The chemical formula of sodium carbonate is
There are
The chemical formula of sodium carbonate is
There are
The chemical formula of sodium carbonate is
Given that
Taking antilog we get
Given that
Taking antilog we get
Given that
Taking antilog we get
And thus
Lead chlorides are pretty insoluble. A solution of this concentration may represent a saturated solution.
We ASSUME that a
And thus
Lead chlorides are pretty insoluble. A solution of this concentration may represent a saturated solution.
We ASSUME that a
And thus
Lead chlorides are pretty insoluble. A solution of this concentration may represent a saturated solution.
The answer is 15.175 moles
To solve this let us find out the molar mass of Argon, it is 39.948 g/mol or roughly 40 g /mol.
It means that one mole of Argon weighs 40g.
if one mole of Argon weighs 40 g , then 607 g of Argon has
607g / 40g moles or 15.175 moles.
15.175 moles.
The answer is 15.175 moles
To solve this let us find out the molar mass of Argon, it is 39.948 g/mol or roughly 40 g /mol.
It means that one mole of Argon weighs 40g.
if one mole of Argon weighs 40 g , then 607 g of Argon has
607g / 40g moles or 15.175 moles.
15.175 moles.
The answer is 15.175 moles
To solve this let us find out the molar mass of Argon, it is 39.948 g/mol or roughly 40 g /mol.
It means that one mole of Argon weighs 40g.
if one mole of Argon weighs 40 g , then 607 g of Argon has
607g / 40g moles or 15.175 moles.
Photosynthesis is the combining of Carbon Dioxide and Water to make Glucose and Oxygen. The equation is:
#12 H_2O + 6 CO_2 -> 6 H_2O + C_6H_12O_6 + 6O_2#
This reaction must occur in the presence of sunlight because light energy is required.
The equation can also be written out in words as:
In the presence of sunlight, six moles of carbon dioxide and six moles of water react to form one molecule of glucose and six moles of oxygen.
Sunlight sends energy in packets of light called photons. These photons activate chlorophyll in the grans to break apart molecules of water. The Hydrogen atoms pick up the electrons from the energy and become energized. The Oxygen gets released as a waste product
The excited
The equation is:
#12 H_2O + 6 CO_2 -> 6 H_2O + C6H_12O_6 + 6O_2#
Photosynthesis is the combining of Carbon Dioxide and Water to make Glucose and Oxygen. The equation is:
#12 H_2O + 6 CO_2 -> 6 H_2O + C_6H_12O_6 + 6O_2#
This reaction must occur in the presence of sunlight because light energy is required.
The equation can also be written out in words as:
In the presence of sunlight, six moles of carbon dioxide and six moles of water react to form one molecule of glucose and six moles of oxygen.
Sunlight sends energy in packets of light called photons. These photons activate chlorophyll in the grans to break apart molecules of water. The Hydrogen atoms pick up the electrons from the energy and become energized. The Oxygen gets released as a waste product
The excited
The equation is:
#12 H_2O + 6 CO_2 -> 6 H_2O + C6H_12O_6 + 6O_2#
Photosynthesis is the combining of Carbon Dioxide and Water to make Glucose and Oxygen. The equation is:
#12 H_2O + 6 CO_2 -> 6 H_2O + C_6H_12O_6 + 6O_2#
This reaction must occur in the presence of sunlight because light energy is required.
The equation can also be written out in words as:
In the presence of sunlight, six moles of carbon dioxide and six moles of water react to form one molecule of glucose and six moles of oxygen.
Sunlight sends energy in packets of light called photons. These photons activate chlorophyll in the grans to break apart molecules of water. The Hydrogen atoms pick up the electrons from the energy and become energized. The Oxygen gets released as a waste product
The excited
The idea here is that lead(II) iodate is considered insoluble in water, so right from the start, you should expect to find a very low concentration of iodate anions,
The dissociation equilibrium that describes the dissociation of lead(II) iodate looks like this
#""Pb""(""IO""_ 3)_ (2(s)) rightleftharpoons ""Pb""_ ((aq))^(2+) + color(red)(2)""IO""_ (3(aq))^(-)#
Notice that every mole of lead(II) iodate that dissociates produces
This means that, at equilibrium, a saturated solution of lead(II) iodate will have
#[""IO""_3^(-)] = color(red)(2) * [""Pb""^(2+)]#
Now, the solubility product constant for this dissociation equilibrium looks like this
#K_(sp) = [""Pb""^(2+)] * [""IO""_3^(-)]^color(red)(2)#
If you take
#K_(sp) = s * (color(red)(2)s)^color(red)(2)#
which is equivalent to
#2.6 * 10^(-13) = 4s^3#
Rearrange to solve for
#s = root(3)( (2.6 * 10^(-13))/4) = 4.02 * 10^(-5)#
This means that a saturated solution of lead(II) iodate will have
#[""Pb""^(2+)] = 4.02 * 10^(-5)# #""M""#
and
#[""IO""_3^(-)] = color(red)(2) * 4.02 * 10^(-5)color(white)(.)""M"" = color(darkgreen)(ul(color(black)(8.0 * 10^(-5)color(white)(.)""M"")))#
I'll leave the answer rounded to two sig figs.
The idea here is that lead(II) iodate is considered insoluble in water, so right from the start, you should expect to find a very low concentration of iodate anions,
The dissociation equilibrium that describes the dissociation of lead(II) iodate looks like this
#""Pb""(""IO""_ 3)_ (2(s)) rightleftharpoons ""Pb""_ ((aq))^(2+) + color(red)(2)""IO""_ (3(aq))^(-)#
Notice that every mole of lead(II) iodate that dissociates produces
This means that, at equilibrium, a saturated solution of lead(II) iodate will have
#[""IO""_3^(-)] = color(red)(2) * [""Pb""^(2+)]#
Now, the solubility product constant for this dissociation equilibrium looks like this
#K_(sp) = [""Pb""^(2+)] * [""IO""_3^(-)]^color(red)(2)#
If you take
#K_(sp) = s * (color(red)(2)s)^color(red)(2)#
which is equivalent to
#2.6 * 10^(-13) = 4s^3#
Rearrange to solve for
#s = root(3)( (2.6 * 10^(-13))/4) = 4.02 * 10^(-5)#
This means that a saturated solution of lead(II) iodate will have
#[""Pb""^(2+)] = 4.02 * 10^(-5)# #""M""#
and
#[""IO""_3^(-)] = color(red)(2) * 4.02 * 10^(-5)color(white)(.)""M"" = color(darkgreen)(ul(color(black)(8.0 * 10^(-5)color(white)(.)""M"")))#
I'll leave the answer rounded to two sig figs.
The idea here is that lead(II) iodate is considered insoluble in water, so right from the start, you should expect to find a very low concentration of iodate anions,
The dissociation equilibrium that describes the dissociation of lead(II) iodate looks like this
#""Pb""(""IO""_ 3)_ (2(s)) rightleftharpoons ""Pb""_ ((aq))^(2+) + color(red)(2)""IO""_ (3(aq))^(-)#
Notice that every mole of lead(II) iodate that dissociates produces
This means that, at equilibrium, a saturated solution of lead(II) iodate will have
#[""IO""_3^(-)] = color(red)(2) * [""Pb""^(2+)]#
Now, the solubility product constant for this dissociation equilibrium looks like this
#K_(sp) = [""Pb""^(2+)] * [""IO""_3^(-)]^color(red)(2)#
If you take
#K_(sp) = s * (color(red)(2)s)^color(red)(2)#
which is equivalent to
#2.6 * 10^(-13) = 4s^3#
Rearrange to solve for
#s = root(3)( (2.6 * 10^(-13))/4) = 4.02 * 10^(-5)#
This means that a saturated solution of lead(II) iodate will have
#[""Pb""^(2+)] = 4.02 * 10^(-5)# #""M""#
and
#[""IO""_3^(-)] = color(red)(2) * 4.02 * 10^(-5)color(white)(.)""M"" = color(darkgreen)(ul(color(black)(8.0 * 10^(-5)color(white)(.)""M"")))#
I'll leave the answer rounded to two sig figs.
Unbalanced Equation
Balance the hydrogen.
Place a coefficient of
Balance the oxygen.
Place a coefficient of
There are now equal numbers of atoms of each element on both sides of the equation, and it is now balanced.
Number of Atoms on Left Side:
Number of Atoms on Right Side:
Unbalanced Equation
Balance the hydrogen.
Place a coefficient of
Balance the oxygen.
Place a coefficient of
There are now equal numbers of atoms of each element on both sides of the equation, and it is now balanced.
Number of Atoms on Left Side:
Number of Atoms on Right Side:
Unbalanced Equation
Balance the hydrogen.
Place a coefficient of
Balance the oxygen.
Place a coefficient of
There are now equal numbers of atoms of each element on both sides of the equation, and it is now balanced.
Number of Atoms on Left Side:
Number of Atoms on Right Side:
In
One mole of methane has a mass of 16.0 g.
In
One mole of methane has a mass of 16.0 g.
In
The molar mass (mass of one mole) of
Convert 3.8 mmol to mol.
Multiply moles times molar mass.
The idea here is that when you're performing a dilution, the concentration of the solution decreases by the same factor (known as the dilution factor) as the volume increases.
#""volume"" uarr ""by a factor""color(white)(.)color(blue)(""DF"") implies ""concentration"" darr ""by the same factor""color(white)(.)color(blue)(""DF"")#
In your case, the volume of the solution increases from
#""DF"" = (500 color(red)(cancel(color(black)(""mL""))))/(25.0color(red)(cancel(color(black)(""mL"")))) = color(blue)(20)#
Now remember, the same dilution factor is applied to the concentration of the solution, i.e. the concentration of the diluted solution must be
This means that the concentration of the diluted solution will be
#""% diluted"" = (10%)/color(blue)(20) = color(darkgreen)(ul(color(black)(0.5%)))#
The answer is rounded to one significant figure.
The idea here is that when you're performing a dilution, the concentration of the solution decreases by the same factor (known as the dilution factor) as the volume increases.
#""volume"" uarr ""by a factor""color(white)(.)color(blue)(""DF"") implies ""concentration"" darr ""by the same factor""color(white)(.)color(blue)(""DF"")#
In your case, the volume of the solution increases from
#""DF"" = (500 color(red)(cancel(color(black)(""mL""))))/(25.0color(red)(cancel(color(black)(""mL"")))) = color(blue)(20)#
Now remember, the same dilution factor is applied to the concentration of the solution, i.e. the concentration of the diluted solution must be
This means that the concentration of the diluted solution will be
#""% diluted"" = (10%)/color(blue)(20) = color(darkgreen)(ul(color(black)(0.5%)))#
The answer is rounded to one significant figure.
The idea here is that when you're performing a dilution, the concentration of the solution decreases by the same factor (known as the dilution factor) as the volume increases.
#""volume"" uarr ""by a factor""color(white)(.)color(blue)(""DF"") implies ""concentration"" darr ""by the same factor""color(white)(.)color(blue)(""DF"")#
In your case, the volume of the solution increases from
#""DF"" = (500 color(red)(cancel(color(black)(""mL""))))/(25.0color(red)(cancel(color(black)(""mL"")))) = color(blue)(20)#
Now remember, the same dilution factor is applied to the concentration of the solution, i.e. the concentration of the diluted solution must be
This means that the concentration of the diluted solution will be
#""% diluted"" = (10%)/color(blue)(20) = color(darkgreen)(ul(color(black)(0.5%)))#
The answer is rounded to one significant figure.
A piece of copper with a mass of 22.6 g initially at a temperature of 16.3 °C is heated to a temperature of 33.1 °C. Assuming the specific heat of copper is 0.386 J/(g°C), how much heat was needed for this temperature change to take place?
146,55 J
A piece of copper with a mass of 22.6 g initially at a temperature of 16.3 °C is heated to a temperature of 33.1 °C. Assuming the specific heat of copper is 0.386 J/(g°C), how much heat was needed for this temperature change to take place?
146,55 J
A piece of copper with a mass of 22.6 g initially at a temperature of 16.3 °C is heated to a temperature of 33.1 °C. Assuming the specific heat of copper is 0.386 J/(g°C), how much heat was needed for this temperature change to take place?
Balancing the equation by balancing number of atoms equal both sides.
Balancing the equation by balancing number of atoms equal both sides.
Balancing the equation by balancing number of atoms equal both sides.
As you know, molarity is defined as the number of moles of solute, which in your case is sodium chloride,
#color(blue)(""molarity"" = ""moles of solute""/""liters of solution"")#
Simply put, dissolving one mole of a solute in one liter of solution will produce a
In your case, the volume of the solution is said to be equal to
Since you have a little less than half of a liter, i.e.
More specifically, the solution will contain - do not forget to convert the volume of the solution from milliliters to liters!
#color(blue)(c = n/V implies n = c * V)#
#n = 0.244 ""moles""/color(red)(cancel(color(black)(""L""))) * 467 * 10^(-3) color(red)(cancel(color(black)(""L""))) = ""0.1139 moles NaCl""#
Rounded to three sig figs, the answer will be
#n_(NaCl) = color(green)(""0.114 moles"")#
SIDE NOTE Because sodium chloride dissociates completely in aqueous solution, you wouldn't actually say that you have a
Instead, you would say that you have a solution that is
As you know, molarity is defined as the number of moles of solute, which in your case is sodium chloride,
#color(blue)(""molarity"" = ""moles of solute""/""liters of solution"")#
Simply put, dissolving one mole of a solute in one liter of solution will produce a
In your case, the volume of the solution is said to be equal to
Since you have a little less than half of a liter, i.e.
More specifically, the solution will contain - do not forget to convert the volume of the solution from milliliters to liters!
#color(blue)(c = n/V implies n = c * V)#
#n = 0.244 ""moles""/color(red)(cancel(color(black)(""L""))) * 467 * 10^(-3) color(red)(cancel(color(black)(""L""))) = ""0.1139 moles NaCl""#
Rounded to three sig figs, the answer will be
#n_(NaCl) = color(green)(""0.114 moles"")#
SIDE NOTE Because sodium chloride dissociates completely in aqueous solution, you wouldn't actually say that you have a
Instead, you would say that you have a solution that is
As you know, molarity is defined as the number of moles of solute, which in your case is sodium chloride,
#color(blue)(""molarity"" = ""moles of solute""/""liters of solution"")#
Simply put, dissolving one mole of a solute in one liter of solution will produce a
In your case, the volume of the solution is said to be equal to
Since you have a little less than half of a liter, i.e.
More specifically, the solution will contain - do not forget to convert the volume of the solution from milliliters to liters!
#color(blue)(c = n/V implies n = c * V)#
#n = 0.244 ""moles""/color(red)(cancel(color(black)(""L""))) * 467 * 10^(-3) color(red)(cancel(color(black)(""L""))) = ""0.1139 moles NaCl""#
Rounded to three sig figs, the answer will be
#n_(NaCl) = color(green)(""0.114 moles"")#
SIDE NOTE Because sodium chloride dissociates completely in aqueous solution, you wouldn't actually say that you have a
Instead, you would say that you have a solution that is
For this kind of questions, we use the formula:
To calculate the mass of rubidium iodide (
We have a 13.9 L solution with a concentration of 0.65 M, using the formula above, rewritten:
So in our 1 L solution, we had 9.035 mol
Since we took 1 L from the original solution, the 9.035 mol/L is the concentration of the original solution. We had 4.6 L of the original solution and we need the amount of mol, therefore, you guessed it, we use again the same formula!
Now we only have to convert mol
Which, when rounded to two significant figures (the lowest number given in the problem), is
The amount of
For this kind of questions, we use the formula:
To calculate the mass of rubidium iodide (
We have a 13.9 L solution with a concentration of 0.65 M, using the formula above, rewritten:
So in our 1 L solution, we had 9.035 mol
Since we took 1 L from the original solution, the 9.035 mol/L is the concentration of the original solution. We had 4.6 L of the original solution and we need the amount of mol, therefore, you guessed it, we use again the same formula!
Now we only have to convert mol
Which, when rounded to two significant figures (the lowest number given in the problem), is
The amount of
For this kind of questions, we use the formula:
To calculate the mass of rubidium iodide (
We have a 13.9 L solution with a concentration of 0.65 M, using the formula above, rewritten:
So in our 1 L solution, we had 9.035 mol
Since we took 1 L from the original solution, the 9.035 mol/L is the concentration of the original solution. We had 4.6 L of the original solution and we need the amount of mol, therefore, you guessed it, we use again the same formula!
Now we only have to convert mol
Which, when rounded to two significant figures (the lowest number given in the problem), is
Here is another way to do this problem using the extensive property of countable numbers. I also get
OVERVIEW:
Think critically through the problem... What did we do? We started with some starting mols of
That was then diluted from
Looking at the end of the question, we were given
#13.9 cancel""L"" xx ""0.65 mols""/cancel""L"" = ""9.035 mols RbI""# for every#""L""# of solution.
The number of mols in the
Mols are an extensive quantity, just like mass. If you have a greater volume of a certain concentration, you have more mols of it.
Therefore, in
#""9.035 mols RbI"" xx (4.6 cancel""L"")/cancel""1 L""#
#=# #""41.561 mols RbI""# in that#""4.6 L""# of solution.
Therefore, you had a mass of:
#color(blue)(m_(RbI)) = 41.561 cancel""mols RbI"" xx ""212.3723 g""/cancel""1 mol RbI""#
#=# #""8826.41 g""#
#-># #color(blue)(""8800 g"")# to two sig figs
Now think back through the problem... What did we do?
We started with some
That was then diluted
As with all these problems, we ASSUME, that there is a
We break this quantity down into atoms.
And, there is
And
If we divide thru by the lowest molar amount, we get an empirical formula of
Note that an examiner would be quite justified at A level to say that
As with all these problems, we ASSUME, that there is a
We break this quantity down into atoms.
And, there is
And
If we divide thru by the lowest molar amount, we get an empirical formula of
Note that an examiner would be quite justified at A level to say that
As with all these problems, we ASSUME, that there is a
We break this quantity down into atoms.
And, there is
And
If we divide thru by the lowest molar amount, we get an empirical formula of
Note that an examiner would be quite justified at A level to say that
This is a fairly straightforward application of the ideal gas law equation, which looks like this
#color(blue)(PV = nRT)"" ""# , where
Rearrange this equation to solve for
#PV = nRT implies T = (PV)/(nR)#
Before plugging in your values, check to make sure that the units given to you match the units used in the expression of the universal gas constant.
As it stands, moles, atm, and liters are all units used for
Plug in your values to get
#T = (2.46 color(red)(cancel(color(black)(""atm""))) * 5.0 color(red)(cancel(color(black)(""L""))))/(2.0 color(red)(cancel(color(black)(""moles""))) * 0.0821 (color(red)(cancel(color(black)(""atm""))) * color(red)(cancel(color(black)(""L""))))/(color(red)(cancel(color(black)(""mol""))) * ""K"")) = ""74.91 K""#
Rounded to two sig figs, the answer will be
#T = color(green)(""75 K"")#
This is a fairly straightforward application of the ideal gas law equation, which looks like this
#color(blue)(PV = nRT)"" ""# , where
Rearrange this equation to solve for
#PV = nRT implies T = (PV)/(nR)#
Before plugging in your values, check to make sure that the units given to you match the units used in the expression of the universal gas constant.
As it stands, moles, atm, and liters are all units used for
Plug in your values to get
#T = (2.46 color(red)(cancel(color(black)(""atm""))) * 5.0 color(red)(cancel(color(black)(""L""))))/(2.0 color(red)(cancel(color(black)(""moles""))) * 0.0821 (color(red)(cancel(color(black)(""atm""))) * color(red)(cancel(color(black)(""L""))))/(color(red)(cancel(color(black)(""mol""))) * ""K"")) = ""74.91 K""#
Rounded to two sig figs, the answer will be
#T = color(green)(""75 K"")#
This is a fairly straightforward application of the ideal gas law equation, which looks like this
#color(blue)(PV = nRT)"" ""# , where
Rearrange this equation to solve for
#PV = nRT implies T = (PV)/(nR)#
Before plugging in your values, check to make sure that the units given to you match the units used in the expression of the universal gas constant.
As it stands, moles, atm, and liters are all units used for
Plug in your values to get
#T = (2.46 color(red)(cancel(color(black)(""atm""))) * 5.0 color(red)(cancel(color(black)(""L""))))/(2.0 color(red)(cancel(color(black)(""moles""))) * 0.0821 (color(red)(cancel(color(black)(""atm""))) * color(red)(cancel(color(black)(""L""))))/(color(red)(cancel(color(black)(""mol""))) * ""K"")) = ""74.91 K""#
Rounded to two sig figs, the answer will be
#T = color(green)(""75 K"")#
Molecular mass HCl=36.4g , Zn=65.4g
n°mol=
In the reaction, we can see HCl doubles the n°mol of Zn, so...
Molecular mass HCl=36.4g , Zn=65.4g
n°mol=
In the reaction, we can see HCl doubles the n°mol of Zn, so...
Molecular mass HCl=36.4g , Zn=65.4g
n°mol=
In the reaction, we can see HCl doubles the n°mol of Zn, so...
The Henderson - Hasselbalch equation allows you to calculate the pH of buffer solution that contains a weak acid and its conjugate base by using the concentrations of these two species and the
#color(blue)(""pH"" = pK_a + log( ([""conjugate base""])/([""weak acid""])))#
In your case, the weak acid is hypochlorous acid,
The acid dissociation constant,
Now, before doing any calculations, try to predict what you expect the solution's pH to be compared with the acid's
Notice that when you have equal concentrations of weak acid and conjugate base, the log term will be equal to zero, since
#log(1) = 0#
This tells you that if you have more conjugate base than weak acid, the log term will be greater than
With this in mind, plug in your values into the H-H equation to get
#""pH"" = -log(3.5 * 10^(-8)) + log( (0.185color(red)(cancel(color(black)(""M""))))/(0.120color(red)(cancel(color(black)(""M"")))))#
#""pH"" = 7.46 + 0.188 = color(green)(7.65)#
The Henderson - Hasselbalch equation allows you to calculate the pH of buffer solution that contains a weak acid and its conjugate base by using the concentrations of these two species and the
#color(blue)(""pH"" = pK_a + log( ([""conjugate base""])/([""weak acid""])))#
In your case, the weak acid is hypochlorous acid,
The acid dissociation constant,
Now, before doing any calculations, try to predict what you expect the solution's pH to be compared with the acid's
Notice that when you have equal concentrations of weak acid and conjugate base, the log term will be equal to zero, since
#log(1) = 0#
This tells you that if you have more conjugate base than weak acid, the log term will be greater than
With this in mind, plug in your values into the H-H equation to get
#""pH"" = -log(3.5 * 10^(-8)) + log( (0.185color(red)(cancel(color(black)(""M""))))/(0.120color(red)(cancel(color(black)(""M"")))))#
#""pH"" = 7.46 + 0.188 = color(green)(7.65)#
The Henderson - Hasselbalch equation allows you to calculate the pH of buffer solution that contains a weak acid and its conjugate base by using the concentrations of these two species and the
#color(blue)(""pH"" = pK_a + log( ([""conjugate base""])/([""weak acid""])))#
In your case, the weak acid is hypochlorous acid,
The acid dissociation constant,
Now, before doing any calculations, try to predict what you expect the solution's pH to be compared with the acid's
Notice that when you have equal concentrations of weak acid and conjugate base, the log term will be equal to zero, since
#log(1) = 0#
This tells you that if you have more conjugate base than weak acid, the log term will be greater than
With this in mind, plug in your values into the H-H equation to get
#""pH"" = -log(3.5 * 10^(-8)) + log( (0.185color(red)(cancel(color(black)(""M""))))/(0.120color(red)(cancel(color(black)(""M"")))))#
#""pH"" = 7.46 + 0.188 = color(green)(7.65)#
In order to determine the number of moles of a given mass of a substance, divide the given mass by its molar mass.
The molar mass of helium is
There are 5.00 moles He in 20.0 g He.
In order to determine the number of moles of a given mass of a substance, divide the given mass by its molar mass.
The molar mass of helium is
There are 5.00 moles He in 20.0 g He.
In order to determine the number of moles of a given mass of a substance, divide the given mass by its molar mass.
The molar mass of helium is
Nitrates are generally soluble in water. In the reaction above both MASS and CHARGE are conserved, as they must be for any chemical reaction. Most of the time, alkali metal cations are along for the ride; the business end of the molecule is the nitrate anion, which can be a potent oxidant.
Nitrates are generally soluble in water. In the reaction above both MASS and CHARGE are conserved, as they must be for any chemical reaction. Most of the time, alkali metal cations are along for the ride; the business end of the molecule is the nitrate anion, which can be a potent oxidant.
Nitrates are generally soluble in water. In the reaction above both MASS and CHARGE are conserved, as they must be for any chemical reaction. Most of the time, alkali metal cations are along for the ride; the business end of the molecule is the nitrate anion, which can be a potent oxidant.
Aluminum is in slight deficiency, and is thus the limiting reagent. So
I should add that there is a very practical application to this reaction (which I think is still used). When setters lay train tracks, obviously they lay steel tracks over wooden or concrete sleepers. In order to join the rails between lengths they use precisely this reaction (of course they have to heat it up to get it to go!). Such welding gives a very good seal and a smooth ride.
See here
Approx.
Aluminum is in slight deficiency, and is thus the limiting reagent. So
I should add that there is a very practical application to this reaction (which I think is still used). When setters lay train tracks, obviously they lay steel tracks over wooden or concrete sleepers. In order to join the rails between lengths they use precisely this reaction (of course they have to heat it up to get it to go!). Such welding gives a very good seal and a smooth ride.
See here
Approx.
Aluminum is in slight deficiency, and is thus the limiting reagent. So
I should add that there is a very practical application to this reaction (which I think is still used). When setters lay train tracks, obviously they lay steel tracks over wooden or concrete sleepers. In order to join the rails between lengths they use precisely this reaction (of course they have to heat it up to get it to go!). Such welding gives a very good seal and a smooth ride.
See here
This question requires you to use Avogadro's number to convert the atoms to moles, and then use the molar mass of gold (
I'm assuming by ""how heavy"" you mean mass, but if you wanted to calculate weight you would need to calculate a force by using
weight of
This question requires you to use Avogadro's number to convert the atoms to moles, and then use the molar mass of gold (
I'm assuming by ""how heavy"" you mean mass, but if you wanted to calculate weight you would need to calculate a force by using
weight of
This question requires you to use Avogadro's number to convert the atoms to moles, and then use the molar mass of gold (
I'm assuming by ""how heavy"" you mean mass, but if you wanted to calculate weight you would need to calculate a force by using
Zinc reacts with hydrochloric acid to produce zinc chloride and hidrogen.
Zinc reacts with hydrochloric acid to produce zinc chloride and hidrogen.
Zinc reacts with hydrochloric acid to produce zinc chloride and hidrogen.
Before doing any calculation, take a look at the values of the equilibrium concentrations for the three chemical species that take part in the reaction.
Notice that the concentrations of the two reactants, nitrogen gas,
This tells you that, at this given temperature, the reaction favors the reactants over the formation of the product. Simply put, the equilibrium lies to the left, so you can expect the equilibrium constant,
For your given reaction
#""N""_text(2(g]) + color(red)(3)""H""_text(2(g]) rightleftharpoons color(blue)(2)""NH""_text(3(g])#
the equilibrium constant takes the form
#K_c = ( [""NH""_3]^color(blue)(2))/([""N""_2] * [""H""_2]^color(red)(3))#
Plug in your values to get
#K_c = (2.00)^color(blue)(2)/(10.00 * 8.00^color(red)(3)) = 4.00/(10.00 * 512)#
#K_c = color(green)(7.81 * 10^(-4))#
The answer is rounded to two sig figs.
Before doing any calculation, take a look at the values of the equilibrium concentrations for the three chemical species that take part in the reaction.
Notice that the concentrations of the two reactants, nitrogen gas,
This tells you that, at this given temperature, the reaction favors the reactants over the formation of the product. Simply put, the equilibrium lies to the left, so you can expect the equilibrium constant,
For your given reaction
#""N""_text(2(g]) + color(red)(3)""H""_text(2(g]) rightleftharpoons color(blue)(2)""NH""_text(3(g])#
the equilibrium constant takes the form
#K_c = ( [""NH""_3]^color(blue)(2))/([""N""_2] * [""H""_2]^color(red)(3))#
Plug in your values to get
#K_c = (2.00)^color(blue)(2)/(10.00 * 8.00^color(red)(3)) = 4.00/(10.00 * 512)#
#K_c = color(green)(7.81 * 10^(-4))#
The answer is rounded to two sig figs.
Before doing any calculation, take a look at the values of the equilibrium concentrations for the three chemical species that take part in the reaction.
Notice that the concentrations of the two reactants, nitrogen gas,
This tells you that, at this given temperature, the reaction favors the reactants over the formation of the product. Simply put, the equilibrium lies to the left, so you can expect the equilibrium constant,
For your given reaction
#""N""_text(2(g]) + color(red)(3)""H""_text(2(g]) rightleftharpoons color(blue)(2)""NH""_text(3(g])#
the equilibrium constant takes the form
#K_c = ( [""NH""_3]^color(blue)(2))/([""N""_2] * [""H""_2]^color(red)(3))#
Plug in your values to get
#K_c = (2.00)^color(blue)(2)/(10.00 * 8.00^color(red)(3)) = 4.00/(10.00 * 512)#
#K_c = color(green)(7.81 * 10^(-4))#
The answer is rounded to two sig figs.
Chemical formula of Calcium chloride=
Molar mass of
Mass of Calcium chloride (given) =
Now, to calculate no. of moles:
No. of moles =
No. of moles=
Converting
Water is solvent while
Now,
Chemical formula of Calcium chloride=
Molar mass of
Mass of Calcium chloride (given) =
Now, to calculate no. of moles:
No. of moles =
No. of moles=
Converting
Water is solvent while
Now,
Chemical formula of Calcium chloride=
Molar mass of
Mass of Calcium chloride (given) =
Now, to calculate no. of moles:
No. of moles =
No. of moles=
Converting
Water is solvent while
Now,
And thus..............................................
Note that at (relatively!) low concentrations, the calculated
Since the pressure and amount of gas is held constant, you can use Charles' law, the volume-temperature law, which states that at constant pressure, the volume of a gas is directly proportional to its Kelvin temperature. This means that as the volume increases, the temperature increases, and vice versa.
The initial calculation will require the conversion of the Celsius temperature to Kelvins. After the calculation is complete, the new temperature will be converted from Kelvins back to degrees Celsius.
Equation
Known
Unknown
Solution
Rearrange the equation to isolate
Final temperature in degrees Celsius
The final temperature in degrees Celsius is
Since the pressure and amount of gas is held constant, you can use Charles' law, the volume-temperature law, which states that at constant pressure, the volume of a gas is directly proportional to its Kelvin temperature. This means that as the volume increases, the temperature increases, and vice versa.
The initial calculation will require the conversion of the Celsius temperature to Kelvins. After the calculation is complete, the new temperature will be converted from Kelvins back to degrees Celsius.
Equation
Known
Unknown
Solution
Rearrange the equation to isolate
Final temperature in degrees Celsius
The final temperature in degrees Celsius is
Since the pressure and amount of gas is held constant, you can use Charles' law, the volume-temperature law, which states that at constant pressure, the volume of a gas is directly proportional to its Kelvin temperature. This means that as the volume increases, the temperature increases, and vice versa.
The initial calculation will require the conversion of the Celsius temperature to Kelvins. After the calculation is complete, the new temperature will be converted from Kelvins back to degrees Celsius.
Equation
Known
Unknown
Solution
Rearrange the equation to isolate
Final temperature in degrees Celsius
The formula mass - which I believe is what you were looking for in the first place - can be found by adding up the relative molecular masses of each species in the formula unit.
The formula of potassium phosphate is
Its relative formula mass is
The formula mass - which I believe is what you were looking for in the first place - can be found by adding up the relative molecular masses of each species in the formula unit.
The formula of potassium phosphate is
Its relative formula mass is
The formula mass - which I believe is what you were looking for in the first place - can be found by adding up the relative molecular masses of each species in the formula unit.
So we approach the equation by the method of half equations:
And
Both half equations conserve mass and conserve charge, as they must if they are to reflect chemical reality.
And we add the individual half reactions together so as to remove the electrons: i.e.
To give (almost finally):
And we can add TWO
Are mass and charge balanced? If they don't, what must you do?
This is a formal redox reaction...........
So we approach the equation by the method of half equations:
And
Both half equations conserve mass and conserve charge, as they must if they are to reflect chemical reality.
And we add the individual half reactions together so as to remove the electrons: i.e.
To give (almost finally):
And we can add TWO
Are mass and charge balanced? If they don't, what must you do?
This is a formal redox reaction...........
So we approach the equation by the method of half equations:
And
Both half equations conserve mass and conserve charge, as they must if they are to reflect chemical reality.
And we add the individual half reactions together so as to remove the electrons: i.e.
To give (almost finally):
And we can add TWO
Are mass and charge balanced? If they don't, what must you do?
First determine the molar mass of fluorine gas
Molar Mass of
Multiply the molar mass of
Moles
Divide the given mass by the molar mass.
I am keeping a couple of guard digits to reduce rounding errors. The final answer will be rounded to three significant figures.
Use the formula for the ideal gas law to determine the volume of the fluorine gas.
https://en.wikipedia.org/w/index.php?title=Gas_constant&oldid=729635962
Rearrange the formula to isolate volume, substitute the known values into the formula and solve for volume.
If your teacher still uses
The volume of
First determine the molar mass of fluorine gas
Molar Mass of
Multiply the molar mass of
Moles
Divide the given mass by the molar mass.
I am keeping a couple of guard digits to reduce rounding errors. The final answer will be rounded to three significant figures.
Use the formula for the ideal gas law to determine the volume of the fluorine gas.
https://en.wikipedia.org/w/index.php?title=Gas_constant&oldid=729635962
Rearrange the formula to isolate volume, substitute the known values into the formula and solve for volume.
If your teacher still uses
The volume of
First determine the molar mass of fluorine gas
Molar Mass of
Multiply the molar mass of
Moles
Divide the given mass by the molar mass.
I am keeping a couple of guard digits to reduce rounding errors. The final answer will be rounded to three significant figures.
Use the formula for the ideal gas law to determine the volume of the fluorine gas.
https://en.wikipedia.org/w/index.php?title=Gas_constant&oldid=729635962
Rearrange the formula to isolate volume, substitute the known values into the formula and solve for volume.
If your teacher still uses
You're looking for the mass of potassium sulfite,
Now, molality is used to express the concentration of a solution in terms of how many moles of solute it contains per kilogram of solvent.
This means that in order to find a solution's molality, you need to know
- the number of moles of solute
- the mass of the solvent expressed in kilograms
In your case, you already know the mass of the solvent in grams, so the very first thing to do here is convert it to kilograms
#872 color(red)(cancel(color(black)(""g""))) * ""1 kg""/(10^3color(red)(cancel(color(black)(""g"")))) = ""0.872 kg""#
Now, a
#0.872 color(red)(cancel(color(black)(""kg solvent""))) * overbrace((""0.128 moles K""_2""SO""_3)/(1color(red)(cancel(color(black)(""kg solvent"")))))^(color(blue)(""= 0.128 m"")) = ""0.1116 moles K""_2""SO""_3#
All you have to do now is use the molar mass of potassium sulfite to figure out how many grams would contain that many moles
#0.1116 color(red)(cancel(color(black)(""moles K""_2""SO""_3))) * ""158.26 g""/(1color(red)(cancel(color(black)(""mole K""_2""SO""_3)))) = color(green)(|bar(ul(color(white)(a/a)color(black)(""17.7 g"")color(white)(a/a)|)))#
The answer is rounded to three sig figs.
You're looking for the mass of potassium sulfite,
Now, molality is used to express the concentration of a solution in terms of how many moles of solute it contains per kilogram of solvent.
This means that in order to find a solution's molality, you need to know
- the number of moles of solute
- the mass of the solvent expressed in kilograms
In your case, you already know the mass of the solvent in grams, so the very first thing to do here is convert it to kilograms
#872 color(red)(cancel(color(black)(""g""))) * ""1 kg""/(10^3color(red)(cancel(color(black)(""g"")))) = ""0.872 kg""#
Now, a
#0.872 color(red)(cancel(color(black)(""kg solvent""))) * overbrace((""0.128 moles K""_2""SO""_3)/(1color(red)(cancel(color(black)(""kg solvent"")))))^(color(blue)(""= 0.128 m"")) = ""0.1116 moles K""_2""SO""_3#
All you have to do now is use the molar mass of potassium sulfite to figure out how many grams would contain that many moles
#0.1116 color(red)(cancel(color(black)(""moles K""_2""SO""_3))) * ""158.26 g""/(1color(red)(cancel(color(black)(""mole K""_2""SO""_3)))) = color(green)(|bar(ul(color(white)(a/a)color(black)(""17.7 g"")color(white)(a/a)|)))#
The answer is rounded to three sig figs.
You're looking for the mass of potassium sulfite,
Now, molality is used to express the concentration of a solution in terms of how many moles of solute it contains per kilogram of solvent.
This means that in order to find a solution's molality, you need to know
- the number of moles of solute
- the mass of the solvent expressed in kilograms
In your case, you already know the mass of the solvent in grams, so the very first thing to do here is convert it to kilograms
#872 color(red)(cancel(color(black)(""g""))) * ""1 kg""/(10^3color(red)(cancel(color(black)(""g"")))) = ""0.872 kg""#
Now, a
#0.872 color(red)(cancel(color(black)(""kg solvent""))) * overbrace((""0.128 moles K""_2""SO""_3)/(1color(red)(cancel(color(black)(""kg solvent"")))))^(color(blue)(""= 0.128 m"")) = ""0.1116 moles K""_2""SO""_3#
All you have to do now is use the molar mass of potassium sulfite to figure out how many grams would contain that many moles
#0.1116 color(red)(cancel(color(black)(""moles K""_2""SO""_3))) * ""158.26 g""/(1color(red)(cancel(color(black)(""mole K""_2""SO""_3)))) = color(green)(|bar(ul(color(white)(a/a)color(black)(""17.7 g"")color(white)(a/a)|)))#
The answer is rounded to three sig figs.
AS is typical in these problems, we ASSUME an
And we divide thru by the SMALLEST molar quantity to get an empirical formula of
AS is typical in these problems, we ASSUME an
And we divide thru by the SMALLEST molar quantity to get an empirical formula of
AS is typical in these problems, we ASSUME an
And we divide thru by the SMALLEST molar quantity to get an empirical formula of
I think you have intuitively realized the correct approach.
The
And thus
And
And
Of course all the individual percentages must sum to 100%. Why?
Alternatively see here.
I think you have intuitively realized the correct approach.
The
And thus
And
And
Of course all the individual percentages must sum to 100%. Why?
Alternatively see here.
I think you have intuitively realized the correct approach.
The
And thus
And
And
Of course all the individual percentages must sum to 100%. Why?
Alternatively see here.
Moles = mass divided by molar mass.
Moles = mass divided by molar mass.
Moles = mass divided by molar mass.
Molar mass:
The idea here is that you need to use the molar mass of cisplatin,
As you can see by looking at the compound's chemical formula, one mole of cisplatin contains
- one mole of platinum,
#1 xx ""Pt""# - two moles of nitrogen,
#2 xx ""N""# - six moles of hydrogen,
#6 xx ""H""# - two moles of chlorine,
#2 xx ""Cl""#
You also know that cisplatin has a molar mass of
Platinum has a molar mass of
Since
#100color(red)(cancel(color(black)(""g cisplatin""))) * ""195.08 g Pt""/(300. color(red)(cancel(color(black)(""g cisplatin"")))) = ""65.03 g Pt""#
Since percent composition tells you the mass of a given element per
This means that your
#25.0color(red)(cancel(color(black)(""g cisplatin""))) * overbrace(""65.03 g Pt""/(100color(red)(cancel(color(black)(""g cisplatin"")))))^(color(purple)(""65.03% Pt"")) = color(green)(|bar(ul(color(white)(a/a)""16.3 g Pt""color(white)(a/a)|)))#
The answer is rounded to three sig figs.
Notice that you can get the same result without calculating the percent composition of platinum in cisplatin. All you have to do is use the mass of platinum you get in one mole of cisplatin
#25.0 color(red)(cancel(color(black)(""g cisplatin""))) * ""195.08 g Pt""/(300. color(red)(cancel(color(black)(""g cisplatin"")))) = color(green)(|bar(ul(color(white)(a/a)""16.3 g Pt""color(white)(a/a)|)))#
The idea here is that you need to use the molar mass of cisplatin,
As you can see by looking at the compound's chemical formula, one mole of cisplatin contains
- one mole of platinum,
#1 xx ""Pt""# - two moles of nitrogen,
#2 xx ""N""# - six moles of hydrogen,
#6 xx ""H""# - two moles of chlorine,
#2 xx ""Cl""#
You also know that cisplatin has a molar mass of
Platinum has a molar mass of
Since
#100color(red)(cancel(color(black)(""g cisplatin""))) * ""195.08 g Pt""/(300. color(red)(cancel(color(black)(""g cisplatin"")))) = ""65.03 g Pt""#
Since percent composition tells you the mass of a given element per
This means that your
#25.0color(red)(cancel(color(black)(""g cisplatin""))) * overbrace(""65.03 g Pt""/(100color(red)(cancel(color(black)(""g cisplatin"")))))^(color(purple)(""65.03% Pt"")) = color(green)(|bar(ul(color(white)(a/a)""16.3 g Pt""color(white)(a/a)|)))#
The answer is rounded to three sig figs.
Notice that you can get the same result without calculating the percent composition of platinum in cisplatin. All you have to do is use the mass of platinum you get in one mole of cisplatin
#25.0 color(red)(cancel(color(black)(""g cisplatin""))) * ""195.08 g Pt""/(300. color(red)(cancel(color(black)(""g cisplatin"")))) = color(green)(|bar(ul(color(white)(a/a)""16.3 g Pt""color(white)(a/a)|)))#
Molar mass:
The idea here is that you need to use the molar mass of cisplatin,
As you can see by looking at the compound's chemical formula, one mole of cisplatin contains
- one mole of platinum,
#1 xx ""Pt""# - two moles of nitrogen,
#2 xx ""N""# - six moles of hydrogen,
#6 xx ""H""# - two moles of chlorine,
#2 xx ""Cl""#
You also know that cisplatin has a molar mass of
Platinum has a molar mass of
Since
#100color(red)(cancel(color(black)(""g cisplatin""))) * ""195.08 g Pt""/(300. color(red)(cancel(color(black)(""g cisplatin"")))) = ""65.03 g Pt""#
Since percent composition tells you the mass of a given element per
This means that your
#25.0color(red)(cancel(color(black)(""g cisplatin""))) * overbrace(""65.03 g Pt""/(100color(red)(cancel(color(black)(""g cisplatin"")))))^(color(purple)(""65.03% Pt"")) = color(green)(|bar(ul(color(white)(a/a)""16.3 g Pt""color(white)(a/a)|)))#
The answer is rounded to three sig figs.
Notice that you can get the same result without calculating the percent composition of platinum in cisplatin. All you have to do is use the mass of platinum you get in one mole of cisplatin
#25.0 color(red)(cancel(color(black)(""g cisplatin""))) * ""195.08 g Pt""/(300. color(red)(cancel(color(black)(""g cisplatin"")))) = color(green)(|bar(ul(color(white)(a/a)""16.3 g Pt""color(white)(a/a)|)))#
And
I look on the Periodic Table, and at
My advice is to look at the Periodic Table, and get to know it; you will soon learn the atomic masses of the common reagents.
By definition, there are
And
I look on the Periodic Table, and at
My advice is to look at the Periodic Table, and get to know it; you will soon learn the atomic masses of the common reagents.
By definition, there are
And
I look on the Periodic Table, and at
My advice is to look at the Periodic Table, and get to know it; you will soon learn the atomic masses of the common reagents.
Or, to be more precise:
The double arrow means that there is an equilibrium between the carbon dioxide and the acid.
It forms carbonic acid
Or, to be more precise:
The double arrow means that there is an equilibrium between the carbon dioxide and the acid.
It forms carbonic acid
Or, to be more precise:
The double arrow means that there is an equilibrium between the carbon dioxide and the acid.
First, it's important to know that the chemical formula for water is
Then, we need to find how many moles of water there are in
To find how many moles of water there are in
In
And, since the chemical formula for water is
Therefore, there will be
First, it's important to know that the chemical formula for water is
Then, we need to find how many moles of water there are in
To find how many moles of water there are in
In
And, since the chemical formula for water is
Therefore, there will be
First, it's important to know that the chemical formula for water is
Then, we need to find how many moles of water there are in
To find how many moles of water there are in
In
And, since the chemical formula for water is
Therefore, there will be
Ethanoic acid is a weak acid which dissociates:
The expression for
So the molar ratio of acetate ion to acetic acid is:
The ratio of conentrations which I have given also corresponds to the molar ratio since the volume is common to both.
We are told that the
Ethanoic acid is a weak acid which dissociates:
The expression for
So the molar ratio of acetate ion to acetic acid is:
The ratio of conentrations which I have given also corresponds to the molar ratio since the volume is common to both.
We are told that the
Ethanoic acid is a weak acid which dissociates:
The expression for
So the molar ratio of acetate ion to acetic acid is:
The ratio of conentrations which I have given also corresponds to the molar ratio since the volume is common to both.
We are told that the
As is the standard with these problems, we assume
In this mass, there are
And,
And,
We divide thru, by the smallest molar quantity (
Because the empirical formula is by definition the smallest WHOLE number ratio that describes constituent atoms in a species, we DOUBLE this formula.
As is the standard with these problems, we assume
In this mass, there are
And,
And,
We divide thru, by the smallest molar quantity (
Because the empirical formula is by definition the smallest WHOLE number ratio that describes constituent atoms in a species, we DOUBLE this formula.
As is the standard with these problems, we assume
In this mass, there are
And,
And,
We divide thru, by the smallest molar quantity (
Because the empirical formula is by definition the smallest WHOLE number ratio that describes constituent atoms in a species, we DOUBLE this formula.
The empirical formula is the simplest whole number ratio defining constituent atoms in a species.
In all of these problems it is useful to assume
So, in
If we divide thru by the smallest molar quantity, we get a formula of
The molecuar mass of fructose is
The empirical formula is the simplest whole number ratio defining constituent atoms in a species.
In all of these problems it is useful to assume
So, in
If we divide thru by the smallest molar quantity, we get a formula of
The molecuar mass of fructose is
The empirical formula is the simplest whole number ratio defining constituent atoms in a species.
In all of these problems it is useful to assume
So, in
If we divide thru by the smallest molar quantity, we get a formula of
The molecuar mass of fructose is
This is an example of Charles' law, which states that the volume of a given amount of gas held at constant pressure varies directly with the temperature in Kelvins. The equation to use is
Given
Unknown
Solution
Rearrange the equation to isolate
The final temperature is 60 K.
This is an example of Charles' law, which states that the volume of a given amount of gas held at constant pressure varies directly with the temperature in Kelvins. The equation to use is
Given
Unknown
Solution
Rearrange the equation to isolate
The final temperature is 60 K.
This is an example of Charles' law, which states that the volume of a given amount of gas held at constant pressure varies directly with the temperature in Kelvins. The equation to use is
Given
Unknown
Solution
Rearrange the equation to isolate
Volume changes due to the addition of
Notice that the solution goes from being acidic at
As you know, the concentration of hydronium cations is equal to
#color(blue)(ul(color(black)([""H""_3""O""^(+)] = 10^(-""pH""))))#
This means that the initial solution contains
#[""H""_3""O""^(+)] = 10^(-3) quad ""M""#
Now, use the volume of the solution to calculate the number of moles of hydronium cations it contains.
#500 color(red)(cancel(color(black)(""mL solution""))) * (10^(-3) quad ""moles H""_3""O""^(+))/(10^3color(red)(cancel(color(black)(""mL solution"")))) = ""0.00050 moles H""_3""O""^(+)#
Sodium hydroxide and hydrochloric acid react in a
#""HCl""_ ((aq)) + ""NaOH""_ ((aq)) -> ""NaCl""_ ((aq)) + ""H""_ 2""O""_ ((l))#
so you know that in order to completely neutralize the acid and make the solution neutral, you need to add
So you can say that at
#""0.0005 moles NaOH added "" -> "" pH"" = 7#
Now, you want the target solution to have
#""pH"" = 10#
which implies that it must have
#""pOH"" = 14 -10 = 4#
This means that the concentration of hydroxide anions in the final solution must be equal to
#[""OH""^(-)] = 10^(-4) quad ""M""#
Since you are told to assume that the volume of the solution does not change upon the addition of the salt, you can use it to find the number of moles of hydroxide anions present in the final solution.
#500 color(red)(cancel(color(black)(""mL solution""))) * (10^(-4) quad ""moles OH""^(-))/(10^3color(red)(cancel(color(black)(""mL solution"")))) = ""0.000050 moles OH""^(-)#
Therefore, you can say that if you add
At that point, adding an additional
The total number of moles of sodium hydroxide that you must add--remember that sodium hydroxide delivers hydroxide anions to the solution in a
#""0.00050 moles + 0.000050 moles = 0.00055 moles NaOH""#
Finally, to convert the number of moles to grams, use the molar mass of the salt.
#0.00055 color(red)(cancel(color(black)(""moles NaOH""))) * ""40 g""/(1color(red)(cancel(color(black)(""mole NaOH"")))) = color(darkgreen)(ul(color(black)(""0.02 g"")))#
The answer is rounded to one significant figure.
Notice that the solution goes from being acidic at
As you know, the concentration of hydronium cations is equal to
#color(blue)(ul(color(black)([""H""_3""O""^(+)] = 10^(-""pH""))))#
This means that the initial solution contains
#[""H""_3""O""^(+)] = 10^(-3) quad ""M""#
Now, use the volume of the solution to calculate the number of moles of hydronium cations it contains.
#500 color(red)(cancel(color(black)(""mL solution""))) * (10^(-3) quad ""moles H""_3""O""^(+))/(10^3color(red)(cancel(color(black)(""mL solution"")))) = ""0.00050 moles H""_3""O""^(+)#
Sodium hydroxide and hydrochloric acid react in a
#""HCl""_ ((aq)) + ""NaOH""_ ((aq)) -> ""NaCl""_ ((aq)) + ""H""_ 2""O""_ ((l))#
so you know that in order to completely neutralize the acid and make the solution neutral, you need to add
So you can say that at
#""0.0005 moles NaOH added "" -> "" pH"" = 7#
Now, you want the target solution to have
#""pH"" = 10#
which implies that it must have
#""pOH"" = 14 -10 = 4#
This means that the concentration of hydroxide anions in the final solution must be equal to
#[""OH""^(-)] = 10^(-4) quad ""M""#
Since you are told to assume that the volume of the solution does not change upon the addition of the salt, you can use it to find the number of moles of hydroxide anions present in the final solution.
#500 color(red)(cancel(color(black)(""mL solution""))) * (10^(-4) quad ""moles OH""^(-))/(10^3color(red)(cancel(color(black)(""mL solution"")))) = ""0.000050 moles OH""^(-)#
Therefore, you can say that if you add
At that point, adding an additional
The total number of moles of sodium hydroxide that you must add--remember that sodium hydroxide delivers hydroxide anions to the solution in a
#""0.00050 moles + 0.000050 moles = 0.00055 moles NaOH""#
Finally, to convert the number of moles to grams, use the molar mass of the salt.
#0.00055 color(red)(cancel(color(black)(""moles NaOH""))) * ""40 g""/(1color(red)(cancel(color(black)(""mole NaOH"")))) = color(darkgreen)(ul(color(black)(""0.02 g"")))#
The answer is rounded to one significant figure.
Volume changes due to the addition of
Notice that the solution goes from being acidic at
As you know, the concentration of hydronium cations is equal to
#color(blue)(ul(color(black)([""H""_3""O""^(+)] = 10^(-""pH""))))#
This means that the initial solution contains
#[""H""_3""O""^(+)] = 10^(-3) quad ""M""#
Now, use the volume of the solution to calculate the number of moles of hydronium cations it contains.
#500 color(red)(cancel(color(black)(""mL solution""))) * (10^(-3) quad ""moles H""_3""O""^(+))/(10^3color(red)(cancel(color(black)(""mL solution"")))) = ""0.00050 moles H""_3""O""^(+)#
Sodium hydroxide and hydrochloric acid react in a
#""HCl""_ ((aq)) + ""NaOH""_ ((aq)) -> ""NaCl""_ ((aq)) + ""H""_ 2""O""_ ((l))#
so you know that in order to completely neutralize the acid and make the solution neutral, you need to add
So you can say that at
#""0.0005 moles NaOH added "" -> "" pH"" = 7#
Now, you want the target solution to have
#""pH"" = 10#
which implies that it must have
#""pOH"" = 14 -10 = 4#
This means that the concentration of hydroxide anions in the final solution must be equal to
#[""OH""^(-)] = 10^(-4) quad ""M""#
Since you are told to assume that the volume of the solution does not change upon the addition of the salt, you can use it to find the number of moles of hydroxide anions present in the final solution.
#500 color(red)(cancel(color(black)(""mL solution""))) * (10^(-4) quad ""moles OH""^(-))/(10^3color(red)(cancel(color(black)(""mL solution"")))) = ""0.000050 moles OH""^(-)#
Therefore, you can say that if you add
At that point, adding an additional
The total number of moles of sodium hydroxide that you must add--remember that sodium hydroxide delivers hydroxide anions to the solution in a
#""0.00050 moles + 0.000050 moles = 0.00055 moles NaOH""#
Finally, to convert the number of moles to grams, use the molar mass of the salt.
#0.00055 color(red)(cancel(color(black)(""moles NaOH""))) * ""40 g""/(1color(red)(cancel(color(black)(""mole NaOH"")))) = color(darkgreen)(ul(color(black)(""0.02 g"")))#
The answer is rounded to one significant figure.
The initial conditions are:
STP means:
The volume of the gas is
Since the volume (
The initial conditions are:
STP means:
The volume of the gas is
Since the volume (
The initial conditions are:
STP means:
The volume of the gas is
Since the volume (
Always start with a balanced equation.
Determine the molar mass of butane.
Molar Mass of
Determine the Mole Ratios for Butane and Carbon dioxide
Divide the given mass of butane by its molar mass (multiply by its reciprocal). This will give the moles of butane. Multiply times the molar ratio that places carbon dioxide in the numerator. This will give the moles of carbon dioxide.
Always start with a balanced equation.
Determine the molar mass of butane.
Molar Mass of
Determine the Mole Ratios for Butane and Carbon dioxide
Divide the given mass of butane by its molar mass (multiply by its reciprocal). This will give the moles of butane. Multiply times the molar ratio that places carbon dioxide in the numerator. This will give the moles of carbon dioxide.
Always start with a balanced equation.
Determine the molar mass of butane.
Molar Mass of
Determine the Mole Ratios for Butane and Carbon dioxide
Divide the given mass of butane by its molar mass (multiply by its reciprocal). This will give the moles of butane. Multiply times the molar ratio that places carbon dioxide in the numerator. This will give the moles of carbon dioxide.
In order to solve this problem, you would need the value of the solubility product constant,
In this case, I'll pick
#K_(sp) = 2.0 * 10^(-19)# .
You could approach this problem by using an ICE table (more here: http://en.wikipedia.org/wiki/RICE_chart) to help you find the molar solubility,
Since you're dealing with an insoluble ionic compound, an equilibrium will be established between the undissolved solid and the dissolved ions.
#"" "" ""LaF""_ (color(red)(3)(s)) "" ""rightleftharpoons"" "" ""La""_ ((aq))^(3+) "" ""+"" "" color(red)(3)""F""_ ((aq))^(-)#
Initially, the concentrations of the
Keep in mind that the solid's concentration is presumed to be either unknown or constant, which is why it's not relevant here.
By definition, the solubility product constant for this equilibrium will be
#K_(sp) = [""La""^(3+)] * [""F""^(-)]^color(red)(3)#
This will be equivalent to
#K_(sp) = s * (color(red)(3)s)^color(red)(3)#
#2.0 * 10^(-19) = 27s^4#
You will thus have
#s = root(4)((2.0 * 10^(-19))/27) = 9.3 * 10^(-6)#
Since
#s = 9.3 * 10^(-6)""mol L""^(-)#
In order to express the solubility in grams per liter,
#9.3 * 10^(-6) color(red)(cancel(color(black)(""mol"")))/""L"" * ""195.9 g""/(1color(red)(cancel(color(black)(""mol"")))) = color(green)(1.8 * 10^(-3) ""g L""^(-1))#
In order to solve this problem, you would need the value of the solubility product constant,
In this case, I'll pick
#K_(sp) = 2.0 * 10^(-19)# .
You could approach this problem by using an ICE table (more here: http://en.wikipedia.org/wiki/RICE_chart) to help you find the molar solubility,
Since you're dealing with an insoluble ionic compound, an equilibrium will be established between the undissolved solid and the dissolved ions.
#"" "" ""LaF""_ (color(red)(3)(s)) "" ""rightleftharpoons"" "" ""La""_ ((aq))^(3+) "" ""+"" "" color(red)(3)""F""_ ((aq))^(-)#
Initially, the concentrations of the
Keep in mind that the solid's concentration is presumed to be either unknown or constant, which is why it's not relevant here.
By definition, the solubility product constant for this equilibrium will be
#K_(sp) = [""La""^(3+)] * [""F""^(-)]^color(red)(3)#
This will be equivalent to
#K_(sp) = s * (color(red)(3)s)^color(red)(3)#
#2.0 * 10^(-19) = 27s^4#
You will thus have
#s = root(4)((2.0 * 10^(-19))/27) = 9.3 * 10^(-6)#
Since
#s = 9.3 * 10^(-6)""mol L""^(-)#
In order to express the solubility in grams per liter,
#9.3 * 10^(-6) color(red)(cancel(color(black)(""mol"")))/""L"" * ""195.9 g""/(1color(red)(cancel(color(black)(""mol"")))) = color(green)(1.8 * 10^(-3) ""g L""^(-1))#
In order to solve this problem, you would need the value of the solubility product constant,
In this case, I'll pick
#K_(sp) = 2.0 * 10^(-19)# .
You could approach this problem by using an ICE table (more here: http://en.wikipedia.org/wiki/RICE_chart) to help you find the molar solubility,
Since you're dealing with an insoluble ionic compound, an equilibrium will be established between the undissolved solid and the dissolved ions.
#"" "" ""LaF""_ (color(red)(3)(s)) "" ""rightleftharpoons"" "" ""La""_ ((aq))^(3+) "" ""+"" "" color(red)(3)""F""_ ((aq))^(-)#
Initially, the concentrations of the
Keep in mind that the solid's concentration is presumed to be either unknown or constant, which is why it's not relevant here.
By definition, the solubility product constant for this equilibrium will be
#K_(sp) = [""La""^(3+)] * [""F""^(-)]^color(red)(3)#
This will be equivalent to
#K_(sp) = s * (color(red)(3)s)^color(red)(3)#
#2.0 * 10^(-19) = 27s^4#
You will thus have
#s = root(4)((2.0 * 10^(-19))/27) = 9.3 * 10^(-6)#
Since
#s = 9.3 * 10^(-6)""mol L""^(-)#
In order to express the solubility in grams per liter,
#9.3 * 10^(-6) color(red)(cancel(color(black)(""mol"")))/""L"" * ""195.9 g""/(1color(red)(cancel(color(black)(""mol"")))) = color(green)(1.8 * 10^(-3) ""g L""^(-1))#
In order to be able to answer this question, you need to have the solubility graph of potassium chloride,
Since the solubility of potassium chloride is given per
Keep in mind that molality is defined as moles of solute, which in your case is potassium chloride, divided by kilograms** of solvent, which in your case is water.
#color(blue)(b = n_""solute""/m_""solvent"")#
This means that you have
#n_""solute"" = b * m_""solvent""#
#n_""solute"" = ""3.5 mol"" color(red)(cancel(color(black)(""kg""^(-1)))) * 100 * 10^(-3)color(red)(cancel(color(black)(""kg""))) = ""0.35 moles KCl""#
Next, use potassium chloride's molar mass to figure out how many grams would contain this many moles
#0.35color(red)(cancel(color(black)(""moles KCl""))) * ""74.5 g""/(1color(red)(cancel(color(black)(""mole KCl"")))) = ""26.1 g""#
Now take a look at the solubility graph and decide at which temperature dissolving
Practically speaking, you're looking for the temperature at which the saturation line, drawn on the graph in
From the look of it, dissolving this much potassium chloride per
In order to be able to answer this question, you need to have the solubility graph of potassium chloride,
Since the solubility of potassium chloride is given per
Keep in mind that molality is defined as moles of solute, which in your case is potassium chloride, divided by kilograms** of solvent, which in your case is water.
#color(blue)(b = n_""solute""/m_""solvent"")#
This means that you have
#n_""solute"" = b * m_""solvent""#
#n_""solute"" = ""3.5 mol"" color(red)(cancel(color(black)(""kg""^(-1)))) * 100 * 10^(-3)color(red)(cancel(color(black)(""kg""))) = ""0.35 moles KCl""#
Next, use potassium chloride's molar mass to figure out how many grams would contain this many moles
#0.35color(red)(cancel(color(black)(""moles KCl""))) * ""74.5 g""/(1color(red)(cancel(color(black)(""mole KCl"")))) = ""26.1 g""#
Now take a look at the solubility graph and decide at which temperature dissolving
Practically speaking, you're looking for the temperature at which the saturation line, drawn on the graph in
From the look of it, dissolving this much potassium chloride per
In order to be able to answer this question, you need to have the solubility graph of potassium chloride,
Since the solubility of potassium chloride is given per
Keep in mind that molality is defined as moles of solute, which in your case is potassium chloride, divided by kilograms** of solvent, which in your case is water.
#color(blue)(b = n_""solute""/m_""solvent"")#
This means that you have
#n_""solute"" = b * m_""solvent""#
#n_""solute"" = ""3.5 mol"" color(red)(cancel(color(black)(""kg""^(-1)))) * 100 * 10^(-3)color(red)(cancel(color(black)(""kg""))) = ""0.35 moles KCl""#
Next, use potassium chloride's molar mass to figure out how many grams would contain this many moles
#0.35color(red)(cancel(color(black)(""moles KCl""))) * ""74.5 g""/(1color(red)(cancel(color(black)(""mole KCl"")))) = ""26.1 g""#
Now take a look at the solubility graph and decide at which temperature dissolving
Practically speaking, you're looking for the temperature at which the saturation line, drawn on the graph in
From the look of it, dissolving this much potassium chloride per
We have:
as a balanced chemical reaction.
I see you want to know the amount
Consider the graphic below:
We have:
as a balanced chemical reaction.
I see you want to know the amount
Consider the graphic below:
We have:
as a balanced chemical reaction.
I see you want to know the amount
Consider the graphic below:
Change the percentages to grams. If there was 100 grams then Ca would have 18.29 grams Chorine would have 32.37 grams and 49.34 grams of water.
The grams can be changed to moles by dividing the number of grams by the grams per mole that you can find from the periodic table.
Now that the moles of each element ( compound) is known it is possible to find a whole number ratio of the elements and compounds giving an empirical formula.
So the empirical formula is
Ca = +2 Cl = -1 so
the empirical formula is also the molecular formula.
Change the percentages to grams. If there was 100 grams then Ca would have 18.29 grams Chorine would have 32.37 grams and 49.34 grams of water.
The grams can be changed to moles by dividing the number of grams by the grams per mole that you can find from the periodic table.
Now that the moles of each element ( compound) is known it is possible to find a whole number ratio of the elements and compounds giving an empirical formula.
So the empirical formula is
Ca = +2 Cl = -1 so
the empirical formula is also the molecular formula.
Change the percentages to grams. If there was 100 grams then Ca would have 18.29 grams Chorine would have 32.37 grams and 49.34 grams of water.
The grams can be changed to moles by dividing the number of grams by the grams per mole that you can find from the periodic table.
Now that the moles of each element ( compound) is known it is possible to find a whole number ratio of the elements and compounds giving an empirical formula.
So the empirical formula is
Ca = +2 Cl = -1 so
the empirical formula is also the molecular formula.
And all we did here was to apply the definition of
Would this solution be slightly acidic or slightly basic........?
And all we did here was to apply the definition of
Would this solution be slightly acidic or slightly basic........?
And all we did here was to apply the definition of
Would this solution be slightly acidic or slightly basic........?
Unbalanced Equation
Balance the Cl
There are 2 Cl atoms on the left side and 1 Cl atom on the right side.
Add a coefficient of 2 in front of KCl on the right side.
There are now 2 Cl atoms on both sides.
Balance the Br
There are 2 Br atoms on the right side and 1 Br atom on the left side. Add a coefficient of 2 in front of KBr.
The equation is now balanced with 2 Cl atoms, 2 K atoms, and 2 Br atoms on both sides of the equation.
Note: You can substitute the words atom or atoms by mole or moles.
Unbalanced Equation
Balance the Cl
There are 2 Cl atoms on the left side and 1 Cl atom on the right side.
Add a coefficient of 2 in front of KCl on the right side.
There are now 2 Cl atoms on both sides.
Balance the Br
There are 2 Br atoms on the right side and 1 Br atom on the left side. Add a coefficient of 2 in front of KBr.
The equation is now balanced with 2 Cl atoms, 2 K atoms, and 2 Br atoms on both sides of the equation.
Note: You can substitute the words atom or atoms by mole or moles.
Unbalanced Equation
Balance the Cl
There are 2 Cl atoms on the left side and 1 Cl atom on the right side.
Add a coefficient of 2 in front of KCl on the right side.
There are now 2 Cl atoms on both sides.
Balance the Br
There are 2 Br atoms on the right side and 1 Br atom on the left side. Add a coefficient of 2 in front of KBr.
The equation is now balanced with 2 Cl atoms, 2 K atoms, and 2 Br atoms on both sides of the equation.
Note: You can substitute the words atom or atoms by mole or moles.
Since you're dealing with a non-volatile solute, the vapor pressure of the solution will depend exclusively on the mole fraction of the solvent and on the vapor pressure of the pure solvent at that temperature.
Mathematically, the relationship between the mole fraction of the solvent and the vapor pressure of the solution looks like this
#color(blue)(P_""sol"" = chi_""solvent"" * P_""solvent""^@)"" ""# , where
So, your goal here is to figure out how many moles of sodium chloride and of water you have in the solution.
To do that, use the two compounds' respective molar masses. For sodium chloride you'll have
#150color(red)(cancel(color(black)(""g""))) * ""1 mole NaCl""/(58.44color(red)(cancel(color(black)(""g"")))) = ""2.567 moles NaCl""#
For water you'll have
#1.00color(red)(cancel(color(black)(""kg""))) * (1000color(red)(cancel(color(black)(""g""))))/(1color(red)(cancel(color(black)(""kg"")))) * (""1 mole H""_2""O"")/(18.015color(red)(cancel(color(black)(""g"")))) = ""55.51 moles H""_2""O""#
Now, the mole fraction of water is equal to the number of moles of water divided by the total number of moles present in the solution.
#n_""total"" = n_""water"" + n_""NaCl""#
#n_""total"" = 55.51 + 2.567 = ""58.08 moles""#
The mole faction of water will thus be
#chi_""water"" = (55.51color(red)(cancel(color(black)(""moles""))))/(58.08color(red)(cancel(color(black)(""moles"")))) = 0.956#
This means that the vapor pressure of the solution will be
#P_""sol"" = 0.956 * ""23.8 torr"" = color(green)(""22.8 torr"")#
I'll leave the answer rounded to three sig figs, despite the fact that you only have two sig figs for the mass of sodium chloride.
Since you're dealing with a non-volatile solute, the vapor pressure of the solution will depend exclusively on the mole fraction of the solvent and on the vapor pressure of the pure solvent at that temperature.
Mathematically, the relationship between the mole fraction of the solvent and the vapor pressure of the solution looks like this
#color(blue)(P_""sol"" = chi_""solvent"" * P_""solvent""^@)"" ""# , where
So, your goal here is to figure out how many moles of sodium chloride and of water you have in the solution.
To do that, use the two compounds' respective molar masses. For sodium chloride you'll have
#150color(red)(cancel(color(black)(""g""))) * ""1 mole NaCl""/(58.44color(red)(cancel(color(black)(""g"")))) = ""2.567 moles NaCl""#
For water you'll have
#1.00color(red)(cancel(color(black)(""kg""))) * (1000color(red)(cancel(color(black)(""g""))))/(1color(red)(cancel(color(black)(""kg"")))) * (""1 mole H""_2""O"")/(18.015color(red)(cancel(color(black)(""g"")))) = ""55.51 moles H""_2""O""#
Now, the mole fraction of water is equal to the number of moles of water divided by the total number of moles present in the solution.
#n_""total"" = n_""water"" + n_""NaCl""#
#n_""total"" = 55.51 + 2.567 = ""58.08 moles""#
The mole faction of water will thus be
#chi_""water"" = (55.51color(red)(cancel(color(black)(""moles""))))/(58.08color(red)(cancel(color(black)(""moles"")))) = 0.956#
This means that the vapor pressure of the solution will be
#P_""sol"" = 0.956 * ""23.8 torr"" = color(green)(""22.8 torr"")#
I'll leave the answer rounded to three sig figs, despite the fact that you only have two sig figs for the mass of sodium chloride.
Since you're dealing with a non-volatile solute, the vapor pressure of the solution will depend exclusively on the mole fraction of the solvent and on the vapor pressure of the pure solvent at that temperature.
Mathematically, the relationship between the mole fraction of the solvent and the vapor pressure of the solution looks like this
#color(blue)(P_""sol"" = chi_""solvent"" * P_""solvent""^@)"" ""# , where
So, your goal here is to figure out how many moles of sodium chloride and of water you have in the solution.
To do that, use the two compounds' respective molar masses. For sodium chloride you'll have
#150color(red)(cancel(color(black)(""g""))) * ""1 mole NaCl""/(58.44color(red)(cancel(color(black)(""g"")))) = ""2.567 moles NaCl""#
For water you'll have
#1.00color(red)(cancel(color(black)(""kg""))) * (1000color(red)(cancel(color(black)(""g""))))/(1color(red)(cancel(color(black)(""kg"")))) * (""1 mole H""_2""O"")/(18.015color(red)(cancel(color(black)(""g"")))) = ""55.51 moles H""_2""O""#
Now, the mole fraction of water is equal to the number of moles of water divided by the total number of moles present in the solution.
#n_""total"" = n_""water"" + n_""NaCl""#
#n_""total"" = 55.51 + 2.567 = ""58.08 moles""#
The mole faction of water will thus be
#chi_""water"" = (55.51color(red)(cancel(color(black)(""moles""))))/(58.08color(red)(cancel(color(black)(""moles"")))) = 0.956#
This means that the vapor pressure of the solution will be
#P_""sol"" = 0.956 * ""23.8 torr"" = color(green)(""22.8 torr"")#
I'll leave the answer rounded to three sig figs, despite the fact that you only have two sig figs for the mass of sodium chloride.
We divide thru by the lowest molar quantity to give an empirical formula of
Now the molecular formula is always a multiple of the the empirical formula:
So using the quoted molecular mass:
Clearly,
Dinitrogen pentoxide,
We divide thru by the lowest molar quantity to give an empirical formula of
Now the molecular formula is always a multiple of the the empirical formula:
So using the quoted molecular mass:
Clearly,
Dinitrogen pentoxide,
We divide thru by the lowest molar quantity to give an empirical formula of
Now the molecular formula is always a multiple of the the empirical formula:
So using the quoted molecular mass:
Clearly,
Lead sulfide occurs in the mineral,
The given reaction illustrates conservation of mass and conservation of charge, as indeed it must if it reflects chemical reality. Why?
Lead sulfide occurs in the mineral,
The given reaction illustrates conservation of mass and conservation of charge, as indeed it must if it reflects chemical reality. Why?
Lead sulfide occurs in the mineral,
The given reaction illustrates conservation of mass and conservation of charge, as indeed it must if it reflects chemical reality. Why?
How did I know this? Well, I have access to a Periodic Table, and you should have access to one as well.
The Periodic Table tells us that the molar mass of oxygen is
What do I mean by molar mass?
How did I know this? Well, I have access to a Periodic Table, and you should have access to one as well.
The Periodic Table tells us that the molar mass of oxygen is
What do I mean by molar mass?
How did I know this? Well, I have access to a Periodic Table, and you should have access to one as well.
The Periodic Table tells us that the molar mass of oxygen is
What do I mean by molar mass?
Let's define those three terms before we get started:
The percent yield of a reaction is the ratio of actual yield to the theoretical yield times one-hundred percent.
The actual yield is the mass of product that was obtained in an experiment.
The theoretical yield is the mass of product calculated from the amounts of reactants used in the experiment; the amount of product that you should get in theory.
Now, to calculate the percent yield we use the equation below:
We know the actual yield and the theoretical yield, so all we have to do is divide the numbers and multiply that value by
Thus, the percent yield is
The percent yield is
Let's define those three terms before we get started:
The percent yield of a reaction is the ratio of actual yield to the theoretical yield times one-hundred percent.
The actual yield is the mass of product that was obtained in an experiment.
The theoretical yield is the mass of product calculated from the amounts of reactants used in the experiment; the amount of product that you should get in theory.
Now, to calculate the percent yield we use the equation below:
We know the actual yield and the theoretical yield, so all we have to do is divide the numbers and multiply that value by
Thus, the percent yield is
The percent yield is
Let's define those three terms before we get started:
The percent yield of a reaction is the ratio of actual yield to the theoretical yield times one-hundred percent.
The actual yield is the mass of product that was obtained in an experiment.
The theoretical yield is the mass of product calculated from the amounts of reactants used in the experiment; the amount of product that you should get in theory.
Now, to calculate the percent yield we use the equation below:
We know the actual yield and the theoretical yield, so all we have to do is divide the numbers and multiply that value by
Thus, the percent yield is
A compound's empirical formula tells you what the smallest whole number ratio between the elements that make up that compound is.
In your case, you know that this ionic compound contains
#0.012# moles of sodium#0.012# moles of sulfur#0.018# moles of oxygen
To get the mole ratio that exists between these elements in the compound, divide these numbers by the smallest one
#""For Na: "" (0.012color(red)(cancel(color(black)(""moles""))))/(0.012color(red)(cancel(color(black)(""moles"")))) = 1#
#""For S: "" (0.012color(red)(cancel(color(black)(""moles""))))/(0.012color(red)(cancel(color(black)(""moles"")))) = 1#
#""For O: "" (0.018color(red)(cancel(color(black)(""moles""))))/(0.012color(red)(cancel(color(black)(""moles"")))) = 1.5#
Now, since you can't have fractional numbers as mole ratios, you will need to find the smallest whole number ratio that satisfies the condition
Notice that if you multiply all the numbers by
#(""Na""_1""S""_1""O""_1.5)_2 implies ""Na""_2""S""_2""O""_3#
For ionic compounds, the empirical formula is always the same as the chemical formula, which is why you cn say that your unknown compound is actually sodium thiosulfate,
A compound's empirical formula tells you what the smallest whole number ratio between the elements that make up that compound is.
In your case, you know that this ionic compound contains
#0.012# moles of sodium#0.012# moles of sulfur#0.018# moles of oxygen
To get the mole ratio that exists between these elements in the compound, divide these numbers by the smallest one
#""For Na: "" (0.012color(red)(cancel(color(black)(""moles""))))/(0.012color(red)(cancel(color(black)(""moles"")))) = 1#
#""For S: "" (0.012color(red)(cancel(color(black)(""moles""))))/(0.012color(red)(cancel(color(black)(""moles"")))) = 1#
#""For O: "" (0.018color(red)(cancel(color(black)(""moles""))))/(0.012color(red)(cancel(color(black)(""moles"")))) = 1.5#
Now, since you can't have fractional numbers as mole ratios, you will need to find the smallest whole number ratio that satisfies the condition
Notice that if you multiply all the numbers by
#(""Na""_1""S""_1""O""_1.5)_2 implies ""Na""_2""S""_2""O""_3#
For ionic compounds, the empirical formula is always the same as the chemical formula, which is why you cn say that your unknown compound is actually sodium thiosulfate,
A compound's empirical formula tells you what the smallest whole number ratio between the elements that make up that compound is.
In your case, you know that this ionic compound contains
#0.012# moles of sodium#0.012# moles of sulfur#0.018# moles of oxygen
To get the mole ratio that exists between these elements in the compound, divide these numbers by the smallest one
#""For Na: "" (0.012color(red)(cancel(color(black)(""moles""))))/(0.012color(red)(cancel(color(black)(""moles"")))) = 1#
#""For S: "" (0.012color(red)(cancel(color(black)(""moles""))))/(0.012color(red)(cancel(color(black)(""moles"")))) = 1#
#""For O: "" (0.018color(red)(cancel(color(black)(""moles""))))/(0.012color(red)(cancel(color(black)(""moles"")))) = 1.5#
Now, since you can't have fractional numbers as mole ratios, you will need to find the smallest whole number ratio that satisfies the condition
Notice that if you multiply all the numbers by
#(""Na""_1""S""_1""O""_1.5)_2 implies ""Na""_2""S""_2""O""_3#
For ionic compounds, the empirical formula is always the same as the chemical formula, which is why you cn say that your unknown compound is actually sodium thiosulfate,
The thing to remember about any solution is that the particles of solute and the particles of solvent are evenly mixed.
This essentially means that if you start with a solution of known molarity and take out a sample of this solution, the molarity of the sample will be the same as the molarity of the initial solution.
In your case, you dissolve
If you take
#100 color(red)(cancel(color(black)(""mL""))) * ""1 L""/(10^3color(red)(cancel(color(black)(""mL"")))) = ""0.1 L""#
of this solution, this sample must have the same concentration as the initial solution, i.e.
Notice that the volume of the sample is
#(1color(red)(cancel(color(black)(""L""))))/(0.1color(red)(cancel(color(black)(""L"")))) = color(blue)(10)#
times smaller than the volume of the initial solution, so it follows that it must contain
Sodium chloride was said to have a molar mass of
Since the initial solution was made by dissolving the equivalent of
#""58.44 g "" = "" 1 mole NaCl""#
in
#""1 mole NaCl""/color(blue)(10) = color(green)(|bar(ul(color(white)(a/a)color(black)(""0.1 moles NaCl"")color(white)(a/a)|)))#
This is equivalent to saying that the
#""58.44 g NaCl""/color(blue)(10) = ""5.844 g NaCl""#
ALTERNATIVELY
You can get the same result by using the formula for molarity, which is
#color(blue)(|bar(ul(color(white)(a/a)c = n_""solute""/V_""solution""color(white)(a/a)|)))#
Rearrange this equation to solve for
#c = n_""solute""/V_""solution"" implies n_""solute"" = c * V_""solution""#
#n_""NaCl"" = ""1.0 mol"" color(red)(cancel(color(black)(""L""^(-1)))) * overbrace(100 * 10^(-3)color(red)(cancel(color(black)(""L""))))^(color(blue)(""volume in liters"")) = color(green)(|bar(ul(color(white)(a/a)color(black)(""0.1 moles NaCl"")color(white)(a/a)|)))#
Keep in mind that the volume must always be expressed in liters when working with molarity.
The thing to remember about any solution is that the particles of solute and the particles of solvent are evenly mixed.
This essentially means that if you start with a solution of known molarity and take out a sample of this solution, the molarity of the sample will be the same as the molarity of the initial solution.
In your case, you dissolve
If you take
#100 color(red)(cancel(color(black)(""mL""))) * ""1 L""/(10^3color(red)(cancel(color(black)(""mL"")))) = ""0.1 L""#
of this solution, this sample must have the same concentration as the initial solution, i.e.
Notice that the volume of the sample is
#(1color(red)(cancel(color(black)(""L""))))/(0.1color(red)(cancel(color(black)(""L"")))) = color(blue)(10)#
times smaller than the volume of the initial solution, so it follows that it must contain
Sodium chloride was said to have a molar mass of
Since the initial solution was made by dissolving the equivalent of
#""58.44 g "" = "" 1 mole NaCl""#
in
#""1 mole NaCl""/color(blue)(10) = color(green)(|bar(ul(color(white)(a/a)color(black)(""0.1 moles NaCl"")color(white)(a/a)|)))#
This is equivalent to saying that the
#""58.44 g NaCl""/color(blue)(10) = ""5.844 g NaCl""#
ALTERNATIVELY
You can get the same result by using the formula for molarity, which is
#color(blue)(|bar(ul(color(white)(a/a)c = n_""solute""/V_""solution""color(white)(a/a)|)))#
Rearrange this equation to solve for
#c = n_""solute""/V_""solution"" implies n_""solute"" = c * V_""solution""#
#n_""NaCl"" = ""1.0 mol"" color(red)(cancel(color(black)(""L""^(-1)))) * overbrace(100 * 10^(-3)color(red)(cancel(color(black)(""L""))))^(color(blue)(""volume in liters"")) = color(green)(|bar(ul(color(white)(a/a)color(black)(""0.1 moles NaCl"")color(white)(a/a)|)))#
Keep in mind that the volume must always be expressed in liters when working with molarity.
The thing to remember about any solution is that the particles of solute and the particles of solvent are evenly mixed.
This essentially means that if you start with a solution of known molarity and take out a sample of this solution, the molarity of the sample will be the same as the molarity of the initial solution.
In your case, you dissolve
If you take
#100 color(red)(cancel(color(black)(""mL""))) * ""1 L""/(10^3color(red)(cancel(color(black)(""mL"")))) = ""0.1 L""#
of this solution, this sample must have the same concentration as the initial solution, i.e.
Notice that the volume of the sample is
#(1color(red)(cancel(color(black)(""L""))))/(0.1color(red)(cancel(color(black)(""L"")))) = color(blue)(10)#
times smaller than the volume of the initial solution, so it follows that it must contain
Sodium chloride was said to have a molar mass of
Since the initial solution was made by dissolving the equivalent of
#""58.44 g "" = "" 1 mole NaCl""#
in
#""1 mole NaCl""/color(blue)(10) = color(green)(|bar(ul(color(white)(a/a)color(black)(""0.1 moles NaCl"")color(white)(a/a)|)))#
This is equivalent to saying that the
#""58.44 g NaCl""/color(blue)(10) = ""5.844 g NaCl""#
ALTERNATIVELY
You can get the same result by using the formula for molarity, which is
#color(blue)(|bar(ul(color(white)(a/a)c = n_""solute""/V_""solution""color(white)(a/a)|)))#
Rearrange this equation to solve for
#c = n_""solute""/V_""solution"" implies n_""solute"" = c * V_""solution""#
#n_""NaCl"" = ""1.0 mol"" color(red)(cancel(color(black)(""L""^(-1)))) * overbrace(100 * 10^(-3)color(red)(cancel(color(black)(""L""))))^(color(blue)(""volume in liters"")) = color(green)(|bar(ul(color(white)(a/a)color(black)(""0.1 moles NaCl"")color(white)(a/a)|)))#
Keep in mind that the volume must always be expressed in liters when working with molarity.
A mole is defined as the amount of a substance whose mass in grams equals the atomic, molecular, or formula-unit mass in atomic mass units.
Example 1: Water (
Example 2: Beryllium has an aromuc mass of
A mole is defined as the amount of a substance whose mass in grams equals the atomic, molecular, or formula-unit mass in atomic mass units.
Example 1: Water (
Example 2: Beryllium has an aromuc mass of
A mole is defined as the amount of a substance whose mass in grams equals the atomic, molecular, or formula-unit mass in atomic mass units.
Example 1: Water (
Example 2: Beryllium has an aromuc mass of
Acetic acid:
To determine the percent composition of carbon in acetic acid, divide the molar mass of acetic acid
The percent composition of carbon in acetic acid is
Acetic acid:
To determine the percent composition of carbon in acetic acid, divide the molar mass of acetic acid
The percent composition of carbon in acetic acid is
Acetic acid:
To determine the percent composition of carbon in acetic acid, divide the molar mass of acetic acid
If the formula of a weak acid be reprsented as HA.then its dissociation in aqueous medium can be written as follows:
#HA"" ""+H_2O""""rightleftharpoons""""H_3O^+""""+"" ""A^-#
Where
If cM be the initial concentration of the weak acid then the concentrations different species at equilibrium will be as follows
#[HA]=c(1-alpha)M#
#[H_3O^+]=calphaM#
#[A^-]=calphaM#
The concentration of
Now the acid dissociation constant
#K_a=([H_3O^+][A^-])/[HA]^2#
#=((calpha)*(calpha))/(c(1-alpha))#
#=(calpha^2)/(1-alpha)#
Given
#K_a=(6.03xx10^-6*(0.2)^2)/(1-0.2)#
#=(6.03xx10^-6xx0.04)/0.8#
#3.015xx10^-7#
The
#=7-log3.015=6.52#
If the formula of a weak acid be reprsented as HA.then its dissociation in aqueous medium can be written as follows:
#HA"" ""+H_2O""""rightleftharpoons""""H_3O^+""""+"" ""A^-#
Where
If cM be the initial concentration of the weak acid then the concentrations different species at equilibrium will be as follows
#[HA]=c(1-alpha)M#
#[H_3O^+]=calphaM#
#[A^-]=calphaM#
The concentration of
Now the acid dissociation constant
#K_a=([H_3O^+][A^-])/[HA]^2#
#=((calpha)*(calpha))/(c(1-alpha))#
#=(calpha^2)/(1-alpha)#
Given
#K_a=(6.03xx10^-6*(0.2)^2)/(1-0.2)#
#=(6.03xx10^-6xx0.04)/0.8#
#3.015xx10^-7#
The
#=7-log3.015=6.52#
If the formula of a weak acid be reprsented as HA.then its dissociation in aqueous medium can be written as follows:
#HA"" ""+H_2O""""rightleftharpoons""""H_3O^+""""+"" ""A^-#
Where
If cM be the initial concentration of the weak acid then the concentrations different species at equilibrium will be as follows
#[HA]=c(1-alpha)M#
#[H_3O^+]=calphaM#
#[A^-]=calphaM#
The concentration of
Now the acid dissociation constant
#K_a=([H_3O^+][A^-])/[HA]^2#
#=((calpha)*(calpha))/(c(1-alpha))#
#=(calpha^2)/(1-alpha)#
Given
#K_a=(6.03xx10^-6*(0.2)^2)/(1-0.2)#
#=(6.03xx10^-6xx0.04)/0.8#
#3.015xx10^-7#
The
#=7-log3.015=6.52#
So, to balance a chemical equation we make the number of each element the same on both sides.
