Gases Workshop Activity - Oct 3rd_4th (Workshop 6)
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131
Subject
Chemistry
Date
Feb 20, 2024
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25
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Gases Workshop Activity - Oct 3rd/4th (Workshop 6)
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Gases Workshop Activity - Oct 3rd/4th (Workshop 6)
Due: 12:00am on Monday, December 12, 2022
You will receive no credit for items you complete after the assignment is due. Grading Policy
Exercise 6.26 - Enhanced - with Feedback
MISSED THIS?
Read Section 6.2 (Page) .
The pressure on top of Mount Everest (29,029 ) averages about 235 .
Part A
Convert this pressure to .
ANSWER:
Correct
The units of and are equivalent. This means that 235 is equal to 235 .
Part B
Convert this pressure to .
ANSWER:
Correct
Converting from to requires two steps. The first is to convert to , and the second is to convert to .
Part C
Convert this pressure to .
ANSWER:
Correct
The pressure in is converted to the pressure in by dividing by 760. The pressure in is then determined by multiplying by
29.92.
Part D
Convert this pressure to .
ANSWER:
Correct
Converting to is a one-step conversion that involves dividing by 760.
Exercise 6.33 - Enhanced - with Feedback and Hints
= 235
= 4.55
= 9.25
= 0.309
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MISSED THIS?
Read Section 6.3 (Page) ; Watch KCV 6.3
.
A 48.5 sample of gas in a cylinder is warmed from 20 to 84 .
Part A
What is its volume at the final temperature? (Assume constant pressure.)
Express your answer in millililters to three significant figures.
Hint 1. Convert between degrees Celsius and kelvins
The temperature values used in the simple gas laws are in kelvins. Convert the initial and final temperatures of the sample to kelvins.
Express your answers in kelvins to three significant figures separated by a comma.
ANSWER:
ANSWER:
Correct
Charles's law is
where and are the initial volume and temperature of the gas and and are the final volume and temperature.
Solving this equation for gives
Before substituting the numerical values to calculate , convert the initial and the final temperatures to kelvins . Substitute , , and to calculate :
Exercise 6.32 - Enhanced - with Feedback
MISSED THIS?
Read Section 6.3 (Page) ; Watch KCV 6.3
.
A sample of gas has an initial volume of 14.3 at a pressure of 1.20 .
Part A
If the sample is compressed to a volume of 10.5 , what is its pressure?
Express the pressure in atmospheres to three significant figures.
ANSWER:
, = 293,357
= 59.1
= 1.63
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Boyle’s law (
) relates the initial and final pressures and volumes of a gas when the number of moles and temperature remain
constant during compression (or expansion). The calculation of the final pressure is summarized as follows:
Exercise 6.35 - Enhanced - with Feedback and Hints
MISSED THIS?
Read Section 6.3 (Page) ; Watch KCV 6.3
.
A balloon contains 0.110 of gas and has a volume of 2.90 .
Part A
If an additional 0.118 of gas is added to the balloon (at the same temperature and pressure), what will its final volume be?
Express your answer in liters to three significant figures.
Hint 1. Calculate the total moles of gas
0.118 of gas is added to the balloon containing 0.110 of gas (at the same temperature and pressure). Calculate the final number of moles
of gas.
Express your answer in moles to three significant figures.
ANSWER:
ANSWER:
Correct
If the number of moles of gas in a balloon increase at a constant temperature and pressure, its volume increases in direct proportion, as the
greater number of gas particles fill more space. This statement can be mathematically described by Avogadro's law:
Solve this equation for the final volume from the initial volume (
), initial number of moles (
) and final number of moles (
):
The total number of moles of gas particles isS
Substituting and the given quantities to calculate gives
Exercise 6.37 - Enhanced - with Feedback and Hints
= 0.228
= 6.01
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MISSED THIS?
Read Section 6.4 (Page) ; Watch KCV 6.3
, IWE 6.5
.
Part A
What volume is occupied by 0.103 of helium gas at a pressure of 0.94 and a temperature of 302 ? Express your answer using two significant figures.
Hint 1. Rearrange the ideal gas law to solve for volume
You are given a temperature (
), pressure (
), and the number of moles (
) of helium and asked to solve for the unknown volume (
). The ideal
gas law relates all the given conditions. Rearrange the ideal gas law to solve for .
Express your answer in terms of , , , and .
ANSWER:
ANSWER:
Correct
Rearrange the ideal gas law to solve for the unknown volume (
).
