chapter 101 gases. 2 homework: 10.12, 10.28, 10.42, 10.48, 10.54, 10.66, 10.72
TRANSCRIPT
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Chapter 10 1
GasesGases
Chapter 10Chapter 10
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Chapter 10 2
Homework:10.12, 10.28, 10.42, 10.48, 10.54, 10.66, 10.72
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Chapter 10 3
Characteristics of GasesCharacteristics of Gases
- Expand to fill a volume (expandability)- Compressible- Readily forms homogeneous mixtures with other gases- These behaviors are due to large distances between the
gas molecules.
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Chapter 10 4
PressurePressurePressure - force acting on an object per unit area.
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Chapter 10 5
PressurePressurePressure - force acting on an object per unit area.
AF
P
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Chapter 10 6
PressurePressurePressure - force acting on an object per unit area.
AF
P
- Atmospheric pressure is measured with a barometer.
- Standard atmospheric pressure is the pressure required to support 760 mm of Hg in a column.
- There are several units used for pressure:- Pascal (Pa), N/m2
- Millimeters of Mercury (mmHg)- Atmospheres (atm)
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Chapter 10 7
PressurePressure- Conversion Factors
- 1 atm = 760 mmHg - 1 atm = 760 torr - 1 atm = 1.01325 105 Pa - 1 atm = 101.325 kPa.
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Chapter 10 8
The Gas LawsThe Gas Laws- There are four variables required to describe a gas:
- Amount of substance: moles- Volume of substance: volume- Pressures of substance: pressure- Temperature of substance: temperature
- The gas laws will hold two of the variables constant and see how the other two vary.
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Chapter 10 9
The Gas LawsThe Gas LawsThe Pressure-Volume Relationship: Boyle’s LawBoyle’s Law - The volume of a fixed quantity of gas is
inversely proportional to its pressure.
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Chapter 10 10
The Gas LawsThe Gas LawsThe Pressure-Volume Relationship: Boyle’s LawBoyle’s Law - The volume of a fixed quantity of gas is
inversely proportional to its pressure.
) and (constant 1
TnV
P
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Chapter 10 11
The Gas LawsThe Gas LawsThe Pressure-Volume Relationship: Boyle’s LawBoyle’s Law - The volume of a fixed quantity of gas is
inversely proportional to its pressure.
2211
) and (constant 1
VPVP
TnV
P
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Chapter 10 12
The Gas LawsThe Gas LawsThe Pressure-Volume Relationship: Boyle’s Law
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Chapter 10 13
The Gas LawsThe Gas Laws
Charles’s Law - the volume of a fixed quantity of gas at constant pressure increases as the temperature increases.
The Temperature-Volume Relationship: Charles’s Law
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Chapter 10 14
The Gas LawsThe Gas Laws
Charles’s Law - the volume of a fixed quantity of gas at constant pressure increases as the temperature increases.
The Temperature-Volume Relationship: Charles’s Law
)and(constant PnTV
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Chapter 10 15
The Gas LawsThe Gas Laws
Charles’s Law - the volume of a fixed quantity of gas at constant pressure increases as the temperature increases.
The Temperature-Volume Relationship: Charles’s Law
2
2
1
1
)and(constant
T
V
T
V
PnTV
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Chapter 10 16
The Gas LawsThe Gas LawsThe Temperature-Volume Relationship:
Charles’s Law
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Chapter 10 17
The Gas LawsThe Gas Laws
Avogadro’s Law - The volume of gas at a given temperature and pressure is directly proportional to the number of moles of gas.
The Quantity-Volume Relationship: Avogadro’s Law
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Chapter 10 18
The Gas LawsThe Gas Laws
Avogadro’s Law - The volume of gas at a given temperature and pressure is directly proportional to the number of moles of gas.
) and (constant TPnV
The Quantity-Volume Relationship: Avogadro’s Law
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Chapter 10 19
The Ideal Gas EquationThe Ideal Gas Equation- Combine the gas laws (Boyle, Charles, Avogadro)
yields a new law or equation.