We have:
LHS
Mg= 1
H= 1
Cl= 1
RHS
Mg= 1
Cl=2
H=2
To make Cl and H have 2 on both sides, we place a 2 in front of the HCl
LHS
Mg=1, H=2, Cl=2
RHS
Mg=1, H=1, Cl=2
Thus the equation is balanced and the coefficient of Mg is 1, which is understood so it is not written in the final equation.
1 is the coefficient. See below.
So, to balance a chemical equation we make the number of each element the same on both sides.
We have:
LHS
Mg= 1
H= 1
Cl= 1
RHS
Mg= 1
Cl=2
H=2
To make Cl and H have 2 on both sides, we place a 2 in front of the HCl
LHS
Mg=1, H=2, Cl=2
RHS
Mg=1, H=1, Cl=2
Thus the equation is balanced and the coefficient of Mg is 1, which is understood so it is not written in the final equation.
1 is the coefficient. See below.
So, to balance a chemical equation we make the number of each element the same on both sides.
We have:
LHS
Mg= 1
H= 1
Cl= 1
RHS
Mg= 1
Cl=2
H=2
To make Cl and H have 2 on both sides, we place a 2 in front of the HCl
LHS
Mg=1, H=2, Cl=2
RHS
Mg=1, H=1, Cl=2
Thus the equation is balanced and the coefficient of Mg is 1, which is understood so it is not written in the final equation.
The zinc ion will have a charge of +2 and the bromide ion has a charge of -1 because it is a halogen.
Combining the ions in a 1:2 ratio produces a compound which is electrically neutral.
Zinc can sometimes have other oxidation states (+1 for example) but these are pretty rare...
This video discusses some more examples of how to figure out a formula for a compound from its name or vice versa.
Hope this helps!
The zinc ion will have a charge of +2 and the bromide ion has a charge of -1 because it is a halogen.
Combining the ions in a 1:2 ratio produces a compound which is electrically neutral.
Zinc can sometimes have other oxidation states (+1 for example) but these are pretty rare...
This video discusses some more examples of how to figure out a formula for a compound from its name or vice versa.
Hope this helps!
The zinc ion will have a charge of +2 and the bromide ion has a charge of -1 because it is a halogen.
Combining the ions in a 1:2 ratio produces a compound which is electrically neutral.
Zinc can sometimes have other oxidation states (+1 for example) but these are pretty rare...
This video discusses some more examples of how to figure out a formula for a compound from its name or vice versa.
Hope this helps!
I am a bit cagy when I quote these values, because there are different standards in each different syllabus......
AS far as I know,
And thus............
Well,
I am a bit cagy when I quote these values, because there are different standards in each different syllabus......
AS far as I know,
And thus............
Well,
I am a bit cagy when I quote these values, because there are different standards in each different syllabus......
AS far as I know,
And thus............
(or
We're asked to find the mass, in
Chemists often use this fact: One mole of an (ideal) gas at standard temperature and pressure occupies a volume of
(The definition of standard temperature and pressure varies slightly depending on usage, but you'll be using this fact most of the time.)
With that being said, we can use the conversion factor
to convert the given
Now that we know the number of moles of
Which, if you wish to round to one significant figure, is simply
If silver ion concentration is
Under
If silver ion concentration is
Under
If silver ion concentration is
You need a stoichiometrically balanced equation to represent the combustion:
So are charge and mass balanced here?
Clearly, the reaction specifies 3 equiv of carbon dioxide per equiv of propane. If we start with 20 molecules of propane, these will necessarily be 60 carbon dioxide product molecules upon complete combustion.
You need a stoichiometrically balanced equation to represent the combustion:
So are charge and mass balanced here?
Clearly, the reaction specifies 3 equiv of carbon dioxide per equiv of propane. If we start with 20 molecules of propane, these will necessarily be 60 carbon dioxide product molecules upon complete combustion.
You need a stoichiometrically balanced equation to represent the combustion:
So are charge and mass balanced here?
Clearly, the reaction specifies 3 equiv of carbon dioxide per equiv of propane. If we start with 20 molecules of propane, these will necessarily be 60 carbon dioxide product molecules upon complete combustion.
The initial strength of the solution is 15% m/m. [Mass/Mass]
That means, the salt present in the solution is =
The water present =
If we evaporate some of the water off the solution, the salt should not change its mass, unless it decomposes or something like that.
Let's assume that we have evaporated
According to the condition,
That means 28.5 kg of water needs to be evaporated.
28.5 kg
The initial strength of the solution is 15% m/m. [Mass/Mass]
That means, the salt present in the solution is =
The water present =
If we evaporate some of the water off the solution, the salt should not change its mass, unless it decomposes or something like that.
Let's assume that we have evaporated
According to the condition,
That means 28.5 kg of water needs to be evaporated.
28.5 kg
The initial strength of the solution is 15% m/m. [Mass/Mass]
That means, the salt present in the solution is =
The water present =
If we evaporate some of the water off the solution, the salt should not change its mass, unless it decomposes or something like that.
Let's assume that we have evaporated
According to the condition,
That means 28.5 kg of water needs to be evaporated.
Convert 200 mg to g.
Convert carats to grams to moles.
Convert moles C to atoms C.
Convert 200 mg to g.
Convert carats to grams to moles.
Convert moles C to atoms C.
Convert 200 mg to g.
Convert carats to grams to moles.
Convert moles C to atoms C.
The 'specific heat of water' is
That means it takes
In this case, we have a mass,
Over all, then, the change in energy of the water,
We use the equation
The 'specific heat of water' is
That means it takes
In this case, we have a mass,
Over all, then, the change in energy of the water,
We use the equation
The 'specific heat of water' is
That means it takes
In this case, we have a mass,
Over all, then, the change in energy of the water,
Convert 10 grams to pounds:
Convert 50 degrees C to Fahrenheit:
Convert 40 degrees C to Fahrenheit:
Calculate the change in temperature:
Heat lost in British Thermal Units (BTU):
0.396 BTU
Convert 10 grams to pounds:
Convert 50 degrees C to Fahrenheit:
Convert 40 degrees C to Fahrenheit:
Calculate the change in temperature:
Heat lost in British Thermal Units (BTU):
0.396 BTU
Convert 10 grams to pounds:
Convert 50 degrees C to Fahrenheit:
Convert 40 degrees C to Fahrenheit:
Calculate the change in temperature:
Heat lost in British Thermal Units (BTU):
The process will give off 420 J of heat.
The energy
#color(blue)(bar(ul(|color(white)(a/a)q = mcΔTcolor(white)(a/a)|)))"" ""#
where
In your problem,
∴
The negative sign tells you that energy is being given off.
If there are
If there are
If there are
From the periodic table, the molar mass of Sb is
Since
Ksp = (Na^+) x (Cl^-) = 38,65
but (Na^+) = (Cl^-) = sqrt (38,65) = 6,2 mol /L
g/L = mol/L x g/mol = 6,2 x 58,45 = 363 g/L
in 2 liters = 726 g
726 g
Ksp = (Na^+) x (Cl^-) = 38,65
but (Na^+) = (Cl^-) = sqrt (38,65) = 6,2 mol /L
g/L = mol/L x g/mol = 6,2 x 58,45 = 363 g/L
in 2 liters = 726 g
726 g
Ksp = (Na^+) x (Cl^-) = 38,65
but (Na^+) = (Cl^-) = sqrt (38,65) = 6,2 mol /L
g/L = mol/L x g/mol = 6,2 x 58,45 = 363 g/L
in 2 liters = 726 g
Use the equation:
where
The specific heat of water,
https://water.usgs.gov/edu/heat-capacity.html
Organize the Information
Given/Known
Unknown:
Solution
Rearrange the equation to isolate
The temperature change was
Use the equation:
where
The specific heat of water,
https://water.usgs.gov/edu/heat-capacity.html
Organize the Information
Given/Known
Unknown:
Solution
Rearrange the equation to isolate
The temperature change was
Use the equation:
where
The specific heat of water,
https://water.usgs.gov/edu/heat-capacity.html
Organize the Information
Given/Known
Unknown:
Solution
Rearrange the equation to isolate
Here's your unbalanced equation:
Begin by fixing the obvious. We need 3
This solves our mercury problem, now on to phosphate. There is one phosphate ion on the left and two on right, so add a coefficient of two to
The only thing left to the balance is the water. On the left side, we have a total of 12
Always double check your answer to make sure all the elements balance. In this case, they do, so we are done.
Here's your unbalanced equation:
Begin by fixing the obvious. We need 3
This solves our mercury problem, now on to phosphate. There is one phosphate ion on the left and two on right, so add a coefficient of two to
The only thing left to the balance is the water. On the left side, we have a total of 12
Always double check your answer to make sure all the elements balance. In this case, they do, so we are done.
Here's your unbalanced equation:
Begin by fixing the obvious. We need 3
This solves our mercury problem, now on to phosphate. There is one phosphate ion on the left and two on right, so add a coefficient of two to
The only thing left to the balance is the water. On the left side, we have a total of 12
Always double check your answer to make sure all the elements balance. In this case, they do, so we are done.
The molecular formula
To calculate the mass % of H atoms in
Mass of
Mass of
Divide and multiply by 100:
The mass % of
The molecular formula
To calculate the mass % of H atoms in
Mass of
Mass of
Divide and multiply by 100:
The mass % of
The molecular formula
To calculate the mass % of H atoms in
Mass of
Mass of
Divide and multiply by 100:
The stoichiometric equation for the combustion of aluminum is:
If
Over
The stoichiometric equation for the combustion of aluminum is:
If
Over
The stoichiometric equation for the combustion of aluminum is:
If
Given mass of
Molar mass of
No of moles of
According to balanced equation;
3 moles of
Thus, 0.084 moles of
= 0.028mol of
So, 0.028mol of
0.084 moles of
Given mass of
Molar mass of
No of moles of
According to balanced equation;
3 moles of
Thus, 0.084 moles of
= 0.028mol of
So, 0.028mol of
0.084 moles of
Given mass of
Molar mass of
No of moles of
According to balanced equation;
3 moles of
Thus, 0.084 moles of
= 0.028mol of
So, 0.028mol of
How many dozens is 72? you just need to divide 72 by 12.
With mols it is exactly the same thing.
1 mol is approximately
Since the molar mass of Sulfur is about 32.06 g/mol you will get:
This number is extraordinarly huge. It is comparable with the mass of the whole Universe, which it's taken as being at least
How many dozens is 72? you just need to divide 72 by 12.
With mols it is exactly the same thing.
1 mol is approximately
Since the molar mass of Sulfur is about 32.06 g/mol you will get:
This number is extraordinarly huge. It is comparable with the mass of the whole Universe, which it's taken as being at least
How many dozens is 72? you just need to divide 72 by 12.
With mols it is exactly the same thing.
1 mol is approximately
Since the molar mass of Sulfur is about 32.06 g/mol you will get:
This number is extraordinarly huge. It is comparable with the mass of the whole Universe, which it's taken as being at least
We're asked to find the pressure of a gas, given its temperature, and volume, and number of moles.
We can use the ideal-gas equation:
#ul(PV = nRT#
where
Let's rearrange the above equation to solve for the pressure,
#P = (nRT)/V#
Plugging in known values:
#color(red)(P) = ((10cancel(""mol""))(0.082057(cancel(""L"")·""atm"")/(cancel(""mol"")·cancel(""K"")))(295cancel(""K"")))/(1.00cancel(""L"")) = color(red)(ulbar(|stackrel("" "")("" ""242color(white)(l)""atm"""" "")|)#
The pressure is thus
We're asked to find the pressure of a gas, given its temperature, and volume, and number of moles.
We can use the ideal-gas equation:
#ul(PV = nRT#
where
Let's rearrange the above equation to solve for the pressure,
#P = (nRT)/V#
Plugging in known values:
#color(red)(P) = ((10cancel(""mol""))(0.082057(cancel(""L"")·""atm"")/(cancel(""mol"")·cancel(""K"")))(295cancel(""K"")))/(1.00cancel(""L"")) = color(red)(ulbar(|stackrel("" "")("" ""242color(white)(l)""atm"""" "")|)#
The pressure is thus
We're asked to find the pressure of a gas, given its temperature, and volume, and number of moles.
We can use the ideal-gas equation:
#ul(PV = nRT#
where
Let's rearrange the above equation to solve for the pressure,
#P = (nRT)/V#
Plugging in known values:
#color(red)(P) = ((10cancel(""mol""))(0.082057(cancel(""L"")·""atm"")/(cancel(""mol"")·cancel(""K"")))(295cancel(""K"")))/(1.00cancel(""L"")) = color(red)(ulbar(|stackrel("" "")("" ""242color(white)(l)""atm"""" "")|)#
The pressure is thus
In aqueous solution,
Now since we interrogate the equilibrium:
On rearrangement,
And this is our defining relationship,
So, we find the
And finally,
For a similar problem, see this site.
In aqueous solution,
Now since we interrogate the equilibrium:
On rearrangement,
And this is our defining relationship,
So, we find the
And finally,
For a similar problem, see this site.
In aqueous solution,
Now since we interrogate the equilibrium:
On rearrangement,
And this is our defining relationship,
So, we find the
And finally,
For a similar problem, see this site.
And thus,
And thus
And thus,
And thus
And thus,
And thus
Anyway you asked the mass of sodium reacted. This is simply
Can you tell me (i) the volume of dihydrogen gas released given
Enquiring minds want to know!
The stoichiometrically balanced equation for this is.....
Anyway you asked the mass of sodium reacted. This is simply
Can you tell me (i) the volume of dihydrogen gas released given
Enquiring minds want to know!
The stoichiometrically balanced equation for this is.....
Anyway you asked the mass of sodium reacted. This is simply
Can you tell me (i) the volume of dihydrogen gas released given
Enquiring minds want to know!
Moles of sodium hydroxide are equivalent to the moles of hydrochloric acid. I know (or can calculate) the moles of sodium hydroxide, and I know the equivalent quantity of
Volume (
So
Correctly! And according to the equation:
Moles of sodium hydroxide are equivalent to the moles of hydrochloric acid. I know (or can calculate) the moles of sodium hydroxide, and I know the equivalent quantity of
Volume (
So
Correctly! And according to the equation:
Moles of sodium hydroxide are equivalent to the moles of hydrochloric acid. I know (or can calculate) the moles of sodium hydroxide, and I know the equivalent quantity of
Volume (
So
Multiply mol
Multiply mol
Multiply mol
First we find the molar mass of the entire molecule. So N is 14 and O is 16. Molar mass of
To find of the percent we take the element's mass in the molecule over the total molar mass. For
It is
First we find the molar mass of the entire molecule. So N is 14 and O is 16. Molar mass of
To find of the percent we take the element's mass in the molecule over the total molar mass. For
It is
First we find the molar mass of the entire molecule. So N is 14 and O is 16. Molar mass of
To find of the percent we take the element's mass in the molecule over the total molar mass. For
% of N =
% of O =
For this type of problem, we want to use the following dilution formula:
* TIP. * Whenever you are diluting a solution (going from a highly concentrated substance to a less concentrated substance by increasing the volume of the solvent) you always use this equation.
We know the initial concentration, final volume, and final concentration. All we have to do rearrange the equation to solve for the initial volume:
62.5 mL of HCl is required.
For this type of problem, we want to use the following dilution formula:
* TIP. * Whenever you are diluting a solution (going from a highly concentrated substance to a less concentrated substance by increasing the volume of the solvent) you always use this equation.
We know the initial concentration, final volume, and final concentration. All we have to do rearrange the equation to solve for the initial volume:
62.5 mL of HCl is required.
For this type of problem, we want to use the following dilution formula:
* TIP. * Whenever you are diluting a solution (going from a highly concentrated substance to a less concentrated substance by increasing the volume of the solvent) you always use this equation.
We know the initial concentration, final volume, and final concentration. All we have to do rearrange the equation to solve for the initial volume:
Molar mass of
Number of moles in
#(20.0 cancel""g"")/(58.4 cancel""g""""/mol"") = ""0.342 mol""#
Molar mass of
Number of moles in
#(20.0 cancel""g"")/(58.4 cancel""g""""/mol"") = ""0.342 mol""#
Molar mass of
Number of moles in
#(20.0 cancel""g"")/(58.4 cancel""g""""/mol"") = ""0.342 mol""#
The above equation is stoichiometrically balanced: garbage out equals garbage in. If you like you can double the equation to remove the non-integral coefficient. I have never seen the need to do so, inasmuch the arithmetic is easier when you use this form with the 1 equivalent of hydrocarbon reactant.
So 2 questions with regard to this reaction: (i) How does energy transfer in this reaction; (ii) does this represent an oxidation reduction reaction?
The above equation is stoichiometrically balanced: garbage out equals garbage in. If you like you can double the equation to remove the non-integral coefficient. I have never seen the need to do so, inasmuch the arithmetic is easier when you use this form with the 1 equivalent of hydrocarbon reactant.
So 2 questions with regard to this reaction: (i) How does energy transfer in this reaction; (ii) does this represent an oxidation reduction reaction?
The above equation is stoichiometrically balanced: garbage out equals garbage in. If you like you can double the equation to remove the non-integral coefficient. I have never seen the need to do so, inasmuch the arithmetic is easier when you use this form with the 1 equivalent of hydrocarbon reactant.
So 2 questions with regard to this reaction: (i) How does energy transfer in this reaction; (ii) does this represent an oxidation reduction reaction?
We need to have the same number of moles of each substance on each side. The balanced equation is:
We start out with the unbalanced equation:
We need to find the values of the coefficients
For the moment, let's leave
There are 18 moles of H on the left so we need 18 on the right, but each
Now we turn our attention to the right side. There are two O in each of 8
On the left, each
If the fraction makes you uncomfortable, another way is to multiply all the coefficients by two:
The unbalanced equation is
One way to balance the equation is to recognize that the polyatomic ions stay together.
Then we can make some substitutions.
Let
Then the equation becomes
We start with the most complicated formula,
Balance
We have fixed 3 atoms of
Balance
We have fixed 4 atoms of
Balance
We have fixed 12 atoms of
Every formula has a coefficient, and the equation is balanced.
Now, we replace the original formulas in the equation.
The balanced equation is
The unbalanced equation is
One way to balance the equation is to recognize that the polyatomic ions stay together.
Then we can make some substitutions.
Let
Then the equation becomes
We start with the most complicated formula,
Balance
We have fixed 3 atoms of
Balance
We have fixed 4 atoms of
Balance
We have fixed 12 atoms of
Every formula has a coefficient, and the equation is balanced.
Now, we replace the original formulas in the equation.
The balanced equation is
The unbalanced equation is
One way to balance the equation is to recognize that the polyatomic ions stay together.
Then we can make some substitutions.
Let
Then the equation becomes
We start with the most complicated formula,
Balance
We have fixed 3 atoms of
Balance
We have fixed 4 atoms of
Balance
We have fixed 12 atoms of
Every formula has a coefficient, and the equation is balanced.
Now, we replace the original formulas in the equation.
The balanced equation is
We divide thru by the lowest molar quantity to give
We divide thru by the lowest molar quantity to give
We divide thru by the lowest molar quantity to give
We need a stoichiometric equation:
Given the stoichiometry of the precipitation reaction, there were
So we have a molar quantity, and a given mass, and the
We need a stoichiometric equation:
Given the stoichiometry of the precipitation reaction, there were
So we have a molar quantity, and a given mass, and the
We need a stoichiometric equation:
Given the stoichiometry of the precipitation reaction, there were
So we have a molar quantity, and a given mass, and the
The first thing to take note of here is that the problem provides you information you don't actually need.
The molarity of the solution is not important in determining how many grams of hydrochloric acid you have in that
So, you know that the solution has a density of
The mass of the sample you have here will thus be
#100.0color(red)(cancel(color(black)(""mL""))) * ""1.191 g""/(1color(red)(cancel(color(black)(""mL"")))) = ""119.1 g""#
Now, you know that the solution has a
This means that your sample will contain
#119.1color(red)(cancel(color(black)(""g solution""))) * ""37 g HCl""/(100color(red)(cancel(color(black)(""g solution"")))) = ""44.067 g HCl""#
Rounded to two sig figs, the number of sig figs you have for the percent concentration by mass, the answer will be
#m_""HCl"" = color(green)(""44 g HCl"")#
The first thing to take note of here is that the problem provides you information you don't actually need.
The molarity of the solution is not important in determining how many grams of hydrochloric acid you have in that
So, you know that the solution has a density of
The mass of the sample you have here will thus be
#100.0color(red)(cancel(color(black)(""mL""))) * ""1.191 g""/(1color(red)(cancel(color(black)(""mL"")))) = ""119.1 g""#
Now, you know that the solution has a
This means that your sample will contain
#119.1color(red)(cancel(color(black)(""g solution""))) * ""37 g HCl""/(100color(red)(cancel(color(black)(""g solution"")))) = ""44.067 g HCl""#
Rounded to two sig figs, the number of sig figs you have for the percent concentration by mass, the answer will be
#m_""HCl"" = color(green)(""44 g HCl"")#
The first thing to take note of here is that the problem provides you information you don't actually need.
The molarity of the solution is not important in determining how many grams of hydrochloric acid you have in that
So, you know that the solution has a density of
The mass of the sample you have here will thus be
#100.0color(red)(cancel(color(black)(""mL""))) * ""1.191 g""/(1color(red)(cancel(color(black)(""mL"")))) = ""119.1 g""#
Now, you know that the solution has a
This means that your sample will contain
#119.1color(red)(cancel(color(black)(""g solution""))) * ""37 g HCl""/(100color(red)(cancel(color(black)(""g solution"")))) = ""44.067 g HCl""#
Rounded to two sig figs, the number of sig figs you have for the percent concentration by mass, the answer will be
#m_""HCl"" = color(green)(""44 g HCl"")#
Thus
The advantage of using Boyle's Law is that I can use any units I like,
From Boyle's Law,
Thus
The advantage of using Boyle's Law is that I can use any units I like,
From Boyle's Law,
Thus
The advantage of using Boyle's Law is that I can use any units I like,
First, we know this is an ionic compound because one of the elements is a metal.
From the periodic table or from the valence shells of the atoms, we can determine that magnesium forms an ion of
The formula is based on combining these ions in the smallest ratio that results in a total charge of zero, because all compounds must be electrically neutral.
So, three
A compound with three magnesium ions and two phosphorus ions would be written
The formula would be
First, we know this is an ionic compound because one of the elements is a metal.
From the periodic table or from the valence shells of the atoms, we can determine that magnesium forms an ion of
The formula is based on combining these ions in the smallest ratio that results in a total charge of zero, because all compounds must be electrically neutral.
So, three
A compound with three magnesium ions and two phosphorus ions would be written
The formula would be
First, we know this is an ionic compound because one of the elements is a metal.
From the periodic table or from the valence shells of the atoms, we can determine that magnesium forms an ion of
The formula is based on combining these ions in the smallest ratio that results in a total charge of zero, because all compounds must be electrically neutral.
So, three
A compound with three magnesium ions and two phosphorus ions would be written
The empirical formula is the simplest while number ratio defining constituent atoms in a chemical species.
We assumes that we got
And thus
And
But you have undoubtedly already noted that the quoted percentages do not add up to 100%. In this scenario IT IS ALWAYS ASSUMED that the BALANCE, the missing percentage, is DUE TO OXYGEN.....
And
And please note that here WE CAN MAKE NO OTHER ASSUMPTION.
And so we divide each molar quantity thru by the SMALLEST molar quantity to get an empirical formula of.....
..........................
We could quote a molecular formula PROVIDED that we get a measurement of molecular mass.
I makes it
The empirical formula is the simplest while number ratio defining constituent atoms in a chemical species.
We assumes that we got
And thus
And
But you have undoubtedly already noted that the quoted percentages do not add up to 100%. In this scenario IT IS ALWAYS ASSUMED that the BALANCE, the missing percentage, is DUE TO OXYGEN.....
And
And please note that here WE CAN MAKE NO OTHER ASSUMPTION.
And so we divide each molar quantity thru by the SMALLEST molar quantity to get an empirical formula of.....
..........................
We could quote a molecular formula PROVIDED that we get a measurement of molecular mass.
I makes it
The empirical formula is the simplest while number ratio defining constituent atoms in a chemical species.
We assumes that we got
And thus
And
But you have undoubtedly already noted that the quoted percentages do not add up to 100%. In this scenario IT IS ALWAYS ASSUMED that the BALANCE, the missing percentage, is DUE TO OXYGEN.....
And
And please note that here WE CAN MAKE NO OTHER ASSUMPTION.
And so we divide each molar quantity thru by the SMALLEST molar quantity to get an empirical formula of.....
..........................
We could quote a molecular formula PROVIDED that we get a measurement of molecular mass.
Can you tell us the molar mass given that carbon has a molar mass of
Well,
Can you tell us the molar mass given that carbon has a molar mass of
Well,
Can you tell us the molar mass given that carbon has a molar mass of
A substance's specific heat tells you how much heat much either be added or removed from
The change in temperature,
#color(blue)(DeltaT = T_""final"" - T_""initial"")#
Now, when the substance absorbs heat, its temperature will increase, which implies that
Your goal here will be to find the change in temperature first, then use it to find the final temperature of the sample.
You will have to use this equation
#color(blue)(q = m * c * DeltaT)"" ""# , where
As you can see, this equation establishes a relationship between the amount of heat added or removed from a sample, the mass of that substance, its specific heat, and the resulting change in temperature.
In your case, adding
#q = m * c * DeltaT implies DeltaT = q/(m * c)#
Plug in your values to get
#DeltaT = (5275 color(red)(cancel(color(black)(""J""))))/(50.0color(red)(cancel(color(black)(""g""))) * 0.50color(red)(cancel(color(black)(""J"")))/(color(red)(cancel(color(black)(""g""))) """"^@""C"")) = 211^@""C""#
So, adding that much heat to your sample will result in a
#T_""final"" = T_""initial"" + DeltaT#
#T_""final"" = 20.0^@""C"" + 211^@""C"" = color(green)(230^@""C"")#
The answer is rounded to two sig figs, the number of sig figs you have for the specific heat of glass.
A substance's specific heat tells you how much heat much either be added or removed from
The change in temperature,
#color(blue)(DeltaT = T_""final"" - T_""initial"")#
Now, when the substance absorbs heat, its temperature will increase, which implies that
Your goal here will be to find the change in temperature first, then use it to find the final temperature of the sample.
You will have to use this equation
#color(blue)(q = m * c * DeltaT)"" ""# , where
As you can see, this equation establishes a relationship between the amount of heat added or removed from a sample, the mass of that substance, its specific heat, and the resulting change in temperature.
In your case, adding
#q = m * c * DeltaT implies DeltaT = q/(m * c)#
Plug in your values to get
#DeltaT = (5275 color(red)(cancel(color(black)(""J""))))/(50.0color(red)(cancel(color(black)(""g""))) * 0.50color(red)(cancel(color(black)(""J"")))/(color(red)(cancel(color(black)(""g""))) """"^@""C"")) = 211^@""C""#
So, adding that much heat to your sample will result in a
#T_""final"" = T_""initial"" + DeltaT#
#T_""final"" = 20.0^@""C"" + 211^@""C"" = color(green)(230^@""C"")#
The answer is rounded to two sig figs, the number of sig figs you have for the specific heat of glass.
A substance's specific heat tells you how much heat much either be added or removed from
The change in temperature,
#color(blue)(DeltaT = T_""final"" - T_""initial"")#
Now, when the substance absorbs heat, its temperature will increase, which implies that
Your goal here will be to find the change in temperature first, then use it to find the final temperature of the sample.
You will have to use this equation
#color(blue)(q = m * c * DeltaT)"" ""# , where
As you can see, this equation establishes a relationship between the amount of heat added or removed from a sample, the mass of that substance, its specific heat, and the resulting change in temperature.
In your case, adding
#q = m * c * DeltaT implies DeltaT = q/(m * c)#
Plug in your values to get
#DeltaT = (5275 color(red)(cancel(color(black)(""J""))))/(50.0color(red)(cancel(color(black)(""g""))) * 0.50color(red)(cancel(color(black)(""J"")))/(color(red)(cancel(color(black)(""g""))) """"^@""C"")) = 211^@""C""#
So, adding that much heat to your sample will result in a
#T_""final"" = T_""initial"" + DeltaT#
#T_""final"" = 20.0^@""C"" + 211^@""C"" = color(green)(230^@""C"")#
The answer is rounded to two sig figs, the number of sig figs you have for the specific heat of glass.
Molar mass NaOH = 40 g/mol#
Molarity is a concentration term in which the no. of moles and the volume are related.
Here
So
# = ""10 g""/(""40 g/mol"" \ xx \ ""0.5 L"")#
# = ""10 mol""/""20 L"" = ""0.5 molar""#
Molarity is a concentration term in which the no. of moles and the volume are related.
Here
So
# = ""10 g""/(""40 g/mol"" \ xx \ ""0.5 L"")#
# = ""10 mol""/""20 L"" = ""0.5 molar""#
Molar mass NaOH = 40 g/mol#
Molarity is a concentration term in which the no. of moles and the volume are related.
Here
So
# = ""10 g""/(""40 g/mol"" \ xx \ ""0.5 L"")#
# = ""10 mol""/""20 L"" = ""0.5 molar""#
Because the magnesium ion has a +2 charge, the formula for magnesium chloride is
To balance the four Cl ions in titanium (IV) chloride, the reaction must produce 2 units of
Because the magnesium ion has a +2 charge, the formula for magnesium chloride is
To balance the four Cl ions in titanium (IV) chloride, the reaction must produce 2 units of
Because the magnesium ion has a +2 charge, the formula for magnesium chloride is
To balance the four Cl ions in titanium (IV) chloride, the reaction must produce 2 units of
We could do this by individual redox steps:
With all chemical reactions, 2 conditions must be absolutely satisfied: (i) mass must be balanced; and (ii) charge must be balanced. Well, is they?
Note that just because I can write the reaction, there is no likelihood that the reaction can be performed. Should I attempt to reduce the chloride with magnesium metal I would likely get a mess. Titanic chloride is unstable with respect to water and dioxygen.
The idea here is that bromophenol blue acts as a weak acid in aqueous solution.
#""HBb""_ ((aq)) + ""H""_ 2""O""_ ((l)) rightleftharpoons ""Bb""_ ((aq))^(-) + ""H""_ 3""O""_ ((aq))^(+)#
As you can see, this equilibrium is influenced by the
In your case, you know that you have
#""pH"" = 4.84#
This implies that you have
#[""H""_3""O""^(+)] = 10^(-4.84)#
Now, by definition, the acid dissociation constant is equal to
#K_a = ([""Bb""^(-)] * [""H""_3""O""^(+)])/([""HBb""])#
Rearrange to find the ratio that exists between the equilibrium concentration of the conjugate base, which is the indicator is its basic form, and the equilibrium concentration of the weak acid.
#([""Bb""^(-)])/([""HBb""]) = K_a/([""H""_3""O""^(+)])#
Plug in your values to find
#([""Bb""^(-)])/([""HBb""]) = (5.84 * 10^(-5))/10^(-4.84)#
#([""Bb""^(-)])/([""HBb""]) = 5.84 * 10^((-5 + 4.84)) = 5.84 * 10^(-0.16)"" """" ""color(darkorange)(""(*)"")#
Now, in order to find the percent dissociation of the weak acid, you need to take into account the fact the initial concentration of the weak acid is equal to
#[""HBb""]_0 = [""Bb""^(-)] + [""HBb""]#
This is the case because the acid dissociates in a
This means that the percent dissociation of the weak acid will be equal to
#""% dissociation"" = ([""Bb""^(-)])/([""Bb""^(-)] + [""HBb""]) * 100%#
Use equation
#[""Bb""^(-)] = 5.84 * 10^(-0.16) * [""HBb""]#
The percent dissociation of the weak acid will thus be equal to
#""% dissociation"" = (5.84 * 10^(-0.16) * color(red)(cancel(color(black)([""HBb""]))))/(5.84 * 10^(-0.16)color(red)(cancel(color(black)([""HBb""]))) + color(red)(cancel(color(black)([""HBb""])))) * 100%#
#""% dissociation"" = (5.84 * 10^(-0.16))/(5.84 * 10^(-0.16) + 1) * 100%#
Now, you know that
#log(5.84) = 0.7664"" ""# and#"" "" 10^0.6064 = 4.04#
This means that you can write
#10^log(5.84) = 10^0.7664#
#5.84 = 10^0.7664#
Plug this into the equation to get
#""% dissociation"" = (10^0.7664 * 10^(-0.16))/(10^0.7664 * 10^(-0.16) + 1) * 100%#
#""% dissociation"" = 10^0.6064/(10^0.6064 + 1) * 100%#
Finally, you will end up with
#""% dissociation"" = 4.04/(4.04 + 1) * 100% = color(darkgreen)(ul(color(black)(80.2%)))#
I'll leave the answer rounded to three sig figs.
The idea here is that bromophenol blue acts as a weak acid in aqueous solution.
#""HBb""_ ((aq)) + ""H""_ 2""O""_ ((l)) rightleftharpoons ""Bb""_ ((aq))^(-) + ""H""_ 3""O""_ ((aq))^(+)#
As you can see, this equilibrium is influenced by the
In your case, you know that you have
#""pH"" = 4.84#
This implies that you have
#[""H""_3""O""^(+)] = 10^(-4.84)#
Now, by definition, the acid dissociation constant is equal to
#K_a = ([""Bb""^(-)] * [""H""_3""O""^(+)])/([""HBb""])#
Rearrange to find the ratio that exists between the equilibrium concentration of the conjugate base, which is the indicator is its basic form, and the equilibrium concentration of the weak acid.
#([""Bb""^(-)])/([""HBb""]) = K_a/([""H""_3""O""^(+)])#
Plug in your values to find
#([""Bb""^(-)])/([""HBb""]) = (5.84 * 10^(-5))/10^(-4.84)#
#([""Bb""^(-)])/([""HBb""]) = 5.84 * 10^((-5 + 4.84)) = 5.84 * 10^(-0.16)"" """" ""color(darkorange)(""(*)"")#
Now, in order to find the percent dissociation of the weak acid, you need to take into account the fact the initial concentration of the weak acid is equal to
#[""HBb""]_0 = [""Bb""^(-)] + [""HBb""]#
This is the case because the acid dissociates in a
This means that the percent dissociation of the weak acid will be equal to
#""% dissociation"" = ([""Bb""^(-)])/([""Bb""^(-)] + [""HBb""]) * 100%#
Use equation
#[""Bb""^(-)] = 5.84 * 10^(-0.16) * [""HBb""]#
The percent dissociation of the weak acid will thus be equal to
#""% dissociation"" = (5.84 * 10^(-0.16) * color(red)(cancel(color(black)([""HBb""]))))/(5.84 * 10^(-0.16)color(red)(cancel(color(black)([""HBb""]))) + color(red)(cancel(color(black)([""HBb""])))) * 100%#
#""% dissociation"" = (5.84 * 10^(-0.16))/(5.84 * 10^(-0.16) + 1) * 100%#
Now, you know that
#log(5.84) = 0.7664"" ""# and#"" "" 10^0.6064 = 4.04#
This means that you can write
#10^log(5.84) = 10^0.7664#
#5.84 = 10^0.7664#
Plug this into the equation to get
#""% dissociation"" = (10^0.7664 * 10^(-0.16))/(10^0.7664 * 10^(-0.16) + 1) * 100%#
#""% dissociation"" = 10^0.6064/(10^0.6064 + 1) * 100%#
Finally, you will end up with
#""% dissociation"" = 4.04/(4.04 + 1) * 100% = color(darkgreen)(ul(color(black)(80.2%)))#
I'll leave the answer rounded to three sig figs.
The idea here is that bromophenol blue acts as a weak acid in aqueous solution.
#""HBb""_ ((aq)) + ""H""_ 2""O""_ ((l)) rightleftharpoons ""Bb""_ ((aq))^(-) + ""H""_ 3""O""_ ((aq))^(+)#
As you can see, this equilibrium is influenced by the
In your case, you know that you have
#""pH"" = 4.84#
This implies that you have
#[""H""_3""O""^(+)] = 10^(-4.84)#
Now, by definition, the acid dissociation constant is equal to
#K_a = ([""Bb""^(-)] * [""H""_3""O""^(+)])/([""HBb""])#
Rearrange to find the ratio that exists between the equilibrium concentration of the conjugate base, which is the indicator is its basic form, and the equilibrium concentration of the weak acid.
#([""Bb""^(-)])/([""HBb""]) = K_a/([""H""_3""O""^(+)])#
Plug in your values to find
#([""Bb""^(-)])/([""HBb""]) = (5.84 * 10^(-5))/10^(-4.84)#
#([""Bb""^(-)])/([""HBb""]) = 5.84 * 10^((-5 + 4.84)) = 5.84 * 10^(-0.16)"" """" ""color(darkorange)(""(*)"")#
Now, in order to find the percent dissociation of the weak acid, you need to take into account the fact the initial concentration of the weak acid is equal to
#[""HBb""]_0 = [""Bb""^(-)] + [""HBb""]#
This is the case because the acid dissociates in a
This means that the percent dissociation of the weak acid will be equal to
#""% dissociation"" = ([""Bb""^(-)])/([""Bb""^(-)] + [""HBb""]) * 100%#
Use equation
#[""Bb""^(-)] = 5.84 * 10^(-0.16) * [""HBb""]#
The percent dissociation of the weak acid will thus be equal to
#""% dissociation"" = (5.84 * 10^(-0.16) * color(red)(cancel(color(black)([""HBb""]))))/(5.84 * 10^(-0.16)color(red)(cancel(color(black)([""HBb""]))) + color(red)(cancel(color(black)([""HBb""])))) * 100%#
#""% dissociation"" = (5.84 * 10^(-0.16))/(5.84 * 10^(-0.16) + 1) * 100%#
Now, you know that
#log(5.84) = 0.7664"" ""# and#"" "" 10^0.6064 = 4.04#
This means that you can write
#10^log(5.84) = 10^0.7664#
#5.84 = 10^0.7664#
Plug this into the equation to get
#""% dissociation"" = (10^0.7664 * 10^(-0.16))/(10^0.7664 * 10^(-0.16) + 1) * 100%#
#""% dissociation"" = 10^0.6064/(10^0.6064 + 1) * 100%#
Finally, you will end up with
#""% dissociation"" = 4.04/(4.04 + 1) * 100% = color(darkgreen)(ul(color(black)(80.2%)))#
I'll leave the answer rounded to three sig figs.
Indicators like bromophenol blue are weak acids where the undissociated (acidic) form is a different colour from the dissociated (basic) form:
We know that
To get the percentage which are dissociated we can find
We know that
This means that the basic, dissociated form is 80.2 %.
How much calcium would be added at the same time?
Let's express the concentration of 2 ppm as grams per litre.
For 40 L of water, you will need
The molar mass of
Of this, 124.01 g comes from the nitrate ions.
Hence, to get 0.08 g of nitrate, you will have to use
The required volume of concentrated salt solution is
There are 40.078 g of
∴ The mass of calcium added is
You must add 0.3 mL of the calcium nitrate solution. This will include 25 mg of calcium.
Let's express the concentration of 2 ppm as grams per litre.
For 40 L of water, you will need
The molar mass of
Of this, 124.01 g comes from the nitrate ions.
Hence, to get 0.08 g of nitrate, you will have to use
The required volume of concentrated salt solution is
There are 40.078 g of
∴ The mass of calcium added is
How much calcium would be added at the same time?
You must add 0.3 mL of the calcium nitrate solution. This will include 25 mg of calcium.
Let's express the concentration of 2 ppm as grams per litre.
For 40 L of water, you will need
The molar mass of
Of this, 124.01 g comes from the nitrate ions.
Hence, to get 0.08 g of nitrate, you will have to use
The required volume of concentrated salt solution is
There are 40.078 g of
∴ The mass of calcium added is
1) calculate mmoles of
these are also the mmoles of carbon in ten milligrams of substance.
We deduce that ten milligrams of substance contain:
We can deduce soon that the molecule contains a number of hydrogen atoms that is the double of carbon atoms; so the empirical formula should be (provisionally)
2) calculate moles of dioxygen (
then:
Therefore the
then we can calculate oxygen moles:
3) calculate number of moles of oxygen in the ten milligrams of substance, applying the conservation law to oxygen alone
The number of mmoles of oxygen atoms in the products is the double of
The mmole number of oxygen atoms in the substance is obtained by subtracting the oxygen number of mmoles used to combust the substance from this total number of atoms in mmoles:
4) Calculate the empirical formula by taking the moles ratios and getting the minimum integer indexes of C, H and O.
We see that carbon mmoles are approximately ten times oxygen mmoles:
Therefore we conclude that putting at least one oxygen atom in the empirical formula, we must put ten times this number in the carbon index an the double in hydrogen index, so we get:
This is also the molecular formula of menthol .
Empirical formula:
1) calculate mmoles of
these are also the mmoles of carbon in ten milligrams of substance.
We deduce that ten milligrams of substance contain:
We can deduce soon that the molecule contains a number of hydrogen atoms that is the double of carbon atoms; so the empirical formula should be (provisionally)
2) calculate moles of dioxygen (
then:
Therefore the
then we can calculate oxygen moles:
3) calculate number of moles of oxygen in the ten milligrams of substance, applying the conservation law to oxygen alone
The number of mmoles of oxygen atoms in the products is the double of
The mmole number of oxygen atoms in the substance is obtained by subtracting the oxygen number of mmoles used to combust the substance from this total number of atoms in mmoles:
4) Calculate the empirical formula by taking the moles ratios and getting the minimum integer indexes of C, H and O.
We see that carbon mmoles are approximately ten times oxygen mmoles:
Therefore we conclude that putting at least one oxygen atom in the empirical formula, we must put ten times this number in the carbon index an the double in hydrogen index, so we get:
This is also the molecular formula of menthol .
Empirical formula:
1) calculate mmoles of
these are also the mmoles of carbon in ten milligrams of substance.
We deduce that ten milligrams of substance contain:
We can deduce soon that the molecule contains a number of hydrogen atoms that is the double of carbon atoms; so the empirical formula should be (provisionally)
2) calculate moles of dioxygen (
then:
Therefore the
then we can calculate oxygen moles:
3) calculate number of moles of oxygen in the ten milligrams of substance, applying the conservation law to oxygen alone
The number of mmoles of oxygen atoms in the products is the double of
The mmole number of oxygen atoms in the substance is obtained by subtracting the oxygen number of mmoles used to combust the substance from this total number of atoms in mmoles:
4) Calculate the empirical formula by taking the moles ratios and getting the minimum integer indexes of C, H and O.
We see that carbon mmoles are approximately ten times oxygen mmoles:
Therefore we conclude that putting at least one oxygen atom in the empirical formula, we must put ten times this number in the carbon index an the double in hydrogen index, so we get:
This is also the molecular formula of menthol .
In aqueous solution, magnesium sulfate speciates to give two moles of ions.......i.e.
Charge and mass are balanced as is absolutely required.
And since magnesium sulfate gives rise to TWO EQUIVS of soluble ions, the osmolarity is equal to TWICE the molarity of
The
In aqueous solution, magnesium sulfate speciates to give two moles of ions.......i.e.
Charge and mass are balanced as is absolutely required.
And since magnesium sulfate gives rise to TWO EQUIVS of soluble ions, the osmolarity is equal to TWICE the molarity of
The
In aqueous solution, magnesium sulfate speciates to give two moles of ions.......i.e.
Charge and mass are balanced as is absolutely required.
And since magnesium sulfate gives rise to TWO EQUIVS of soluble ions, the osmolarity is equal to TWICE the molarity of
For this problem, we can use Gay-Lussac's Law of Combining Volumes:
If pressure and temperature are constant, the ratio between the volumes of the reactant gases and the products can be expressed in simple whole numbers.
The balanced equation for the combustion is
#color(white)(l)""CH""_4 + ""2O""_2 → ""CO""_2 + ""2H""_2""O""#
#""1 dm""^3color(white)(ll)""2 dm""^3#
According to Gay-Lussac,
∴
The volume of oxygen required is
For this problem, we can use Gay-Lussac's Law of Combining Volumes:
If pressure and temperature are constant, the ratio between the volumes of the reactant gases and the products can be expressed in simple whole numbers.
The balanced equation for the combustion is
#color(white)(l)""CH""_4 + ""2O""_2 → ""CO""_2 + ""2H""_2""O""#
#""1 dm""^3color(white)(ll)""2 dm""^3#
According to Gay-Lussac,
∴
The volume of oxygen required is
For this problem, we can use Gay-Lussac's Law of Combining Volumes:
If pressure and temperature are constant, the ratio between the volumes of the reactant gases and the products can be expressed in simple whole numbers.
The balanced equation for the combustion is
#color(white)(l)""CH""_4 + ""2O""_2 → ""CO""_2 + ""2H""_2""O""#
#""1 dm""^3color(white)(ll)""2 dm""^3#
According to Gay-Lussac,
∴
Percent yield is defined as the ratio between the actual yield and the theoretical yield of the reaction, multiplied by
#color(blue)(|bar(ul(color(white)(a/a)""% yield"" = ""what you actually get""/""what you should theoretically get"" xx 100 color(white)(a/a)|)))#
So, you know that your reaction has a theoretical yield of
In other words, the theoretical yield tells you how much product is produced for a
Now, the reaction is said to produce
#""actual yield"" = ""82.5 g"" - ""12.3 g"" = ""70.2 g""#
The reaction's percent yield will thus be
#""% yield"" = (70.2 color(red)(cancel(color(black)(""g""))))/(82.5color(red)(cancel(color(black)(""g"")))) xx 100 = color(green)(|bar(ul(color(white)(a/a)""85.1%""color(white)(a/a)|)))#
The answer is rounded to three sig figs.
Percent yield is defined as the ratio between the actual yield and the theoretical yield of the reaction, multiplied by
#color(blue)(|bar(ul(color(white)(a/a)""% yield"" = ""what you actually get""/""what you should theoretically get"" xx 100 color(white)(a/a)|)))#
So, you know that your reaction has a theoretical yield of
In other words, the theoretical yield tells you how much product is produced for a
Now, the reaction is said to produce
#""actual yield"" = ""82.5 g"" - ""12.3 g"" = ""70.2 g""#
The reaction's percent yield will thus be
#""% yield"" = (70.2 color(red)(cancel(color(black)(""g""))))/(82.5color(red)(cancel(color(black)(""g"")))) xx 100 = color(green)(|bar(ul(color(white)(a/a)""85.1%""color(white)(a/a)|)))#
The answer is rounded to three sig figs.
Percent yield is defined as the ratio between the actual yield and the theoretical yield of the reaction, multiplied by
#color(blue)(|bar(ul(color(white)(a/a)""% yield"" = ""what you actually get""/""what you should theoretically get"" xx 100 color(white)(a/a)|)))#
So, you know that your reaction has a theoretical yield of
In other words, the theoretical yield tells you how much product is produced for a
Now, the reaction is said to produce
#""actual yield"" = ""82.5 g"" - ""12.3 g"" = ""70.2 g""#
The reaction's percent yield will thus be
#""% yield"" = (70.2 color(red)(cancel(color(black)(""g""))))/(82.5color(red)(cancel(color(black)(""g"")))) xx 100 = color(green)(|bar(ul(color(white)(a/a)""85.1%""color(white)(a/a)|)))#
The answer is rounded to three sig figs.
The first thing to do here is make sure that you have a balanced chemical equation. The equation given to you is actually unbalanced, so focus on writing a balanced version
#color(red)(2)""C""_2""H""_text(6(g]) + 7""O""_text(2(g]) -> color(purple)(4)""CO""_text(2(g]) + 6""H""_2""O""_text((g])#
Now, the problem provides you with a mass of ethane,
As you know, STP conditions are defined as a temperature of
So, what this means is that if you know how many moles of carbon dioxide are produced by the reaction, you can use the molar volume of the gas to find the requested volume.
Use the molar mass of ethane to determine how many moles you'd get in that
#240.0 color(red)(cancel(color(black)(""g""))) * (""1 mole C""_2""H""_6)/(30.07color(red)(cancel(color(black)(""g"")))) = ""7.9814 moles C""_2""H""_6#
Now, notice that you have a
This means that the reaction will produce
#7.9814 color(red)(cancel(color(black)(""moles C""_2""H""_6))) * (color(purple)(4)"" moles CO""_2)/(color(red)(2)color(red)(cancel(color(black)(""moles C""_2""H""_6)))) = ""15.9628 moles CO""_2#
So, if one mole of any ideal gas occupies
#15.9628color(red)(cancel(color(black)(""moles CO""_2))) * ""22.7 L""/(1color(red)(cancel(color(black)(""mole CO""_2)))) = color(green)(""362.4 L"")#
The answer is rounded to four sig figs, the number of sig figs you have for the mass of ethane.
The first thing to do here is make sure that you have a balanced chemical equation. The equation given to you is actually unbalanced, so focus on writing a balanced version
#color(red)(2)""C""_2""H""_text(6(g]) + 7""O""_text(2(g]) -> color(purple)(4)""CO""_text(2(g]) + 6""H""_2""O""_text((g])#
Now, the problem provides you with a mass of ethane,
As you know, STP conditions are defined as a temperature of
So, what this means is that if you know how many moles of carbon dioxide are produced by the reaction, you can use the molar volume of the gas to find the requested volume.
Use the molar mass of ethane to determine how many moles you'd get in that
#240.0 color(red)(cancel(color(black)(""g""))) * (""1 mole C""_2""H""_6)/(30.07color(red)(cancel(color(black)(""g"")))) = ""7.9814 moles C""_2""H""_6#
Now, notice that you have a
This means that the reaction will produce
#7.9814 color(red)(cancel(color(black)(""moles C""_2""H""_6))) * (color(purple)(4)"" moles CO""_2)/(color(red)(2)color(red)(cancel(color(black)(""moles C""_2""H""_6)))) = ""15.9628 moles CO""_2#
So, if one mole of any ideal gas occupies
#15.9628color(red)(cancel(color(black)(""moles CO""_2))) * ""22.7 L""/(1color(red)(cancel(color(black)(""mole CO""_2)))) = color(green)(""362.4 L"")#
The answer is rounded to four sig figs, the number of sig figs you have for the mass of ethane.
The first thing to do here is make sure that you have a balanced chemical equation. The equation given to you is actually unbalanced, so focus on writing a balanced version
#color(red)(2)""C""_2""H""_text(6(g]) + 7""O""_text(2(g]) -> color(purple)(4)""CO""_text(2(g]) + 6""H""_2""O""_text((g])#
Now, the problem provides you with a mass of ethane,
As you know, STP conditions are defined as a temperature of
So, what this means is that if you know how many moles of carbon dioxide are produced by the reaction, you can use the molar volume of the gas to find the requested volume.
Use the molar mass of ethane to determine how many moles you'd get in that
#240.0 color(red)(cancel(color(black)(""g""))) * (""1 mole C""_2""H""_6)/(30.07color(red)(cancel(color(black)(""g"")))) = ""7.9814 moles C""_2""H""_6#
Now, notice that you have a
This means that the reaction will produce
#7.9814 color(red)(cancel(color(black)(""moles C""_2""H""_6))) * (color(purple)(4)"" moles CO""_2)/(color(red)(2)color(red)(cancel(color(black)(""moles C""_2""H""_6)))) = ""15.9628 moles CO""_2#
So, if one mole of any ideal gas occupies
#15.9628color(red)(cancel(color(black)(""moles CO""_2))) * ""22.7 L""/(1color(red)(cancel(color(black)(""mole CO""_2)))) = color(green)(""362.4 L"")#
The answer is rounded to four sig figs, the number of sig figs you have for the mass of ethane.
You have a little mor econverting to do to solve this one.
You know that the drug dosage is set to
To start, you can convert the dosage from miligrams per kilogram to miligrams per pound by using the conversion factor
#""1 kg"" ~= ""2.2046 lbs""#
This means that you have
#5.00""mg""/color(red)(cancel(color(black)(""kg""))) * (1color(red)(cancel(color(black)(""kg""))))/""2.2046 lbs"" = ""2.268 mg/lbs""#
So, your patient needs 2.268 mg of the drug per pound of body weight. This means that if you supply 425 mg, his body weight must be
#425color(red)(cancel(color(black)(""mg""))) * ""1 lbs""/(2.268color(red)(cancel(color(black)(""mg"")))) = ""187.39 lbs""#
Rounded to three sig figs, the number of sig figs you gave for the dosage and mass of drug administered, the answer will be
#color(green)(""187 lbs"")#
You have a little mor econverting to do to solve this one.
You know that the drug dosage is set to
To start, you can convert the dosage from miligrams per kilogram to miligrams per pound by using the conversion factor
#""1 kg"" ~= ""2.2046 lbs""#
This means that you have
#5.00""mg""/color(red)(cancel(color(black)(""kg""))) * (1color(red)(cancel(color(black)(""kg""))))/""2.2046 lbs"" = ""2.268 mg/lbs""#
So, your patient needs 2.268 mg of the drug per pound of body weight. This means that if you supply 425 mg, his body weight must be
#425color(red)(cancel(color(black)(""mg""))) * ""1 lbs""/(2.268color(red)(cancel(color(black)(""mg"")))) = ""187.39 lbs""#
Rounded to three sig figs, the number of sig figs you gave for the dosage and mass of drug administered, the answer will be
#color(green)(""187 lbs"")#
You have a little mor econverting to do to solve this one.
You know that the drug dosage is set to
To start, you can convert the dosage from miligrams per kilogram to miligrams per pound by using the conversion factor
#""1 kg"" ~= ""2.2046 lbs""#
This means that you have
#5.00""mg""/color(red)(cancel(color(black)(""kg""))) * (1color(red)(cancel(color(black)(""kg""))))/""2.2046 lbs"" = ""2.268 mg/lbs""#
So, your patient needs 2.268 mg of the drug per pound of body weight. This means that if you supply 425 mg, his body weight must be
#425color(red)(cancel(color(black)(""mg""))) * ""1 lbs""/(2.268color(red)(cancel(color(black)(""mg"")))) = ""187.39 lbs""#
Rounded to three sig figs, the number of sig figs you gave for the dosage and mass of drug administered, the answer will be
#color(green)(""187 lbs"")#
The patient weighs
Given/Known
Medication mass/body mass:
Dosage:
Just over 187 lb
Therefore
But you want the weight of the patient in lb, so you need to build in the conversion
Therefore the final equation is:
Plug in the numbers and you get 187.43 lb.
You can assume that the volume of the tire and the amount of air it contains remain constant, which means that you can focus solely on the relationship that exists between pressure and temperature under these conditions.
More specifically, you should know that when volume and number of moles are kept constant, temperature and pressure have a direct relationship - this is known as Gay Lussac's Law.
So, when temperature increases, pressure increases as well. Likewise, when temperature decreases, pressure decreases as well.
In your case, the pressure in the tire increased from
Mathematically, you can write this as
#color(blue)(|bar(ul(color(white)(a/a)P_1/T_1 = P_2/T_2color(white)(a/a)|)))"" ""# , where
Rearrange the equation to solve for
#P_1/T_1 = P_2/T_2 implies T_2 = P_2/P_1 * T_1#
Plug in your values to get
#T_2 = (238color(red)(cancel(color(black)(""kPa""))))/(205color(red)(cancel(color(black)(""kPa"")))) * ""303 K"" = ""351.78 K""#
Rounded to three sig figs, the answer will be
#T_2 = color(green)(|bar(ul(color(white)(a/a)""352 K""color(white)(a/a)|)))#
As predicted, the temperature of the air increased.
You can assume that the volume of the tire and the amount of air it contains remain constant, which means that you can focus solely on the relationship that exists between pressure and temperature under these conditions.
More specifically, you should know that when volume and number of moles are kept constant, temperature and pressure have a direct relationship - this is known as Gay Lussac's Law.
So, when temperature increases, pressure increases as well. Likewise, when temperature decreases, pressure decreases as well.
In your case, the pressure in the tire increased from
Mathematically, you can write this as
#color(blue)(|bar(ul(color(white)(a/a)P_1/T_1 = P_2/T_2color(white)(a/a)|)))"" ""# , where
Rearrange the equation to solve for
#P_1/T_1 = P_2/T_2 implies T_2 = P_2/P_1 * T_1#
Plug in your values to get
#T_2 = (238color(red)(cancel(color(black)(""kPa""))))/(205color(red)(cancel(color(black)(""kPa"")))) * ""303 K"" = ""351.78 K""#
Rounded to three sig figs, the answer will be
#T_2 = color(green)(|bar(ul(color(white)(a/a)""352 K""color(white)(a/a)|)))#
As predicted, the temperature of the air increased.
You can assume that the volume of the tire and the amount of air it contains remain constant, which means that you can focus solely on the relationship that exists between pressure and temperature under these conditions.
More specifically, you should know that when volume and number of moles are kept constant, temperature and pressure have a direct relationship - this is known as Gay Lussac's Law.
So, when temperature increases, pressure increases as well. Likewise, when temperature decreases, pressure decreases as well.
In your case, the pressure in the tire increased from
Mathematically, you can write this as
#color(blue)(|bar(ul(color(white)(a/a)P_1/T_1 = P_2/T_2color(white)(a/a)|)))"" ""# , where
Rearrange the equation to solve for
#P_1/T_1 = P_2/T_2 implies T_2 = P_2/P_1 * T_1#
Plug in your values to get
#T_2 = (238color(red)(cancel(color(black)(""kPa""))))/(205color(red)(cancel(color(black)(""kPa"")))) * ""303 K"" = ""351.78 K""#
Rounded to three sig figs, the answer will be
#T_2 = color(green)(|bar(ul(color(white)(a/a)""352 K""color(white)(a/a)|)))#
As predicted, the temperature of the air increased.
Here's how I go about doing this.
What we can do is use the ideal gas equation for both
I won't show the unit conversions here, as I predict you already know how; the moles of each substance is
We'll now use the ideal gas equation to solve for the pressure at the new conditions:
Here's how I go about doing this.
What we can do is use the ideal gas equation for both
I won't show the unit conversions here, as I predict you already know how; the moles of each substance is
We'll now use the ideal gas equation to solve for the pressure at the new conditions:
Here's how I go about doing this.
What we can do is use the ideal gas equation for both
I won't show the unit conversions here, as I predict you already know how; the moles of each substance is
We'll now use the ideal gas equation to solve for the pressure at the new conditions:
The first thing that you need to do here is to use the density of the solution to determine the mass of
The density of the solution is said to be equal to
This means that your sample will have a mass of
#150 color(red)(cancel(color(black)(""mL""))) * ""1.1 g""/(1color(red)(cancel(color(black)(""mL"")))) = ""165 g""#
Now, in order to find the solution's percent concentration by mass,
Since you know that
#100 color(red)(cancel(color(black)(""g solution""))) * ""30. g HCl""/(165color(red)(cancel(color(black)(""g solution"")))) = ""18.18 g HCl""#
You can thus say that the solution has a percent concentration by mass equal to
#color(darkgreen)(ul(color(black)(""% m/m = 18% HCl"")))#
The answer is rounded to two sig figs.
The first thing that you need to do here is to use the density of the solution to determine the mass of
The density of the solution is said to be equal to
This means that your sample will have a mass of
#150 color(red)(cancel(color(black)(""mL""))) * ""1.1 g""/(1color(red)(cancel(color(black)(""mL"")))) = ""165 g""#
Now, in order to find the solution's percent concentration by mass,
Since you know that
#100 color(red)(cancel(color(black)(""g solution""))) * ""30. g HCl""/(165color(red)(cancel(color(black)(""g solution"")))) = ""18.18 g HCl""#
You can thus say that the solution has a percent concentration by mass equal to
#color(darkgreen)(ul(color(black)(""% m/m = 18% HCl"")))#
The answer is rounded to two sig figs.
The first thing that you need to do here is to use the density of the solution to determine the mass of
The density of the solution is said to be equal to
This means that your sample will have a mass of
#150 color(red)(cancel(color(black)(""mL""))) * ""1.1 g""/(1color(red)(cancel(color(black)(""mL"")))) = ""165 g""#
Now, in order to find the solution's percent concentration by mass,
Since you know that
#100 color(red)(cancel(color(black)(""g solution""))) * ""30. g HCl""/(165color(red)(cancel(color(black)(""g solution"")))) = ""18.18 g HCl""#
You can thus say that the solution has a percent concentration by mass equal to
#color(darkgreen)(ul(color(black)(""% m/m = 18% HCl"")))#
The answer is rounded to two sig figs.
We know the molality (m), but not the number of moles of solute. The mass of solute can be converted into moles using the molar mass of HgCl (236.04 g/mol) as a conversion factor:
Since the moles of solute and molality are known, we can rearrange the equation to solve for mass of solvent :
Lastly, the mass of solvent has to be converted from kg to g using the following conversion factor:
Therefore,
4,980 g
4,980 g
We know the molality (m), but not the number of moles of solute. The mass of solute can be converted into moles using the molar mass of HgCl (236.04 g/mol) as a conversion factor:
Since the moles of solute and molality are known, we can rearrange the equation to solve for mass of solvent :
Lastly, the mass of solvent has to be converted from kg to g using the following conversion factor:
Therefore,
4,980 g
4,980 g
We know the molality (m), but not the number of moles of solute. The mass of solute can be converted into moles using the molar mass of HgCl (236.04 g/mol) as a conversion factor:
Since the moles of solute and molality are known, we can rearrange the equation to solve for mass of solvent :
Lastly, the mass of solvent has to be converted from kg to g using the following conversion factor:
Therefore,
4,980 g
The product
Anyway, as to your problem,
Note that we typically buy conc.
The product
Anyway, as to your problem,
Note that we typically buy conc.
The product
Anyway, as to your problem,
Note that we typically buy conc.
H2(g) + I2(g) --> 2HI(g)
deltaH(rxn) = +53 kJ
I happen to know the answer is 5.2 kJ, but I'm not sure how to get to it. I'm studying for finals and need help. Thank you!
The thermochemical equation given to you tells you how much heat is needed in order to produce
In other words, you know that in order to produce
#DeltaH_""rxn"" = + ""53 kJ""#
The plus sign tells you that this reaction taken is
#color(blue)(ul(color(black)(""1 mole I""_2 color(white)(.)""and 1 mole H""_2 -> ""53 kJ of heat consumed"")))#
So, convert the samples of hydrogen gas and iodine to moles by using the molar masses of the two reactants.
#5.0 color(red)(cancel(color(black)(""g""))) * ""1 mole H""_2/(2.016color(red)(cancel(color(black)(""g"")))) = ""2.48 moles H""_2#
#25 color(red)(cancel(color(black)(""g""))) * ""1 mole I""_2/(253.81color(red)(cancel(color(black)(""g"")))) = ""0.0985 moles I""_2#
The reaction consumes hydrogen gas and iodine in a
Therefore, the reaction will consume
This means that the reaction will require
#0.0985 color(red)(cancel(color(black)(""moles I""_2))) * ""53 kJ""/(1color(red)(cancel(color(black)(""mole I""_2)))) = color(darkgreen)(ul(color(black)(""5.2 kJ"")))#
of heat. The answer is rounded to two sig figs. You can thus say that you have
#DeltaH_ (""rxn for 0.0985 moles I""_2 color(white)(.)""and H""_2) = + ""5.2 kJ""#
The thermochemical equation given to you tells you how much heat is needed in order to produce
In other words, you know that in order to produce
#DeltaH_""rxn"" = + ""53 kJ""#
The plus sign tells you that this reaction taken is
#color(blue)(ul(color(black)(""1 mole I""_2 color(white)(.)""and 1 mole H""_2 -> ""53 kJ of heat consumed"")))#
So, convert the samples of hydrogen gas and iodine to moles by using the molar masses of the two reactants.
#5.0 color(red)(cancel(color(black)(""g""))) * ""1 mole H""_2/(2.016color(red)(cancel(color(black)(""g"")))) = ""2.48 moles H""_2#
#25 color(red)(cancel(color(black)(""g""))) * ""1 mole I""_2/(253.81color(red)(cancel(color(black)(""g"")))) = ""0.0985 moles I""_2#
The reaction consumes hydrogen gas and iodine in a
Therefore, the reaction will consume
This means that the reaction will require
#0.0985 color(red)(cancel(color(black)(""moles I""_2))) * ""53 kJ""/(1color(red)(cancel(color(black)(""mole I""_2)))) = color(darkgreen)(ul(color(black)(""5.2 kJ"")))#
of heat. The answer is rounded to two sig figs. You can thus say that you have
#DeltaH_ (""rxn for 0.0985 moles I""_2 color(white)(.)""and H""_2) = + ""5.2 kJ""#
H2(g) + I2(g) --> 2HI(g)
deltaH(rxn) = +53 kJ
I happen to know the answer is 5.2 kJ, but I'm not sure how to get to it. I'm studying for finals and need help. Thank you!
The thermochemical equation given to you tells you how much heat is needed in order to produce
In other words, you know that in order to produce
#DeltaH_""rxn"" = + ""53 kJ""#
The plus sign tells you that this reaction taken is
#color(blue)(ul(color(black)(""1 mole I""_2 color(white)(.)""and 1 mole H""_2 -> ""53 kJ of heat consumed"")))#
So, convert the samples of hydrogen gas and iodine to moles by using the molar masses of the two reactants.
#5.0 color(red)(cancel(color(black)(""g""))) * ""1 mole H""_2/(2.016color(red)(cancel(color(black)(""g"")))) = ""2.48 moles H""_2#
#25 color(red)(cancel(color(black)(""g""))) * ""1 mole I""_2/(253.81color(red)(cancel(color(black)(""g"")))) = ""0.0985 moles I""_2#
The reaction consumes hydrogen gas and iodine in a
Therefore, the reaction will consume
This means that the reaction will require
#0.0985 color(red)(cancel(color(black)(""moles I""_2))) * ""53 kJ""/(1color(red)(cancel(color(black)(""mole I""_2)))) = color(darkgreen)(ul(color(black)(""5.2 kJ"")))#
of heat. The answer is rounded to two sig figs. You can thus say that you have
#DeltaH_ (""rxn for 0.0985 moles I""_2 color(white)(.)""and H""_2) = + ""5.2 kJ""#
The important thing to notice here is that the reaction takes place at STP conditions, which are defined as a pressure of
Moreover, at STP one mole of any ideal gas occupies exactly
Since all the gases are at the same conditions for pressure and temperature, the mole ratios become volume ratios.
To prove this, use the ideal gas law equation to write the number of moles of hydrogen gas and of chlorine gas as
#PV = nRT implies n = (PV)/(RT)#
For hydrogen, you would have
#n_""hydrogen"" = (P * V_""hydrogen"")/(RT)#
and for chlorine you have
#n_""chlorine"" = (P * V_""chlorine"")/(RT)#
Thus, the mole ratio between hydrogen and chlorine will be
#n_""hydrogen""/n_""chlorine""= (color(red)(cancel(color(black)(P))) V_""hydronge"")/color(red)(cancel(color(black)(RT))) * color(red)(cancel(color(black)(RT)))/(color(red)(cancel(color(black)(P))) * V_""chlorine"") = V_""hydrogen""/V_""chlorine""#
The same principle applies to the mole ratio that exists between hydrogen and hydrogen chloride.
So, the balanced chemical equation for this reaction is
#""H""_text(2(g]) + ""Cl""_text(2(g]) -> color(blue)(2)""HCl""_text((g])#
Notice that you have a
This means that the reaction will produce twice as many moles as you the number of moles of hydrogen gas that reacts.
Use the volume ratio to find what volume of hydrogen chloride will be produced by the reaction
#4.9color(red)(cancel(color(black)("" L H""_2))) * (color(blue)(2)"" L HCl"")/(1color(red)(cancel(color(black)("" L H""_2)))) = ""9.8 L HCl""#
Now use the molar volume to find how many moles you'd get in this volume of gas at STP
#9.8color(red)(cancel(color(black)("" L HCl""))) * ""1 mole HCl""/(22.7color(red)(cancel(color(black)("" L HCl"")))) = ""0.4317 moles HCl""#
Finally, use hydrogen chloride's molar mass to find how many grams would contain this many moles
#0.4317color(red)(cancel(color(black)(""moles HCl""))) * ""36.461 g""/(1color(red)(cancel(color(black)(""mole HCl"")))) = ""15.74 g""#
Rounded to two sig figs, the answer will be
#m_""HCl"" = color(green)(""16 g"")#
The important thing to notice here is that the reaction takes place at STP conditions, which are defined as a pressure of
Moreover, at STP one mole of any ideal gas occupies exactly
Since all the gases are at the same conditions for pressure and temperature, the mole ratios become volume ratios.
To prove this, use the ideal gas law equation to write the number of moles of hydrogen gas and of chlorine gas as
#PV = nRT implies n = (PV)/(RT)#
For hydrogen, you would have
#n_""hydrogen"" = (P * V_""hydrogen"")/(RT)#
and for chlorine you have
#n_""chlorine"" = (P * V_""chlorine"")/(RT)#
Thus, the mole ratio between hydrogen and chlorine will be
#n_""hydrogen""/n_""chlorine""= (color(red)(cancel(color(black)(P))) V_""hydronge"")/color(red)(cancel(color(black)(RT))) * color(red)(cancel(color(black)(RT)))/(color(red)(cancel(color(black)(P))) * V_""chlorine"") = V_""hydrogen""/V_""chlorine""#
The same principle applies to the mole ratio that exists between hydrogen and hydrogen chloride.
So, the balanced chemical equation for this reaction is
#""H""_text(2(g]) + ""Cl""_text(2(g]) -> color(blue)(2)""HCl""_text((g])#
Notice that you have a
This means that the reaction will produce twice as many moles as you the number of moles of hydrogen gas that reacts.
Use the volume ratio to find what volume of hydrogen chloride will be produced by the reaction
#4.9color(red)(cancel(color(black)("" L H""_2))) * (color(blue)(2)"" L HCl"")/(1color(red)(cancel(color(black)("" L H""_2)))) = ""9.8 L HCl""#
Now use the molar volume to find how many moles you'd get in this volume of gas at STP
#9.8color(red)(cancel(color(black)("" L HCl""))) * ""1 mole HCl""/(22.7color(red)(cancel(color(black)("" L HCl"")))) = ""0.4317 moles HCl""#
Finally, use hydrogen chloride's molar mass to find how many grams would contain this many moles
#0.4317color(red)(cancel(color(black)(""moles HCl""))) * ""36.461 g""/(1color(red)(cancel(color(black)(""mole HCl"")))) = ""15.74 g""#
Rounded to two sig figs, the answer will be
#m_""HCl"" = color(green)(""16 g"")#
The important thing to notice here is that the reaction takes place at STP conditions, which are defined as a pressure of
Moreover, at STP one mole of any ideal gas occupies exactly
Since all the gases are at the same conditions for pressure and temperature, the mole ratios become volume ratios.
To prove this, use the ideal gas law equation to write the number of moles of hydrogen gas and of chlorine gas as
#PV = nRT implies n = (PV)/(RT)#
For hydrogen, you would have
#n_""hydrogen"" = (P * V_""hydrogen"")/(RT)#
and for chlorine you have
#n_""chlorine"" = (P * V_""chlorine"")/(RT)#
Thus, the mole ratio between hydrogen and chlorine will be
#n_""hydrogen""/n_""chlorine""= (color(red)(cancel(color(black)(P))) V_""hydronge"")/color(red)(cancel(color(black)(RT))) * color(red)(cancel(color(black)(RT)))/(color(red)(cancel(color(black)(P))) * V_""chlorine"") = V_""hydrogen""/V_""chlorine""#
The same principle applies to the mole ratio that exists between hydrogen and hydrogen chloride.
So, the balanced chemical equation for this reaction is
#""H""_text(2(g]) + ""Cl""_text(2(g]) -> color(blue)(2)""HCl""_text((g])#
Notice that you have a
This means that the reaction will produce twice as many moles as you the number of moles of hydrogen gas that reacts.
Use the volume ratio to find what volume of hydrogen chloride will be produced by the reaction
#4.9color(red)(cancel(color(black)("" L H""_2))) * (color(blue)(2)"" L HCl"")/(1color(red)(cancel(color(black)("" L H""_2)))) = ""9.8 L HCl""#
Now use the molar volume to find how many moles you'd get in this volume of gas at STP
#9.8color(red)(cancel(color(black)("" L HCl""))) * ""1 mole HCl""/(22.7color(red)(cancel(color(black)("" L HCl"")))) = ""0.4317 moles HCl""#
Finally, use hydrogen chloride's molar mass to find how many grams would contain this many moles
#0.4317color(red)(cancel(color(black)(""moles HCl""))) * ""36.461 g""/(1color(red)(cancel(color(black)(""mole HCl"")))) = ""15.74 g""#
Rounded to two sig figs, the answer will be
#m_""HCl"" = color(green)(""16 g"")#
Concentration of HCl is
as
[
pH =
Concentration of HCl is
as
[
pH =
Concentration of HCl is
as
[
pH =
First, let's rewrite
Now we can see how many of each element/polyatomic ion we have on each side of the equation.
Left:
Na - 1
H - 1
OH - 1
Right:
Na - 1
OH -1
H -2
Looking at what we have, there is an unequal amount of hydrogen on the left side and the right side. We have only 1 on the left and 2 on the right. Let's double everything on the left. Now we get:
Since we have 2 of everything on the left, we should make sure that the right side is accounted for. As mentioned in our original count, the
Changing back
First, let's rewrite
Now we can see how many of each element/polyatomic ion we have on each side of the equation.
Left:
Na - 1
H - 1
OH - 1
Right:
Na - 1
OH -1
H -2
Looking at what we have, there is an unequal amount of hydrogen on the left side and the right side. We have only 1 on the left and 2 on the right. Let's double everything on the left. Now we get:
Since we have 2 of everything on the left, we should make sure that the right side is accounted for. As mentioned in our original count, the
Changing back
First, let's rewrite
Now we can see how many of each element/polyatomic ion we have on each side of the equation.
Left:
Na - 1
H - 1
OH - 1
Right:
Na - 1
OH -1
H -2
Looking at what we have, there is an unequal amount of hydrogen on the left side and the right side. We have only 1 on the left and 2 on the right. Let's double everything on the left. Now we get:
Since we have 2 of everything on the left, we should make sure that the right side is accounted for. As mentioned in our original count, the
Changing back
We use the Ideal Gas equation,
We are left with units of
We use the Ideal Gas equation,
We are left with units of
We use the Ideal Gas equation,
We are left with units of
The stoichiometry of the equation is crucial; inasmuch as the stoichiometry, the chemical proportion, shows that hydrogen fluoride and sodium fluoride are formed in equal amounts, and all I need are the molecular weights of hydrogen fluoride (
So moles of
Moles of
The stoichiometry of the equation is crucial; inasmuch as the stoichiometry, the chemical proportion, shows that hydrogen fluoride and sodium fluoride are formed in equal amounts, and all I need are the molecular weights of hydrogen fluoride (
So moles of
Moles of
The stoichiometry of the equation is crucial; inasmuch as the stoichiometry, the chemical proportion, shows that hydrogen fluoride and sodium fluoride are formed in equal amounts, and all I need are the molecular weights of hydrogen fluoride (
So moles of
Moles of
I would use the relationship describing heat exchanged
With our data:
Rearranging:
I found
I would use the relationship describing heat exchanged
With our data:
Rearranging:
I found
I would use the relationship describing heat exchanged
With our data:
Rearranging:
Gay-Lussac's law states that the pressure of a gas held at constant volume is directly proportional to its temperature in Kelvins.
The equation for Charles' law is
Rearrange the equation to isolate
The temperature at which the gas pressure inside the tank will equal 825 kPa is 508 K.
Gay-Lussac's law states that the pressure of a gas held at constant volume is directly proportional to its temperature in Kelvins.
The equation for Charles' law is
Rearrange the equation to isolate
The temperature at which the gas pressure inside the tank will equal 825 kPa is 508 K.
Gay-Lussac's law states that the pressure of a gas held at constant volume is directly proportional to its temperature in Kelvins.
The equation for Charles' law is
Rearrange the equation to isolate
Specific heat is given by
#""S"" = ""Q""/""mΔT""#
Where
Specific heat is given by
#""S"" = ""Q""/""mΔT""#
Where
Specific heat is given by
#""S"" = ""Q""/""mΔT""#
Where
57.45 g
57.45 g
Given substance is Sodium,
Therefore,
There is 1:1 equivalence between acid and base.
Thus
There is 1:1 equivalence between acid and base.
Thus
There is 1:1 equivalence between acid and base.
Thus
Even without doing any calculation, you could predict that the empirical formula for this compound will come out to be
So if you're familair with ionic compounds and polyatomic ions, you can predict that the answer will be
Let's do some calculations to see if the prediction turns out to be correct.
Your strategy here will be to pick a sample of this compound and use the given percent composition to find how many grams of each element you'd get in this sample.
To make calculations easier, you can pic ka
#""18.8 g""# of lithium#""16.3 g""# of carbon#""64.9 g""# of oxygen
Next, use the molar mass of each element to determine how many moles of each you'd get in this sample
#""For Li: "" 18.8 color(red)(cancel(color(black)(""g""))) * ""1 mole Li""/(6.941color(red)(cancel(color(black)(""g"")))) = ""2.709 moles Li""#
#""For C: "" 16.3 color(red)(cancel(color(black)(""g""))) * ""1 mole C""/(12.011color(red)(cancel(color(black)(""g"")))) = ""1.357 moles C""#
#""For O: "" 64.9 color(red)(cancel(color(black)(""g""))) * ""1 mole O""/(15.9994 color(red)(cancel(color(black)(""g"")))) = ""4.056 moles O""#
Now, a compound's empirical formula gives you the smallest whole number ratio that exists between the elements that make up the compound.
To get the mole ratio that exists between these three elements, divide these three values by the smallest one.
#""For Li: "" (2.709 color(red)(cancel(color(black)(""moles""))))/(1.357color(red)(cancel(color(black)(""moles"")))) = 1.996 ~~ 2#
#""For C: "" (1.357 color(red)(cancel(color(black)(""moles""))))/(1.357color(red)(cancel(color(black)(""moles"")))) = 1#
#""For O: "" (4.056 color(red)(cancel(color(black)(""moles""))))/(1.357color(red)(cancel(color(black)(""moles"")))) = 2.989 ~~ 3#
Since this ratio, i.e.
#""Li""_2""CO""_3 -># lithium carbonate
Even without doing any calculation, you could predict that the empirical formula for this compound will come out to be
So if you're familair with ionic compounds and polyatomic ions, you can predict that the answer will be
Let's do some calculations to see if the prediction turns out to be correct.
Your strategy here will be to pick a sample of this compound and use the given percent composition to find how many grams of each element you'd get in this sample.
To make calculations easier, you can pic ka
#""18.8 g""# of lithium#""16.3 g""# of carbon#""64.9 g""# of oxygen
Next, use the molar mass of each element to determine how many moles of each you'd get in this sample
#""For Li: "" 18.8 color(red)(cancel(color(black)(""g""))) * ""1 mole Li""/(6.941color(red)(cancel(color(black)(""g"")))) = ""2.709 moles Li""#
#""For C: "" 16.3 color(red)(cancel(color(black)(""g""))) * ""1 mole C""/(12.011color(red)(cancel(color(black)(""g"")))) = ""1.357 moles C""#
#""For O: "" 64.9 color(red)(cancel(color(black)(""g""))) * ""1 mole O""/(15.9994 color(red)(cancel(color(black)(""g"")))) = ""4.056 moles O""#
Now, a compound's empirical formula gives you the smallest whole number ratio that exists between the elements that make up the compound.
To get the mole ratio that exists between these three elements, divide these three values by the smallest one.
#""For Li: "" (2.709 color(red)(cancel(color(black)(""moles""))))/(1.357color(red)(cancel(color(black)(""moles"")))) = 1.996 ~~ 2#
#""For C: "" (1.357 color(red)(cancel(color(black)(""moles""))))/(1.357color(red)(cancel(color(black)(""moles"")))) = 1#
#""For O: "" (4.056 color(red)(cancel(color(black)(""moles""))))/(1.357color(red)(cancel(color(black)(""moles"")))) = 2.989 ~~ 3#
Since this ratio, i.e.
#""Li""_2""CO""_3 -># lithium carbonate
Even without doing any calculation, you could predict that the empirical formula for this compound will come out to be
So if you're familair with ionic compounds and polyatomic ions, you can predict that the answer will be
Let's do some calculations to see if the prediction turns out to be correct.
Your strategy here will be to pick a sample of this compound and use the given percent composition to find how many grams of each element you'd get in this sample.
To make calculations easier, you can pic ka
#""18.8 g""# of lithium#""16.3 g""# of carbon#""64.9 g""# of oxygen
Next, use the molar mass of each element to determine how many moles of each you'd get in this sample
#""For Li: "" 18.8 color(red)(cancel(color(black)(""g""))) * ""1 mole Li""/(6.941color(red)(cancel(color(black)(""g"")))) = ""2.709 moles Li""#
#""For C: "" 16.3 color(red)(cancel(color(black)(""g""))) * ""1 mole C""/(12.011color(red)(cancel(color(black)(""g"")))) = ""1.357 moles C""#
#""For O: "" 64.9 color(red)(cancel(color(black)(""g""))) * ""1 mole O""/(15.9994 color(red)(cancel(color(black)(""g"")))) = ""4.056 moles O""#
Now, a compound's empirical formula gives you the smallest whole number ratio that exists between the elements that make up the compound.
To get the mole ratio that exists between these three elements, divide these three values by the smallest one.
#""For Li: "" (2.709 color(red)(cancel(color(black)(""moles""))))/(1.357color(red)(cancel(color(black)(""moles"")))) = 1.996 ~~ 2#
#""For C: "" (1.357 color(red)(cancel(color(black)(""moles""))))/(1.357color(red)(cancel(color(black)(""moles"")))) = 1#
#""For O: "" (4.056 color(red)(cancel(color(black)(""moles""))))/(1.357color(red)(cancel(color(black)(""moles"")))) = 2.989 ~~ 3#
Since this ratio, i.e.
#""Li""_2""CO""_3 -># lithium carbonate
This is so since sodium is in group
Overall nett charge of molecule must be zero.
The ions will be held together by strong ionic bonds and the molecules by strong coulombic forces of attraction.
This is so since sodium is in group
Overall nett charge of molecule must be zero.
The ions will be held together by strong ionic bonds and the molecules by strong coulombic forces of attraction.
This is so since sodium is in group
Overall nett charge of molecule must be zero.
The ions will be held together by strong ionic bonds and the molecules by strong coulombic forces of attraction.
From the reaction equation coefficients, we can understand that for every
However,
So
From the reaction equation coefficients, we can understand that for every
However,
So
From the reaction equation coefficients, we can understand that for every
However,
So
The conjugate acid of a base is the original species PLUS a proton,
What are the conjugate bases of
The conjugate acid of a base is the original species PLUS a proton,
What are the conjugate bases of
The conjugate acid of a base is the original species PLUS a proton,
What are the conjugate bases of
yet again, i have no idea what im doing
Start with a balanced equation. It will be used to determine mol ratios between the product and the reactant.
Multiply the given mol oxygen gas by the mol ratio
The given amount of oxygen gas will produce
Start with a balanced equation. It will be used to determine mol ratios between the product and the reactant.
Multiply the given mol oxygen gas by the mol ratio
yet again, i have no idea what im doing
The given amount of oxygen gas will produce
Start with a balanced equation. It will be used to determine mol ratios between the product and the reactant.
Multiply the given mol oxygen gas by the mol ratio
This illustrates old Dalton's Law of Partial Pressures. In a gaseous mixture, the partial pressure exerted by a component gas is the same as if it alone occupied the container.....
And thus
Where
And here the container is conveniently the atmosphere, which to a first approx. is
And so what is
This illustrates old Dalton's Law of Partial Pressures. In a gaseous mixture, the partial pressure exerted by a component gas is the same as if it alone occupied the container.....
And thus
Where
And here the container is conveniently the atmosphere, which to a first approx. is
And so what is
This illustrates old Dalton's Law of Partial Pressures. In a gaseous mixture, the partial pressure exerted by a component gas is the same as if it alone occupied the container.....
And thus
Where
And here the container is conveniently the atmosphere, which to a first approx. is
And so what is
So,
The hexahydrate has a beautiful purple colour. The anhydrous salt,
So,
The hexahydrate has a beautiful purple colour. The anhydrous salt,
So,
The hexahydrate has a beautiful purple colour. The anhydrous salt,
Now, the basic equation is
So here,
Certainly, this expression is dimensionally consistent. I wanted an answer with units of mass, and the calculation did in fact give such units.
Now, the basic equation is
So here,
Certainly, this expression is dimensionally consistent. I wanted an answer with units of mass, and the calculation did in fact give such units.
Now, the basic equation is
So here,
Certainly, this expression is dimensionally consistent. I wanted an answer with units of mass, and the calculation did in fact give such units.
The mass of
Multiply moles by molar mass.
The molar mass of acetic acid
https://www.ncbi.nlm.nih.gov/pccompound?term=%22acetic+acid%22
The thing to keep in mind about formula units is that you need
In this case, you know that
Moreover, you know that calcium chloride has a molar mass of
So if
This means that your sample has a mass of
#1.2 * 10^8 color(red)(cancel(color(black)(""f. units CaCl""_2))) * ""110.98 g""/(6.022 * 10^(23)color(red)(cancel(color(black)(""f. units CaCl""_2)))) = color(darkgreen)(ul(color(black)(2.2 * 10^(-4) quad ""g"")))#
The answer is rounded to two sig figs, the number of sig figs you have for the number of formula units.
The thing to keep in mind about formula units is that you need
In this case, you know that
Moreover, you know that calcium chloride has a molar mass of
So if
This means that your sample has a mass of
#1.2 * 10^8 color(red)(cancel(color(black)(""f. units CaCl""_2))) * ""110.98 g""/(6.022 * 10^(23)color(red)(cancel(color(black)(""f. units CaCl""_2)))) = color(darkgreen)(ul(color(black)(2.2 * 10^(-4) quad ""g"")))#
The answer is rounded to two sig figs, the number of sig figs you have for the number of formula units.