The temperature ( ) and pressure ( ) are given in the correct units of kelvin ( ) and atmospheres ( ). The moles ( ) of helium gas
are given. Use the gas constant so that the units of atmospheres, kelvins, and moles cancel, leaving only
the desired units of liters ( ). Substitute the given values into the ideal gas law and solve for volume. Part B
Would the volume be different if the gas was argon (under the same conditions)?
Match the words in the left column to the appropriate blanks in the sentences on the right.
ANSWER:
=
= 2.7
Reset
Help
different
The volume would be the same
if the gas was argon because the ideal gas law is dependent on the
number of moles
of the gas and not the mass
.
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Correct
The ideal gas law given below is used to calculate the volume of a gas. The terms , , , and represent the pressure, volume, number of moles, and temperature, respectively. The terms represents the gas
constant. Notice that the moles of a gas are what is used in the calculation for volume, not the mass of the gas. Therefore, as long as the number
of moles of gas remains constant, along with the other given conditions, the volume also remains the same. Thus, the gas law is dependent on
the number of moles of gas not the mass of a gas.
Exercise 6.44 - Enhanced - with Feedback
MISSED THIS?
Read Section 6.4 (Page) ; Watch KCV 6.3
, IWE 6.5
.
Part A
A weather balloon is inflated to a volume of 29.7 at a pressure of 733 and a temperature of 26.3 . The balloon rises in the atmosphere to an
altitude where the pressure is 385 and the temperature is -15.7 .
Assuming the balloon can freely expand, calculate the volume of the balloon at this altitude.
ANSWER:
Correct
The number of moles in the weather balloon will remain constant. According to the ideal gas law, the changes in pressure and temperature will
cause a change in volume. Volume and pressure are inversely related, while volume and temperature are directly related.
Exercise 6.40 - Enhanced - with Feedback
MISSED THIS?
Read Section 6.4 (Page) ; Watch KCV 6.3
, IWE 6.5
.
Part A
What is the pressure in a 23.0-
cylinder filled with 49.9 of oxygen gas at a temperature of 331 ?
Express your answer to three significant figures with the appropriate units.
ANSWER:
Correct
The pressure can be solved for using the ideal gas equation (
). The moles of oxygen must be determined using the molar mass of
molecular oxygen, (
).
Exercise 6.54 - Enhanced - with Feedback
MISSED THIS?
Read Section 6.5 (Page) .
Part A
=
48.6
= 1.84
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Use the molar volume of a gas at STP to calculate the density (in ) of oxygen gas at STP.
Express the density in grams per liter to three significant figures.
ANSWER:
Correct
The standard molar volume of a gas is 22.414 , and the molar mass of is 32.00 . The density is found by dividing the mass of
a gas by its volume: . Therefore, you can divide the molar mass of a gas (in grams per mole) by its molar volume (in liters per mole) to
determine its density (in grams per liter).
Exercise 6.55 - Enhanced - with Feedback and Hints
MISSED THIS?
Read Section 6.5 (Page) ; Watch IWE 6.7
.
Part A
What is the density (in ) of hydrogen gas at 24 and a pressure of 1645 ? Express your answer in grams per liter to three significant figures.
Hint 1. Derive the equation for density from the ideal gas law
Because density is a mass per unit volume, and because molar mass is the mass per mole, the density of a gas
is its molar mass (
) divided by its
molar volume (moles (
) per volume (
)):
By substituting these into the ideal gas law
where is pressure, is the ideal gas constant, and is temperature, derive the general equation for the density of a gas (
) that can be used to
calculate the density under any conditions.
Express your answer in terms of , , , and .
ANSWER:
ANSWER:
=
1.43
=
=
9.25
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The density of a gas can be calculated under any conditions by using the ideal gas law. Since the density is the molar mass divided by the molar
volume and the molar volume equals (from the ideal gas law), the equation for density (
) is the following (and you can review the hint for
this part for more details):
where is the pressure of a gas, is its molar mass, is the ideal gas constant, and is the temperature.
First, convert all quantities to the appropriate units given that the units for are . Convert the temperature from degrees
Celsius ) to kelvins and the pressure from pounds per square inch to atmospheres :
Then substitute the quantities into the equation to calculate density:
Exercise 6.59 - Enhanced - with Feedback and Hints
MISSED THIS?
Read Section 6.5 (Page) ; Watch IWE 6.8
.
Part A
A sample of gas has a mass of 38.9 . Its volume is 224 at a temperature of 55 and a pressure of 867 . Find the molar mass of the gas. Express your answer in grams per mole to three significant figures.