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Chapter 10 20
The Ideal Gas EquationThe Ideal Gas Equation- Combine the gas laws (Boyle, Charles, Avogadro)
yields a new law or equation.
Ideal gas equation:
PV = nRT
R = gas constant = 0.08206 L.atm/mol-K
P = pressure (atm) V = volume (L)
n = moles T = temperature (K)
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Chapter 10 21
The Ideal Gas EquationThe Ideal Gas Equation- We define STP (standard temperature and pressure)
as 0C, 273.15 K, 1 atm.- Volume of 1 mol of gas at STP is 22.4 L.
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Chapter 10 22
Gas Densities and Molar Mass- Rearranging the ideal-gas equation with M as molar
mass yields
Further Applications of The Ideal-Gas Further Applications of The Ideal-Gas EquationEquation
RT
Pd
M
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Chapter 10 23
Gas Mixtures and Partial PressuresGas Mixtures and Partial Pressures
Dalton’s Law - In a gas mixture the total pressure is given by the sum of partial pressures of each component:
Pt = P1 + P2 + P3 + …
- The pressure due to an individual gas is called a partial pressure.
Dalton’s Law
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Chapter 10 24
Gas Mixtures and Partial PressuresGas Mixtures and Partial Pressures
- The partial pressure of a gas can be determined if you know the mole fraction of the gas of interest and the total pressure of the system.
- Let ni be the number of moles of gas i exerting a partial pressure Pi, then
Pi = iPt
where i is the mole fraction (ni/nt).
Partial Pressures and Mole Fractions
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Chapter 10 25
Kinetic-Molecular TheoryKinetic-Molecular Theory- Theory developed to explain gas behavior- To describe the behavior of a gas, we must first
describe what a gas is:– Gases consist of a large number of molecules in constant
random motion.
– Volume of individual molecules negligible compared to volume of container.
– Intermolecular forces (forces between gas molecules) negligible.
– Energy can be transferred between molecules, but total kinetic energy is constant at constant temperature.
– Average kinetic energy of molecules is proportional to temperature.
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Chapter 10 26
Kinetic-Molecular TheoryKinetic-Molecular Theory
• Pressure exerted by a gas is the result of bombardment of the walls of the container by the gas molecules.
• Pressure varies directly with the number of molecules hitting the wall per unit time.– If the volume is reduced the number of impacts is
increased; therefore, the pressure is increased.
Boyle’s Law
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Chapter 10 27
Kinetic-Molecular TheoryKinetic-Molecular Theory
• The force that a gas molecule strikes the side of a container is directly proportional to the temperature.
• As the temperature is increased the kinetic energy (force) of the gas particles increases.
• For the pressure to be constant, the area the force is applied to must be increased (the volume is increased).
Charles’ Law
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Chapter 10 28
Kinetic-Molecular TheoryKinetic-Molecular Theory
• There are large distances between the gas molecules.• The components of a mixture will bombard the walls
of the container with the same frequency in the presence of a mixture as it would by itself.
Dalton’s Law
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Chapter 10 29
Kinetic-Molecular TheoryKinetic-Molecular Theory
• As kinetic energy increases, the velocity of the gas molecules increases.
• Root mean square speed, u, is the speed of a gas molecule having average kinetic energy.
• Average kinetic energy, , is related to root mean square speed:
= ½mu2
Molecular Speed
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Chapter 10 30
Kinetic-Molecular TheoryKinetic-Molecular TheoryMolecular Speed
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Chapter 10 31
Molecular Effusion and DiffusionMolecular Effusion and Diffusion
• Consider two gases at the same temperature: the lighter gas has a higher u than the heavier gas.
• Mathematically:
Molecular Speed
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Chapter 10 32
Molecular Effusion and DiffusionMolecular Effusion and Diffusion
• Consider two gases at the same temperature: the lighter gas has a higher u than the heavier gas.