The thing to keep in mind about formula units is that you need
In this case, you know that
Moreover, you know that calcium chloride has a molar mass of
So if
This means that your sample has a mass of
#1.2 * 10^8 color(red)(cancel(color(black)(""f. units CaCl""_2))) * ""110.98 g""/(6.022 * 10^(23)color(red)(cancel(color(black)(""f. units CaCl""_2)))) = color(darkgreen)(ul(color(black)(2.2 * 10^(-4) quad ""g"")))#
The answer is rounded to two sig figs, the number of sig figs you have for the number of formula units.
To make this problem more interesting, let's calculate the mass of a single atom of nitrogen first, then use that value as a conversion factor to determine the mass of
The starting point here will be the molar mass of nitrogen, which is listed as
#M_(""M N"") = ""14.00674 g mol""^(-1)#
This tells you that one mole of nitrogen has a mass of
#1 color(red)(cancel(color(black)(""atom N""))) * ""14.00674 g""/(6.022 * 10^(23)color(red)(cancel(color(black)(""atoms N"")))) = 2.326 * 10^(-23)""g""#
Now all you have to do is multiply this value by the number of atoms given to you to find
#9.76 * 10^(12) color(red)(cancel(color(black)(""atoms N""))) * (2.326 * 10^(-23)""g"")/(1color(red)(cancel(color(black)(""atom N"")))) = color(green)(|bar(ul(color(white)(a/a)color(black)(2.27 * 10^(-10)""g"")color(white)(a/a)|)))#
The answer is rounded to three sig figs.
To make this problem more interesting, let's calculate the mass of a single atom of nitrogen first, then use that value as a conversion factor to determine the mass of
The starting point here will be the molar mass of nitrogen, which is listed as
#M_(""M N"") = ""14.00674 g mol""^(-1)#
This tells you that one mole of nitrogen has a mass of
#1 color(red)(cancel(color(black)(""atom N""))) * ""14.00674 g""/(6.022 * 10^(23)color(red)(cancel(color(black)(""atoms N"")))) = 2.326 * 10^(-23)""g""#
Now all you have to do is multiply this value by the number of atoms given to you to find
#9.76 * 10^(12) color(red)(cancel(color(black)(""atoms N""))) * (2.326 * 10^(-23)""g"")/(1color(red)(cancel(color(black)(""atom N"")))) = color(green)(|bar(ul(color(white)(a/a)color(black)(2.27 * 10^(-10)""g"")color(white)(a/a)|)))#
The answer is rounded to three sig figs.
To make this problem more interesting, let's calculate the mass of a single atom of nitrogen first, then use that value as a conversion factor to determine the mass of
The starting point here will be the molar mass of nitrogen, which is listed as
#M_(""M N"") = ""14.00674 g mol""^(-1)#
This tells you that one mole of nitrogen has a mass of
#1 color(red)(cancel(color(black)(""atom N""))) * ""14.00674 g""/(6.022 * 10^(23)color(red)(cancel(color(black)(""atoms N"")))) = 2.326 * 10^(-23)""g""#
Now all you have to do is multiply this value by the number of atoms given to you to find
#9.76 * 10^(12) color(red)(cancel(color(black)(""atoms N""))) * (2.326 * 10^(-23)""g"")/(1color(red)(cancel(color(black)(""atom N"")))) = color(green)(|bar(ul(color(white)(a/a)color(black)(2.27 * 10^(-10)""g"")color(white)(a/a)|)))#
The answer is rounded to three sig figs.
So
So
So
The heat absorbed is
The mass is
The specific heat of copper is
The change in temperature is
Therefore,
The heat absorbed is
The heat absorbed is
The mass is
The specific heat of copper is
The change in temperature is
Therefore,
The heat absorbed is
The heat absorbed is
The mass is
The specific heat of copper is
The change in temperature is
Therefore,
Here are all informations that we have :
Then we start
Firstly the formula,
SHC
Then we replace the values and calculate it
Therefore, heat absorbed
(I put the answer at 3 significant figures (s.f) but for further calculations use the first value obtained)
As it stands, the problem fails to provide you with an essential piece of information - the value of aluminium's specific heat.
This means that you're going to have to track it yourself. You can find it listed here
http://www.engineeringtoolbox.com/specific-heat-metals-d_152.html
as being equal to
#c = ""0.91 J g""^(-1)""""^@""C""^(-1)#
So, a substance's specific heat tells you how much heat is needed in order to increase the temperature of
In this case, you need to provide aluminium with
You know that the block of aluminium has a mass
#42.7 color(red)(cancel(color(black)(""g""))) xx ""0.91 J"" color(red)(cancel(color(black)(""g""^(-1))))""""^@""C""^(-1) = ""38.86 J"" """"^@""C""^(-1)#
This much heat would increase the temperature of
#15.2color(red)(cancel(color(black)(""""^@""C""))) xx ""38.86 J"" color(red)(cancel(color(black)(""""^@""C""^(-1)))) = ""590.67 J""#
Rounded to three sig figs, the answer will be
#""heat needed"" = color(green)(|bar(ul(color(white)(a/a)""591 J""color(white)(a/a)|)))#
Alternatively, you can use the formula
#color(blue)(|bar(ul(color(white)(a/a)q = m * c * DeltaTcolor(white)(a/a)|)))"" ""# , where
Plug in your values to get
#q = 42.7 color(red)(cancel(color(black)(""g""))) * ""0.91 J"" color(red)(cancel(color(black)(""g""^(-1))))color(red)(cancel(color(black)(""""^@""C""^(-1)))) * 15.2color(red)(cancel(color(black)(""""^@""C"")))#
#q = color(green)(|bar(ul(color(white)(a/a)""591 J""color(white)(a/a)|)))#
As it stands, the problem fails to provide you with an essential piece of information - the value of aluminium's specific heat.
This means that you're going to have to track it yourself. You can find it listed here
http://www.engineeringtoolbox.com/specific-heat-metals-d_152.html
as being equal to
#c = ""0.91 J g""^(-1)""""^@""C""^(-1)#
So, a substance's specific heat tells you how much heat is needed in order to increase the temperature of
In this case, you need to provide aluminium with
You know that the block of aluminium has a mass
#42.7 color(red)(cancel(color(black)(""g""))) xx ""0.91 J"" color(red)(cancel(color(black)(""g""^(-1))))""""^@""C""^(-1) = ""38.86 J"" """"^@""C""^(-1)#
This much heat would increase the temperature of
#15.2color(red)(cancel(color(black)(""""^@""C""))) xx ""38.86 J"" color(red)(cancel(color(black)(""""^@""C""^(-1)))) = ""590.67 J""#
Rounded to three sig figs, the answer will be
#""heat needed"" = color(green)(|bar(ul(color(white)(a/a)""591 J""color(white)(a/a)|)))#
Alternatively, you can use the formula
#color(blue)(|bar(ul(color(white)(a/a)q = m * c * DeltaTcolor(white)(a/a)|)))"" ""# , where
Plug in your values to get
#q = 42.7 color(red)(cancel(color(black)(""g""))) * ""0.91 J"" color(red)(cancel(color(black)(""g""^(-1))))color(red)(cancel(color(black)(""""^@""C""^(-1)))) * 15.2color(red)(cancel(color(black)(""""^@""C"")))#
#q = color(green)(|bar(ul(color(white)(a/a)""591 J""color(white)(a/a)|)))#
As it stands, the problem fails to provide you with an essential piece of information - the value of aluminium's specific heat.
This means that you're going to have to track it yourself. You can find it listed here
http://www.engineeringtoolbox.com/specific-heat-metals-d_152.html
as being equal to
#c = ""0.91 J g""^(-1)""""^@""C""^(-1)#
So, a substance's specific heat tells you how much heat is needed in order to increase the temperature of
In this case, you need to provide aluminium with
You know that the block of aluminium has a mass
#42.7 color(red)(cancel(color(black)(""g""))) xx ""0.91 J"" color(red)(cancel(color(black)(""g""^(-1))))""""^@""C""^(-1) = ""38.86 J"" """"^@""C""^(-1)#
This much heat would increase the temperature of
#15.2color(red)(cancel(color(black)(""""^@""C""))) xx ""38.86 J"" color(red)(cancel(color(black)(""""^@""C""^(-1)))) = ""590.67 J""#
Rounded to three sig figs, the answer will be
#""heat needed"" = color(green)(|bar(ul(color(white)(a/a)""591 J""color(white)(a/a)|)))#
Alternatively, you can use the formula
#color(blue)(|bar(ul(color(white)(a/a)q = m * c * DeltaTcolor(white)(a/a)|)))"" ""# , where
Plug in your values to get
#q = 42.7 color(red)(cancel(color(black)(""g""))) * ""0.91 J"" color(red)(cancel(color(black)(""g""^(-1))))color(red)(cancel(color(black)(""""^@""C""^(-1)))) * 15.2color(red)(cancel(color(black)(""""^@""C"")))#
#q = color(green)(|bar(ul(color(white)(a/a)""591 J""color(white)(a/a)|)))#
First, you need to write out the balanced chemical reaction equation. Then you convert the given mass to moles to see which compound “limits” the reaction (the one completely used up before the other one is fully reacted). Finally, you again use the balanced equation to convert the actual moles of reagents into the respective moles of product, and then convert those moles back into masses.
From our equation, see that our 0.025 mole
A mass balance will confirm the correct calculations.
Original Reagent mass: 2.5g + 2.5g = 5.0g
Final Reaction Mass: 1.1g + 2.78g + 0.45g + 0.66g = 5.0g
First, you need to write out the balanced chemical reaction equation. Then you convert the given mass to moles to see which compound “limits” the reaction (the one completely used up before the other one is fully reacted). Finally, you again use the balanced equation to convert the actual moles of reagents into the respective moles of product, and then convert those moles back into masses.
From our equation, see that our 0.025 mole
A mass balance will confirm the correct calculations.
Original Reagent mass: 2.5g + 2.5g = 5.0g
Final Reaction Mass: 1.1g + 2.78g + 0.45g + 0.66g = 5.0g
First, you need to write out the balanced chemical reaction equation. Then you convert the given mass to moles to see which compound “limits” the reaction (the one completely used up before the other one is fully reacted). Finally, you again use the balanced equation to convert the actual moles of reagents into the respective moles of product, and then convert those moles back into masses.
From our equation, see that our 0.025 mole
A mass balance will confirm the correct calculations.
Original Reagent mass: 2.5g + 2.5g = 5.0g
Final Reaction Mass: 1.1g + 2.78g + 0.45g + 0.66g = 5.0g
Balanced Equation
The basic process will be:
First convert the masses of calcium carbonate
Theoretical Mass of
Multiply mol
Determine the mass in grams of
Theoretical Mass of
Multiply the mol
Determine the mass in grams of
Theoretical Mass of
Multiply the mol
Determine the mass in grams of
Theoretical Mass of
Determine the mol
Multiply the mol
Determine the mass in grams of
Summary
We need (i) a stoichiometric equation.......
And (ii) we need equivalent quantities of metal salt, and hydrogen chloride.
And thus the acid is in VAST stoichiometric excess, and thus calcium carbonate is the LIMITING reagent.
And so we gets.................
And likewise we gets...........
An extension of this question would be to ask the VOLUME gas produced under standard conditions of
For such mass related problems we use average mass of elements or compounds. A verge mass of element is dependent on its isotopes and respective fraction found naturally.
In the given question we need to know average mass of Phosphorus and Oxygen.
average mass of Phosphorus P = 30.973762 amu
average mass of Oxygen O = 15.9994 amu
Molar mass of
For such mass related problems we use average mass of elements or compounds. A verge mass of element is dependent on its isotopes and respective fraction found naturally.
In the given question we need to know average mass of Phosphorus and Oxygen.
average mass of Phosphorus P = 30.973762 amu
average mass of Oxygen O = 15.9994 amu
Molar mass of
For such mass related problems we use average mass of elements or compounds. A verge mass of element is dependent on its isotopes and respective fraction found naturally.
In the given question we need to know average mass of Phosphorus and Oxygen.
average mass of Phosphorus P = 30.973762 amu
average mass of Oxygen O = 15.9994 amu
Molar mass of
This is an example of the combined gas law, which combines Boyle's, Charles', and Gay-Lussac's laws. It shows the relationship between the pressure, volume, and temperature when the quantity of ideal gas is constant.
The equation to use is:
https://en.wikipedia.org/wiki/Gas_laws
Housekeeping Issues
Current STP
Standard temperature is
The Kelvin temperature scale must be used in gas problems. To convert temperature in degrees Celsius to Kelvins, add
The pressure in
I'm giving the pressure to three sig figs to reduce rounding errors.
Organize the data:
Known
Unknown
Solution
Rearrange the combined gas law equation to isolate
The volume at STP is
This is an example of the combined gas law, which combines Boyle's, Charles', and Gay-Lussac's laws. It shows the relationship between the pressure, volume, and temperature when the quantity of ideal gas is constant.
The equation to use is:
https://en.wikipedia.org/wiki/Gas_laws
Housekeeping Issues
Current STP
Standard temperature is
The Kelvin temperature scale must be used in gas problems. To convert temperature in degrees Celsius to Kelvins, add
The pressure in
I'm giving the pressure to three sig figs to reduce rounding errors.
Organize the data:
Known
Unknown
Solution
Rearrange the combined gas law equation to isolate
The volume at STP is
This is an example of the combined gas law, which combines Boyle's, Charles', and Gay-Lussac's laws. It shows the relationship between the pressure, volume, and temperature when the quantity of ideal gas is constant.
The equation to use is:
https://en.wikipedia.org/wiki/Gas_laws
Housekeeping Issues
Current STP
Standard temperature is
The Kelvin temperature scale must be used in gas problems. To convert temperature in degrees Celsius to Kelvins, add
The pressure in
I'm giving the pressure to three sig figs to reduce rounding errors.
Organize the data:
Known
Unknown
Solution
Rearrange the combined gas law equation to isolate
Calcium hydroxide is very sparingly soluble. I don't think you could make a
Calcium hydroxide is very sparingly soluble. I don't think you could make a
Calcium hydroxide is very sparingly soluble. I don't think you could make a
If the molar mass of 138.02 g/mol has been provided for you,
divide it by the molar mass of
Nitrogen has a molar mass of 14.0 g/mol, whereas oxygen is 16.0 g/mol
138.02 g/mol
138.02 g/mol
Since 3.00 is the whole number, multiply it by the empirical formula.
(
138.02 g/mol
then
Molecular Formulae = Empirical Formulae
If the molar mass of 138.02 g/mol has been provided for you,
divide it by the molar mass of
Nitrogen has a molar mass of 14.0 g/mol, whereas oxygen is 16.0 g/mol
138.02 g/mol
138.02 g/mol
Since 3.00 is the whole number, multiply it by the empirical formula.
(
138.02 g/mol
then
Molecular Formulae = Empirical Formulae
If the molar mass of 138.02 g/mol has been provided for you,
divide it by the molar mass of
Nitrogen has a molar mass of 14.0 g/mol, whereas oxygen is 16.0 g/mol
138.02 g/mol
138.02 g/mol
Since 3.00 is the whole number, multiply it by the empirical formula.
(
If you have 10.81 grams of
This is your answer: 14.188 grams of nitrogen gan
From 64 grams of
If you have 10.81 grams of
This is your answer: 14.188 grams of nitrogen gan
From 64 grams of
If you have 10.81 grams of
This is your answer: 14.188 grams of nitrogen gan
We need (i)
And (ii) we need a stoichiometric equation.........
And by definition, this equilibrium is governed by the quotient.....
Now if initially,
This is a quadratic in
And thus
Because the approximations have converged, we are willing to accept this value. But
And since we know (or should know) that
We need (i)
And (ii) we need a stoichiometric equation.........
And by definition, this equilibrium is governed by the quotient.....
Now if initially,
This is a quadratic in
And thus
Because the approximations have converged, we are willing to accept this value. But
And since we know (or should know) that
We need (i)
And (ii) we need a stoichiometric equation.........
And by definition, this equilibrium is governed by the quotient.....
Now if initially,
This is a quadratic in
And thus
Because the approximations have converged, we are willing to accept this value. But
And since we know (or should know) that
According to Dalton's law of partial pressures,
So the total pressure of the system is merely the partial pressures of each component added together:
According to Dalton's law of partial pressures,
So the total pressure of the system is merely the partial pressures of each component added together:
According to Dalton's law of partial pressures,
So the total pressure of the system is merely the partial pressures of each component added together:
The idea here is that a solution's mass by volume percent concentration,
In your case, the solution is said to have a mass by volume percent concentration of
Now, you know that your target solution must contain
#5.00 color(red)(cancel(color(black)(""g solute""))) * ""100 mL solution""/(10.4color(red)(cancel(color(black)(""g solute"")))) = color(darkgreen)(ul(color(black)(""48.1 mL solution"")))#
The answer is rounded to three sig figs, the number of sig figs you have for your values.
So, in order to make this
The idea here is that a solution's mass by volume percent concentration,
In your case, the solution is said to have a mass by volume percent concentration of
Now, you know that your target solution must contain
#5.00 color(red)(cancel(color(black)(""g solute""))) * ""100 mL solution""/(10.4color(red)(cancel(color(black)(""g solute"")))) = color(darkgreen)(ul(color(black)(""48.1 mL solution"")))#
The answer is rounded to three sig figs, the number of sig figs you have for your values.
So, in order to make this
The idea here is that a solution's mass by volume percent concentration,
In your case, the solution is said to have a mass by volume percent concentration of
Now, you know that your target solution must contain
#5.00 color(red)(cancel(color(black)(""g solute""))) * ""100 mL solution""/(10.4color(red)(cancel(color(black)(""g solute"")))) = color(darkgreen)(ul(color(black)(""48.1 mL solution"")))#
The answer is rounded to three sig figs, the number of sig figs you have for your values.
So, in order to make this
In a mole of aluminium atoms, there exist
In a mole of aluminium atoms, there exist
In a mole of aluminium atoms, there exist
As always with these problems, it is useful to assume
And we divide thru by the LOWEST molar quantity to give an empirical formula of.....
But we gots a molecular mass, and we know that the molecular formula is a whole number multiple of the empirical formula.
And so
And thus
I prefer questions that quote actual microanalytical data.
As always with these problems, it is useful to assume
And we divide thru by the LOWEST molar quantity to give an empirical formula of.....
But we gots a molecular mass, and we know that the molecular formula is a whole number multiple of the empirical formula.
And so
And thus
I prefer questions that quote actual microanalytical data.
As always with these problems, it is useful to assume
And we divide thru by the LOWEST molar quantity to give an empirical formula of.....
But we gots a molecular mass, and we know that the molecular formula is a whole number multiple of the empirical formula.
And so
And thus
I prefer questions that quote actual microanalytical data.
Your strategy when dealing with mass percentages is to pick a sample of your compound and determine how many moles of each element it contains.
To make the calculations easier, you can pick a
#""89.1 g "" -># chlorine#""0.84 g "" -># hydrogen#""10.06 g "" -># carbon
Next, use the molar masses of the three elements to find how many moles of each you have in this sample
#""For Cl: "" 89.1 color(red)(cancel(color(black)(""g""))) * ""1 mole Cl""/(35.453 color(red)(cancel(color(black)(""g"")))) = ""2.513 moles C""#
#""For H: "" 0.84 color(red)(cancel(color(black)(""g""))) * ""1 mole H""/(1.00794color(red)(cancel(color(black)(""g"")))) = ""0.8334 moles H""#
#""For C: "" 10.06 color(red)(cancel(color(black)(""g""))) * ""1 mole C""/(12.011 color(red)(cancel(color(black)(""g"")))) = ""0.8376 moles C""#
To find the mole ratios that exist between the elements in the compound, divided these values by the smallest one
#""For Cl: "" (2.513 color(red)(cancel(color(black)(""moles""))))/(0.8334 color(red)(cancel(color(black)(""moles"")))) = 3.0154 ~~ 3#
#""For H: "" (0.8334 color(red)(cancel(color(black)(""moles""))))/(0.8334color(red)(cancel(color(black)(""moles"")))) = 1#
#""For C: "" (0.8376 color(red)(cancel(color(black)(""moles""))))/(0.8334color(red)(cancel(color(black)(""moles"")))) = 1.005 ~~ 1#
The compound's empirical formula, which tells you what the smallest whole number ratio that exists between the elements that make up a compound is, will thus be
#""C""_1""H""_1""Cl""_3 implies color(green)(""CHCl""_3)#
Your strategy when dealing with mass percentages is to pick a sample of your compound and determine how many moles of each element it contains.
To make the calculations easier, you can pick a
#""89.1 g "" -># chlorine#""0.84 g "" -># hydrogen#""10.06 g "" -># carbon
Next, use the molar masses of the three elements to find how many moles of each you have in this sample
#""For Cl: "" 89.1 color(red)(cancel(color(black)(""g""))) * ""1 mole Cl""/(35.453 color(red)(cancel(color(black)(""g"")))) = ""2.513 moles C""#
#""For H: "" 0.84 color(red)(cancel(color(black)(""g""))) * ""1 mole H""/(1.00794color(red)(cancel(color(black)(""g"")))) = ""0.8334 moles H""#
#""For C: "" 10.06 color(red)(cancel(color(black)(""g""))) * ""1 mole C""/(12.011 color(red)(cancel(color(black)(""g"")))) = ""0.8376 moles C""#
To find the mole ratios that exist between the elements in the compound, divided these values by the smallest one
#""For Cl: "" (2.513 color(red)(cancel(color(black)(""moles""))))/(0.8334 color(red)(cancel(color(black)(""moles"")))) = 3.0154 ~~ 3#
#""For H: "" (0.8334 color(red)(cancel(color(black)(""moles""))))/(0.8334color(red)(cancel(color(black)(""moles"")))) = 1#
#""For C: "" (0.8376 color(red)(cancel(color(black)(""moles""))))/(0.8334color(red)(cancel(color(black)(""moles"")))) = 1.005 ~~ 1#
The compound's empirical formula, which tells you what the smallest whole number ratio that exists between the elements that make up a compound is, will thus be
#""C""_1""H""_1""Cl""_3 implies color(green)(""CHCl""_3)#
Your strategy when dealing with mass percentages is to pick a sample of your compound and determine how many moles of each element it contains.
To make the calculations easier, you can pick a
#""89.1 g "" -># chlorine#""0.84 g "" -># hydrogen#""10.06 g "" -># carbon
Next, use the molar masses of the three elements to find how many moles of each you have in this sample
#""For Cl: "" 89.1 color(red)(cancel(color(black)(""g""))) * ""1 mole Cl""/(35.453 color(red)(cancel(color(black)(""g"")))) = ""2.513 moles C""#
#""For H: "" 0.84 color(red)(cancel(color(black)(""g""))) * ""1 mole H""/(1.00794color(red)(cancel(color(black)(""g"")))) = ""0.8334 moles H""#
#""For C: "" 10.06 color(red)(cancel(color(black)(""g""))) * ""1 mole C""/(12.011 color(red)(cancel(color(black)(""g"")))) = ""0.8376 moles C""#
To find the mole ratios that exist between the elements in the compound, divided these values by the smallest one
#""For Cl: "" (2.513 color(red)(cancel(color(black)(""moles""))))/(0.8334 color(red)(cancel(color(black)(""moles"")))) = 3.0154 ~~ 3#
#""For H: "" (0.8334 color(red)(cancel(color(black)(""moles""))))/(0.8334color(red)(cancel(color(black)(""moles"")))) = 1#
#""For C: "" (0.8376 color(red)(cancel(color(black)(""moles""))))/(0.8334color(red)(cancel(color(black)(""moles"")))) = 1.005 ~~ 1#
The compound's empirical formula, which tells you what the smallest whole number ratio that exists between the elements that make up a compound is, will thus be
#""C""_1""H""_1""Cl""_3 implies color(green)(""CHCl""_3)#
We have:
LHS
C= 1
S= 1
O=2
RHS
C=2
S=2
O=2
We balance the equation by making the number of each element the same on both sides.
So:
LHS
C=
S=
O=
RHS
C= 1 (from
S= 2
O=
Then we rewrite the equation:
Hope this isn't too confusing!
We have:
LHS
C= 1
S= 1
O=2
RHS
C=2
S=2
O=2
We balance the equation by making the number of each element the same on both sides.
So:
LHS
C=
S=
O=
RHS
C= 1 (from
S= 2
O=
Then we rewrite the equation:
Hope this isn't too confusing!
We have:
LHS
C= 1
S= 1
O=2
RHS
C=2
S=2
O=2
We balance the equation by making the number of each element the same on both sides.
So:
LHS
C=
S=
O=
RHS
C= 1 (from
S= 2
O=
Then we rewrite the equation:
Hope this isn't too confusing!
Given:
According to the Law of Conservation of mass, the number of atoms of each type on the left side of the equation must be equal to the number of atoms on the right side of the equation:
You can't change formulas, only add a different number of molecules (coefficient). Since the Carbon atoms are in two different molecules on the right, start with balancing the Sulfur:
Add a
Finally, put a
Balance!
In ammoniacal solution, sodium metal eventually produces stoichiometric quantities sodium amide and dihydrogen gas. When the metal is introduced to the ammonia, a deep blue colour developes that is attributed to the solvated electron that eventually reduces the hydrogen of ammonia. Sometimes electrons couple to give a bronze colour. Preparative reactions in ammonia generally use a bit of iron salts to facilitate the electron transfer.
Note that the given equation is precisely equivalent to the reaction of water with sodium metal (in water, the lifetime of the solvated electron is much, much shorter!).
In ammoniacal solution, sodium metal eventually produces stoichiometric quantities sodium amide and dihydrogen gas. When the metal is introduced to the ammonia, a deep blue colour developes that is attributed to the solvated electron that eventually reduces the hydrogen of ammonia. Sometimes electrons couple to give a bronze colour. Preparative reactions in ammonia generally use a bit of iron salts to facilitate the electron transfer.
Note that the given equation is precisely equivalent to the reaction of water with sodium metal (in water, the lifetime of the solvated electron is much, much shorter!).
In ammoniacal solution, sodium metal eventually produces stoichiometric quantities sodium amide and dihydrogen gas. When the metal is introduced to the ammonia, a deep blue colour developes that is attributed to the solvated electron that eventually reduces the hydrogen of ammonia. Sometimes electrons couple to give a bronze colour. Preparative reactions in ammonia generally use a bit of iron salts to facilitate the electron transfer.
Note that the given equation is precisely equivalent to the reaction of water with sodium metal (in water, the lifetime of the solvated electron is much, much shorter!).
A compound's solubility in water is usually expressed in grams per
In your case, you can use the information given to you to find the solubility of silver nitrate,
So, you know that you can dissolve
#100color(red)(cancel(color(black)(""g H""_2""O""))) * ""11.1 g AgNO""_3/(5.0color(red)(cancel(color(black)(""g H""_2""O"")))) = ""222 g""#
This means that the solubility of the salt will be
#""solubility AgNO""_3 = color(green)(|bar(ul(color(white)(a/a)color(black)(""220 g/100 g H""_2""O"")color(white)(a/a)|)))#
The answer is rounded to two sig figs, the number of sig figs you have for the mass of water that can dissolve
A compound's solubility in water is usually expressed in grams per
In your case, you can use the information given to you to find the solubility of silver nitrate,
So, you know that you can dissolve
#100color(red)(cancel(color(black)(""g H""_2""O""))) * ""11.1 g AgNO""_3/(5.0color(red)(cancel(color(black)(""g H""_2""O"")))) = ""222 g""#
This means that the solubility of the salt will be
#""solubility AgNO""_3 = color(green)(|bar(ul(color(white)(a/a)color(black)(""220 g/100 g H""_2""O"")color(white)(a/a)|)))#
The answer is rounded to two sig figs, the number of sig figs you have for the mass of water that can dissolve
A compound's solubility in water is usually expressed in grams per
In your case, you can use the information given to you to find the solubility of silver nitrate,
So, you know that you can dissolve
#100color(red)(cancel(color(black)(""g H""_2""O""))) * ""11.1 g AgNO""_3/(5.0color(red)(cancel(color(black)(""g H""_2""O"")))) = ""222 g""#
This means that the solubility of the salt will be
#""solubility AgNO""_3 = color(green)(|bar(ul(color(white)(a/a)color(black)(""220 g/100 g H""_2""O"")color(white)(a/a)|)))#
The answer is rounded to two sig figs, the number of sig figs you have for the mass of water that can dissolve
And I didn't even need a calculator.........
By definition,
And if
And we know that hydroxide/hydronium ions obey the following equilibrium in aqueous solution.......
Why do we use such an absurd definition? Well, back in the day, approx. 30-40 years ago BEFORE the proliferation of hand-held electronic calculators, students and engineers would routinely use log tables for lengthy calculation involving multiplication and division. because the product
These days with a simple electronic calculator (and I bought one for a quid in a discounter's shop a couple of weeks ago) we have access to a computational power that would have astonished both Newton and Gauss (they say that Gauss in particular, a mathematical prodigy, had memorized the log tables so that he could do his calculations).
Well,
And I didn't even need a calculator.........
By definition,
And if
And we know that hydroxide/hydronium ions obey the following equilibrium in aqueous solution.......
Why do we use such an absurd definition? Well, back in the day, approx. 30-40 years ago BEFORE the proliferation of hand-held electronic calculators, students and engineers would routinely use log tables for lengthy calculation involving multiplication and division. because the product
These days with a simple electronic calculator (and I bought one for a quid in a discounter's shop a couple of weeks ago) we have access to a computational power that would have astonished both Newton and Gauss (they say that Gauss in particular, a mathematical prodigy, had memorized the log tables so that he could do his calculations).
Well,
And I didn't even need a calculator.........
By definition,
And if
And we know that hydroxide/hydronium ions obey the following equilibrium in aqueous solution.......
Why do we use such an absurd definition? Well, back in the day, approx. 30-40 years ago BEFORE the proliferation of hand-held electronic calculators, students and engineers would routinely use log tables for lengthy calculation involving multiplication and division. because the product
These days with a simple electronic calculator (and I bought one for a quid in a discounter's shop a couple of weeks ago) we have access to a computational power that would have astonished both Newton and Gauss (they say that Gauss in particular, a mathematical prodigy, had memorized the log tables so that he could do his calculations).
And
And
And
The molar volume at
And thus the
What mass does this volume represent?
Approx.
The molar volume at
And thus the
What mass does this volume represent?
Approx.
The molar volume at
And thus the
What mass does this volume represent?
At Standard Temperature and Pressure,
Knowing this, we can think about this for a second without involving any math or calculations.
If we know
This means, there is
Answer: 39.87 L
A useful piece of information to have when looking for the mass of
Once you know the mass of
So, the mass of
#M_(""M H""_2""SO""_4) = ""98.08 g mol""^(-1)#
This implies that every mole of sulfuric acid has a mass of
#2.25 color(red)(cancel(color(black)(""moles H""_2""SO""_4))) * overbrace(""98.08 g""/(1color(red)(cancel(color(black)(""mole H""_2""SO""_4)))))^(color(blue)(""molar mass of H""_2""SO""_4)) = color(green)(|bar(ul(color(white)(a/a)color(black)(""221 g"")color(white)(a/a)|)))#
The answer is rounded to three sig figs.
A useful piece of information to have when looking for the mass of
Once you know the mass of
So, the mass of
#M_(""M H""_2""SO""_4) = ""98.08 g mol""^(-1)#
This implies that every mole of sulfuric acid has a mass of
#2.25 color(red)(cancel(color(black)(""moles H""_2""SO""_4))) * overbrace(""98.08 g""/(1color(red)(cancel(color(black)(""mole H""_2""SO""_4)))))^(color(blue)(""molar mass of H""_2""SO""_4)) = color(green)(|bar(ul(color(white)(a/a)color(black)(""221 g"")color(white)(a/a)|)))#
The answer is rounded to three sig figs.
A useful piece of information to have when looking for the mass of
Once you know the mass of
So, the mass of
#M_(""M H""_2""SO""_4) = ""98.08 g mol""^(-1)#
This implies that every mole of sulfuric acid has a mass of
#2.25 color(red)(cancel(color(black)(""moles H""_2""SO""_4))) * overbrace(""98.08 g""/(1color(red)(cancel(color(black)(""mole H""_2""SO""_4)))))^(color(blue)(""molar mass of H""_2""SO""_4)) = color(green)(|bar(ul(color(white)(a/a)color(black)(""221 g"")color(white)(a/a)|)))#
The answer is rounded to three sig figs.
Use the equality above to determine the moles of carbon dioxide as shown below.
The answer is
Use the equality above to determine the moles of carbon dioxide as shown below.
The answer is
Use the equality above to determine the moles of carbon dioxide as shown below.
..........or at least you need
From this site, we learn that
Now in aqueous solution, methylamine undergoes the acid base reaction:
And we write the equilibrium equation in the usual way:
So if
As is normal, this is a quadratic in
And so
So,
But
Now I admit this does seem like a lot of work (it was more work for me, because I had to format the equations on this editor!). But I can assure you that you can get very proficient at these sorts of problems, especially if you can work your calculator competently. Remember the approach, approximate and then justify. And then recycle the approximation. Good luck.
Well, for a start you need
..........or at least you need
From this site, we learn that
Now in aqueous solution, methylamine undergoes the acid base reaction:
And we write the equilibrium equation in the usual way:
So if
As is normal, this is a quadratic in
And so
So,
But
Now I admit this does seem like a lot of work (it was more work for me, because I had to format the equations on this editor!). But I can assure you that you can get very proficient at these sorts of problems, especially if you can work your calculator competently. Remember the approach, approximate and then justify. And then recycle the approximation. Good luck.
Well, for a start you need
..........or at least you need
From this site, we learn that
Now in aqueous solution, methylamine undergoes the acid base reaction:
And we write the equilibrium equation in the usual way:
So if
As is normal, this is a quadratic in
And so
So,
But
Now I admit this does seem like a lot of work (it was more work for me, because I had to format the equations on this editor!). But I can assure you that you can get very proficient at these sorts of problems, especially if you can work your calculator competently. Remember the approach, approximate and then justify. And then recycle the approximation. Good luck.
It is a 0.815 M solution, so that means that in 1 litre you have 0.815 moles of solute.
Therefore in 497.2 litres of the solution you have 497.2 x 0.815 = 405.218 moles of solute.
(By the way, the correct terminology is ""0.815 M sodium cyanide solution"" - saying ""0.815 M sodium cyanide"" is meaningless).
405.218 moles
It is a 0.815 M solution, so that means that in 1 litre you have 0.815 moles of solute.
Therefore in 497.2 litres of the solution you have 497.2 x 0.815 = 405.218 moles of solute.
(By the way, the correct terminology is ""0.815 M sodium cyanide solution"" - saying ""0.815 M sodium cyanide"" is meaningless).
405.218 moles
It is a 0.815 M solution, so that means that in 1 litre you have 0.815 moles of solute.
Therefore in 497.2 litres of the solution you have 497.2 x 0.815 = 405.218 moles of solute.
(By the way, the correct terminology is ""0.815 M sodium cyanide solution"" - saying ""0.815 M sodium cyanide"" is meaningless).
Molarity is represented by the following equation:
In our case, we already have the molarity and the volume of solution. However, the volume does not have the proper units since it is given in terms of milliliters instead of liters. We can convert 500 mL into liters by using the conversion factor 1000mL = 1L. When 500 mL is divided by 1000 mL/L, we obtain a volume of 0.50 L.
Now we can rearrange the equation to solve for the number of moles, which will allow us to determine the mass. We can do this by multiplying by liters of solution on both sides of the equation. The liters of solution will cancel out on the right side, leaving the number of moles being equal to the molarity times volume like so:
Moles of solute = (liters of solution) * (Molarity)
Moles of solute = (0.50 L) (6.0 M) = 3.0 mol of HCl
Now we have to convert the 3.0 mol of HCl into grams of HCl. This can be done by multiplying 3.0 mol by the molecular weight of HCl, which is 36.46 g/mol.
(3.0 mol)(36.46 g/mol) = 109 g HCl
109 g of HCl
Molarity is represented by the following equation:
In our case, we already have the molarity and the volume of solution. However, the volume does not have the proper units since it is given in terms of milliliters instead of liters. We can convert 500 mL into liters by using the conversion factor 1000mL = 1L. When 500 mL is divided by 1000 mL/L, we obtain a volume of 0.50 L.
Now we can rearrange the equation to solve for the number of moles, which will allow us to determine the mass. We can do this by multiplying by liters of solution on both sides of the equation. The liters of solution will cancel out on the right side, leaving the number of moles being equal to the molarity times volume like so:
Moles of solute = (liters of solution) * (Molarity)
Moles of solute = (0.50 L) (6.0 M) = 3.0 mol of HCl
Now we have to convert the 3.0 mol of HCl into grams of HCl. This can be done by multiplying 3.0 mol by the molecular weight of HCl, which is 36.46 g/mol.
(3.0 mol)(36.46 g/mol) = 109 g HCl
109 g of HCl
Molarity is represented by the following equation:
In our case, we already have the molarity and the volume of solution. However, the volume does not have the proper units since it is given in terms of milliliters instead of liters. We can convert 500 mL into liters by using the conversion factor 1000mL = 1L. When 500 mL is divided by 1000 mL/L, we obtain a volume of 0.50 L.
Now we can rearrange the equation to solve for the number of moles, which will allow us to determine the mass. We can do this by multiplying by liters of solution on both sides of the equation. The liters of solution will cancel out on the right side, leaving the number of moles being equal to the molarity times volume like so:
Moles of solute = (liters of solution) * (Molarity)
Moles of solute = (0.50 L) (6.0 M) = 3.0 mol of HCl
Now we have to convert the 3.0 mol of HCl into grams of HCl. This can be done by multiplying 3.0 mol by the molecular weight of HCl, which is 36.46 g/mol.
(3.0 mol)(36.46 g/mol) = 109 g HCl
........And so the metal oxidation state would be
And thus its perhalide formula would be
You might get a
Well, the parent phosphate ion is
........And so the metal oxidation state would be
And thus its perhalide formula would be
You might get a
Well, the parent phosphate ion is
........And so the metal oxidation state would be
And thus its perhalide formula would be
You might get a
The formula for the heat absorbed by a substance is
#color(blue)(|bar(ul(color(white)(a/a) q = mcΔT color(white)(a/a)|)))"" ""#
where
You can rearrange the formula to calculate the specific heat capacity:
The specific heat capacity of mercury is
The formula for the heat absorbed by a substance is
#color(blue)(|bar(ul(color(white)(a/a) q = mcΔT color(white)(a/a)|)))"" ""#
where
You can rearrange the formula to calculate the specific heat capacity:
The specific heat capacity of mercury is
The formula for the heat absorbed by a substance is
#color(blue)(|bar(ul(color(white)(a/a) q = mcΔT color(white)(a/a)|)))"" ""#
where
You can rearrange the formula to calculate the specific heat capacity:
Magnesium metal lies in Group II of the Periodic Table; it readily loses 2 electrons to give rise to
What are the formulae for calcium hydroxide and barium hydroxides?
Magnesium metal lies in Group II of the Periodic Table; it readily loses 2 electrons to give rise to
What are the formulae for calcium hydroxide and barium hydroxides?
Magnesium metal lies in Group II of the Periodic Table; it readily loses 2 electrons to give rise to
What are the formulae for calcium hydroxide and barium hydroxides?
The thing to remember when dealing with ionic compounds is that their empirical formula is equivalent to their chemical formula .
In other words, if you know an ionic compound's chemical formula, you also know its empirical formula.
A compound's empirical formula will tell you what the smallest whole number ratio is for the atoms of the elements that make up that compound.
In the case of ionic compounds, you know that their chemical formula is given by the formula unit, which tells you the smallest whole number ratio that exist between the cations and anions that form said compound.
You know that magnesium hydroxide is composed of magnesium cations,
In order to blance the positive charge of the cations, the formula unit of magnesium hydroxide will consist of two hydroxide anions.
#""Mg""^(2+) + 2""OH""^(-) -> ""Mg""(""OH"")_2#
This is also the compound's empirical formula.
How would you remove the half coefficient on the oxygen? What does the
How would you remove the half coefficient on the oxygen? What does the
How would you remove the half coefficient on the oxygen? What does the
Formula P1 x V1 / T1 = P2 x V2 / T2
Fill in what you know
Pressure is constant so no need to put that in making the formula
V1 / T1 = V2 / T2
Voulme 1= 950 mL
Volume 2= ?
Temperature 1 = 25 C
Temperature 2 = 50 C
Convert your temperature to Kelvin
C+273=K
Temperature 1 = 25 C + 273 = 298 K
Temperature 2 = 50 C + 273 = 323 K
Plug in to the Formula
950 mL/298 K = ? / 323 K
Rearrange the formula to make one to solve for what is missing.
To get 323 K out of the denominator multiply by it.
Making it
950 mL x 323 K / 298 K = ?
Plug it in
950 mL x 323 K / 298 K = 1027.9 mL
1027.9 mL
Formula P1 x V1 / T1 = P2 x V2 / T2
Fill in what you know
Pressure is constant so no need to put that in making the formula
V1 / T1 = V2 / T2
Voulme 1= 950 mL
Volume 2= ?
Temperature 1 = 25 C
Temperature 2 = 50 C
Convert your temperature to Kelvin
C+273=K
Temperature 1 = 25 C + 273 = 298 K
Temperature 2 = 50 C + 273 = 323 K
Plug in to the Formula
950 mL/298 K = ? / 323 K
Rearrange the formula to make one to solve for what is missing.
To get 323 K out of the denominator multiply by it.
Making it
950 mL x 323 K / 298 K = ?
Plug it in
950 mL x 323 K / 298 K = 1027.9 mL
1027.9 mL
Formula P1 x V1 / T1 = P2 x V2 / T2
Fill in what you know
Pressure is constant so no need to put that in making the formula
V1 / T1 = V2 / T2
Voulme 1= 950 mL
Volume 2= ?
Temperature 1 = 25 C
Temperature 2 = 50 C
Convert your temperature to Kelvin
C+273=K
Temperature 1 = 25 C + 273 = 298 K
Temperature 2 = 50 C + 273 = 323 K
Plug in to the Formula
950 mL/298 K = ? / 323 K
Rearrange the formula to make one to solve for what is missing.
To get 323 K out of the denominator multiply by it.
Making it
950 mL x 323 K / 298 K = ?
Plug it in
950 mL x 323 K / 298 K = 1027.9 mL
The idea here is that you're looking for the mass of this unknown metal that can combine with
To make the calculations easier, pick a
#60%color(white)(.)""metal "" stackrel(color(white)(color(blue)(""100 g sample"")aaa))(rarr) "" 60 g metal""# #40%color(white)(.)""oxygen "" stackrel(color(white)(color(blue)(""100 g sample"")aaa))(rarr) "" 40 g oxygen""#
So, you know that
#8 color(red)(cancel(color(black)(""g oxygen""))) * ""60 g metal""/(40color(red)(cancel(color(black)(""g oxygen"")))) = ""12 g metal""#
Therefore, you can say that the equivalent mass of the metal is equal to
#color(darkgreen)(ul(color(black)(""equivalent mass of metal in 60% metal oxide = 12 g"")))#
The idea here is that you're looking for the mass of this unknown metal that can combine with
To make the calculations easier, pick a
#60%color(white)(.)""metal "" stackrel(color(white)(color(blue)(""100 g sample"")aaa))(rarr) "" 60 g metal""# #40%color(white)(.)""oxygen "" stackrel(color(white)(color(blue)(""100 g sample"")aaa))(rarr) "" 40 g oxygen""#
So, you know that
#8 color(red)(cancel(color(black)(""g oxygen""))) * ""60 g metal""/(40color(red)(cancel(color(black)(""g oxygen"")))) = ""12 g metal""#
Therefore, you can say that the equivalent mass of the metal is equal to
#color(darkgreen)(ul(color(black)(""equivalent mass of metal in 60% metal oxide = 12 g"")))#
The idea here is that you're looking for the mass of this unknown metal that can combine with
To make the calculations easier, pick a
#60%color(white)(.)""metal "" stackrel(color(white)(color(blue)(""100 g sample"")aaa))(rarr) "" 60 g metal""# #40%color(white)(.)""oxygen "" stackrel(color(white)(color(blue)(""100 g sample"")aaa))(rarr) "" 40 g oxygen""#
So, you know that
#8 color(red)(cancel(color(black)(""g oxygen""))) * ""60 g metal""/(40color(red)(cancel(color(black)(""g oxygen"")))) = ""12 g metal""#
Therefore, you can say that the equivalent mass of the metal is equal to
#color(darkgreen)(ul(color(black)(""equivalent mass of metal in 60% metal oxide = 12 g"")))#
This is a pretty straightforward example of how to sue Avogadro's number to figure out the number of molecules present in a given sample.
In your case, the sample is said to contain
#color(blue)(|bar(ul(color(white)(a/a)""1 mole"" = 6.022 * 10^(23)""molecules"" color(white)(a/a)|))) -># Avogadro's number
This is what Avogadro's number is all about -- the number of molecules needed to form
So, now that you know how many molecules are needed to have
#2.50 color(red)(cancel(color(black)(""moles H""_2""O""))) * (6.022 * 10^(23)""molec. H""_2""O"")/(1color(red)(cancel(color(black)(""mole H""_2""O"")))) = color(green)(|bar(ul(color(white)(a/a)color(black)(1.51 * 10^(24)""molec. H""_2""O"")color(white)(a/a)|)))#
The answer is rounded to three sig figs.
This is a pretty straightforward example of how to sue Avogadro's number to figure out the number of molecules present in a given sample.
In your case, the sample is said to contain
#color(blue)(|bar(ul(color(white)(a/a)""1 mole"" = 6.022 * 10^(23)""molecules"" color(white)(a/a)|))) -># Avogadro's number
This is what Avogadro's number is all about -- the number of molecules needed to form
So, now that you know how many molecules are needed to have
#2.50 color(red)(cancel(color(black)(""moles H""_2""O""))) * (6.022 * 10^(23)""molec. H""_2""O"")/(1color(red)(cancel(color(black)(""mole H""_2""O"")))) = color(green)(|bar(ul(color(white)(a/a)color(black)(1.51 * 10^(24)""molec. H""_2""O"")color(white)(a/a)|)))#
The answer is rounded to three sig figs.
This is a pretty straightforward example of how to sue Avogadro's number to figure out the number of molecules present in a given sample.
In your case, the sample is said to contain
#color(blue)(|bar(ul(color(white)(a/a)""1 mole"" = 6.022 * 10^(23)""molecules"" color(white)(a/a)|))) -># Avogadro's number
This is what Avogadro's number is all about -- the number of molecules needed to form
So, now that you know how many molecules are needed to have
#2.50 color(red)(cancel(color(black)(""moles H""_2""O""))) * (6.022 * 10^(23)""molec. H""_2""O"")/(1color(red)(cancel(color(black)(""mole H""_2""O"")))) = color(green)(|bar(ul(color(white)(a/a)color(black)(1.51 * 10^(24)""molec. H""_2""O"")color(white)(a/a)|)))#
The answer is rounded to three sig figs.
The formula for molarity is:
#color(blue)(bar(ul(|color(white)(a/a) ""Molarity"" = ""moles""/""litres""color(white)(a/a)|)))"" ""#
or
#color(blue)(bar(ul(|color(white)(a/a) M = n/Vcolor(white)(a/a)|)))"" ""#
In your problem,
∴
Now, we can use Avogadro's number to calculate the number of molecules.
There are
The formula for molarity is:
#color(blue)(bar(ul(|color(white)(a/a) ""Molarity"" = ""moles""/""litres""color(white)(a/a)|)))"" ""#
or
#color(blue)(bar(ul(|color(white)(a/a) M = n/Vcolor(white)(a/a)|)))"" ""#
In your problem,
∴
Now, we can use Avogadro's number to calculate the number of molecules.
There are
The formula for molarity is:
#color(blue)(bar(ul(|color(white)(a/a) ""Molarity"" = ""moles""/""litres""color(white)(a/a)|)))"" ""#
or
#color(blue)(bar(ul(|color(white)(a/a) M = n/Vcolor(white)(a/a)|)))"" ""#
In your problem,
∴
Now, we can use Avogadro's number to calculate the number of molecules.
So if there is a quantity of
Barium metal has a molar mass of
So if there is a quantity of
Barium metal has a molar mass of
So if there is a quantity of
The product
In this problem,
The old equation is
But I think you have worded the question improperly......
The product
In this problem,
The old equation is
But I think you have worded the question improperly......
The product
In this problem,
Because we are at STP, we must use the ideal gas law equation
Next, list your known and unknown variables. Our only unknown is the volume of
At STP, the temperature is 273K and the pressure is 1 atm. The proportionality constant, R, is equal to 0.0821
The only issue is the mass of
Now all we have to do is rearrange the equation and solve for V like so:
Because we are at STP, we must use the ideal gas law equation
Next, list your known and unknown variables. Our only unknown is the volume of
At STP, the temperature is 273K and the pressure is 1 atm. The proportionality constant, R, is equal to 0.0821
The only issue is the mass of
Now all we have to do is rearrange the equation and solve for V like so:
Because we are at STP, we must use the ideal gas law equation
Next, list your known and unknown variables. Our only unknown is the volume of
At STP, the temperature is 273K and the pressure is 1 atm. The proportionality constant, R, is equal to 0.0821
The only issue is the mass of
Now all we have to do is rearrange the equation and solve for V like so:
And thus
And thus
And thus
I'll show you hot to solve this one without using the Henderson - Hasselbalch equation. This method is a bit long, but I think that it's a great help to understanding the general idea behind how buffers work.
So, your solution is said to contain
You know that this buffer is prepared by dissolving
#color(blue)(c = n/V)#
#[""HA""] = [""A""^(-)] = ""1.0 moles""/""1.0 L"" = ""1.0 M""#
Keep in mind that the salt dissociates in a
Now, you take
#color(blue)(c = n/V implies n = c * V)#
#n_(HA) = n_(A^(-)) = ""1.0 M"" * 500 * 10^(-3)""L"" = ""0.50 moles""#
To find the pH of this buffer solution, you need to use an ICE table - remember to use the molarities of the weak acid and conjugate base!
#"" """"HA""_text((aq]) + ""H""_2""O""_text((l]) "" ""rightleftharpoons"" "" ""H""_3""O""_text((aq])^(+) "" ""+"" "" ""A""_text((aq])^(-)#
By definition, the acid dissociation constant,
#K_a = ( [""H""_3""O""^(+)] * [""A""^(-)])/([""HA""]) = (x * (1+x))/(1-x) = 1.75 * 10^(-5)#
Because the value of
#1.0 +- x ~~ 1.0#
This will get you
#1.75 * 10^(-5) = (x * 1)/(1) = x#
Here
#color(blue)(""pH"" = - log([""H""_3""O""^(+)]))#
#""pH""_0 = - log(1.75 * 10^(-5)) = 4.757#
SIDE NOTE This is why the pH of a buffer solution is said to be equal to the acid's
Now, sodium hydroxide is a strong base, which means that it will neutralize the weak acid to produce water and the conjugate base of the acid.
This means that you can expect the pH of the solution to increase a little after the addition of the strong base.
The balanced chemical equation for this reaction looks like this - I'll use the hydroxide anion to represent the strong base
#""HA""_text((aq]) + ""OH""_text((aq])^(-) -> ""A""_text((aq])^(-) + ""H""_2""O""_text((l])#
Notice that the hydroxide anions react in a
Use the molarity and volume of the sodium hydroxide solution to find how many moles of strong base are added to the buffer
#color(blue)(c = n/V implies n = c * V)#
#n_(OH^(-)) = ""0.20 M"" * 25.0 * 10^(-3)""L"" = ""0.0050 moles OH""^(-)#
This means that after the neutralization reaction takes place, you will be left with
#n_(OH^(-)) = ""0 moles"" -># completely consumed
#n_(HA) = ""0.50 moles"" - ""0.0050 moles"" = ""0.495 moles HA""#
#n_(A^(-)) = ""0.50 moles"" + ""0.0050 moles"" = ""0.505 moles A""^(-)#
The new volume of the buffer will be
#V_""total"" = V_""initial"" + V_""base""#
#V_""total"" = ""500 mL"" + ""25.0 mL"" = ""525 mL""#
Now calculate the new molarities of the weak acid and conjugate base
#[""HA""] = ""0.495 moles""/(525 * 10^(-3)""L"") = ""0.9429 M""#
#[""A""^(-)] = ""0.505 moles""/(525 * 10^(-3)""L"") = ""0.9619 M""#
Now it's back to the ICE table. To keep the answer at a reasonable length, I won't add it again. This time, the initial concentrations of the weak acid and conjugate base will no longer be
Once again, you will have
#K_a = ([""H""_3""O""^(+)] * [""A""^(-)])/([""HA""])#
In this case, this will be equal to
#K_a = (x * (0.9619 + x))/(0.9429 + x) = 1.75 * 10^(-5)#
Using the same approximation, you can say that
#1.75 * 10^(-5) = x * 0.9619/0.9429#
Therefore,
#x = (1.75 * 0.9429)/0.9619 * 10^(-5) = 1.7154 * 10^(-5)#
The pH of the solution will now be
#""pH""_1 = - log(1.7154 * 10^(-5)) = 4.766#
The change in pH can be calculated as
#Delta_""pH"" = ""pH""_1 - ""pH""_0#
#Delta_""pH"" = 4.766 - 4.757 = color(green)(0.009)#
Here's what I got.
I'll show you hot to solve this one without using the Henderson - Hasselbalch equation. This method is a bit long, but I think that it's a great help to understanding the general idea behind how buffers work.
So, your solution is said to contain
You know that this buffer is prepared by dissolving
#color(blue)(c = n/V)#
#[""HA""] = [""A""^(-)] = ""1.0 moles""/""1.0 L"" = ""1.0 M""#
Keep in mind that the salt dissociates in a
Now, you take
#color(blue)(c = n/V implies n = c * V)#
#n_(HA) = n_(A^(-)) = ""1.0 M"" * 500 * 10^(-3)""L"" = ""0.50 moles""#
To find the pH of this buffer solution, you need to use an ICE table - remember to use the molarities of the weak acid and conjugate base!
#"" """"HA""_text((aq]) + ""H""_2""O""_text((l]) "" ""rightleftharpoons"" "" ""H""_3""O""_text((aq])^(+) "" ""+"" "" ""A""_text((aq])^(-)#
By definition, the acid dissociation constant,
#K_a = ( [""H""_3""O""^(+)] * [""A""^(-)])/([""HA""]) = (x * (1+x))/(1-x) = 1.75 * 10^(-5)#
Because the value of
#1.0 +- x ~~ 1.0#
This will get you
#1.75 * 10^(-5) = (x * 1)/(1) = x#
Here
#color(blue)(""pH"" = - log([""H""_3""O""^(+)]))#
#""pH""_0 = - log(1.75 * 10^(-5)) = 4.757#
SIDE NOTE This is why the pH of a buffer solution is said to be equal to the acid's
Now, sodium hydroxide is a strong base, which means that it will neutralize the weak acid to produce water and the conjugate base of the acid.
This means that you can expect the pH of the solution to increase a little after the addition of the strong base.
The balanced chemical equation for this reaction looks like this - I'll use the hydroxide anion to represent the strong base
#""HA""_text((aq]) + ""OH""_text((aq])^(-) -> ""A""_text((aq])^(-) + ""H""_2""O""_text((l])#
Notice that the hydroxide anions react in a
Use the molarity and volume of the sodium hydroxide solution to find how many moles of strong base are added to the buffer
#color(blue)(c = n/V implies n = c * V)#
#n_(OH^(-)) = ""0.20 M"" * 25.0 * 10^(-3)""L"" = ""0.0050 moles OH""^(-)#
This means that after the neutralization reaction takes place, you will be left with
#n_(OH^(-)) = ""0 moles"" -># completely consumed
#n_(HA) = ""0.50 moles"" - ""0.0050 moles"" = ""0.495 moles HA""#
#n_(A^(-)) = ""0.50 moles"" + ""0.0050 moles"" = ""0.505 moles A""^(-)#
The new volume of the buffer will be
#V_""total"" = V_""initial"" + V_""base""#
#V_""total"" = ""500 mL"" + ""25.0 mL"" = ""525 mL""#
Now calculate the new molarities of the weak acid and conjugate base
#[""HA""] = ""0.495 moles""/(525 * 10^(-3)""L"") = ""0.9429 M""#
#[""A""^(-)] = ""0.505 moles""/(525 * 10^(-3)""L"") = ""0.9619 M""#
Now it's back to the ICE table. To keep the answer at a reasonable length, I won't add it again. This time, the initial concentrations of the weak acid and conjugate base will no longer be
Once again, you will have
#K_a = ([""H""_3""O""^(+)] * [""A""^(-)])/([""HA""])#
In this case, this will be equal to
#K_a = (x * (0.9619 + x))/(0.9429 + x) = 1.75 * 10^(-5)#
Using the same approximation, you can say that
#1.75 * 10^(-5) = x * 0.9619/0.9429#
Therefore,
#x = (1.75 * 0.9429)/0.9619 * 10^(-5) = 1.7154 * 10^(-5)#
The pH of the solution will now be
#""pH""_1 = - log(1.7154 * 10^(-5)) = 4.766#
The change in pH can be calculated as
#Delta_""pH"" = ""pH""_1 - ""pH""_0#
#Delta_""pH"" = 4.766 - 4.757 = color(green)(0.009)#
Here's what I got.
I'll show you hot to solve this one without using the Henderson - Hasselbalch equation. This method is a bit long, but I think that it's a great help to understanding the general idea behind how buffers work.
So, your solution is said to contain
You know that this buffer is prepared by dissolving
#color(blue)(c = n/V)#
#[""HA""] = [""A""^(-)] = ""1.0 moles""/""1.0 L"" = ""1.0 M""#
Keep in mind that the salt dissociates in a
Now, you take
#color(blue)(c = n/V implies n = c * V)#
#n_(HA) = n_(A^(-)) = ""1.0 M"" * 500 * 10^(-3)""L"" = ""0.50 moles""#
To find the pH of this buffer solution, you need to use an ICE table - remember to use the molarities of the weak acid and conjugate base!
#"" """"HA""_text((aq]) + ""H""_2""O""_text((l]) "" ""rightleftharpoons"" "" ""H""_3""O""_text((aq])^(+) "" ""+"" "" ""A""_text((aq])^(-)#
By definition, the acid dissociation constant,
#K_a = ( [""H""_3""O""^(+)] * [""A""^(-)])/([""HA""]) = (x * (1+x))/(1-x) = 1.75 * 10^(-5)#
Because the value of
#1.0 +- x ~~ 1.0#
This will get you
#1.75 * 10^(-5) = (x * 1)/(1) = x#
Here
#color(blue)(""pH"" = - log([""H""_3""O""^(+)]))#
#""pH""_0 = - log(1.75 * 10^(-5)) = 4.757#
SIDE NOTE This is why the pH of a buffer solution is said to be equal to the acid's
Now, sodium hydroxide is a strong base, which means that it will neutralize the weak acid to produce water and the conjugate base of the acid.
This means that you can expect the pH of the solution to increase a little after the addition of the strong base.
The balanced chemical equation for this reaction looks like this - I'll use the hydroxide anion to represent the strong base
#""HA""_text((aq]) + ""OH""_text((aq])^(-) -> ""A""_text((aq])^(-) + ""H""_2""O""_text((l])#
Notice that the hydroxide anions react in a
Use the molarity and volume of the sodium hydroxide solution to find how many moles of strong base are added to the buffer
#color(blue)(c = n/V implies n = c * V)#
#n_(OH^(-)) = ""0.20 M"" * 25.0 * 10^(-3)""L"" = ""0.0050 moles OH""^(-)#
This means that after the neutralization reaction takes place, you will be left with
#n_(OH^(-)) = ""0 moles"" -># completely consumed
#n_(HA) = ""0.50 moles"" - ""0.0050 moles"" = ""0.495 moles HA""#
#n_(A^(-)) = ""0.50 moles"" + ""0.0050 moles"" = ""0.505 moles A""^(-)#
The new volume of the buffer will be
#V_""total"" = V_""initial"" + V_""base""#
#V_""total"" = ""500 mL"" + ""25.0 mL"" = ""525 mL""#
Now calculate the new molarities of the weak acid and conjugate base
#[""HA""] = ""0.495 moles""/(525 * 10^(-3)""L"") = ""0.9429 M""#
#[""A""^(-)] = ""0.505 moles""/(525 * 10^(-3)""L"") = ""0.9619 M""#
Now it's back to the ICE table. To keep the answer at a reasonable length, I won't add it again. This time, the initial concentrations of the weak acid and conjugate base will no longer be
Once again, you will have
#K_a = ([""H""_3""O""^(+)] * [""A""^(-)])/([""HA""])#
In this case, this will be equal to
#K_a = (x * (0.9619 + x))/(0.9429 + x) = 1.75 * 10^(-5)#
Using the same approximation, you can say that
#1.75 * 10^(-5) = x * 0.9619/0.9429#
Therefore,
#x = (1.75 * 0.9429)/0.9619 * 10^(-5) = 1.7154 * 10^(-5)#
The pH of the solution will now be
#""pH""_1 = - log(1.7154 * 10^(-5)) = 4.766#
The change in pH can be calculated as
#Delta_""pH"" = ""pH""_1 - ""pH""_0#
#Delta_""pH"" = 4.766 - 4.757 = color(green)(0.009)#
The strategy to follow is
1. Chemical Equation
2.
We can use the Henderson-Hasselbalch equation to calculate the
3. Moles of base added
4. New moles of
You are using only half of the original buffer, so you are starting with 0.50 mol each of
The added
5.
Since
6. Change in pH
We usually express
The MOLARITY of a solution is determined by the ratio of moles of solute, (the material being dissolved), compared to the liters of solution. the equation is
For this
Therefore
M = 0.75 M
V = 50 mL or 0.050 L
moles =
Use the molar mass of
The MOLARITY of a solution is determined by the ratio of moles of solute, (the material being dissolved), compared to the liters of solution. the equation is
For this
Therefore
M = 0.75 M
V = 50 mL or 0.050 L
moles =
Use the molar mass of
The MOLARITY of a solution is determined by the ratio of moles of solute, (the material being dissolved), compared to the liters of solution. the equation is
For this
Therefore
M = 0.75 M
V = 50 mL or 0.050 L
moles =
Use the molar mass of
Moles of calcium carbide:
Given the equation,
Approx.
Moles of calcium carbide:
Given the equation,
Approx.
Moles of calcium carbide:
Given the equation,
You will need
Balanced Equation
We will make the following conversions to answer this question.
In order to do this, we need to determine the molar mass of each compound. We can calculate it or look it up on a reputable source. I like to use The PubChem Project.
Molar Masses
https://pubchem.ncbi.nlm.nih.gov/compound/6352#section=Top
https://pubchem.ncbi.nlm.nih.gov/compound/962#section=Top
We also need the mole ratio between the two compounds.
From the balanced equation we can see that the mole ratio between
Solution
Convert
Convert
Convert
We can put all of the steps together like this:
This way you can do the calculations all at once and round at the end, which reduces rounding errors.
Any substance containing as many particles as there are in 1 mole of carbon-12.
1 mole of Carbon-12 contains
Therefore, Avogadro's constant contains
Avogadro's constant contains
Any substance containing as many particles as there are in 1 mole of carbon-12.
1 mole of Carbon-12 contains
Therefore, Avogadro's constant contains
Avogadro's constant contains
Any substance containing as many particles as there are in 1 mole of carbon-12.
1 mole of Carbon-12 contains
Therefore, Avogadro's constant contains
What does the
What does the
What does the
As with all these problems we assume a
And we divide thru the individual molar quantities by the smallest such quantity, that of nitrogen.
Because, by definition, empirical formulae are the simplest WHOLE number ratio definining constituent atoms in a species, we double this ratio to get:
As with all these problems we assume a
And we divide thru the individual molar quantities by the smallest such quantity, that of nitrogen.
Because, by definition, empirical formulae are the simplest WHOLE number ratio definining constituent atoms in a species, we double this ratio to get:
As with all these problems we assume a
And we divide thru the individual molar quantities by the smallest such quantity, that of nitrogen.
Because, by definition, empirical formulae are the simplest WHOLE number ratio definining constituent atoms in a species, we double this ratio to get:
acetic acid (ethanoic acid):
sodium hydroxide:
neutralisation reactions always produce a salt and water.
in this case, the salt is sodium ethanoate
(
equation:
carbon:
sodium:
hydrogen:
oxygen:
this means that the equation is balanced, with
acetic acid (ethanoic acid):
sodium hydroxide:
neutralisation reactions always produce a salt and water.
in this case, the salt is sodium ethanoate
(
equation:
carbon:
sodium:
hydrogen:
oxygen:
this means that the equation is balanced, with
acetic acid (ethanoic acid):
sodium hydroxide:
neutralisation reactions always produce a salt and water.
in this case, the salt is sodium ethanoate
(
equation:
carbon:
sodium:
hydrogen:
oxygen:
this means that the equation is balanced, with
The total pressure is the sum of the pressures of the three gases in the flask.
The total pressure is 25.7 atm.
The total pressure is the sum of the pressures of the three gases in the flask.
The total pressure is 25.7 atm.
The total pressure is the sum of the pressures of the three gases in the flask.
First, let's write the equation for this titration reaction:
The solubility of
For any titration problem like this, where you try and find the concentration of one substance when you know the other concentration and volumes of both, the next step is to convert the standard solution (the one with known volume and concentration) to moles:
Now we'll use the stoichiometrically equivalent values to find the moles of the substance with unknown concentration,
Finally, we'll use this value and its known volume to find its molar concentration:
First, let's write the equation for this titration reaction:
The solubility of
For any titration problem like this, where you try and find the concentration of one substance when you know the other concentration and volumes of both, the next step is to convert the standard solution (the one with known volume and concentration) to moles:
Now we'll use the stoichiometrically equivalent values to find the moles of the substance with unknown concentration,
Finally, we'll use this value and its known volume to find its molar concentration:
First, let's write the equation for this titration reaction:
The solubility of
For any titration problem like this, where you try and find the concentration of one substance when you know the other concentration and volumes of both, the next step is to convert the standard solution (the one with known volume and concentration) to moles:
Now we'll use the stoichiometrically equivalent values to find the moles of the substance with unknown concentration,
Finally, we'll use this value and its known volume to find its molar concentration:
You first balance the metals, namely Sodium (
Next you do the non-metals, starting with anything but Oxygen (
Next is Boron (
The you balance the Hydrogen (
So you add a 5 to the water molecule on the left such that it doesn't effect the Sulphate ion, and in doing so balances the amount of Hydrogens (12 each side).
Everything should now balance, including Oxygen (16 on left and 16 on the right), if it doesn't, just keep trying by putting different values of coefficients before each molecule until it works out.
Remember that the coefficients should be the lowest value possible, and must not be fractions.
Hope I helped :)
You first balance the metals, namely Sodium (
Next you do the non-metals, starting with anything but Oxygen (
Next is Boron (
The you balance the Hydrogen (
So you add a 5 to the water molecule on the left such that it doesn't effect the Sulphate ion, and in doing so balances the amount of Hydrogens (12 each side).
Everything should now balance, including Oxygen (16 on left and 16 on the right), if it doesn't, just keep trying by putting different values of coefficients before each molecule until it works out.
Remember that the coefficients should be the lowest value possible, and must not be fractions.
Hope I helped :)
You first balance the metals, namely Sodium (
Next you do the non-metals, starting with anything but Oxygen (
Next is Boron (
The you balance the Hydrogen (
So you add a 5 to the water molecule on the left such that it doesn't effect the Sulphate ion, and in doing so balances the amount of Hydrogens (12 each side).
Everything should now balance, including Oxygen (16 on left and 16 on the right), if it doesn't, just keep trying by putting different values of coefficients before each molecule until it works out.
Remember that the coefficients should be the lowest value possible, and must not be fractions.
Hope I helped :)
It does not matter how long the equation is. You need to always start with tallying the number of atoms (based on the subscripts).
Left side:
Na = 2
B = 4
O = 7 + 4 + 1 (DO NOT ADD IT UP YET)
H = 2 + 2 (DO NOT ADD IT UP YET)
S= 1
Right side:
Na = 2
B = 1
O = 3 + 4 (DO NOT ADD IT UP YET)
H = 3
S= 1
Start with the element easiest to balance, in this case boron (B). Since B is part of the substance
Left side:
Na = 2
B = 4
O = 7 + 4 + 1
H = 2 + 2
S= 1
Right side:
Na = 2
B = 1 x 4 = 4
O = (3 x 4) + 4
H = (3 x 4) = 12
S= 1
Now, the number of H on your right side is 12. There are two H atoms on your left side that you can choose to multiply by 5 (since both values are 2) to balance. Choose the least complicated substance or the one with least number of other atoms attached to it. In this case, it is
Left side:
Na = 2
B = 4
O = 7 + 4 + (1 x 5)
H = 2 + (2 x 5) = 12
S= 1
Right side:
Na = 2
B = 1 x 4 = 4
O = (3 x 4) + 4
H = (3 x 4) = 12
S= 1
Now, look at your tally sheet and see it everything is balanced out.
Left side:
Na = 2
B = 4
O = 7 + 4 + (1 x 5) = 16
H = 2 + (2 x 5) = 12
S= 1
Right side:
Na = 2
B = 1 x 4 = 4
O = (3 x 4) + 4 = 16
H = (3 x 4) = 12
S= 1
The equation is now balanced.
The formula for molality
#color(blue)(|bar(ul(color(white)(a/a) b = ""moles of solute""/""kilograms of solvent""color(white)(a/a)|)))"" ""#
∴
The molality is 2.50 mol/kg.
The formula for molality
#color(blue)(|bar(ul(color(white)(a/a) b = ""moles of solute""/""kilograms of solvent""color(white)(a/a)|)))"" ""#
∴
The molality is 2.50 mol/kg.
The formula for molality
#color(blue)(|bar(ul(color(white)(a/a) b = ""moles of solute""/""kilograms of solvent""color(white)(a/a)|)))"" ""#
∴
First write balanced chemical equation for this ion-exchange reaction :
This balanced chemical equation represents the mole ratio in which the chemicals combine and form products.
So clearly 1 mole of each reactant combines to form 3 moles potassium nitrate and 1 mole aluminium phosphate.
Hence, by ratio and proportion, 2 moles of each reactant will produce 6 moles potassium nitrate and 2 moles aluminium phosphate.
First write balanced chemical equation for this ion-exchange reaction :
This balanced chemical equation represents the mole ratio in which the chemicals combine and form products.
So clearly 1 mole of each reactant combines to form 3 moles potassium nitrate and 1 mole aluminium phosphate.
Hence, by ratio and proportion, 2 moles of each reactant will produce 6 moles potassium nitrate and 2 moles aluminium phosphate.
First write balanced chemical equation for this ion-exchange reaction :
This balanced chemical equation represents the mole ratio in which the chemicals combine and form products.
So clearly 1 mole of each reactant combines to form 3 moles potassium nitrate and 1 mole aluminium phosphate.
Hence, by ratio and proportion, 2 moles of each reactant will produce 6 moles potassium nitrate and 2 moles aluminium phosphate.
Molarity is represented by the equation below:
We know the molarity and the volume of solution. The volume does not have good units since it is given in terms of milliliters instead of liters.
225 mL can be converted into liters by using the conversion factor 1000mL = 1L.
When 225 mL is divided by 1000 mL/L, you obtain a volume of 0.225 L.
Now we can rearrange the equation to solve for the number of moles. This can be accomplished by multiplying by liters of solution on both sides of the equation. The liters of solution will cancel out on the right side, leaving the number of moles being equal to the molarity times volume like so:
Moles of solute =
Moles of solute =
0.180 moles of
Molarity is represented by the equation below:
We know the molarity and the volume of solution. The volume does not have good units since it is given in terms of milliliters instead of liters.
225 mL can be converted into liters by using the conversion factor 1000mL = 1L.
When 225 mL is divided by 1000 mL/L, you obtain a volume of 0.225 L.
Now we can rearrange the equation to solve for the number of moles. This can be accomplished by multiplying by liters of solution on both sides of the equation. The liters of solution will cancel out on the right side, leaving the number of moles being equal to the molarity times volume like so:
Moles of solute =
Moles of solute =
0.180 moles of
Molarity is represented by the equation below:
We know the molarity and the volume of solution. The volume does not have good units since it is given in terms of milliliters instead of liters.
225 mL can be converted into liters by using the conversion factor 1000mL = 1L.
When 225 mL is divided by 1000 mL/L, you obtain a volume of 0.225 L.
Now we can rearrange the equation to solve for the number of moles. This can be accomplished by multiplying by liters of solution on both sides of the equation. The liters of solution will cancel out on the right side, leaving the number of moles being equal to the molarity times volume like so:
Moles of solute =
Moles of solute =
Final concentration
Final concentration
The balanced equation clearly specifies a 1:1 equivalence.
Alternatively, we could write:
And the fashion seems to favour this representation.
The given equations are equivalent they represent the acid/base equilibrium that occurs in water, which at
The balanced equation clearly specifies a 1:1 equivalence.
Alternatively, we could write:
And the fashion seems to favour this representation.
The given equations are equivalent they represent the acid/base equilibrium that occurs in water, which at
The balanced equation clearly specifies a 1:1 equivalence.
Alternatively, we could write:
And the fashion seems to favour this representation.
The given equations are equivalent they represent the acid/base equilibrium that occurs in water, which at
The idea here is that the
Formic acid is a weak acid, which means that it only partially ionizes to produce formate anions, the conjugate base of formic acid, and hydronium cations. For the sake of simplicity, I'll use
#""HA""_ ((aq)) + ""H""_ 2""O""_ ((l)) rightleftharpoons ""H""_ 3""O""_ ((aq))^(+) + ""A""_ ((aq))^(-)#
As you know, the
#""pH"" = - log([""H""_3""O""^(+)])#
This implies that the concentration of hydronium cations is equal to
#[""H""_3""O""^(+)] = 10^(-""pH"")color(white)(.)""M""#
Now, notice that in order for the ionization of the weak acid to produce
If you take
#[""HA""]_ 0 = [""HA""] + [""H""_ 3""O""^(+)]#
This is equivalent to saying that in order to get an equilibrium concentration of
#[""HA""] = [""HA""]_ 0 - [""H""_ 3""O""^(+)]#
Now, by definition, the acid dissociation constant is equal to
#K_a = ([""H""_3""O""^(+)] * [""A""^(-)])/([""HA""])#
Since you know that, at equilibrium, you have
#[""H""_3""O""^(+)] = [""A""^(-)] = 10^(-""pH"")#
you can rewrite the expression you have for the acid dissociation constant as
#K_a = ([""H""_3""O""^(+)]^2)/([""HA""]_0 - [""H""_3""O""^(+)])#
which is, of course, equivalent to
#K_a = (10^(-""pH""))^2/([""HA""]_0 - 10^(-""pH""))#
Rearrange to solve for the initial concentration of the formic acid
#[""HA""]_0 * K_a = 10^(-2""pH"") + 10^(-""pH"") * K_a#
#[""HA""]_0 = 10^(-2""pH"")/K_a + 10^(-""pH"")#
Finally, plug in your values to get
#[""HA""]_ 0 = (10^(-2 * 1.90))/(1.8 * 10^(-4)) + 10^(-1.90) = color(darkgreen)(ul(color(black)(""0.89 M"")))#
The answer is rounded to two sig figs, the number of decimal places you have for the
The idea here is that the
Formic acid is a weak acid, which means that it only partially ionizes to produce formate anions, the conjugate base of formic acid, and hydronium cations. For the sake of simplicity, I'll use
#""HA""_ ((aq)) + ""H""_ 2""O""_ ((l)) rightleftharpoons ""H""_ 3""O""_ ((aq))^(+) + ""A""_ ((aq))^(-)#
As you know, the
#""pH"" = - log([""H""_3""O""^(+)])#
This implies that the concentration of hydronium cations is equal to
#[""H""_3""O""^(+)] = 10^(-""pH"")color(white)(.)""M""#
Now, notice that in order for the ionization of the weak acid to produce
If you take
#[""HA""]_ 0 = [""HA""] + [""H""_ 3""O""^(+)]#
This is equivalent to saying that in order to get an equilibrium concentration of
#[""HA""] = [""HA""]_ 0 - [""H""_ 3""O""^(+)]#
Now, by definition, the acid dissociation constant is equal to
#K_a = ([""H""_3""O""^(+)] * [""A""^(-)])/([""HA""])#
Since you know that, at equilibrium, you have
#[""H""_3""O""^(+)] = [""A""^(-)] = 10^(-""pH"")#
you can rewrite the expression you have for the acid dissociation constant as
#K_a = ([""H""_3""O""^(+)]^2)/([""HA""]_0 - [""H""_3""O""^(+)])#
which is, of course, equivalent to
#K_a = (10^(-""pH""))^2/([""HA""]_0 - 10^(-""pH""))#
Rearrange to solve for the initial concentration of the formic acid
#[""HA""]_0 * K_a = 10^(-2""pH"") + 10^(-""pH"") * K_a#
#[""HA""]_0 = 10^(-2""pH"")/K_a + 10^(-""pH"")#
Finally, plug in your values to get
#[""HA""]_ 0 = (10^(-2 * 1.90))/(1.8 * 10^(-4)) + 10^(-1.90) = color(darkgreen)(ul(color(black)(""0.89 M"")))#
The answer is rounded to two sig figs, the number of decimal places you have for the
The idea here is that the
Formic acid is a weak acid, which means that it only partially ionizes to produce formate anions, the conjugate base of formic acid, and hydronium cations. For the sake of simplicity, I'll use
#""HA""_ ((aq)) + ""H""_ 2""O""_ ((l)) rightleftharpoons ""H""_ 3""O""_ ((aq))^(+) + ""A""_ ((aq))^(-)#
As you know, the
#""pH"" = - log([""H""_3""O""^(+)])#
This implies that the concentration of hydronium cations is equal to
#[""H""_3""O""^(+)] = 10^(-""pH"")color(white)(.)""M""#
Now, notice that in order for the ionization of the weak acid to produce
If you take
#[""HA""]_ 0 = [""HA""] + [""H""_ 3""O""^(+)]#
This is equivalent to saying that in order to get an equilibrium concentration of
#[""HA""] = [""HA""]_ 0 - [""H""_ 3""O""^(+)]#
Now, by definition, the acid dissociation constant is equal to
#K_a = ([""H""_3""O""^(+)] * [""A""^(-)])/([""HA""])#
Since you know that, at equilibrium, you have
#[""H""_3""O""^(+)] = [""A""^(-)] = 10^(-""pH"")#
you can rewrite the expression you have for the acid dissociation constant as
#K_a = ([""H""_3""O""^(+)]^2)/([""HA""]_0 - [""H""_3""O""^(+)])#
which is, of course, equivalent to
#K_a = (10^(-""pH""))^2/([""HA""]_0 - 10^(-""pH""))#
Rearrange to solve for the initial concentration of the formic acid
#[""HA""]_0 * K_a = 10^(-2""pH"") + 10^(-""pH"") * K_a#
#[""HA""]_0 = 10^(-2""pH"")/K_a + 10^(-""pH"")#
Finally, plug in your values to get
#[""HA""]_ 0 = (10^(-2 * 1.90))/(1.8 * 10^(-4)) + 10^(-1.90) = color(darkgreen)(ul(color(black)(""0.89 M"")))#
The answer is rounded to two sig figs, the number of decimal places you have for the
The concentration is 0.89 mol/L.
Step 1. Calculate
Step 2. Calculate the concentration of formic acid
We can use an ICE table to organize our calculations.
Let
In this problem,
So, our ICE table becomes
Check:
It checks!
We assume that the volumes are additive, and we work out the molar concentration of the new solution.
And thus.......
What are
We use the relationship
We assume that the volumes are additive, and we work out the molar concentration of the new solution.
And thus.......
What are
We use the relationship
We assume that the volumes are additive, and we work out the molar concentration of the new solution.
And thus.......
What are
Step 1. Calculate the total moles of
(a) Moles of
(b) Moles of
(c) Total moles of
Step 2. Calculate the total volume of the solution
Step 3. Calculate the total concentration of
To find the empirical formula for caffeine we begin with the Molecular (true) Formula
We can then reduce the Molecular Formula to the Empirical (simple) Formula by dividing each of the subscripts by the greatest common factor. In this case we divide by 2.
This is the Empirical Formula.
I hope this was beneficial.
SMARTERTEACHER
To find the empirical formula for caffeine we begin with the Molecular (true) Formula
We can then reduce the Molecular Formula to the Empirical (simple) Formula by dividing each of the subscripts by the greatest common factor. In this case we divide by 2.
This is the Empirical Formula.
I hope this was beneficial.
SMARTERTEACHER
To find the empirical formula for caffeine we begin with the Molecular (true) Formula
We can then reduce the Molecular Formula to the Empirical (simple) Formula by dividing each of the subscripts by the greatest common factor. In this case we divide by 2.
This is the Empirical Formula.
I hope this was beneficial.
SMARTERTEACHER
The conjugate base of any Brønsted-Lowry acid can be found by removing a proton (
#HNO_2 (aq) + H_2O (l) rightleftharpoons NO_2^(-) (aq) + H_3O^(+) (aq)#
Here, the Brønsted-Lowry acid,
The conjugate base of any Brønsted-Lowry acid can be found by removing a proton (
#HNO_2 (aq) + H_2O (l) rightleftharpoons NO_2^(-) (aq) + H_3O^(+) (aq)#
Here, the Brønsted-Lowry acid,
The conjugate base of any Brønsted-Lowry acid can be found by removing a proton (
#HNO_2 (aq) + H_2O (l) rightleftharpoons NO_2^(-) (aq) + H_3O^(+) (aq)#
Here, the Brønsted-Lowry acid,
The term 'conjugate base' in the chemical community is typically used in association with the term 'conjugate acid' and comes from the Bronsted-Lowry theory of Acids and Bases. The textbook definition is based upon the proton transfer relationship between an acidic substance that can donate a hydrogen ion (proton) to an alkaline substance that can accept the hydrogen ion forming a stronger bond with the accepting substance. The Bronsted-Acid is strictly defined as a proton donor and the Bronsted-Base a proton acceptior. These definitions imply the need for a chemical reaction in order to show 'donator' substance and 'acceptor' substance. The application origin is found in the 'Bronsted-Lowry' theory of acids and bases.
The following can be classified as a molecular form Bronsted-Lowry Acid-Base Rxn.
HCl(aq) + NaOH(aq) => NaCl(aq) + HOH(l)
In aqueous media, the the proton from the Hydrochloric Acid molecule is transferred to the Hydroxide ion of the Sodium Hydroxide molecule producing Sodium Chloride and molecular water. The essential driving force of the reaction is the tendency for the transferring Hydrogen to form a stronger, more stable bond with the Hydroxide ion and thus a weaker acid.
The products, once formed, are now themselves acids and bases but are referred to as 'Conjugate Acid' and 'Conjugate Base'. The water ( H-OH ) which now has the transferred Hydrogen is now a proton donor, or 'Conjugate Acid' of the reactant base (
Now, it should be noted that anions of negative charge are generally, and collectively referred to in the chemical community linguistically as 'Conjugate Bases' if the species are negative and as a 'Conjugate Acids' if the species are positive. The association is referred to as a 'Conjugate Acid-Base Pair'.
Some Examples:
The 'Conjugate Acid-Base Pair' for ...
The difference between a Conjugate Base and Acid of the conjugate base is the presence of 1 Hydrogen. That is, to any anion, adding 1 Hydrogen, gives the acid of that conjugate base.
Generally:
There are two steps to this thermochemistry process
First we are going to calculate the condensation process at
Then calculate the cooling of the liquid from
Step 1
Step 2
There are two steps to this thermochemistry process
First we are going to calculate the condensation process at
Then calculate the cooling of the liquid from
Step 1
Step 2
There are two steps to this thermochemistry process
First we are going to calculate the condensation process at
Then calculate the cooling of the liquid from
Step 1
Step 2
Let's start by writing the chemical reaction for the dissociation of silver phosphate:
Now, set the Ksp value equal to the products (you don't care about the reactants because it's a solid). Ag is raised to the 3rd power because the coefficient is 3.
Replace each reactant with the letter x because that's what you're trying to find. Since the coefficient in front of silver is 3, a 3 must be placed in front of the x and it must be raised to the 3rd power.
Now take
Now you're left with
(
The value of x that we just obtained is our molar solubility.
Let's start by writing the chemical reaction for the dissociation of silver phosphate:
Now, set the Ksp value equal to the products (you don't care about the reactants because it's a solid). Ag is raised to the 3rd power because the coefficient is 3.
Replace each reactant with the letter x because that's what you're trying to find. Since the coefficient in front of silver is 3, a 3 must be placed in front of the x and it must be raised to the 3rd power.
Now take
Now you're left with
(
The value of x that we just obtained is our molar solubility.
Let's start by writing the chemical reaction for the dissociation of silver phosphate:
Now, set the Ksp value equal to the products (you don't care about the reactants because it's a solid). Ag is raised to the 3rd power because the coefficient is 3.
Replace each reactant with the letter x because that's what you're trying to find. Since the coefficient in front of silver is 3, a 3 must be placed in front of the x and it must be raised to the 3rd power.
Now take
Now you're left with
(
The value of x that we just obtained is our molar solubility.
You have already down the hard yards: the mass of oxygen is the mass of the sample LESS the mass of the metals.
I make that about
Well, clearly it is
You have already down the hard yards: the mass of oxygen is the mass of the sample LESS the mass of the metals.
I make that about
Well, clearly it is
You have already down the hard yards: the mass of oxygen is the mass of the sample LESS the mass of the metals.
I make that about
Aluminum nitrite consists of the aluminum cation
Since the aluminum ion has a charge of 3+, and the nitrite ion has a charge of 1-, there need to be three nitrite ions for every aluminum ion. Therefore, the chemical formula for aluminum nitrite is
The formula for aluminum nitrite is
Aluminum nitrite consists of the aluminum cation
Since the aluminum ion has a charge of 3+, and the nitrite ion has a charge of 1-, there need to be three nitrite ions for every aluminum ion. Therefore, the chemical formula for aluminum nitrite is
The formula for aluminum nitrite is
Aluminum nitrite consists of the aluminum cation
Since the aluminum ion has a charge of 3+, and the nitrite ion has a charge of 1-, there need to be three nitrite ions for every aluminum ion. Therefore, the chemical formula for aluminum nitrite is
Simply, the reaction is
From 1 mol of
Hence,
We know,1mol of any ideal gas at STP (
Produced ammonia is
Ans:
Simply, the reaction is
From 1 mol of
Hence,
We know,1mol of any ideal gas at STP (
Produced ammonia is
Ans:
Simply, the reaction is
From 1 mol of
Hence,
We know,1mol of any ideal gas at STP (
Produced ammonia is
Magnesium is
Sulphate is
Hence, magnesium sulphate is
""Hepta-"" refers to ""seven"", ""hydrate"" refers to ""water"", so we have to add
Formula:
Magnesium is
Sulphate is
Hence, magnesium sulphate is
""Hepta-"" refers to ""seven"", ""hydrate"" refers to ""water"", so we have to add
Formula:
Magnesium is
Sulphate is
Hence, magnesium sulphate is
""Hepta-"" refers to ""seven"", ""hydrate"" refers to ""water"", so we have to add
Formula:
Mg(OH)2 has a solubility of 9.0x10^-5 M. What is the value of Ksp for Mg(OH)2?
The chemical equation for the equilibrium is
∴
The chemical equation for the equilibrium is
∴
Mg(OH)2 has a solubility of 9.0x10^-5 M. What is the value of Ksp for Mg(OH)2?
The chemical equation for the equilibrium is
∴
The Reaction:
Data given:
Pressure
Volume
Temperature
Work :
At 303 K,
∴
Apply the Ideal Gas Law
The constant
The Reaction:
Data given:
Pressure
Volume
Temperature
Work :
At 303 K,
∴
Apply the Ideal Gas Law
The constant
The Reaction:
Data given:
Pressure
Volume
Temperature
Work :
At 303 K,
∴
Apply the Ideal Gas Law
The constant
A solution's osmolarity basically tells you the number of moles of particles that contribute to the solution's osmotic pressure present in one liter of solution.
A particle that contributes to a solution's osmotic pressure is called an osmole.
Now, urea,
This implies that every mole of urea dissolved in solution will produce one osmole of particles of solute.
Use the molar mass of urea to calculate how many moles you have in that sample
#25 color(red)(cancel(color(black)(""g""))) * ""1 mole urea""/(60.055color(red)(cancel(color(black)(""g"")))) = ""0.4163 moles urea""#
Assuming that the volume of the solutionThis means that the solution's osmolarity will be equal to
#0.4163 color(white)(a)color(red)(cancel(color(black)(""moles"")))/""1 L"" * ""1 Osmol""/(1color(red)(cancel(color(black)(""mole"")))) = color(green)(|bar(ul(color(white)(a/a)color(black)(""0.42 Osmol L""^(-1))color(white)(a/a)|)))#
I'll leave the answer rounded to two sig figs, but keep in mind that you only have one sig fig for the volume of the solution.
A solution's osmolarity basically tells you the number of moles of particles that contribute to the solution's osmotic pressure present in one liter of solution.
A particle that contributes to a solution's osmotic pressure is called an osmole.
Now, urea,
This implies that every mole of urea dissolved in solution will produce one osmole of particles of solute.
Use the molar mass of urea to calculate how many moles you have in that sample
#25 color(red)(cancel(color(black)(""g""))) * ""1 mole urea""/(60.055color(red)(cancel(color(black)(""g"")))) = ""0.4163 moles urea""#
Assuming that the volume of the solutionThis means that the solution's osmolarity will be equal to
#0.4163 color(white)(a)color(red)(cancel(color(black)(""moles"")))/""1 L"" * ""1 Osmol""/(1color(red)(cancel(color(black)(""mole"")))) = color(green)(|bar(ul(color(white)(a/a)color(black)(""0.42 Osmol L""^(-1))color(white)(a/a)|)))#
I'll leave the answer rounded to two sig figs, but keep in mind that you only have one sig fig for the volume of the solution.
A solution's osmolarity basically tells you the number of moles of particles that contribute to the solution's osmotic pressure present in one liter of solution.
A particle that contributes to a solution's osmotic pressure is called an osmole.
Now, urea,
This implies that every mole of urea dissolved in solution will produce one osmole of particles of solute.
Use the molar mass of urea to calculate how many moles you have in that sample
#25 color(red)(cancel(color(black)(""g""))) * ""1 mole urea""/(60.055color(red)(cancel(color(black)(""g"")))) = ""0.4163 moles urea""#
Assuming that the volume of the solutionThis means that the solution's osmolarity will be equal to
#0.4163 color(white)(a)color(red)(cancel(color(black)(""moles"")))/""1 L"" * ""1 Osmol""/(1color(red)(cancel(color(black)(""mole"")))) = color(green)(|bar(ul(color(white)(a/a)color(black)(""0.42 Osmol L""^(-1))color(white)(a/a)|)))#
I'll leave the answer rounded to two sig figs, but keep in mind that you only have one sig fig for the volume of the solution.
The purpose of using oxalic acid dihydrate,
In order to determine the normality of the oxalic acid solution, you need to first figure out how many electrons will the reducing agent lose.
These electrons will help you determine the equivalents needed to calculate the solution's normality.
Now, I won't show you exactly how to balance redox reactions because I don't want the answer to become too long. The balanced chemical equation for the redox titration of potassium permanganate with oxalic acid looks like this
I will show you how to balance the oxidation half-reaction, in which the oxalate ion,
This half-reaction will tell you how many electrons are being lost by the reducing agent.
The carbon atoms are going from an oxidation state of +3 on the reactants' side, to an oxidation state of +4 on the products' side.
However, keep in mind that you have 2 carbon atoms on the reactants' side, so multiply the carbon dioxide by 2 to get
This means that a total of 2 electrons are being lost, one from each of the two carbon atoms.
This tells you that 1 mole of oxalic acid produces 2 moles of electrons for the redox reaction. These are your equivalents. You thus have
Now, in order to determine how many moles of oxalic acid you have, you need to use the percent composition of the dihydrate form.
Oxalic acid dihydrate contains
by mass. This means that your 3.6-g sample of dihydrate actually contains
The number of moles of oxalic acid will be
This means that you get
The solution's normality will thus be
Normality: 0.38 N.
The purpose of using oxalic acid dihydrate,
In order to determine the normality of the oxalic acid solution, you need to first figure out how many electrons will the reducing agent lose.
These electrons will help you determine the equivalents needed to calculate the solution's normality.
Now, I won't show you exactly how to balance redox reactions because I don't want the answer to become too long. The balanced chemical equation for the redox titration of potassium permanganate with oxalic acid looks like this
I will show you how to balance the oxidation half-reaction, in which the oxalate ion,
This half-reaction will tell you how many electrons are being lost by the reducing agent.
The carbon atoms are going from an oxidation state of +3 on the reactants' side, to an oxidation state of +4 on the products' side.
However, keep in mind that you have 2 carbon atoms on the reactants' side, so multiply the carbon dioxide by 2 to get
This means that a total of 2 electrons are being lost, one from each of the two carbon atoms.
This tells you that 1 mole of oxalic acid produces 2 moles of electrons for the redox reaction. These are your equivalents. You thus have
Now, in order to determine how many moles of oxalic acid you have, you need to use the percent composition of the dihydrate form.
Oxalic acid dihydrate contains
by mass. This means that your 3.6-g sample of dihydrate actually contains
The number of moles of oxalic acid will be
This means that you get
The solution's normality will thus be
Normality: 0.38 N.
The purpose of using oxalic acid dihydrate,
In order to determine the normality of the oxalic acid solution, you need to first figure out how many electrons will the reducing agent lose.
These electrons will help you determine the equivalents needed to calculate the solution's normality.
Now, I won't show you exactly how to balance redox reactions because I don't want the answer to become too long. The balanced chemical equation for the redox titration of potassium permanganate with oxalic acid looks like this
I will show you how to balance the oxidation half-reaction, in which the oxalate ion,
This half-reaction will tell you how many electrons are being lost by the reducing agent.
The carbon atoms are going from an oxidation state of +3 on the reactants' side, to an oxidation state of +4 on the products' side.
However, keep in mind that you have 2 carbon atoms on the reactants' side, so multiply the carbon dioxide by 2 to get
This means that a total of 2 electrons are being lost, one from each of the two carbon atoms.
This tells you that 1 mole of oxalic acid produces 2 moles of electrons for the redox reaction. These are your equivalents. You thus have
Now, in order to determine how many moles of oxalic acid you have, you need to use the percent composition of the dihydrate form.
Oxalic acid dihydrate contains
by mass. This means that your 3.6-g sample of dihydrate actually contains
The number of moles of oxalic acid will be
This means that you get
The solution's normality will thus be
This is NOT the stoichiometric equation:
How would you modify it so that the left had side is equivalent to the right hand side? You can certainly use half-coeffcients,
You balance it stoichiometrically.
This is NOT the stoichiometric equation:
How would you modify it so that the left had side is equivalent to the right hand side? You can certainly use half-coeffcients,
You balance it stoichiometrically.
This is NOT the stoichiometric equation:
How would you modify it so that the left had side is equivalent to the right hand side? You can certainly use half-coeffcients,
Balance Cl.
There are 3 Cl atoms on the product side and 1 on the reactant side. Add a coefficient of 3 in front of
There are now 3 Cl atoms on both sides.
Balance H.
There are 3 H atoms on the reactant side and 2 on the product side. 3 and 2 are multiples of 6. So change the coefficient in front of HCl from 3 to 6, and add a coefficient of 3 in front of the
There are now 6 H atoms on both sides.
Go back to the Cl. There are now 6 Cl on the reactant side and 3 on the product side. Add a coefficient of 2 in front of
There are now 6 Cl atoms on both sides.
Balance the Al
There is 1 Al on the reactant side and 2 Al on the product side. Add a coefficient in front of Al on the reactant side.
There are now 2 Al on both sides and the equation is balanced.
Reactants:
Products:
And we need a stoichiometric equation that represents the complete combustion of methanol:
Is it balanced? It looks like it to me.
And so we combust
Given the stoichiometry, at most, we can get
If you are unsatisfied, I am willing to try again.
I get
And we need a stoichiometric equation that represents the complete combustion of methanol:
Is it balanced? It looks like it to me.
And so we combust
Given the stoichiometry, at most, we can get
If you are unsatisfied, I am willing to try again.
I get
And we need a stoichiometric equation that represents the complete combustion of methanol:
Is it balanced? It looks like it to me.
And so we combust
Given the stoichiometry, at most, we can get
If you are unsatisfied, I am willing to try again.
And thus
And thus
And thus
Before we begin let me introduce Dalton's Law of Partial Pressures equation:
Where
Based on what you've given me, we know the total pressure,
Therefore,
The partial pressure of the other gas is
Before we begin let me introduce Dalton's Law of Partial Pressures equation:
Where
Based on what you've given me, we know the total pressure,
Therefore,
The partial pressure of the other gas is
Before we begin let me introduce Dalton's Law of Partial Pressures equation:
Where
Based on what you've given me, we know the total pressure,
Therefore,
I think this is the primary component of
All I have done here is made sure that garbage out equals garbage in (have I?). I have assumed that SOME of the hydrocarbon reactant is incompletely combusted. What do I mean by this? The actual degree of combustion would have to be measured by some means of analysis. We could certainly represent it.
Well, for a start,
I think this is the primary component of
All I have done here is made sure that garbage out equals garbage in (have I?). I have assumed that SOME of the hydrocarbon reactant is incompletely combusted. What do I mean by this? The actual degree of combustion would have to be measured by some means of analysis. We could certainly represent it.
Well, for a start,
I think this is the primary component of
All I have done here is made sure that garbage out equals garbage in (have I?). I have assumed that SOME of the hydrocarbon reactant is incompletely combusted. What do I mean by this? The actual degree of combustion would have to be measured by some means of analysis. We could certainly represent it.
An alternative approach is to simply use the mole ratios that exist between the species that take part in the reaction, then use the molar mass of copper(II) hydroxide,
So, your sodium hydroxide solution contains
#C = n/V implies n = C * V#
#n_(OH^(-)) = ""0.450 M"" * 100 * 10^(-3)""L"" = ""0.0450 moles OH""""""^(-)#
The net ionic equation for this double replacement reaction looks like this
#""Cu""_text((aq])^(2+) + color(red)(2)""OH""_text((aq])^(-) -> ""Cu""(""OH"")_text(2(s]) darr#
So,
This means that the reaction will produce
#0.0450color(red)(cancel(color(black)(""moles OH""""""^(-)))) * (""1 mole Cu""(""OH"")_2)/(color(red)(2)color(red)(cancel(color(black)(""moles OH""""""^(-))))) = ""0.0225 moles Cu""(""OH"")_2#
To find how many grams of copper(II) hydroxide will contain this many moles, use the compound's molar mass
#0.0225color(red)(cancel(color(black)(""moles OH""""""^(-)))) * ""97.56 g""/(1color(red)(cancel(color(black)(""mole OH""""""^(-))))) = ""2.195 g Cu""(""OH"")_2#
Rounded to three sig figs, the answer will be
#m_(Cu(OH)_2) = color(green)(""2.20 g"")#
An alternative approach is to simply use the mole ratios that exist between the species that take part in the reaction, then use the molar mass of copper(II) hydroxide,
So, your sodium hydroxide solution contains
#C = n/V implies n = C * V#
#n_(OH^(-)) = ""0.450 M"" * 100 * 10^(-3)""L"" = ""0.0450 moles OH""""""^(-)#
The net ionic equation for this double replacement reaction looks like this
#""Cu""_text((aq])^(2+) + color(red)(2)""OH""_text((aq])^(-) -> ""Cu""(""OH"")_text(2(s]) darr#
So,
This means that the reaction will produce
#0.0450color(red)(cancel(color(black)(""moles OH""""""^(-)))) * (""1 mole Cu""(""OH"")_2)/(color(red)(2)color(red)(cancel(color(black)(""moles OH""""""^(-))))) = ""0.0225 moles Cu""(""OH"")_2#
To find how many grams of copper(II) hydroxide will contain this many moles, use the compound's molar mass
#0.0225color(red)(cancel(color(black)(""moles OH""""""^(-)))) * ""97.56 g""/(1color(red)(cancel(color(black)(""mole OH""""""^(-))))) = ""2.195 g Cu""(""OH"")_2#
Rounded to three sig figs, the answer will be
#m_(Cu(OH)_2) = color(green)(""2.20 g"")#
An alternative approach is to simply use the mole ratios that exist between the species that take part in the reaction, then use the molar mass of copper(II) hydroxide,
So, your sodium hydroxide solution contains
#C = n/V implies n = C * V#
#n_(OH^(-)) = ""0.450 M"" * 100 * 10^(-3)""L"" = ""0.0450 moles OH""""""^(-)#
The net ionic equation for this double replacement reaction looks like this
#""Cu""_text((aq])^(2+) + color(red)(2)""OH""_text((aq])^(-) -> ""Cu""(""OH"")_text(2(s]) darr#
So,
This means that the reaction will produce
#0.0450color(red)(cancel(color(black)(""moles OH""""""^(-)))) * (""1 mole Cu""(""OH"")_2)/(color(red)(2)color(red)(cancel(color(black)(""moles OH""""""^(-))))) = ""0.0225 moles Cu""(""OH"")_2#
To find how many grams of copper(II) hydroxide will contain this many moles, use the compound's molar mass
#0.0225color(red)(cancel(color(black)(""moles OH""""""^(-)))) * ""97.56 g""/(1color(red)(cancel(color(black)(""mole OH""""""^(-))))) = ""2.195 g Cu""(""OH"")_2#
Rounded to three sig figs, the answer will be
#m_(Cu(OH)_2) = color(green)(""2.20 g"")#
You know that you're diluting
#0.820 color(red)(cancel(color(black)(""L""))) * (10^3 quad ""mL"")/(1color(red)(cancel(color(black)(""L"")))) = ""820. mL""#
Now, when you're diluting a solution, you must keep in mind that the ratio that exists between the volume of the diluted solution and the volume of the concentrated solution is equal to the dilution factor,
#""DF"" = V_""diluted""/V_""stock""#
Moreover, the dilution factor is also equal to the ratio that exists between the concentration of the concentrated solution and the concentration of the diluted solution.
#""DF"" = c_""stock""/c_""diluted""#
In your case, the dilution factor is equal to
#""DF"" = (820. color(red)(cancel(color(black)(""mL""))))/(1.00 * 10^2color(red)(cancel(color(black)(""L"")))) = color(blue)(8.20)#
You can thus say that the concentration of the initial solution, i.e. the concentrated solution, was
#c_""diluted"" = c_""stock""/color(blue)(8.20)#
#c_""diluted"" = ""12.4 M""/color(blue)(8.20) = color(darkgreen)(ul(color(black)(""1.51 M"")))#
The answer is rounded to three sig figs.
So, if you start with
Don't forget that when diluting strong acids, you must always add the acid to the water and not the water to the acid!
You know that you're diluting
#0.820 color(red)(cancel(color(black)(""L""))) * (10^3 quad ""mL"")/(1color(red)(cancel(color(black)(""L"")))) = ""820. mL""#
Now, when you're diluting a solution, you must keep in mind that the ratio that exists between the volume of the diluted solution and the volume of the concentrated solution is equal to the dilution factor,
#""DF"" = V_""diluted""/V_""stock""#
Moreover, the dilution factor is also equal to the ratio that exists between the concentration of the concentrated solution and the concentration of the diluted solution.
#""DF"" = c_""stock""/c_""diluted""#
In your case, the dilution factor is equal to
#""DF"" = (820. color(red)(cancel(color(black)(""mL""))))/(1.00 * 10^2color(red)(cancel(color(black)(""L"")))) = color(blue)(8.20)#
You can thus say that the concentration of the initial solution, i.e. the concentrated solution, was
#c_""diluted"" = c_""stock""/color(blue)(8.20)#
#c_""diluted"" = ""12.4 M""/color(blue)(8.20) = color(darkgreen)(ul(color(black)(""1.51 M"")))#
The answer is rounded to three sig figs.
So, if you start with
Don't forget that when diluting strong acids, you must always add the acid to the water and not the water to the acid!
You know that you're diluting
#0.820 color(red)(cancel(color(black)(""L""))) * (10^3 quad ""mL"")/(1color(red)(cancel(color(black)(""L"")))) = ""820. mL""#
Now, when you're diluting a solution, you must keep in mind that the ratio that exists between the volume of the diluted solution and the volume of the concentrated solution is equal to the dilution factor,
#""DF"" = V_""diluted""/V_""stock""#
Moreover, the dilution factor is also equal to the ratio that exists between the concentration of the concentrated solution and the concentration of the diluted solution.
#""DF"" = c_""stock""/c_""diluted""#
In your case, the dilution factor is equal to
#""DF"" = (820. color(red)(cancel(color(black)(""mL""))))/(1.00 * 10^2color(red)(cancel(color(black)(""L"")))) = color(blue)(8.20)#
You can thus say that the concentration of the initial solution, i.e. the concentrated solution, was
#c_""diluted"" = c_""stock""/color(blue)(8.20)#
#c_""diluted"" = ""12.4 M""/color(blue)(8.20) = color(darkgreen)(ul(color(black)(""1.51 M"")))#
The answer is rounded to three sig figs.
So, if you start with
Don't forget that when diluting strong acids, you must always add the acid to the water and not the water to the acid!
The relationship between work (
where,
since the gas is expanding, then the work is done by the system and it is of a negative value .
Note that work, in this case, should be expressed in
Since work is done by the system:
Pressure should then be expressed in
Thus, replacing every term in its value in the expression
Note that
Here is a video that further explains this topic:
Thermochemistry | The Nature of Energy.
The relationship between work (
where,
since the gas is expanding, then the work is done by the system and it is of a negative value .
Note that work, in this case, should be expressed in
Since work is done by the system:
Pressure should then be expressed in
Thus, replacing every term in its value in the expression
Note that
Here is a video that further explains this topic:
Thermochemistry | The Nature of Energy.
The relationship between work (
where,
since the gas is expanding, then the work is done by the system and it is of a negative value .
Note that work, in this case, should be expressed in
Since work is done by the system:
Pressure should then be expressed in
Thus, replacing every term in its value in the expression
Note that
Here is a video that further explains this topic:
Thermochemistry | The Nature of Energy.
Start by calculating the number of moles of ammonia present in the
As you know, a
This means that your sample will contain
#25.0 color(red)(cancel(color(black)(""mL solution""))) * ""1.5 g NH""_3/(100color(red)(cancel(color(black)(""mL solution"")))) = ""0.375 g NH""_3#
Use the molar mass of ammonia to convert this to moles
#0.375 color(red)(cancel(color(black)(""g""))) * ""1 mole NH""_3/(17.031color(red)(cancel(color(black)(""g"")))) = ""0.02202 moles NH""_3#
Now, you're diluting this sample by adding
#""25.0 mL + 250 mL = 275 mL""#
Since you only added water, the number of moles of ammonia will remain unchanged, i.e. the diluted solution contains the same number of moles as the
As you know, the molarity of a solution tells you the number of moles of solute present for every
In your case, the number of moles of ammonia present in
#10^3 color(red)(cancel(color(black)(""mL solution""))) * overbrace(""0.02202 moles H""_3/(275 color(red)(cancel(color(black)(""mL solution"")))))^(color(blue)(""the known composition of the diluted solution"")) = ""0.080 moles NH""_3#
So, if you have
#color(darkgreen)(ul(color(black)(""molarity = 0.080 mol L""^(-1))))#
The answer is rounded to two sig figs.
Start by calculating the number of moles of ammonia present in the
As you know, a
This means that your sample will contain
#25.0 color(red)(cancel(color(black)(""mL solution""))) * ""1.5 g NH""_3/(100color(red)(cancel(color(black)(""mL solution"")))) = ""0.375 g NH""_3#
Use the molar mass of ammonia to convert this to moles
#0.375 color(red)(cancel(color(black)(""g""))) * ""1 mole NH""_3/(17.031color(red)(cancel(color(black)(""g"")))) = ""0.02202 moles NH""_3#
Now, you're diluting this sample by adding
#""25.0 mL + 250 mL = 275 mL""#
Since you only added water, the number of moles of ammonia will remain unchanged, i.e. the diluted solution contains the same number of moles as the
As you know, the molarity of a solution tells you the number of moles of solute present for every
In your case, the number of moles of ammonia present in
#10^3 color(red)(cancel(color(black)(""mL solution""))) * overbrace(""0.02202 moles H""_3/(275 color(red)(cancel(color(black)(""mL solution"")))))^(color(blue)(""the known composition of the diluted solution"")) = ""0.080 moles NH""_3#
So, if you have
#color(darkgreen)(ul(color(black)(""molarity = 0.080 mol L""^(-1))))#
The answer is rounded to two sig figs.
Start by calculating the number of moles of ammonia present in the
As you know, a
This means that your sample will contain
#25.0 color(red)(cancel(color(black)(""mL solution""))) * ""1.5 g NH""_3/(100color(red)(cancel(color(black)(""mL solution"")))) = ""0.375 g NH""_3#
Use the molar mass of ammonia to convert this to moles
#0.375 color(red)(cancel(color(black)(""g""))) * ""1 mole NH""_3/(17.031color(red)(cancel(color(black)(""g"")))) = ""0.02202 moles NH""_3#
Now, you're diluting this sample by adding
#""25.0 mL + 250 mL = 275 mL""#
Since you only added water, the number of moles of ammonia will remain unchanged, i.e. the diluted solution contains the same number of moles as the
As you know, the molarity of a solution tells you the number of moles of solute present for every
In your case, the number of moles of ammonia present in
#10^3 color(red)(cancel(color(black)(""mL solution""))) * overbrace(""0.02202 moles H""_3/(275 color(red)(cancel(color(black)(""mL solution"")))))^(color(blue)(""the known composition of the diluted solution"")) = ""0.080 moles NH""_3#
So, if you have
#color(darkgreen)(ul(color(black)(""molarity = 0.080 mol L""^(-1))))#
The answer is rounded to two sig figs.
So, I have 0.35 moles of ethanol:
thus I have
Now
If there are 0.35 moles of something, then there is the number
So, I have 0.35 moles of ethanol:
thus I have
Now
If there are 0.35 moles of something, then there is the number
So, I have 0.35 moles of ethanol:
thus I have
Now
It can be titrated as a base, or here as an acid. It is air stable, non-hygroscopic, and thus useful as a primary standard.
And thus moles of
It can be titrated as a base, or here as an acid. It is air stable, non-hygroscopic, and thus useful as a primary standard.
And thus moles of
It can be titrated as a base, or here as an acid. It is air stable, non-hygroscopic, and thus useful as a primary standard.
And thus moles of
The balanced chemical equation that describes this double replacement reaction
#2""AgNO""_ (3(aq)) + ""BaCl""_ (2(aq)) -> 2""AgCl""_ ((s)) darr + ""Ba""(""NO""_ 3)_ (2(aq))#
tells you that when
In other words, you have
#(""moles of BaCl""_2 quad ""consumed"")/(""moles of AgCl produced"") = 1/2#
Now, in order to be able to use this
#4.62 color(red)(cancel(color(black)(""g""))) * ""1 mole BaCl""_2/(208.23 color(red)(cancel(color(black)(""g"")))) = ""0.02219 moles BaCl""_2#
You can now say that the reaction will produce
#0.02219 color(red)(cancel(color(black)(""moles BaCl""_2))) * ""2 moles AgCl""/(1color(red)(cancel(color(black)(""mole BaCl""_2)))) = ""0.04438 moles AgCl""#
Finally, to convert the number of moles of silver chloride to grams, use the molar mass of the compound.
#0.04438 color(red)(cancel(color(black)(""moles AgCl""))) * ""143.32 g""/(1color(red)(cancel(color(black)(""mole AgCl"")))) = color(darkgreen)(ul(color(black)(""6.36 g"")))#
The answer is rounded to three sig figs.
The balanced chemical equation that describes this double replacement reaction
#2""AgNO""_ (3(aq)) + ""BaCl""_ (2(aq)) -> 2""AgCl""_ ((s)) darr + ""Ba""(""NO""_ 3)_ (2(aq))#
tells you that when
In other words, you have
#(""moles of BaCl""_2 quad ""consumed"")/(""moles of AgCl produced"") = 1/2#
Now, in order to be able to use this
#4.62 color(red)(cancel(color(black)(""g""))) * ""1 mole BaCl""_2/(208.23 color(red)(cancel(color(black)(""g"")))) = ""0.02219 moles BaCl""_2#
You can now say that the reaction will produce
#0.02219 color(red)(cancel(color(black)(""moles BaCl""_2))) * ""2 moles AgCl""/(1color(red)(cancel(color(black)(""mole BaCl""_2)))) = ""0.04438 moles AgCl""#
Finally, to convert the number of moles of silver chloride to grams, use the molar mass of the compound.
#0.04438 color(red)(cancel(color(black)(""moles AgCl""))) * ""143.32 g""/(1color(red)(cancel(color(black)(""mole AgCl"")))) = color(darkgreen)(ul(color(black)(""6.36 g"")))#
The answer is rounded to three sig figs.
The balanced chemical equation that describes this double replacement reaction
#2""AgNO""_ (3(aq)) + ""BaCl""_ (2(aq)) -> 2""AgCl""_ ((s)) darr + ""Ba""(""NO""_ 3)_ (2(aq))#
tells you that when
In other words, you have
#(""moles of BaCl""_2 quad ""consumed"")/(""moles of AgCl produced"") = 1/2#
Now, in order to be able to use this
#4.62 color(red)(cancel(color(black)(""g""))) * ""1 mole BaCl""_2/(208.23 color(red)(cancel(color(black)(""g"")))) = ""0.02219 moles BaCl""_2#
You can now say that the reaction will produce
#0.02219 color(red)(cancel(color(black)(""moles BaCl""_2))) * ""2 moles AgCl""/(1color(red)(cancel(color(black)(""mole BaCl""_2)))) = ""0.04438 moles AgCl""#
Finally, to convert the number of moles of silver chloride to grams, use the molar mass of the compound.
#0.04438 color(red)(cancel(color(black)(""moles AgCl""))) * ""143.32 g""/(1color(red)(cancel(color(black)(""mole AgCl"")))) = color(darkgreen)(ul(color(black)(""6.36 g"")))#
The answer is rounded to three sig figs.
The unbalanced equation is:
Since there is already one mole of carbon
This is a correct, balanced chemical equation.
If fractional coefficients make you uncomfortable, though, it would be possible to multiply all coefficients by 2:
The balanced equation is:
The unbalanced equation is:
Since there is already one mole of carbon
This is a correct, balanced chemical equation.
If fractional coefficients make you uncomfortable, though, it would be possible to multiply all coefficients by 2:
The balanced equation is:
The unbalanced equation is:
Since there is already one mole of carbon
This is a correct, balanced chemical equation.
If fractional coefficients make you uncomfortable, though, it would be possible to multiply all coefficients by 2:
Your starting point here will be the balanced chemical equation for this neutralization reaction.
Calcium carbonate,
#""CaCO""_ (3(s)) + color(red)(2)""HCl""_ ((aq)) -> ""CaCl""_ (2(aq)) + ""H""_ 2""O""_ ((l)) + ""CO""_ (2(g)) uarr#
Notice that it takes
The problem provides you with the molarity and volume of the hydrochloric acid solution. This means that you can use the definition of molarity to find how many moles of acid it contained
#color(blue)(c = n/V implies n = c * V)#
#n_(HCl) = ""0.547 M"" * 24.65 * 10^(-3)""L"" = ""0.013484 moles HCl""#
Use the aforementioned mole ratio to find the number of moles of calcium carbonate that must have been present in the Rolaids table
#0.013484 color(red)(cancel(color(black)(""moles HCl""))) * (""1 mole CaCO""_3)/(color(red)(2)color(red)(cancel(color(black)(""moles HCl"")))) = ""0.0067420 moles CaCO""_3#
Finally, to find the mass of calcium carbonate that contains this many moles, use the compound's molar mass
#0.0067420 color(red)(cancel(color(black)(""moles CaCO""_3))) * ""100.1 g""/(1color(red)(cancel(color(black)(""mole CaCO""_3)))) = ""0.645 g""#
Expressed in milligrams, the answer will be
#0.645 color(red)(cancel(color(black)(""g""))) * (10^3""mg"")/(1color(red)(cancel(color(black)(""g"")))) = color(green)(""645 mg"")#
The answer is rounded to three sig figs.
Your starting point here will be the balanced chemical equation for this neutralization reaction.
Calcium carbonate,
#""CaCO""_ (3(s)) + color(red)(2)""HCl""_ ((aq)) -> ""CaCl""_ (2(aq)) + ""H""_ 2""O""_ ((l)) + ""CO""_ (2(g)) uarr#
Notice that it takes
The problem provides you with the molarity and volume of the hydrochloric acid solution. This means that you can use the definition of molarity to find how many moles of acid it contained
#color(blue)(c = n/V implies n = c * V)#
#n_(HCl) = ""0.547 M"" * 24.65 * 10^(-3)""L"" = ""0.013484 moles HCl""#
Use the aforementioned mole ratio to find the number of moles of calcium carbonate that must have been present in the Rolaids table
#0.013484 color(red)(cancel(color(black)(""moles HCl""))) * (""1 mole CaCO""_3)/(color(red)(2)color(red)(cancel(color(black)(""moles HCl"")))) = ""0.0067420 moles CaCO""_3#
Finally, to find the mass of calcium carbonate that contains this many moles, use the compound's molar mass
#0.0067420 color(red)(cancel(color(black)(""moles CaCO""_3))) * ""100.1 g""/(1color(red)(cancel(color(black)(""mole CaCO""_3)))) = ""0.645 g""#
Expressed in milligrams, the answer will be
#0.645 color(red)(cancel(color(black)(""g""))) * (10^3""mg"")/(1color(red)(cancel(color(black)(""g"")))) = color(green)(""645 mg"")#
The answer is rounded to three sig figs.
Your starting point here will be the balanced chemical equation for this neutralization reaction.
Calcium carbonate,
#""CaCO""_ (3(s)) + color(red)(2)""HCl""_ ((aq)) -> ""CaCl""_ (2(aq)) + ""H""_ 2""O""_ ((l)) + ""CO""_ (2(g)) uarr#
Notice that it takes
The problem provides you with the molarity and volume of the hydrochloric acid solution. This means that you can use the definition of molarity to find how many moles of acid it contained
#color(blue)(c = n/V implies n = c * V)#
#n_(HCl) = ""0.547 M"" * 24.65 * 10^(-3)""L"" = ""0.013484 moles HCl""#
Use the aforementioned mole ratio to find the number of moles of calcium carbonate that must have been present in the Rolaids table
#0.013484 color(red)(cancel(color(black)(""moles HCl""))) * (""1 mole CaCO""_3)/(color(red)(2)color(red)(cancel(color(black)(""moles HCl"")))) = ""0.0067420 moles CaCO""_3#
Finally, to find the mass of calcium carbonate that contains this many moles, use the compound's molar mass
#0.0067420 color(red)(cancel(color(black)(""moles CaCO""_3))) * ""100.1 g""/(1color(red)(cancel(color(black)(""mole CaCO""_3)))) = ""0.645 g""#
Expressed in milligrams, the answer will be
#0.645 color(red)(cancel(color(black)(""g""))) * (10^3""mg"")/(1color(red)(cancel(color(black)(""g"")))) = color(green)(""645 mg"")#
The answer is rounded to three sig figs.
You know that one molecule of nitrogen gas,
Now, a mole is simply a very, very large collection of particles. In order to have one mole of things, let's say particles, you need to have
So, in one mole of nitrogen gas you have
#6.022 * 10^(23)color(red)(cancel(color(black)(""molecules N""_2))) * (color(blue)(2)color(white)(a)""atoms of N"")/(1color(red)(cancel(color(black)(""molecule N""_2)))) #
# = 1.2044 * 10^(24)""taoms of N""#
Alternatively, you can express this as
You know that one molecule of nitrogen gas,
Now, a mole is simply a very, very large collection of particles. In order to have one mole of things, let's say particles, you need to have
So, in one mole of nitrogen gas you have
#6.022 * 10^(23)color(red)(cancel(color(black)(""molecules N""_2))) * (color(blue)(2)color(white)(a)""atoms of N"")/(1color(red)(cancel(color(black)(""molecule N""_2)))) #
# = 1.2044 * 10^(24)""taoms of N""#
Alternatively, you can express this as
You know that one molecule of nitrogen gas,
Now, a mole is simply a very, very large collection of particles. In order to have one mole of things, let's say particles, you need to have
So, in one mole of nitrogen gas you have
#6.022 * 10^(23)color(red)(cancel(color(black)(""molecules N""_2))) * (color(blue)(2)color(white)(a)""atoms of N"")/(1color(red)(cancel(color(black)(""molecule N""_2)))) #
# = 1.2044 * 10^(24)""taoms of N""#
Alternatively, you can express this as
And thus
Why did I double the molar quantity with respect to
Well I make
And thus
Why did I double the molar quantity with respect to
Well I make
And thus
Why did I double the molar quantity with respect to
We can use an ICE table to calculate the concentrations of the ions in solution.
The chemical equation is
Let's rewrite this as
Check for negligibility:
∴
Then
pH = 12.03
We can use an ICE table to calculate the concentrations of the ions in solution.
The chemical equation is
Let's rewrite this as
Check for negligibility:
∴
Then
pH = 12.03
We can use an ICE table to calculate the concentrations of the ions in solution.
The chemical equation is
Let's rewrite this as
Check for negligibility:
∴
Then
A
We know that
How many atoms in an
A water molecule contains
A
We know that
How many atoms in an
A water molecule contains
A
We know that
How many atoms in an
Judging by the values you have, the most helpful concentration that can be calculated here is that in parts per million, ppm.
A solution's ppm concentration tells you the number of grams of solute present for every
#10^6 = 1,000,000#
grams of solution. Now, notice that you can approximate the mass of the solution, which is equal to
#""mass solution"" = ""490 g"" + ""0.0085 g"" #
#""mass solution = 490.0085 g""#
to be equal to the mass of water, the solvent.
#""490.0085 g "" ~~ "" 490 g""#
You can thus say that, in your case, a
Use the known composition of the solution as a conversion factor to calculate the number of grams of calcium cations present in
#10^6 color(red)(cancel(color(black)(""g solution""))) * (8.5 * 10^(-3)color(white)(.)""g Ca""^(2+))/(490color(red)(cancel(color(black)(""g solution"")))) = ""17.35 g Ca""^(2+)#
Since this represents the number of grams of calcium cations present in
#color(darkgreen)(ul(color(black)(""concentration"" = ""17 ppm Ca""^(2+))))#
The answer is rounded to two sig figs, the number of sig figs you have for your values.
Judging by the values you have, the most helpful concentration that can be calculated here is that in parts per million, ppm.
A solution's ppm concentration tells you the number of grams of solute present for every
#10^6 = 1,000,000#
grams of solution. Now, notice that you can approximate the mass of the solution, which is equal to
#""mass solution"" = ""490 g"" + ""0.0085 g"" #
#""mass solution = 490.0085 g""#
to be equal to the mass of water, the solvent.
#""490.0085 g "" ~~ "" 490 g""#
You can thus say that, in your case, a
Use the known composition of the solution as a conversion factor to calculate the number of grams of calcium cations present in
#10^6 color(red)(cancel(color(black)(""g solution""))) * (8.5 * 10^(-3)color(white)(.)""g Ca""^(2+))/(490color(red)(cancel(color(black)(""g solution"")))) = ""17.35 g Ca""^(2+)#
Since this represents the number of grams of calcium cations present in
#color(darkgreen)(ul(color(black)(""concentration"" = ""17 ppm Ca""^(2+))))#
The answer is rounded to two sig figs, the number of sig figs you have for your values.
Judging by the values you have, the most helpful concentration that can be calculated here is that in parts per million, ppm.
A solution's ppm concentration tells you the number of grams of solute present for every
#10^6 = 1,000,000#
grams of solution. Now, notice that you can approximate the mass of the solution, which is equal to
#""mass solution"" = ""490 g"" + ""0.0085 g"" #
#""mass solution = 490.0085 g""#
to be equal to the mass of water, the solvent.
#""490.0085 g "" ~~ "" 490 g""#
You can thus say that, in your case, a
Use the known composition of the solution as a conversion factor to calculate the number of grams of calcium cations present in
#10^6 color(red)(cancel(color(black)(""g solution""))) * (8.5 * 10^(-3)color(white)(.)""g Ca""^(2+))/(490color(red)(cancel(color(black)(""g solution"")))) = ""17.35 g Ca""^(2+)#
Since this represents the number of grams of calcium cations present in
#color(darkgreen)(ul(color(black)(""concentration"" = ""17 ppm Ca""^(2+))))#
The answer is rounded to two sig figs, the number of sig figs you have for your values.
M = 0.5 mol/L
WE have the stoichiometric equation:
Which tells us unequivocally that the combustion of 1 mole of propane gives 3 moles of carbon dioxide.
And then we simply access the molar quantity of propane:
And since, per the stoichiometric equation, if
At normal pressure and temperature,
Approx.
WE have the stoichiometric equation:
Which tells us unequivocally that the combustion of 1 mole of propane gives 3 moles of carbon dioxide.
And then we simply access the molar quantity of propane:
And since, per the stoichiometric equation, if
At normal pressure and temperature,
Approx.
WE have the stoichiometric equation:
Which tells us unequivocally that the combustion of 1 mole of propane gives 3 moles of carbon dioxide.
And then we simply access the molar quantity of propane:
And since, per the stoichiometric equation, if
At normal pressure and temperature,
24 litres at RTP (298K) or 22.4 litres at STP (273K)
1) The balanced equation gives us the mole ratio propane:oxygen which is 1:5 - this means that every mole of propane will require 5 moles of oxygen for complete combustion.
2) However we don't have one mole of propane, we have 8.8g which, using moles =mass/molar mass, equals 8.8/44 = 0.2 moles. (RMM of propane = 12x3 + 1x8 = 44, therefore molar mass = 44g/mol).
3) Applying the mole ratio, it follows that 0.2 moles of propane will react with 0.2x5 = 1 mole of oxygen.
4) One mole of any gas (assuming ideal behaviour, which is a perfectly reasonable assumption, especially with a gas like oxygen with small and non-polar molecules) occupies 24 litres at RTP and 22.4 at STP. Presto!
Your strategy here will be to
use ethanol's molar mass to determine how many grams would contain the given number of moles
use the given percent concentration by mass to find the volume of solution that would contain that many grams of ethanol
do a quick search for the density of an ethanol solution that has that percent concentration of ethanol
So, ethanol's molar mass is equal to
#3.150 color(red)(cancel(color(black)(""moles C""_2""H""_6""O""))) * ""46.07 g""/(1color(red)(cancel(color(black)(""mole C""_2""H""_6""O"")))) = ""145.12 g""#
As you know, a solution's percent concentration by mass is defined as the ratio between the mass of the solute and the total mass of the solution, multiplied by
#color(blue)(""% w/w"" = ""mass of solute""/""mass of solution"" xx 100)#
Now, a
#145.12 color(red)(cancel(color(black)(""g C""_2""H""_6""O""))) * ""100 g solution""/(7.700color(red)(cancel(color(black)(""g C""_2""H""_6""O"")))) = ""1884.7 g solution""#
Now, to get the volume of this solution, you need to use its density, which you can find here
http://wissen.science-and-fun.de/chemistry/chemistry/density-tables/ethanol-water-mixtures/
The density of a
#1884.7 color(red)(cancel(color(black)(""g""))) * "" 1mL""/(0.98725 color(red)(cancel(color(black)(""g"")))) = color(green)(""1909 mL"")#
The answer is rounded to four sig figs.
Your strategy here will be to
use ethanol's molar mass to determine how many grams would contain the given number of moles
use the given percent concentration by mass to find the volume of solution that would contain that many grams of ethanol
do a quick search for the density of an ethanol solution that has that percent concentration of ethanol
So, ethanol's molar mass is equal to
#3.150 color(red)(cancel(color(black)(""moles C""_2""H""_6""O""))) * ""46.07 g""/(1color(red)(cancel(color(black)(""mole C""_2""H""_6""O"")))) = ""145.12 g""#
As you know, a solution's percent concentration by mass is defined as the ratio between the mass of the solute and the total mass of the solution, multiplied by
#color(blue)(""% w/w"" = ""mass of solute""/""mass of solution"" xx 100)#
Now, a
#145.12 color(red)(cancel(color(black)(""g C""_2""H""_6""O""))) * ""100 g solution""/(7.700color(red)(cancel(color(black)(""g C""_2""H""_6""O"")))) = ""1884.7 g solution""#
Now, to get the volume of this solution, you need to use its density, which you can find here
http://wissen.science-and-fun.de/chemistry/chemistry/density-tables/ethanol-water-mixtures/
The density of a
#1884.7 color(red)(cancel(color(black)(""g""))) * "" 1mL""/(0.98725 color(red)(cancel(color(black)(""g"")))) = color(green)(""1909 mL"")#
The answer is rounded to four sig figs.
Your strategy here will be to
use ethanol's molar mass to determine how many grams would contain the given number of moles
use the given percent concentration by mass to find the volume of solution that would contain that many grams of ethanol
do a quick search for the density of an ethanol solution that has that percent concentration of ethanol
So, ethanol's molar mass is equal to
#3.150 color(red)(cancel(color(black)(""moles C""_2""H""_6""O""))) * ""46.07 g""/(1color(red)(cancel(color(black)(""mole C""_2""H""_6""O"")))) = ""145.12 g""#
As you know, a solution's percent concentration by mass is defined as the ratio between the mass of the solute and the total mass of the solution, multiplied by
#color(blue)(""% w/w"" = ""mass of solute""/""mass of solution"" xx 100)#
Now, a
#145.12 color(red)(cancel(color(black)(""g C""_2""H""_6""O""))) * ""100 g solution""/(7.700color(red)(cancel(color(black)(""g C""_2""H""_6""O"")))) = ""1884.7 g solution""#
Now, to get the volume of this solution, you need to use its density, which you can find here
http://wissen.science-and-fun.de/chemistry/chemistry/density-tables/ethanol-water-mixtures/
The density of a
#1884.7 color(red)(cancel(color(black)(""g""))) * "" 1mL""/(0.98725 color(red)(cancel(color(black)(""g"")))) = color(green)(""1909 mL"")#
The answer is rounded to four sig figs.
This is a combustion reaction of an alkane with oxygen. It is exothermic (
This is a combustion reaction of an alkane with oxygen. It is exothermic (
This is a combustion reaction of an alkane with oxygen. It is exothermic (
Let me start by writing a simple equation
x stands for amount of oxygen in grams
When you solve this equation,
Now in percent form
Mass percent of oxygen
27.8% oxygen
Let me start by writing a simple equation
x stands for amount of oxygen in grams
When you solve this equation,
Now in percent form
Mass percent of oxygen
27.8% oxygen
Let me start by writing a simple equation
x stands for amount of oxygen in grams
When you solve this equation,
Now in percent form
Mass percent of oxygen
QUICK ANSWER
The temperature of the solution will increase by
Basically, when you increase the quantities of strong acid and strong base that you're mixing, the heat given off and the volume of the resulting solution will increase by the same factor.
#""more acid + base"" => {(""heat given off "" uarr color(blue)("" by a factor f"")), (""total volume of the solution "" uarr color(blue)("" by the same factor f"")) :}#
The reaction will give off more heat, but there will be more of the resulting solution to heat, so you will end up with
#color(darkgreen)(ul(color(black)(DeltaT_""15 mL"" = DeltaT_""5 mL"" = 2^@""C"")))#
DETAILED EXPLANATION
The idea here is that the heat given off by the reaction will depend on how many moles of each reactant you mix, i.e. on the volume of each solution, so right from the start, you should expect to have
#""heat given off for 5 mL "" < "" heat given off for 15 mL""#
because you're mixing a smaller number of moles of the strong acid and of the strong base, provided, of course, that the concentrations of the two reactants do not change.
However, you will also have
#""volume of solution for 5 mL "" < "" volume of solution for 15 mL""#
so look for these two factors to cancel each other out and result in the same
A generic thermochemical equation that can describe this neutralization reaction looks like this
#""HA""_ ((aq)) + ""BOH""_ ((aq)) -> ""H""_ 2""O""_ ((l)) + ""BA""_ ((aq)),"" ""DeltaH_""neut"" = +xcolor(white)(.)""kJ mol""^(-1)#
This tells you that when
So, assuming that the strong acid solution and the strong base solution have concentrations equal to
#((c * 15)/10^3) color(white)(.)""moles HA"" "" ""# and#"" "" ((c * 15)/10^3)color(white)(.)""moles BOH""#
This means that when you mix
#((c * 15)/10^3) color(red)(cancel(color(black)(""moles HA""))) * (x color(white)(.)""kJ"")/(1color(red)(cancel(color(black)(""mole HA"")))) = ((c * 15 * x)/10^3)color(white)(.)""kJ""#
of heat, the equivalent of
Similarly, when you mix
#((3 * 5)/10^3) color(red)(cancel(color(black)(""moles HA""))) * (x color(white)(.)""kJ"")/(1color(red)(cancel(color(black)(""mole HA"")))) = ( (c * 5 * x)/10^3)color(white)(.)""kJ""#
of heat, the equivalent of
So, just as you would expect, the reaction gives of more heat when you're mixing a bigger number of moles of each reactant.
Now, the heat given off by the neutralization reaction will be absorbed by the solution, which is why its temperature increases.
More specifically, you know that
#color(blue)(ul(color(black)(q_""absorbed"" = m_""sol"" * c_""sol"" * DeltaT)))#
Here
#q_""absorbed""# is the heat absorbed by the solution#m_""sol""# is the mass of the solution#c_""sol""# is the specific heat of the solution#DeltaT# is the change in temperature
When you mix
#V_""total"" = ""15 mL"" + ""15 mL""#
#V_""total"" = ""30 mL""#
If you take
#30 color(red)(cancel(color(black)(""mL solution""))) * (rho color(white)(.)""g"")/(1color(red)(cancel(color(black)(""mL solution"")))) = (30 * rho)color(white)(.)""g""#
This means that you have
#(c * 15 * x) color(red)(cancel(color(black)(""J""))) = (30 * rho) color(red)(cancel(color(black)(""g""))) * c_""sol"" color(red)(cancel(color(black)(""J""))) color(red)(cancel(color(black)(""g"")))^(-1) """"^@""C""^(-1) * DeltaT_""15 mL""#
which gets you
#DeltaT_""15 mL"" = ((c * 15 * x)/(30 * rho * c_""sol"")) """"^@""C""#
#color(blue)(ul(color(black)(DeltaT_""15 mL"" = (1/2 * (c * x)/(rho * c_""sol""))""""^@""C"")))#
Similarly, you can say that when you mix
#V_""total"" = ""5 mL"" + ""5 mL"" = ""10 mL""#
This time, you will have
#(c * 5 * x) color(red)(cancel(color(black)(""J""))) = (10 * rho) color(red)(cancel(color(black)(""g""))) * c_""sol"" color(red)(cancel(color(black)(""J""))) color(red)(cancel(color(black)(""g"")))^(-1) """"^@""C""^(-1) * DeltaT_""5 mL""#
which gets you
#DeltaT_""5 mL"" = ((c * 5 * x)/(10 * rho * c_""sol""))""""^@""C""#
#color(blue)(ul(color(black)(DeltaT_""5 mL"" = (1/2 * (c * x)/(rho * c_""sol""))""""^@""C"")))#
As you can see, you have
#color(darkgreen)(ul(color(black)(DeltaT_""15 mL"" = DeltaT_""5 mL"" = 2^@""C"")))#
The temperature of the solution will increase by
QUICK ANSWER
The temperature of the solution will increase by
Basically, when you increase the quantities of strong acid and strong base that you're mixing, the heat given off and the volume of the resulting solution will increase by the same factor.
#""more acid + base"" => {(""heat given off "" uarr color(blue)("" by a factor f"")), (""total volume of the solution "" uarr color(blue)("" by the same factor f"")) :}#
The reaction will give off more heat, but there will be more of the resulting solution to heat, so you will end up with
#color(darkgreen)(ul(color(black)(DeltaT_""15 mL"" = DeltaT_""5 mL"" = 2^@""C"")))#
DETAILED EXPLANATION
The idea here is that the heat given off by the reaction will depend on how many moles of each reactant you mix, i.e. on the volume of each solution, so right from the start, you should expect to have
#""heat given off for 5 mL "" < "" heat given off for 15 mL""#
because you're mixing a smaller number of moles of the strong acid and of the strong base, provided, of course, that the concentrations of the two reactants do not change.
However, you will also have
#""volume of solution for 5 mL "" < "" volume of solution for 15 mL""#
so look for these two factors to cancel each other out and result in the same
A generic thermochemical equation that can describe this neutralization reaction looks like this
#""HA""_ ((aq)) + ""BOH""_ ((aq)) -> ""H""_ 2""O""_ ((l)) + ""BA""_ ((aq)),"" ""DeltaH_""neut"" = +xcolor(white)(.)""kJ mol""^(-1)#
This tells you that when
So, assuming that the strong acid solution and the strong base solution have concentrations equal to
#((c * 15)/10^3) color(white)(.)""moles HA"" "" ""# and#"" "" ((c * 15)/10^3)color(white)(.)""moles BOH""#
This means that when you mix
#((c * 15)/10^3) color(red)(cancel(color(black)(""moles HA""))) * (x color(white)(.)""kJ"")/(1color(red)(cancel(color(black)(""mole HA"")))) = ((c * 15 * x)/10^3)color(white)(.)""kJ""#
of heat, the equivalent of
Similarly, when you mix
#((3 * 5)/10^3) color(red)(cancel(color(black)(""moles HA""))) * (x color(white)(.)""kJ"")/(1color(red)(cancel(color(black)(""mole HA"")))) = ( (c * 5 * x)/10^3)color(white)(.)""kJ""#
of heat, the equivalent of
So, just as you would expect, the reaction gives of more heat when you're mixing a bigger number of moles of each reactant.
Now, the heat given off by the neutralization reaction will be absorbed by the solution, which is why its temperature increases.
More specifically, you know that
#color(blue)(ul(color(black)(q_""absorbed"" = m_""sol"" * c_""sol"" * DeltaT)))#
Here
#q_""absorbed""# is the heat absorbed by the solution#m_""sol""# is the mass of the solution#c_""sol""# is the specific heat of the solution#DeltaT# is the change in temperature
When you mix
#V_""total"" = ""15 mL"" + ""15 mL""#
#V_""total"" = ""30 mL""#
If you take
#30 color(red)(cancel(color(black)(""mL solution""))) * (rho color(white)(.)""g"")/(1color(red)(cancel(color(black)(""mL solution"")))) = (30 * rho)color(white)(.)""g""#
This means that you have
#(c * 15 * x) color(red)(cancel(color(black)(""J""))) = (30 * rho) color(red)(cancel(color(black)(""g""))) * c_""sol"" color(red)(cancel(color(black)(""J""))) color(red)(cancel(color(black)(""g"")))^(-1) """"^@""C""^(-1) * DeltaT_""15 mL""#
which gets you
#DeltaT_""15 mL"" = ((c * 15 * x)/(30 * rho * c_""sol"")) """"^@""C""#
#color(blue)(ul(color(black)(DeltaT_""15 mL"" = (1/2 * (c * x)/(rho * c_""sol""))""""^@""C"")))#
Similarly, you can say that when you mix
#V_""total"" = ""5 mL"" + ""5 mL"" = ""10 mL""#
This time, you will have
#(c * 5 * x) color(red)(cancel(color(black)(""J""))) = (10 * rho) color(red)(cancel(color(black)(""g""))) * c_""sol"" color(red)(cancel(color(black)(""J""))) color(red)(cancel(color(black)(""g"")))^(-1) """"^@""C""^(-1) * DeltaT_""5 mL""#
which gets you
#DeltaT_""5 mL"" = ((c * 5 * x)/(10 * rho * c_""sol""))""""^@""C""#
#color(blue)(ul(color(black)(DeltaT_""5 mL"" = (1/2 * (c * x)/(rho * c_""sol""))""""^@""C"")))#
As you can see, you have
#color(darkgreen)(ul(color(black)(DeltaT_""15 mL"" = DeltaT_""5 mL"" = 2^@""C"")))#
The temperature of the solution will increase by
QUICK ANSWER
The temperature of the solution will increase by
Basically, when you increase the quantities of strong acid and strong base that you're mixing, the heat given off and the volume of the resulting solution will increase by the same factor.
#""more acid + base"" => {(""heat given off "" uarr color(blue)("" by a factor f"")), (""total volume of the solution "" uarr color(blue)("" by the same factor f"")) :}#
The reaction will give off more heat, but there will be more of the resulting solution to heat, so you will end up with
#color(darkgreen)(ul(color(black)(DeltaT_""15 mL"" = DeltaT_""5 mL"" = 2^@""C"")))#
DETAILED EXPLANATION
The idea here is that the heat given off by the reaction will depend on how many moles of each reactant you mix, i.e. on the volume of each solution, so right from the start, you should expect to have
#""heat given off for 5 mL "" < "" heat given off for 15 mL""#
because you're mixing a smaller number of moles of the strong acid and of the strong base, provided, of course, that the concentrations of the two reactants do not change.
However, you will also have
#""volume of solution for 5 mL "" < "" volume of solution for 15 mL""#
so look for these two factors to cancel each other out and result in the same
A generic thermochemical equation that can describe this neutralization reaction looks like this
#""HA""_ ((aq)) + ""BOH""_ ((aq)) -> ""H""_ 2""O""_ ((l)) + ""BA""_ ((aq)),"" ""DeltaH_""neut"" = +xcolor(white)(.)""kJ mol""^(-1)#
This tells you that when
So, assuming that the strong acid solution and the strong base solution have concentrations equal to
#((c * 15)/10^3) color(white)(.)""moles HA"" "" ""# and#"" "" ((c * 15)/10^3)color(white)(.)""moles BOH""#
This means that when you mix
#((c * 15)/10^3) color(red)(cancel(color(black)(""moles HA""))) * (x color(white)(.)""kJ"")/(1color(red)(cancel(color(black)(""mole HA"")))) = ((c * 15 * x)/10^3)color(white)(.)""kJ""#
of heat, the equivalent of
Similarly, when you mix
#((3 * 5)/10^3) color(red)(cancel(color(black)(""moles HA""))) * (x color(white)(.)""kJ"")/(1color(red)(cancel(color(black)(""mole HA"")))) = ( (c * 5 * x)/10^3)color(white)(.)""kJ""#
of heat, the equivalent of
So, just as you would expect, the reaction gives of more heat when you're mixing a bigger number of moles of each reactant.
Now, the heat given off by the neutralization reaction will be absorbed by the solution, which is why its temperature increases.
More specifically, you know that
#color(blue)(ul(color(black)(q_""absorbed"" = m_""sol"" * c_""sol"" * DeltaT)))#
Here
#q_""absorbed""# is the heat absorbed by the solution#m_""sol""# is the mass of the solution#c_""sol""# is the specific heat of the solution#DeltaT# is the change in temperature
When you mix
#V_""total"" = ""15 mL"" + ""15 mL""#
#V_""total"" = ""30 mL""#
If you take
#30 color(red)(cancel(color(black)(""mL solution""))) * (rho color(white)(.)""g"")/(1color(red)(cancel(color(black)(""mL solution"")))) = (30 * rho)color(white)(.)""g""#
This means that you have
#(c * 15 * x) color(red)(cancel(color(black)(""J""))) = (30 * rho) color(red)(cancel(color(black)(""g""))) * c_""sol"" color(red)(cancel(color(black)(""J""))) color(red)(cancel(color(black)(""g"")))^(-1) """"^@""C""^(-1) * DeltaT_""15 mL""#
which gets you
#DeltaT_""15 mL"" = ((c * 15 * x)/(30 * rho * c_""sol"")) """"^@""C""#
#color(blue)(ul(color(black)(DeltaT_""15 mL"" = (1/2 * (c * x)/(rho * c_""sol""))""""^@""C"")))#
Similarly, you can say that when you mix
#V_""total"" = ""5 mL"" + ""5 mL"" = ""10 mL""#
This time, you will have
#(c * 5 * x) color(red)(cancel(color(black)(""J""))) = (10 * rho) color(red)(cancel(color(black)(""g""))) * c_""sol"" color(red)(cancel(color(black)(""J""))) color(red)(cancel(color(black)(""g"")))^(-1) """"^@""C""^(-1) * DeltaT_""5 mL""#
which gets you
#DeltaT_""5 mL"" = ((c * 5 * x)/(10 * rho * c_""sol""))""""^@""C""#
#color(blue)(ul(color(black)(DeltaT_""5 mL"" = (1/2 * (c * x)/(rho * c_""sol""))""""^@""C"")))#
As you can see, you have
#color(darkgreen)(ul(color(black)(DeltaT_""15 mL"" = DeltaT_""5 mL"" = 2^@""C"")))#
The alkali neutralises the acid:
The moles reacting (n) of each species is given by:
This shows that they are added in the same mole ratio as defined by the equation.
We can, therefore, say that the no. moles of
A salt formed from a weak acid and a strong base is slightly alkaline due to hydrolysis:
From an ICE table we can use this expression which applies to a weak base:
Where
The total volume is now 200 ml + 100 ml = 300 ml = 0.3 L
You need to look up the
To get
Putting in the numbers:
As expected the solution is slightly alkaline, even though the acid and base have been added in the same molar ratio.
The alkali neutralises the acid:
The moles reacting (n) of each species is given by:
This shows that they are added in the same mole ratio as defined by the equation.
We can, therefore, say that the no. moles of
A salt formed from a weak acid and a strong base is slightly alkaline due to hydrolysis:
From an ICE table we can use this expression which applies to a weak base:
Where
The total volume is now 200 ml + 100 ml = 300 ml = 0.3 L
You need to look up the
To get
Putting in the numbers:
As expected the solution is slightly alkaline, even though the acid and base have been added in the same molar ratio.
The alkali neutralises the acid:
The moles reacting (n) of each species is given by:
This shows that they are added in the same mole ratio as defined by the equation.
We can, therefore, say that the no. moles of
A salt formed from a weak acid and a strong base is slightly alkaline due to hydrolysis:
From an ICE table we can use this expression which applies to a weak base:
Where
The total volume is now 200 ml + 100 ml = 300 ml = 0.3 L
You need to look up the
To get
Putting in the numbers:
As expected the solution is slightly alkaline, even though the acid and base have been added in the same molar ratio.
This is really a two-part question:
What's in the mixture?
I like to use an ICE table to keep track of the stoichiometry calculations.
We have 300 mL of a solution that contains 0.020 mol of
2. What's the
The solution will be basic.
Now, we set up another ICE table (with molarities) to calculate the
Check if
At STP of 273.15 K
In order to answer this question, divide the given volume by the molar volume of 27.710980 L/mol.
https://en.m.wikipedia.org/wiki/Molar_volume
Just in case your teacher uses the older values for STP, 273.15 K and 1 atm. The molar volume of a gas at this STP is 22.414 mol/L.
https://en.m.wikipedia.org/wiki/Molar_volume
Just as with the previous calculation, divide the given volume by the molar volume of 22.414 mol/L.
At STP of 273.15 K and 100 kPa, 63.3 L of an ideal gas would contain 2.33 moles.
At STP of 273.15 K and 1 atm, 63.3 L of an ideal gas would contain 2.82 moles.
At STP of 273.15 K
In order to answer this question, divide the given volume by the molar volume of 27.710980 L/mol.
https://en.m.wikipedia.org/wiki/Molar_volume
Just in case your teacher uses the older values for STP, 273.15 K and 1 atm. The molar volume of a gas at this STP is 22.414 mol/L.
https://en.m.wikipedia.org/wiki/Molar_volume
Just as with the previous calculation, divide the given volume by the molar volume of 22.414 mol/L.
At STP of 273.15 K and 100 kPa, 63.3 L of an ideal gas would contain 2.33 moles.
At STP of 273.15 K and 1 atm, 63.3 L of an ideal gas would contain 2.82 moles.
At STP of 273.15 K
In order to answer this question, divide the given volume by the molar volume of 27.710980 L/mol.
https://en.m.wikipedia.org/wiki/Molar_volume
Just in case your teacher uses the older values for STP, 273.15 K and 1 atm. The molar volume of a gas at this STP is 22.414 mol/L.
https://en.m.wikipedia.org/wiki/Molar_volume
Just as with the previous calculation, divide the given volume by the molar volume of 22.414 mol/L.
And (ii) we need equivalent quantities of dioxygen and pentane. We know that
Clearly, there is INSUFFICIENT dioxygen gas for complete combustion. And thus
By the way the question proposed that oxygen and pentane were mixed together.........this is not something that I would be happy doing, and I would run a long way away if I saw someone doing this....
We need (i) a stoichiometric equation:
And (ii) we need equivalent quantities of dioxygen and pentane. We know that
Clearly, there is INSUFFICIENT dioxygen gas for complete combustion. And thus
By the way the question proposed that oxygen and pentane were mixed together.........this is not something that I would be happy doing, and I would run a long way away if I saw someone doing this....
We need (i) a stoichiometric equation:
And (ii) we need equivalent quantities of dioxygen and pentane. We know that
Clearly, there is INSUFFICIENT dioxygen gas for complete combustion. And thus
By the way the question proposed that oxygen and pentane were mixed together.........this is not something that I would be happy doing, and I would run a long way away if I saw someone doing this....
The limiting reactant is oxygen.
Here's another way to identify the limiting reactant.
We know that we will need a balanced equation with masses, moles, and molar masses of the compounds involved.
1. Gather all the information in one place with molar masses above the formulas and everything else below the formulas.
Note: We do not have to stick with gram-moles. We can use pound-moles because the numbers will still be in the same ratio.
2. Identify the limiting reactant
An easy way to identify the limiting reactant is to calculate the ""moles of reaction"" each will give:
You divide the moles of each reactant by its corresponding coefficient in the balanced equation.
I did that for you in the table above.
Write the reaction which has a heat of reaction equal to heat of formation for HCl(g) ?
A compound's standard heat of formation, or standard enthalpy of formation,
In your case, the chemical reaction that has a change in enthalpy equal to
Now, one molecule of hydrogen chloride contains one atom of hydrogen,
The key now is to remember that hydrogen and chlorine exist as diatomic elements in their standard state.
This means that you can write
#""H""_ (2(g)) + ""Cl""_ (2(g)) -> 2""HCl""_((g))#
This represents one form of the balanced chemical equation that describes the synthesis of hydrogen chloride.
In order for the enthalpy change of reaction to be equal to that of the standard enthalpy change of formation, you must divide everything by
#color(green)(|bar(ul(color(white)(a/a)color(black)(1/2""H""_ (2(g)) + 1/2""Cl""_ (2(g)) -> ""HCl""_((g)))color(white)(a/a)|)))#
This equation will have
#DeltaH_""rxn""^@ = DeltaH_f^@#
because it describes the formation of one mole of hydrogen chloride from its constituent elements in their standard state.
A compound's standard heat of formation, or standard enthalpy of formation,
In your case, the chemical reaction that has a change in enthalpy equal to
Now, one molecule of hydrogen chloride contains one atom of hydrogen,
The key now is to remember that hydrogen and chlorine exist as diatomic elements in their standard state.
This means that you can write
#""H""_ (2(g)) + ""Cl""_ (2(g)) -> 2""HCl""_((g))#
This represents one form of the balanced chemical equation that describes the synthesis of hydrogen chloride.
In order for the enthalpy change of reaction to be equal to that of the standard enthalpy change of formation, you must divide everything by
#color(green)(|bar(ul(color(white)(a/a)color(black)(1/2""H""_ (2(g)) + 1/2""Cl""_ (2(g)) -> ""HCl""_((g)))color(white)(a/a)|)))#
This equation will have
#DeltaH_""rxn""^@ = DeltaH_f^@#
because it describes the formation of one mole of hydrogen chloride from its constituent elements in their standard state.
Write the reaction which has a heat of reaction equal to heat of formation for HCl(g) ?
A compound's standard heat of formation, or standard enthalpy of formation,
In your case, the chemical reaction that has a change in enthalpy equal to
Now, one molecule of hydrogen chloride contains one atom of hydrogen,
The key now is to remember that hydrogen and chlorine exist as diatomic elements in their standard state.
This means that you can write
#""H""_ (2(g)) + ""Cl""_ (2(g)) -> 2""HCl""_((g))#
This represents one form of the balanced chemical equation that describes the synthesis of hydrogen chloride.
In order for the enthalpy change of reaction to be equal to that of the standard enthalpy change of formation, you must divide everything by
#color(green)(|bar(ul(color(white)(a/a)color(black)(1/2""H""_ (2(g)) + 1/2""Cl""_ (2(g)) -> ""HCl""_((g)))color(white)(a/a)|)))#
This equation will have
#DeltaH_""rxn""^@ = DeltaH_f^@#
because it describes the formation of one mole of hydrogen chloride from its constituent elements in their standard state.
And if there are
And
And if there are
And
And if there are
And
The important thing to remember about gases that are being collected over water is that the total pressure measured includes the vapor pressure of water at that given temperature.
In this case, the collection tube will hold a gaseous mixture that contains molecules of carbon dioxide,
This means that you can use Dalton's Law of Partial Pressures to determine the partial pressure of carbon dioxide in this mixture.
#color(blue)(P_""total"" = sum_i P_i)"" ""# , where
In this case, the vapor pressure of water at
http://www.endmemo.com/chem/vaporpressurewater.php
This means that you have
#P_""total"" = P_(CO_2) + P_(H_2O)#
#P_(CO_2) = ""101.3 kPa"" - ""31.09 kPa"" = ""70.21 kPa""#
Now, STP conditions imply a temperature of
#color(blue)((P_1V_1)/T_1 = (P_2V_2)/T_2)"" ""# , where
Plug in your values and solve for
#V_2 = P_1/P_2 * T_2/T_1 * V_1#
#V_2 = (70.21 color(red)(cancel(color(black)(""kPa""))))/(100.0color(red)(cancel(color(black)(""kPa"")))) * ( (273.15 + 0)color(red)(cancel(color(black)(""K""))))/((273.15 + 70.0)color(red)(cancel(color(black)(""K"")))) * ""60.0 mL""#
#V_2 = color(green)(""33.5 mL"")#
The answer is rounded to three sig figs.
The important thing to remember about gases that are being collected over water is that the total pressure measured includes the vapor pressure of water at that given temperature.
In this case, the collection tube will hold a gaseous mixture that contains molecules of carbon dioxide,
This means that you can use Dalton's Law of Partial Pressures to determine the partial pressure of carbon dioxide in this mixture.
#color(blue)(P_""total"" = sum_i P_i)"" ""# , where
In this case, the vapor pressure of water at
http://www.endmemo.com/chem/vaporpressurewater.php
This means that you have
#P_""total"" = P_(CO_2) + P_(H_2O)#
#P_(CO_2) = ""101.3 kPa"" - ""31.09 kPa"" = ""70.21 kPa""#
Now, STP conditions imply a temperature of
#color(blue)((P_1V_1)/T_1 = (P_2V_2)/T_2)"" ""# , where
Plug in your values and solve for
#V_2 = P_1/P_2 * T_2/T_1 * V_1#
#V_2 = (70.21 color(red)(cancel(color(black)(""kPa""))))/(100.0color(red)(cancel(color(black)(""kPa"")))) * ( (273.15 + 0)color(red)(cancel(color(black)(""K""))))/((273.15 + 70.0)color(red)(cancel(color(black)(""K"")))) * ""60.0 mL""#
#V_2 = color(green)(""33.5 mL"")#
The answer is rounded to three sig figs.
The important thing to remember about gases that are being collected over water is that the total pressure measured includes the vapor pressure of water at that given temperature.
In this case, the collection tube will hold a gaseous mixture that contains molecules of carbon dioxide,
This means that you can use Dalton's Law of Partial Pressures to determine the partial pressure of carbon dioxide in this mixture.
#color(blue)(P_""total"" = sum_i P_i)"" ""# , where
In this case, the vapor pressure of water at
http://www.endmemo.com/chem/vaporpressurewater.php
This means that you have
#P_""total"" = P_(CO_2) + P_(H_2O)#
#P_(CO_2) = ""101.3 kPa"" - ""31.09 kPa"" = ""70.21 kPa""#
Now, STP conditions imply a temperature of
#color(blue)((P_1V_1)/T_1 = (P_2V_2)/T_2)"" ""# , where
Plug in your values and solve for
#V_2 = P_1/P_2 * T_2/T_1 * V_1#
#V_2 = (70.21 color(red)(cancel(color(black)(""kPa""))))/(100.0color(red)(cancel(color(black)(""kPa"")))) * ( (273.15 + 0)color(red)(cancel(color(black)(""K""))))/((273.15 + 70.0)color(red)(cancel(color(black)(""K"")))) * ""60.0 mL""#
#V_2 = color(green)(""33.5 mL"")#
The answer is rounded to three sig figs.
First we need a chemical equation:
Given the equation, the
The equivalent weight of the metal is
First we need a chemical equation:
Given the equation, the
The equivalent weight of the metal is
First we need a chemical equation:
Given the equation, the
Equivalent weight of metal is 20 g/eq
For HCl
Molarity = Normality
Gram equivalents of HCl = Normality × Volume in litres
For complete neutralisation
Gram equivalents of HCl = Gram equivalents of metal carbonate
Equivalent weight of metal carbonate
Equivalent weight of carbonate
Equivalent weight of metal carbonate = Equivalent weight of metal + Equivalent weight of carbonate
50 g/eq = x + 30 g/eq
x = 20 g/eq
∴ Equivalent weight of metal is 20 g/eq
The number of moles of
So the mass of
Now we know from the law of equivalent proportion that the ratio of masses of two reacting substances for neutralization is equal to the ratio of their equivalent masses.
Hence we can say
Please note
Equivalent mass of
The vapor pressure for a solution that contains a non-volatile solute will depend on the vapor pressure of the pure solvent at that temperature and on the mole faction of the solvent.
#color(blue)(P_""sol"" = chi_""solvent"" * P_""solvent""^@)"" ""# , where
This means that in order to be able to calculate the mole fraction of sodium chloride, you need to know what the vapor pressure of pure water is at
You can use an online calculator to find that the vapor pressure of pure water at
http://www.endmemo.com/chem/vaporpressurewater.php
So, plug in your values into the above equation and solve for
#chi_""water"" = P_""sol""/P_""water""^@#
#chi_""water"" = (19.6color(red)(cancel(color(black)(""torr""))))/(23.7color(red)(cancel(color(black)(""torr"")))) = 0.827#
Since the solution only contains water and sodium chloride, you can say that
#chi_""water"" + chi_""NaCl"" = 1#
This means that the mole fraction of sodium chloride is
#chi_""NaCl"" = 1 - chi_""water""#
#chi_""NaCl"" = 1 - 0.827 = color(green)(0.173)#
The vapor pressure for a solution that contains a non-volatile solute will depend on the vapor pressure of the pure solvent at that temperature and on the mole faction of the solvent.
#color(blue)(P_""sol"" = chi_""solvent"" * P_""solvent""^@)"" ""# , where
This means that in order to be able to calculate the mole fraction of sodium chloride, you need to know what the vapor pressure of pure water is at
You can use an online calculator to find that the vapor pressure of pure water at
http://www.endmemo.com/chem/vaporpressurewater.php
So, plug in your values into the above equation and solve for
#chi_""water"" = P_""sol""/P_""water""^@#
#chi_""water"" = (19.6color(red)(cancel(color(black)(""torr""))))/(23.7color(red)(cancel(color(black)(""torr"")))) = 0.827#
Since the solution only contains water and sodium chloride, you can say that
#chi_""water"" + chi_""NaCl"" = 1#
This means that the mole fraction of sodium chloride is
#chi_""NaCl"" = 1 - chi_""water""#
#chi_""NaCl"" = 1 - 0.827 = color(green)(0.173)#
The vapor pressure for a solution that contains a non-volatile solute will depend on the vapor pressure of the pure solvent at that temperature and on the mole faction of the solvent.
#color(blue)(P_""sol"" = chi_""solvent"" * P_""solvent""^@)"" ""# , where
This means that in order to be able to calculate the mole fraction of sodium chloride, you need to know what the vapor pressure of pure water is at
You can use an online calculator to find that the vapor pressure of pure water at
http://www.endmemo.com/chem/vaporpressurewater.php
So, plug in your values into the above equation and solve for
#chi_""water"" = P_""sol""/P_""water""^@#
#chi_""water"" = (19.6color(red)(cancel(color(black)(""torr""))))/(23.7color(red)(cancel(color(black)(""torr"")))) = 0.827#
Since the solution only contains water and sodium chloride, you can say that
#chi_""water"" + chi_""NaCl"" = 1#
This means that the mole fraction of sodium chloride is
#chi_""NaCl"" = 1 - chi_""water""#
#chi_""NaCl"" = 1 - 0.827 = color(green)(0.173)#
And thus the mass of
Well I know that
And thus the mass of
Well I know that
And thus the mass of
1 molar
The key to this problem is the balanced chemical equation for this neutralization reaction.
Acetic acid,
#""CH""_3""COOH""_text((aq]) + ""OH""_text((aq])^(-) -> ""CH""_3""COO""_text((aq])^(-) + ""H""_2""O""_text((l])#
As you can see, the acid and the base react in a
The problem provides you with the molarity and volume of the sodium hydroxide solution, which means that you can use the definition of molarity to find how many moles of hydroxide anions were needed to completely neutralize the acid.
#color(blue)(|bar(ul(color(white)(a/a)c = n_""solute""/V_""solution"" implies n_""solute"" = c * V_""solution""color(white)(a/a)|)))#
In this case, you will have
#n_(OH^(-)) = ""0.232 mol"" color(red)(cancel(color(black)(""L""^(-1)))) * 53.2 * 10^(-3)color(red)(cancel(color(black)(""L""))) = ""0.01234 moles OH""^(-)#
This means that the acid solution must have contained
The molarity of the acetic acid solution was
#[""CH""_3""COOH""] = ""0.01234 moles""/(25.0 * 10^(-3)""L"") = color(green)(|bar(ul(color(white)(a/a)""0.494 mol L""^(-1)color(white)(a/a)|)))#
The answer is rounded to three sig figs.
It's worth noting that the pH of the solution at equivalence point will not be equal to
Instead, the pH of the solution at equivalence point will be greater than
The key to this problem is the balanced chemical equation for this neutralization reaction.
Acetic acid,
#""CH""_3""COOH""_text((aq]) + ""OH""_text((aq])^(-) -> ""CH""_3""COO""_text((aq])^(-) + ""H""_2""O""_text((l])#
As you can see, the acid and the base react in a
The problem provides you with the molarity and volume of the sodium hydroxide solution, which means that you can use the definition of molarity to find how many moles of hydroxide anions were needed to completely neutralize the acid.
#color(blue)(|bar(ul(color(white)(a/a)c = n_""solute""/V_""solution"" implies n_""solute"" = c * V_""solution""color(white)(a/a)|)))#
In this case, you will have
#n_(OH^(-)) = ""0.232 mol"" color(red)(cancel(color(black)(""L""^(-1)))) * 53.2 * 10^(-3)color(red)(cancel(color(black)(""L""))) = ""0.01234 moles OH""^(-)#
This means that the acid solution must have contained
The molarity of the acetic acid solution was
#[""CH""_3""COOH""] = ""0.01234 moles""/(25.0 * 10^(-3)""L"") = color(green)(|bar(ul(color(white)(a/a)""0.494 mol L""^(-1)color(white)(a/a)|)))#
The answer is rounded to three sig figs.
It's worth noting that the pH of the solution at equivalence point will not be equal to
Instead, the pH of the solution at equivalence point will be greater than
The key to this problem is the balanced chemical equation for this neutralization reaction.
Acetic acid,
#""CH""_3""COOH""_text((aq]) + ""OH""_text((aq])^(-) -> ""CH""_3""COO""_text((aq])^(-) + ""H""_2""O""_text((l])#
As you can see, the acid and the base react in a
The problem provides you with the molarity and volume of the sodium hydroxide solution, which means that you can use the definition of molarity to find how many moles of hydroxide anions were needed to completely neutralize the acid.
#color(blue)(|bar(ul(color(white)(a/a)c = n_""solute""/V_""solution"" implies n_""solute"" = c * V_""solution""color(white)(a/a)|)))#
In this case, you will have
#n_(OH^(-)) = ""0.232 mol"" color(red)(cancel(color(black)(""L""^(-1)))) * 53.2 * 10^(-3)color(red)(cancel(color(black)(""L""))) = ""0.01234 moles OH""^(-)#
This means that the acid solution must have contained
The molarity of the acetic acid solution was
#[""CH""_3""COOH""] = ""0.01234 moles""/(25.0 * 10^(-3)""L"") = color(green)(|bar(ul(color(white)(a/a)""0.494 mol L""^(-1)color(white)(a/a)|)))#
The answer is rounded to three sig figs.
It's worth noting that the pH of the solution at equivalence point will not be equal to
Instead, the pH of the solution at equivalence point will be greater than
The formation of ammonia is probably one of the best studied equilibrium reactions. This is an equilibrium reaction that requires high pressures, and high temperatures to provide acceptable rates. The one thing this reaction has got going for it is that the ammonia product is condensable (perhaps on a cold finger), whereas the reactants are essentially non-condensable.
Well clearly stoichiometry dictates that 2 equiv of ammonia are formed in the stoichiometric reaction. But there is a catch......
The formation of ammonia is probably one of the best studied equilibrium reactions. This is an equilibrium reaction that requires high pressures, and high temperatures to provide acceptable rates. The one thing this reaction has got going for it is that the ammonia product is condensable (perhaps on a cold finger), whereas the reactants are essentially non-condensable.
Well clearly stoichiometry dictates that 2 equiv of ammonia are formed in the stoichiometric reaction. But there is a catch......
The formation of ammonia is probably one of the best studied equilibrium reactions. This is an equilibrium reaction that requires high pressures, and high temperatures to provide acceptable rates. The one thing this reaction has got going for it is that the ammonia product is condensable (perhaps on a cold finger), whereas the reactants are essentially non-condensable.
There is a slight excess of lead nitrate:
Note that if we did the actual experiment, we would probably get
There is a slight excess of lead nitrate:
Note that if we did the actual experiment, we would probably get
There is a slight excess of lead nitrate:
Note that if we did the actual experiment, we would probably get
Your strategy here will be to
- pick a sample of this ionic compound
- use the molar masses of iron and of sulfur to find how many moles of each it contains
- find the smallest whole number ratio that exists between iron and sulfur in the compound
To make the calculations easier, pick a
#""53.73 g Fe""# #""46.27 g S""#
Convert these masses to moles
#""For Fe: "" 53.73 color(red)(cancel(color(black)(""g""))) * ""1 mole Fe""/(55.845color(red)(cancel(color(black)(""g"")))) = 0.962#
#""For S: "" 46.27 color(red)(cancel(color(black)(""g""))) * ""1 mole S""/(32.066color(red)(cancel(color(black)(""g"")))) = 1.44#
Next, divide both values by the smallest one
#""For Fe: "" (0.95 color(red)(cancel(color(black)(""moles""))))/(0.95color(red)(cancel(color(black)(""moles"")))) = 1#
#""For S: "" (1.44 color(red)(cancel(color(black)(""moles""))))/(0.95color(red)(cancel(color(black)(""moles"")))) = 1.52#
Now, to find the smallest whole number ratio that exists between the two elements in the compound, multiply both values by
#""For Fe: "" 2 xx 1 = 2#
#""For S: "" 2 xx 1.52 = 3.04 ~~ 3#
The empirical formula of the unknown compound will thus be
#color(green)(bar(ul(|color(white)(a/a)color(black)(""Fe""_2""S""_3)color(white)(a/a)|)))#
Your strategy here will be to
- pick a sample of this ionic compound
- use the molar masses of iron and of sulfur to find how many moles of each it contains
- find the smallest whole number ratio that exists between iron and sulfur in the compound
To make the calculations easier, pick a
#""53.73 g Fe""# #""46.27 g S""#
Convert these masses to moles
#""For Fe: "" 53.73 color(red)(cancel(color(black)(""g""))) * ""1 mole Fe""/(55.845color(red)(cancel(color(black)(""g"")))) = 0.962#
#""For S: "" 46.27 color(red)(cancel(color(black)(""g""))) * ""1 mole S""/(32.066color(red)(cancel(color(black)(""g"")))) = 1.44#
Next, divide both values by the smallest one
#""For Fe: "" (0.95 color(red)(cancel(color(black)(""moles""))))/(0.95color(red)(cancel(color(black)(""moles"")))) = 1#
#""For S: "" (1.44 color(red)(cancel(color(black)(""moles""))))/(0.95color(red)(cancel(color(black)(""moles"")))) = 1.52#
Now, to find the smallest whole number ratio that exists between the two elements in the compound, multiply both values by
#""For Fe: "" 2 xx 1 = 2#
#""For S: "" 2 xx 1.52 = 3.04 ~~ 3#
The empirical formula of the unknown compound will thus be
#color(green)(bar(ul(|color(white)(a/a)color(black)(""Fe""_2""S""_3)color(white)(a/a)|)))#
Your strategy here will be to
- pick a sample of this ionic compound
- use the molar masses of iron and of sulfur to find how many moles of each it contains
- find the smallest whole number ratio that exists between iron and sulfur in the compound
To make the calculations easier, pick a
#""53.73 g Fe""# #""46.27 g S""#
Convert these masses to moles
#""For Fe: "" 53.73 color(red)(cancel(color(black)(""g""))) * ""1 mole Fe""/(55.845color(red)(cancel(color(black)(""g"")))) = 0.962#
#""For S: "" 46.27 color(red)(cancel(color(black)(""g""))) * ""1 mole S""/(32.066color(red)(cancel(color(black)(""g"")))) = 1.44#
Next, divide both values by the smallest one
#""For Fe: "" (0.95 color(red)(cancel(color(black)(""moles""))))/(0.95color(red)(cancel(color(black)(""moles"")))) = 1#
#""For S: "" (1.44 color(red)(cancel(color(black)(""moles""))))/(0.95color(red)(cancel(color(black)(""moles"")))) = 1.52#
Now, to find the smallest whole number ratio that exists between the two elements in the compound, multiply both values by
#""For Fe: "" 2 xx 1 = 2#
#""For S: "" 2 xx 1.52 = 3.04 ~~ 3#
The empirical formula of the unknown compound will thus be
#color(green)(bar(ul(|color(white)(a/a)color(black)(""Fe""_2""S""_3)color(white)(a/a)|)))#
This is a great example of how a molality practice problem looks like. Before doing any calculations, make sure that you have a clear understanding of what molality means.
As you know, molality is defined as moles of solute, which in your case is sodium chloride, divided by the mass of solvent, always expressed in kilograms!
#color(blue)(""molality"" = ""moles of solute""/""kilograms of solvent"")#
So, a
Now, your solution is said to have a molality of
However, you are told that your solution contains less than
#806 color(red)(cancel(color(black)(""g""))) * "" 1 kg""/(10^3color(red)(cancel(color(black)(""g"")))) = ""0.806 kg""#
of water. This means that you can expect it to contain fewer than
#0.806 color(red)(cancel(color(black)(""kg water""))) * overbrace(""2.48 moles NaCl""/(1color(red)(cancel(color(black)(""kg water"")))))^(color(purple)(""the given molality"")) = ""1.9989 moles NaCl""#
Now, if you need to express this in grams of sodium chloride, simply use the compound's molar mass to go from moles to grams
#1.9989 color(red)(cancel(color(black)(""moles NaCl""))) * ""58.443 g""/(1color(red)(cancel(color(black)(""mole NaCl"")))) = ""116.82 g""#
Rounded to three sig figs, the answer will be
#m_(NaCl) = color(green)(""117 g"")#
This is a great example of how a molality practice problem looks like. Before doing any calculations, make sure that you have a clear understanding of what molality means.
As you know, molality is defined as moles of solute, which in your case is sodium chloride, divided by the mass of solvent, always expressed in kilograms!