• Mathematically:
MRT
u3
Molecular Speed
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Chapter 10 33
Molecular Effusion and DiffusionMolecular Effusion and Diffusion
• Consider two gases at the same temperature: the lighter gas has a higher u than the heavier gas.
• Mathematically:
• The lower the molar mass, M, the higher the u for that gas at a constant temperature.
MRT
u3
Molecular Speed
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Chapter 10 34
Molecular Effusion and DiffusionMolecular Effusion and Diffusion
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Chapter 10 35
Molecular Effusion and DiffusionMolecular Effusion and DiffusionGraham’s Law of Effusion
Effusion – The escape of gas through a small opening.
Diffusion – The spreading of one substance through another.
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Chapter 10 36
Molecular Effusion and DiffusionMolecular Effusion and Diffusion
Graham’s Law of Effusion – The rate of effusion of a gas is inversely proportional to the square root of its molecular mass.
Graham’s Law of Effusion
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Chapter 10 37
Molecular Effusion and DiffusionMolecular Effusion and Diffusion
Graham’s Law of Effusion – The rate of effusion of a gas is inversely proportional to the square root of its molecular mass.
Graham’s Law of Effusion
1
2
2
1MM
rr
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Chapter 10 38
Molecular Effusion and DiffusionMolecular Effusion and Diffusion
Graham’s Law of Effusion – The rate of effusion of a gas is inversely proportional to the square root of its molecular mass.
• Gas escaping from a balloon is a good example.
Graham’s Law of Effusion
1
2
2
1MM
rr
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Chapter 10 39
Molecular Effusion and DiffusionMolecular Effusion and DiffusionGraham’s Law of Effusion
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Chapter 10 40
Molecular Effusion and DiffusionMolecular Effusion and Diffusion
• Diffusion of a gas is the spread of the gas through space.
• Diffusion is faster for light gas molecules.• Diffusion is slowed by gas molecules colliding with
each other.• Average distance of a gas molecule between collisions
is called mean free path.
Diffusion and Mean Free Path
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Chapter 10 41
Molecular Effusion and DiffusionMolecular Effusion and DiffusionDiffusion and Mean Free Path• At sea level, mean free path is about 6 10-6 cm.
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Chapter 10 42
Real Gases: Deviations from Ideal Real Gases: Deviations from Ideal BehaviorBehavior• From the ideal gas equation, we have
PV = nRT• This equation breaks-down at
– High pressure• At high pressure, the attractive and repulsive forces between gas
molecules becomes significant.
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Chapter 10 43
Real Gases: Deviations from Ideal Real Gases: Deviations from Ideal BehaviorBehavior
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Chapter 10 44
Real Gases: Deviations from Ideal Real Gases: Deviations from Ideal BehaviorBehavior• From the ideal gas equation, we have
PV = nRT• This equation breaks-down at
– High pressure• At high pressure, the attractive and repulsive forces between gas
molecules becomes significant.
– Small volume
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Chapter 10 45
Real Gases: Deviations from Ideal Real Gases: Deviations from Ideal BehaviorBehavior• From the ideal gas equation, we have
PV = nRT• This equation breaks-down at
– High pressure• At high pressure, the attractive and repulsive forces between gas
molecules becomes significant.
– Small volume• At small volumes, the volume due to the gas molecules is a source of
error.
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Chapter 10 46
Real Gases: Deviations from Ideal Real Gases: Deviations from Ideal BehaviorBehavior
• Two terms are added to the ideal gas equation to correct for volume of molecules and one to correct for intermolecular attractions.
The Van der Waals Equation
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Chapter 10 47
Real Gases: Deviations from Ideal Real Gases: Deviations from Ideal BehaviorBehavior
• Two terms are added to the ideal gas equation to correct for volume of molecules and one to correct for intermolecular attractions.
The Van der Waals Equation
2
2
V
annbV
nRTP
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Chapter 10 48
Real Gases: Deviations from Ideal Real Gases: Deviations from Ideal BehaviorBehavior
• Two terms are added to the ideal gas equation to correct for volume of molecules and one to correct for intermolecular attractions.