#color(blue)(""molality"" = ""moles of solute""/""kilograms of solvent"")#
So, a
Now, your solution is said to have a molality of
However, you are told that your solution contains less than
#806 color(red)(cancel(color(black)(""g""))) * "" 1 kg""/(10^3color(red)(cancel(color(black)(""g"")))) = ""0.806 kg""#
of water. This means that you can expect it to contain fewer than
#0.806 color(red)(cancel(color(black)(""kg water""))) * overbrace(""2.48 moles NaCl""/(1color(red)(cancel(color(black)(""kg water"")))))^(color(purple)(""the given molality"")) = ""1.9989 moles NaCl""#
Now, if you need to express this in grams of sodium chloride, simply use the compound's molar mass to go from moles to grams
#1.9989 color(red)(cancel(color(black)(""moles NaCl""))) * ""58.443 g""/(1color(red)(cancel(color(black)(""mole NaCl"")))) = ""116.82 g""#
Rounded to three sig figs, the answer will be
#m_(NaCl) = color(green)(""117 g"")#
This is a great example of how a molality practice problem looks like. Before doing any calculations, make sure that you have a clear understanding of what molality means.
As you know, molality is defined as moles of solute, which in your case is sodium chloride, divided by the mass of solvent, always expressed in kilograms!
#color(blue)(""molality"" = ""moles of solute""/""kilograms of solvent"")#
So, a
Now, your solution is said to have a molality of
However, you are told that your solution contains less than
#806 color(red)(cancel(color(black)(""g""))) * "" 1 kg""/(10^3color(red)(cancel(color(black)(""g"")))) = ""0.806 kg""#
of water. This means that you can expect it to contain fewer than
#0.806 color(red)(cancel(color(black)(""kg water""))) * overbrace(""2.48 moles NaCl""/(1color(red)(cancel(color(black)(""kg water"")))))^(color(purple)(""the given molality"")) = ""1.9989 moles NaCl""#
Now, if you need to express this in grams of sodium chloride, simply use the compound's molar mass to go from moles to grams
#1.9989 color(red)(cancel(color(black)(""moles NaCl""))) * ""58.443 g""/(1color(red)(cancel(color(black)(""mole NaCl"")))) = ""116.82 g""#
Rounded to three sig figs, the answer will be
#m_(NaCl) = color(green)(""117 g"")#
We need to find the limiting reactant (also called limiting reagent).
Iron(III) Oxide
First divide the given mass of
Next multiply times the mole ratio of
Then multiply times the molar mass of
This will give you the amount of pure
Aluminum
First divide the given mass of
Next multiply times the mole ratio of
Then multiply times the molar mass of
This will give you the amount of pure
Aluminum is the limiting reactant. The maximum amount of pure
Aluminum is the limiting reactant. The maximum amount of pure
We need to find the limiting reactant (also called limiting reagent).
Iron(III) Oxide
First divide the given mass of
Next multiply times the mole ratio of
Then multiply times the molar mass of
This will give you the amount of pure
Aluminum
First divide the given mass of
Next multiply times the mole ratio of
Then multiply times the molar mass of
This will give you the amount of pure
Aluminum is the limiting reactant. The maximum amount of pure
Aluminum is the limiting reactant. The maximum amount of pure
We need to find the limiting reactant (also called limiting reagent).
Iron(III) Oxide
First divide the given mass of
Next multiply times the mole ratio of
Then multiply times the molar mass of
This will give you the amount of pure
Aluminum
First divide the given mass of
Next multiply times the mole ratio of
Then multiply times the molar mass of
This will give you the amount of pure
Aluminum is the limiting reactant. The maximum amount of pure
The first thing to mention here is that water's specific heat at around
http://www.engineeringtoolbox.com/water-thermal-properties-d_162.html
However, since this is the value given to you in the problem, you must use it.
So, the idea here is that the heat gained by the metal will be equal to the heat lost by the water. The final temperature of the water will thus be equal to the final temperature of the ring.
Mathematically, this is written as
#color(blue)(-q_""water"" = q_""metal"")#
Here the negative sign is used because
The equation that establishes a relationship between heat lost/gained and temperature change looks like this
#color(blue)(q = - m * c * DeltaT)"" ""# , where
In your case, you would have
#- overbrace(m_""water"" * c_""water"" * DeltaT_""water"")^(color(blue)(=q_""water"")) = overbrace(m_""metal"" * c_""metal"" * DeltaT_""metal"")^(color(red)(=q_""metal""))#
Rearrange this equation to solve for
#c_""metal"" = -m_""water""/m_""metal"" * (DeltaT_""water"")/(DeltaT_""metal"") * c_""water""#
Plug in your values to get
#c_""metal"" = (15color(red)(cancel(color(black)(""g""))))/(20color(red)(cancel(color(black)(""g"")))) * ((97 - 100)color(red)(cancel(color(black)(""""^@""C""))))/((97 - 25)color(red)(cancel(color(black)(""""^@""C"")))) * 4.184""J""/(""g"" """"^@""C"")#
#c_""metal"" = 0.13075""J""/(""g"" """"^@""C"")#
I'll leave the answer rounded to two sig figs, despite the fact that you only gave one sig fig for the mass of the ring
#c_""metal"" = color(green)(0.13""J""/(""g"" """"^@""C""))#
The first thing to mention here is that water's specific heat at around
http://www.engineeringtoolbox.com/water-thermal-properties-d_162.html
However, since this is the value given to you in the problem, you must use it.
So, the idea here is that the heat gained by the metal will be equal to the heat lost by the water. The final temperature of the water will thus be equal to the final temperature of the ring.
Mathematically, this is written as
#color(blue)(-q_""water"" = q_""metal"")#
Here the negative sign is used because
The equation that establishes a relationship between heat lost/gained and temperature change looks like this
#color(blue)(q = - m * c * DeltaT)"" ""# , where
In your case, you would have
#- overbrace(m_""water"" * c_""water"" * DeltaT_""water"")^(color(blue)(=q_""water"")) = overbrace(m_""metal"" * c_""metal"" * DeltaT_""metal"")^(color(red)(=q_""metal""))#
Rearrange this equation to solve for
#c_""metal"" = -m_""water""/m_""metal"" * (DeltaT_""water"")/(DeltaT_""metal"") * c_""water""#
Plug in your values to get
#c_""metal"" = (15color(red)(cancel(color(black)(""g""))))/(20color(red)(cancel(color(black)(""g"")))) * ((97 - 100)color(red)(cancel(color(black)(""""^@""C""))))/((97 - 25)color(red)(cancel(color(black)(""""^@""C"")))) * 4.184""J""/(""g"" """"^@""C"")#
#c_""metal"" = 0.13075""J""/(""g"" """"^@""C"")#
I'll leave the answer rounded to two sig figs, despite the fact that you only gave one sig fig for the mass of the ring
#c_""metal"" = color(green)(0.13""J""/(""g"" """"^@""C""))#
The first thing to mention here is that water's specific heat at around
http://www.engineeringtoolbox.com/water-thermal-properties-d_162.html
However, since this is the value given to you in the problem, you must use it.
So, the idea here is that the heat gained by the metal will be equal to the heat lost by the water. The final temperature of the water will thus be equal to the final temperature of the ring.
Mathematically, this is written as
#color(blue)(-q_""water"" = q_""metal"")#
Here the negative sign is used because
The equation that establishes a relationship between heat lost/gained and temperature change looks like this
#color(blue)(q = - m * c * DeltaT)"" ""# , where
In your case, you would have
#- overbrace(m_""water"" * c_""water"" * DeltaT_""water"")^(color(blue)(=q_""water"")) = overbrace(m_""metal"" * c_""metal"" * DeltaT_""metal"")^(color(red)(=q_""metal""))#
Rearrange this equation to solve for
#c_""metal"" = -m_""water""/m_""metal"" * (DeltaT_""water"")/(DeltaT_""metal"") * c_""water""#
Plug in your values to get
#c_""metal"" = (15color(red)(cancel(color(black)(""g""))))/(20color(red)(cancel(color(black)(""g"")))) * ((97 - 100)color(red)(cancel(color(black)(""""^@""C""))))/((97 - 25)color(red)(cancel(color(black)(""""^@""C"")))) * 4.184""J""/(""g"" """"^@""C"")#
#c_""metal"" = 0.13075""J""/(""g"" """"^@""C"")#
I'll leave the answer rounded to two sig figs, despite the fact that you only gave one sig fig for the mass of the ring
#c_""metal"" = color(green)(0.13""J""/(""g"" """"^@""C""))#
20.0 g of calcium carbonate is heated to produce carbon dioxide according to
the equation
CaCO3(s) → CaO(s) + CO2(g)
If the reaction is carried out at a temperature of 800.0 degrees C and a constant pressure of
1.0 atm is maintained, what volume of carbon dioxide gas is recovered?
There are
Given the stoichiometry, an equivalent molar quantity of carbon dioxide gas will be evolved.
At
There are
Given the stoichiometry, an equivalent molar quantity of carbon dioxide gas will be evolved.
At
20.0 g of calcium carbonate is heated to produce carbon dioxide according to
the equation
CaCO3(s) → CaO(s) + CO2(g)
If the reaction is carried out at a temperature of 800.0 degrees C and a constant pressure of
1.0 atm is maintained, what volume of carbon dioxide gas is recovered?
There are
Given the stoichiometry, an equivalent molar quantity of carbon dioxide gas will be evolved.
At
The first thing that you need to do here is to pick a sample of this solution and use its molality to determine how many moles of solute it contains.
To make the calculations easier, let's pick a sample that contains exactly
Next, use the molar mass of oxalic acid to calculate how many grams of solute are present in this sample.
#0.585 color(red)(cancel(color(black)(""moles H""_2""C""_2""O""_4))) * ""90.03 g""/(1color(red)(cancel(color(black)(""mole H""_2""C""_2""O""_4)))) = ""52.68 g""#
This means that the total mass of the sample, which includes the mass of the solute and the mass of the solvent, which is equal to
#10^3 quad ""g"" + ""52.68 g"" = ""1052.68 g""#
Now, in order to find the molarity of the solution, you need to find the number of moles of solute present in exactly
Use the density of the solution to find the total volume of the sample
#1052.68 color(red)(cancel(color(black)(""g""))) * ""1 mL""/(1.022color(red)(cancel(color(black)(""g"")))) = ""1030.0 mL""#
Since you know that this sample contains
#10^3 color(red)(cancel(color(black)(""mL solution""))) * (""0.585 moles H""_2""C""_2""O""_4)/(1030.0color(red)(cancel(color(black)(""mL solution"")))) = ""0.568 moles H""_2""C""_2""O""_4#
This means that the solution has a molarity of
#color(darkgreen)(ul(color(black)(""molarity"" = ""0.568 mol L""^(-1))))#
The answer is rounded to three sig figs, the number of sig figs you have for the molality of the solution.
The first thing that you need to do here is to pick a sample of this solution and use its molality to determine how many moles of solute it contains.
To make the calculations easier, let's pick a sample that contains exactly
Next, use the molar mass of oxalic acid to calculate how many grams of solute are present in this sample.
#0.585 color(red)(cancel(color(black)(""moles H""_2""C""_2""O""_4))) * ""90.03 g""/(1color(red)(cancel(color(black)(""mole H""_2""C""_2""O""_4)))) = ""52.68 g""#
This means that the total mass of the sample, which includes the mass of the solute and the mass of the solvent, which is equal to
#10^3 quad ""g"" + ""52.68 g"" = ""1052.68 g""#
Now, in order to find the molarity of the solution, you need to find the number of moles of solute present in exactly
Use the density of the solution to find the total volume of the sample
#1052.68 color(red)(cancel(color(black)(""g""))) * ""1 mL""/(1.022color(red)(cancel(color(black)(""g"")))) = ""1030.0 mL""#
Since you know that this sample contains
#10^3 color(red)(cancel(color(black)(""mL solution""))) * (""0.585 moles H""_2""C""_2""O""_4)/(1030.0color(red)(cancel(color(black)(""mL solution"")))) = ""0.568 moles H""_2""C""_2""O""_4#
This means that the solution has a molarity of
#color(darkgreen)(ul(color(black)(""molarity"" = ""0.568 mol L""^(-1))))#
The answer is rounded to three sig figs, the number of sig figs you have for the molality of the solution.
The first thing that you need to do here is to pick a sample of this solution and use its molality to determine how many moles of solute it contains.
To make the calculations easier, let's pick a sample that contains exactly
Next, use the molar mass of oxalic acid to calculate how many grams of solute are present in this sample.
#0.585 color(red)(cancel(color(black)(""moles H""_2""C""_2""O""_4))) * ""90.03 g""/(1color(red)(cancel(color(black)(""mole H""_2""C""_2""O""_4)))) = ""52.68 g""#
This means that the total mass of the sample, which includes the mass of the solute and the mass of the solvent, which is equal to
#10^3 quad ""g"" + ""52.68 g"" = ""1052.68 g""#
Now, in order to find the molarity of the solution, you need to find the number of moles of solute present in exactly
Use the density of the solution to find the total volume of the sample
#1052.68 color(red)(cancel(color(black)(""g""))) * ""1 mL""/(1.022color(red)(cancel(color(black)(""g"")))) = ""1030.0 mL""#
Since you know that this sample contains
#10^3 color(red)(cancel(color(black)(""mL solution""))) * (""0.585 moles H""_2""C""_2""O""_4)/(1030.0color(red)(cancel(color(black)(""mL solution"")))) = ""0.568 moles H""_2""C""_2""O""_4#
This means that the solution has a molarity of
#color(darkgreen)(ul(color(black)(""molarity"" = ""0.568 mol L""^(-1))))#
The answer is rounded to three sig figs, the number of sig figs you have for the molality of the solution.
The answer can be determined by using Boyle's Law:
Let's identify our known and unknown variables.
Rearrange the equation to solve for the final pressure by dividing both sides by
Plug in your given values to obtain the final pressure:
That sample of gas has a new pressure of
The answer can be determined by using Boyle's Law:
Let's identify our known and unknown variables.
Rearrange the equation to solve for the final pressure by dividing both sides by
Plug in your given values to obtain the final pressure:
That sample of gas has a new pressure of
The answer can be determined by using Boyle's Law:
Let's identify our known and unknown variables.
Rearrange the equation to solve for the final pressure by dividing both sides by
Plug in your given values to obtain the final pressure:
A
And thus we need to find the amount of substance of both
We use the relationship,
Note that I converted the
Clearly, the hydrobromic acid is in excess. And given 1:1 stoichiometry, there are
So
A stoichiometric equation is required:
We get (finally)
And thus we need to find the amount of substance of both
We use the relationship,
Note that I converted the
Clearly, the hydrobromic acid is in excess. And given 1:1 stoichiometry, there are
So
A
A stoichiometric equation is required:
We get (finally)
And thus we need to find the amount of substance of both
We use the relationship,
Note that I converted the
Clearly, the hydrobromic acid is in excess. And given 1:1 stoichiometry, there are
So
So if you have
The molecular mass of octane is
So if you have
The molecular mass of octane is
So if you have
Thus
This site tells me that vapour pressure of water at
Thus
This site tells me that vapour pressure of water at
Thus
1) Use PV = nRT to determine moles of gas present in gas:
(1.10 atm) (2.55 L) = (n) (0.08206) (328 K)
n = 0.104 mol
2) Get molar mass of gas using calculated moles:
0.104 mol x M g/mol = 3.45g
M = 3.45 g / 0.104 mol = 33.17 g/mol
Density = Molar Mass x P / RT
= 33.17 g/mol x 1.10 atm / 0.0821 L mol / atm .K x 328 K
= 1.35 g /L
1.35 g/L
1) Use PV = nRT to determine moles of gas present in gas:
(1.10 atm) (2.55 L) = (n) (0.08206) (328 K)
n = 0.104 mol
2) Get molar mass of gas using calculated moles:
0.104 mol x M g/mol = 3.45g
M = 3.45 g / 0.104 mol = 33.17 g/mol
Density = Molar Mass x P / RT
= 33.17 g/mol x 1.10 atm / 0.0821 L mol / atm .K x 328 K
= 1.35 g /L
1.35 g/L
1) Use PV = nRT to determine moles of gas present in gas:
(1.10 atm) (2.55 L) = (n) (0.08206) (328 K)
n = 0.104 mol
2) Get molar mass of gas using calculated moles:
0.104 mol x M g/mol = 3.45g
M = 3.45 g / 0.104 mol = 33.17 g/mol
Density = Molar Mass x P / RT
= 33.17 g/mol x 1.10 atm / 0.0821 L mol / atm .K x 328 K
= 1.35 g /L
Introduced into a 1.30 −L flask is 0.120 mol of
What is the total pressure of the gases in the flask at this point? [Hint: Use data from Appendix D in the textbook and appropriate relationships from this chapter.]
From what I know, I need to apply
Here's the information I got from the book...
for
for
Yes, you need
Calculate
You can calculate it from the values tabulated at 298 K.
∴ At 227 °C (500 K),
Calculate
Calculate equilibrium concentrations
Finally, we can set up an ICE table to calculate the equilibrium concentrations.
Test for negligibility:
Calculate total pressure
Warning! Long Answer.
Yes, you need
Calculate
You can calculate it from the values tabulated at 298 K.
∴ At 227 °C (500 K),
Calculate
Calculate equilibrium concentrations
Finally, we can set up an ICE table to calculate the equilibrium concentrations.
Test for negligibility:
Calculate total pressure
Introduced into a 1.30 −L flask is 0.120 mol of
What is the total pressure of the gases in the flask at this point? [Hint: Use data from Appendix D in the textbook and appropriate relationships from this chapter.]
From what I know, I need to apply
Here's the information I got from the book...
for
for
Warning! Long Answer.
Yes, you need
Calculate
You can calculate it from the values tabulated at 298 K.
∴ At 227 °C (500 K),
Calculate
Calculate equilibrium concentrations
Finally, we can set up an ICE table to calculate the equilibrium concentrations.
Test for negligibility:
Calculate total pressure
0.28g
0.28g
You are presumed to have
By the stoichiometry of the reaction listed, for each mole of diborane that combusts,
Assuming ideality,
Why did I use
Please note that this answer rests on the assumption that you mean diborane. There are many other borohydrides.
I assume that you mean diborane,
You are presumed to have
By the stoichiometry of the reaction listed, for each mole of diborane that combusts,
Assuming ideality,
Why did I use
Please note that this answer rests on the assumption that you mean diborane. There are many other borohydrides.
I assume that you mean diborane,
You are presumed to have
By the stoichiometry of the reaction listed, for each mole of diborane that combusts,
Assuming ideality,
Why did I use
Please note that this answer rests on the assumption that you mean diborane. There are many other borohydrides.
Because bromine exists in the form of a diatomic molecule, the equation must be based on one molecule of
Because bromine exists in the form of a diatomic molecule, the equation must be based on one molecule of
Because bromine exists in the form of a diatomic molecule, the equation must be based on one molecule of
This is not a reaction I would care to do. Bromine is one of the most corrosive substances you use in a lab, and demands a great deal of respect.
Is this a redox equation? How do you know?
So we know that
Thus
There are
So we know that
Thus
There are
So we know that
Thus
This is a limiting reactant problem.
We know that we will need a balanced equation and moles of each reactant.
1. Gather all the information in one place with the experimental number of moles below the formulas.
2. Identify the limiting reactant
An easy way to identify the limiting reactant is to calculate the ""moles of reaction"" each will give:
You divide the moles of each reactant by its corresponding coefficient in the balanced equation.
I did that for you in the table above.
3. Calculate the theoretical yield of
4. Calculate the percent yield
The percent yield is 50 %.
This is a limiting reactant problem.
We know that we will need a balanced equation and moles of each reactant.
1. Gather all the information in one place with the experimental number of moles below the formulas.
2. Identify the limiting reactant
An easy way to identify the limiting reactant is to calculate the ""moles of reaction"" each will give:
You divide the moles of each reactant by its corresponding coefficient in the balanced equation.
I did that for you in the table above.
3. Calculate the theoretical yield of
4. Calculate the percent yield
The percent yield is 50 %.
This is a limiting reactant problem.
We know that we will need a balanced equation and moles of each reactant.
1. Gather all the information in one place with the experimental number of moles below the formulas.
2. Identify the limiting reactant
An easy way to identify the limiting reactant is to calculate the ""moles of reaction"" each will give:
You divide the moles of each reactant by its corresponding coefficient in the balanced equation.
I did that for you in the table above.
3. Calculate the theoretical yield of
4. Calculate the percent yield
Calcium is a metal, so its formula will simply be
Nitrogen is a diatomic molecular compound, making it
Since calcium nitride is an ionic compound, by evaluating its constituent ions we can determine its formula. The calcium ion is a 2+ ion, or
Balancing the equation, then, you will notice that the number of nitrogens on each side balances However, on the left side of the equation we have
I have also provided the states of each reactant and the product at room temperature, though without further information on reaction conditions these cannot be confirmed as correct and so I recommend leaving them out unless there is further advice given.
It may seem unnatural to refer to these species as ""calciums"" and the like, but it is important to be aware that we are not always talking in terms of atoms: in this case, we may also be referring to the ions on the right hand side of the equation.
Calcium is a metal, so its formula will simply be
Nitrogen is a diatomic molecular compound, making it
Since calcium nitride is an ionic compound, by evaluating its constituent ions we can determine its formula. The calcium ion is a 2+ ion, or
Balancing the equation, then, you will notice that the number of nitrogens on each side balances However, on the left side of the equation we have
I have also provided the states of each reactant and the product at room temperature, though without further information on reaction conditions these cannot be confirmed as correct and so I recommend leaving them out unless there is further advice given.
It may seem unnatural to refer to these species as ""calciums"" and the like, but it is important to be aware that we are not always talking in terms of atoms: in this case, we may also be referring to the ions on the right hand side of the equation.
Calcium is a metal, so its formula will simply be
Nitrogen is a diatomic molecular compound, making it
Since calcium nitride is an ionic compound, by evaluating its constituent ions we can determine its formula. The calcium ion is a 2+ ion, or
Balancing the equation, then, you will notice that the number of nitrogens on each side balances However, on the left side of the equation we have
I have also provided the states of each reactant and the product at room temperature, though without further information on reaction conditions these cannot be confirmed as correct and so I recommend leaving them out unless there is further advice given.
It may seem unnatural to refer to these species as ""calciums"" and the like, but it is important to be aware that we are not always talking in terms of atoms: in this case, we may also be referring to the ions on the right hand side of the equation.
In order for an aqueous solution to be neutral, you need it to contain equal concentrations of hydronium ions,
Now, a solution's pH is calculated by taking the negative common logarithm (this is simply a base 10 log) of the concentration of hydronium ions.
#""pH"" = - log( [""H""_3""O""^(+)])#
Likewise, a solution's pOH is calculated by taking the negative common logarithm of the concentration of hydroxide ions
#""pOH"" = - log( [""OH""^(-)])#
For aqueous solutions, you can say that
#color(blue)(""pH"" + ""pOH"" = 14)#
Here
So, take a look at your solution. You know that its pH is equal to
#""pOH"" = 14 - ""pH""#
#""pOH"" = 14 - 2 = 12#
Now, a lower pH is equivalent to a higher concentration of hydronium ions, and implicitly a lower concentration of hydroxide ions.
Use the log definitions of the pH and pOH to get
#[""H""_3""O""^(+)] = 10^(-""pH"") = 10^(-2)""M""#
and
#[""OH""^(-)] = 10^(-""pOH"") = 10^(-12)""M""#
This means that a solution that has a pH equal to
#([""H""_3""O""^(+)])/([""OH""^(-)]) = (10^(-2)color(red)(cancel(color(black)(""M""))))/(10^(-12)color(red)(cancel(color(black)(""M"")))) = 10^10#
You can say that it contains
In order for an aqueous solution to be neutral, you need it to contain equal concentrations of hydronium ions,
Now, a solution's pH is calculated by taking the negative common logarithm (this is simply a base 10 log) of the concentration of hydronium ions.
#""pH"" = - log( [""H""_3""O""^(+)])#
Likewise, a solution's pOH is calculated by taking the negative common logarithm of the concentration of hydroxide ions
#""pOH"" = - log( [""OH""^(-)])#
For aqueous solutions, you can say that
#color(blue)(""pH"" + ""pOH"" = 14)#
Here
So, take a look at your solution. You know that its pH is equal to
#""pOH"" = 14 - ""pH""#
#""pOH"" = 14 - 2 = 12#
Now, a lower pH is equivalent to a higher concentration of hydronium ions, and implicitly a lower concentration of hydroxide ions.
Use the log definitions of the pH and pOH to get
#[""H""_3""O""^(+)] = 10^(-""pH"") = 10^(-2)""M""#
and
#[""OH""^(-)] = 10^(-""pOH"") = 10^(-12)""M""#
This means that a solution that has a pH equal to
#([""H""_3""O""^(+)])/([""OH""^(-)]) = (10^(-2)color(red)(cancel(color(black)(""M""))))/(10^(-12)color(red)(cancel(color(black)(""M"")))) = 10^10#
You can say that it contains
In order for an aqueous solution to be neutral, you need it to contain equal concentrations of hydronium ions,
Now, a solution's pH is calculated by taking the negative common logarithm (this is simply a base 10 log) of the concentration of hydronium ions.
#""pH"" = - log( [""H""_3""O""^(+)])#
Likewise, a solution's pOH is calculated by taking the negative common logarithm of the concentration of hydroxide ions
#""pOH"" = - log( [""OH""^(-)])#
For aqueous solutions, you can say that
#color(blue)(""pH"" + ""pOH"" = 14)#
Here
So, take a look at your solution. You know that its pH is equal to
#""pOH"" = 14 - ""pH""#
#""pOH"" = 14 - 2 = 12#
Now, a lower pH is equivalent to a higher concentration of hydronium ions, and implicitly a lower concentration of hydroxide ions.
Use the log definitions of the pH and pOH to get
#[""H""_3""O""^(+)] = 10^(-""pH"") = 10^(-2)""M""#
and
#[""OH""^(-)] = 10^(-""pOH"") = 10^(-12)""M""#
This means that a solution that has a pH equal to
#([""H""_3""O""^(+)])/([""OH""^(-)]) = (10^(-2)color(red)(cancel(color(black)(""M""))))/(10^(-12)color(red)(cancel(color(black)(""M"")))) = 10^10#
The oxidation number of an atom in a chemical bond, is the charge left on the atom of interest when all the bonding pairs of electrons are broken, with the charge devolving to the most electronegative atom.
When we do this for
If it is elemental gas, the oxidation number is
The oxidation number of an atom in a chemical bond, is the charge left on the atom of interest when all the bonding pairs of electrons are broken, with the charge devolving to the most electronegative atom.
When we do this for
If it is elemental gas, the oxidation number is
The oxidation number of an atom in a chemical bond, is the charge left on the atom of interest when all the bonding pairs of electrons are broken, with the charge devolving to the most electronegative atom.
When we do this for
From Gay Lussac’s Law
#color(white)(...)""P ∝ T""#
#""P""_1/""T""_1 = ""P""_2/""T""_2#
From Gay Lussac’s Law
#color(white)(...)""P ∝ T""#
#""P""_1/""T""_1 = ""P""_2/""T""_2#
From Gay Lussac’s Law
#color(white)(...)""P ∝ T""#
#""P""_1/""T""_1 = ""P""_2/""T""_2#
We interrogate the decomposition reaction....
And given complete decomposition we have a molar quantity of
And the volume expressed by this quantity is simply given by the old Ideal Gas equation..
And the number of molecules is simply
Well,
We interrogate the decomposition reaction....
And given complete decomposition we have a molar quantity of
And the volume expressed by this quantity is simply given by the old Ideal Gas equation..
And the number of molecules is simply
Well,
We interrogate the decomposition reaction....
And given complete decomposition we have a molar quantity of
And the volume expressed by this quantity is simply given by the old Ideal Gas equation..
And the number of molecules is simply
Magnesium is
Sulphate is
Hence, magnesium sulphate is
""Hepta-"" refers to ""seven"", ""hydrate"" refers to ""water"", so we have to add
Formula:
Magnesium is
Sulphate is
Hence, magnesium sulphate is
""Hepta-"" refers to ""seven"", ""hydrate"" refers to ""water"", so we have to add
Formula:
Magnesium is
Sulphate is
Hence, magnesium sulphate is
""Hepta-"" refers to ""seven"", ""hydrate"" refers to ""water"", so we have to add
Formula:
The idea here is that you need to use the fact that a solution is electrically neutral to determine how many milliequivalents of sodium cations are needed to balance the milliequivalents of the two anions.
The two anions present in solution are
As you know, an equivalent for an ion is calculated by multiplying the number of moles of that ion by its valence.
In this case, every liter of solution will contain
#""no. of mEq anions"" = ""40. mEq"" + ""15 mEq"" = ""55 mEq""#
You know that a solution is electrically neutral, so
#color(blue)(""no. of mEq anions "" = "" no. of mEq cations"")#
Because sodium is the only cation present in this solution, its number of milliequivalents must be equal to the total number of milliequivalents for the anions.
This means that the solution will contain
#[""Na""^(+)] = color(green)(""55 mEq/L"")#
The idea here is that you need to use the fact that a solution is electrically neutral to determine how many milliequivalents of sodium cations are needed to balance the milliequivalents of the two anions.
The two anions present in solution are
As you know, an equivalent for an ion is calculated by multiplying the number of moles of that ion by its valence.
In this case, every liter of solution will contain
#""no. of mEq anions"" = ""40. mEq"" + ""15 mEq"" = ""55 mEq""#
You know that a solution is electrically neutral, so
#color(blue)(""no. of mEq anions "" = "" no. of mEq cations"")#
Because sodium is the only cation present in this solution, its number of milliequivalents must be equal to the total number of milliequivalents for the anions.
This means that the solution will contain
#[""Na""^(+)] = color(green)(""55 mEq/L"")#
The idea here is that you need to use the fact that a solution is electrically neutral to determine how many milliequivalents of sodium cations are needed to balance the milliequivalents of the two anions.
The two anions present in solution are
As you know, an equivalent for an ion is calculated by multiplying the number of moles of that ion by its valence.
In this case, every liter of solution will contain
#""no. of mEq anions"" = ""40. mEq"" + ""15 mEq"" = ""55 mEq""#
You know that a solution is electrically neutral, so
#color(blue)(""no. of mEq anions "" = "" no. of mEq cations"")#
Because sodium is the only cation present in this solution, its number of milliequivalents must be equal to the total number of milliequivalents for the anions.
This means that the solution will contain
#[""Na""^(+)] = color(green)(""55 mEq/L"")#
Assuming near ideal gas behavior where
Rearranging the equation we have
V = 40.0 L
R = 0.0820575 L-atm/K-mol
T = 22'C = 295'K
n = (2500g/32.0 g/mol) = 78.125 moles of
P = 47.3atm
47.3 atm
Assuming near ideal gas behavior where
Rearranging the equation we have
V = 40.0 L
R = 0.0820575 L-atm/K-mol
T = 22'C = 295'K
n = (2500g/32.0 g/mol) = 78.125 moles of
P = 47.3atm
47.3 atm
Assuming near ideal gas behavior where
Rearranging the equation we have
V = 40.0 L
R = 0.0820575 L-atm/K-mol
T = 22'C = 295'K
n = (2500g/32.0 g/mol) = 78.125 moles of
P = 47.3atm
We can calculate the masses of
Now, we must convert these masses to moles and find their ratios.
From here on, I like to summarize the calculations in a table.
The empirical formula is
The empirical formula is
We can calculate the masses of
Now, we must convert these masses to moles and find their ratios.
From here on, I like to summarize the calculations in a table.
The empirical formula is
The empirical formula is
We can calculate the masses of
Now, we must convert these masses to moles and find their ratios.
From here on, I like to summarize the calculations in a table.
The empirical formula is
Right from the start, you should be able to predict that the pH of the solution will be higher than
You could look at the concentration of the base and say that the pH will only be slightly higher than
The key here is the self-ionization of water, which at room temperature has an equilibrium constant equal to
#K_W = 10^(-14) -># water's ionization constant
In pure water, water undergoes self-ionization to form hydronium cations,
#2""H""_ 2""O""_ ((l)) rightleftharpoons ""H""_ 3""O""_ ((aq))^(+) + ""OH""_ ((aq))^(-)#
By definition, the ionization constant is equal to
#K_W = [""H""_3""O""^(+)] * [""OH""^(-)]#
If you take
#K_w = x * x = x^2#
This gets you
#x = sqrt(K_W) = sqrt(10^(-14)) = 10^(-7)#
So, the self-ionization of water produces
#[""H""_3""O""^(+)] = 10^(-7)""M ""# and#"" "" [""OH""^(-)] = 10^(-7)""M""#
at room temperature. Now, your solution contains sodium hydroxide,
Therefore, you're adding
#[""OH""^(-)] = [""NaOH""] = 10^(-8)""M""#
to pure water. The key now is the fact that the self-ionization of water will still take place, but because the concentration of hydroxide anions has increased, the equilibrium will produce fewer moles of hydronium and hydroxide ions.
If you take
#"" ""2""H""_ 2""O""_ ((l)) rightleftharpoons"" "" ""H""_ 3""O""_ ((aq))^(+) "" ""+"" "" "" """"OH""_ ((aq))^(-)#
This time, the ionization constant will be equal to
#K_W = y * (10^(-8) + y)#
#K_w = y^2 + 10^(-8) * y#
This will get you
#y^2 + 10^(-8) * y - 10^(-14) = 0#
This quadratic equation will produce two solutions, one positive and one negative. Since
#y = 9.51 * 10^(-8)#
This means that the equilibrium concentration of hydronium cations will be
#[""H""_3""O""^(+)] = 9.51 * 10^(-8)""M""#
The pH of the solution is given by
#color(blue)(|bar(ul(color(white)(a/a)""pH"" = - log([""H""_3""O""^(+)])color(white)(a/a)|)))#
Plug in your value to find
#""pH"" = - log(9.51 * 10^(-8)) = color(green)(|bar(ul(color(white)(a/a)color(black)(7.02)color(white)(a/a)|)))#
As you can see, the pH of the solution is consistent with the fact that your solution contains a strong base, albeit in a very small concentration.
Right from the start, you should be able to predict that the pH of the solution will be higher than
You could look at the concentration of the base and say that the pH will only be slightly higher than
The key here is the self-ionization of water, which at room temperature has an equilibrium constant equal to
#K_W = 10^(-14) -># water's ionization constant
In pure water, water undergoes self-ionization to form hydronium cations,
#2""H""_ 2""O""_ ((l)) rightleftharpoons ""H""_ 3""O""_ ((aq))^(+) + ""OH""_ ((aq))^(-)#
By definition, the ionization constant is equal to
#K_W = [""H""_3""O""^(+)] * [""OH""^(-)]#
If you take
#K_w = x * x = x^2#
This gets you
#x = sqrt(K_W) = sqrt(10^(-14)) = 10^(-7)#
So, the self-ionization of water produces
#[""H""_3""O""^(+)] = 10^(-7)""M ""# and#"" "" [""OH""^(-)] = 10^(-7)""M""#
at room temperature. Now, your solution contains sodium hydroxide,
Therefore, you're adding
#[""OH""^(-)] = [""NaOH""] = 10^(-8)""M""#
to pure water. The key now is the fact that the self-ionization of water will still take place, but because the concentration of hydroxide anions has increased, the equilibrium will produce fewer moles of hydronium and hydroxide ions.
If you take
#"" ""2""H""_ 2""O""_ ((l)) rightleftharpoons"" "" ""H""_ 3""O""_ ((aq))^(+) "" ""+"" "" "" """"OH""_ ((aq))^(-)#
This time, the ionization constant will be equal to
#K_W = y * (10^(-8) + y)#
#K_w = y^2 + 10^(-8) * y#
This will get you
#y^2 + 10^(-8) * y - 10^(-14) = 0#
This quadratic equation will produce two solutions, one positive and one negative. Since
#y = 9.51 * 10^(-8)#
This means that the equilibrium concentration of hydronium cations will be
#[""H""_3""O""^(+)] = 9.51 * 10^(-8)""M""#
The pH of the solution is given by
#color(blue)(|bar(ul(color(white)(a/a)""pH"" = - log([""H""_3""O""^(+)])color(white)(a/a)|)))#
Plug in your value to find
#""pH"" = - log(9.51 * 10^(-8)) = color(green)(|bar(ul(color(white)(a/a)color(black)(7.02)color(white)(a/a)|)))#
As you can see, the pH of the solution is consistent with the fact that your solution contains a strong base, albeit in a very small concentration.
Right from the start, you should be able to predict that the pH of the solution will be higher than
You could look at the concentration of the base and say that the pH will only be slightly higher than
The key here is the self-ionization of water, which at room temperature has an equilibrium constant equal to
#K_W = 10^(-14) -># water's ionization constant
In pure water, water undergoes self-ionization to form hydronium cations,
#2""H""_ 2""O""_ ((l)) rightleftharpoons ""H""_ 3""O""_ ((aq))^(+) + ""OH""_ ((aq))^(-)#
By definition, the ionization constant is equal to
#K_W = [""H""_3""O""^(+)] * [""OH""^(-)]#
If you take
#K_w = x * x = x^2#
This gets you
#x = sqrt(K_W) = sqrt(10^(-14)) = 10^(-7)#
So, the self-ionization of water produces
#[""H""_3""O""^(+)] = 10^(-7)""M ""# and#"" "" [""OH""^(-)] = 10^(-7)""M""#
at room temperature. Now, your solution contains sodium hydroxide,
Therefore, you're adding
#[""OH""^(-)] = [""NaOH""] = 10^(-8)""M""#
to pure water. The key now is the fact that the self-ionization of water will still take place, but because the concentration of hydroxide anions has increased, the equilibrium will produce fewer moles of hydronium and hydroxide ions.
If you take
#"" ""2""H""_ 2""O""_ ((l)) rightleftharpoons"" "" ""H""_ 3""O""_ ((aq))^(+) "" ""+"" "" "" """"OH""_ ((aq))^(-)#
This time, the ionization constant will be equal to
#K_W = y * (10^(-8) + y)#
#K_w = y^2 + 10^(-8) * y#
This will get you
#y^2 + 10^(-8) * y - 10^(-14) = 0#
This quadratic equation will produce two solutions, one positive and one negative. Since
#y = 9.51 * 10^(-8)#
This means that the equilibrium concentration of hydronium cations will be
#[""H""_3""O""^(+)] = 9.51 * 10^(-8)""M""#
The pH of the solution is given by
#color(blue)(|bar(ul(color(white)(a/a)""pH"" = - log([""H""_3""O""^(+)])color(white)(a/a)|)))#
Plug in your value to find
#""pH"" = - log(9.51 * 10^(-8)) = color(green)(|bar(ul(color(white)(a/a)color(black)(7.02)color(white)(a/a)|)))#
As you can see, the pH of the solution is consistent with the fact that your solution contains a strong base, albeit in a very small concentration.
This is a tiny concentration for NaOH so I would expect the pH to be close to 7 or slightly greater.
Because the concentration is so small, we must take into account the ions that are produced from the auto - ionisation of water:
For which
This tells us that, in pure water, the hydrogen and hydroxide ion concentrations are
To get the total hydroxide concentration you might think you simply add
We need to apply Le Chatelier's Principle.
If we do a thought experiment we can imagine some pure water to which a tiny amount of sodium hydroxide is added such that the concentration of NaOH is
We have now disturbed a system at equilibrium by adding extra
The reaction quotient
We can set up an ICE table using equilibrium concentrations to show this:
This gives us:
If we multiply this out we get a quadratic equation so the quadratic formula can be used to solve for
We can now get the equilibrium concentration of
This is an example of ""The Common Ion Effect"", hydroxide being the common ion in question.
The molar enthalpy of vaporization is simply the enthalpy change needed in order for one mole of a given substance to undergo a liquid
Now, the molar enthalpy of vaporization will carry a positive sign, since it's used with the notion of energy needed, or energy absorbed, by a system in order to cause its molecules to go from the liquid state into the gaseous state.
Phase changes always occur at constant temperature, which is why you'll sometimes see the molar enthlpy of vaporization being referred to as the latent molar enthalpy of vaporization.
So, you know that you have
Mathematically, this can be expressed as
#color(blue)(q = n * DeltaH_""vap"") "" ""# , where
Plug your values into this equation and rearrange to solve for
#q = n * DeltaH_""vap"" implies DeltaH_""vap"" = q/n#
#DeltaH_""vap"" = ""28.4 kJ""/""3.21 moles"" = ""8.8474 kJ/mol""#
Rounded to thre sig figs, the answer will be
#DeltaH_""vap"" = color(green)(""8.85 kJ/mol"")#
The molar enthalpy of vaporization is simply the enthalpy change needed in order for one mole of a given substance to undergo a liquid
Now, the molar enthalpy of vaporization will carry a positive sign, since it's used with the notion of energy needed, or energy absorbed, by a system in order to cause its molecules to go from the liquid state into the gaseous state.
Phase changes always occur at constant temperature, which is why you'll sometimes see the molar enthlpy of vaporization being referred to as the latent molar enthalpy of vaporization.
So, you know that you have
Mathematically, this can be expressed as
#color(blue)(q = n * DeltaH_""vap"") "" ""# , where
Plug your values into this equation and rearrange to solve for
#q = n * DeltaH_""vap"" implies DeltaH_""vap"" = q/n#
#DeltaH_""vap"" = ""28.4 kJ""/""3.21 moles"" = ""8.8474 kJ/mol""#
Rounded to thre sig figs, the answer will be
#DeltaH_""vap"" = color(green)(""8.85 kJ/mol"")#
The molar enthalpy of vaporization is simply the enthalpy change needed in order for one mole of a given substance to undergo a liquid
Now, the molar enthalpy of vaporization will carry a positive sign, since it's used with the notion of energy needed, or energy absorbed, by a system in order to cause its molecules to go from the liquid state into the gaseous state.
Phase changes always occur at constant temperature, which is why you'll sometimes see the molar enthlpy of vaporization being referred to as the latent molar enthalpy of vaporization.
So, you know that you have
Mathematically, this can be expressed as
#color(blue)(q = n * DeltaH_""vap"") "" ""# , where
Plug your values into this equation and rearrange to solve for
#q = n * DeltaH_""vap"" implies DeltaH_""vap"" = q/n#
#DeltaH_""vap"" = ""28.4 kJ""/""3.21 moles"" = ""8.8474 kJ/mol""#
Rounded to thre sig figs, the answer will be
#DeltaH_""vap"" = color(green)(""8.85 kJ/mol"")#
We assume a
And thus there are
And
And
We divide thru by the lowest molar quantity, that of oxygen to get a trial empirical formula.....of
This organic formula is a bit whack, in that I think it contains a peroxo linkage (or it's a diether or diol); it must do if its MOLECULAR formula is
We assume a
And thus there are
And
And
We divide thru by the lowest molar quantity, that of oxygen to get a trial empirical formula.....of
This organic formula is a bit whack, in that I think it contains a peroxo linkage (or it's a diether or diol); it must do if its MOLECULAR formula is
We assume a
And thus there are
And
And
We divide thru by the lowest molar quantity, that of oxygen to get a trial empirical formula.....of
This organic formula is a bit whack, in that I think it contains a peroxo linkage (or it's a diether or diol); it must do if its MOLECULAR formula is
Here's my thought process:
- From name: phosphorus
#-># #""P""# - ""di""
#-># #2# - From name: oxide
#-># #""O""# - Anion stem: ""ox""
- ""pent"" (instead of ""penta"", since ""ox"" starts with a vowel)
#-># #5#
#-># #""P""_2""O""_5# (empirical formula)
To find the concentration of something that dissociates into
So this becomes:
To find the concentration of something that dissociates into
So this becomes:
To find the concentration of something that dissociates into
So this becomes:
The
#[H^+]# is the hydrogen ion concentration in terms of molarity
The dissociation equation for hydrochloric acid is:
So, one mole of hydrochloric acid contains one mole of hydrogen ions.
Therefore, the hydrogen ion concentration will be,
Start by writing out the balanced chemical equation that describes your reaction
#""CaO""_ ((s)) + color(red)(2)""HCl""_ ((aq)) -> ""CaCl""_ (2(aq)) + ""H""_ 2""O""_ ((l))#
You need
You an convert this mole ratio to a gram ratio by using the molar masses of the two compounds
#M_(""M CaO"") = ""56.08 g mol""^(-1)#
#M_(""M HCl"") = ""36.46 g mol""^(-1)#
A
#(56.08 color(red)(cancel(color(black)(""g mol""^(-1)))))/(color(red)(2) * 36.46color(red)(cancel(color(black)(""g mol""^(-1))))) = 56.08/72.92 -># gram ratio
So, you need
You know that you have a sample of
#60.4 color(red)(cancel(color(black)(""g CaO""))) * ""72.92 g HCl""/(56.08color(red)(cancel(color(black)(""g CaO"")))) = ""78.54 g HCl""#
Since you have less hydrochloric acid than you would need in order for all the mass of calcium oxide to react
#overbrace(""69.0 g"")^(color(blue)(""what you have"")) < overbrace(""78.54 g"")^(color(darkgreen)(""what you need""))#
you can conclude that hydrochloric acid will act as a limiting reagent, i.e. it will be completely consumed before all the grams of calcium oxide get a chance to react.
Use the molar mass of calcium chloride to convert the
#M_(""M CaCl""_2) = ""111.0 g mol""^(-1)#
You will have
#(color(red)(2) * 36.46 color(red)(cancel(color(black)(""g mol""^(-1)))))/(111.0color(red)(cancel(color(black)(""g mol""^(-1))))) = 72.92/111.0 -># gram ratio
Since all the hydrochloric acid will react, you can say that the reaction will produce
#69.0 color(red)(cancel(color(black)(""g HCl""))) * ""111.0 g CaCl""_2/(72.92color(red)(cancel(color(black)(""g HCl"")))) = color(green)(|bar(ul(color(white)(a/a)color(black)(""105 g CaCl""_2)color(white)(a/a)|)))#
The answer is rounded to three sig figs.
Start by writing out the balanced chemical equation that describes your reaction
#""CaO""_ ((s)) + color(red)(2)""HCl""_ ((aq)) -> ""CaCl""_ (2(aq)) + ""H""_ 2""O""_ ((l))#
You need
You an convert this mole ratio to a gram ratio by using the molar masses of the two compounds
#M_(""M CaO"") = ""56.08 g mol""^(-1)#
#M_(""M HCl"") = ""36.46 g mol""^(-1)#
A
#(56.08 color(red)(cancel(color(black)(""g mol""^(-1)))))/(color(red)(2) * 36.46color(red)(cancel(color(black)(""g mol""^(-1))))) = 56.08/72.92 -># gram ratio
So, you need
You know that you have a sample of
#60.4 color(red)(cancel(color(black)(""g CaO""))) * ""72.92 g HCl""/(56.08color(red)(cancel(color(black)(""g CaO"")))) = ""78.54 g HCl""#
Since you have less hydrochloric acid than you would need in order for all the mass of calcium oxide to react
#overbrace(""69.0 g"")^(color(blue)(""what you have"")) < overbrace(""78.54 g"")^(color(darkgreen)(""what you need""))#
you can conclude that hydrochloric acid will act as a limiting reagent, i.e. it will be completely consumed before all the grams of calcium oxide get a chance to react.
Use the molar mass of calcium chloride to convert the
#M_(""M CaCl""_2) = ""111.0 g mol""^(-1)#
You will have
#(color(red)(2) * 36.46 color(red)(cancel(color(black)(""g mol""^(-1)))))/(111.0color(red)(cancel(color(black)(""g mol""^(-1))))) = 72.92/111.0 -># gram ratio
Since all the hydrochloric acid will react, you can say that the reaction will produce
#69.0 color(red)(cancel(color(black)(""g HCl""))) * ""111.0 g CaCl""_2/(72.92color(red)(cancel(color(black)(""g HCl"")))) = color(green)(|bar(ul(color(white)(a/a)color(black)(""105 g CaCl""_2)color(white)(a/a)|)))#
The answer is rounded to three sig figs.
Start by writing out the balanced chemical equation that describes your reaction
#""CaO""_ ((s)) + color(red)(2)""HCl""_ ((aq)) -> ""CaCl""_ (2(aq)) + ""H""_ 2""O""_ ((l))#
You need
You an convert this mole ratio to a gram ratio by using the molar masses of the two compounds
#M_(""M CaO"") = ""56.08 g mol""^(-1)#
#M_(""M HCl"") = ""36.46 g mol""^(-1)#
A
#(56.08 color(red)(cancel(color(black)(""g mol""^(-1)))))/(color(red)(2) * 36.46color(red)(cancel(color(black)(""g mol""^(-1))))) = 56.08/72.92 -># gram ratio
So, you need
You know that you have a sample of
#60.4 color(red)(cancel(color(black)(""g CaO""))) * ""72.92 g HCl""/(56.08color(red)(cancel(color(black)(""g CaO"")))) = ""78.54 g HCl""#
Since you have less hydrochloric acid than you would need in order for all the mass of calcium oxide to react
#overbrace(""69.0 g"")^(color(blue)(""what you have"")) < overbrace(""78.54 g"")^(color(darkgreen)(""what you need""))#
you can conclude that hydrochloric acid will act as a limiting reagent, i.e. it will be completely consumed before all the grams of calcium oxide get a chance to react.
Use the molar mass of calcium chloride to convert the
#M_(""M CaCl""_2) = ""111.0 g mol""^(-1)#
You will have
#(color(red)(2) * 36.46 color(red)(cancel(color(black)(""g mol""^(-1)))))/(111.0color(red)(cancel(color(black)(""g mol""^(-1))))) = 72.92/111.0 -># gram ratio
Since all the hydrochloric acid will react, you can say that the reaction will produce
#69.0 color(red)(cancel(color(black)(""g HCl""))) * ""111.0 g CaCl""_2/(72.92color(red)(cancel(color(black)(""g HCl"")))) = color(green)(|bar(ul(color(white)(a/a)color(black)(""105 g CaCl""_2)color(white)(a/a)|)))#
The answer is rounded to three sig figs.
But given the formulation of the salt,
But given the formulation of the salt,
But given the formulation of the salt,
For this problem, we can use Gay-Lussac's Law of Combining Volumes:
If pressure and temperature are constant, the ratio between the volumes of the reactant gases and the products can be expressed in simple whole numbers.
The balanced equation for the combustion is
#color(white)(l)""CH""_4 + ""2O""_2 → ""CO""_2 + ""2H""_2""O""#
#""1 dm""^3color(white)(ll)""2 dm""^3#
According to Gay-Lussac,
∴
The volume of oxygen required is
For this problem, we can use Gay-Lussac's Law of Combining Volumes:
If pressure and temperature are constant, the ratio between the volumes of the reactant gases and the products can be expressed in simple whole numbers.
The balanced equation for the combustion is
#color(white)(l)""CH""_4 + ""2O""_2 → ""CO""_2 + ""2H""_2""O""#
#""1 dm""^3color(white)(ll)""2 dm""^3#
According to Gay-Lussac,
∴
The volume of oxygen required is
For this problem, we can use Gay-Lussac's Law of Combining Volumes:
If pressure and temperature are constant, the ratio between the volumes of the reactant gases and the products can be expressed in simple whole numbers.
The balanced equation for the combustion is
#color(white)(l)""CH""_4 + ""2O""_2 → ""CO""_2 + ""2H""_2""O""#
#""1 dm""^3color(white)(ll)""2 dm""^3#
According to Gay-Lussac,
∴
Since water is required for the complete reaction or if NaOH is taken in liquid [Lye form], H2O already exists. Hence reaction is as under :-
2NaOH + 2Al + 2H2O = 2NaAlO2 [ Sodium aluminate] + 3 H2
2NaOH + 2Al + 2H2O = 2NaAlO2 [ Sodium aluminate] + 3 H2
Since water is required for the complete reaction or if NaOH is taken in liquid [Lye form], H2O already exists. Hence reaction is as under :-
2NaOH + 2Al + 2H2O = 2NaAlO2 [ Sodium aluminate] + 3 H2
2NaOH + 2Al + 2H2O = 2NaAlO2 [ Sodium aluminate] + 3 H2
Since water is required for the complete reaction or if NaOH is taken in liquid [Lye form], H2O already exists. Hence reaction is as under :-
2NaOH + 2Al + 2H2O = 2NaAlO2 [ Sodium aluminate] + 3 H2
We're asked to find the molar concentration of the
Let's first write the chemical equation for this reaction:
Using the molarity equation, we can find the number of moles of
(volume converted to liters)
Now, using the coefficients of the chemical reaction, we can determine the number of moles of
Lastly, we'll use the molarity equation (using given volume of
(volume converted to liters)
We're asked to find the molar concentration of the
Let's first write the chemical equation for this reaction:
Using the molarity equation, we can find the number of moles of
(volume converted to liters)
Now, using the coefficients of the chemical reaction, we can determine the number of moles of
Lastly, we'll use the molarity equation (using given volume of
(volume converted to liters)
We're asked to find the molar concentration of the
Let's first write the chemical equation for this reaction:
Using the molarity equation, we can find the number of moles of
(volume converted to liters)
Now, using the coefficients of the chemical reaction, we can determine the number of moles of
Lastly, we'll use the molarity equation (using given volume of
(volume converted to liters)
We first need to define
http://goldbook.iupac.org/S05910.html
Most people use
At
http://jfi.uchicago.edu/~ntt/formula/node197.html
https://en.m.wikipedia.org/wiki/Molar_volume
To determine the moles of a gas, divide the given volume by the molar volume.
At
We first need to define
http://goldbook.iupac.org/S05910.html
Most people use
At
http://jfi.uchicago.edu/~ntt/formula/node197.html
https://en.m.wikipedia.org/wiki/Molar_volume
To determine the moles of a gas, divide the given volume by the molar volume.
At
We first need to define
http://goldbook.iupac.org/S05910.html
Most people use
At
http://jfi.uchicago.edu/~ntt/formula/node197.html
https://en.m.wikipedia.org/wiki/Molar_volume
To determine the moles of a gas, divide the given volume by the molar volume.
Now
And so
All I have done is to use a
Now
And so
All I have done is to use a
Now
And so
All I have done is to use a
What we're doing here is calculating basic mole-mole relationships, something that you'll be doing quite a bit!
The steps to solving mole-mole problems like this are
write the balanced chemical equation for the reaction (this is given)
divide the number of moles of the given known substance (
multiply that number by the coefficient of the substance you're trying to find (
Using simple dimensional analysis, it looks like this:
rounded to
Thus, if the reaction goes to completion,
What we're doing here is calculating basic mole-mole relationships, something that you'll be doing quite a bit!
The steps to solving mole-mole problems like this are
write the balanced chemical equation for the reaction (this is given)
divide the number of moles of the given known substance (
multiply that number by the coefficient of the substance you're trying to find (
Using simple dimensional analysis, it looks like this:
rounded to
Thus, if the reaction goes to completion,
What we're doing here is calculating basic mole-mole relationships, something that you'll be doing quite a bit!
The steps to solving mole-mole problems like this are
write the balanced chemical equation for the reaction (this is given)
divide the number of moles of the given known substance (
multiply that number by the coefficient of the substance you're trying to find (
Using simple dimensional analysis, it looks like this:
rounded to
Thus, if the reaction goes to completion,
Convert 2.35 mol/kg RbNO3 (aq) solution to percent by mass.
Your goal here is to figure out the number of grams of solute present for every
Now, you know that the solution has a molality equal to
To make the calculations easier, pick a sample of this solution that contains exactly
Use the molar mass of the solute to convert the number of moles to grams
#2.35 color(red)(cancel(color(black)(""moles RbNO""_3))) * ""147.473 g""/(1color(red)(cancel(color(black)(""mole RbNO""_3)))) = ""346.56 g""#
This means that the total mass of the sample is equal to
#""346.56 g "" + quad 10^3 quad ""g"" = ""1346.56 g""#
So, you know that you have
#100 color(red)(cancel(color(black)(""g solution""))) * ""346.56 g RbNO""_3/(1346.56 color(red)(cancel(color(black)(""g solution"")))) = ""25.7 g RbNO""_3#
This means that the solution's percent concentration by mass is equal to
#color(darkgreen)(ul(color(black)(""% m/m"" = ""25.7% RbNO""_3)))# This tells you that you get
#""25.7 g""# of rubidium nitrate for every#""100 g""# of the solution.
The answer is rounded to three sig figs.
Your goal here is to figure out the number of grams of solute present for every
Now, you know that the solution has a molality equal to
To make the calculations easier, pick a sample of this solution that contains exactly
Use the molar mass of the solute to convert the number of moles to grams
#2.35 color(red)(cancel(color(black)(""moles RbNO""_3))) * ""147.473 g""/(1color(red)(cancel(color(black)(""mole RbNO""_3)))) = ""346.56 g""#
This means that the total mass of the sample is equal to
#""346.56 g "" + quad 10^3 quad ""g"" = ""1346.56 g""#
So, you know that you have
#100 color(red)(cancel(color(black)(""g solution""))) * ""346.56 g RbNO""_3/(1346.56 color(red)(cancel(color(black)(""g solution"")))) = ""25.7 g RbNO""_3#
This means that the solution's percent concentration by mass is equal to
#color(darkgreen)(ul(color(black)(""% m/m"" = ""25.7% RbNO""_3)))# This tells you that you get
#""25.7 g""# of rubidium nitrate for every#""100 g""# of the solution.
The answer is rounded to three sig figs.
Convert 2.35 mol/kg RbNO3 (aq) solution to percent by mass.
Your goal here is to figure out the number of grams of solute present for every
Now, you know that the solution has a molality equal to
To make the calculations easier, pick a sample of this solution that contains exactly
Use the molar mass of the solute to convert the number of moles to grams
#2.35 color(red)(cancel(color(black)(""moles RbNO""_3))) * ""147.473 g""/(1color(red)(cancel(color(black)(""mole RbNO""_3)))) = ""346.56 g""#
This means that the total mass of the sample is equal to
#""346.56 g "" + quad 10^3 quad ""g"" = ""1346.56 g""#
So, you know that you have
#100 color(red)(cancel(color(black)(""g solution""))) * ""346.56 g RbNO""_3/(1346.56 color(red)(cancel(color(black)(""g solution"")))) = ""25.7 g RbNO""_3#
This means that the solution's percent concentration by mass is equal to
#color(darkgreen)(ul(color(black)(""% m/m"" = ""25.7% RbNO""_3)))# This tells you that you get
#""25.7 g""# of rubidium nitrate for every#""100 g""# of the solution.
The answer is rounded to three sig figs.
The specific heat of a substance tells you how much heat is needed in order to increase the temperature of one unit of mass of that substance, usually
In your case, the specific heat of foam is given as
Now, you must figure out how much heat is needed to increase the temperature of
In this case, you know that you need
In other words, you will need
#2 color(red)(cancel(color(black)(""K""))) * overbrace(""1200 J""/(1color(red)(cancel(color(black)(""K"")))))^(color(blue)(""needed for 1 kg of foam"")) = color(darkgreen)(ul(color(black)(""2400 J"")))#
I'll leave the answer rounded to two sig figs.
The specific heat of a substance tells you how much heat is needed in order to increase the temperature of one unit of mass of that substance, usually
In your case, the specific heat of foam is given as
Now, you must figure out how much heat is needed to increase the temperature of
In this case, you know that you need
In other words, you will need
#2 color(red)(cancel(color(black)(""K""))) * overbrace(""1200 J""/(1color(red)(cancel(color(black)(""K"")))))^(color(blue)(""needed for 1 kg of foam"")) = color(darkgreen)(ul(color(black)(""2400 J"")))#
I'll leave the answer rounded to two sig figs.
The specific heat of a substance tells you how much heat is needed in order to increase the temperature of one unit of mass of that substance, usually
In your case, the specific heat of foam is given as
Now, you must figure out how much heat is needed to increase the temperature of
In this case, you know that you need
In other words, you will need
#2 color(red)(cancel(color(black)(""K""))) * overbrace(""1200 J""/(1color(red)(cancel(color(black)(""K"")))))^(color(blue)(""needed for 1 kg of foam"")) = color(darkgreen)(ul(color(black)(""2400 J"")))#
I'll leave the answer rounded to two sig figs.
For your problem,
The general aqueous solubility of nitrate species allows such a high concentration to be achieved.
For your problem,
The general aqueous solubility of nitrate species allows such a high concentration to be achieved.
For your problem,
The general aqueous solubility of nitrate species allows such a high concentration to be achieved.
In
In
There is a total of
In
In
There is a total of
In
In
There is a total of
In all problems of this sort, it is convenient to assume a
We divide thru by the SMALLEST molar quantity, that of the metal, to get an empirical formula of
We assume a
In all problems of this sort, it is convenient to assume a
We divide thru by the SMALLEST molar quantity, that of the metal, to get an empirical formula of
We assume a
In all problems of this sort, it is convenient to assume a
We divide thru by the SMALLEST molar quantity, that of the metal, to get an empirical formula of
Here I have
This question is the same as if your were asked,
Can you answer that?
Here I have
This question is the same as if your were asked,
Can you answer that?
Here I have
The formula for osmotic pressure
#color(blue)(bar(ul(|color(white)(a/a)Pi = cRTcolor(white)(a/a)|)))"" ""#
where
Since
#Pi = (nRT)/V#
Since
#Pi = (nRT)/(MV)#
We can solve this equation for the molar mass and get
#color(blue)(bar(ul(|color(white)(a/a)M=(mRT)/(PiV)color(white)(a/a)|)))"" ""#
In this problem
∴
The molar mass is
The formula for osmotic pressure
#color(blue)(bar(ul(|color(white)(a/a)Pi = cRTcolor(white)(a/a)|)))"" ""#
where
Since
#Pi = (nRT)/V#
Since
#Pi = (nRT)/(MV)#
We can solve this equation for the molar mass and get
#color(blue)(bar(ul(|color(white)(a/a)M=(mRT)/(PiV)color(white)(a/a)|)))"" ""#
In this problem
∴
The molar mass is
The formula for osmotic pressure
#color(blue)(bar(ul(|color(white)(a/a)Pi = cRTcolor(white)(a/a)|)))"" ""#
where
Since
#Pi = (nRT)/V#
Since
#Pi = (nRT)/(MV)#
We can solve this equation for the molar mass and get
#color(blue)(bar(ul(|color(white)(a/a)M=(mRT)/(PiV)color(white)(a/a)|)))"" ""#
In this problem
∴
Water specific heat is 4.178 J/g-°K.
The temperature change is 28.5 – 22.0 = 6.5 (because it is a difference, °C or °K doesn't matter)
41840J = 4.178 J/g-°K * Xg * deltaT
Xg = 41840J / (4.178 * 6.5) = 1541g water
1541g
Water specific heat is 4.178 J/g-°K.
The temperature change is 28.5 – 22.0 = 6.5 (because it is a difference, °C or °K doesn't matter)
41840J = 4.178 J/g-°K * Xg * deltaT
Xg = 41840J / (4.178 * 6.5) = 1541g water
1541g
Water specific heat is 4.178 J/g-°K.
The temperature change is 28.5 – 22.0 = 6.5 (because it is a difference, °C or °K doesn't matter)
41840J = 4.178 J/g-°K * Xg * deltaT
Xg = 41840J / (4.178 * 6.5) = 1541g water
Concentration here is simply (amount of substance)/(volume of solution).
So
Concentration here is simply (amount of substance)/(volume of solution).
So
Concentration here is simply (amount of substance)/(volume of solution).
So
For starters, we know that a
This implies that a
So all you have to do now is figure out how many moles of magnesium chloride would be equivalent in
#(color(red)(?)color(white)(.)""moles MgCl""_2)/""2.0 L solution"" = overbrace(""4.5 moles MgCl""_2/(""1.0 L solution""))^(color(blue)(""= 4.5 M solution""))#
Use cross multiplication to get
#color(red)(?) = (2.0 color(red)(cancel(color(black)(""L solution""))))/(1.0color(red)(cancel(color(black)(""L solution"")))) * ""4.5 moles MgCl""_2#
This will be equal to
#color(red)(?) = ""9.0 moles MgCl""_2#
So, you know that you can get a
To convert this to grams, use the compound's molar mass
#9.0 color(red)(cancel(color(black)(""moles MgCl""_2))) * ""95.211 g""/(1color(red)(cancel(color(black)(""mole MgCl""_2)))) = color(darkgreen)(ul(color(black)(""860 g"")))#
The answer is rounded to two sig figs, the number of sig figs you have for your values.
For starters, we know that a
This implies that a
So all you have to do now is figure out how many moles of magnesium chloride would be equivalent in
#(color(red)(?)color(white)(.)""moles MgCl""_2)/""2.0 L solution"" = overbrace(""4.5 moles MgCl""_2/(""1.0 L solution""))^(color(blue)(""= 4.5 M solution""))#
Use cross multiplication to get
#color(red)(?) = (2.0 color(red)(cancel(color(black)(""L solution""))))/(1.0color(red)(cancel(color(black)(""L solution"")))) * ""4.5 moles MgCl""_2#
This will be equal to
#color(red)(?) = ""9.0 moles MgCl""_2#
So, you know that you can get a
To convert this to grams, use the compound's molar mass
#9.0 color(red)(cancel(color(black)(""moles MgCl""_2))) * ""95.211 g""/(1color(red)(cancel(color(black)(""mole MgCl""_2)))) = color(darkgreen)(ul(color(black)(""860 g"")))#
The answer is rounded to two sig figs, the number of sig figs you have for your values.
For starters, we know that a
This implies that a
So all you have to do now is figure out how many moles of magnesium chloride would be equivalent in
#(color(red)(?)color(white)(.)""moles MgCl""_2)/""2.0 L solution"" = overbrace(""4.5 moles MgCl""_2/(""1.0 L solution""))^(color(blue)(""= 4.5 M solution""))#
Use cross multiplication to get
#color(red)(?) = (2.0 color(red)(cancel(color(black)(""L solution""))))/(1.0color(red)(cancel(color(black)(""L solution"")))) * ""4.5 moles MgCl""_2#
This will be equal to
#color(red)(?) = ""9.0 moles MgCl""_2#
So, you know that you can get a
To convert this to grams, use the compound's molar mass
#9.0 color(red)(cancel(color(black)(""moles MgCl""_2))) * ""95.211 g""/(1color(red)(cancel(color(black)(""mole MgCl""_2)))) = color(darkgreen)(ul(color(black)(""860 g"")))#
The answer is rounded to two sig figs, the number of sig figs you have for your values.
Use the combined gas law.
Given
Unknown
Equation
Solution
Rearrange the equation to isolate
The initial volume is
Use the combined gas law.
Given
Unknown
Equation
Solution
Rearrange the equation to isolate
The initial volume is
Use the combined gas law.
Given
Unknown
Equation
Solution
Rearrange the equation to isolate
Since we have a multiple of
If we have
Then we need
Hence the balanced form of the equation is:
#3MnO_2+4Al -> 3Mn+2Al_2O_3#
Since we have a multiple of
If we have
Then we need
Hence the balanced form of the equation is:
#3MnO_2+4Al -> 3Mn+2Al_2O_3#
Since we have a multiple of
If we have
Then we need
Hence the balanced form of the equation is:
#3MnO_2+4Al -> 3Mn+2Al_2O_3#
The idea here is that the partial pressure exerted by a gas that's part of a gaseous mixture is proportional to its mole fraction.
This means that in order to find the partial pressure of carbon monoxide, you need to find
- the number of moles of carbon dioxide found in the mixture
- the total number of moles present in the mixture
So, you know that the percent composition of the mixture is
#100% - (29.1% + 61.8%) = 9.1%#
Your strategy now will be to pick a sample of this mixture and determine the mass of each gas it contains. To make the calculations easier, let's say that you have a
Using the percentages given, you can say that this sample will contain
#""29.1 g SO""_2# #""61.8 g PF""_3# #""9.1 g CO""#
Use the respective molar masses of the three gases to find their number of moles
#""For SO""_2:"" "" 29.1 color(red)(cancel(color(black)(""g""))) * ""1 mole SO""_2/(64.06color(red)(cancel(color(black)(""g"")))) = ""0.4543 moles SO""_2#
#""For PF""_3: "" "" 61.8 color(red)(cancel(color(black)(""g""))) * ""1 mole PF""_3/(87.97color(red)(cancel(color(black)(""g"")))) = ""0.7025 moles PF""_3#
#""For CO"": "" "" 9.1color(red)(cancel(color(black)(""g""))) * ""1 mole CO""/(28.01color(red)(cancel(color(black)(""g"")))) = ""0.3249 moles CO""#
The total number of moles present in the mixture will be
#n_""total"" = 0.4543 + 0.7025 + 0.3249 = ""1.4817 moles""#
Now, the mole fraction of carbon monoxide will be equal to the number of moles of carbon monoxide divided by the total number of moles present in the mixture
#chi_(CO) = n_(CO)/n_""total""#
#chi_(CO) = ( 0.3249 color(red)(cancel(color(black)(""moles""))))/(1.4817 color(red)(cancel(color(black)(""moles"")))) = 0.2193#
The partial pressure of carbon monoxide will thus be
#P_(CO) = chi_(CO) xx P_""total""#
#P_(CO) = 0.2193 * ""684 torr"" = color(green)(""150. torr"")#
The answer is rounded to three sig figs.
The idea here is that the partial pressure exerted by a gas that's part of a gaseous mixture is proportional to its mole fraction.
This means that in order to find the partial pressure of carbon monoxide, you need to find
- the number of moles of carbon dioxide found in the mixture
- the total number of moles present in the mixture
So, you know that the percent composition of the mixture is
#100% - (29.1% + 61.8%) = 9.1%#
Your strategy now will be to pick a sample of this mixture and determine the mass of each gas it contains. To make the calculations easier, let's say that you have a
Using the percentages given, you can say that this sample will contain
#""29.1 g SO""_2# #""61.8 g PF""_3# #""9.1 g CO""#
Use the respective molar masses of the three gases to find their number of moles
#""For SO""_2:"" "" 29.1 color(red)(cancel(color(black)(""g""))) * ""1 mole SO""_2/(64.06color(red)(cancel(color(black)(""g"")))) = ""0.4543 moles SO""_2#
#""For PF""_3: "" "" 61.8 color(red)(cancel(color(black)(""g""))) * ""1 mole PF""_3/(87.97color(red)(cancel(color(black)(""g"")))) = ""0.7025 moles PF""_3#
#""For CO"": "" "" 9.1color(red)(cancel(color(black)(""g""))) * ""1 mole CO""/(28.01color(red)(cancel(color(black)(""g"")))) = ""0.3249 moles CO""#
The total number of moles present in the mixture will be
#n_""total"" = 0.4543 + 0.7025 + 0.3249 = ""1.4817 moles""#
Now, the mole fraction of carbon monoxide will be equal to the number of moles of carbon monoxide divided by the total number of moles present in the mixture
#chi_(CO) = n_(CO)/n_""total""#
#chi_(CO) = ( 0.3249 color(red)(cancel(color(black)(""moles""))))/(1.4817 color(red)(cancel(color(black)(""moles"")))) = 0.2193#
The partial pressure of carbon monoxide will thus be
#P_(CO) = chi_(CO) xx P_""total""#
#P_(CO) = 0.2193 * ""684 torr"" = color(green)(""150. torr"")#
The answer is rounded to three sig figs.
The idea here is that the partial pressure exerted by a gas that's part of a gaseous mixture is proportional to its mole fraction.
This means that in order to find the partial pressure of carbon monoxide, you need to find
- the number of moles of carbon dioxide found in the mixture
- the total number of moles present in the mixture
So, you know that the percent composition of the mixture is
#100% - (29.1% + 61.8%) = 9.1%#
Your strategy now will be to pick a sample of this mixture and determine the mass of each gas it contains. To make the calculations easier, let's say that you have a
Using the percentages given, you can say that this sample will contain
#""29.1 g SO""_2# #""61.8 g PF""_3# #""9.1 g CO""#
Use the respective molar masses of the three gases to find their number of moles
#""For SO""_2:"" "" 29.1 color(red)(cancel(color(black)(""g""))) * ""1 mole SO""_2/(64.06color(red)(cancel(color(black)(""g"")))) = ""0.4543 moles SO""_2#
#""For PF""_3: "" "" 61.8 color(red)(cancel(color(black)(""g""))) * ""1 mole PF""_3/(87.97color(red)(cancel(color(black)(""g"")))) = ""0.7025 moles PF""_3#
#""For CO"": "" "" 9.1color(red)(cancel(color(black)(""g""))) * ""1 mole CO""/(28.01color(red)(cancel(color(black)(""g"")))) = ""0.3249 moles CO""#
The total number of moles present in the mixture will be
#n_""total"" = 0.4543 + 0.7025 + 0.3249 = ""1.4817 moles""#
Now, the mole fraction of carbon monoxide will be equal to the number of moles of carbon monoxide divided by the total number of moles present in the mixture
#chi_(CO) = n_(CO)/n_""total""#
#chi_(CO) = ( 0.3249 color(red)(cancel(color(black)(""moles""))))/(1.4817 color(red)(cancel(color(black)(""moles"")))) = 0.2193#
The partial pressure of carbon monoxide will thus be
#P_(CO) = chi_(CO) xx P_""total""#
#P_(CO) = 0.2193 * ""684 torr"" = color(green)(""150. torr"")#
The answer is rounded to three sig figs.
A reaction's percent yield essentially tells you many grams of a given product will actually be produced instead of a theoretical
Simply put, you can calculate a reaction's percent yield by dividing the actual yield, which is what the reaction actually produces, by the theoretical yield, which is what the reaction should theoretically produce, and multiplying the result by
#color(blue)(|bar(ul(color(blue)(""% yield"" = ""what you actually get""/""what you should theoretically get"" xx 100color(white)(a/a)|)))#
In your case, the reaction is said to have a theoretical yield of
The actual yield of the reaction is said to
The percent yield of the reaction will be
#""% yield"" = (0.104 color(red)(cancel(color(black)(""g""))))/(0.145color(red)(cancel(color(black)(""g"")))) xx 100 = color(green)(|bar(ul(color(white)(a/a)""71.7%color(white)(a/a)|))) -># rounded to three sig figs
This means that for every
A reaction's percent yield essentially tells you many grams of a given product will actually be produced instead of a theoretical
Simply put, you can calculate a reaction's percent yield by dividing the actual yield, which is what the reaction actually produces, by the theoretical yield, which is what the reaction should theoretically produce, and multiplying the result by
#color(blue)(|bar(ul(color(blue)(""% yield"" = ""what you actually get""/""what you should theoretically get"" xx 100color(white)(a/a)|)))#
In your case, the reaction is said to have a theoretical yield of
The actual yield of the reaction is said to
The percent yield of the reaction will be
#""% yield"" = (0.104 color(red)(cancel(color(black)(""g""))))/(0.145color(red)(cancel(color(black)(""g"")))) xx 100 = color(green)(|bar(ul(color(white)(a/a)""71.7%color(white)(a/a)|))) -># rounded to three sig figs
This means that for every
A reaction's percent yield essentially tells you many grams of a given product will actually be produced instead of a theoretical
Simply put, you can calculate a reaction's percent yield by dividing the actual yield, which is what the reaction actually produces, by the theoretical yield, which is what the reaction should theoretically produce, and multiplying the result by
#color(blue)(|bar(ul(color(blue)(""% yield"" = ""what you actually get""/""what you should theoretically get"" xx 100color(white)(a/a)|)))#
In your case, the reaction is said to have a theoretical yield of
The actual yield of the reaction is said to
The percent yield of the reaction will be
#""% yield"" = (0.104 color(red)(cancel(color(black)(""g""))))/(0.145color(red)(cancel(color(black)(""g"")))) xx 100 = color(green)(|bar(ul(color(white)(a/a)""71.7%color(white)(a/a)|))) -># rounded to three sig figs
This means that for every
Each formula unit of magnesium chloride,
Multiply the number of formula units by the conversion factor
Use Avogadro’s number to calculate the number of moles of
Therefore, there are 12 moles of Cl.
Use Avogadro’s number to calculate the number of moles of
Therefore, there are 12 moles of Cl.
There are
Each formula unit of magnesium chloride,
Multiply the number of formula units by the conversion factor
The idea here is that you need to use the mass of water, the heat added to the sample, and the specific heat of water to find the resulting change in temperature.
The equation that establishes a relationship between heat absorbed and change in temperature looks like this
#color(blue)(q = m * c * DeltaT)"" ""# , where
Plug in your values and solve for
#q = m * c * DeltaT implies DeltaT = q/(m * c)#
#DeltaT = (1.0 * 10^(3)color(red)(cancel(color(black)(""J""))))/(50.0color(red)(cancel(color(black)(""g""))) * 4.18color(red)(cancel(color(black)(""J"")))/(color(red)(cancel(color(black)(""g""))) """"^@""C"")) = 4.785^@""C""#
So, if the temperature of the water changed by
#DeltaT = T_""f"" - T_""i"" implies T_""i"" = T_""f"" - DeltaT#
#T_""i"" = 50.0^@""C"" - 4.785^@""C"" = color(green)(45.2^@""C"")#
The idea here is that you need to use the mass of water, the heat added to the sample, and the specific heat of water to find the resulting change in temperature.
The equation that establishes a relationship between heat absorbed and change in temperature looks like this
#color(blue)(q = m * c * DeltaT)"" ""# , where
Plug in your values and solve for
#q = m * c * DeltaT implies DeltaT = q/(m * c)#
#DeltaT = (1.0 * 10^(3)color(red)(cancel(color(black)(""J""))))/(50.0color(red)(cancel(color(black)(""g""))) * 4.18color(red)(cancel(color(black)(""J"")))/(color(red)(cancel(color(black)(""g""))) """"^@""C"")) = 4.785^@""C""#
So, if the temperature of the water changed by
#DeltaT = T_""f"" - T_""i"" implies T_""i"" = T_""f"" - DeltaT#
#T_""i"" = 50.0^@""C"" - 4.785^@""C"" = color(green)(45.2^@""C"")#
The idea here is that you need to use the mass of water, the heat added to the sample, and the specific heat of water to find the resulting change in temperature.
The equation that establishes a relationship between heat absorbed and change in temperature looks like this
#color(blue)(q = m * c * DeltaT)"" ""# , where
Plug in your values and solve for
#q = m * c * DeltaT implies DeltaT = q/(m * c)#
#DeltaT = (1.0 * 10^(3)color(red)(cancel(color(black)(""J""))))/(50.0color(red)(cancel(color(black)(""g""))) * 4.18color(red)(cancel(color(black)(""J"")))/(color(red)(cancel(color(black)(""g""))) """"^@""C"")) = 4.785^@""C""#
So, if the temperature of the water changed by
#DeltaT = T_""f"" - T_""i"" implies T_""i"" = T_""f"" - DeltaT#
#T_""i"" = 50.0^@""C"" - 4.785^@""C"" = color(green)(45.2^@""C"")#
A monoprotic acid can contribute one proton to the neutralization reaction that takes place when sodium hydroxide,
If you take
#""HA""_ ((aq)) + ""NaOH""_ ((aq)) -> ""NaA""_ ((aq)) + ""H""_ 2""O""_ ((l))#
Iin order to have a complete neutralization, you need to add equal numbers of acid and of base.
As you know, molarity is defined as the number of moles of solute present in one liter of solution. You already know the molarity of the sodium hydroxide solution.
Now, let's assume that the monoprotic acid solution has a molarity equal to
However, you know that the acid solution has a volume of
#(45.5 color(red)(cancel(color(black)(""mL""))))/(10.0 color(red)(cancel(color(black)(""mL"")))) = color(blue)(4.55)#
smaller than what you'd need if the acid solution had the same molarity as the sodium hydroxide solution. This can only mean that the acid solution is
You thus have
#c_(""HA"") = color(blue)(4.55) * ""0.200 M"" = color(green)(|bar(ul(color(white)(a/a)color(black)(""0.910 M"")color(white)(a/a)|)))#
Therefore, you can say for a fact that
A monoprotic acid can contribute one proton to the neutralization reaction that takes place when sodium hydroxide,
If you take
#""HA""_ ((aq)) + ""NaOH""_ ((aq)) -> ""NaA""_ ((aq)) + ""H""_ 2""O""_ ((l))#
Iin order to have a complete neutralization, you need to add equal numbers of acid and of base.
As you know, molarity is defined as the number of moles of solute present in one liter of solution. You already know the molarity of the sodium hydroxide solution.
Now, let's assume that the monoprotic acid solution has a molarity equal to
However, you know that the acid solution has a volume of
#(45.5 color(red)(cancel(color(black)(""mL""))))/(10.0 color(red)(cancel(color(black)(""mL"")))) = color(blue)(4.55)#
smaller than what you'd need if the acid solution had the same molarity as the sodium hydroxide solution. This can only mean that the acid solution is
You thus have
#c_(""HA"") = color(blue)(4.55) * ""0.200 M"" = color(green)(|bar(ul(color(white)(a/a)color(black)(""0.910 M"")color(white)(a/a)|)))#
Therefore, you can say for a fact that
A monoprotic acid can contribute one proton to the neutralization reaction that takes place when sodium hydroxide,
If you take
#""HA""_ ((aq)) + ""NaOH""_ ((aq)) -> ""NaA""_ ((aq)) + ""H""_ 2""O""_ ((l))#
Iin order to have a complete neutralization, you need to add equal numbers of acid and of base.
As you know, molarity is defined as the number of moles of solute present in one liter of solution. You already know the molarity of the sodium hydroxide solution.
Now, let's assume that the monoprotic acid solution has a molarity equal to
However, you know that the acid solution has a volume of
#(45.5 color(red)(cancel(color(black)(""mL""))))/(10.0 color(red)(cancel(color(black)(""mL"")))) = color(blue)(4.55)#
smaller than what you'd need if the acid solution had the same molarity as the sodium hydroxide solution. This can only mean that the acid solution is
You thus have
#c_(""HA"") = color(blue)(4.55) * ""0.200 M"" = color(green)(|bar(ul(color(white)(a/a)color(black)(""0.910 M"")color(white)(a/a)|)))#
Therefore, you can say for a fact that
And thus
So
And thus
So
And thus
So
The molecule iron (III) sulfide is an ionic molecule between the metal Iron and the non-metal sulfur.
In an ionic molecule the formula is determined by balancing the charges on the ions.
For this case the roman numeral (III) tells us the iron has a charge of
The the element sulfur has a charge of
To balance these ion charges it takes two
The molecule formula is
The molecule formula is
The molecule iron (III) sulfide is an ionic molecule between the metal Iron and the non-metal sulfur.
In an ionic molecule the formula is determined by balancing the charges on the ions.
For this case the roman numeral (III) tells us the iron has a charge of
The the element sulfur has a charge of
To balance these ion charges it takes two
The molecule formula is
The molecule formula is
The molecule iron (III) sulfide is an ionic molecule between the metal Iron and the non-metal sulfur.
In an ionic molecule the formula is determined by balancing the charges on the ions.
For this case the roman numeral (III) tells us the iron has a charge of
The the element sulfur has a charge of
To balance these ion charges it takes two
The molecule formula is
As for any chemical reaction, it is balanced stoichiometrically. What does ""
As for any chemical reaction, it is balanced stoichiometrically. What does ""
As for any chemical reaction, it is balanced stoichiometrically. What does ""
The idea here is that you need to use the number of moles of anhydrous barium iodide,
So, divide both values by the smallest one to find
#""For BaI""_2: color(white)(a)(0.0243 color(red)(cancel(color(black)(""moles""))))/(0.0243color(red)(cancel(color(black)(""moles"")))) = 1#
#""For H""_2""O: "" (0.098 color(red)(cancel(color(black)(""moles""))))/(0.0243color(red)(cancel(color(black)(""moles"")))) = 4.033 ~~ 4#
This means that one formula unit of this hydrate will contain
#color(green)(|bar(ul(color(white)(a/a)color(black)(""BaI""_2 * 4""H""_2""O"")color(white)(a/a)|))) -># barium iodide tetrahydrate
The idea here is that you need to use the number of moles of anhydrous barium iodide,
So, divide both values by the smallest one to find
#""For BaI""_2: color(white)(a)(0.0243 color(red)(cancel(color(black)(""moles""))))/(0.0243color(red)(cancel(color(black)(""moles"")))) = 1#
#""For H""_2""O: "" (0.098 color(red)(cancel(color(black)(""moles""))))/(0.0243color(red)(cancel(color(black)(""moles"")))) = 4.033 ~~ 4#
This means that one formula unit of this hydrate will contain
#color(green)(|bar(ul(color(white)(a/a)color(black)(""BaI""_2 * 4""H""_2""O"")color(white)(a/a)|))) -># barium iodide tetrahydrate
The idea here is that you need to use the number of moles of anhydrous barium iodide,
So, divide both values by the smallest one to find
#""For BaI""_2: color(white)(a)(0.0243 color(red)(cancel(color(black)(""moles""))))/(0.0243color(red)(cancel(color(black)(""moles"")))) = 1#
#""For H""_2""O: "" (0.098 color(red)(cancel(color(black)(""moles""))))/(0.0243color(red)(cancel(color(black)(""moles"")))) = 4.033 ~~ 4#
This means that one formula unit of this hydrate will contain
#color(green)(|bar(ul(color(white)(a/a)color(black)(""BaI""_2 * 4""H""_2""O"")color(white)(a/a)|))) -># barium iodide tetrahydrate
The idea here is that you need to find the number of moles of particles of solute present in your solution.
Knowing this will allow you to find the mole fraction of the solvent, which will then get you the vapor pressure of the solution.
Notice that a formula unit of calcium nitrate,
- one mole of calcium cations,
#1 xx ""Ca""^(2+)# - two moles of nitrate anions,
#2 xx ""NO""_3^(-)#
This means that every mole of calcium nitrate that dissociates in aqueous solution will produce one mole of calcium cations and two moles of nitrate anions
#""Ca""(""NO""_ 3)_ (2(aq)) -> ""Ca""_ ((aq))^(2+) + 2""NO""_(3(aq))^(-)#
In other words, for every mole of calcium nitrate that you dissolve in solution, you get three moles of solute particles, i.e .of ions.