• a and b are constants, determined by the particular gas.
The Van der Waals Equation
2
2
V
annbV
nRTP
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Chapter 10 49
Homework:10.12, 10.28, 10.42, 10.48, 10.54, 10.66, 10.72
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Chapter 10 50
ProblemProblem
A sample of gas occupies a volume of 7.50L at 0.988 atm and 28.0oC. Calculate the pressure of the gas if its volume is decreased to 4.89L while its temperature is held constant.
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Chapter 10 51
ProblemProblem
A sample of gas occupies a volume of 7.50L at 0.988 atm and 28.0oC. Calculate the pressure of the gas if its volume is decreased to 4.89L while its temperature is held constant.
--BOYLES LAW
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Chapter 10 52
ProblemProblem
A sample of gas occupies a volume of 7.50L at 0.988 atm and 28.0oC. Calculate the pressure of the gas if its volume is decreased to 4.89L while its temperature is held constant.
--BOYLES LAW
2211 VPVP
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Chapter 10 53
ProblemProblem
A sample of gas occupies a volume of 7.50L at 0.988 atm and 28.0oC. Calculate the pressure of the gas if its volume is decreased to 4.89L while its temperature is held constant.
--BOYLES LAW
)89.4()50.7(988.0 2
2221
LPLatm
VPVP
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Chapter 10 54
ProblemProblem
A sample of gas occupies a volume of 7.50L at 0.988 atm and 28.0oC. Calculate the pressure of the gas if its volume is decreased to 4.89L while its temperature is held constant.
--BOYLES LAW
atmL
LatmP
LPLatm
VPVP
52.189.4
)50.7(988.0
)89.4()50.7(988.0
2
2
2221
![Page 55: Chapter 101 Gases. 2 Homework: 10.12, 10.28, 10.42, 10.48, 10.54, 10.66, 10.72](https://reader035.vdocuments.site/reader035/viewer/2022081506/5697c0011a28abf838cc29e7/html5/thumbnails/55.jpg)
Chapter 10 55
ProblemProblem
A sample of gas occupies a volume of 7.50L at 0.988 atm and 28.0oC. At what temperature in degrees Celsius is the volume of the gas 4.00L if the pressure is kept constant.
![Page 56: Chapter 101 Gases. 2 Homework: 10.12, 10.28, 10.42, 10.48, 10.54, 10.66, 10.72](https://reader035.vdocuments.site/reader035/viewer/2022081506/5697c0011a28abf838cc29e7/html5/thumbnails/56.jpg)
Chapter 10 56
ProblemProblem
A sample of gas occupies a volume of 7.50L at 0.988 atm and 28.0oC. At what temperature in degrees Celsius is the volume of the gas 4.00L if the pressure is kept constant.
--Charles’ Law
![Page 57: Chapter 101 Gases. 2 Homework: 10.12, 10.28, 10.42, 10.48, 10.54, 10.66, 10.72](https://reader035.vdocuments.site/reader035/viewer/2022081506/5697c0011a28abf838cc29e7/html5/thumbnails/57.jpg)
Chapter 10 57
ProblemProblem
A sample of gas occupies a volume of 7.50L at 0.988 atm and 28.0oC. At what temperature in degrees Celsius is the volume of the gas 4.00L if the pressure is kept constant.
--Charles’ Law
2
2
1
1
T
V
T
V
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Chapter 10 58
ProblemProblem
A sample of gas occupies a volume of 7.50L at 0.988 atm and 28.0oC. At what temperature in degrees Celsius is the volume of the gas 4.00L if the pressure is kept constant.
--Charles’ Law
KK
V
P
V
P
30128273
kelvintoetemperaturconvert2
2
2
1
![Page 59: Chapter 101 Gases. 2 Homework: 10.12, 10.28, 10.42, 10.48, 10.54, 10.66, 10.72](https://reader035.vdocuments.site/reader035/viewer/2022081506/5697c0011a28abf838cc29e7/html5/thumbnails/59.jpg)
Chapter 10 59
ProblemProblem
A sample of gas occupies a volume of 7.50L at 0.988 atm and 28.0oC. At what temperature in degrees Celsius is the volume of the gas 4.00L if the pressure is kept constant.
--Charles’ Law
x
L
K
L
V
P
V
P
00.4
301
50.72
2
2
1
![Page 60: Chapter 101 Gases. 2 Homework: 10.12, 10.28, 10.42, 10.48, 10.54, 10.66, 10.72](https://reader035.vdocuments.site/reader035/viewer/2022081506/5697c0011a28abf838cc29e7/html5/thumbnails/60.jpg)
Chapter 10 60
ProblemProblem
A sample of gas occupies a volume of 7.50L at 0.988 atm and 28.0oC. At what temperature in degrees Celsius is the volume of the gas 4.00L if the pressure is kept constant.
--Charles’ Law
5.1125.160
00.4
301
50.72
2
2
1
Ctoconvert o
Kx
x
L
K
L
V
P
V
P
![Page 61: Chapter 101 Gases. 2 Homework: 10.12, 10.28, 10.42, 10.48, 10.54, 10.66, 10.72](https://reader035.vdocuments.site/reader035/viewer/2022081506/5697c0011a28abf838cc29e7/html5/thumbnails/61.jpg)
Chapter 10 61
ProblemProblemCalcium hydride, CaH2, reacts with water to form hydrogen gas:
CaH2(s) + 2 H2O(l) Ca(OH)2(aq) + 2 H2(g)
How many grams of CaH2 are needed to generate 10.0L of H2 gas if the pressure of H2 is 740 torr at 23oC?
![Page 62: Chapter 101 Gases. 2 Homework: 10.12, 10.28, 10.42, 10.48, 10.54, 10.66, 10.72](https://reader035.vdocuments.site/reader035/viewer/2022081506/5697c0011a28abf838cc29e7/html5/thumbnails/62.jpg)
Chapter 10 62
ProblemProblemCaH2(s) + 2 H2O(l) Ca(OH)2(aq) + 2 H2(g)
-Calculate moles of H2 formed
-Calculate moles of CaH2 needed
-Convert moles CaH2 to grams
![Page 63: Chapter 101 Gases. 2 Homework: 10.12, 10.28, 10.42, 10.48, 10.54, 10.66, 10.72](https://reader035.vdocuments.site/reader035/viewer/2022081506/5697c0011a28abf838cc29e7/html5/thumbnails/63.jpg)
Chapter 10 63
ProblemProblemCaH2(s) + 2 H2O(l) Ca(OH)2(aq) + 2 H2(g)
)(/)(08206.0
29627323
0.10
974.0760/740
KmolatmLr
KCT
LV
torrP
o
![Page 64: Chapter 101 Gases. 2 Homework: 10.12, 10.28, 10.42, 10.48, 10.54, 10.66, 10.72](https://reader035.vdocuments.site/reader035/viewer/2022081506/5697c0011a28abf838cc29e7/html5/thumbnails/64.jpg)
Chapter 10 64
ProblemProblemCaH2(s) + 2 H2O(l) Ca(OH)2(aq) + 2 H2(g)
)296(08206.0
)0.10(974.0
)(/)(08206.0
29627323
0.10
974.0760/740
2 K
Latmn
KmolatmLr
KCT
LV
torrP
H
o