Now, use calcium nitrate's molar mass to determine how many moles you're adding to the solution
#14 color(red)(cancel(color(black)(""g""))) * (""1 mole Ca""(""NO""_3)_2)/(164.09color(red)(cancel(color(black)(""g"")))) = ""0.0853 moles Ca""(""NO""_3)_2#
At
#0.0853color(red)(cancel(color(black)(""moles Ca""(""NO""_3)_2))) * ""3 moles ions""/(1color(red)(cancel(color(black)(""mole Ca""(""NO""_3)_2)))) = ""0.256 moles ions""#
However, you know that the salt has a
#0.256 color(red)(cancel(color(black)(""moles ions""))) * overbrace(""70 moles ions""/(100color(red)(cancel(color(black)(""moles ions"")))))^(color(purple)(""= 70% dissociation"")) = ""0.179 moles ions""#
Use water's molar mass to determine how many moles of solvent you have present in the solution
#200color(red)(cancel(color(black)(""g""))) * (""1 mole H""_2""O"")/(18.015color(red)(cancel(color(black)(""g"")))) = ""11.1 moles H""_2""O""#
Now, according to Raoult's Law, the vapor pressure of a solution that contains a non-volatile solute will depend on the mole fraction of the solvent,
#color(blue)(|bar(ul(color(white)(a/a)P_""solution"" = chi_""solvent"" xx P_""solvent""^@color(white)(a/a)|)))#
The mole fraction of water will be equal to the number of moles of water divided by the total number of moles present in solution
#chi_""water"" = (11.1 color(red)(cancel(color(black)(""moles""))))/((11.1 + 0.179color(red)(cancel(color(black)(""moles""))))) = 0.984#
The vapor pressure of the solution will thus be
#P_""solution"" = 0.984 xx ""760 mmHg"" = color(green)(|bar(ul(color(white)(a/a)""745 mmHg""color(white)(a/a)|)))#
I'll leave the answer rounded to three sig figs, despite the fact that your values do not justify using this many sig figs.
The idea here is that you need to find the number of moles of particles of solute present in your solution.
Knowing this will allow you to find the mole fraction of the solvent, which will then get you the vapor pressure of the solution.
Notice that a formula unit of calcium nitrate,
- one mole of calcium cations,
#1 xx ""Ca""^(2+)# - two moles of nitrate anions,
#2 xx ""NO""_3^(-)#
This means that every mole of calcium nitrate that dissociates in aqueous solution will produce one mole of calcium cations and two moles of nitrate anions
#""Ca""(""NO""_ 3)_ (2(aq)) -> ""Ca""_ ((aq))^(2+) + 2""NO""_(3(aq))^(-)#
In other words, for every mole of calcium nitrate that you dissolve in solution, you get three moles of solute particles, i.e .of ions.
Now, use calcium nitrate's molar mass to determine how many moles you're adding to the solution
#14 color(red)(cancel(color(black)(""g""))) * (""1 mole Ca""(""NO""_3)_2)/(164.09color(red)(cancel(color(black)(""g"")))) = ""0.0853 moles Ca""(""NO""_3)_2#
At
#0.0853color(red)(cancel(color(black)(""moles Ca""(""NO""_3)_2))) * ""3 moles ions""/(1color(red)(cancel(color(black)(""mole Ca""(""NO""_3)_2)))) = ""0.256 moles ions""#
However, you know that the salt has a
#0.256 color(red)(cancel(color(black)(""moles ions""))) * overbrace(""70 moles ions""/(100color(red)(cancel(color(black)(""moles ions"")))))^(color(purple)(""= 70% dissociation"")) = ""0.179 moles ions""#
Use water's molar mass to determine how many moles of solvent you have present in the solution
#200color(red)(cancel(color(black)(""g""))) * (""1 mole H""_2""O"")/(18.015color(red)(cancel(color(black)(""g"")))) = ""11.1 moles H""_2""O""#
Now, according to Raoult's Law, the vapor pressure of a solution that contains a non-volatile solute will depend on the mole fraction of the solvent,
#color(blue)(|bar(ul(color(white)(a/a)P_""solution"" = chi_""solvent"" xx P_""solvent""^@color(white)(a/a)|)))#
The mole fraction of water will be equal to the number of moles of water divided by the total number of moles present in solution
#chi_""water"" = (11.1 color(red)(cancel(color(black)(""moles""))))/((11.1 + 0.179color(red)(cancel(color(black)(""moles""))))) = 0.984#
The vapor pressure of the solution will thus be
#P_""solution"" = 0.984 xx ""760 mmHg"" = color(green)(|bar(ul(color(white)(a/a)""745 mmHg""color(white)(a/a)|)))#
I'll leave the answer rounded to three sig figs, despite the fact that your values do not justify using this many sig figs.
The idea here is that you need to find the number of moles of particles of solute present in your solution.
Knowing this will allow you to find the mole fraction of the solvent, which will then get you the vapor pressure of the solution.
Notice that a formula unit of calcium nitrate,
- one mole of calcium cations,
#1 xx ""Ca""^(2+)# - two moles of nitrate anions,
#2 xx ""NO""_3^(-)#
This means that every mole of calcium nitrate that dissociates in aqueous solution will produce one mole of calcium cations and two moles of nitrate anions
#""Ca""(""NO""_ 3)_ (2(aq)) -> ""Ca""_ ((aq))^(2+) + 2""NO""_(3(aq))^(-)#
In other words, for every mole of calcium nitrate that you dissolve in solution, you get three moles of solute particles, i.e .of ions.
Now, use calcium nitrate's molar mass to determine how many moles you're adding to the solution
#14 color(red)(cancel(color(black)(""g""))) * (""1 mole Ca""(""NO""_3)_2)/(164.09color(red)(cancel(color(black)(""g"")))) = ""0.0853 moles Ca""(""NO""_3)_2#
At
#0.0853color(red)(cancel(color(black)(""moles Ca""(""NO""_3)_2))) * ""3 moles ions""/(1color(red)(cancel(color(black)(""mole Ca""(""NO""_3)_2)))) = ""0.256 moles ions""#
However, you know that the salt has a
#0.256 color(red)(cancel(color(black)(""moles ions""))) * overbrace(""70 moles ions""/(100color(red)(cancel(color(black)(""moles ions"")))))^(color(purple)(""= 70% dissociation"")) = ""0.179 moles ions""#
Use water's molar mass to determine how many moles of solvent you have present in the solution
#200color(red)(cancel(color(black)(""g""))) * (""1 mole H""_2""O"")/(18.015color(red)(cancel(color(black)(""g"")))) = ""11.1 moles H""_2""O""#
Now, according to Raoult's Law, the vapor pressure of a solution that contains a non-volatile solute will depend on the mole fraction of the solvent,
#color(blue)(|bar(ul(color(white)(a/a)P_""solution"" = chi_""solvent"" xx P_""solvent""^@color(white)(a/a)|)))#
The mole fraction of water will be equal to the number of moles of water divided by the total number of moles present in solution
#chi_""water"" = (11.1 color(red)(cancel(color(black)(""moles""))))/((11.1 + 0.179color(red)(cancel(color(black)(""moles""))))) = 0.984#
The vapor pressure of the solution will thus be
#P_""solution"" = 0.984 xx ""760 mmHg"" = color(green)(|bar(ul(color(white)(a/a)""745 mmHg""color(white)(a/a)|)))#
I'll leave the answer rounded to three sig figs, despite the fact that your values do not justify using this many sig figs.
And so the oxides are reduced to iron...
And aluminim metal is oxidized to alumina...
And we add
And now we cancel common reagents...
And finally....
And this looks balanced to me... but please check my 'rithmetic; all care taken but no responsibility admitted. And this is an important reaction practically. Where there is a join in the track of a railway line, they fill it with the iron oxide and alumina, and have at it with a blow torch. The aluminum oxidizes to alumina (and is expelled from the join), and the molten iron welds the track together....
Well...
And so the oxides are reduced to iron...
And aluminim metal is oxidized to alumina...
And we add
And now we cancel common reagents...
And finally....
And this looks balanced to me... but please check my 'rithmetic; all care taken but no responsibility admitted. And this is an important reaction practically. Where there is a join in the track of a railway line, they fill it with the iron oxide and alumina, and have at it with a blow torch. The aluminum oxidizes to alumina (and is expelled from the join), and the molten iron welds the track together....
Well...
And so the oxides are reduced to iron...
And aluminim metal is oxidized to alumina...
And we add
And now we cancel common reagents...
And finally....
And this looks balanced to me... but please check my 'rithmetic; all care taken but no responsibility admitted. And this is an important reaction practically. Where there is a join in the track of a railway line, they fill it with the iron oxide and alumina, and have at it with a blow torch. The aluminum oxidizes to alumina (and is expelled from the join), and the molten iron welds the track together....
To balance the reaction, if it is happening at the solid state (decomposition):
However, to balance the reaction, if it is happening in aqueous medium:
We should recognize that this is a reduction half equation since the oxidation number of oxygen is increasing from
The oxidation numbers for
To balance this reaction then:
Therefore, in order for this reaction to occur, you will need a reducing agent which will reduce the oxidizing agent
The oxidation half equation will then be:
The redox reaction will thus be:
Oxidation:
Reduction:
RedOx:
or
Here is a video that explains how to balance a redox reaction in acidic medium:
Balancing Redox Reactions | Acidic Medium.
And here is another one on how to balance a redox reaction in basic medium:
Balancing Redox Reactions | Basic Medium.
or
To balance the reaction, if it is happening at the solid state (decomposition):
However, to balance the reaction, if it is happening in aqueous medium:
We should recognize that this is a reduction half equation since the oxidation number of oxygen is increasing from
The oxidation numbers for
To balance this reaction then:
Therefore, in order for this reaction to occur, you will need a reducing agent which will reduce the oxidizing agent
The oxidation half equation will then be:
The redox reaction will thus be:
Oxidation:
Reduction:
RedOx:
or
Here is a video that explains how to balance a redox reaction in acidic medium:
Balancing Redox Reactions | Acidic Medium.
And here is another one on how to balance a redox reaction in basic medium:
Balancing Redox Reactions | Basic Medium.
or
To balance the reaction, if it is happening at the solid state (decomposition):
However, to balance the reaction, if it is happening in aqueous medium:
We should recognize that this is a reduction half equation since the oxidation number of oxygen is increasing from
The oxidation numbers for
To balance this reaction then:
Therefore, in order for this reaction to occur, you will need a reducing agent which will reduce the oxidizing agent
The oxidation half equation will then be:
The redox reaction will thus be:
Oxidation:
Reduction:
RedOx:
or
Here is a video that explains how to balance a redox reaction in acidic medium:
Balancing Redox Reactions | Acidic Medium.
And here is another one on how to balance a redox reaction in basic medium:
Balancing Redox Reactions | Basic Medium.
Start off with what you are given.
If you know your gas laws, you have to utilise a certain gas law called Charles' Law:
Remember to convert Celsius values to Kelvin whenever you are dealing with gas problems. This can be done by adding 273 to whatever value in Celsius you have.
Normally in these types of problems (gas law problems), you are given all the variables but one to solve. In this case, the full setup would look like this:
By cross multiplying, we have...
291
Dividing both sides by 291 to isolate
In my school, we learnt that we use the Kelvin value in temperature to count significant figures, so in this case, the answer should have 3 sigfigs.
Therefore,
3.01 L
Start off with what you are given.
If you know your gas laws, you have to utilise a certain gas law called Charles' Law:
Remember to convert Celsius values to Kelvin whenever you are dealing with gas problems. This can be done by adding 273 to whatever value in Celsius you have.
Normally in these types of problems (gas law problems), you are given all the variables but one to solve. In this case, the full setup would look like this:
By cross multiplying, we have...
291
Dividing both sides by 291 to isolate
In my school, we learnt that we use the Kelvin value in temperature to count significant figures, so in this case, the answer should have 3 sigfigs.
Therefore,
3.01 L
Start off with what you are given.
If you know your gas laws, you have to utilise a certain gas law called Charles' Law:
Remember to convert Celsius values to Kelvin whenever you are dealing with gas problems. This can be done by adding 273 to whatever value in Celsius you have.
Normally in these types of problems (gas law problems), you are given all the variables but one to solve. In this case, the full setup would look like this:
By cross multiplying, we have...
291
Dividing both sides by 291 to isolate
In my school, we learnt that we use the Kelvin value in temperature to count significant figures, so in this case, the answer should have 3 sigfigs.
Therefore,
Your strategy here will be to
use the ideal gas law equation to determine how many moles of oxygen gas you have in that sample
use oxygen gas' molar mass to find how many grams would contain that many moles
The ideal gas law equation looks like this
#color(blue)(PV = nRT)"" ""# , where
Now, before plugging in your values, you need to make sure that the units given to you match those used in the expression of the universal gas constant.
More specifically, you need to have a pressure in atm, a temperature in Kelvin, and a volume in liters, so don't forget to keep track of the unit conversions needed to get you to these units.
So, rearrange the ideal gas law equation to solve for
#PV = nRT implies n = (PV)/(RT)#
This means that you have
#n = (755/760color(red)(cancel(color(black)(""atm""))) * 125 * 10^(-3)color(red)(cancel(color(black)(""L""))))/(0.0821(color(red)(cancel(color(black)(""atm""))) * color(red)(cancel(color(black)(""L""))))/(""mol"" * color(red)(cancel(color(black)(""K"")))) * (273.15 + 30.0)color(red)(cancel(color(black)(""K"")))) = ""0.004989 moles O""_2#
As you know, a substance's molar mass tells you what the mass of one mole of that substance is. In this case, oxygen has a molar mass of
This means that you sample will have a mass of
#0.004989 color(red)(cancel(color(black)(""moles""))) * ""31.9988 g""/(1color(red)(cancel(color(black)(""mole"")))) = color(green)(""0.160 g O""_2)#
Your strategy here will be to
use the ideal gas law equation to determine how many moles of oxygen gas you have in that sample
use oxygen gas' molar mass to find how many grams would contain that many moles
The ideal gas law equation looks like this
#color(blue)(PV = nRT)"" ""# , where
Now, before plugging in your values, you need to make sure that the units given to you match those used in the expression of the universal gas constant.
More specifically, you need to have a pressure in atm, a temperature in Kelvin, and a volume in liters, so don't forget to keep track of the unit conversions needed to get you to these units.
So, rearrange the ideal gas law equation to solve for
#PV = nRT implies n = (PV)/(RT)#
This means that you have
#n = (755/760color(red)(cancel(color(black)(""atm""))) * 125 * 10^(-3)color(red)(cancel(color(black)(""L""))))/(0.0821(color(red)(cancel(color(black)(""atm""))) * color(red)(cancel(color(black)(""L""))))/(""mol"" * color(red)(cancel(color(black)(""K"")))) * (273.15 + 30.0)color(red)(cancel(color(black)(""K"")))) = ""0.004989 moles O""_2#
As you know, a substance's molar mass tells you what the mass of one mole of that substance is. In this case, oxygen has a molar mass of
This means that you sample will have a mass of
#0.004989 color(red)(cancel(color(black)(""moles""))) * ""31.9988 g""/(1color(red)(cancel(color(black)(""mole"")))) = color(green)(""0.160 g O""_2)#
Your strategy here will be to
use the ideal gas law equation to determine how many moles of oxygen gas you have in that sample
use oxygen gas' molar mass to find how many grams would contain that many moles
The ideal gas law equation looks like this
#color(blue)(PV = nRT)"" ""# , where
Now, before plugging in your values, you need to make sure that the units given to you match those used in the expression of the universal gas constant.
More specifically, you need to have a pressure in atm, a temperature in Kelvin, and a volume in liters, so don't forget to keep track of the unit conversions needed to get you to these units.
So, rearrange the ideal gas law equation to solve for
#PV = nRT implies n = (PV)/(RT)#
This means that you have
#n = (755/760color(red)(cancel(color(black)(""atm""))) * 125 * 10^(-3)color(red)(cancel(color(black)(""L""))))/(0.0821(color(red)(cancel(color(black)(""atm""))) * color(red)(cancel(color(black)(""L""))))/(""mol"" * color(red)(cancel(color(black)(""K"")))) * (273.15 + 30.0)color(red)(cancel(color(black)(""K"")))) = ""0.004989 moles O""_2#
As you know, a substance's molar mass tells you what the mass of one mole of that substance is. In this case, oxygen has a molar mass of
This means that you sample will have a mass of
#0.004989 color(red)(cancel(color(black)(""moles""))) * ""31.9988 g""/(1color(red)(cancel(color(black)(""mole"")))) = color(green)(""0.160 g O""_2)#
To obtain the mass of water, let's use the equation below:
Based on the information you've provided, we know the following variables:
All we have to do is rearrange the equation to solve for m. This can be accomplished by dividing both sides by
Now, we just plug in the known values:
To obtain the mass of water, let's use the equation below:
Based on the information you've provided, we know the following variables:
All we have to do is rearrange the equation to solve for m. This can be accomplished by dividing both sides by
Now, we just plug in the known values:
To obtain the mass of water, let's use the equation below:
Based on the information you've provided, we know the following variables:
All we have to do is rearrange the equation to solve for m. This can be accomplished by dividing both sides by
Now, we just plug in the known values:
The answer to this problem depends on whether or not you should approximate the density of water to be equal to
Since no information about density was provide, I assume that this is what you must do. However, it's important to note that water's density varies with temperature, and that the value
The idea here is that the heat lost by the hot water sample will be equal to the heat absorbed by the room-temperature water sample.
#color(blue)(-q_""lost"" = q_""absorbed"")"" "" "" ""color(orange)(""(*)"")#
The minus sign is used here because heat lost carries a negative sign.
Your go-to equation here will be
#color(blue)(|bar(ul(color(white)(a/a)q = m * c * DeltaTcolor(white)(a/a)|)))"" ""# , where
Since you're dealing with two samples of water, you don't need to know the value of water's specific heat to solve for the final temperature of the mixture,
So, the change in temperature for the two samples will be
#""For the hot sample: "" DeltaT_""hot"" = T_f - 95.00^@""C""#
#""For the warm sample: "" DeltaT_""warm"" = T_f - 25.00^@""C""#
If you take the density to be equal to
#360.0 color(red)(cancel(color(black)(""mL""))) * ""1.0 g""/(1color(red)(cancel(color(black)(""mL"")))) = ""360.0 g"" "" ""# and#"" ""120.0color(red)(cancel(color(black)(""mL""))) * ""1.0 g""/(1color(red)(cancel(color(black)(""mL"")))) = ""120.0 g""#
Use equation
#overbrace(-120.0color(red)(cancel(color(black)(""g""))) * color(red)(cancel(color(black)(c_""water""))) * (T_f - 95.00^@""C""))^(color(purple)(""heat lost by the hot sample"")) = overbrace(360.0color(red)(cancel(color(black)(""g""))) * color(red)(cancel(color(black)(c_""water""))) * (T_f - 25.00)^@""C"")^(color(blue)(""heat gained by warm sample""))#
This will get you
#-120.0 * T_f + 11400^@""C"" = 360.0 * T_f - 9000^@""C""#
#480.0 * T_f = 20400^@""C""#
#T_f = (20400^@""C"")/480.0 = color(green)(|bar(ul(color(white)(a/a)42.50^@""C""color(white)(a/a)|)))#
The answer is rounded to four sig figs.
SIDE NOTE If you use the actual densities of water at
#T_f = 38.76^@""C""#
As practice, you should try redoing the calculations using the actual densities of water at those two temperatures. You can find info on that here
http://antoine.frostburg.edu/chem/senese/javascript/water-density.html
The answer to this problem depends on whether or not you should approximate the density of water to be equal to
Since no information about density was provide, I assume that this is what you must do. However, it's important to note that water's density varies with temperature, and that the value
The idea here is that the heat lost by the hot water sample will be equal to the heat absorbed by the room-temperature water sample.
#color(blue)(-q_""lost"" = q_""absorbed"")"" "" "" ""color(orange)(""(*)"")#
The minus sign is used here because heat lost carries a negative sign.
Your go-to equation here will be
#color(blue)(|bar(ul(color(white)(a/a)q = m * c * DeltaTcolor(white)(a/a)|)))"" ""# , where
Since you're dealing with two samples of water, you don't need to know the value of water's specific heat to solve for the final temperature of the mixture,
So, the change in temperature for the two samples will be
#""For the hot sample: "" DeltaT_""hot"" = T_f - 95.00^@""C""#
#""For the warm sample: "" DeltaT_""warm"" = T_f - 25.00^@""C""#
If you take the density to be equal to
#360.0 color(red)(cancel(color(black)(""mL""))) * ""1.0 g""/(1color(red)(cancel(color(black)(""mL"")))) = ""360.0 g"" "" ""# and#"" ""120.0color(red)(cancel(color(black)(""mL""))) * ""1.0 g""/(1color(red)(cancel(color(black)(""mL"")))) = ""120.0 g""#
Use equation
#overbrace(-120.0color(red)(cancel(color(black)(""g""))) * color(red)(cancel(color(black)(c_""water""))) * (T_f - 95.00^@""C""))^(color(purple)(""heat lost by the hot sample"")) = overbrace(360.0color(red)(cancel(color(black)(""g""))) * color(red)(cancel(color(black)(c_""water""))) * (T_f - 25.00)^@""C"")^(color(blue)(""heat gained by warm sample""))#
This will get you
#-120.0 * T_f + 11400^@""C"" = 360.0 * T_f - 9000^@""C""#
#480.0 * T_f = 20400^@""C""#
#T_f = (20400^@""C"")/480.0 = color(green)(|bar(ul(color(white)(a/a)42.50^@""C""color(white)(a/a)|)))#
The answer is rounded to four sig figs.
SIDE NOTE If you use the actual densities of water at
#T_f = 38.76^@""C""#
As practice, you should try redoing the calculations using the actual densities of water at those two temperatures. You can find info on that here
http://antoine.frostburg.edu/chem/senese/javascript/water-density.html
The answer to this problem depends on whether or not you should approximate the density of water to be equal to
Since no information about density was provide, I assume that this is what you must do. However, it's important to note that water's density varies with temperature, and that the value
The idea here is that the heat lost by the hot water sample will be equal to the heat absorbed by the room-temperature water sample.
#color(blue)(-q_""lost"" = q_""absorbed"")"" "" "" ""color(orange)(""(*)"")#
The minus sign is used here because heat lost carries a negative sign.
Your go-to equation here will be
#color(blue)(|bar(ul(color(white)(a/a)q = m * c * DeltaTcolor(white)(a/a)|)))"" ""# , where
Since you're dealing with two samples of water, you don't need to know the value of water's specific heat to solve for the final temperature of the mixture,
So, the change in temperature for the two samples will be
#""For the hot sample: "" DeltaT_""hot"" = T_f - 95.00^@""C""#
#""For the warm sample: "" DeltaT_""warm"" = T_f - 25.00^@""C""#
If you take the density to be equal to
#360.0 color(red)(cancel(color(black)(""mL""))) * ""1.0 g""/(1color(red)(cancel(color(black)(""mL"")))) = ""360.0 g"" "" ""# and#"" ""120.0color(red)(cancel(color(black)(""mL""))) * ""1.0 g""/(1color(red)(cancel(color(black)(""mL"")))) = ""120.0 g""#
Use equation
#overbrace(-120.0color(red)(cancel(color(black)(""g""))) * color(red)(cancel(color(black)(c_""water""))) * (T_f - 95.00^@""C""))^(color(purple)(""heat lost by the hot sample"")) = overbrace(360.0color(red)(cancel(color(black)(""g""))) * color(red)(cancel(color(black)(c_""water""))) * (T_f - 25.00)^@""C"")^(color(blue)(""heat gained by warm sample""))#
This will get you
#-120.0 * T_f + 11400^@""C"" = 360.0 * T_f - 9000^@""C""#
#480.0 * T_f = 20400^@""C""#
#T_f = (20400^@""C"")/480.0 = color(green)(|bar(ul(color(white)(a/a)42.50^@""C""color(white)(a/a)|)))#
The answer is rounded to four sig figs.
SIDE NOTE If you use the actual densities of water at
#T_f = 38.76^@""C""#
As practice, you should try redoing the calculations using the actual densities of water at those two temperatures. You can find info on that here
http://antoine.frostburg.edu/chem/senese/javascript/water-density.html
Molarity is simply a measure of the concentration of a solution in terms of how many moles of solute are present per liter of solution.
#color(blue)(|bar(ul(""molarity"" = ""moles of solute""/""liters of solution""))|)#
This means that in order to find a solution's molarity you must know two things
- the number of moles of solute present
- the volume of the solution
In order to find the number of moles of silver nitrate,
#10.0 color(red)(cancel(color(black)(""g""))) * ""1 mole AgNO""_3/(169.87color(red)(cancel(color(black)(""g"")))) = ""0.05887 moles AgNO""_3#
Now, do not forget that molarity is expressed per liter of solution. This means that you're going to have to convert the volume from milliliters to liters by using the conversion factor
#""1 L"" = 10^3""mL""#
Plug in your values to get
#color(blue)( |bar(ul(c = n_""solute""/V_""solution""))|)#
#c = ""0.05887 moles""/(500. * 10^(-3)""L"") = color(green)(|bar(ul(""0.118 M""))|)#
The answer is rounded to three sig figs.
Molarity is simply a measure of the concentration of a solution in terms of how many moles of solute are present per liter of solution.
#color(blue)(|bar(ul(""molarity"" = ""moles of solute""/""liters of solution""))|)#
This means that in order to find a solution's molarity you must know two things
- the number of moles of solute present
- the volume of the solution
In order to find the number of moles of silver nitrate,
#10.0 color(red)(cancel(color(black)(""g""))) * ""1 mole AgNO""_3/(169.87color(red)(cancel(color(black)(""g"")))) = ""0.05887 moles AgNO""_3#
Now, do not forget that molarity is expressed per liter of solution. This means that you're going to have to convert the volume from milliliters to liters by using the conversion factor
#""1 L"" = 10^3""mL""#
Plug in your values to get
#color(blue)( |bar(ul(c = n_""solute""/V_""solution""))|)#
#c = ""0.05887 moles""/(500. * 10^(-3)""L"") = color(green)(|bar(ul(""0.118 M""))|)#
The answer is rounded to three sig figs.
Molarity is simply a measure of the concentration of a solution in terms of how many moles of solute are present per liter of solution.
#color(blue)(|bar(ul(""molarity"" = ""moles of solute""/""liters of solution""))|)#
This means that in order to find a solution's molarity you must know two things
- the number of moles of solute present
- the volume of the solution
In order to find the number of moles of silver nitrate,
#10.0 color(red)(cancel(color(black)(""g""))) * ""1 mole AgNO""_3/(169.87color(red)(cancel(color(black)(""g"")))) = ""0.05887 moles AgNO""_3#
Now, do not forget that molarity is expressed per liter of solution. This means that you're going to have to convert the volume from milliliters to liters by using the conversion factor
#""1 L"" = 10^3""mL""#
Plug in your values to get
#color(blue)( |bar(ul(c = n_""solute""/V_""solution""))|)#
#c = ""0.05887 moles""/(500. * 10^(-3)""L"") = color(green)(|bar(ul(""0.118 M""))|)#
The answer is rounded to three sig figs.
The oxdiation number of manganese in the anion
This is the manganate ion, where manganese has a
The oxdiation number of manganese in the anion
This is the manganate ion, where manganese has a
The oxdiation number of manganese in the anion
The first thing you need to realize is there are two bonds in this formula: the ionic bond between
With this kind of substances, it would be helpful if you separate the ions first since it would break down your equation into a more simpler one.
Hence,
Now that you separated the ions, you can safely ignore the ammonium ion for the time being, as it would not affect your calculations for the oxidation state of
Thus,
Thus we have the equation:
Therefore, the oxidation state of
Still in doubt? Let's try solving
[(+1) (2)] +
(+2) +
Same answer. :)
I would use fractions for
The volume increases of
the percentual increase will be:
I got
I would use fractions for
The volume increases of
the percentual increase will be:
I got
I would use fractions for
The volume increases of
the percentual increase will be:
The general pattern for solving these types of problems is
So to solve the problem:
Molecular mass of
= 122.55
Moles of
mol ratio:
so
mol of KCl =
mol of KCl =
mol of KCl = 0.204 mol
4.. molecular mass of KCl = 39.10 + 35.45 = 74.55
mass KCl =
Therefore 15.2 g of KCl would be formed.
15.2 g of KCl would be produced
The general pattern for solving these types of problems is
So to solve the problem:
Molecular mass of
= 122.55
Moles of
mol ratio:
so
mol of KCl =
mol of KCl =
mol of KCl = 0.204 mol
4.. molecular mass of KCl = 39.10 + 35.45 = 74.55
mass KCl =
Therefore 15.2 g of KCl would be formed.
15.2 g of KCl would be produced
The general pattern for solving these types of problems is
So to solve the problem:
Molecular mass of
= 122.55
Moles of
mol ratio:
so
mol of KCl =
mol of KCl =
mol of KCl = 0.204 mol
4.. molecular mass of KCl = 39.10 + 35.45 = 74.55
mass KCl =
Therefore 15.2 g of KCl would be formed.
I include the units, even tho it is a bit a pfaff, to ensure that I have done the right order of operation. It is all too easy to divide when we should have multiplied, and vice versa. We want an answer with units of concentration, i.e.
Remember
I include the units, even tho it is a bit a pfaff, to ensure that I have done the right order of operation. It is all too easy to divide when we should have multiplied, and vice versa. We want an answer with units of concentration, i.e.
Remember
I include the units, even tho it is a bit a pfaff, to ensure that I have done the right order of operation. It is all too easy to divide when we should have multiplied, and vice versa. We want an answer with units of concentration, i.e.
Remember
Your strategy here will be to use the volume and molarity of the solution to determine how many moles of ammonium sulfate,
Once you know the number of moles of solute needed to make that solution, use ammonium sulfate's molar mass to determine how many grams would contain that many moles.
So, a solution's molarity tells you how many moles of solute you get per liter of solution.
In this case, the solution must have a molarity of
Use this value as a conversion factor to find the number of moles that you'd get in
#0.25 color(red)(cancel(color(black)(""L solution""))) * (""6 moles""color(white)(a) (""NH""_4)_2""SO""_4)/(1color(red)(cancel(color(black)(""L solution"")))) = ""1.5 moles"" color(white)(a)(""NH""_4)_2""SO""_4#
Now use ammonium sulfate's molar mass, which is listed as being equal to
#1.5color(red)(cancel(color(black)(""moles""color(white)(a)(""NH""_4)_2""SO""_4))) * ""132.14 g""/(1color(red)(cancel(color(black)(""mole""color(white)(a)(""NH""_4)_2""SO""_4)))) = ""198.21 g""#
Rounded to one significant figure, the answer will be
#""mass of ammonium sulfate"" = color(green)(|bar(ul(color(white)(a/a)""200 g""color(white)(a/a)|)))#
Your strategy here will be to use the volume and molarity of the solution to determine how many moles of ammonium sulfate,
Once you know the number of moles of solute needed to make that solution, use ammonium sulfate's molar mass to determine how many grams would contain that many moles.
So, a solution's molarity tells you how many moles of solute you get per liter of solution.
In this case, the solution must have a molarity of
Use this value as a conversion factor to find the number of moles that you'd get in
#0.25 color(red)(cancel(color(black)(""L solution""))) * (""6 moles""color(white)(a) (""NH""_4)_2""SO""_4)/(1color(red)(cancel(color(black)(""L solution"")))) = ""1.5 moles"" color(white)(a)(""NH""_4)_2""SO""_4#
Now use ammonium sulfate's molar mass, which is listed as being equal to
#1.5color(red)(cancel(color(black)(""moles""color(white)(a)(""NH""_4)_2""SO""_4))) * ""132.14 g""/(1color(red)(cancel(color(black)(""mole""color(white)(a)(""NH""_4)_2""SO""_4)))) = ""198.21 g""#
Rounded to one significant figure, the answer will be
#""mass of ammonium sulfate"" = color(green)(|bar(ul(color(white)(a/a)""200 g""color(white)(a/a)|)))#
Your strategy here will be to use the volume and molarity of the solution to determine how many moles of ammonium sulfate,
Once you know the number of moles of solute needed to make that solution, use ammonium sulfate's molar mass to determine how many grams would contain that many moles.
So, a solution's molarity tells you how many moles of solute you get per liter of solution.
In this case, the solution must have a molarity of
Use this value as a conversion factor to find the number of moles that you'd get in
#0.25 color(red)(cancel(color(black)(""L solution""))) * (""6 moles""color(white)(a) (""NH""_4)_2""SO""_4)/(1color(red)(cancel(color(black)(""L solution"")))) = ""1.5 moles"" color(white)(a)(""NH""_4)_2""SO""_4#
Now use ammonium sulfate's molar mass, which is listed as being equal to
#1.5color(red)(cancel(color(black)(""moles""color(white)(a)(""NH""_4)_2""SO""_4))) * ""132.14 g""/(1color(red)(cancel(color(black)(""mole""color(white)(a)(""NH""_4)_2""SO""_4)))) = ""198.21 g""#
Rounded to one significant figure, the answer will be
#""mass of ammonium sulfate"" = color(green)(|bar(ul(color(white)(a/a)""200 g""color(white)(a/a)|)))#
You have to search the limiting reagent between Na and
You have
As 1 mole occupies 22,4 L at STP you have
The limiting reagent is
since from the balanced reaction you have 2 mol of NaCl from one of
67,9 g
You have to search the limiting reagent between Na and
You have
As 1 mole occupies 22,4 L at STP you have
The limiting reagent is
since from the balanced reaction you have 2 mol of NaCl from one of
67,9 g
You have to search the limiting reagent between Na and
You have
As 1 mole occupies 22,4 L at STP you have
The limiting reagent is
since from the balanced reaction you have 2 mol of NaCl from one of
The specific heat capacity of ice is equal to
Note that this is not the same as the ones of water (
Here is a video that explains this topic in much more details:
Thermochemistry | Enthalpy and Calorimetry.
The specific heat capacity of ice is equal to
Note that this is not the same as the ones of water (
Here is a video that explains this topic in much more details:
Thermochemistry | Enthalpy and Calorimetry.
The specific heat capacity of ice is equal to
Note that this is not the same as the ones of water (
Here is a video that explains this topic in much more details:
Thermochemistry | Enthalpy and Calorimetry.
Is the reaction above a redox reaction? What is oxidized, and what is reduced? What are the oxidation states? Nitrous oxide is capable of being further oxidized to
As to your question, we start with
Rxn:
Is the reaction above a redox reaction? What is oxidized, and what is reduced? What are the oxidation states? Nitrous oxide is capable of being further oxidized to
As to your question, we start with
Rxn:
Is the reaction above a redox reaction? What is oxidized, and what is reduced? What are the oxidation states? Nitrous oxide is capable of being further oxidized to
As to your question, we start with
The balanced equation for the reaction is
Phosphoric acid
Thus, one molecule of phosphoric acid will need three formula units of sodium hydroxide, and
One mole of phosphoric acid will need three moles of sodium hydroxide.
This is what the balanced equation tells us:
3 mol of
You need 3 mol of sodium hydroxide to neutralize 1 mol of phosphoric acid.
The balanced equation for the reaction is
Phosphoric acid
Thus, one molecule of phosphoric acid will need three formula units of sodium hydroxide, and
One mole of phosphoric acid will need three moles of sodium hydroxide.
This is what the balanced equation tells us:
3 mol of
You need 3 mol of sodium hydroxide to neutralize 1 mol of phosphoric acid.
The balanced equation for the reaction is
Phosphoric acid
Thus, one molecule of phosphoric acid will need three formula units of sodium hydroxide, and
One mole of phosphoric acid will need three moles of sodium hydroxide.
This is what the balanced equation tells us:
3 mol of
H2(g) + I2(g) --> 2HI(g)
deltaH(rxn) = +53 kJ
I happen to know the answer is 5.2 kJ, but I'm not sure how to get to it. I'm studying for finals and need help. Thank you!
The thermochemical equation given to you tells you how much heat is needed in order to produce
In other words, you know that in order to produce
#DeltaH_""rxn"" = + ""53 kJ""#
The plus sign tells you that this reaction taken is
#color(blue)(ul(color(black)(""1 mole I""_2 color(white)(.)""and 1 mole H""_2 -> ""53 kJ of heat consumed"")))#
So, convert the samples of hydrogen gas and iodine to moles by using the molar masses of the two reactants.
#5.0 color(red)(cancel(color(black)(""g""))) * ""1 mole H""_2/(2.016color(red)(cancel(color(black)(""g"")))) = ""2.48 moles H""_2#
#25 color(red)(cancel(color(black)(""g""))) * ""1 mole I""_2/(253.81color(red)(cancel(color(black)(""g"")))) = ""0.0985 moles I""_2#
The reaction consumes hydrogen gas and iodine in a
Therefore, the reaction will consume
This means that the reaction will require
#0.0985 color(red)(cancel(color(black)(""moles I""_2))) * ""53 kJ""/(1color(red)(cancel(color(black)(""mole I""_2)))) = color(darkgreen)(ul(color(black)(""5.2 kJ"")))#
of heat. The answer is rounded to two sig figs. You can thus say that you have
#DeltaH_ (""rxn for 0.0985 moles I""_2 color(white)(.)""and H""_2) = + ""5.2 kJ""#
The thermochemical equation given to you tells you how much heat is needed in order to produce
In other words, you know that in order to produce
#DeltaH_""rxn"" = + ""53 kJ""#
The plus sign tells you that this reaction taken is
#color(blue)(ul(color(black)(""1 mole I""_2 color(white)(.)""and 1 mole H""_2 -> ""53 kJ of heat consumed"")))#
So, convert the samples of hydrogen gas and iodine to moles by using the molar masses of the two reactants.
#5.0 color(red)(cancel(color(black)(""g""))) * ""1 mole H""_2/(2.016color(red)(cancel(color(black)(""g"")))) = ""2.48 moles H""_2#
#25 color(red)(cancel(color(black)(""g""))) * ""1 mole I""_2/(253.81color(red)(cancel(color(black)(""g"")))) = ""0.0985 moles I""_2#
The reaction consumes hydrogen gas and iodine in a
Therefore, the reaction will consume
This means that the reaction will require
#0.0985 color(red)(cancel(color(black)(""moles I""_2))) * ""53 kJ""/(1color(red)(cancel(color(black)(""mole I""_2)))) = color(darkgreen)(ul(color(black)(""5.2 kJ"")))#
of heat. The answer is rounded to two sig figs. You can thus say that you have
#DeltaH_ (""rxn for 0.0985 moles I""_2 color(white)(.)""and H""_2) = + ""5.2 kJ""#
H2(g) + I2(g) --> 2HI(g)
deltaH(rxn) = +53 kJ
I happen to know the answer is 5.2 kJ, but I'm not sure how to get to it. I'm studying for finals and need help. Thank you!
The thermochemical equation given to you tells you how much heat is needed in order to produce
In other words, you know that in order to produce
#DeltaH_""rxn"" = + ""53 kJ""#
The plus sign tells you that this reaction taken is
#color(blue)(ul(color(black)(""1 mole I""_2 color(white)(.)""and 1 mole H""_2 -> ""53 kJ of heat consumed"")))#
So, convert the samples of hydrogen gas and iodine to moles by using the molar masses of the two reactants.
#5.0 color(red)(cancel(color(black)(""g""))) * ""1 mole H""_2/(2.016color(red)(cancel(color(black)(""g"")))) = ""2.48 moles H""_2#
#25 color(red)(cancel(color(black)(""g""))) * ""1 mole I""_2/(253.81color(red)(cancel(color(black)(""g"")))) = ""0.0985 moles I""_2#
The reaction consumes hydrogen gas and iodine in a
Therefore, the reaction will consume
This means that the reaction will require
#0.0985 color(red)(cancel(color(black)(""moles I""_2))) * ""53 kJ""/(1color(red)(cancel(color(black)(""mole I""_2)))) = color(darkgreen)(ul(color(black)(""5.2 kJ"")))#
of heat. The answer is rounded to two sig figs. You can thus say that you have
#DeltaH_ (""rxn for 0.0985 moles I""_2 color(white)(.)""and H""_2) = + ""5.2 kJ""#
We work out (i) the number of moles of
We work out (i) the number of moles of
Of course, we need to find the appropriate gas constant.
This site gives me
Given this, we plug in the numbers:
We should get an answer in
The temperature is so low because there is a large molar quantity of dihydrogen. At this temperature we would anticipate some deviation from ideality. We are in no position to assess this.
Of course, we need to find the appropriate gas constant.
This site gives me
Given this, we plug in the numbers:
We should get an answer in
The temperature is so low because there is a large molar quantity of dihydrogen. At this temperature we would anticipate some deviation from ideality. We are in no position to assess this.
Of course, we need to find the appropriate gas constant.
This site gives me
Given this, we plug in the numbers:
We should get an answer in
The temperature is so low because there is a large molar quantity of dihydrogen. At this temperature we would anticipate some deviation from ideality. We are in no position to assess this.
The key to this problem is the specific heat of iron, which is said to be equal to
#c_""iron"" = ""0.449 J g""^(-1)""""^@""C""^(-1)#
As you know, the specific heat of a substance tells you how much heat is needed to raise the temperature of a sample of
Keep in mind that this represents the amount of heat needed per gram, per degree Celsius!
So, how much heat would be needed to increase the temperature of a
#50.0 color(red)(cancel(color(black)(""g""))) * overbrace(""0.449 J"" /(1color(red)(cancel(color(black)(""g""))) """"^@""C""))^(color(blue)(""the specific heat of iron"")) = ""22.45 J"" """"^@""C""^(-1)#
This means that in order to increase the temperature of
You can thus say that in order to increase the temperature of the block by
#10.0 color(red)(cancel(color(black)(""""^@""C""))) * overbrace(""22.45 J""/(1color(red)(cancel(color(black)(""""^@""C"")))))^(color(blue)(""for 50.0 g of iron"")) = color(green)(bar(ul(|color(white)(a/a)color(black)(""225 J"")color(white)(a/a)|)))#
The answer is rounded to three sig figs.
The key to this problem is the specific heat of iron, which is said to be equal to
#c_""iron"" = ""0.449 J g""^(-1)""""^@""C""^(-1)#
As you know, the specific heat of a substance tells you how much heat is needed to raise the temperature of a sample of
Keep in mind that this represents the amount of heat needed per gram, per degree Celsius!
So, how much heat would be needed to increase the temperature of a
#50.0 color(red)(cancel(color(black)(""g""))) * overbrace(""0.449 J"" /(1color(red)(cancel(color(black)(""g""))) """"^@""C""))^(color(blue)(""the specific heat of iron"")) = ""22.45 J"" """"^@""C""^(-1)#
This means that in order to increase the temperature of
You can thus say that in order to increase the temperature of the block by
#10.0 color(red)(cancel(color(black)(""""^@""C""))) * overbrace(""22.45 J""/(1color(red)(cancel(color(black)(""""^@""C"")))))^(color(blue)(""for 50.0 g of iron"")) = color(green)(bar(ul(|color(white)(a/a)color(black)(""225 J"")color(white)(a/a)|)))#
The answer is rounded to three sig figs.
The key to this problem is the specific heat of iron, which is said to be equal to
#c_""iron"" = ""0.449 J g""^(-1)""""^@""C""^(-1)#
As you know, the specific heat of a substance tells you how much heat is needed to raise the temperature of a sample of
Keep in mind that this represents the amount of heat needed per gram, per degree Celsius!
So, how much heat would be needed to increase the temperature of a
#50.0 color(red)(cancel(color(black)(""g""))) * overbrace(""0.449 J"" /(1color(red)(cancel(color(black)(""g""))) """"^@""C""))^(color(blue)(""the specific heat of iron"")) = ""22.45 J"" """"^@""C""^(-1)#
This means that in order to increase the temperature of
You can thus say that in order to increase the temperature of the block by
#10.0 color(red)(cancel(color(black)(""""^@""C""))) * overbrace(""22.45 J""/(1color(red)(cancel(color(black)(""""^@""C"")))))^(color(blue)(""for 50.0 g of iron"")) = color(green)(bar(ul(|color(white)(a/a)color(black)(""225 J"")color(white)(a/a)|)))#
The answer is rounded to three sig figs.
find the number of co2 that exert a pressure of 785torrs at a volume of 32.5L and a temperature 32c ?
I'm assuming that the question wants you to determine the number of moles of carbon dioxide,
Your tool of choice here will be the ideal gas law equation
#color(blue)(PV = nRT)"" ""# , where
Now, the problem provides you with everything that you need in order to solve for the number of moles of gas. However, notice that some of the units used in the expression of
This means that you're going to have to do a couple of units conversions to get your units to match those used in expression of
#""1 atm "" = "" 760 torr""#
#t [""K""] = 273.15 + t [""""^@""C""]#
Rearrange the ideal gas law equation to solve for
#PV = nRT implies n = (PV)/(RT)#
Plug in your values to get
#n = (785/760color(red)(cancel(color(black)(""atm""))) * 32.5color(red)(cancel(color(black)(""L""))))/(0.0821(color(red)(cancel(color(black)(""atm""))) * color(red)(cancel(color(black)(""L""))))/(""mol"" * color(red)(cancel(color(black)(""K"")))) * (273.15 + 32)color(red)(cancel(color(black)(""K"")))) = ""1.3399 moles""#
Rounded to two sig figs, the number of sig figs you have for the temperature of the gas, the answer will be
#n = color(green)(""1.3 moles CO""_2)#
I'm assuming that the question wants you to determine the number of moles of carbon dioxide,
Your tool of choice here will be the ideal gas law equation
#color(blue)(PV = nRT)"" ""# , where
Now, the problem provides you with everything that you need in order to solve for the number of moles of gas. However, notice that some of the units used in the expression of
This means that you're going to have to do a couple of units conversions to get your units to match those used in expression of
#""1 atm "" = "" 760 torr""#
#t [""K""] = 273.15 + t [""""^@""C""]#
Rearrange the ideal gas law equation to solve for
#PV = nRT implies n = (PV)/(RT)#
Plug in your values to get
#n = (785/760color(red)(cancel(color(black)(""atm""))) * 32.5color(red)(cancel(color(black)(""L""))))/(0.0821(color(red)(cancel(color(black)(""atm""))) * color(red)(cancel(color(black)(""L""))))/(""mol"" * color(red)(cancel(color(black)(""K"")))) * (273.15 + 32)color(red)(cancel(color(black)(""K"")))) = ""1.3399 moles""#
Rounded to two sig figs, the number of sig figs you have for the temperature of the gas, the answer will be
#n = color(green)(""1.3 moles CO""_2)#
find the number of co2 that exert a pressure of 785torrs at a volume of 32.5L and a temperature 32c ?
I'm assuming that the question wants you to determine the number of moles of carbon dioxide,
Your tool of choice here will be the ideal gas law equation
#color(blue)(PV = nRT)"" ""# , where
Now, the problem provides you with everything that you need in order to solve for the number of moles of gas. However, notice that some of the units used in the expression of
This means that you're going to have to do a couple of units conversions to get your units to match those used in expression of
#""1 atm "" = "" 760 torr""#
#t [""K""] = 273.15 + t [""""^@""C""]#
Rearrange the ideal gas law equation to solve for
#PV = nRT implies n = (PV)/(RT)#
Plug in your values to get
#n = (785/760color(red)(cancel(color(black)(""atm""))) * 32.5color(red)(cancel(color(black)(""L""))))/(0.0821(color(red)(cancel(color(black)(""atm""))) * color(red)(cancel(color(black)(""L""))))/(""mol"" * color(red)(cancel(color(black)(""K"")))) * (273.15 + 32)color(red)(cancel(color(black)(""K"")))) = ""1.3399 moles""#
Rounded to two sig figs, the number of sig figs you have for the temperature of the gas, the answer will be
#n = color(green)(""1.3 moles CO""_2)#
The equation relating moles and volume is
Given values are
Substitute the given values into the equation.
The molarity of the solution is
The equation relating moles and volume is
Given values are
Substitute the given values into the equation.
The molarity of the solution is
The equation relating moles and volume is
Given values are
Substitute the given values into the equation.
Normally, this rxn must be calalyzed with a little
We spit out a
This represents a mass of
This is not a reaction that you do at home.
Normally, this rxn must be calalyzed with a little
We spit out a
This represents a mass of
This is not a reaction that you do at home.
Normally, this rxn must be calalyzed with a little
We spit out a
This represents a mass of
This is not a reaction that you do at home.
The molar mass of sodium chloride is
The molar mass of sodium chloride is
The molar mass of sodium chloride is
Number of moles of
There are thus
Number of moles of
There are thus
Number of moles of
There are thus
The idea here is that you need to find the moles of citric acid by using the molarity of the sodium hydroxide solution, then use the mass of the orange juice to find the percent concentration by mass of citric acid.
So, start with the balanced chemical equation for this neutralization reaction
#""C""_5""H""_8""O""_text(7(aq]) + color(red)(3)""NaOH""_text((aq]) -> ""Na""_3""C""_6""H""_5""O""_text(7(aq]) + ""H""_2""O""_text((l])#
Notice that your reaction needs
#color(blue)(c = n/V implies n = c * V)#
#n_""NaOH"" = ""0.090 M"" * 20.4 * 10^(-3)""L"" = ""0.001836 moles NaOH""#
This means that the orange juice contained
#0.001836color(red)(cancel(color(black)(""moles NaOH""))) * ""1 mole citric acid""/(color(red)(3)color(red)(cancel(color(black)(""moles NaOH"")))) = ""0.0006120 moles C""_3""H""_8""O""_7#
Now that you know how many moles of citric acid you had in the orange juice, use its molar mass to determine how many moles of citric acid would contain this many moles
#0.000612color(red)(cancel(color(black)(""moles""))) * ""192.124 g""/(1color(red)(cancel(color(black)(""mole"")))) = ""0.11758 g C""_3""H""_8""O""_7#
Now, a solution's percent concentration by mass is defined as the ratio between the mass of the solute, which in your case is citric acid, and the total mass of the solution, multiplied by
#color(blue)(""%m/m"" = ""mass of solute""/""mass of soluteion"" xx 100)#
In your case, the percent concentration by mass of the orange juice will be
#""%m/m"" = (0.11758color(red)(cancel(color(black)(""g""))))/(20.31color(red)(cancel(color(black)(""g"")))) xx 100 = color(green)(""0.579% C""_3""H""_8""O""_7)#
The idea here is that you need to find the moles of citric acid by using the molarity of the sodium hydroxide solution, then use the mass of the orange juice to find the percent concentration by mass of citric acid.
So, start with the balanced chemical equation for this neutralization reaction
#""C""_5""H""_8""O""_text(7(aq]) + color(red)(3)""NaOH""_text((aq]) -> ""Na""_3""C""_6""H""_5""O""_text(7(aq]) + ""H""_2""O""_text((l])#
Notice that your reaction needs
#color(blue)(c = n/V implies n = c * V)#
#n_""NaOH"" = ""0.090 M"" * 20.4 * 10^(-3)""L"" = ""0.001836 moles NaOH""#
This means that the orange juice contained
#0.001836color(red)(cancel(color(black)(""moles NaOH""))) * ""1 mole citric acid""/(color(red)(3)color(red)(cancel(color(black)(""moles NaOH"")))) = ""0.0006120 moles C""_3""H""_8""O""_7#
Now that you know how many moles of citric acid you had in the orange juice, use its molar mass to determine how many moles of citric acid would contain this many moles
#0.000612color(red)(cancel(color(black)(""moles""))) * ""192.124 g""/(1color(red)(cancel(color(black)(""mole"")))) = ""0.11758 g C""_3""H""_8""O""_7#
Now, a solution's percent concentration by mass is defined as the ratio between the mass of the solute, which in your case is citric acid, and the total mass of the solution, multiplied by
#color(blue)(""%m/m"" = ""mass of solute""/""mass of soluteion"" xx 100)#
In your case, the percent concentration by mass of the orange juice will be
#""%m/m"" = (0.11758color(red)(cancel(color(black)(""g""))))/(20.31color(red)(cancel(color(black)(""g"")))) xx 100 = color(green)(""0.579% C""_3""H""_8""O""_7)#
The idea here is that you need to find the moles of citric acid by using the molarity of the sodium hydroxide solution, then use the mass of the orange juice to find the percent concentration by mass of citric acid.
So, start with the balanced chemical equation for this neutralization reaction
#""C""_5""H""_8""O""_text(7(aq]) + color(red)(3)""NaOH""_text((aq]) -> ""Na""_3""C""_6""H""_5""O""_text(7(aq]) + ""H""_2""O""_text((l])#
Notice that your reaction needs
#color(blue)(c = n/V implies n = c * V)#
#n_""NaOH"" = ""0.090 M"" * 20.4 * 10^(-3)""L"" = ""0.001836 moles NaOH""#
This means that the orange juice contained
#0.001836color(red)(cancel(color(black)(""moles NaOH""))) * ""1 mole citric acid""/(color(red)(3)color(red)(cancel(color(black)(""moles NaOH"")))) = ""0.0006120 moles C""_3""H""_8""O""_7#
Now that you know how many moles of citric acid you had in the orange juice, use its molar mass to determine how many moles of citric acid would contain this many moles
#0.000612color(red)(cancel(color(black)(""moles""))) * ""192.124 g""/(1color(red)(cancel(color(black)(""mole"")))) = ""0.11758 g C""_3""H""_8""O""_7#
Now, a solution's percent concentration by mass is defined as the ratio between the mass of the solute, which in your case is citric acid, and the total mass of the solution, multiplied by
#color(blue)(""%m/m"" = ""mass of solute""/""mass of soluteion"" xx 100)#
In your case, the percent concentration by mass of the orange juice will be
#""%m/m"" = (0.11758color(red)(cancel(color(black)(""g""))))/(20.31color(red)(cancel(color(black)(""g"")))) xx 100 = color(green)(""0.579% C""_3""H""_8""O""_7)#
Manganese is oxidized:
Chromium is reduced:
And so we take
Charge is balanced, and mass is balanced, and so this is a reasonable representation of reality. Whether the reaction actually works, I don't know, and cannot be bothered to look up redox tables.
Manganese is oxidized:
Chromium is reduced:
And so we take
Charge is balanced, and mass is balanced, and so this is a reasonable representation of reality. Whether the reaction actually works, I don't know, and cannot be bothered to look up redox tables.
Manganese is oxidized:
Chromium is reduced:
And so we take
Charge is balanced, and mass is balanced, and so this is a reasonable representation of reality. Whether the reaction actually works, I don't know, and cannot be bothered to look up redox tables.
There are
There are
........And so the metal oxidation state would be
And thus its perhalide formula would be
You might get a
Well, the parent phosphate ion is
........And so the metal oxidation state would be
And thus its perhalide formula would be
You might get a
Well, the parent phosphate ion is
........And so the metal oxidation state would be
And thus its perhalide formula would be
You might get a
Because we are given the pressure, number of moles, and temperature, we will have to use the ideal gas law equation:
Based on the units that are associated with each variable, the number of moles have good units. The temperature has to be converted to Kelvins; we can do this by adding
Therefore,
The next issue is that pressure has units of torr instead of atmospheres; we can use the following relationship to give the correct units:
Thus,
R has the same value no matter what chemical species you are dealing with.
Now we know P,T, n, and R.
All we have to do is rearrange the equation to solve for V. We can do this by dividing by the pressure on both sides of the equation:
Finally, plug in your known values like so:
The gas has a volume of
Because we are given the pressure, number of moles, and temperature, we will have to use the ideal gas law equation:
Based on the units that are associated with each variable, the number of moles have good units. The temperature has to be converted to Kelvins; we can do this by adding
Therefore,
The next issue is that pressure has units of torr instead of atmospheres; we can use the following relationship to give the correct units:
Thus,
R has the same value no matter what chemical species you are dealing with.
Now we know P,T, n, and R.
All we have to do is rearrange the equation to solve for V. We can do this by dividing by the pressure on both sides of the equation:
Finally, plug in your known values like so:
The gas has a volume of
Because we are given the pressure, number of moles, and temperature, we will have to use the ideal gas law equation:
Based on the units that are associated with each variable, the number of moles have good units. The temperature has to be converted to Kelvins; we can do this by adding
Therefore,
The next issue is that pressure has units of torr instead of atmospheres; we can use the following relationship to give the correct units:
Thus,
R has the same value no matter what chemical species you are dealing with.
Now we know P,T, n, and R.
All we have to do is rearrange the equation to solve for V. We can do this by dividing by the pressure on both sides of the equation:
Finally, plug in your known values like so:
For
Water is an exceptionally concentrated solvent.
Around about
For
Water is an exceptionally concentrated solvent.
Around about
For
Water is an exceptionally concentrated solvent.
We can use a heat equilibrium equation to find this value.
Since the density of water is
Cold
Cold
We can use a heat equilibrium equation to find this value.
Since the density of water is
Cold
Cold
We can use a heat equilibrium equation to find this value.
Since the density of water is
Cold
Cold
To begin, using a
Your values will look like this :
Next, convert each value into moles by using molar mass.
(Divide your original value by the molar mass to find moles)
Take your smallest value of moles and divide every numerical value by it. In this case, the smallest value is
To find the smallest whole number ratio, multiply each value by
Your formula is now:
Remember, empirical formulas are related by a whole number ratios or the least common multiple of all numbers.
The empirical formula is:
(Remember that if the subscript of an element is
The empirical formula is
To begin, using a
Your values will look like this :
Next, convert each value into moles by using molar mass.
(Divide your original value by the molar mass to find moles)
Take your smallest value of moles and divide every numerical value by it. In this case, the smallest value is
To find the smallest whole number ratio, multiply each value by
Your formula is now:
Remember, empirical formulas are related by a whole number ratios or the least common multiple of all numbers.
The empirical formula is:
(Remember that if the subscript of an element is
The empirical formula is
To begin, using a
Your values will look like this :
Next, convert each value into moles by using molar mass.
(Divide your original value by the molar mass to find moles)
Take your smallest value of moles and divide every numerical value by it. In this case, the smallest value is
To find the smallest whole number ratio, multiply each value by
Your formula is now:
Remember, empirical formulas are related by a whole number ratios or the least common multiple of all numbers.
The empirical formula is:
(Remember that if the subscript of an element is
As always we ASSUME a
And so we divide thru by the SMALLEST molar quantity to get a trial empirical formula of...
And of course this is zinc nitrate...
Moles of water cancel, leaving you with moles of Oxygen.
Factor out 18/18=1.
25 x 1 = 25
You will need 25 moles, but this is not based on the equation alone.
Moles of water cancel, leaving you with moles of Oxygen.
Factor out 18/18=1.
25 x 1 = 25
You will need 25 moles, but this is not based on the equation alone.
Moles of water cancel, leaving you with moles of Oxygen.
Factor out 18/18=1.
25 x 1 = 25
The equation above represents probably the most important inorganic reaction on the planet in that it allows the formation of ammonia and nitrates for nitrogenous fertilizer.
Your question specified 2 moles of nitrogen. I would be quite justified in assuming that the questioner meant 2 moles of dinitrogen gas, i.e. the equivalent of
Thus, if
The balanced equation for dinitrogen reduction is:
The equation above represents probably the most important inorganic reaction on the planet in that it allows the formation of ammonia and nitrates for nitrogenous fertilizer.
Your question specified 2 moles of nitrogen. I would be quite justified in assuming that the questioner meant 2 moles of dinitrogen gas, i.e. the equivalent of
Thus, if
The balanced equation for dinitrogen reduction is:
The equation above represents probably the most important inorganic reaction on the planet in that it allows the formation of ammonia and nitrates for nitrogenous fertilizer.
Your question specified 2 moles of nitrogen. I would be quite justified in assuming that the questioner meant 2 moles of dinitrogen gas, i.e. the equivalent of
Thus, if
First, convert the grams of lead dioxide into moles.
Since for every mole of lead dioxide you need 2 moles of nitric acid, your total moles of needed nitric acid would be:
Because
Seeing that you only have 3 sig figs, the answer would be:
First, convert the grams of lead dioxide into moles.
Since for every mole of lead dioxide you need 2 moles of nitric acid, your total moles of needed nitric acid would be:
Because
Seeing that you only have 3 sig figs, the answer would be:
First, convert the grams of lead dioxide into moles.
Since for every mole of lead dioxide you need 2 moles of nitric acid, your total moles of needed nitric acid would be:
Because
Seeing that you only have 3 sig figs, the answer would be:
The thing to remember about a solution's molarity is that you can express it as a fraction that has
In your case, a
#""2.0 M"" = ""2.0 moles KCl""/""1 L solution""#
Now, you should know that
#""1 L"" = 10^3# #""mL""#
This means that you can rewrite the molarity of the solution as
#""2.0 M"" = ""2.0 moles KCl""/(10^3color(white)(.)""mL solution"")#
So, you need to figure out how many moles of potassium chloride must be dissolved in water to make
In other words, you must find the number of moles that when dissolved in
#(color(blue)(?)color(white)(.)""moles KCl"")/""100.0 mL solution"" = ""2.0 moles KCl""/(10^3color(white)(.)""mL solution"")#
Solve this equation to find
#color(blue)(?) = (100.0 color(red)(cancel(color(black)(""mL solution""))))/(10^3color(red)(cancel(color(black)(""mL solution"")))) * ""2.0 moles""#
#color(blue)(?) = ""0.20 moles"" -># rounded to two sig figs
Therefore, you can say that if you dissolve
The thing to remember about a solution's molarity is that you can express it as a fraction that has
In your case, a
#""2.0 M"" = ""2.0 moles KCl""/""1 L solution""#
Now, you should know that
#""1 L"" = 10^3# #""mL""#
This means that you can rewrite the molarity of the solution as
#""2.0 M"" = ""2.0 moles KCl""/(10^3color(white)(.)""mL solution"")#
So, you need to figure out how many moles of potassium chloride must be dissolved in water to make
In other words, you must find the number of moles that when dissolved in
#(color(blue)(?)color(white)(.)""moles KCl"")/""100.0 mL solution"" = ""2.0 moles KCl""/(10^3color(white)(.)""mL solution"")#
Solve this equation to find
#color(blue)(?) = (100.0 color(red)(cancel(color(black)(""mL solution""))))/(10^3color(red)(cancel(color(black)(""mL solution"")))) * ""2.0 moles""#
#color(blue)(?) = ""0.20 moles"" -># rounded to two sig figs
Therefore, you can say that if you dissolve
The thing to remember about a solution's molarity is that you can express it as a fraction that has
In your case, a
#""2.0 M"" = ""2.0 moles KCl""/""1 L solution""#
Now, you should know that
#""1 L"" = 10^3# #""mL""#
This means that you can rewrite the molarity of the solution as
#""2.0 M"" = ""2.0 moles KCl""/(10^3color(white)(.)""mL solution"")#
So, you need to figure out how many moles of potassium chloride must be dissolved in water to make
In other words, you must find the number of moles that when dissolved in
#(color(blue)(?)color(white)(.)""moles KCl"")/""100.0 mL solution"" = ""2.0 moles KCl""/(10^3color(white)(.)""mL solution"")#
Solve this equation to find
#color(blue)(?) = (100.0 color(red)(cancel(color(black)(""mL solution""))))/(10^3color(red)(cancel(color(black)(""mL solution"")))) * ""2.0 moles""#
#color(blue)(?) = ""0.20 moles"" -># rounded to two sig figs
Therefore, you can say that if you dissolve
Find the number of moles for each element by dividing the mass present with the relative mass of the atom.
Now divide each number of moles by the lowest one, in this case
Round to the nearest whole number
Find the number of moles for each element by dividing the mass present with the relative mass of the atom.
Now divide each number of moles by the lowest one, in this case
Round to the nearest whole number
Find the number of moles for each element by dividing the mass present with the relative mass of the atom.
Now divide each number of moles by the lowest one, in this case
Round to the nearest whole number
We got
Note that NORMALLY you would NEVER be given the percentage oxygen. Why not? Because there are very few ways to measure the proportion of this gas. At a 1st year undergrad level this question would have proposed an oxide of chromium that contained
When we divide thru by the lowest number of moles (
But by definition, the empirical formula is the simplest WHOLE number ratio defining constituent atoms in a species, so we must double this provisional formula to give:
And this is a crystalline, orange powder, that is widely used in oxidations of organic materials....
Since the chemical equation,
#color(red)2H_2O->2H_2+color(blue)1O_2#
requires a specific number of moles of the reactant, then a specific number of moles of the products are created.
With this knowledge, you can create a mole ratio.
The general format of a mole ratio is as follows:
#color(blue)(|bar(ul(color(white)(a/a)color(black)((""required based on balanced equation"")/(""product based on balanced equation"")=(""required"")/(""product""))color(white)(a/a)|)))#
In your case, you're looking for the moles of
Thus, your mole ratio would be:
#(color(red)2color(white)(i)molcolor(white)(i)H_2O)/(color(blue)1color(white)(i)molcolor(white)(i)O_2)=x/(2.5color(white)(i)molcolor(white)(i)O_2)#
#color(darkorange)(rArr)# where#x# represents the moles of#H_2O# required
From this point on, your goal is to solve for
#x=2.5color(purple)cancelcolor(black)(molcolor(white)(i)O_2)xx(2color(white)(i)molcolor(white)(i)H_2O)/(1color(purple)cancelcolor(black)(molcolor(white)(i)O_2))#
#x=color(green)(|bar(ul(color(white)(a/a)5color(white)(i)molcolor(white)(i)H_2Ocolor(white)(a/a)|)))#
Since the chemical equation,
#color(red)2H_2O->2H_2+color(blue)1O_2#
requires a specific number of moles of the reactant, then a specific number of moles of the products are created.
With this knowledge, you can create a mole ratio.
The general format of a mole ratio is as follows:
#color(blue)(|bar(ul(color(white)(a/a)color(black)((""required based on balanced equation"")/(""product based on balanced equation"")=(""required"")/(""product""))color(white)(a/a)|)))#
In your case, you're looking for the moles of
Thus, your mole ratio would be:
#(color(red)2color(white)(i)molcolor(white)(i)H_2O)/(color(blue)1color(white)(i)molcolor(white)(i)O_2)=x/(2.5color(white)(i)molcolor(white)(i)O_2)#
#color(darkorange)(rArr)# where#x# represents the moles of#H_2O# required
From this point on, your goal is to solve for
#x=2.5color(purple)cancelcolor(black)(molcolor(white)(i)O_2)xx(2color(white)(i)molcolor(white)(i)H_2O)/(1color(purple)cancelcolor(black)(molcolor(white)(i)O_2))#
#x=color(green)(|bar(ul(color(white)(a/a)5color(white)(i)molcolor(white)(i)H_2Ocolor(white)(a/a)|)))#
Since the chemical equation,
#color(red)2H_2O->2H_2+color(blue)1O_2#
requires a specific number of moles of the reactant, then a specific number of moles of the products are created.
With this knowledge, you can create a mole ratio.
The general format of a mole ratio is as follows:
#color(blue)(|bar(ul(color(white)(a/a)color(black)((""required based on balanced equation"")/(""product based on balanced equation"")=(""required"")/(""product""))color(white)(a/a)|)))#
In your case, you're looking for the moles of
Thus, your mole ratio would be:
#(color(red)2color(white)(i)molcolor(white)(i)H_2O)/(color(blue)1color(white)(i)molcolor(white)(i)O_2)=x/(2.5color(white)(i)molcolor(white)(i)O_2)#
#color(darkorange)(rArr)# where#x# represents the moles of#H_2O# required
From this point on, your goal is to solve for
#x=2.5color(purple)cancelcolor(black)(molcolor(white)(i)O_2)xx(2color(white)(i)molcolor(white)(i)H_2O)/(1color(purple)cancelcolor(black)(molcolor(white)(i)O_2))#
#x=color(green)(|bar(ul(color(white)(a/a)5color(white)(i)molcolor(white)(i)H_2Ocolor(white)(a/a)|)))#
Balanced Equation
This at first glance looks like a double replacement (double displacement) reaction. However, a double replacement reaction does not occur unless one of the products is a solid precipitate, an insoluble gas, or water. In this case, none of those is a product. All of the compounds are aqueous.
This reaction will not actually occur. All you will have is a mixture of
However, I will work it out so you can see how a problem of this type is to be done.
Step 1: You need the molar masses of lithium nitrate and lithium sulfate.
You can calculate it or find it in a resource.
This type of problem follows the pattern:
Step 2: Convert mass of
Divide the mass of lithium sulfate by its molar mass.
I am including a couple of guard units to reduce rounding errors. I will round the final answer to two significant figures.
Step 3: Convert moles of
Multiply moles of
Step 4: Convert moles
Multiply the moles
This problem can be worked out all at once as follows.
If this reaction were to occur,
Balanced Equation
This at first glance looks like a double replacement (double displacement) reaction. However, a double replacement reaction does not occur unless one of the products is a solid precipitate, an insoluble gas, or water. In this case, none of those is a product. All of the compounds are aqueous.
This reaction will not actually occur. All you will have is a mixture of
However, I will work it out so you can see how a problem of this type is to be done.
Step 1: You need the molar masses of lithium nitrate and lithium sulfate.
You can calculate it or find it in a resource.
This type of problem follows the pattern:
Step 2: Convert mass of
Divide the mass of lithium sulfate by its molar mass.
I am including a couple of guard units to reduce rounding errors. I will round the final answer to two significant figures.
Step 3: Convert moles of
Multiply moles of
Step 4: Convert moles
Multiply the moles
This problem can be worked out all at once as follows.
If this reaction were to occur,
Balanced Equation
This at first glance looks like a double replacement (double displacement) reaction. However, a double replacement reaction does not occur unless one of the products is a solid precipitate, an insoluble gas, or water. In this case, none of those is a product. All of the compounds are aqueous.
This reaction will not actually occur. All you will have is a mixture of
However, I will work it out so you can see how a problem of this type is to be done.
Step 1: You need the molar masses of lithium nitrate and lithium sulfate.
You can calculate it or find it in a resource.
This type of problem follows the pattern:
Step 2: Convert mass of
Divide the mass of lithium sulfate by its molar mass.
I am including a couple of guard units to reduce rounding errors. I will round the final answer to two significant figures.
Step 3: Convert moles of
Multiply moles of
Step 4: Convert moles
Multiply the moles
This problem can be worked out all at once as follows.
Specific heat is the heat capacity per gram of the substance of interest. Heat capacity is the amount of heat energy a body can cold.
The specific heat of water,
Specific heat is the heat capacity per gram of the substance of interest. Heat capacity is the amount of heat energy a body can cold.
The specific heat of water,
Specific heat is the heat capacity per gram of the substance of interest. Heat capacity is the amount of heat energy a body can cold.
Barium is listed under Group 2A of the Periodic Table. Looking at the column, Barium is Ba. The elements listed in Group 2A have a charge of +2.
So we have Barium as
Now let's move to Chloride. Don't bother looking at the table, because it is not listed. Chloride is an ion of Chlorine, which is from Group 7A and we know anything in that column has a charge of -1. So Chlorine has gained an electron and becomes Chloride. Note that this is an anion and has a negative charge. You can see how easy it was to change the name, you just change the suffix to -ide.
So we have Chloride as
We're not finished though! The charges do not balance!
To balance it we simply add another Chloride so that you have two negatives balancing the two positives. You'll write it as
Now they balance; you have one barium
The final formula is written as
Hope that helps!
Barium is listed under Group 2A of the Periodic Table. Looking at the column, Barium is Ba. The elements listed in Group 2A have a charge of +2.
So we have Barium as
Now let's move to Chloride. Don't bother looking at the table, because it is not listed. Chloride is an ion of Chlorine, which is from Group 7A and we know anything in that column has a charge of -1. So Chlorine has gained an electron and becomes Chloride. Note that this is an anion and has a negative charge. You can see how easy it was to change the name, you just change the suffix to -ide.
So we have Chloride as
We're not finished though! The charges do not balance!
To balance it we simply add another Chloride so that you have two negatives balancing the two positives. You'll write it as
Now they balance; you have one barium
The final formula is written as
Hope that helps!
Barium is listed under Group 2A of the Periodic Table. Looking at the column, Barium is Ba. The elements listed in Group 2A have a charge of +2.
So we have Barium as
Now let's move to Chloride. Don't bother looking at the table, because it is not listed. Chloride is an ion of Chlorine, which is from Group 7A and we know anything in that column has a charge of -1. So Chlorine has gained an electron and becomes Chloride. Note that this is an anion and has a negative charge. You can see how easy it was to change the name, you just change the suffix to -ide.
So we have Chloride as
We're not finished though! The charges do not balance!
To balance it we simply add another Chloride so that you have two negatives balancing the two positives. You'll write it as
Now they balance; you have one barium
The final formula is written as
Hope that helps!
We can work out the concentration of the hydrochloric acid:
There is no such beast as ammonium hydroxide. There is
We can work out the concentration of the hydrochloric acid:
There is no such beast as ammonium hydroxide. There is
We can work out the concentration of the hydrochloric acid:
Given what I have said:
Clearly,
Of course this is reasonable inasmuch that we know that the styrene monomer is
The molecular formula is always a whole number multiple of the empirical formula. Of course, the multiple might be
Given what I have said:
Clearly,
Of course this is reasonable inasmuch that we know that the styrene monomer is
The molecular formula is always a whole number multiple of the empirical formula. Of course, the multiple might be
Given what I have said:
Clearly,
Of course this is reasonable inasmuch that we know that the styrene monomer is
One way to approach this problem would be to use the volume of the initial solution and the volume of the target solution to find the dilution factor.
This will then allow you to find the molarity of the diluted solution.
So, the dilution factor, which tells you how concentrated the starting solution is compared with the dilution solution, is calculated like this
#color(blue)(|bar(ul(color(white)(a/a)""D.F."" = V_""final""/V_""initial""color(white)(a/a)|)))#
Here you have
In your case, you have
#""D.F."" = (800.0 color(red)(cancel(color(black)(""mL""))))/(250.0color(red)(cancel(color(black)(""mL"")))) = 3.2#
This means that the initial solution was
#color(blue)(|bar(ul(color(white)(a/a)color(black)(c_""concentrated"" = ""D.F."" xx c_""diluted"")color(white)(a/a)|)))#
which gets you
#c_""diluted"" = ""0.96 M""/3.2 = color(green)(|bar(ul(color(white)(a/a)color(black)(""0.30 M"")color(white)(a/a)|)))#
The answer is rounded to two sig figs, the number of sig figs you have for the molarity of the concentrated solution.
One way to approach this problem would be to use the volume of the initial solution and the volume of the target solution to find the dilution factor.
This will then allow you to find the molarity of the diluted solution.
So, the dilution factor, which tells you how concentrated the starting solution is compared with the dilution solution, is calculated like this
#color(blue)(|bar(ul(color(white)(a/a)""D.F."" = V_""final""/V_""initial""color(white)(a/a)|)))#
Here you have
In your case, you have
#""D.F."" = (800.0 color(red)(cancel(color(black)(""mL""))))/(250.0color(red)(cancel(color(black)(""mL"")))) = 3.2#
This means that the initial solution was
#color(blue)(|bar(ul(color(white)(a/a)color(black)(c_""concentrated"" = ""D.F."" xx c_""diluted"")color(white)(a/a)|)))#
which gets you
#c_""diluted"" = ""0.96 M""/3.2 = color(green)(|bar(ul(color(white)(a/a)color(black)(""0.30 M"")color(white)(a/a)|)))#
The answer is rounded to two sig figs, the number of sig figs you have for the molarity of the concentrated solution.
One way to approach this problem would be to use the volume of the initial solution and the volume of the target solution to find the dilution factor.
This will then allow you to find the molarity of the diluted solution.
So, the dilution factor, which tells you how concentrated the starting solution is compared with the dilution solution, is calculated like this
#color(blue)(|bar(ul(color(white)(a/a)""D.F."" = V_""final""/V_""initial""color(white)(a/a)|)))#
Here you have
In your case, you have
#""D.F."" = (800.0 color(red)(cancel(color(black)(""mL""))))/(250.0color(red)(cancel(color(black)(""mL"")))) = 3.2#
This means that the initial solution was
#color(blue)(|bar(ul(color(white)(a/a)color(black)(c_""concentrated"" = ""D.F."" xx c_""diluted"")color(white)(a/a)|)))#
which gets you
#c_""diluted"" = ""0.96 M""/3.2 = color(green)(|bar(ul(color(white)(a/a)color(black)(""0.30 M"")color(white)(a/a)|)))#
The answer is rounded to two sig figs, the number of sig figs you have for the molarity of the concentrated solution.
The empirical formula tells you what the smallest integer ratio between the atoms that make up a substance is.
In essence, the empirical formula can be thought of as the building block of a substance. The molecular formula will thus tell you how many building blocks are needed to form that substance.
So, for a given molecule, the empirical formula will tell you what is the minimum number of atoms of each element needed to form the building block for the molecular formula.
In your case, the molecular formula is given to be
- two atoms of carbon
- four atoms of hydrogen
- two atoms of oxygen
But what is the minimum number of atoms of each element that you need in order to be able to start building the molecule?
The minimum number of atoms an element can contriubte to a building block is
With this in mind, notice that if you divide all these values by
- one atom of carbon
- two atoms of hydrogen
- one atom of oxygen
This tells you that the building block for this particular molecule is
In other words, if you take two such building blocks,
The empirical formula tells you what the smallest integer ratio between the atoms that make up a substance is.
In essence, the empirical formula can be thought of as the building block of a substance. The molecular formula will thus tell you how many building blocks are needed to form that substance.
So, for a given molecule, the empirical formula will tell you what is the minimum number of atoms of each element needed to form the building block for the molecular formula.
In your case, the molecular formula is given to be
- two atoms of carbon
- four atoms of hydrogen
- two atoms of oxygen
But what is the minimum number of atoms of each element that you need in order to be able to start building the molecule?
The minimum number of atoms an element can contriubte to a building block is
With this in mind, notice that if you divide all these values by
- one atom of carbon
- two atoms of hydrogen
- one atom of oxygen
This tells you that the building block for this particular molecule is
In other words, if you take two such building blocks,
The empirical formula tells you what the smallest integer ratio between the atoms that make up a substance is.
In essence, the empirical formula can be thought of as the building block of a substance. The molecular formula will thus tell you how many building blocks are needed to form that substance.
So, for a given molecule, the empirical formula will tell you what is the minimum number of atoms of each element needed to form the building block for the molecular formula.
In your case, the molecular formula is given to be
- two atoms of carbon
- four atoms of hydrogen
- two atoms of oxygen
But what is the minimum number of atoms of each element that you need in order to be able to start building the molecule?
The minimum number of atoms an element can contriubte to a building block is
With this in mind, notice that if you divide all these values by
- one atom of carbon
- two atoms of hydrogen
- one atom of oxygen
This tells you that the building block for this particular molecule is
In other words, if you take two such building blocks,
We must use Avogadro's number:
We set up the problem like this:
Then we just compute
We must use Avogadro's number:
We set up the problem like this:
Then we just compute
We must use Avogadro's number:
We set up the problem like this:
Then we just compute
And the ammine ligands may be removed as neutral entities, whose removal causes no change to the charge of the complex.......
to leave
So
Note that all I have done here is to conserve CHARGE, and MASS. All of stoichiometry depends on these conservations.
Capisce?
Note, that looking at this question again, it was clearly formulated by someone who has not formally trained as a chemist. The
Well because the complex as written has a single chloride counterion, we can write..........
And the ammine ligands may be removed as neutral entities, whose removal causes no change to the charge of the complex.......
to leave
So
Note that all I have done here is to conserve CHARGE, and MASS. All of stoichiometry depends on these conservations.
Capisce?
Note, that looking at this question again, it was clearly formulated by someone who has not formally trained as a chemist. The
Well because the complex as written has a single chloride counterion, we can write..........
And the ammine ligands may be removed as neutral entities, whose removal causes no change to the charge of the complex.......
to leave
So
Note that all I have done here is to conserve CHARGE, and MASS. All of stoichiometry depends on these conservations.
Capisce?
Note, that looking at this question again, it was clearly formulated by someone who has not formally trained as a chemist. The
The empirical formula is the simplest whole number molar ratio of the elements in the compound.
We must convert the masses of
Step 1. Calculate the mass of
Step 2. Calculate the moles of
From this point on, I like to summarize the calculations in a table.
The molar ratio is
The empirical formula is
Here is a video that illustrates how to determine an empirical formula.
The empirical formula is
The empirical formula is the simplest whole number molar ratio of the elements in the compound.
We must convert the masses of
Step 1. Calculate the mass of
Step 2. Calculate the moles of
From this point on, I like to summarize the calculations in a table.
The molar ratio is
The empirical formula is
Here is a video that illustrates how to determine an empirical formula.
The empirical formula is
The empirical formula is the simplest whole number molar ratio of the elements in the compound.
We must convert the masses of
Step 1. Calculate the mass of
Step 2. Calculate the moles of
From this point on, I like to summarize the calculations in a table.
The molar ratio is
The empirical formula is
Here is a video that illustrates how to determine an empirical formula.
Magnesium hydroxide,
Magnesium hydroxide dissociates only partially to form magnesium cations,
#""Mg""(""OH"")_text(2(s]) rightleftharpoons ""Mg""_text((aq])^(2+) + color(red)(2)""OH""_text((aq])^(-)#
For an insoluble compound, its molar solubility tells you how many moles of the compound can be dissolved per liter of aqueous solution before reaching saturation.
In your case, a molar solubility of
#s = 1.6 * 10^(-4)#
means that you can only dissolve
Take a look at the dissociation equilibrium. Notice that every mole of magnesium hydroxide that dissociates produces
This tells you that if you successfully dissolve
#n_(Mg^(2+)) = 1 xx 1.6 * 10^(-4) = 1.6 * 10^(-4)""moles Mg""^(2+)#
and
#n_(OH^(-)) = color(red)(2) xx 1.6 * 10^(-4) = 3.2 * 10^(-4)""moles""#
Since we're working with one liter of solution, you can cay that
#[""Mg""^(2+)] = 1.6 * 10^(-4)""M""#
#[""OH""^(-)] = 3.2 * 10^(-4)""M""#
By definition, the solubility product constant,
#K_(sp) = [""Mg""^(2+)] * [""OH""^(-)]^color(red)(2)#
Plug in these values to get
#K_(sp) = 1.6 * 10^(-4) * (3.2 * 10^(-4))^color(red)(2)#
#K_(sp) = color(green)(1.6 * 10^(-11)) -># rounded to two sig figs
The listed value for magnesium hydroxide's solubility product is
http://www.wiredchemist.com/chemistry/data/solubility-product-constants
Magnesium hydroxide,
Magnesium hydroxide dissociates only partially to form magnesium cations,
#""Mg""(""OH"")_text(2(s]) rightleftharpoons ""Mg""_text((aq])^(2+) + color(red)(2)""OH""_text((aq])^(-)#
For an insoluble compound, its molar solubility tells you how many moles of the compound can be dissolved per liter of aqueous solution before reaching saturation.
In your case, a molar solubility of
#s = 1.6 * 10^(-4)#
means that you can only dissolve
Take a look at the dissociation equilibrium. Notice that every mole of magnesium hydroxide that dissociates produces
This tells you that if you successfully dissolve
#n_(Mg^(2+)) = 1 xx 1.6 * 10^(-4) = 1.6 * 10^(-4)""moles Mg""^(2+)#
and
#n_(OH^(-)) = color(red)(2) xx 1.6 * 10^(-4) = 3.2 * 10^(-4)""moles""#
Since we're working with one liter of solution, you can cay that
#[""Mg""^(2+)] = 1.6 * 10^(-4)""M""#
#[""OH""^(-)] = 3.2 * 10^(-4)""M""#
By definition, the solubility product constant,
#K_(sp) = [""Mg""^(2+)] * [""OH""^(-)]^color(red)(2)#
Plug in these values to get
#K_(sp) = 1.6 * 10^(-4) * (3.2 * 10^(-4))^color(red)(2)#
#K_(sp) = color(green)(1.6 * 10^(-11)) -># rounded to two sig figs
The listed value for magnesium hydroxide's solubility product is
http://www.wiredchemist.com/chemistry/data/solubility-product-constants
Magnesium hydroxide,
Magnesium hydroxide dissociates only partially to form magnesium cations,
#""Mg""(""OH"")_text(2(s]) rightleftharpoons ""Mg""_text((aq])^(2+) + color(red)(2)""OH""_text((aq])^(-)#
For an insoluble compound, its molar solubility tells you how many moles of the compound can be dissolved per liter of aqueous solution before reaching saturation.
In your case, a molar solubility of
#s = 1.6 * 10^(-4)#
means that you can only dissolve
Take a look at the dissociation equilibrium. Notice that every mole of magnesium hydroxide that dissociates produces
This tells you that if you successfully dissolve
#n_(Mg^(2+)) = 1 xx 1.6 * 10^(-4) = 1.6 * 10^(-4)""moles Mg""^(2+)#
and
#n_(OH^(-)) = color(red)(2) xx 1.6 * 10^(-4) = 3.2 * 10^(-4)""moles""#
Since we're working with one liter of solution, you can cay that
#[""Mg""^(2+)] = 1.6 * 10^(-4)""M""#
#[""OH""^(-)] = 3.2 * 10^(-4)""M""#
By definition, the solubility product constant,
#K_(sp) = [""Mg""^(2+)] * [""OH""^(-)]^color(red)(2)#
Plug in these values to get
#K_(sp) = 1.6 * 10^(-4) * (3.2 * 10^(-4))^color(red)(2)#
#K_(sp) = color(green)(1.6 * 10^(-11)) -># rounded to two sig figs
The listed value for magnesium hydroxide's solubility product is
http://www.wiredchemist.com/chemistry/data/solubility-product-constants
For this type of problem, we want to use the following dilution formula:
* TIP. * Whenever you are diluting a solution (going from a highly concentrated substance to a less concentrated substance by increasing the volume of the solvent) you always use this equation.
We know the initial concentration, final volume, and final concentration. All we have to do rearrange the equation to solve for the initial volume:
62.5 mL of HCl is required.
For this type of problem, we want to use the following dilution formula:
* TIP. * Whenever you are diluting a solution (going from a highly concentrated substance to a less concentrated substance by increasing the volume of the solvent) you always use this equation.
We know the initial concentration, final volume, and final concentration. All we have to do rearrange the equation to solve for the initial volume:
62.5 mL of HCl is required.
For this type of problem, we want to use the following dilution formula:
* TIP. * Whenever you are diluting a solution (going from a highly concentrated substance to a less concentrated substance by increasing the volume of the solvent) you always use this equation.