![Page 65: Chapter 101 Gases. 2 Homework: 10.12, 10.28, 10.42, 10.48, 10.54, 10.66, 10.72](https://reader035.vdocuments.site/reader035/viewer/2022081506/5697c0011a28abf838cc29e7/html5/thumbnails/65.jpg)
Chapter 10 65
ProblemProblemCaH2(s) + 2 H2O(l) Ca(OH)2(aq) + 2 H2(g)
moln
K
Latmn
KmolatmLr
KCT
LV
torrP
H
H
o
401.0
)296(08206.0
)0.10(974.0
)(/)(08206.0
29627323
0.10
974.0760/740
2
2
![Page 66: Chapter 101 Gases. 2 Homework: 10.12, 10.28, 10.42, 10.48, 10.54, 10.66, 10.72](https://reader035.vdocuments.site/reader035/viewer/2022081506/5697c0011a28abf838cc29e7/html5/thumbnails/66.jpg)
Chapter 10 66
ProblemProblemCaH2(s) + 2 H2O(l) Ca(OH)2(aq) + 2 H2(g)
mol
x
H
CaH
moln
K
Latmn
KmolatmLr
KCT
LV
torrP
H
H
o
401.02
1
401.0
)296(08206.0
)0.10(974.0
)(/)(08206.0
29627323
0.10
974.0760/740
2
2
2
2
![Page 67: Chapter 101 Gases. 2 Homework: 10.12, 10.28, 10.42, 10.48, 10.54, 10.66, 10.72](https://reader035.vdocuments.site/reader035/viewer/2022081506/5697c0011a28abf838cc29e7/html5/thumbnails/67.jpg)
Chapter 10 67
ProblemProblemCaH2(s) + 2 H2O(l) Ca(OH)2(aq) + 2 H2(g)
molx
mol
x
H
CaH
moln
K
Latmn
KmolatmLr
KCT
LV
torrP
H
H
o
2005.0
401.02
1
401.0
)296(08206.0
)0.10(974.0
)(/)(08206.0
29627323
0.10
974.0760/740
2
2
2
2
![Page 68: Chapter 101 Gases. 2 Homework: 10.12, 10.28, 10.42, 10.48, 10.54, 10.66, 10.72](https://reader035.vdocuments.site/reader035/viewer/2022081506/5697c0011a28abf838cc29e7/html5/thumbnails/68.jpg)
Chapter 10 68
ProblemProblemCaH2(s) + 2 H2O(l) Ca(OH)2(aq) + 2 H2(g)
)/10.42(2005.0
2005.0
401.02
1
401.0
)296(08206.0
)0.10(974.0
)(/)(08206.0
29627323
0.10
974.0760/740
2
2
2
2
2
molgmolgCaH
molx
mol
x
H
CaH
moln
K
Latmn
KmolatmLr
KCT
LV
torrP
H
H
o
![Page 69: Chapter 101 Gases. 2 Homework: 10.12, 10.28, 10.42, 10.48, 10.54, 10.66, 10.72](https://reader035.vdocuments.site/reader035/viewer/2022081506/5697c0011a28abf838cc29e7/html5/thumbnails/69.jpg)
Chapter 10 69
ProblemProblemCaH2(s) + 2 H2O(l) Ca(OH)2(aq) + 2 H2(g)
g
molgmolgCaH
molx
mol
x
H
CaH
moln
K
Latmn
KmolatmLr
KCT
LV
torrP
H
H
o
44.8
)/10.42(2005.0
2005.0
401.02
1
401.0
)296(08206.0
)0.10(974.0
)(/)(08206.0
29627323
0.10
974.0760/740
2
2
2
2
2
![Page 70: Chapter 101 Gases. 2 Homework: 10.12, 10.28, 10.42, 10.48, 10.54, 10.66, 10.72](https://reader035.vdocuments.site/reader035/viewer/2022081506/5697c0011a28abf838cc29e7/html5/thumbnails/70.jpg)
Chapter 10 70
ProblemProblemCaH2(s) + 2 H2O(l) Ca(OH)2(aq) + 2 H2(g)
g
molgmolgCaH
molx
mol
x
H
CaH
moln
K
Latmn
KmolatmLr
KCT
LV
torrP
H
H
o
44.8
)/10.42(2005.0
2005.0
401.02
1
401.0
)296(08206.0
)0.10(974.0
)(/)(08206.0
29627323
0.10
974.0760/740
2
2
2
2
2