We know the initial concentration, final volume, and final concentration. All we have to do rearrange the equation to solve for the initial volume:
Is charge balanced? Tick. Is mass balanced? Tick.
The
Most such combustion reactions, certainly in the
Is charge balanced? Tick. Is mass balanced? Tick.
The
Most such combustion reactions, certainly in the
Is charge balanced? Tick. Is mass balanced? Tick.
The
Most such combustion reactions, certainly in the
Write the (first unbalanced) chemical equation:
Balanced:
1 mole of the organic carboxylic acid ethanoic acid (acetic acid) has a mass equal to its molar mass
But 1 mole contains an Avagadro number of molecules, ie
Therefore 1 molecule of this weak acid has a mass of
1 mole of the organic carboxylic acid ethanoic acid (acetic acid) has a mass equal to its molar mass
But 1 mole contains an Avagadro number of molecules, ie
Therefore 1 molecule of this weak acid has a mass of
1 mole of the organic carboxylic acid ethanoic acid (acetic acid) has a mass equal to its molar mass
But 1 mole contains an Avagadro number of molecules, ie
Therefore 1 molecule of this weak acid has a mass of
Now, the first thing is to monitor each element as an individual and count the number of moles present in each on the left side and equate it to exactly the same number of moles to the same element on the right side.
For example, we have
On the left side, there are 3 moles of
Now, the first thing is to monitor each element as an individual and count the number of moles present in each on the left side and equate it to exactly the same number of moles to the same element on the right side.
For example, we have
On the left side, there are 3 moles of
Now, the first thing is to monitor each element as an individual and count the number of moles present in each on the left side and equate it to exactly the same number of moles to the same element on the right side.
For example, we have
On the left side, there are 3 moles of
And the question you ask yourself...is mass balanced, and is charge balanced...and so you count up the atoms on each side.... There are three metals atoms, 2 nitrogens, 12 hydrogens, and 6 oxygens ON BOTH SIDES of the equation....so the given reaction at least conforms to the principle of conservation of mass. It also conforms to experiment....
The problem that can arise in trying to balance a combustion equation is the fact that water has only one O atom, but oxygen gas supplies them in pairs. The ""trick"" to getting the job done is to make certain the coefficient of water is an even number, even if this means you don't start with the usual technique of placing a ""1"" in front of a compound.
Start with a ""2"" in front of the benzene
The problem that can arise in trying to balance a combustion equation is the fact that water has only one O atom, but oxygen gas supplies them in pairs. The ""trick"" to getting the job done is to make certain the coefficient of water is an even number, even if this means you don't start with the usual technique of placing a ""1"" in front of a compound.
Start with a ""2"" in front of the benzene
The problem that can arise in trying to balance a combustion equation is the fact that water has only one O atom, but oxygen gas supplies them in pairs. The ""trick"" to getting the job done is to make certain the coefficient of water is an even number, even if this means you don't start with the usual technique of placing a ""1"" in front of a compound.
Since the type of iron is not specified then it is assumed to be 'Cast Iron' which has a melting point of 1204 deg-Celsius*.
*http://www.onlinemetals.com/meltpt.cfm
The total heat transfer would be ,,,
..............................................................................................................................
=>
=
(Specific Heat of Molten Iron) http://www.engineeringtoolbox.com/liquid-metal-boiling-points-specific-heat-d_1893.html
..............................................................................................................................
=>
=
(Heat of Fusion of Molten Iron) http://www.engineeringtoolbox.com/fusion-heat-metals-d_1266.html
...............................................................................................................................
=>
=
(Specific Heat of Iron (s) = 0.46
..............................................................................................................................
900 Kj
Since the type of iron is not specified then it is assumed to be 'Cast Iron' which has a melting point of 1204 deg-Celsius*.
*http://www.onlinemetals.com/meltpt.cfm
The total heat transfer would be ,,,
..............................................................................................................................
=>
=
(Specific Heat of Molten Iron) http://www.engineeringtoolbox.com/liquid-metal-boiling-points-specific-heat-d_1893.html
..............................................................................................................................
=>
=
(Heat of Fusion of Molten Iron) http://www.engineeringtoolbox.com/fusion-heat-metals-d_1266.html
...............................................................................................................................
=>
=
(Specific Heat of Iron (s) = 0.46
..............................................................................................................................
900 Kj
Since the type of iron is not specified then it is assumed to be 'Cast Iron' which has a melting point of 1204 deg-Celsius*.
*http://www.onlinemetals.com/meltpt.cfm
The total heat transfer would be ,,,
..............................................................................................................................
=>
=
(Specific Heat of Molten Iron) http://www.engineeringtoolbox.com/liquid-metal-boiling-points-specific-heat-d_1893.html
..............................................................................................................................
=>
=
(Heat of Fusion of Molten Iron) http://www.engineeringtoolbox.com/fusion-heat-metals-d_1266.html
...............................................................................................................................
=>
=
(Specific Heat of Iron (s) = 0.46
..............................................................................................................................
I got
You can get more context here:
https://en.wikipedia.org/wiki/Iron#Phase_diagram_and_allotropes
and you can examine the specific heat capacity variations more closely here:
http://webbook.nist.gov/cgi/cbook.cgi?ID=C7439896&Mask=2&Type=JANAFS&Plot=on#JANAFS
On another note, this
If you simply assume a
There is a HUGE assumption here that iron's specific heat capacity doesn't change from
Since all these phases at
However, the specific heat capacity
[
The wonky curve is the
Aren't you glad we aren't doing phase changes? :-)
So, we would have the heat of cooling as the negative of the heat of heating:
#q_""cool"" = -(q_1 + . . . + q_7)#
#= -m(C_(P1)DeltaT_(0->1) + . . . + C_(P7)DeltaT_(6->7))#
I'll leave the units out, but you know that they are
#= -1000 cdot [0.531(700 - 298.15) + 0.719(935 - 700) + 1.064(1042 - 935) + 1.177(1100 - 1042) + 0.770(1183.15 - 1100) + 0.642(1667.15 - 1183.15) + 0.748(1773.15 - 1667.15)]#
Each phase then approximately contributes:
#= overbrace(-""628487 J"")^(alpha"" phase"") + overbrace(-""310728 J"")^(gamma"" phase"") + overbrace(-""79288 J"")^(delta"" phase"")#
#~~# #-1.020 xx 10^(6)# #""J""# ,
or about
Thermal history of cooling cast iron
Thermal history of cooling cast iron from
Propane has a chemical formula of
Combustion is when a hydro-carbon like propane containing carbon and hydrogen is burned in oxygen, releasing carbon dioxide and water.
Propane has a chemical formula of
Combustion is when a hydro-carbon like propane containing carbon and hydrogen is burned in oxygen, releasing carbon dioxide and water.
Propane has a chemical formula of
Combustion is when a hydro-carbon like propane containing carbon and hydrogen is burned in oxygen, releasing carbon dioxide and water.
One last question from me. I thought it would be Ksp=[2Ag^+]^2[SO3^2-], but plugging in 1.80×10^-3 for Ag and doing algebra doesn't seem to get me the correct answer when I put it on the online homework tool we use. Thanks for your help! I still can't seem to understand concentrations.
The expression of the solubility product constant for silver sulfite looks like this
#K_(sp) = [""Ag""^(+)]^color(red)(2) * [""SO""_3^(2-)]#
This is the case because silver sulfite will partially ionize to produce silver(I) cations and sulfite anions in aqueous solution.
#""Ag""_ 2""SO""_ (3(s)) rightleftharpoons color(red)(2)""Ag""_ ((aq))^(+) + ""SO""_ (3(aq))^(2-)#
Notice that the fact that every mole of silver sulfite that dissociates in aqueous solution produce
Mind you, the coefficients of the ions are used as expoents in the expression of the solubility product constant, not as coefficients.
So something like
#K_(sp) = [color(red)(2)""Ag""^(+)]^color(red)(2) * [""SO""_3^(2-)]#
is not correct because the coefficient must only be used as an exponent in the expression of the solubility proeduct constant.
In your case, you have--I'll do the calculations with the added units for the solubility product constant!
#1.50 * 10^(-14) quad ""M""^3 = (1.80 * 10^(-3))^color(red)(2) quad ""M""^color(red)(2) * [""SO""_3^(2-)]#
So you can say that the equilibrium concentration of the sulfite anions will be equal to
#[S""O""_3^(2-)] = (1.50 * 10^(-14) quad ""M""^color(red)(cancel(color(black)(3))))/((1.80 * 10^(-3))^color(red)(2) quad color(red)(cancel(color(black)(""M""^2)))) = color(darkgreen)(ul(color(black)(4.63 * 10^(-9) quad ""M"")))#
The answer is rounded to three sig figs.
So, a quick recap
#""Ag""_ 2""SO""_ (3(s)) rightleftharpoons color(red)(2)""Ag""_ ((aq))^(+) + ""SO""_ (3(aq))^(2-)# Here
#color(red)(2)# is a coefficient!
#K_(sp) = [""Ag""^(+)]^color(red)(2) * [""SO""_3^(2-)]# Here
#color(red)(2)# must be an exponent, so don't use it as a coefficient!
The expression of the solubility product constant for silver sulfite looks like this
#K_(sp) = [""Ag""^(+)]^color(red)(2) * [""SO""_3^(2-)]#
This is the case because silver sulfite will partially ionize to produce silver(I) cations and sulfite anions in aqueous solution.
#""Ag""_ 2""SO""_ (3(s)) rightleftharpoons color(red)(2)""Ag""_ ((aq))^(+) + ""SO""_ (3(aq))^(2-)#
Notice that the fact that every mole of silver sulfite that dissociates in aqueous solution produce
Mind you, the coefficients of the ions are used as expoents in the expression of the solubility product constant, not as coefficients.
So something like
#K_(sp) = [color(red)(2)""Ag""^(+)]^color(red)(2) * [""SO""_3^(2-)]#
is not correct because the coefficient must only be used as an exponent in the expression of the solubility proeduct constant.
In your case, you have--I'll do the calculations with the added units for the solubility product constant!
#1.50 * 10^(-14) quad ""M""^3 = (1.80 * 10^(-3))^color(red)(2) quad ""M""^color(red)(2) * [""SO""_3^(2-)]#
So you can say that the equilibrium concentration of the sulfite anions will be equal to
#[S""O""_3^(2-)] = (1.50 * 10^(-14) quad ""M""^color(red)(cancel(color(black)(3))))/((1.80 * 10^(-3))^color(red)(2) quad color(red)(cancel(color(black)(""M""^2)))) = color(darkgreen)(ul(color(black)(4.63 * 10^(-9) quad ""M"")))#
The answer is rounded to three sig figs.
So, a quick recap
#""Ag""_ 2""SO""_ (3(s)) rightleftharpoons color(red)(2)""Ag""_ ((aq))^(+) + ""SO""_ (3(aq))^(2-)# Here
#color(red)(2)# is a coefficient!
#K_(sp) = [""Ag""^(+)]^color(red)(2) * [""SO""_3^(2-)]# Here
#color(red)(2)# must be an exponent, so don't use it as a coefficient!
One last question from me. I thought it would be Ksp=[2Ag^+]^2[SO3^2-], but plugging in 1.80×10^-3 for Ag and doing algebra doesn't seem to get me the correct answer when I put it on the online homework tool we use. Thanks for your help! I still can't seem to understand concentrations.
The expression of the solubility product constant for silver sulfite looks like this
#K_(sp) = [""Ag""^(+)]^color(red)(2) * [""SO""_3^(2-)]#
This is the case because silver sulfite will partially ionize to produce silver(I) cations and sulfite anions in aqueous solution.
#""Ag""_ 2""SO""_ (3(s)) rightleftharpoons color(red)(2)""Ag""_ ((aq))^(+) + ""SO""_ (3(aq))^(2-)#
Notice that the fact that every mole of silver sulfite that dissociates in aqueous solution produce
Mind you, the coefficients of the ions are used as expoents in the expression of the solubility product constant, not as coefficients.
So something like
#K_(sp) = [color(red)(2)""Ag""^(+)]^color(red)(2) * [""SO""_3^(2-)]#
is not correct because the coefficient must only be used as an exponent in the expression of the solubility proeduct constant.
In your case, you have--I'll do the calculations with the added units for the solubility product constant!
#1.50 * 10^(-14) quad ""M""^3 = (1.80 * 10^(-3))^color(red)(2) quad ""M""^color(red)(2) * [""SO""_3^(2-)]#
So you can say that the equilibrium concentration of the sulfite anions will be equal to
#[S""O""_3^(2-)] = (1.50 * 10^(-14) quad ""M""^color(red)(cancel(color(black)(3))))/((1.80 * 10^(-3))^color(red)(2) quad color(red)(cancel(color(black)(""M""^2)))) = color(darkgreen)(ul(color(black)(4.63 * 10^(-9) quad ""M"")))#
The answer is rounded to three sig figs.
So, a quick recap
#""Ag""_ 2""SO""_ (3(s)) rightleftharpoons color(red)(2)""Ag""_ ((aq))^(+) + ""SO""_ (3(aq))^(2-)# Here
#color(red)(2)# is a coefficient!
#K_(sp) = [""Ag""^(+)]^color(red)(2) * [""SO""_3^(2-)]# Here
#color(red)(2)# must be an exponent, so don't use it as a coefficient!
We know that aluminum has a valency of
So in writing the formula, there is a cross of the two valencies because three chlorine atoms will have to react with one aluminium atom to gain one electron each.
#[""Al""]^(3+) + 3[""Cl""]^(-) -> ""AlCl""_3#
We know that aluminum has a valency of
So in writing the formula, there is a cross of the two valencies because three chlorine atoms will have to react with one aluminium atom to gain one electron each.
#[""Al""]^(3+) + 3[""Cl""]^(-) -> ""AlCl""_3#
We know that aluminum has a valency of
So in writing the formula, there is a cross of the two valencies because three chlorine atoms will have to react with one aluminium atom to gain one electron each.
#[""Al""]^(3+) + 3[""Cl""]^(-) -> ""AlCl""_3#
Oxidation states with ""n+"" mean there are n number of electrons lost or removed from the total number of electrons of the ground state element . ""n-"" on the other hand means there are n number of electrons gained or added to the total number of electrons of the ground state element.
3 electrons are gained.
Oxidation states with ""n+"" mean there are n number of electrons lost or removed from the total number of electrons of the ground state element . ""n-"" on the other hand means there are n number of electrons gained or added to the total number of electrons of the ground state element.
3 electrons are gained.
Oxidation states with ""n+"" mean there are n number of electrons lost or removed from the total number of electrons of the ground state element . ""n-"" on the other hand means there are n number of electrons gained or added to the total number of electrons of the ground state element.
First, let's rewrite
Now we can see how many of each element/polyatomic ion we have on each side of the equation.
Left:
Na - 1
H - 1
OH - 1
Right:
Na - 1
OH -1
H -2
Looking at what we have, there is an unequal amount of hydrogen on the left side and the right side. We have only 1 on the left and 2 on the right. Let's double everything on the left. Now we get:
Since we have 2 of everything on the left, we should make sure that the right side is accounted for. As mentioned in our original count, the
Changing back
First, let's rewrite
Now we can see how many of each element/polyatomic ion we have on each side of the equation.
Left:
Na - 1
H - 1
OH - 1
Right:
Na - 1
OH -1
H -2
Looking at what we have, there is an unequal amount of hydrogen on the left side and the right side. We have only 1 on the left and 2 on the right. Let's double everything on the left. Now we get:
Since we have 2 of everything on the left, we should make sure that the right side is accounted for. As mentioned in our original count, the
Changing back
First, let's rewrite
Now we can see how many of each element/polyatomic ion we have on each side of the equation.
Left:
Na - 1
H - 1
OH - 1
Right:
Na - 1
OH -1
H -2
Looking at what we have, there is an unequal amount of hydrogen on the left side and the right side. We have only 1 on the left and 2 on the right. Let's double everything on the left. Now we get:
Since we have 2 of everything on the left, we should make sure that the right side is accounted for. As mentioned in our original count, the
Changing back
This question is an example of the combined gas law. The formula is:
Organize the data:
Known
Unknown
Solution
Rearrange the formula to isolate
(
This question is an example of the combined gas law. The formula is:
Organize the data:
Known
Unknown
Solution
Rearrange the formula to isolate
(
This question is an example of the combined gas law. The formula is:
Organize the data:
Known
Unknown
Solution
Rearrange the formula to isolate
(
We use Charles' Law:
Plug in the values:
And don't forget to convert degrees Celsius to Kelvin! Just add 273.
Solve the algebra:
You can convert it back to degrees Celsius by subtracting 273.
391 K or 118 degrees Celsius
We use Charles' Law:
Plug in the values:
And don't forget to convert degrees Celsius to Kelvin! Just add 273.
Solve the algebra:
You can convert it back to degrees Celsius by subtracting 273.
391 K or 118 degrees Celsius
We use Charles' Law:
Plug in the values:
And don't forget to convert degrees Celsius to Kelvin! Just add 273.
Solve the algebra:
You can convert it back to degrees Celsius by subtracting 273.
Calcium sulfate dihydrate is an ionic substance composed of
Calcium sulfate dihydrate is an ionic substance composed of
Calcium sulfate dihydrate is an ionic substance composed of
And thus ONE equiv of dihydrogen results from each equiv of sulfuric acid (and also from each equiv of zinc metal).........
As far as I know
And thus
And so volume at
We interrogate the redox reaction......
And thus ONE equiv of dihydrogen results from each equiv of sulfuric acid (and also from each equiv of zinc metal).........
As far as I know
And thus
And so volume at
We interrogate the redox reaction......
And thus ONE equiv of dihydrogen results from each equiv of sulfuric acid (and also from each equiv of zinc metal).........
As far as I know
And thus
And so volume at
We calculate (i) a molar quantity of iron oxide:
And then (ii) convert this molar quantity into a mass of iron:
Why did I multiply the molar quantity by 2?
We calculate (i) a molar quantity of iron oxide:
And then (ii) convert this molar quantity into a mass of iron:
Why did I multiply the molar quantity by 2?
We calculate (i) a molar quantity of iron oxide:
And then (ii) convert this molar quantity into a mass of iron:
Why did I multiply the molar quantity by 2?
And thus for
Given that
Well, I gets a
And thus for
Given that
Well, I gets a
And thus for
Given that
Required Heat to raise the given temperature by
We get,
1125 calories
Required Heat to raise the given temperature by
We get,
1125 calories
Required Heat to raise the given temperature by
We get,
We were given
Use
to get
Now use
to get
So
We were given
Use
to get
Now use
to get
So
We were given
Use
to get
Now use
to get
So
The
We were given data for the reaction........
Since this reaction results in the formation of 2 mol of hydrogen radicals, i.e.
Agreed?
The
We were given data for the reaction........
Since this reaction results in the formation of 2 mol of hydrogen radicals, i.e.
Agreed?
The
We were given data for the reaction........
Since this reaction results in the formation of 2 mol of hydrogen radicals, i.e.
Agreed?
And thus...............................................
Well,
And thus...............................................
Well,
And thus...............................................
Notice that the number of atoms of
Now the
The equation is now balanced with 2
Notice that the number of atoms of
Now the
The equation is now balanced with 2
Notice that the number of atoms of
Now the
The equation is now balanced with 2
The empirical formula of a compound represents the lowest whole number ratio of elements in the compound. This ratio is represented by subscripts in the formula. In order to determine the empirical formula, we first must determine the number of moles of each element and then determine the mole ratio of each element. The mole ratio will give us the subscripts for the empirical formula.
Since the percentages of each element add up to 100, we can assume a 100 g sample of the compound and convert the percentages to grams.
Masses
Molar Masses
Moles
Divide the mass of each element by its molar mass (atomic weight on the periodic table in g/mol).
Mole Ratios
Divide the number of moles of each element by the lowest number of moles.
The empirical formula for this compound is
The empirical formula for the compound containing uranium and fluorine is
The empirical formula of a compound represents the lowest whole number ratio of elements in the compound. This ratio is represented by subscripts in the formula. In order to determine the empirical formula, we first must determine the number of moles of each element and then determine the mole ratio of each element. The mole ratio will give us the subscripts for the empirical formula.
Since the percentages of each element add up to 100, we can assume a 100 g sample of the compound and convert the percentages to grams.
Masses
Molar Masses
Moles
Divide the mass of each element by its molar mass (atomic weight on the periodic table in g/mol).
Mole Ratios
Divide the number of moles of each element by the lowest number of moles.
The empirical formula for this compound is
The empirical formula for the compound containing uranium and fluorine is
The empirical formula of a compound represents the lowest whole number ratio of elements in the compound. This ratio is represented by subscripts in the formula. In order to determine the empirical formula, we first must determine the number of moles of each element and then determine the mole ratio of each element. The mole ratio will give us the subscripts for the empirical formula.
Since the percentages of each element add up to 100, we can assume a 100 g sample of the compound and convert the percentages to grams.
Masses
Molar Masses
Moles
Divide the mass of each element by its molar mass (atomic weight on the periodic table in g/mol).
Mole Ratios
Divide the number of moles of each element by the lowest number of moles.
The empirical formula for this compound is
From the information given for this question, we can see that this kind of situation is involving Boyle's Law.
Boyle's Law states that the pressure of a fixed mass of gas at a constant temperature is inversely proportional to its volume .
In which from the definition, the equation is derived as;
When there are two situations , given initial and final value of both pressure and volume, the equation is derived as;
From the information given in this question;
Calculating
From the information given for this question, we can see that this kind of situation is involving Boyle's Law.
Boyle's Law states that the pressure of a fixed mass of gas at a constant temperature is inversely proportional to its volume .
In which from the definition, the equation is derived as;
When there are two situations , given initial and final value of both pressure and volume, the equation is derived as;
From the information given in this question;
Calculating
From the information given for this question, we can see that this kind of situation is involving Boyle's Law.
Boyle's Law states that the pressure of a fixed mass of gas at a constant temperature is inversely proportional to its volume .
In which from the definition, the equation is derived as;
When there are two situations , given initial and final value of both pressure and volume, the equation is derived as;
From the information given in this question;
Calculating
We want
And already we see from the quotient we ARE going to get an answer with
We use the combined gas law, which states for a GIVEN quantity of gas..........
We want
And already we see from the quotient we ARE going to get an answer with
We use the combined gas law, which states for a GIVEN quantity of gas..........
We want
And already we see from the quotient we ARE going to get an answer with
The pKa of acetic acid is 4.74.
Your strategy here will be to use the Henderson - Hasselbalch equation to find the ratio that must exist between the concentrations of the weak acid and of the conjugate in order to have a solution that has a pH of
The Henderson - Hasselbalch equation looks like this
#color(blue)(|bar(ul(color(white)(a/a)""pH"" = pK_a + log( ([""conjugate base""])/([""weak acid""]))color(white)(a/a)|)))#
Your buffer solution contains acetic acid,
This means that you have
#5.06 = 4.74 + log( ([""CH""_3""COO""^(-)])/([""CH""_3""COOH""]))#
Now, before doing any calculations, try to predict how many mmoles of acetate anions you'd need relative to the number of mmoles of acetic acid.
Notice that the pH of the target solution is higher than the
#5.06 > pK_a#
This tells you that the target solution must contain more conjugate base than weak acid. Implicitly, this solution must contain more moles of acetate anions than of acetic acid.
Rearrange the H - H equation to isolate the log term on one side
#0.32 = log( ([""CH""_3""COO""^(-)])/([""CH""_3""COOH""]))#
This will be equivalent to
#10^0.32 = 10^log( ([""CH""_3""COO""^(-)])/([""CH""_3""COOH""]))#
which will give you
#([""CH""_3""COO""^(-)])/([""CH""_3""COOH""]) = 10^0.32 = 2.09#
As predicted, the concentration of the conjugate base must be higher than that of the weak acid
#color(green)(|bar(ul(color(white)(a/a)color(black)([""CH""_3""COO""^(-)] = 2.09 xx [""CH""_3""COOH""])color(white)(a/a)|)))#
Now, the volume of the solution is the same for both chemical species, let's say
#n_(CH_3COO^(-))/color(red)(cancel(color(black)(V))) = 2.09 xx n_(CH_3COOH)/color(red)(cancel(color(black)(V)))#
Since the solution is said to contain
#n_(CH_3COO^(-)) = 2.09 xx ""10 mmoles"" = color(green)(|bar(ul(color(white)(a/a)""21 mmoles CH""_3""COO""^(-)color(white)(a/a)|)))#
I'll leave the answer rounded to two sig figs.
Your strategy here will be to use the Henderson - Hasselbalch equation to find the ratio that must exist between the concentrations of the weak acid and of the conjugate in order to have a solution that has a pH of
The Henderson - Hasselbalch equation looks like this
#color(blue)(|bar(ul(color(white)(a/a)""pH"" = pK_a + log( ([""conjugate base""])/([""weak acid""]))color(white)(a/a)|)))#
Your buffer solution contains acetic acid,
This means that you have
#5.06 = 4.74 + log( ([""CH""_3""COO""^(-)])/([""CH""_3""COOH""]))#
Now, before doing any calculations, try to predict how many mmoles of acetate anions you'd need relative to the number of mmoles of acetic acid.
Notice that the pH of the target solution is higher than the
#5.06 > pK_a#
This tells you that the target solution must contain more conjugate base than weak acid. Implicitly, this solution must contain more moles of acetate anions than of acetic acid.
Rearrange the H - H equation to isolate the log term on one side
#0.32 = log( ([""CH""_3""COO""^(-)])/([""CH""_3""COOH""]))#
This will be equivalent to
#10^0.32 = 10^log( ([""CH""_3""COO""^(-)])/([""CH""_3""COOH""]))#
which will give you
#([""CH""_3""COO""^(-)])/([""CH""_3""COOH""]) = 10^0.32 = 2.09#
As predicted, the concentration of the conjugate base must be higher than that of the weak acid
#color(green)(|bar(ul(color(white)(a/a)color(black)([""CH""_3""COO""^(-)] = 2.09 xx [""CH""_3""COOH""])color(white)(a/a)|)))#
Now, the volume of the solution is the same for both chemical species, let's say
#n_(CH_3COO^(-))/color(red)(cancel(color(black)(V))) = 2.09 xx n_(CH_3COOH)/color(red)(cancel(color(black)(V)))#
Since the solution is said to contain
#n_(CH_3COO^(-)) = 2.09 xx ""10 mmoles"" = color(green)(|bar(ul(color(white)(a/a)""21 mmoles CH""_3""COO""^(-)color(white)(a/a)|)))#
I'll leave the answer rounded to two sig figs.
The pKa of acetic acid is 4.74.
Your strategy here will be to use the Henderson - Hasselbalch equation to find the ratio that must exist between the concentrations of the weak acid and of the conjugate in order to have a solution that has a pH of
The Henderson - Hasselbalch equation looks like this
#color(blue)(|bar(ul(color(white)(a/a)""pH"" = pK_a + log( ([""conjugate base""])/([""weak acid""]))color(white)(a/a)|)))#
Your buffer solution contains acetic acid,
This means that you have
#5.06 = 4.74 + log( ([""CH""_3""COO""^(-)])/([""CH""_3""COOH""]))#
Now, before doing any calculations, try to predict how many mmoles of acetate anions you'd need relative to the number of mmoles of acetic acid.
Notice that the pH of the target solution is higher than the
#5.06 > pK_a#
This tells you that the target solution must contain more conjugate base than weak acid. Implicitly, this solution must contain more moles of acetate anions than of acetic acid.
Rearrange the H - H equation to isolate the log term on one side
#0.32 = log( ([""CH""_3""COO""^(-)])/([""CH""_3""COOH""]))#
This will be equivalent to
#10^0.32 = 10^log( ([""CH""_3""COO""^(-)])/([""CH""_3""COOH""]))#
which will give you
#([""CH""_3""COO""^(-)])/([""CH""_3""COOH""]) = 10^0.32 = 2.09#
As predicted, the concentration of the conjugate base must be higher than that of the weak acid
#color(green)(|bar(ul(color(white)(a/a)color(black)([""CH""_3""COO""^(-)] = 2.09 xx [""CH""_3""COOH""])color(white)(a/a)|)))#
Now, the volume of the solution is the same for both chemical species, let's say
#n_(CH_3COO^(-))/color(red)(cancel(color(black)(V))) = 2.09 xx n_(CH_3COOH)/color(red)(cancel(color(black)(V)))#
Since the solution is said to contain
#n_(CH_3COO^(-)) = 2.09 xx ""10 mmoles"" = color(green)(|bar(ul(color(white)(a/a)""21 mmoles CH""_3""COO""^(-)color(white)(a/a)|)))#
I'll leave the answer rounded to two sig figs.
Based on
So for each
Based on
So for each
Based on
So for each
Well, clearly we gots an
We gots
On the other hand (if you are an undergrad), phosphoric acid is ONLY a diacid in water...tritation with sodium hydroxide yields a stoichiometric endpoint at
And so we might have a biphosphate species of the form
You can use the ideal gas law to answer this question. The equation is:
where
We will use the ideal gas law to determine moles of diacetyl gas, then divide the given mass by the moles of diacetyl gas to determine its molar mass.
Known
https://chem.libretexts.org/Core/Physical_and_Theoretical_Chemistry/Physical_Properties_of_Matter/States_of_Matter/Properties_of_Gases/Gas_Laws/The_Ideal_Gas_Law
Unknown
Solve for
Rearrange the equation to isolate
Now that you have moles, divide the mass of diacetyl given in the question by the moles.
The molecular formula for diacetyl is
The molar mass of diacetyl,
You can use the ideal gas law to answer this question. The equation is:
where
We will use the ideal gas law to determine moles of diacetyl gas, then divide the given mass by the moles of diacetyl gas to determine its molar mass.
Known
https://chem.libretexts.org/Core/Physical_and_Theoretical_Chemistry/Physical_Properties_of_Matter/States_of_Matter/Properties_of_Gases/Gas_Laws/The_Ideal_Gas_Law
Unknown
Solve for
Rearrange the equation to isolate
Now that you have moles, divide the mass of diacetyl given in the question by the moles.
The molecular formula for diacetyl is
The molar mass of diacetyl,
You can use the ideal gas law to answer this question. The equation is:
where
We will use the ideal gas law to determine moles of diacetyl gas, then divide the given mass by the moles of diacetyl gas to determine its molar mass.
Known
https://chem.libretexts.org/Core/Physical_and_Theoretical_Chemistry/Physical_Properties_of_Matter/States_of_Matter/Properties_of_Gases/Gas_Laws/The_Ideal_Gas_Law
Unknown
Solve for
Rearrange the equation to isolate
Now that you have moles, divide the mass of diacetyl given in the question by the moles.
The molecular formula for diacetyl is
What we can do here is calculate the density of the diacetyl, and use that to directly calculate the molar mass. We will use the equation
where
Where is this equation derived from? Read the steps below if you would like to know, otherwise, skip to the next step.
Well, let's recall our ideal-gas equation, and rearrange it to solve for units similar to that of density,
#""mol""/""L""# , which is#n/V# :
#PV = nRT#
#P = (nRT)/V#
#P/(RT) = n/V# Now, let's multiply both sides of the equation by
#M# , the molar mass with units#""g""/""mol""# :
#(PM)/(RT) = (nM)/V# If we list the right side of the equation in terms of units, we have
#cancel(""mol"")/""L"" xx ""g""/cancel(""mol"") = ""g""/""L""# Which is the units for density. Thus, the value
#(nM)/V# is the density of the gas, and if we plug this back into our equation:
#(PM)/(RT) = (nM)/V = d# Thus,
#d = (MP)/(RT)# , and rearranging to solve for the molar mass yields our original equation,#M = (dRT)/P# .
The density of the diacetyl is
The temperature, in Kelvin, is
and the pressure, in atmospheres, is
Finally, plugging in all our known variables, we have
To check this, the molecular formula of diacetyl is
The equation for the combustion of methanol is:
Firstly you need to work out how many moles of carbon dioxide are in 5 litres at STP. If you assume that it is an idea gas (it isn't, but it's close enough for this sort of purpose) then 1 mole at STP occupies 22.4 litres. Therefore, at STP 5 litres is
Looking at the equation, it tells you that 2 moles of carbon dioxide are produced by complete combustion of 2 moles of methanol. Therefore, if you have 0.223 moles of carbon dioxide, you would need 0.223 moles of methanol.
Molar mass of methanol is 32.04 g/mol so you would need 0.223 x 32.04 = 7.14 g of methanol.
0.223 moles
The equation for the combustion of methanol is:
Firstly you need to work out how many moles of carbon dioxide are in 5 litres at STP. If you assume that it is an idea gas (it isn't, but it's close enough for this sort of purpose) then 1 mole at STP occupies 22.4 litres. Therefore, at STP 5 litres is
Looking at the equation, it tells you that 2 moles of carbon dioxide are produced by complete combustion of 2 moles of methanol. Therefore, if you have 0.223 moles of carbon dioxide, you would need 0.223 moles of methanol.
Molar mass of methanol is 32.04 g/mol so you would need 0.223 x 32.04 = 7.14 g of methanol.
0.223 moles
The equation for the combustion of methanol is:
Firstly you need to work out how many moles of carbon dioxide are in 5 litres at STP. If you assume that it is an idea gas (it isn't, but it's close enough for this sort of purpose) then 1 mole at STP occupies 22.4 litres. Therefore, at STP 5 litres is
Looking at the equation, it tells you that 2 moles of carbon dioxide are produced by complete combustion of 2 moles of methanol. Therefore, if you have 0.223 moles of carbon dioxide, you would need 0.223 moles of methanol.
Molar mass of methanol is 32.04 g/mol so you would need 0.223 x 32.04 = 7.14 g of methanol.
In order to find a solution's molar concentration, or molarity, you need to determine how many moles of solute, which in your case is sodium sulfate,
That is how molarity was defined -- the number of moles of solute in one liter of solution.
So, you know that you have
#""1 L"" = 10^3""mL""#
of solution. Convert the volume from milliliters to liters first
#12 color(red)(cancel(color(black)(""mL""))) * ""1 L""/(10^3color(red)(cancel(color(black)(""mL"")))) = ""0.012 L""#
If
#1 color(red)(cancel(color(black)(""L solution""))) * (""0.09 moles Na""_2""SO""_4)/(0.012color(red)(cancel(color(black)(""L solution"")))) = ""7.5 moles Na""_2""SO""_4#
So, if
#""molarity"" = color(green)(|bar(ul(color(white)(a/a)color(black)(""7.5 mol L""^(-1) = ""7.5 molar"" = ""7.5 M"")color(white)(a/a)|)))#
The answer is rounded to two sig figs.
In order to find a solution's molar concentration, or molarity, you need to determine how many moles of solute, which in your case is sodium sulfate,
That is how molarity was defined -- the number of moles of solute in one liter of solution.
So, you know that you have
#""1 L"" = 10^3""mL""#
of solution. Convert the volume from milliliters to liters first
#12 color(red)(cancel(color(black)(""mL""))) * ""1 L""/(10^3color(red)(cancel(color(black)(""mL"")))) = ""0.012 L""#
If
#1 color(red)(cancel(color(black)(""L solution""))) * (""0.09 moles Na""_2""SO""_4)/(0.012color(red)(cancel(color(black)(""L solution"")))) = ""7.5 moles Na""_2""SO""_4#
So, if
#""molarity"" = color(green)(|bar(ul(color(white)(a/a)color(black)(""7.5 mol L""^(-1) = ""7.5 molar"" = ""7.5 M"")color(white)(a/a)|)))#
The answer is rounded to two sig figs.
In order to find a solution's molar concentration, or molarity, you need to determine how many moles of solute, which in your case is sodium sulfate,
That is how molarity was defined -- the number of moles of solute in one liter of solution.
So, you know that you have
#""1 L"" = 10^3""mL""#
of solution. Convert the volume from milliliters to liters first
#12 color(red)(cancel(color(black)(""mL""))) * ""1 L""/(10^3color(red)(cancel(color(black)(""mL"")))) = ""0.012 L""#
If
#1 color(red)(cancel(color(black)(""L solution""))) * (""0.09 moles Na""_2""SO""_4)/(0.012color(red)(cancel(color(black)(""L solution"")))) = ""7.5 moles Na""_2""SO""_4#
So, if
#""molarity"" = color(green)(|bar(ul(color(white)(a/a)color(black)(""7.5 mol L""^(-1) = ""7.5 molar"" = ""7.5 M"")color(white)(a/a)|)))#
The answer is rounded to two sig figs.
#pH = -log[H^+]#
#pH = -log[H^+]#
#pH = -log[H^+]#
The is the arsenical analogue of phosphoric acid,
Perhaps
The is the arsenical analogue of phosphoric acid,
Perhaps
The is the arsenical analogue of phosphoric acid,
The salt is composed of
The salt is composed of
The salt is composed of
As you know, the molality of a solution tells you the number of moles of solute present for every
This means that the first thing that you need to do here is to figure out how many grams of water are present in your sample. To do that, use the density of water.
#500. color(red)(cancel(color(black)(""mL""))) * ""1.00 g""/(1color(red)(cancel(color(black)(""mL"")))) = ""500. g""#
Next, use the molar mass of the solute to determine how many moles are present in the sample.
#115 color(red)(cancel(color(black)(""g""))) * ""1 mole NanO""_3/(85.0color(red)(cancel(color(black)(""g"")))) = ""1.353 moles NaNO""_3#
So, you know that this solution will contain
In order to find the molality of the solution, you must figure out how many moles of solute would be present for
#10^3 color(red)(cancel(color(black)(""g water""))) * ""1.353 moles NaNO""_3/(500. color(red)(cancel(color(black)(""g water"")))) = ""2.706 moles NaNO""_3#
You can thus say that the molality of the solution is equal to
#color(darkgreen)(ul(color(black)(""molality""))) = ""2.706 mol kg""^(-1) ~~ color(darkgreen)(ul(color(black)(""2.71 mol kg""^(-1)))#
The answer is rounded to three sig figs.
So, you know that this solution will contain
As you know, the molality of a solution tells you the number of moles of solute present for every
This means that the first thing that you need to do here is to figure out how many grams of water are present in your sample. To do that, use the density of water.
#500. color(red)(cancel(color(black)(""mL""))) * ""1.00 g""/(1color(red)(cancel(color(black)(""mL"")))) = ""500. g""#
Next, use the molar mass of the solute to determine how many moles are present in the sample.
#115 color(red)(cancel(color(black)(""g""))) * ""1 mole NanO""_3/(85.0color(red)(cancel(color(black)(""g"")))) = ""1.353 moles NaNO""_3#
So, you know that this solution will contain
In order to find the molality of the solution, you must figure out how many moles of solute would be present for
#10^3 color(red)(cancel(color(black)(""g water""))) * ""1.353 moles NaNO""_3/(500. color(red)(cancel(color(black)(""g water"")))) = ""2.706 moles NaNO""_3#
You can thus say that the molality of the solution is equal to
#color(darkgreen)(ul(color(black)(""molality""))) = ""2.706 mol kg""^(-1) ~~ color(darkgreen)(ul(color(black)(""2.71 mol kg""^(-1)))#
The answer is rounded to three sig figs.
So, you know that this solution will contain
As you know, the molality of a solution tells you the number of moles of solute present for every
This means that the first thing that you need to do here is to figure out how many grams of water are present in your sample. To do that, use the density of water.
#500. color(red)(cancel(color(black)(""mL""))) * ""1.00 g""/(1color(red)(cancel(color(black)(""mL"")))) = ""500. g""#
Next, use the molar mass of the solute to determine how many moles are present in the sample.
#115 color(red)(cancel(color(black)(""g""))) * ""1 mole NanO""_3/(85.0color(red)(cancel(color(black)(""g"")))) = ""1.353 moles NaNO""_3#
So, you know that this solution will contain
In order to find the molality of the solution, you must figure out how many moles of solute would be present for
#10^3 color(red)(cancel(color(black)(""g water""))) * ""1.353 moles NaNO""_3/(500. color(red)(cancel(color(black)(""g water"")))) = ""2.706 moles NaNO""_3#
You can thus say that the molality of the solution is equal to
#color(darkgreen)(ul(color(black)(""molality""))) = ""2.706 mol kg""^(-1) ~~ color(darkgreen)(ul(color(black)(""2.71 mol kg""^(-1)))#
The answer is rounded to three sig figs.
So, you know that this solution will contain
We interrogate the equilibrium...............
Now we know that
So at equilibrium,
This is quadratic in
Thus
Now that we an approximation for
And thus..........................................................
Since the values have converged, we can take this value as the true value.
And so
But we need
And so
We interrogate the equilibrium...............
Now we know that
So at equilibrium,
This is quadratic in
Thus
Now that we an approximation for
And thus..........................................................
Since the values have converged, we can take this value as the true value.
And so
But we need
And so
We interrogate the equilibrium...............
Now we know that
So at equilibrium,
This is quadratic in
Thus
Now that we an approximation for
And thus..........................................................
Since the values have converged, we can take this value as the true value.
And so
But we need
And so
Carbon is oxidized from
Where do the electrons go? To oxygen:
We add (i) and (ii) to get:
Now on the right side we have
Thus:
To give (finally!):
As the balanced redox equation.
Carbon has been oxidized, and dioxygen has been reduced. Mass and charge are balanced as required. This is a lot of work for a simple oxidation reaction. And of course you would never do this formally. You would (i) balance the carbons, (ii) then the hydrogens, and (iii) finally the oxygens, all directly, without all that tedious mucking about with electrons
Carbon is oxidized from
Where do the electrons go? To oxygen:
We add (i) and (ii) to get:
Now on the right side we have
Thus:
To give (finally!):
As the balanced redox equation.
Carbon has been oxidized, and dioxygen has been reduced. Mass and charge are balanced as required. This is a lot of work for a simple oxidation reaction. And of course you would never do this formally. You would (i) balance the carbons, (ii) then the hydrogens, and (iii) finally the oxygens, all directly, without all that tedious mucking about with electrons
Carbon is oxidized from
Where do the electrons go? To oxygen:
We add (i) and (ii) to get:
Now on the right side we have
Thus:
To give (finally!):
As the balanced redox equation.
Carbon has been oxidized, and dioxygen has been reduced. Mass and charge are balanced as required. This is a lot of work for a simple oxidation reaction. And of course you would never do this formally. You would (i) balance the carbons, (ii) then the hydrogens, and (iii) finally the oxygens, all directly, without all that tedious mucking about with electrons
Is the given equation balanced with respect to mass and charge? If the answer is NON, it cannot be accepted as a representation of chemical reality.
Lithium is an active metal, and a strong enuff reductant to reduce the water molecule to give dihydrogen gas and the metal hydroxide......
Is the given equation balanced with respect to mass and charge? If the answer is NON, it cannot be accepted as a representation of chemical reality.
Lithium is an active metal, and a strong enuff reductant to reduce the water molecule to give dihydrogen gas and the metal hydroxide......
Is the given equation balanced with respect to mass and charge? If the answer is NON, it cannot be accepted as a representation of chemical reality.
Lithium is an active metal, and a strong enuff reductant to reduce the water molecule to give dihydrogen gas and the metal hydroxide......
Your starting point here will be the balanced chemical equation for this synthesis reaction
#2""Na""_text((s]) + ""F""_text(2(g]) -> 2""NaF""_text((s])#
Now, the standard enthalpy of formation is always given for the formation of one mole of a substance. In this case, the chemical equation that describes the formation of one mole of sodium fluoride looks like this
#""Na""_text((s]) + 1/2""F""_text(2(g]) -> ""NaF""_text((s])#
Now, notice that you have a
You know that when
#0.560 color(red)(cancel(color(black)(""g""))) * ""1 mole Na""/(23.0color(red)(cancel(color(black)(""g"")))) = ""0.02435 moles Na""#
So, if
#1 color(red)(cancel(color(black)(""mole Na""))) * ""13.8 kJ""/(0.02435color(red)(cancel(color(black)(""moles Na"")))) = ""566.7 kJ""#
Now, when heat is given off, the standard enthalpy change of formation carries a negative sign. This means that the enthalpy change of formation for sodium fluoride will be
#DeltaH_""f""^@ = - color(green)(""567 kJ/mol"")#
The answer is rounded to three sig figs.
The listed value for sodium fluoride's standard enthalpy of formation is
http://nshs-science.net/chemistry/common/pdf/R-standard_enthalpy_of_formation.pdf
Your starting point here will be the balanced chemical equation for this synthesis reaction
#2""Na""_text((s]) + ""F""_text(2(g]) -> 2""NaF""_text((s])#
Now, the standard enthalpy of formation is always given for the formation of one mole of a substance. In this case, the chemical equation that describes the formation of one mole of sodium fluoride looks like this
#""Na""_text((s]) + 1/2""F""_text(2(g]) -> ""NaF""_text((s])#
Now, notice that you have a
You know that when
#0.560 color(red)(cancel(color(black)(""g""))) * ""1 mole Na""/(23.0color(red)(cancel(color(black)(""g"")))) = ""0.02435 moles Na""#
So, if
#1 color(red)(cancel(color(black)(""mole Na""))) * ""13.8 kJ""/(0.02435color(red)(cancel(color(black)(""moles Na"")))) = ""566.7 kJ""#
Now, when heat is given off, the standard enthalpy change of formation carries a negative sign. This means that the enthalpy change of formation for sodium fluoride will be
#DeltaH_""f""^@ = - color(green)(""567 kJ/mol"")#
The answer is rounded to three sig figs.
The listed value for sodium fluoride's standard enthalpy of formation is
http://nshs-science.net/chemistry/common/pdf/R-standard_enthalpy_of_formation.pdf
Your starting point here will be the balanced chemical equation for this synthesis reaction
#2""Na""_text((s]) + ""F""_text(2(g]) -> 2""NaF""_text((s])#
Now, the standard enthalpy of formation is always given for the formation of one mole of a substance. In this case, the chemical equation that describes the formation of one mole of sodium fluoride looks like this
#""Na""_text((s]) + 1/2""F""_text(2(g]) -> ""NaF""_text((s])#
Now, notice that you have a
You know that when
#0.560 color(red)(cancel(color(black)(""g""))) * ""1 mole Na""/(23.0color(red)(cancel(color(black)(""g"")))) = ""0.02435 moles Na""#
So, if
#1 color(red)(cancel(color(black)(""mole Na""))) * ""13.8 kJ""/(0.02435color(red)(cancel(color(black)(""moles Na"")))) = ""566.7 kJ""#
Now, when heat is given off, the standard enthalpy change of formation carries a negative sign. This means that the enthalpy change of formation for sodium fluoride will be
#DeltaH_""f""^@ = - color(green)(""567 kJ/mol"")#
The answer is rounded to three sig figs.
The listed value for sodium fluoride's standard enthalpy of formation is
http://nshs-science.net/chemistry/common/pdf/R-standard_enthalpy_of_formation.pdf
No. of valence electrons:
So around each atom (from left to right above), there are 6, and 5, and 7 valence electrons. Because there are 7 valence electrons around the rightmost oxygen (i.e. 9 electrons in total including the inner shell electrons), this oxygen has a formal negative charge.
Of course there is a resonance isomer:
The neutral nitrogen atom has a formal lone pair in each mesomer.
How would you describe the structure of this ion based on VSEPR?
No. of valence electrons:
So around each atom (from left to right above), there are 6, and 5, and 7 valence electrons. Because there are 7 valence electrons around the rightmost oxygen (i.e. 9 electrons in total including the inner shell electrons), this oxygen has a formal negative charge.
Of course there is a resonance isomer:
The neutral nitrogen atom has a formal lone pair in each mesomer.
How would you describe the structure of this ion based on VSEPR?
No. of valence electrons:
So around each atom (from left to right above), there are 6, and 5, and 7 valence electrons. Because there are 7 valence electrons around the rightmost oxygen (i.e. 9 electrons in total including the inner shell electrons), this oxygen has a formal negative charge.
Of course there is a resonance isomer:
The neutral nitrogen atom has a formal lone pair in each mesomer.
How would you describe the structure of this ion based on VSEPR?
gives 0.893 moles.
An ideal gas at STP occupies 22.4 liters. Calculating Oxygen as if it were an ideal gas there are .893 moles of Oxygen in 20.0 liters.
gives 0.893 moles.
An ideal gas at STP occupies 22.4 liters. Calculating Oxygen as if it were an ideal gas there are .893 moles of Oxygen in 20.0 liters.
gives 0.893 moles.
And so we takes one of the former and six of the latter......
And so we takes one of the former and six of the latter......
And so we takes one of the former and six of the latter......
The change in water's freezing point is simply the freezing-point depression that is produced by the dissolving of sodium chloride,
As its name suggests, the freezing-point depression tells you by how many degrees the freezing point of the solution will decrease compared with that of the pure solvent.
The equation that allows you to calculate freezing-point depression looks like this
#color(blue)(|bar(ul(color(white)(a/a)DeltaT_f = i * K_f * bcolor(white)(a/a)|)))#
Here
Water's cryoscopic constant is listed as
#K_f = 1.86^@""C kg mol""^(-1)#
http://www.vaxasoftware.com/doc_eduen/qui/tcriosebu.pdf
Now, sodium chloride is a strong electrolyte, which means that it dissociates completely in aqueous solution to form sodium cations,
#""NaCl""_ ((aq)) -> ""Na""_ ((aq))^(+) + ""Cl""_ ((aq))^(-)#
Notice that every mole of sodium chloride that is dissolved in aqueous solution produces
This means that the van't Hoff factor, which tells you the ratio that exists between the number of moles of solute dissolved and the number of particles of solute produced in solution, will be equal to
#i = 2 -># one mole dissolved, two moles of ions produced
The next thing to do here is calculate the solution's molality, which is equal to the number of moles of solute present in
Use sodium chloride's molar mass to convert the grams of solute to moles
#3 color(red)(cancel(color(black)(""g""))) * ""1 mole NaCl""/(58.443color(red)(cancel(color(black)(""g"")))) = ""0.08555 moles NaCl""#
Convert the mass of water from grams to kilograms
#75.25 color(red)(cancel(color(black)(""g""))) * ""1 kg""/(10^3color(red)(cancel(color(black)(""kg"")))) = ""0.7525 kg""#
The molality of the solution will be equal to
#b = ""0.08555 moles""/""0.7525 kg"" = ""0.1137 mol kg""^(-1)#
Plug in your values to calculate the value of
#DeltaT_f = 2 * 1.86^@""C"" color(red)(cancel(color(black)(""kg""))) color(red)(cancel(color(black)(""mol""^(-1)))) * 0.1137color(red)(cancel(color(black)(""mol""))) color(red)(cancel(color(black)(""kg"")))#
#DeltaT_f = color(green)(|bar(ul(color(white)(a/a)color(black)(0.42^@""C"")color(white)(a/a)|)))#
I'll leave the answer rounded to two sig figs, but don't forget that you have one sig fig for the mass of sodium chloride.
So, this value tells you that the freezing point of the solution will be
The change in water's freezing point is simply the freezing-point depression that is produced by the dissolving of sodium chloride,
As its name suggests, the freezing-point depression tells you by how many degrees the freezing point of the solution will decrease compared with that of the pure solvent.
The equation that allows you to calculate freezing-point depression looks like this
#color(blue)(|bar(ul(color(white)(a/a)DeltaT_f = i * K_f * bcolor(white)(a/a)|)))#
Here
Water's cryoscopic constant is listed as
#K_f = 1.86^@""C kg mol""^(-1)#
http://www.vaxasoftware.com/doc_eduen/qui/tcriosebu.pdf
Now, sodium chloride is a strong electrolyte, which means that it dissociates completely in aqueous solution to form sodium cations,
#""NaCl""_ ((aq)) -> ""Na""_ ((aq))^(+) + ""Cl""_ ((aq))^(-)#
Notice that every mole of sodium chloride that is dissolved in aqueous solution produces
This means that the van't Hoff factor, which tells you the ratio that exists between the number of moles of solute dissolved and the number of particles of solute produced in solution, will be equal to
#i = 2 -># one mole dissolved, two moles of ions produced
The next thing to do here is calculate the solution's molality, which is equal to the number of moles of solute present in
Use sodium chloride's molar mass to convert the grams of solute to moles
#3 color(red)(cancel(color(black)(""g""))) * ""1 mole NaCl""/(58.443color(red)(cancel(color(black)(""g"")))) = ""0.08555 moles NaCl""#
Convert the mass of water from grams to kilograms
#75.25 color(red)(cancel(color(black)(""g""))) * ""1 kg""/(10^3color(red)(cancel(color(black)(""kg"")))) = ""0.7525 kg""#
The molality of the solution will be equal to
#b = ""0.08555 moles""/""0.7525 kg"" = ""0.1137 mol kg""^(-1)#
Plug in your values to calculate the value of
#DeltaT_f = 2 * 1.86^@""C"" color(red)(cancel(color(black)(""kg""))) color(red)(cancel(color(black)(""mol""^(-1)))) * 0.1137color(red)(cancel(color(black)(""mol""))) color(red)(cancel(color(black)(""kg"")))#
#DeltaT_f = color(green)(|bar(ul(color(white)(a/a)color(black)(0.42^@""C"")color(white)(a/a)|)))#
I'll leave the answer rounded to two sig figs, but don't forget that you have one sig fig for the mass of sodium chloride.
So, this value tells you that the freezing point of the solution will be
The change in water's freezing point is simply the freezing-point depression that is produced by the dissolving of sodium chloride,
As its name suggests, the freezing-point depression tells you by how many degrees the freezing point of the solution will decrease compared with that of the pure solvent.
The equation that allows you to calculate freezing-point depression looks like this
#color(blue)(|bar(ul(color(white)(a/a)DeltaT_f = i * K_f * bcolor(white)(a/a)|)))#
Here
Water's cryoscopic constant is listed as
#K_f = 1.86^@""C kg mol""^(-1)#
http://www.vaxasoftware.com/doc_eduen/qui/tcriosebu.pdf
Now, sodium chloride is a strong electrolyte, which means that it dissociates completely in aqueous solution to form sodium cations,
#""NaCl""_ ((aq)) -> ""Na""_ ((aq))^(+) + ""Cl""_ ((aq))^(-)#
Notice that every mole of sodium chloride that is dissolved in aqueous solution produces
This means that the van't Hoff factor, which tells you the ratio that exists between the number of moles of solute dissolved and the number of particles of solute produced in solution, will be equal to
#i = 2 -># one mole dissolved, two moles of ions produced
The next thing to do here is calculate the solution's molality, which is equal to the number of moles of solute present in
Use sodium chloride's molar mass to convert the grams of solute to moles
#3 color(red)(cancel(color(black)(""g""))) * ""1 mole NaCl""/(58.443color(red)(cancel(color(black)(""g"")))) = ""0.08555 moles NaCl""#
Convert the mass of water from grams to kilograms
#75.25 color(red)(cancel(color(black)(""g""))) * ""1 kg""/(10^3color(red)(cancel(color(black)(""kg"")))) = ""0.7525 kg""#
The molality of the solution will be equal to
#b = ""0.08555 moles""/""0.7525 kg"" = ""0.1137 mol kg""^(-1)#
Plug in your values to calculate the value of
#DeltaT_f = 2 * 1.86^@""C"" color(red)(cancel(color(black)(""kg""))) color(red)(cancel(color(black)(""mol""^(-1)))) * 0.1137color(red)(cancel(color(black)(""mol""))) color(red)(cancel(color(black)(""kg"")))#
#DeltaT_f = color(green)(|bar(ul(color(white)(a/a)color(black)(0.42^@""C"")color(white)(a/a)|)))#
I'll leave the answer rounded to two sig figs, but don't forget that you have one sig fig for the mass of sodium chloride.
So, this value tells you that the freezing point of the solution will be
You have been given the empirical formula for a hydrocarbon. You need to determine the empirical formula mass. Then divide the molecular mass by the empirical mass. Then multiply the subscripts of the empirical formula by the result.
Multiply the subscript of each element by its atomic mass and add.
Divide the molecular mass by the empirical mass.
Multiply the subscripts in the empirical formula by 2 to get the molecular formula.
The molecular formula is
The molecular formula is
Refer to the explanation for the process.
You have been given the empirical formula for a hydrocarbon. You need to determine the empirical formula mass. Then divide the molecular mass by the empirical mass. Then multiply the subscripts of the empirical formula by the result.
Multiply the subscript of each element by its atomic mass and add.
Divide the molecular mass by the empirical mass.
Multiply the subscripts in the empirical formula by 2 to get the molecular formula.
The molecular formula is
The molecular formula is
Refer to the explanation for the process.
You have been given the empirical formula for a hydrocarbon. You need to determine the empirical formula mass. Then divide the molecular mass by the empirical mass. Then multiply the subscripts of the empirical formula by the result.
Multiply the subscript of each element by its atomic mass and add.
Divide the molecular mass by the empirical mass.
Multiply the subscripts in the empirical formula by 2 to get the molecular formula.
The molecular formula is
This simply an application of the Ideal Gas Equation. Units of pressure are always a problem, because these units dictate the choice of gas constant,
This simply an application of the Ideal Gas Equation. Units of pressure are always a problem, because these units dictate the choice of gas constant,
This simply an application of the Ideal Gas Equation. Units of pressure are always a problem, because these units dictate the choice of gas constant,
And so we completely COMBUST propane,
Now this works well for odd-numbered alkanes....for EVEN-NUMBERED alkanes...we reach a problem...
...OR....
Sometimes the latter equation is preferred because if you use a half-integral coefficient, you might die. I tend to find the stoichiometry of the FORMER reaction a bit easier to use when calculating stoichiometric equivalence. In either scenario CHARGE and MASS are balanced ABSOLUTELY....as indeed they must be if we purport to represent an actual chemical reaction.
The typical rigmarole is to....
And so we completely COMBUST propane,
Now this works well for odd-numbered alkanes....for EVEN-NUMBERED alkanes...we reach a problem...
...OR....
Sometimes the latter equation is preferred because if you use a half-integral coefficient, you might die. I tend to find the stoichiometry of the FORMER reaction a bit easier to use when calculating stoichiometric equivalence. In either scenario CHARGE and MASS are balanced ABSOLUTELY....as indeed they must be if we purport to represent an actual chemical reaction.
The typical rigmarole is to....
And so we completely COMBUST propane,
Now this works well for odd-numbered alkanes....for EVEN-NUMBERED alkanes...we reach a problem...
...OR....
Sometimes the latter equation is preferred because if you use a half-integral coefficient, you might die. I tend to find the stoichiometry of the FORMER reaction a bit easier to use when calculating stoichiometric equivalence. In either scenario CHARGE and MASS are balanced ABSOLUTELY....as indeed they must be if we purport to represent an actual chemical reaction.
The chemical equation for this reaction is
then
and finally
The balanced chemical equation would be:
HOWEVER,
if too much or too little of
or
Source: https://en.wikipedia.org/wiki/Propane#Properties_and_reactions
For such mass related problems we use average mass of elements or compounds. A verge mass of element is dependent on its isotopes and respective fraction found naturally.
In the given question we need to know average mass of Phosphorus and Oxygen.
average mass of Phosphorus P = 30.973762 amu
average mass of Oxygen O = 15.9994 amu
Molar mass of
For such mass related problems we use average mass of elements or compounds. A verge mass of element is dependent on its isotopes and respective fraction found naturally.
In the given question we need to know average mass of Phosphorus and Oxygen.
average mass of Phosphorus P = 30.973762 amu
average mass of Oxygen O = 15.9994 amu
Molar mass of
For such mass related problems we use average mass of elements or compounds. A verge mass of element is dependent on its isotopes and respective fraction found naturally.
In the given question we need to know average mass of Phosphorus and Oxygen.
average mass of Phosphorus P = 30.973762 amu
average mass of Oxygen O = 15.9994 amu
Molar mass of
So, we convert the temperatures to the absolute scale, and solve the quotient:
You should convert this temperature back to the
With a constant amount of gas,
So, we convert the temperatures to the absolute scale, and solve the quotient:
You should convert this temperature back to the
With a constant amount of gas,
So, we convert the temperatures to the absolute scale, and solve the quotient:
You should convert this temperature back to the
We must first identify the limiting reactant, and then we calculate the theoretical yield.
We start with the balanced equation.
(a) Identify the limiting reactant
We calculate the amount of
Calculate the moles of
Calculate moles of
Calculate the moles of
Calculate moles of
The limiting reactant is
(b) Calculate the theoretical yield of
(c) Calculate the actual yield of
If % yield = 75 %, then
The mass of ammonia will be 13 g.
We must first identify the limiting reactant, and then we calculate the theoretical yield.
We start with the balanced equation.
(a) Identify the limiting reactant
We calculate the amount of
Calculate the moles of
Calculate moles of
Calculate the moles of
Calculate moles of
The limiting reactant is
(b) Calculate the theoretical yield of
(c) Calculate the actual yield of
If % yield = 75 %, then
The mass of ammonia will be 13 g.
We must first identify the limiting reactant, and then we calculate the theoretical yield.
We start with the balanced equation.
(a) Identify the limiting reactant
We calculate the amount of
Calculate the moles of
Calculate moles of
Calculate the moles of
Calculate moles of
The limiting reactant is
(b) Calculate the theoretical yield of
(c) Calculate the actual yield of
If % yield = 75 %, then
We have
We have
We have
Now the mole is simply a number, i.e.
ANd so we multiply this molar quantity by the molar mass of
And so you want the MASS of
Now the mole is simply a number, i.e.
ANd so we multiply this molar quantity by the molar mass of
And so you want the MASS of
Now the mole is simply a number, i.e.
ANd so we multiply this molar quantity by the molar mass of
And so
We put in some numbers, and we assume that
And we plug in the numbers to our equilibrium expression..
And this is a quadratic in
And thus
And now that we have an approximation for
This method is as exact as if we used the quadratic equation, and so.....
We know that in aqueous solution,
Well, we interrogate the equilibrium.......and gets
And so
We put in some numbers, and we assume that
And we plug in the numbers to our equilibrium expression..
And this is a quadratic in
And thus
And now that we have an approximation for
This method is as exact as if we used the quadratic equation, and so.....
We know that in aqueous solution,
Well, we interrogate the equilibrium.......and gets
And so
We put in some numbers, and we assume that
And we plug in the numbers to our equilibrium expression..
And this is a quadratic in
And thus
And now that we have an approximation for
This method is as exact as if we used the quadratic equation, and so.....
We know that in aqueous solution,
Well, molarity is temperature-dependent, so I will assume
and I got
And at these conditions,
#400 cancel(""g H""_2""O"") xx ""1 mL""/(0.9970749 cancel""g"")#
#=# #""401.17 mL""#
And so, the molarity is given by:
#""M"" = ""mols solute""/""L solution""# (and NOT solvent!)
#= [""20 g NaOH"" xx (""1 mol NaOH"")/(""(22.989 + 15.999 + 1.0079 g) NaOH"")]/(401.17 xx 10^(-3) ""L solvent"" + V_""solute"")#
We could assume that the solvent volume does not differ from the solution volume, but that is a lie... so let's use the density of
#20 cancel""g NaOH"" xx (cancel""1 mL"")/(2.13 cancel""g"") xx ""1 L""/(1000 cancel""mL"")#
#=# #""0.009390 L""#
And so, the solvent volume will rise by about
#color(blue)([""NaOH""]) = [""0.5001 mols NaOH""]/(401.17 xx 10^(-3) ""L solvent"" + ""0.009390 L NaOH"")#
#=# #ulcolor(blue)(""1.218 M"")# (Had you assumed
#V_(""soln"") ~~ V_""solvent""# , you would have gotten about#""1.25 M""# .)
But you have only allowed yourself one measly significant figure, so I guess you only have a
Well, molarity is temperature-dependent, so I will assume
and I got
And at these conditions,
#400 cancel(""g H""_2""O"") xx ""1 mL""/(0.9970749 cancel""g"")#
#=# #""401.17 mL""#
And so, the molarity is given by:
#""M"" = ""mols solute""/""L solution""# (and NOT solvent!)
#= [""20 g NaOH"" xx (""1 mol NaOH"")/(""(22.989 + 15.999 + 1.0079 g) NaOH"")]/(401.17 xx 10^(-3) ""L solvent"" + V_""solute"")#
We could assume that the solvent volume does not differ from the solution volume, but that is a lie... so let's use the density of
#20 cancel""g NaOH"" xx (cancel""1 mL"")/(2.13 cancel""g"") xx ""1 L""/(1000 cancel""mL"")#
#=# #""0.009390 L""#
And so, the solvent volume will rise by about
#color(blue)([""NaOH""]) = [""0.5001 mols NaOH""]/(401.17 xx 10^(-3) ""L solvent"" + ""0.009390 L NaOH"")#
#=# #ulcolor(blue)(""1.218 M"")# (Had you assumed
#V_(""soln"") ~~ V_""solvent""# , you would have gotten about#""1.25 M""# .)
But you have only allowed yourself one measly significant figure, so I guess you only have a
Well, molarity is temperature-dependent, so I will assume
and I got
And at these conditions,
#400 cancel(""g H""_2""O"") xx ""1 mL""/(0.9970749 cancel""g"")#
#=# #""401.17 mL""#
And so, the molarity is given by:
#""M"" = ""mols solute""/""L solution""# (and NOT solvent!)
#= [""20 g NaOH"" xx (""1 mol NaOH"")/(""(22.989 + 15.999 + 1.0079 g) NaOH"")]/(401.17 xx 10^(-3) ""L solvent"" + V_""solute"")#
We could assume that the solvent volume does not differ from the solution volume, but that is a lie... so let's use the density of
#20 cancel""g NaOH"" xx (cancel""1 mL"")/(2.13 cancel""g"") xx ""1 L""/(1000 cancel""mL"")#
#=# #""0.009390 L""#
And so, the solvent volume will rise by about
#color(blue)([""NaOH""]) = [""0.5001 mols NaOH""]/(401.17 xx 10^(-3) ""L solvent"" + ""0.009390 L NaOH"")#
#=# #ulcolor(blue)(""1.218 M"")# (Had you assumed
#V_(""soln"") ~~ V_""solvent""# , you would have gotten about#""1.25 M""# .)
But you have only allowed yourself one measly significant figure, so I guess you only have a
1.25M
The other posted solution is detailed and accurate, but possibly ""over kill"" for this venue. While 20 technically does have only one significant figure without a designated decimal point, I will also assume it to be 20. for calculations.
Also assuming STP for a general chemistry question of this sort, I calculate the moles of NaOH as 0.5 and use the (estimated) density of water at 1g/mL to get 400mL of solvent.
0.5M/400mL = 1.25M solution.
Right on, or certainly within any of the error probabilities, or assumptions of the more ""rigorous"" answer. You see, chemistry doesn't have to be intimidating.
The first thing to do here is write a balanced chemical equation that describes this neutralization reaction
#""KOH""_ ((aq)) + ""HCl""_ ((aq)) -> ""KCl""_ ((aq)) + ""H""_ 2""O""_ ((l))#
Potassium hydroxide,
This tells you that in order to have a complete neutralization, you need to mix equal numbers of moles of each reactant.
Now, the problem provides you with the molarity and volume of the hydrochloric acid solution, which you can use to determine how many moles of acid were needed for the reaction
#color(purple)(|bar(ul(color(white)(a/a)color(black)(c = n_""solute""/V_""solution"" implies n_""solute"" = c * V_""solution"")color(white)(a/a)|)))#
You will have
#n_""HCl"" = ""2.50 mol"" color(red)(cancel(color(black)(""L""^(-1)))) * overbrace(100.0 * 10^(-3)color(red)(cancel(color(black)(""L""))))^(color(blue)(""volume in liters""))#
#n_""HCl"" = ""0.250 moles HCl""#
This is exactly how many moles of potassium hydroxide you must add to the reaction. All you have to do now is use the molarity of the potassium hydroxide solution to figure out what volume would contain that many moles
#color(purple)(|bar(ul(color(white)(a/a)color(black)(c = n_""solute""/V_""solution"" implies V_""solution"" = n_""solute""/c)color(white)(a/a)|)))#
You will have
#V_""KOH"" = (0.250color(red)(cancel(color(black)(""moles""))))/(1.6color(red)(cancel(color(black)(""mol"")))""L""^(-1)) = ""0.1563 mL""#
I'll leave the answer rounded to three sig figs and express it in milliliters
#""volume of KOH solution"" = color(green)(|bar(ul(color(white)(a/a)color(black)(""156 mL"")color(white)(a/a)|)))#
The first thing to do here is write a balanced chemical equation that describes this neutralization reaction
#""KOH""_ ((aq)) + ""HCl""_ ((aq)) -> ""KCl""_ ((aq)) + ""H""_ 2""O""_ ((l))#
Potassium hydroxide,
This tells you that in order to have a complete neutralization, you need to mix equal numbers of moles of each reactant.
Now, the problem provides you with the molarity and volume of the hydrochloric acid solution, which you can use to determine how many moles of acid were needed for the reaction
#color(purple)(|bar(ul(color(white)(a/a)color(black)(c = n_""solute""/V_""solution"" implies n_""solute"" = c * V_""solution"")color(white)(a/a)|)))#
You will have
#n_""HCl"" = ""2.50 mol"" color(red)(cancel(color(black)(""L""^(-1)))) * overbrace(100.0 * 10^(-3)color(red)(cancel(color(black)(""L""))))^(color(blue)(""volume in liters""))#
#n_""HCl"" = ""0.250 moles HCl""#
This is exactly how many moles of potassium hydroxide you must add to the reaction. All you have to do now is use the molarity of the potassium hydroxide solution to figure out what volume would contain that many moles
#color(purple)(|bar(ul(color(white)(a/a)color(black)(c = n_""solute""/V_""solution"" implies V_""solution"" = n_""solute""/c)color(white)(a/a)|)))#
You will have
#V_""KOH"" = (0.250color(red)(cancel(color(black)(""moles""))))/(1.6color(red)(cancel(color(black)(""mol"")))""L""^(-1)) = ""0.1563 mL""#
I'll leave the answer rounded to three sig figs and express it in milliliters
#""volume of KOH solution"" = color(green)(|bar(ul(color(white)(a/a)color(black)(""156 mL"")color(white)(a/a)|)))#
The first thing to do here is write a balanced chemical equation that describes this neutralization reaction
#""KOH""_ ((aq)) + ""HCl""_ ((aq)) -> ""KCl""_ ((aq)) + ""H""_ 2""O""_ ((l))#
Potassium hydroxide,
This tells you that in order to have a complete neutralization, you need to mix equal numbers of moles of each reactant.
Now, the problem provides you with the molarity and volume of the hydrochloric acid solution, which you can use to determine how many moles of acid were needed for the reaction
#color(purple)(|bar(ul(color(white)(a/a)color(black)(c = n_""solute""/V_""solution"" implies n_""solute"" = c * V_""solution"")color(white)(a/a)|)))#
You will have
#n_""HCl"" = ""2.50 mol"" color(red)(cancel(color(black)(""L""^(-1)))) * overbrace(100.0 * 10^(-3)color(red)(cancel(color(black)(""L""))))^(color(blue)(""volume in liters""))#
#n_""HCl"" = ""0.250 moles HCl""#
This is exactly how many moles of potassium hydroxide you must add to the reaction. All you have to do now is use the molarity of the potassium hydroxide solution to figure out what volume would contain that many moles
#color(purple)(|bar(ul(color(white)(a/a)color(black)(c = n_""solute""/V_""solution"" implies V_""solution"" = n_""solute""/c)color(white)(a/a)|)))#
You will have
#V_""KOH"" = (0.250color(red)(cancel(color(black)(""moles""))))/(1.6color(red)(cancel(color(black)(""mol"")))""L""^(-1)) = ""0.1563 mL""#
I'll leave the answer rounded to three sig figs and express it in milliliters
#""volume of KOH solution"" = color(green)(|bar(ul(color(white)(a/a)color(black)(""156 mL"")color(white)(a/a)|)))#
You have quoted a solution of
Because this a
You have quoted a solution of
Because this a
You have quoted a solution of
Because this a
We need a stoichiometric equation that represents the oxidation of aluminum metal.....
And thus each mole of metal generates
A bit over
We need a stoichiometric equation that represents the oxidation of aluminum metal.....
And thus each mole of metal generates
A bit over
We need a stoichiometric equation that represents the oxidation of aluminum metal.....
And thus each mole of metal generates
Using Dimensional Analysis we can set up the correct equation to find the desired result. If we want mass (grams) from a concentration and volume it is this:
Using Dimensional Analysis we can set up the correct equation to find the desired result. If we want mass (grams) from a concentration and volume it is this:
Using Dimensional Analysis we can set up the correct equation to find the desired result. If we want mass (grams) from a concentration and volume it is this:
This is basically a density problem, where the density is known:
In this question, the mass of
To find mass when density and volume are known, we can rearrange the density formula to isolate mass.
Plug in the known values and solve for mass.
Here's how I go about doing this.
What we can do is use the ideal gas equation for both
I won't show the unit conversions here, as I predict you already know how; the moles of each substance is
We'll now use the ideal gas equation to solve for the pressure at the new conditions:
Here's how I go about doing this.
What we can do is use the ideal gas equation for both
I won't show the unit conversions here, as I predict you already know how; the moles of each substance is
We'll now use the ideal gas equation to solve for the pressure at the new conditions:
Here's how I go about doing this.
What we can do is use the ideal gas equation for both
I won't show the unit conversions here, as I predict you already know how; the moles of each substance is
We'll now use the ideal gas equation to solve for the pressure at the new conditions:
Given unbalanced equation is
As a thumb rule
We start with
Let the balanced equation be: (notice that number of N atoms have been kept balanced)
To balance hydrogen atoms we multiply ammonia molecule with 2 (set
The equation looks like
For the remaining unbalanced element:
Total number of oxygen atoms on products side becomes
We know that
Given unbalanced equation is
As a thumb rule
We start with
Let the balanced equation be: (notice that number of N atoms have been kept balanced)
To balance hydrogen atoms we multiply ammonia molecule with 2 (set
The equation looks like
For the remaining unbalanced element:
Total number of oxygen atoms on products side becomes
We know that
Given unbalanced equation is
As a thumb rule
We start with
Let the balanced equation be: (notice that number of N atoms have been kept balanced)
To balance hydrogen atoms we multiply ammonia molecule with 2 (set
The equation looks like
For the remaining unbalanced element:
Total number of oxygen atoms on products side becomes
We know that
The number of atoms in a single mole of any substance is defined by Avogadro's Number, a constant which is equal to
To find the number of atoms in a given amount of substance, simply multiply the amount in moles of the substance by Avogadro's Number:
Number of
The number of atoms in a single mole of any substance is defined by Avogadro's Number, a constant which is equal to
To find the number of atoms in a given amount of substance, simply multiply the amount in moles of the substance by Avogadro's Number:
Number of
The number of atoms in a single mole of any substance is defined by Avogadro's Number, a constant which is equal to
To find the number of atoms in a given amount of substance, simply multiply the amount in moles of the substance by Avogadro's Number:
Number of
Notice the imbalance of atoms. On the left side, the reactant side, there are
An important thing to remember when balancing equations is that the molecules themselves may not be changed—only their coefficients can. For example, we can change
In order to deal with the original imbalance, the
Now we have the same amount of oxygens on each side,
Notice the imbalance of atoms. On the left side, the reactant side, there are
An important thing to remember when balancing equations is that the molecules themselves may not be changed—only their coefficients can. For example, we can change
In order to deal with the original imbalance, the
Now we have the same amount of oxygens on each side,
Notice the imbalance of atoms. On the left side, the reactant side, there are
An important thing to remember when balancing equations is that the molecules themselves may not be changed—only their coefficients can. For example, we can change
In order to deal with the original imbalance, the
Now we have the same amount of oxygens on each side,
The starting species is
Fluorine is an excellent oxidant and gives
The starting species is
Fluorine is an excellent oxidant and gives
The starting species is
Fluorine is an excellent oxidant and gives
Benzoic acid is a weak acid that has antimicrobial properties. Its sodium salt, sodium benzoate, is a preservative found in foods, medications, and personal hygiene products. Benzoic acid dissociates in water:
The
Express the percentage numerically using two significant figures.
( explanation under construction by question owner )
Benzoic acid is a weak acid that has antimicrobial properties. Its sodium salt, sodium benzoate, is a preservative found in foods, medications, and personal hygiene products. Benzoic acid dissociates in water:
The
Express the percentage numerically using two significant figures.
( explanation under construction by question owner )
The heat is
mass,
specific heat,
Heat liberated is
The heat is
The heat is
mass,
specific heat,
Heat liberated is
The heat is
The heat is
mass,
specific heat,
Heat liberated is
Since we are at STP and are given only one set of conditions, we have to use the ideal gas law equation:
I should mention that the pressure does not always have units of atm, it depends on the units of pressure given in the gas constant.
List your known and unknown variables. Our only unknown is the volume of
At STP, the temperature is 273K and the pressure is 1 atm.
Now we have to rearrange the equation to solve for V:
At STP, 2.895 moles of
Since we are at STP and are given only one set of conditions, we have to use the ideal gas law equation:
I should mention that the pressure does not always have units of atm, it depends on the units of pressure given in the gas constant.
List your known and unknown variables. Our only unknown is the volume of
At STP, the temperature is 273K and the pressure is 1 atm.
Now we have to rearrange the equation to solve for V:
At STP, 2.895 moles of
Since we are at STP and are given only one set of conditions, we have to use the ideal gas law equation:
I should mention that the pressure does not always have units of atm, it depends on the units of pressure given in the gas constant.
List your known and unknown variables. Our only unknown is the volume of
At STP, the temperature is 273K and the pressure is 1 atm.
Now we have to rearrange the equation to solve for V:
We take the quotient,
Note the dimensional consistency of the answer....
What are the concentrations with respect to
What is the pH of this solution?
We take the quotient,
Note the dimensional consistency of the answer....
What are the concentrations with respect to
What is the pH of this solution?
We take the quotient,
Note the dimensional consistency of the answer....
What are the concentrations with respect to
What is the pH of this solution?
If you are not confident with the logarithmic function please voice your concern. See here for the relationship between
If you are not confident with the logarithmic function please voice your concern. See here for the relationship between
If you are not confident with the logarithmic function please voice your concern. See here for the relationship between
=>
Given
Substitute into HH Equation and solve for
=>
=>
Given
Substitute into HH Equation and solve for
=>
=>
Given
Substitute into HH Equation and solve for
=>
We interrogate the molar quantities of metal and oxygen.....
We divide thru by the smallest molar quantity, that of the metal, to get a trial empirical formula of.....
But by specification, the empirical formula is the simplest whole number ratio defining constituent atoms in a species...and so....
This is not a good question inasmuch as
We interrogate the molar quantities of metal and oxygen.....
We divide thru by the smallest molar quantity, that of the metal, to get a trial empirical formula of.....
But by specification, the empirical formula is the simplest whole number ratio defining constituent atoms in a species...and so....
This is not a good question inasmuch as
We interrogate the molar quantities of metal and oxygen.....
We divide thru by the smallest molar quantity, that of the metal, to get a trial empirical formula of.....
But by specification, the empirical formula is the simplest whole number ratio defining constituent atoms in a species...and so....
This is not a good question inasmuch as
Assuming complete reaction of both, convert the masses into moles and normalize. I’ll use a singlet oxygen because we don’t know the ratio yet, even though the actual gas would be diatomic.
To calculate the mass percent composition of aluminum in aluminum oxide, you divide the given mass of aluminum by the given mass of aluminum oxide and multiply times 100.
The sample of aluminum oxide contains 71.7% aluminum.
The sample of aluminum oxide contains 71.7% aluminum.
To calculate the mass percent composition of aluminum in aluminum oxide, you divide the given mass of aluminum by the given mass of aluminum oxide and multiply times 100.
The sample of aluminum oxide contains 71.7% aluminum.
The sample of aluminum oxide contains 71.7% aluminum.
To calculate the mass percent composition of aluminum in aluminum oxide, you divide the given mass of aluminum by the given mass of aluminum oxide and multiply times 100.
The sample of aluminum oxide contains 71.7% aluminum.
Given the value of Ksp 2.9x10^-12, what would be the solubility of Mg(OH)2 in a 0.345 M NaOH solution?
Your strategy here will be to use an ICE table to find the solubility of magnesium hydroxide,
Sodium hydroxide dissociates in a
#""NaOH""_ ((aq)) -> ""Na""_ ((aq))^(+) + ""OH""_ ((aq))^(-)#
Since every mole of sodium hydroxide produces one mole of hydroxide anions, your solution will contain
#[""OH""^(-)] = [""NaOH""] = ""0.345 M""#
Set up your ICE table
#"" "" ""Mg""(""OH"")_ (color(red)(2)(s)) "" ""rightleftharpoons "" "" ""Mg""_ ((aq))^(2+) "" ""+"" "" color(red)(2)""OH""_ ((aq))^(-)#
By definition, the solubility product constant,
#K_(sp) = [""Mg""^(2+)] * [""OH""^(-)]^color(red)(2)#
In your case, this will be equal to
#2.9 * 10^(-12) = s * (0.345 + color(red)(2)s)^color(red)(2)#
#2.9 * 10^(-12) = s * (0.119025 + 1.38s + 4s^2)#
Rearrange to get
#4s^3 + 1.38s^2 + 0.119025s - 2.9 * 10^(-12) = 0#
This cubic equation will produce one real solution
#s = 2.44 * 10^(-11)#
This represents the molar solubility of magnesium hydroxide in a solution that contains
#""molar solubility""_(""in 0.345M OH""^(-)) = color(green)(|bar(ul(color(white)(a/a)color(black)(2.44 * 10^(-11)""M"")color(white)(a/a)|)))#
I'll leave the answer rounded to three sig figs.
SIDE NOTE You should be able to predict that the solubility of magnesium hydroxide in this solution is significantly lower than the solubility of the salt in pure water.
In this case, you would have
#"" "" ""Mg""(""OH"")_ (color(red)(2)(s)) "" ""rightleftharpoons "" "" ""Mg""_ ((aq))^(2+) "" ""+"" "" color(red)(2)""OH""_ ((aq))^(-)#
This time, the solubility product constant would be
#K_(sp) = s * (color(red)(2)s)^color(red)(2)#
#2.9 * 10^(-12) = 4s^3 implies s = root(3)( (2.9 * 10^(-12))/4) = 9.0 * 10^(-5)#
The solubility of the salt decreases in a solution that contains hydroxide anions because of the common-ion effect.
Your strategy here will be to use an ICE table to find the solubility of magnesium hydroxide,
Sodium hydroxide dissociates in a
#""NaOH""_ ((aq)) -> ""Na""_ ((aq))^(+) + ""OH""_ ((aq))^(-)#
Since every mole of sodium hydroxide produces one mole of hydroxide anions, your solution will contain
#[""OH""^(-)] = [""NaOH""] = ""0.345 M""#
Set up your ICE table
#"" "" ""Mg""(""OH"")_ (color(red)(2)(s)) "" ""rightleftharpoons "" "" ""Mg""_ ((aq))^(2+) "" ""+"" "" color(red)(2)""OH""_ ((aq))^(-)#
By definition, the solubility product constant,
#K_(sp) = [""Mg""^(2+)] * [""OH""^(-)]^color(red)(2)#
In your case, this will be equal to
#2.9 * 10^(-12) = s * (0.345 + color(red)(2)s)^color(red)(2)#
#2.9 * 10^(-12) = s * (0.119025 + 1.38s + 4s^2)#
Rearrange to get
#4s^3 + 1.38s^2 + 0.119025s - 2.9 * 10^(-12) = 0#
This cubic equation will produce one real solution
#s = 2.44 * 10^(-11)#
This represents the molar solubility of magnesium hydroxide in a solution that contains
#""molar solubility""_(""in 0.345M OH""^(-)) = color(green)(|bar(ul(color(white)(a/a)color(black)(2.44 * 10^(-11)""M"")color(white)(a/a)|)))#
I'll leave the answer rounded to three sig figs.
SIDE NOTE You should be able to predict that the solubility of magnesium hydroxide in this solution is significantly lower than the solubility of the salt in pure water.
In this case, you would have
#"" "" ""Mg""(""OH"")_ (color(red)(2)(s)) "" ""rightleftharpoons "" "" ""Mg""_ ((aq))^(2+) "" ""+"" "" color(red)(2)""OH""_ ((aq))^(-)#
This time, the solubility product constant would be
#K_(sp) = s * (color(red)(2)s)^color(red)(2)#
#2.9 * 10^(-12) = 4s^3 implies s = root(3)( (2.9 * 10^(-12))/4) = 9.0 * 10^(-5)#
The solubility of the salt decreases in a solution that contains hydroxide anions because of the common-ion effect.
Given the value of Ksp 2.9x10^-12, what would be the solubility of Mg(OH)2 in a 0.345 M NaOH solution?
Your strategy here will be to use an ICE table to find the solubility of magnesium hydroxide,
Sodium hydroxide dissociates in a
#""NaOH""_ ((aq)) -> ""Na""_ ((aq))^(+) + ""OH""_ ((aq))^(-)#
Since every mole of sodium hydroxide produces one mole of hydroxide anions, your solution will contain
#[""OH""^(-)] = [""NaOH""] = ""0.345 M""#
Set up your ICE table
#"" "" ""Mg""(""OH"")_ (color(red)(2)(s)) "" ""rightleftharpoons "" "" ""Mg""_ ((aq))^(2+) "" ""+"" "" color(red)(2)""OH""_ ((aq))^(-)#
By definition, the solubility product constant,
#K_(sp) = [""Mg""^(2+)] * [""OH""^(-)]^color(red)(2)#
In your case, this will be equal to
#2.9 * 10^(-12) = s * (0.345 + color(red)(2)s)^color(red)(2)#
#2.9 * 10^(-12) = s * (0.119025 + 1.38s + 4s^2)#
Rearrange to get
#4s^3 + 1.38s^2 + 0.119025s - 2.9 * 10^(-12) = 0#
This cubic equation will produce one real solution
#s = 2.44 * 10^(-11)#
This represents the molar solubility of magnesium hydroxide in a solution that contains
#""molar solubility""_(""in 0.345M OH""^(-)) = color(green)(|bar(ul(color(white)(a/a)color(black)(2.44 * 10^(-11)""M"")color(white)(a/a)|)))#
I'll leave the answer rounded to three sig figs.
SIDE NOTE You should be able to predict that the solubility of magnesium hydroxide in this solution is significantly lower than the solubility of the salt in pure water.
In this case, you would have
#"" "" ""Mg""(""OH"")_ (color(red)(2)(s)) "" ""rightleftharpoons "" "" ""Mg""_ ((aq))^(2+) "" ""+"" "" color(red)(2)""OH""_ ((aq))^(-)#
This time, the solubility product constant would be
#K_(sp) = s * (color(red)(2)s)^color(red)(2)#
#2.9 * 10^(-12) = 4s^3 implies s = root(3)( (2.9 * 10^(-12))/4) = 9.0 * 10^(-5)#
The solubility of the salt decreases in a solution that contains hydroxide anions because of the common-ion effect.
Is the synthesis a redox reaction? Why or why not?
Is the synthesis a redox reaction? Why or why not?
Is the synthesis a redox reaction? Why or why not?
The equation above is stoichiometrically balanced. What does this mean? It means that for every reactant particle, there is a corresponding product particle. Stoichiometry is also practised in commerce: for every debit item (each charge made to an account), there must be a corresponding credit item (a deposit to made to another account). This principle of ""rob Peter, and pay Paul"" is fundamental to stoichiometry.
You started with
Capisce?
The equation above is stoichiometrically balanced. What does this mean? It means that for every reactant particle, there is a corresponding product particle. Stoichiometry is also practised in commerce: for every debit item (each charge made to an account), there must be a corresponding credit item (a deposit to made to another account). This principle of ""rob Peter, and pay Paul"" is fundamental to stoichiometry.
You started with
Capisce?
The equation above is stoichiometrically balanced. What does this mean? It means that for every reactant particle, there is a corresponding product particle. Stoichiometry is also practised in commerce: for every debit item (each charge made to an account), there must be a corresponding credit item (a deposit to made to another account). This principle of ""rob Peter, and pay Paul"" is fundamental to stoichiometry.
You started with
Capisce?
The give quotient is
The give quotient is
The give quotient is
You need to do a little algebra on this one.
You have the substance
Converting this into numbers based on their known oxidation states, you will have:
Na = 2 (+1) = +2
C = 2 (x) = 2x, where x is the unknown
O = 4 (-2) = -8
Hence,
(+2) + 2x + (-8) = 0
2x + (-6) = 0
2x = +6
x = +3
Therefore, the oxidation state of one Carbon atom in the substance
You need to do a little algebra on this one.
You have the substance
Converting this into numbers based on their known oxidation states, you will have:
Na = 2 (+1) = +2
C = 2 (x) = 2x, where x is the unknown
O = 4 (-2) = -8
Hence,
(+2) + 2x + (-8) = 0
2x + (-6) = 0
2x = +6
x = +3
Therefore, the oxidation state of one Carbon atom in the substance
You need to do a little algebra on this one.
You have the substance
Converting this into numbers based on their known oxidation states, you will have:
Na = 2 (+1) = +2
C = 2 (x) = 2x, where x is the unknown
O = 4 (-2) = -8
Hence,
(+2) + 2x + (-8) = 0
2x + (-6) = 0
2x = +6
x = +3
Therefore, the oxidation state of one Carbon atom in the substance
The oxidation number of each carbon atom in
The sum of the oxidation numbers of the elements in a compound is zero.
The oxidation number for sodium is pretty much always
Since there are two sodium atoms, the total oxidation number for sodium is
#+2# .
The oxidation number for oxygen is
Since there are four oxygen atoms, the total oxidation number for the oxygen atoms is
#-8# .The sum of the oxidation numbers for sodium and oxygen is
#+2 - 8 = -6# . Therefore, the total oxidation number for carbon must be#+6# in order for the sum of the oxidation numbers to equal zero.Divide
#+6# by two to get the oxidation number of each carbon atom, which is#bb(+3)# .
Oxidation numbers for all elements in the compound sodium oxalate:
#stackrel(2 xx +1)(""Na""_2)"" ""stackrel(2 xx +3)(""C""_2)"" ""stackrel(4 xx -2)(""O""_4)#
#2xx(+1)+2 xx (+3)+4xx(-2) = 0#
The idea here is that iron(III) chloride,
Now, notice that one formula unit of iron(III) chloride contains
- one iron(III) cation,
#1 xx ""Fe""^(3+)# - three chloride anions,
#3 xx ""Cl""^(-)#
This means that when one mole of iron(III) chloride dissolves in water, you get
- one mole of iron(III) cations
- three moles of chloride anions
Therefore, you can say that the concentration of chloride anions in an aqueous solution of iron(III) chloride will be three times higher than the concentration of the salt itself.
In your case, you can say that the concentration of chloride anions will be
#color(green)(bar(ul(|color(white)(a/a)color(black)([""Cl""^(-)] = 3 xx ""2.5 M"" = ""7.5 M"")color(white)(a/a)|)))#
The idea here is that iron(III) chloride,
Now, notice that one formula unit of iron(III) chloride contains
- one iron(III) cation,
#1 xx ""Fe""^(3+)# - three chloride anions,
#3 xx ""Cl""^(-)#
This means that when one mole of iron(III) chloride dissolves in water, you get
- one mole of iron(III) cations
- three moles of chloride anions
Therefore, you can say that the concentration of chloride anions in an aqueous solution of iron(III) chloride will be three times higher than the concentration of the salt itself.
In your case, you can say that the concentration of chloride anions will be
#color(green)(bar(ul(|color(white)(a/a)color(black)([""Cl""^(-)] = 3 xx ""2.5 M"" = ""7.5 M"")color(white)(a/a)|)))#
The idea here is that iron(III) chloride,
Now, notice that one formula unit of iron(III) chloride contains
- one iron(III) cation,
#1 xx ""Fe""^(3+)# - three chloride anions,
#3 xx ""Cl""^(-)#
This means that when one mole of iron(III) chloride dissolves in water, you get
- one mole of iron(III) cations
- three moles of chloride anions
Therefore, you can say that the concentration of chloride anions in an aqueous solution of iron(III) chloride will be three times higher than the concentration of the salt itself.
In your case, you can say that the concentration of chloride anions will be
#color(green)(bar(ul(|color(white)(a/a)color(black)([""Cl""^(-)] = 3 xx ""2.5 M"" = ""7.5 M"")color(white)(a/a)|)))#
Sodium chloride, also known as table salt, is an ionic compound composed of sodium cations,
The molar mass of an anionic compound tells you what the mass of one mole of formula units of that respective compound is.
You know that sodium chloride's formula unit contains
- one sodium atom
- one chlorine atom
This means that sodium chloride's molar mass will be the sum of the molar masses of those two atoms.
#M_""M NaCl"" = M_"" MNaCl"" + M_""M Cl""#
A quick look in the periodic table and you'll see that the molar masses of sodium and chlorine, respectively, are
Therefore, the molar mass of sodium chloride will be
#M_""M"" = ""22.989770 g/mol"" + ""35.453 g/mol"" = color(green)(""58.44277 g/mol"")#
In stoichiometric calculations, this value is often used as
Sodium chloride, also known as table salt, is an ionic compound composed of sodium cations,
The molar mass of an anionic compound tells you what the mass of one mole of formula units of that respective compound is.
You know that sodium chloride's formula unit contains
- one sodium atom
- one chlorine atom
This means that sodium chloride's molar mass will be the sum of the molar masses of those two atoms.
#M_""M NaCl"" = M_"" MNaCl"" + M_""M Cl""#
A quick look in the periodic table and you'll see that the molar masses of sodium and chlorine, respectively, are
Therefore, the molar mass of sodium chloride will be
#M_""M"" = ""22.989770 g/mol"" + ""35.453 g/mol"" = color(green)(""58.44277 g/mol"")#
In stoichiometric calculations, this value is often used as
Sodium chloride, also known as table salt, is an ionic compound composed of sodium cations,
The molar mass of an anionic compound tells you what the mass of one mole of formula units of that respective compound is.
You know that sodium chloride's formula unit contains
- one sodium atom
- one chlorine atom
This means that sodium chloride's molar mass will be the sum of the molar masses of those two atoms.
#M_""M NaCl"" = M_"" MNaCl"" + M_""M Cl""#
A quick look in the periodic table and you'll see that the molar masses of sodium and chlorine, respectively, are
Therefore, the molar mass of sodium chloride will be
#M_""M"" = ""22.989770 g/mol"" + ""35.453 g/mol"" = color(green)(""58.44277 g/mol"")#
In stoichiometric calculations, this value is often used as
The molar mass of sodium chloride is 58.44 g/mol.
The molar mass of sodium is 22.990 g/mol. The molar mass of chlorine is 35.45 g/mol.
To calculate the molar mass of NaCl, multiply the subscript of each element times its molar mass, then add them together. If there is no subscript, it is understood to be
The problem provides you with the temperature of the solution because at
#color(blue)(ul(color(black)(""pH + pOH = 14"")))#
This means that you can express the
#""pOH"" = - log([""OH""^(-)])#
You can thus say that the
#""pH"" = 14 - ""pOH""#
#""pH"" = 14 - [ - log([""OH""^(-)])]#
#""pH"" = 14 + log([""OH""^(-)])#
So instead of calculating the
#""pH"" = - log([""H""_3""O""^(+)])#
you can do so indirectly by using the concentration of hydroxide anions.
Now, sodium hydroxide is a strong base, which means that it dissociates completely in aqueous solution to produce sodium cations, which are of no interest to you here, and hydroxide anions.
Both ions are produced in
#[""OH""^(-)] = [""NaOH""] = ""0.001 M""#
Now that you know the concentration of hydroxide anions in this solution, you can say that its
#""pH"" = 14 + log(0.001) = color(darkgreen)(ul(color(black)(""11.0"")))#
The answer is rounded to one decimal place, the number of sig figs you have for the concentration of the strong base.
The problem provides you with the temperature of the solution because at
#color(blue)(ul(color(black)(""pH + pOH = 14"")))#
This means that you can express the
#""pOH"" = - log([""OH""^(-)])#
You can thus say that the
#""pH"" = 14 - ""pOH""#
#""pH"" = 14 - [ - log([""OH""^(-)])]#
#""pH"" = 14 + log([""OH""^(-)])#
So instead of calculating the
#""pH"" = - log([""H""_3""O""^(+)])#
you can do so indirectly by using the concentration of hydroxide anions.
Now, sodium hydroxide is a strong base, which means that it dissociates completely in aqueous solution to produce sodium cations, which are of no interest to you here, and hydroxide anions.
Both ions are produced in
#[""OH""^(-)] = [""NaOH""] = ""0.001 M""#
Now that you know the concentration of hydroxide anions in this solution, you can say that its
#""pH"" = 14 + log(0.001) = color(darkgreen)(ul(color(black)(""11.0"")))#
The answer is rounded to one decimal place, the number of sig figs you have for the concentration of the strong base.
The problem provides you with the temperature of the solution because at
#color(blue)(ul(color(black)(""pH + pOH = 14"")))#
This means that you can express the
#""pOH"" = - log([""OH""^(-)])#
You can thus say that the
#""pH"" = 14 - ""pOH""#
#""pH"" = 14 - [ - log([""OH""^(-)])]#
#""pH"" = 14 + log([""OH""^(-)])#
So instead of calculating the
#""pH"" = - log([""H""_3""O""^(+)])#
you can do so indirectly by using the concentration of hydroxide anions.
Now, sodium hydroxide is a strong base, which means that it dissociates completely in aqueous solution to produce sodium cations, which are of no interest to you here, and hydroxide anions.
Both ions are produced in
#[""OH""^(-)] = [""NaOH""] = ""0.001 M""#
Now that you know the concentration of hydroxide anions in this solution, you can say that its
#""pH"" = 14 + log(0.001) = color(darkgreen)(ul(color(black)(""11.0"")))#
The answer is rounded to one decimal place, the number of sig figs you have for the concentration of the strong base.
We can convert the mass of carbon monoxide to moles and then use the Ideal Gas Law to calculate the volume at STP.
Moles of
Volume at STP
The Ideal Gas Law is:
#color(blue)(|bar(ul(PV = nRT)|)# ,
where
We can rearrange the Ideal Gas Law to get
#V = (nRT)/P#
STP is 1 bar and 0 °C.
The volume is 20.3 L.
We can convert the mass of carbon monoxide to moles and then use the Ideal Gas Law to calculate the volume at STP.
Moles of
Volume at STP
The Ideal Gas Law is:
#color(blue)(|bar(ul(PV = nRT)|)# ,
where
We can rearrange the Ideal Gas Law to get
#V = (nRT)/P#
STP is 1 bar and 0 °C.
The volume is 20.3 L.
We can convert the mass of carbon monoxide to moles and then use the Ideal Gas Law to calculate the volume at STP.
Moles of
Volume at STP
The Ideal Gas Law is:
#color(blue)(|bar(ul(PV = nRT)|)# ,
where
We can rearrange the Ideal Gas Law to get
#V = (nRT)/P#
STP is 1 bar and 0 °C.
The formula for
#color(blue)(|bar(ul(color(white)(a/a) ""pH"" = -log[""H""^+]color(white)(a/a)|)))"" ""#
We can rearrange this equation to get
#color(blue)(|bar(ul(color(white)(a/a)[""H""^+] = 10^""-pH""color(white)(a/a)|)))"" ""#
If
The formula for
#color(blue)(|bar(ul(color(white)(a/a) ""pH"" = -log[""H""^+]color(white)(a/a)|)))"" ""#
We can rearrange this equation to get
#color(blue)(|bar(ul(color(white)(a/a)[""H""^+] = 10^""-pH""color(white)(a/a)|)))"" ""#
If
The formula for
#color(blue)(|bar(ul(color(white)(a/a) ""pH"" = -log[""H""^+]color(white)(a/a)|)))"" ""#
We can rearrange this equation to get
#color(blue)(|bar(ul(color(white)(a/a)[""H""^+] = 10^""-pH""color(white)(a/a)|)))"" ""#
If
The correct answer is
Let us see how we got the answer; Look at the electronic arrangement of Ca and N atom.
Ca ( Z= 20) has 20 electrons with following electronic configuration. 1
It loses two electron in its 4s subshell to achieve stability and forms ion
N ( Z=7) on the other hand has seven electrons and wants to gain three electrons to achieve stable noble gas configuration.Nitrogen atom on gaining three electrons forms negative nitride ion,
N ( Z=7) = 1
Two Nitrogen atoms gains three electrons each ( total of six) from three Ca atoms, each Ca atom loses two electrons (total of six) to two Nitrogen atom , in this process each Ca atom becomes
so in all we have three
so the formula becomes
The correct answer is
Let us see how we got the answer; Look at the electronic arrangement of Ca and N atom.
Ca ( Z= 20) has 20 electrons with following electronic configuration. 1
It loses two electron in its 4s subshell to achieve stability and forms ion
N ( Z=7) on the other hand has seven electrons and wants to gain three electrons to achieve stable noble gas configuration.Nitrogen atom on gaining three electrons forms negative nitride ion,
N ( Z=7) = 1
Two Nitrogen atoms gains three electrons each ( total of six) from three Ca atoms, each Ca atom loses two electrons (total of six) to two Nitrogen atom , in this process each Ca atom becomes
so in all we have three
so the formula becomes
The correct answer is
Let us see how we got the answer; Look at the electronic arrangement of Ca and N atom.
Ca ( Z= 20) has 20 electrons with following electronic configuration. 1
It loses two electron in its 4s subshell to achieve stability and forms ion
N ( Z=7) on the other hand has seven electrons and wants to gain three electrons to achieve stable noble gas configuration.Nitrogen atom on gaining three electrons forms negative nitride ion,
N ( Z=7) = 1
Two Nitrogen atoms gains three electrons each ( total of six) from three Ca atoms, each Ca atom loses two electrons (total of six) to two Nitrogen atom , in this process each Ca atom becomes
so in all we have three
so the formula becomes
And thus.......
So what's unusual about this problem? It expresses volume in units of
And so when we are quoted a volume of
Such operations are known as
And thus.......
So what's unusual about this problem? It expresses volume in units of
And so when we are quoted a volume of
Such operations are known as
And thus.......
So what's unusual about this problem? It expresses volume in units of
And so when we are quoted a volume of
Such operations are known as
Concentration of
(one mole of
So
Now we plug in what we know:
If you have the
So you'll need
Extra :
One drop more and the solution will turn cloudy white, and turn dark on exposure to sunlight -- that's what classic photography is based upon.
because the 'sp' suffix stands for 'solution product'.
Concentration of
(one mole of
So
Now we plug in what we know:
If you have the
So you'll need
Extra :
One drop more and the solution will turn cloudy white, and turn dark on exposure to sunlight -- that's what classic photography is based upon.
because the 'sp' suffix stands for 'solution product'.
Concentration of
(one mole of
So
Now we plug in what we know:
If you have the
So you'll need
Extra :
One drop more and the solution will turn cloudy white, and turn dark on exposure to sunlight -- that's what classic photography is based upon.
Your task here is to find the specific heat of iron, so take a second to make sure that you understand what it is you're looking for.
A substance's specific heat tells you how much heat is needed in order to increase the temperature of
So, in essence, you're looking for amount of heat per unit of mass and per unit of temperature.
#color(blue)(""specific heat"" = ""heat""/(""unit of mass "" xx "" unit of temperature""))#
Now, let's assume that you don't know the equation that establishes a relationship between heat added, the mass of the sample, the specific heat of the substance, and the resulting increase in temperature.
Here's how you could play around with the information provided by the problem to understand how to find the specific heat of iron.
For instance, let's assume that adding
#""1086.75 J""/""15.75 g"" = ""69 J/g""#
In this scenario, adding
Now let's assume that this much heat would increase the temperature of
#DeltaT = 175^@""C"" - 25^@""C"" = 150^@""C""#
In this case, the amount of heat needed per degree Celsius will be
#""1086.75 J""/(150^@""C"") = ""7.245 J/""""""^@""C""#
In this scenario, you will get a
But since you know that adding that much heat will increase the temperature of
#""1086.75 J""/(""15.75 g"" * 150^@""C"") = 0.46""J""/(""g"" """"^@""C"")#
And this is the substance's specific heat. So, to find the specific heat of a substance, you need to divide the amount of heat used to produce that increase in temperature for the given sample.
The equation that you'll be using from now on looks like this
#color(blue)(q = m * c * DeltaT)"" ""# , where
Your task here is to find the specific heat of iron, so take a second to make sure that you understand what it is you're looking for.
A substance's specific heat tells you how much heat is needed in order to increase the temperature of
So, in essence, you're looking for amount of heat per unit of mass and per unit of temperature.
#color(blue)(""specific heat"" = ""heat""/(""unit of mass "" xx "" unit of temperature""))#
Now, let's assume that you don't know the equation that establishes a relationship between heat added, the mass of the sample, the specific heat of the substance, and the resulting increase in temperature.
Here's how you could play around with the information provided by the problem to understand how to find the specific heat of iron.
For instance, let's assume that adding
#""1086.75 J""/""15.75 g"" = ""69 J/g""#
In this scenario, adding
Now let's assume that this much heat would increase the temperature of
#DeltaT = 175^@""C"" - 25^@""C"" = 150^@""C""#
In this case, the amount of heat needed per degree Celsius will be
#""1086.75 J""/(150^@""C"") = ""7.245 J/""""""^@""C""#
In this scenario, you will get a
But since you know that adding that much heat will increase the temperature of
#""1086.75 J""/(""15.75 g"" * 150^@""C"") = 0.46""J""/(""g"" """"^@""C"")#
And this is the substance's specific heat. So, to find the specific heat of a substance, you need to divide the amount of heat used to produce that increase in temperature for the given sample.
The equation that you'll be using from now on looks like this
#color(blue)(q = m * c * DeltaT)"" ""# , where
Your task here is to find the specific heat of iron, so take a second to make sure that you understand what it is you're looking for.
A substance's specific heat tells you how much heat is needed in order to increase the temperature of
So, in essence, you're looking for amount of heat per unit of mass and per unit of temperature.
#color(blue)(""specific heat"" = ""heat""/(""unit of mass "" xx "" unit of temperature""))#
Now, let's assume that you don't know the equation that establishes a relationship between heat added, the mass of the sample, the specific heat of the substance, and the resulting increase in temperature.
Here's how you could play around with the information provided by the problem to understand how to find the specific heat of iron.
For instance, let's assume that adding
#""1086.75 J""/""15.75 g"" = ""69 J/g""#
In this scenario, adding
Now let's assume that this much heat would increase the temperature of
#DeltaT = 175^@""C"" - 25^@""C"" = 150^@""C""#
In this case, the amount of heat needed per degree Celsius will be
#""1086.75 J""/(150^@""C"") = ""7.245 J/""""""^@""C""#
In this scenario, you will get a
But since you know that adding that much heat will increase the temperature of
#""1086.75 J""/(""15.75 g"" * 150^@""C"") = 0.46""J""/(""g"" """"^@""C"")#
And this is the substance's specific heat. So, to find the specific heat of a substance, you need to divide the amount of heat used to produce that increase in temperature for the given sample.
The equation that you'll be using from now on looks like this
#color(blue)(q = m * c * DeltaT)"" ""# , where
The volume of the container, we must assume, is static. Of course, the pressure will increase, but this is not the question you asked. The gas could have been enclosed in a piston, expanding against external pressure. Please review your question.
Your question is not well-proposed. The final volume is
The volume of the container, we must assume, is static. Of course, the pressure will increase, but this is not the question you asked. The gas could have been enclosed in a piston, expanding against external pressure. Please review your question.
Your question is not well-proposed. The final volume is
The volume of the container, we must assume, is static. Of course, the pressure will increase, but this is not the question you asked. The gas could have been enclosed in a piston, expanding against external pressure. Please review your question.
Note that the
Note that the
Note that the
The former would be all you needed to write in an exam. But as to background, we KNOW from classic experiments that water undergoes autoprotolysis, i.e. self-ionization.
We could represent this reaction by
OR by
Note that
The equilibrium constant for the reaction, under standard conditions, is..........
And to make the arithmetic a bit easier we can use the
And since we are directly given
How do you think
pH is defined as 'the logarithm (to base 10) of the [H+] with the sign changed'.
pH is defined as 'the logarithm (to base 10) of the [H+] with the sign changed'.
The former would be all you needed to write in an exam. But as to background, we KNOW from classic experiments that water undergoes autoprotolysis, i.e. self-ionization.
We could represent this reaction by
OR by
Note that
The equilibrium constant for the reaction, under standard conditions, is..........
And to make the arithmetic a bit easier we can use the
And since we are directly given
How do you think
A carbon atom has an average mass of
Recall that one mole of atoms is
So here, we got:
And so, the total mass will be
A carbon atom has an average mass of
Recall that one mole of atoms is
So here, we got:
And so, the total mass will be
A carbon atom has an average mass of
Recall that one mole of atoms is
So here, we got:
And so, the total mass will be
Is this reaction exothermic? Could I use this reaction to drive a locomotive?
Is this reaction exothermic? Could I use this reaction to drive a locomotive?
Is this reaction exothermic? Could I use this reaction to drive a locomotive?
Balancing a chemical equation consists of changing the coefficient of molecules in order to make the number of atoms equal on both side.
Consider the given equation:
#Ag_2OrarrAg+O_2#
Notice how there is only one
#color(red)(2)Ag_2OrarrAg+O_2#
Now that the oxygen atoms are balanced, all that remains is the silver. There are now four
#2Ag_2Orarrcolor(blue)(4)Ag+O_2#
This is completely balanced.
Balancing a chemical equation consists of changing the coefficient of molecules in order to make the number of atoms equal on both side.
Consider the given equation:
#Ag_2OrarrAg+O_2#
Notice how there is only one
#color(red)(2)Ag_2OrarrAg+O_2#
Now that the oxygen atoms are balanced, all that remains is the silver. There are now four
#2Ag_2Orarrcolor(blue)(4)Ag+O_2#
This is completely balanced.
Balancing a chemical equation consists of changing the coefficient of molecules in order to make the number of atoms equal on both side.
Consider the given equation:
#Ag_2OrarrAg+O_2#
Notice how there is only one
#color(red)(2)Ag_2OrarrAg+O_2#
Now that the oxygen atoms are balanced, all that remains is the silver. There are now four
#2Ag_2Orarrcolor(blue)(4)Ag+O_2#
This is completely balanced.
If the amount of dissociation is
If
Since the approximations have converged, we are willing to accept this value as the true value (i.e. the same as if we exactly solved the quadratic in
Given
What is
If the amount of dissociation is
If
Since the approximations have converged, we are willing to accept this value as the true value (i.e. the same as if we exactly solved the quadratic in
Given
What is
If the amount of dissociation is
If
Since the approximations have converged, we are willing to accept this value as the true value (i.e. the same as if we exactly solved the quadratic in
Given
What is
Consider the balanced chemical equation, which we need to obtain the molar ratios:
I see you would like to know the grams
Let's use the molar ratios to calculate the amount of nitrogen dioxide required.
Consider the graphic below:
11.46 grams
Consider the balanced chemical equation, which we need to obtain the molar ratios:
I see you would like to know the grams
Let's use the molar ratios to calculate the amount of nitrogen dioxide required.
Consider the graphic below:
11.46 grams
Consider the balanced chemical equation, which we need to obtain the molar ratios:
I see you would like to know the grams
Let's use the molar ratios to calculate the amount of nitrogen dioxide required.
Consider the graphic below:
You follow a systematic procedure to balance the equation.
Start with the unbalanced equation:
A method that often works is to balance everything other than
Another useful procedure is to start with what looks like the most complicated formula.
The most complicated formula looks like
Balance
We have
Balance
We have
Most instructors don't allow fractions in balanced chemical equations, because you can't have a fraction of an atom or a molecule.
To get rid of the fractions, you multiply all the coefficients by 2 and get
Now you can balance the equation by putting a 1 in front of the
We should now have a balanced equation.
Check to make sure:
The balanced equation is
The balanced equation is
You follow a systematic procedure to balance the equation.
Start with the unbalanced equation:
A method that often works is to balance everything other than
Another useful procedure is to start with what looks like the most complicated formula.
The most complicated formula looks like
Balance
We have
Balance
We have
Most instructors don't allow fractions in balanced chemical equations, because you can't have a fraction of an atom or a molecule.
To get rid of the fractions, you multiply all the coefficients by 2 and get
Now you can balance the equation by putting a 1 in front of the
We should now have a balanced equation.
Check to make sure:
The balanced equation is
The balanced equation is
You follow a systematic procedure to balance the equation.
Start with the unbalanced equation:
A method that often works is to balance everything other than
Another useful procedure is to start with what looks like the most complicated formula.
The most complicated formula looks like
Balance
We have
Balance
We have
Most instructors don't allow fractions in balanced chemical equations, because you can't have a fraction of an atom or a molecule.
To get rid of the fractions, you multiply all the coefficients by 2 and get
Now you can balance the equation by putting a 1 in front of the
We should now have a balanced equation.
Check to make sure:
The balanced equation is
The oxidation number is the charge assigned when the bonding electrons are distributed to the most electronegative atom. Clearly, for a homonuclear diatomic molecule, the bound atoms have EQUAL electronegativity, and we conceive that the electrons are shared between the two atoms to give
We has zerovalent hydrogen, i.e.
The oxidation number is the charge assigned when the bonding electrons are distributed to the most electronegative atom. Clearly, for a homonuclear diatomic molecule, the bound atoms have EQUAL electronegativity, and we conceive that the electrons are shared between the two atoms to give
We has zerovalent hydrogen, i.e.
The oxidation number is the charge assigned when the bonding electrons are distributed to the most electronegative atom. Clearly, for a homonuclear diatomic molecule, the bound atoms have EQUAL electronegativity, and we conceive that the electrons are shared between the two atoms to give
One atmosphere of pressure will support of column of mercury that is
Given your problem, I suppose we could state that
And thus your question has not been consistently proposed. The pressure is LOWER at
I do not know why we seem to be getting a lot of questions that quote pressures of
One atmosphere of pressure will support of column of mercury that is
Given your problem, I suppose we could state that
And thus your question has not been consistently proposed. The pressure is LOWER at
I do not know why we seem to be getting a lot of questions that quote pressures of
One atmosphere of pressure will support of column of mercury that is
Given your problem, I suppose we could state that
And thus your question has not been consistently proposed. The pressure is LOWER at
The standard reduction potentials for zinc and aluminum are listed below:
In a galvanic cell, the component with lower standard reduction potential gets oxidized and that it is added to the anode compartment. The second is therefore, forms the cathode compartment.
Since almuninum has the lowest standard reduction potential
Note that when the reduction equation is reversed, the sign of the standard reduction potential is reversed as well.
Since zinc will be in the cathode compartment, it will get reduced as follows:
The overall reaction in the galvanic cell is the sum of the two half equations; oxidation and reduction:
Therefore, the standard cell potential is :
Note that the half equations where multiplied by the corresponding integers (2 and 3) in order to cancel the number of electrons from the overall equation.
Here is a video that further explains this topic:
Electrochemistry | The Standard Reduction Potential.
The standard reduction potentials for zinc and aluminum are listed below:
In a galvanic cell, the component with lower standard reduction potential gets oxidized and that it is added to the anode compartment. The second is therefore, forms the cathode compartment.
Since almuninum has the lowest standard reduction potential
Note that when the reduction equation is reversed, the sign of the standard reduction potential is reversed as well.
Since zinc will be in the cathode compartment, it will get reduced as follows:
The overall reaction in the galvanic cell is the sum of the two half equations; oxidation and reduction:
Therefore, the standard cell potential is :
Note that the half equations where multiplied by the corresponding integers (2 and 3) in order to cancel the number of electrons from the overall equation.
Here is a video that further explains this topic:
Electrochemistry | The Standard Reduction Potential.
The standard reduction potentials for zinc and aluminum are listed below:
In a galvanic cell, the component with lower standard reduction potential gets oxidized and that it is added to the anode compartment. The second is therefore, forms the cathode compartment.
Since almuninum has the lowest standard reduction potential
Note that when the reduction equation is reversed, the sign of the standard reduction potential is reversed as well.
Since zinc will be in the cathode compartment, it will get reduced as follows:
The overall reaction in the galvanic cell is the sum of the two half equations; oxidation and reduction:
Therefore, the standard cell potential is :
Note that the half equations where multiplied by the corresponding integers (2 and 3) in order to cancel the number of electrons from the overall equation.
Here is a video that further explains this topic:
Electrochemistry | The Standard Reduction Potential.
We apply Boyle's Law
The initial volume is
The initial pressure is
The final pressure is
The final volume is
The volume is
We apply Boyle's Law
The initial volume is
The initial pressure is
The final pressure is
The final volume is
The volume is
We apply Boyle's Law
The initial volume is
The initial pressure is
The final pressure is
The final volume is
We got...
Which tells that the formation of
But here we have made much more than 2 mol ammonia, and the enthalpy changes appropriately.
And thus
Why is
How? We use the enthalpy as a variable in the reaction..........and get
We got...
Which tells that the formation of
But here we have made much more than 2 mol ammonia, and the enthalpy changes appropriately.
And thus
Why is
How? We use the enthalpy as a variable in the reaction..........and get
We got...
Which tells that the formation of
But here we have made much more than 2 mol ammonia, and the enthalpy changes appropriately.
And thus
Why is
Start by writing the unbalanced chemical equation that describes this double displacement reaction
#""FeCl""_ (3(aq)) + ""NaOH""_ ((aq)) -> ""Fe""(""OH"")_ (3(s)) darr + ""NaCl""#
In order to balance this chemical equation, you can use the fact that the two reactants and sodium chloride are soluble in water, which implies that they exist as ions in aqueous solution.
#""FeCl""_ (3(aq)) -> ""Fe""_ ((aq))^(3+) + 3""Cl""_ ((aq))^(-)#
#""NaOH""_ ((aq)) -> ""Na""_ ((aq))^(+) + ""OH""_ ((aq))^(-)#
#""NaCl""_ ((aq)) -> ""Na""_ ((aq))^(+) + ""Cl""_ ((aq))^(-)#
You can rewrite the unbalanced chemical equation as
#""Fe""_ ((aq))^(3+) + 3""Cl""_ ((aq))^(-) + ""Na""_ ((aq))^(+) + ""OH""_ ((aq))^(-) -> ""Fe""(""OH"")_ (3(s)) darr + ""Na""_ ((aq))^(+) + ""Cl""_ ((aq))^(-)#
The first thing that stands out here is that you need to balance the chloride anions by multiplying the chloride anions present on the products' side by
Keep in mind that the chloride anions are part of the sodium chloride, so if you multiply the chloride anions by
#""Fe""_ ((aq))^(3+) + 3""Cl""_ ((aq))^(-) + ""Na""_ ((aq))^(+) + ""OH""_ ((aq))^(-) -> ""Fe""(""OH"")_ (3(s)) darr + color(blue)(3)""Na""_ ((aq))^(+) + color(blue)(3)""Cl""_ ((aq))^(-)#
Now balance the sodium cations present on the reactants' side by multiplying them by
#""Fe""_ ((aq))^(3+) + 3""Cl""_ ((aq))^(-) + color(red)(3)""Na""_ ((aq))^(+) + color(red)(3)""OH""_ ((aq))^(-) -> ""Fe""(""OH"")_ (3(s)) darr + color(blue)(3)""Na""_ ((aq))^(+) + color(blue)(3)""Cl""_ ((aq))^(-)#
The hydroxide anions are now balanced because you have
#[""Fe""_ ((aq))^(3+) + 3""Cl""_ ((aq))^(-)] + color(red)(3) xx [""Na""_ ((aq))^(+) + ""OH""_ ((aq))^(-)] -> ""Fe""(""OH"")_ (3(s)) darr + color(blue)(3)xx[""Na""_ ((aq))^(+) + ""Cl""_ ((aq))^(-)]#
to get the balanced chemical equation
#""FeCl""_ (3(aq)) + color(red)(3)""NaOH""_ ((aq)) -> ""Fe""(""OH"")_ (3(s)) darr + color(blue)(3)""NaCl""_ ((aq))#
You could also write the net ionic equation, which does not include the spectator ions, i.e. the ions that are present on both sides of the balanced chemical equation.
#""Fe""_ ((aq))^(3+) + color(red)(3)""OH""_ ((aq))^(-) -> ""Fe""(""OH"")_ (3(s)) darr#
Start by writing the unbalanced chemical equation that describes this double displacement reaction
#""FeCl""_ (3(aq)) + ""NaOH""_ ((aq)) -> ""Fe""(""OH"")_ (3(s)) darr + ""NaCl""#
In order to balance this chemical equation, you can use the fact that the two reactants and sodium chloride are soluble in water, which implies that they exist as ions in aqueous solution.
#""FeCl""_ (3(aq)) -> ""Fe""_ ((aq))^(3+) + 3""Cl""_ ((aq))^(-)#
#""NaOH""_ ((aq)) -> ""Na""_ ((aq))^(+) + ""OH""_ ((aq))^(-)#
#""NaCl""_ ((aq)) -> ""Na""_ ((aq))^(+) + ""Cl""_ ((aq))^(-)#
You can rewrite the unbalanced chemical equation as
#""Fe""_ ((aq))^(3+) + 3""Cl""_ ((aq))^(-) + ""Na""_ ((aq))^(+) + ""OH""_ ((aq))^(-) -> ""Fe""(""OH"")_ (3(s)) darr + ""Na""_ ((aq))^(+) + ""Cl""_ ((aq))^(-)#
The first thing that stands out here is that you need to balance the chloride anions by multiplying the chloride anions present on the products' side by
Keep in mind that the chloride anions are part of the sodium chloride, so if you multiply the chloride anions by
#""Fe""_ ((aq))^(3+) + 3""Cl""_ ((aq))^(-) + ""Na""_ ((aq))^(+) + ""OH""_ ((aq))^(-) -> ""Fe""(""OH"")_ (3(s)) darr + color(blue)(3)""Na""_ ((aq))^(+) + color(blue)(3)""Cl""_ ((aq))^(-)#
Now balance the sodium cations present on the reactants' side by multiplying them by
#""Fe""_ ((aq))^(3+) + 3""Cl""_ ((aq))^(-) + color(red)(3)""Na""_ ((aq))^(+) + color(red)(3)""OH""_ ((aq))^(-) -> ""Fe""(""OH"")_ (3(s)) darr + color(blue)(3)""Na""_ ((aq))^(+) + color(blue)(3)""Cl""_ ((aq))^(-)#
The hydroxide anions are now balanced because you have
#[""Fe""_ ((aq))^(3+) + 3""Cl""_ ((aq))^(-)] + color(red)(3) xx [""Na""_ ((aq))^(+) + ""OH""_ ((aq))^(-)] -> ""Fe""(""OH"")_ (3(s)) darr + color(blue)(3)xx[""Na""_ ((aq))^(+) + ""Cl""_ ((aq))^(-)]#
to get the balanced chemical equation
#""FeCl""_ (3(aq)) + color(red)(3)""NaOH""_ ((aq)) -> ""Fe""(""OH"")_ (3(s)) darr + color(blue)(3)""NaCl""_ ((aq))#
You could also write the net ionic equation, which does not include the spectator ions, i.e. the ions that are present on both sides of the balanced chemical equation.
#""Fe""_ ((aq))^(3+) + color(red)(3)""OH""_ ((aq))^(-) -> ""Fe""(""OH"")_ (3(s)) darr#
Start by writing the unbalanced chemical equation that describes this double displacement reaction
#""FeCl""_ (3(aq)) + ""NaOH""_ ((aq)) -> ""Fe""(""OH"")_ (3(s)) darr + ""NaCl""#
In order to balance this chemical equation, you can use the fact that the two reactants and sodium chloride are soluble in water, which implies that they exist as ions in aqueous solution.
#""FeCl""_ (3(aq)) -> ""Fe""_ ((aq))^(3+) + 3""Cl""_ ((aq))^(-)#
#""NaOH""_ ((aq)) -> ""Na""_ ((aq))^(+) + ""OH""_ ((aq))^(-)#
#""NaCl""_ ((aq)) -> ""Na""_ ((aq))^(+) + ""Cl""_ ((aq))^(-)#
You can rewrite the unbalanced chemical equation as
#""Fe""_ ((aq))^(3+) + 3""Cl""_ ((aq))^(-) + ""Na""_ ((aq))^(+) + ""OH""_ ((aq))^(-) -> ""Fe""(""OH"")_ (3(s)) darr + ""Na""_ ((aq))^(+) + ""Cl""_ ((aq))^(-)#
The first thing that stands out here is that you need to balance the chloride anions by multiplying the chloride anions present on the products' side by
Keep in mind that the chloride anions are part of the sodium chloride, so if you multiply the chloride anions by
#""Fe""_ ((aq))^(3+) + 3""Cl""_ ((aq))^(-) + ""Na""_ ((aq))^(+) + ""OH""_ ((aq))^(-) -> ""Fe""(""OH"")_ (3(s)) darr + color(blue)(3)""Na""_ ((aq))^(+) + color(blue)(3)""Cl""_ ((aq))^(-)#
Now balance the sodium cations present on the reactants' side by multiplying them by
#""Fe""_ ((aq))^(3+) + 3""Cl""_ ((aq))^(-) + color(red)(3)""Na""_ ((aq))^(+) + color(red)(3)""OH""_ ((aq))^(-) -> ""Fe""(""OH"")_ (3(s)) darr + color(blue)(3)""Na""_ ((aq))^(+) + color(blue)(3)""Cl""_ ((aq))^(-)#
The hydroxide anions are now balanced because you have
#[""Fe""_ ((aq))^(3+) + 3""Cl""_ ((aq))^(-)] + color(red)(3) xx [""Na""_ ((aq))^(+) + ""OH""_ ((aq))^(-)] -> ""Fe""(""OH"")_ (3(s)) darr + color(blue)(3)xx[""Na""_ ((aq))^(+) + ""Cl""_ ((aq))^(-)]#
to get the balanced chemical equation
#""FeCl""_ (3(aq)) + color(red)(3)""NaOH""_ ((aq)) -> ""Fe""(""OH"")_ (3(s)) darr + color(blue)(3)""NaCl""_ ((aq))#
You could also write the net ionic equation, which does not include the spectator ions, i.e. the ions that are present on both sides of the balanced chemical equation.
#""Fe""_ ((aq))^(3+) + color(red)(3)""OH""_ ((aq))^(-) -> ""Fe""(""OH"")_ (3(s)) darr#
For this problem, we can use the relation
#color(blue)(bar(ul(|color(white)(a/a) c_1V_1 = c_2V_2 = ""amount of solute"" color(white)(a/a)|)))"" ""#
Let the
Then, after mixing, we have
This is made up of
So, we add
Check:
It checks!
You must add 3 L of the 90 % acid.
For this problem, we can use the relation
#color(blue)(bar(ul(|color(white)(a/a) c_1V_1 = c_2V_2 = ""amount of solute"" color(white)(a/a)|)))"" ""#
Let the
Then, after mixing, we have
This is made up of
So, we add
Check:
It checks!
You must add 3 L of the 90 % acid.
For this problem, we can use the relation
#color(blue)(bar(ul(|color(white)(a/a) c_1V_1 = c_2V_2 = ""amount of solute"" color(white)(a/a)|)))"" ""#
Let the
Then, after mixing, we have
This is made up of
So, we add
Check:
It checks!
For pure water at room temperature, the concentration of hydronium cations,
#color(blue)(|bar(ul(color(white)(a/a)[""H""_3""O""^(+)] * [""OH""^(-)] = 10^(-14)""M""^2color(white)(a/a)|)))#
This relationship is based on the self-ionization of water, which at room temperature produces equal concentrations,
So, you're dealing with an aqueous solution that has
#[""H""_3""O""^(+)] = 2 * 10^(-5)""M""#
Right from the start, you can tell that this solution is acidic, since the concentration of hydronium ions increased compared with that of pure water.
Despite the fact that you have more hydronium ions present, the relationship between the hydronium and hydroxide anions remains valid.
This of course implies that the concentration of hydroxide anions will be lower than
More specifically, the concentration of hydroxide anions will be
#[""OH""^(-)] = (10^(-14)""M""^color(red)(cancel(color(black)(2))))/(2 * 10^(-5)color(red)(cancel(color(black)(""M"")))) = color(green)(|bar(ul(color(white)(a/a)5 * 10^(-10)""M""color(white)(a/a)|)))#
As practice, you can find the pH of this solution by using
#color(blue)(|bar(ul(color(white)(a/a)""pH"" = - log([""H""_3""O""^(+)])color(white)(a/a)|)))#
In this case, you will have
#""pH"" = - log(2 * 10^(-5)) = 4.7#
For pure water at room temperature, the concentration of hydronium cations,
#color(blue)(|bar(ul(color(white)(a/a)[""H""_3""O""^(+)] * [""OH""^(-)] = 10^(-14)""M""^2color(white)(a/a)|)))#
This relationship is based on the self-ionization of water, which at room temperature produces equal concentrations,
So, you're dealing with an aqueous solution that has
#[""H""_3""O""^(+)] = 2 * 10^(-5)""M""#
Right from the start, you can tell that this solution is acidic, since the concentration of hydronium ions increased compared with that of pure water.
Despite the fact that you have more hydronium ions present, the relationship between the hydronium and hydroxide anions remains valid.
This of course implies that the concentration of hydroxide anions will be lower than
More specifically, the concentration of hydroxide anions will be
#[""OH""^(-)] = (10^(-14)""M""^color(red)(cancel(color(black)(2))))/(2 * 10^(-5)color(red)(cancel(color(black)(""M"")))) = color(green)(|bar(ul(color(white)(a/a)5 * 10^(-10)""M""color(white)(a/a)|)))#
As practice, you can find the pH of this solution by using
#color(blue)(|bar(ul(color(white)(a/a)""pH"" = - log([""H""_3""O""^(+)])color(white)(a/a)|)))#
In this case, you will have
#""pH"" = - log(2 * 10^(-5)) = 4.7#
For pure water at room temperature, the concentration of hydronium cations,
#color(blue)(|bar(ul(color(white)(a/a)[""H""_3""O""^(+)] * [""OH""^(-)] = 10^(-14)""M""^2color(white)(a/a)|)))#
This relationship is based on the self-ionization of water, which at room temperature produces equal concentrations,
So, you're dealing with an aqueous solution that has
#[""H""_3""O""^(+)] = 2 * 10^(-5)""M""#
Right from the start, you can tell that this solution is acidic, since the concentration of hydronium ions increased compared with that of pure water.
Despite the fact that you have more hydronium ions present, the relationship between the hydronium and hydroxide anions remains valid.
This of course implies that the concentration of hydroxide anions will be lower than
More specifically, the concentration of hydroxide anions will be
#[""OH""^(-)] = (10^(-14)""M""^color(red)(cancel(color(black)(2))))/(2 * 10^(-5)color(red)(cancel(color(black)(""M"")))) = color(green)(|bar(ul(color(white)(a/a)5 * 10^(-10)""M""color(white)(a/a)|)))#
As practice, you can find the pH of this solution by using
#color(blue)(|bar(ul(color(white)(a/a)""pH"" = - log([""H""_3""O""^(+)])color(white)(a/a)|)))#
In this case, you will have
#""pH"" = - log(2 * 10^(-5)) = 4.7#
With all these problems, a stoichiometricall balanced equation that represents the reaction is a prerequisite.
And then we work out the number of moles of the reagents.
Given that 2 equiv of nitric acid are required to neutralize each equiv sodium carbonate, we can work out the concentration of the nitric acid as:
We work out the number of moles of each reagent...............and calculate
With all these problems, a stoichiometricall balanced equation that represents the reaction is a prerequisite.
And then we work out the number of moles of the reagents.
Given that 2 equiv of nitric acid are required to neutralize each equiv sodium carbonate, we can work out the concentration of the nitric acid as:
We work out the number of moles of each reagent...............and calculate
With all these problems, a stoichiometricall balanced equation that represents the reaction is a prerequisite.
And then we work out the number of moles of the reagents.
Given that 2 equiv of nitric acid are required to neutralize each equiv sodium carbonate, we can work out the concentration of the nitric acid as:
As you know, oxidation numbers are assigned by treating all bonds an atom has as ionic.
The idea behind that is that the more electronegative atom will take all the bonding electrons from the less electronegative atom.
The bond-line notation for ethyl-4-methylpentanoate looks like this
A carbonyl group is simply a carbon atom double bonded to an oxygen atom. In the case of ethyl-4-methylpentanoate, the carbon that's part of the carbonyl group is also bonded to another carbon atom and to another oxygen atom, th oxygen from the ether group..
When a carbon atom is bonded to another carbon atom, the difference in electronegativity between these two atoms is equal to zero.
So the carbonyl carbon will neither take, nor lose electrons from its bond to the other carbon atom. Remember that the ""taking"" or ""losing"" of electrons is hypothetical.
Since oxygen is more electronegative than carbon, it will take the electrons carbon is contributing to the bond. In the case of the double bond, oxygen will take all four bonding electrons, i.e. both electrons carbon shares to the bond.
This will leave carbon deprived of two electrons, giving it a +2 oxidation state.
The ether oxygen that is single bonded to the carbonyl atom will take both bonding electrons, i.e. the electron carbon shares to the bond.
This means that the carbonyl carbon has ""lost"" an additional 1 electron, bringing the total number of electrons it ""lost"" to 3. As a result, the carbonyl carbon's oxidation state is +3.
The oxidation number of the carbonyl carbon is +3.
As you know, oxidation numbers are assigned by treating all bonds an atom has as ionic.
The idea behind that is that the more electronegative atom will take all the bonding electrons from the less electronegative atom.
The bond-line notation for ethyl-4-methylpentanoate looks like this
A carbonyl group is simply a carbon atom double bonded to an oxygen atom. In the case of ethyl-4-methylpentanoate, the carbon that's part of the carbonyl group is also bonded to another carbon atom and to another oxygen atom, th oxygen from the ether group..
When a carbon atom is bonded to another carbon atom, the difference in electronegativity between these two atoms is equal to zero.
So the carbonyl carbon will neither take, nor lose electrons from its bond to the other carbon atom. Remember that the ""taking"" or ""losing"" of electrons is hypothetical.
Since oxygen is more electronegative than carbon, it will take the electrons carbon is contributing to the bond. In the case of the double bond, oxygen will take all four bonding electrons, i.e. both electrons carbon shares to the bond.
This will leave carbon deprived of two electrons, giving it a +2 oxidation state.
The ether oxygen that is single bonded to the carbonyl atom will take both bonding electrons, i.e. the electron carbon shares to the bond.
This means that the carbonyl carbon has ""lost"" an additional 1 electron, bringing the total number of electrons it ""lost"" to 3. As a result, the carbonyl carbon's oxidation state is +3.
The oxidation number of the carbonyl carbon is +3.
As you know, oxidation numbers are assigned by treating all bonds an atom has as ionic.
The idea behind that is that the more electronegative atom will take all the bonding electrons from the less electronegative atom.
The bond-line notation for ethyl-4-methylpentanoate looks like this
A carbonyl group is simply a carbon atom double bonded to an oxygen atom. In the case of ethyl-4-methylpentanoate, the carbon that's part of the carbonyl group is also bonded to another carbon atom and to another oxygen atom, th oxygen from the ether group..
When a carbon atom is bonded to another carbon atom, the difference in electronegativity between these two atoms is equal to zero.
So the carbonyl carbon will neither take, nor lose electrons from its bond to the other carbon atom. Remember that the ""taking"" or ""losing"" of electrons is hypothetical.
Since oxygen is more electronegative than carbon, it will take the electrons carbon is contributing to the bond. In the case of the double bond, oxygen will take all four bonding electrons, i.e. both electrons carbon shares to the bond.
This will leave carbon deprived of two electrons, giving it a +2 oxidation state.
The ether oxygen that is single bonded to the carbonyl atom will take both bonding electrons, i.e. the electron carbon shares to the bond.
This means that the carbonyl carbon has ""lost"" an additional 1 electron, bringing the total number of electrons it ""lost"" to 3. As a result, the carbonyl carbon's oxidation state is +3.
Every chemical reaction conserves the component masses; of course such mass may be lost to the experimenter by evolution of gas or thru handling.
Here
In another experiment a punter used the same quantities of material but used
Why
Every chemical reaction conserves the component masses; of course such mass may be lost to the experimenter by evolution of gas or thru handling.
Here
In another experiment a punter used the same quantities of material but used
Why
Every chemical reaction conserves the component masses; of course such mass may be lost to the experimenter by evolution of gas or thru handling.
Here
In another experiment a punter used the same quantities of material but used
As its name suggests, barium chloride dihydrate,
The water that's part of the solid's crystal structure is called water of crystallization.
In the case of barium chloride dihydrate, you get
In other words, if you were to drive the water of crystallization off by heating the hydrate, you would be left with
The find the hydrate's percent composition of water in barium chloride dihydrate, use the mass of one mole of hydrate and of two moles of water.
Barium chloride dihydrate has a molar mass of
Since one mole of hydrate contains
This means that the percent composition of water is
#(color(red)(2) xx 18.015color(red)(cancel(color(black)(""g""))))/(244.26color(red)(cancel(color(black)(""g"")))) xx 100 = color(green)(|bar(ul(color(white)(a/a)""14.751%""color(white)(a/a)|)))#
This means that you get
Here is a similar lab with analysis conducted using copper (II) sulfate.
Hope this helps!
As its name suggests, barium chloride dihydrate,
The water that's part of the solid's crystal structure is called water of crystallization.
In the case of barium chloride dihydrate, you get
In other words, if you were to drive the water of crystallization off by heating the hydrate, you would be left with
The find the hydrate's percent composition of water in barium chloride dihydrate, use the mass of one mole of hydrate and of two moles of water.
Barium chloride dihydrate has a molar mass of
Since one mole of hydrate contains
This means that the percent composition of water is
#(color(red)(2) xx 18.015color(red)(cancel(color(black)(""g""))))/(244.26color(red)(cancel(color(black)(""g"")))) xx 100 = color(green)(|bar(ul(color(white)(a/a)""14.751%""color(white)(a/a)|)))#
This means that you get
Here is a similar lab with analysis conducted using copper (II) sulfate.
Hope this helps!
As its name suggests, barium chloride dihydrate,
The water that's part of the solid's crystal structure is called water of crystallization.
In the case of barium chloride dihydrate, you get
In other words, if you were to drive the water of crystallization off by heating the hydrate, you would be left with
The find the hydrate's percent composition of water in barium chloride dihydrate, use the mass of one mole of hydrate and of two moles of water.
Barium chloride dihydrate has a molar mass of
Since one mole of hydrate contains
This means that the percent composition of water is
#(color(red)(2) xx 18.015color(red)(cancel(color(black)(""g""))))/(244.26color(red)(cancel(color(black)(""g"")))) xx 100 = color(green)(|bar(ul(color(white)(a/a)""14.751%""color(white)(a/a)|)))#
This means that you get
Here is a similar lab with analysis conducted using copper (II) sulfate.
Hope this helps!
The relationship between work (
where,
since the gas is expanding, then the work is done by the system and it is of a negative value .
Note that work, in this case, should be expressed in
Since work is done by the system:
Pressure should then be expressed in
Thus, replacing every term in its value in the expression
Note that
Here is a video that further explains this topic:
Thermochemistry | The Nature of Energy.
The relationship between work (
where,
since the gas is expanding, then the work is done by the system and it is of a negative value .
Note that work, in this case, should be expressed in
Since work is done by the system:
Pressure should then be expressed in
Thus, replacing every term in its value in the expression
Note that
Here is a video that further explains this topic:
Thermochemistry | The Nature of Energy.
The relationship between work (
where,
since the gas is expanding, then the work is done by the system and it is of a negative value .
Note that work, in this case, should be expressed in
Since work is done by the system:
Pressure should then be expressed in
Thus, replacing every term in its value in the expression
Note that
Here is a video that further explains this topic:
Thermochemistry | The Nature of Energy.
Writing calcium hydroxide + phosphoric acid yield calcium phosphate + water in chemical equation would be,
but it still needs to be balanced. To balance this, you just have to ensure that there is an equal amount of elements in both the reactant and product side.
Balancing this would result to,
Writing calcium hydroxide + phosphoric acid yield calcium phosphate + water in chemical equation would be,
but it still needs to be balanced. To balance this, you just have to ensure that there is an equal amount of elements in both the reactant and product side.
Balancing this would result to,
Writing calcium hydroxide + phosphoric acid yield calcium phosphate + water in chemical equation would be,
but it still needs to be balanced. To balance this, you just have to ensure that there is an equal amount of elements in both the reactant and product side.
Balancing this would result to,
When you balance, start of with the most complicated molecules, and leave the simpler molecules to last. In this case, we want to balance
We need two sodiums on the reactants side, so we place a two in front of sodium chloride. Then we need two chlorines as we just added two to sodium chloride, so we place a two in front of hydrogen chloride.
Now count your elements and see what's missing. We have 1 sulfur on both sides, 2 hydrogens, but we have 5 oxygens on the reactants side and only 4 on the products. Since the number of oxygens on the products can only be even, as it is in
When you balance, start of with the most complicated molecules, and leave the simpler molecules to last. In this case, we want to balance
We need two sodiums on the reactants side, so we place a two in front of sodium chloride. Then we need two chlorines as we just added two to sodium chloride, so we place a two in front of hydrogen chloride.
Now count your elements and see what's missing. We have 1 sulfur on both sides, 2 hydrogens, but we have 5 oxygens on the reactants side and only 4 on the products. Since the number of oxygens on the products can only be even, as it is in
When you balance, start of with the most complicated molecules, and leave the simpler molecules to last. In this case, we want to balance
We need two sodiums on the reactants side, so we place a two in front of sodium chloride. Then we need two chlorines as we just added two to sodium chloride, so we place a two in front of hydrogen chloride.
Now count your elements and see what's missing. We have 1 sulfur on both sides, 2 hydrogens, but we have 5 oxygens on the reactants side and only 4 on the products. Since the number of oxygens on the products can only be even, as it is in
This answer is incorrect because it does not take into account the fact that you're dealing with a basic buffer solution.
The trick here was to recognize the fact that the solution contains ammonium hydroxide, which is actually a solution of ammonia,
A buffer solution is able to resist changes to its pH upon the addition of small amounts of strong acid or strong base.
This means that, even without doing any calculation, you should have recognized that the pH of the solution will virtually remain unchanged by the addition of the sodium hydroxide solution.
Correct answer:
Since I didn't recognize the fact that this was a buffer solution, I will leave my inaccurate answer below to serve as an example of how NOT to answer this question.
The trick here is that the pH of the solution is the only information you actually need here.
Don't worry about what solution
First thing first, try to predict if you expect the pH of the solution to increase or to decrease upon the addition of the sodium hydroxide,
Well, since sodium hydroxide is a strong base, it will dissociate completely to form sodium cations, which are of no interest to you, and hydroxide anions,
So, adding the sodium hydroxide will Increase the concentration of the hydroxide anions, which can only mean that the pH of the solution will increase, i.e. the solution will become even more basic.
Prediction:
#""pH"" > 10.0#
Use the pH of the solution to find the concentration of the hydroxide anions before adding the sodium hydroxide
#color(blue)(""pH"" + ""pOH"" = 14)#
Therefore, you have
#""pOH"" = 14 - 10.0 = 4.0#
This means that the concentration of hydroxide anions is
#""pOH""= - log([""OH""^(-)]) implies [""OH""^(-)] = 10^(-""pOH"")#
#[""OH""^(-)] = 10^(-4)""M""#
Now, you take a
#color(blue)(c = n/V implies n = c * V)#
#n_(OH^(-)) = 10^(-4)""mol"" color(red)(cancel(color(black)(""dm""^(-3)))) * 1 color(red)(cancel(color(black)(""dm""^3))) = 10^(-4)""moles OH""^(-)#
Now use the molarity and volume of the sodium hydroxide solution to determine how many moles of hydroxide anions you're adding to that sample of solution
#n_(OH^(-)""added"") = ""0.1 mol"" color(red)(cancel(color(black)(""dm""^(-3)))) * 1 * 10^(-3)color(red)(cancel(color(black)(""dm""^3))) = 10^(-4)""mol OH""^(-)#
Now, the total number of moles of hydroxide anions will be
#n_(OH^(-)""total"") = n_(OH^(-)) + n_(OH^(-)""added"")#
#n_(OH^(-)""total"") = 10^(-4)""moles"" + 10^(-4)""moles"" = 2 * 10^(-4)""moles OH""^(-)#
The total volume of the resulting solution will be
#V_""total"" = V_""sample P"" + V_""added""#
#V_""total"" = ""1 dm""^3 + 1 * 10^(-3)""dm""^3 = ""1.001 dm""^3#
The molarity of the hydroxide anions after you add the sodium hydroxide solution will be
#[""OH""^(-)] = (2 * 10^(-4)""moles"")/""1.001 dm""^3 = 1.998 * 10^(-4)""M""#
The pOH of the resulting solution will be
#""pOH"" = -log(1.998 * 10^(-4)) = 3.70#
The pH of this solution will be
#""pH"" = 14 - 3.70 = color(green)(10.3)#
As predicted, the pH of the solution increased upon the addition of the sodium hydroxide solution.
See explanation.
This answer is incorrect because it does not take into account the fact that you're dealing with a basic buffer solution.
The trick here was to recognize the fact that the solution contains ammonium hydroxide, which is actually a solution of ammonia,
A buffer solution is able to resist changes to its pH upon the addition of small amounts of strong acid or strong base.
This means that, even without doing any calculation, you should have recognized that the pH of the solution will virtually remain unchanged by the addition of the sodium hydroxide solution.
Correct answer:
Since I didn't recognize the fact that this was a buffer solution, I will leave my inaccurate answer below to serve as an example of how NOT to answer this question.
The trick here is that the pH of the solution is the only information you actually need here.
Don't worry about what solution
First thing first, try to predict if you expect the pH of the solution to increase or to decrease upon the addition of the sodium hydroxide,
Well, since sodium hydroxide is a strong base, it will dissociate completely to form sodium cations, which are of no interest to you, and hydroxide anions,
So, adding the sodium hydroxide will Increase the concentration of the hydroxide anions, which can only mean that the pH of the solution will increase, i.e. the solution will become even more basic.
Prediction:
#""pH"" > 10.0#
Use the pH of the solution to find the concentration of the hydroxide anions before adding the sodium hydroxide
#color(blue)(""pH"" + ""pOH"" = 14)#
Therefore, you have
#""pOH"" = 14 - 10.0 = 4.0#
This means that the concentration of hydroxide anions is
#""pOH""= - log([""OH""^(-)]) implies [""OH""^(-)] = 10^(-""pOH"")#
#[""OH""^(-)] = 10^(-4)""M""#
Now, you take a
#color(blue)(c = n/V implies n = c * V)#
#n_(OH^(-)) = 10^(-4)""mol"" color(red)(cancel(color(black)(""dm""^(-3)))) * 1 color(red)(cancel(color(black)(""dm""^3))) = 10^(-4)""moles OH""^(-)#
Now use the molarity and volume of the sodium hydroxide solution to determine how many moles of hydroxide anions you're adding to that sample of solution
#n_(OH^(-)""added"") = ""0.1 mol"" color(red)(cancel(color(black)(""dm""^(-3)))) * 1 * 10^(-3)color(red)(cancel(color(black)(""dm""^3))) = 10^(-4)""mol OH""^(-)#
Now, the total number of moles of hydroxide anions will be
#n_(OH^(-)""total"") = n_(OH^(-)) + n_(OH^(-)""added"")#
#n_(OH^(-)""total"") = 10^(-4)""moles"" + 10^(-4)""moles"" = 2 * 10^(-4)""moles OH""^(-)#
The total volume of the resulting solution will be
#V_""total"" = V_""sample P"" + V_""added""#
#V_""total"" = ""1 dm""^3 + 1 * 10^(-3)""dm""^3 = ""1.001 dm""^3#
The molarity of the hydroxide anions after you add the sodium hydroxide solution will be
#[""OH""^(-)] = (2 * 10^(-4)""moles"")/""1.001 dm""^3 = 1.998 * 10^(-4)""M""#
The pOH of the resulting solution will be
#""pOH"" = -log(1.998 * 10^(-4)) = 3.70#
The pH of this solution will be
#""pH"" = 14 - 3.70 = color(green)(10.3)#
As predicted, the pH of the solution increased upon the addition of the sodium hydroxide solution.
See explanation.
This answer is incorrect because it does not take into account the fact that you're dealing with a basic buffer solution.
The trick here was to recognize the fact that the solution contains ammonium hydroxide, which is actually a solution of ammonia,
A buffer solution is able to resist changes to its pH upon the addition of small amounts of strong acid or strong base.
This means that, even without doing any calculation, you should have recognized that the pH of the solution will virtually remain unchanged by the addition of the sodium hydroxide solution.
Correct answer:
Since I didn't recognize the fact that this was a buffer solution, I will leave my inaccurate answer below to serve as an example of how NOT to answer this question.
The trick here is that the pH of the solution is the only information you actually need here.
Don't worry about what solution
First thing first, try to predict if you expect the pH of the solution to increase or to decrease upon the addition of the sodium hydroxide,
Well, since sodium hydroxide is a strong base, it will dissociate completely to form sodium cations, which are of no interest to you, and hydroxide anions,
So, adding the sodium hydroxide will Increase the concentration of the hydroxide anions, which can only mean that the pH of the solution will increase, i.e. the solution will become even more basic.
Prediction:
#""pH"" > 10.0#
Use the pH of the solution to find the concentration of the hydroxide anions before adding the sodium hydroxide
#color(blue)(""pH"" + ""pOH"" = 14)#
Therefore, you have
#""pOH"" = 14 - 10.0 = 4.0#
This means that the concentration of hydroxide anions is
#""pOH""= - log([""OH""^(-)]) implies [""OH""^(-)] = 10^(-""pOH"")#
#[""OH""^(-)] = 10^(-4)""M""#
Now, you take a
#color(blue)(c = n/V implies n = c * V)#
#n_(OH^(-)) = 10^(-4)""mol"" color(red)(cancel(color(black)(""dm""^(-3)))) * 1 color(red)(cancel(color(black)(""dm""^3))) = 10^(-4)""moles OH""^(-)#
Now use the molarity and volume of the sodium hydroxide solution to determine how many moles of hydroxide anions you're adding to that sample of solution
#n_(OH^(-)""added"") = ""0.1 mol"" color(red)(cancel(color(black)(""dm""^(-3)))) * 1 * 10^(-3)color(red)(cancel(color(black)(""dm""^3))) = 10^(-4)""mol OH""^(-)#
Now, the total number of moles of hydroxide anions will be
#n_(OH^(-)""total"") = n_(OH^(-)) + n_(OH^(-)""added"")#
#n_(OH^(-)""total"") = 10^(-4)""moles"" + 10^(-4)""moles"" = 2 * 10^(-4)""moles OH""^(-)#
The total volume of the resulting solution will be
#V_""total"" = V_""sample P"" + V_""added""#
#V_""total"" = ""1 dm""^3 + 1 * 10^(-3)""dm""^3 = ""1.001 dm""^3#
The molarity of the hydroxide anions after you add the sodium hydroxide solution will be
#[""OH""^(-)] = (2 * 10^(-4)""moles"")/""1.001 dm""^3 = 1.998 * 10^(-4)""M""#
The pOH of the resulting solution will be
#""pOH"" = -log(1.998 * 10^(-4)) = 3.70#
The pH of this solution will be
#""pH"" = 14 - 3.70 = color(green)(10.3)#
As predicted, the pH of the solution increased upon the addition of the sodium hydroxide solution.
i.e unchanged.
This is a recipe for an alkaline buffer. It resists the addition of small amounts of acid and alkali keeping the pH constant.
How does it work?
You need a solution of a weak base and its salt. In this case we have ammonia solution and the salt ammonium chloride.
Ammonium chloride is an ionic salt and dissociates completely in water:
If a small amount of
In this question a small amount of
A calculation can show this:
Ammonium ions dissociate:
For which:
The number of moles of
These will combine with the
In 1 litre of P we have 0.1 mol of
So we are left with
By the same reasoning you can see that the number of moles of ammonia must have increased by this amount.
This gives
To find the pH of a buffer we rearrange
Concentration = moles/volume. We can ignore the total volume of the solution since it is common to both
So the pH does not change.
This is for illustration only. As you state, this is a multi - choice question so my take on this is that you are required recognise it is the recipe for an alkali buffer so reason that the pH won't change.
The salt (NaCl) content of this water is 11.6%. 20% salty water has 216 degrees Fahrenheit boiling point.
Pure water has a boiling point of 212 degrees Fahrenheit. When you build an equation Delta t = 4 degrees F when salt has a concentration of 20%, you can calculate the boiling point of Juan's water. The answer is 214.32 degrees Fahrenheit.
Approximately 214.32 degrees Fahrenheit
The salt (NaCl) content of this water is 11.6%. 20% salty water has 216 degrees Fahrenheit boiling point.
Pure water has a boiling point of 212 degrees Fahrenheit. When you build an equation Delta t = 4 degrees F when salt has a concentration of 20%, you can calculate the boiling point of Juan's water. The answer is 214.32 degrees Fahrenheit.
Approximately 214.32 degrees Fahrenheit
The salt (NaCl) content of this water is 11.6%. 20% salty water has 216 degrees Fahrenheit boiling point.
Pure water has a boiling point of 212 degrees Fahrenheit. When you build an equation Delta t = 4 degrees F when salt has a concentration of 20%, you can calculate the boiling point of Juan's water. The answer is 214.32 degrees Fahrenheit.
100.50 degrees celcius.
If you dissolve 29.2 g of sodium chloride in 1 kg water, you create what is known as a ""0.5 molal"" solution (as opposed to 0.5 molar solution, which is 29.2 g of sodium chloride dissolved in 1 litre of solution).
The boiling point of water is raised by 0.5 degrees celcius for each 0.5 molality of the solution, so instead of boiling at 100 degrees C it would boil at 100.50 degrees C.
Oxidation number is the charge left on the central atom when each of the bonding pairs are broken, with the charge going to the most electronegative atom. If I break the
What are the oxidation states of nitrogen in
We could represent the oxidation of ammonia
Note that we often speak of dinitrogen reduction to ammonia, the which represents a formal six electron reduction with respect to dinitrogen...
The oxidation state of
Oxidation number is the charge left on the central atom when each of the bonding pairs are broken, with the charge going to the most electronegative atom. If I break the
What are the oxidation states of nitrogen in
We could represent the oxidation of ammonia
Note that we often speak of dinitrogen reduction to ammonia, the which represents a formal six electron reduction with respect to dinitrogen...
The oxidation state of
Oxidation number is the charge left on the central atom when each of the bonding pairs are broken, with the charge going to the most electronegative atom. If I break the
What are the oxidation states of nitrogen in
We could represent the oxidation of ammonia
Note that we often speak of dinitrogen reduction to ammonia, the which represents a formal six electron reduction with respect to dinitrogen...
The oxidation state of Nitrogen in
Nitrogen has a higher electronegativity than Hydrogen. This means that the electron density of the electrons that form the bond between the Nitrogen and the Hydrogen will be pulled closer to the Nitrogen than the Hydrogen.
There are three sets of bonds between Nitrogen and Hydrogen in
The Ideal Gas Law explicitly states that:
Note that the only problem in solving this problem is choosing an appropriate gas constant,
The Ideal Gas Law explicitly states that:
Note that the only problem in solving this problem is choosing an appropriate gas constant,
The Ideal Gas Law explicitly states that:
Note that the only problem in solving this problem is choosing an appropriate gas constant,
The trick here is to recognize the fact that argon will not react with the hot copper and the hot magnesium, but the oxygen and the nitrogen gas present in the sample will.
Basically, the fact that argon is a noble gas tells you that you shouldn't expect to see a reaction take place when the sample of air is passed over the two hot metals.
On the other hand, oxygen gas will react with the hot copper to form copper(II) oxide,
So if you pass the sample over the two hot metals until no reduction in volume takes place, you can assume that all the oxygen gas and all the nitrogen gas have reacted.
This means that the sample will only contain argon. Since this gas makes up
#1 color(red)(cancel(color(black)(""%""))) * overbrace(""1000 mL""/(100color(red)(cancel(color(black)(""%"")))))^(color(blue)(""1 L is 100% of the volume"")) = ""10 mL""#
The answer is rounded to one significant figure.
The trick here is to recognize the fact that argon will not react with the hot copper and the hot magnesium, but the oxygen and the nitrogen gas present in the sample will.
Basically, the fact that argon is a noble gas tells you that you shouldn't expect to see a reaction take place when the sample of air is passed over the two hot metals.
On the other hand, oxygen gas will react with the hot copper to form copper(II) oxide,
So if you pass the sample over the two hot metals until no reduction in volume takes place, you can assume that all the oxygen gas and all the nitrogen gas have reacted.
This means that the sample will only contain argon. Since this gas makes up
#1 color(red)(cancel(color(black)(""%""))) * overbrace(""1000 mL""/(100color(red)(cancel(color(black)(""%"")))))^(color(blue)(""1 L is 100% of the volume"")) = ""10 mL""#
The answer is rounded to one significant figure.
The trick here is to recognize the fact that argon will not react with the hot copper and the hot magnesium, but the oxygen and the nitrogen gas present in the sample will.
Basically, the fact that argon is a noble gas tells you that you shouldn't expect to see a reaction take place when the sample of air is passed over the two hot metals.
On the other hand, oxygen gas will react with the hot copper to form copper(II) oxide,
So if you pass the sample over the two hot metals until no reduction in volume takes place, you can assume that all the oxygen gas and all the nitrogen gas have reacted.
This means that the sample will only contain argon. Since this gas makes up
#1 color(red)(cancel(color(black)(""%""))) * overbrace(""1000 mL""/(100color(red)(cancel(color(black)(""%"")))))^(color(blue)(""1 L is 100% of the volume"")) = ""10 mL""#
The answer is rounded to one significant figure.
In this case, where the mass of the solute is known, and the mass of the solution is desired, the basic equation for percent concentration is the mass of solute divided by the mass of the solution multiplied times 100%.
Given:
mass of solute =
percent concentration =
Unknown:
mass of solution
Solution: Rearrange the equation for percent concentration to isolate mass of solution. Plug in the known values and solve.
The mass of the solution is
In this case, where the mass of the solute is known, and the mass of the solution is desired, the basic equation for percent concentration is the mass of solute divided by the mass of the solution multiplied times 100%.
Given:
mass of solute =
percent concentration =
Unknown:
mass of solution
Solution: Rearrange the equation for percent concentration to isolate mass of solution. Plug in the known values and solve.
The mass of the solution is
In this case, where the mass of the solute is known, and the mass of the solution is desired, the basic equation for percent concentration is the mass of solute divided by the mass of the solution multiplied times 100%.
Given:
mass of solute =
percent concentration =
Unknown:
mass of solution
Solution: Rearrange the equation for percent concentration to isolate mass of solution. Plug in the known values and solve.
Nitrogen molecule is
Flourine molecule is
Nitrogen triflouride is
(tri- means 3, and
so 3
write unbalanced eqn for reaction
now balance the number of atoms on each side:
at least 2 atoms of nitrogen so at least 2 molecules of
2 molecules of
To answer this question, we need two things:
We are given the molality and mass of solvent. The mass of solvent is expressed in terms of grams and we need kilograms so the following relationship must be used:
1000g = 1kg
Divide 156g by 1000 to obtain 0.156 kg of solvent.
Next, we determine the number of moles of solute (barium hydroxide), by rearranging the equation:
Moles of solute = Molality x Mass of solvent
Therefore,
Since we want the mass of
g
6.60g of
To answer this question, we need two things:
We are given the molality and mass of solvent. The mass of solvent is expressed in terms of grams and we need kilograms so the following relationship must be used:
1000g = 1kg
Divide 156g by 1000 to obtain 0.156 kg of solvent.
Next, we determine the number of moles of solute (barium hydroxide), by rearranging the equation:
Moles of solute = Molality x Mass of solvent
Therefore,
Since we want the mass of
g
6.60g of
To answer this question, we need two things:
We are given the molality and mass of solvent. The mass of solvent is expressed in terms of grams and we need kilograms so the following relationship must be used:
1000g = 1kg
Divide 156g by 1000 to obtain 0.156 kg of solvent.
Next, we determine the number of moles of solute (barium hydroxide), by rearranging the equation:
Moles of solute = Molality x Mass of solvent
Therefore,
Since we want the mass of
g
Where
And what is the mass of this number of molecules? And how many ATOMS? You should be able to answer pdq.
There are
Where
And what is the mass of this number of molecules? And how many ATOMS? You should be able to answer pdq.
There are
Where
And what is the mass of this number of molecules? And how many ATOMS? You should be able to answer pdq.
So the mass of .0642 mol ammonium phosphate is
=
So the mass of .0642 mol ammonium phosphate is
=
So the mass of .0642 mol ammonium phosphate is
=
And for stoichiometric reaction we need,
And for stoichiometric reaction we need,
And for stoichiometric reaction we need,
Now
And
Given that it is a binary system, what is
Note that the information given with respect to
Now
And
Given that it is a binary system, what is
Note that the information given with respect to
Now
And
Given that it is a binary system, what is
Note that the information given with respect to
Zinc metal will react with dilute sulfuric acid to produce zinc sulfate,
The balanced chemical equation for this single replacement reaction looks like this
#""Zn""_text((s]) + ""H""_2""SO""_text(4(aq]) -> ""ZnSO""_text(4(aq]) + ""H""_text(2(g]) uarr#
Notice that you have a
In your case, you know that the sulfuric acid is in excess, which means that zinc metal acts as a limiting reagent, i.e. it will be completely consumed by the reaction.
Use zinc metal's molar mass to determine how many moles you have in that
#4.33 color(red)(cancel(color(black)(""g""))) * ""1 mole Zn""/(65.38color(red)(cancel(color(black)(""g"")))) = ""0.06623 moles Zn""#
This means that the reaction produced
Now, in order to determine the volume this much hydrogen gas occupies under those conditions for pressure and temperature, you must use the ideal gas law equation, which looks like this
#color(blue)(PV = nRT)"" ""# , where
It's important to notice here that the units given to you for pressure and temperature do not match those used in the expression of the universal gas constant.
This means that you have to convert these units from torr to atm, and from degrees Celsius to Kelvin, respectively.
So, rearrange the ideal gas law equation to solve for
#V = (nRT)/P#
#V = (0.06623 color(red)(cancel(color(black)(""moles""))) * 0.0821(color(red)(cancel(color(black)(""atm""))) * ""L"")/(color(red)(cancel(color(black)(""moles""))) * color(red)(cancel(color(black)(""K"")))) * (273.15 + 38)color(red)(cancel(color(black)(""K""))))/(760/760color(red)(cancel(color(black)(""atm""))))#
#V = ""1.692 L""#
Rounded to two sig figs, the number of sig figs you have for the pressure and temperature of the gas, the answer will be
#V = color(green)(""1.7 L"")#
Zinc metal will react with dilute sulfuric acid to produce zinc sulfate,
The balanced chemical equation for this single replacement reaction looks like this
#""Zn""_text((s]) + ""H""_2""SO""_text(4(aq]) -> ""ZnSO""_text(4(aq]) + ""H""_text(2(g]) uarr#
Notice that you have a
In your case, you know that the sulfuric acid is in excess, which means that zinc metal acts as a limiting reagent, i.e. it will be completely consumed by the reaction.
Use zinc metal's molar mass to determine how many moles you have in that
#4.33 color(red)(cancel(color(black)(""g""))) * ""1 mole Zn""/(65.38color(red)(cancel(color(black)(""g"")))) = ""0.06623 moles Zn""#
This means that the reaction produced
Now, in order to determine the volume this much hydrogen gas occupies under those conditions for pressure and temperature, you must use the ideal gas law equation, which looks like this
#color(blue)(PV = nRT)"" ""# , where
It's important to notice here that the units given to you for pressure and temperature do not match those used in the expression of the universal gas constant.
This means that you have to convert these units from torr to atm, and from degrees Celsius to Kelvin, respectively.
So, rearrange the ideal gas law equation to solve for
#V = (nRT)/P#
#V = (0.06623 color(red)(cancel(color(black)(""moles""))) * 0.0821(color(red)(cancel(color(black)(""atm""))) * ""L"")/(color(red)(cancel(color(black)(""moles""))) * color(red)(cancel(color(black)(""K"")))) * (273.15 + 38)color(red)(cancel(color(black)(""K""))))/(760/760color(red)(cancel(color(black)(""atm""))))#
#V = ""1.692 L""#
Rounded to two sig figs, the number of sig figs you have for the pressure and temperature of the gas, the answer will be
#V = color(green)(""1.7 L"")#
Zinc metal will react with dilute sulfuric acid to produce zinc sulfate,
The balanced chemical equation for this single replacement reaction looks like this
#""Zn""_text((s]) + ""H""_2""SO""_text(4(aq]) -> ""ZnSO""_text(4(aq]) + ""H""_text(2(g]) uarr#
Notice that you have a
In your case, you know that the sulfuric acid is in excess, which means that zinc metal acts as a limiting reagent, i.e. it will be completely consumed by the reaction.
Use zinc metal's molar mass to determine how many moles you have in that
#4.33 color(red)(cancel(color(black)(""g""))) * ""1 mole Zn""/(65.38color(red)(cancel(color(black)(""g"")))) = ""0.06623 moles Zn""#
This means that the reaction produced
Now, in order to determine the volume this much hydrogen gas occupies under those conditions for pressure and temperature, you must use the ideal gas law equation, which looks like this
#color(blue)(PV = nRT)"" ""# , where
It's important to notice here that the units given to you for pressure and temperature do not match those used in the expression of the universal gas constant.
This means that you have to convert these units from torr to atm, and from degrees Celsius to Kelvin, respectively.
So, rearrange the ideal gas law equation to solve for
#V = (nRT)/P#
#V = (0.06623 color(red)(cancel(color(black)(""moles""))) * 0.0821(color(red)(cancel(color(black)(""atm""))) * ""L"")/(color(red)(cancel(color(black)(""moles""))) * color(red)(cancel(color(black)(""K"")))) * (273.15 + 38)color(red)(cancel(color(black)(""K""))))/(760/760color(red)(cancel(color(black)(""atm""))))#
#V = ""1.692 L""#
Rounded to two sig figs, the number of sig figs you have for the pressure and temperature of the gas, the answer will be
#V = color(green)(""1.7 L"")#
Zinc's only common oxidation state is
First, calculate the moles of
Next, calculate the moles of
In order to find the volume of space occupied by the calculated number of moles of hydrogen gas at the stated conditions, we must consult the ideal gas equation:
Where:
First, convert the given data into workable units:
Next, rearrange the equation to solve for volume:
Finally, substitute in your values and solve the equation:
Another useful unit that is probably suitable to use in this situation is the cubic decimetre (
Since the type of iron is not specified then it is assumed to be 'Cast Iron' which has a melting point of 1204 deg-Celsius*.
*http://www.onlinemetals.com/meltpt.cfm
The total heat transfer would be ,,,
..............................................................................................................................
=>
=
(Specific Heat of Molten Iron) http://www.engineeringtoolbox.com/liquid-metal-boiling-points-specific-heat-d_1893.html
..............................................................................................................................
=>
=
(Heat of Fusion of Molten Iron) http://www.engineeringtoolbox.com/fusion-heat-metals-d_1266.html
...............................................................................................................................
=>
=
(Specific Heat of Iron (s) = 0.46
..............................................................................................................................
900 Kj
Since the type of iron is not specified then it is assumed to be 'Cast Iron' which has a melting point of 1204 deg-Celsius*.
*http://www.onlinemetals.com/meltpt.cfm
The total heat transfer would be ,,,
..............................................................................................................................
=>
=
(Specific Heat of Molten Iron) http://www.engineeringtoolbox.com/liquid-metal-boiling-points-specific-heat-d_1893.html
..............................................................................................................................
=>
=
(Heat of Fusion of Molten Iron) http://www.engineeringtoolbox.com/fusion-heat-metals-d_1266.html
...............................................................................................................................
=>
=
(Specific Heat of Iron (s) = 0.46
..............................................................................................................................
900 Kj
Since the type of iron is not specified then it is assumed to be 'Cast Iron' which has a melting point of 1204 deg-Celsius*.
*http://www.onlinemetals.com/meltpt.cfm
The total heat transfer would be ,,,
..............................................................................................................................
=>
=
(Specific Heat of Molten Iron) http://www.engineeringtoolbox.com/liquid-metal-boiling-points-specific-heat-d_1893.html
..............................................................................................................................
=>
=
(Heat of Fusion of Molten Iron) http://www.engineeringtoolbox.com/fusion-heat-metals-d_1266.html
...............................................................................................................................
=>
=
(Specific Heat of Iron (s) = 0.46
..............................................................................................................................
I got
You can get more context here:
https://en.wikipedia.org/wiki/Iron#Phase_diagram_and_allotropes
and you can examine the specific heat capacity variations more closely here:
http://webbook.nist.gov/cgi/cbook.cgi?ID=C7439896&Mask=2&Type=JANAFS&Plot=on#JANAFS
On another note, this
If you simply assume a
There is a HUGE assumption here that iron's specific heat capacity doesn't change from
Since all these phases at
However, the specific heat capacity
[
The wonky curve is the
Aren't you glad we aren't doing phase changes? :-)
So, we would have the heat of cooling as the negative of the heat of heating:
#q_""cool"" = -(q_1 + . . . + q_7)#
#= -m(C_(P1)DeltaT_(0->1) + . . . + C_(P7)DeltaT_(6->7))#
I'll leave the units out, but you know that they are
#= -1000 cdot [0.531(700 - 298.15) + 0.719(935 - 700) + 1.064(1042 - 935) + 1.177(1100 - 1042) + 0.770(1183.15 - 1100) + 0.642(1667.15 - 1183.15) + 0.748(1773.15 - 1667.15)]#
Each phase then approximately contributes:
#= overbrace(-""628487 J"")^(alpha"" phase"") + overbrace(-""310728 J"")^(gamma"" phase"") + overbrace(-""79288 J"")^(delta"" phase"")#
#~~# #-1.020 xx 10^(6)# #""J""# ,
or about
Thermal history of cooling cast iron
Thermal history of cooling cast iron from
The balanced equation for the reaction is
Phosphoric acid
Thus, one molecule of phosphoric acid will need three formula units of sodium hydroxide, and
One mole of phosphoric acid will need three moles of sodium hydroxide.
This is what the balanced equation tells us:
3 mol of
You need 3 mol of sodium hydroxide to neutralize 1 mol of phosphoric acid.
The balanced equation for the reaction is
Phosphoric acid
Thus, one molecule of phosphoric acid will need three formula units of sodium hydroxide, and
One mole of phosphoric acid will need three moles of sodium hydroxide.
This is what the balanced equation tells us:
3 mol of
You need 3 mol of sodium hydroxide to neutralize 1 mol of phosphoric acid.
The balanced equation for the reaction is
Phosphoric acid
Thus, one molecule of phosphoric acid will need three formula units of sodium hydroxide, and
One mole of phosphoric acid will need three moles of sodium hydroxide.
This is what the balanced equation tells us:
3 mol of
To begin, using a
Your values will look like this :
Next, convert each value into moles by using molar mass.
(Divide your original value by the molar mass to find moles)
Take your smallest value of moles and divide every numerical value by it. In this case, the smallest value is
To find the smallest whole number ratio, multiply each value by
Your formula is now:
Remember, empirical formulas are related by a whole number ratios or the least common multiple of all numbers.
The empirical formula is:
(Remember that if the subscript of an element is
The empirical formula is
To begin, using a
Your values will look like this :
Next, convert each value into moles by using molar mass.
(Divide your original value by the molar mass to find moles)
Take your smallest value of moles and divide every numerical value by it. In this case, the smallest value is
To find the smallest whole number ratio, multiply each value by
Your formula is now:
Remember, empirical formulas are related by a whole number ratios or the least common multiple of all numbers.
The empirical formula is:
(Remember that if the subscript of an element is
The empirical formula is
To begin, using a
Your values will look like this :
Next, convert each value into moles by using molar mass.
(Divide your original value by the molar mass to find moles)
Take your smallest value of moles and divide every numerical value by it. In this case, the smallest value is
To find the smallest whole number ratio, multiply each value by
Your formula is now:
Remember, empirical formulas are related by a whole number ratios or the least common multiple of all numbers.
The empirical formula is:
(Remember that if the subscript of an element is
As always we ASSUME a
And so we divide thru by the SMALLEST molar quantity to get a trial empirical formula of...
And of course this is zinc nitrate...