Hint 1. Calculate the density of the gas
The density of a gas (
) is directly proportional to its molar mass (
) through an equation derived from the ideal gas law. Knowing the density of a
gas and its temperature and pressure, you can determine the molar mass of this gas.
Calculate the density of the gas given a mass of 38.9 and a volume of 224 .
Express your answer in grams per liter to three significant figures.
ANSWER:
ANSWER:
=
0.174
4.10
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To solve the problem, first convert all quantities to the appropriate units given that the units for are . Density will need to
have the units . The complete calculation for density (
) is given in the hint for this part, and the value was found to be . The
conversions for temperature and pressure are
Rearrange the equation
where is pressure, is molar mass, is the ideal gas constant, and is temperature, to solve for the molar mass, then substitute all the
quantities to calculate the molar mass:
Exercise 6.61 - Enhanced - with Feedback and Hints
MISSED THIS?
Read Section 6.6 (Page) ; Watch KCV 6.6
.
A gas mixture contains each of the following gases at the indicated partial pressures: , 236 ; , 105 ; and , 113 .
Part A
What is the total pressure of the mixture? Hint 1. Determine the relationship between the partial pressures and the total pressure of a mixture of gases
You are given the partial pressures of three gases and asked to determine the total pressure of the mixture. Complete the following
sentences describing Dalton's law of partial pressures to determine the relationship between the partial pressures and the total pressure of a
mixture of gases.
Match the words in the left column to the appropriate blanks in the sentences on the right.
ANSWER:
ANSWER:
Reset
Help
product
gas
The sum
of the partial
pressures of the components in a gas mixture
equals the total
pressure.
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Correct
According to Dalton's law of partial pressures, the total pressure of a mixture of gases is the sum of the pressures of the individual gases in the
mixture. Use the given partial pressures of each of the gases to calculate the total pressure.
Part B
What mass of each gas is present in a 1.40 sample of this mixture at 25.0 ? Enter your answers numerically separated by commas.
Hint 1. Rearrange the ideal gas law to solve for moles
You are given a temperature (
), pressure (
), and volume (
) of three gases and asked to solve for the mass of each. To solve for the individual
masses, the moles of each gas must first be calculated. The ideal gas law relates all the given conditions and the molar amount (
). Rearrange the
ideal gas law to solve for .
Express your answer in terms of , , , and .
ANSWER:
Hint 2. Calculate the number of moles of \rm N_2 gas
Given the rearranged ideal gas law from the first hint of this part calculate the number of moles of gas with a partial pressure of 236 .
Enter the quantity in moles to three significant figures.
ANSWER:
ANSWER:
= 454
=
1.78×10
−2
=
0.498,0.253,3.41×10
−2
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Correct
To find the mass in grams (
) of each gas in the mixture, use the ideal gas law to solve for the moles (
) of each gas. Then, convert the moles of
gas to the mass of gas using the molar mass (
) of each gas. First, rearrange the ideal gas law to solve for moles:
where is pressure, is volume, is the ideal gas constant, and is temperature. Calculate the moles of each gas by substituting the given
partial pressures of each gas into the ideal gas law. The partial pressures are given in units of torr. Convert the units of pressure from torr to
atmospheres (
) using the equality . Next, convert the given temperature (
) to kelvins.
Use the gas constant, , so that the units of atmospheres, kelvins, and liters (
) cancel, leaving only the
desired units of moles (
). Substitute the given values and partial pressure for each gas into the ideal gas law. Then, solve for moles.
For ,
For ,
For ,
Finally, convert the calculated moles of each gas to grams using the molar mass of each gas.
For ,
For ,
For ,
Exercise 6.64 - Enhanced - with Feedback
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MISSED THIS?
Read Section 6.6 (Page) ; Watch KCV 6.6
.
A 275-
flask contains pure helium at a pressure of 755 . A second flask with a volume of 465 contains pure argon at a pressure of 729 .
Part A
If we connect the two flasks through a stopcock and we open the stopcock, what is the partial pressure of helium?
Express the partial pressure in to three significant figures.
ANSWER:
Correct
The partial pressure of helium can be calculated using the pressure of helium in the first flask and the volume of the two flasks:
This expression is derived using Boyle's law where the volume of the two flasks is the sum of the two volumes.
These can then be plugged into Boyle's law and can be solved for.
Part B
If we connect the two flasks through a stopcock and we open the stopcock, what is the partial pressure of argon?
Express the partial pressure in to three significant figures.
ANSWER:
Correct
The partial pressure of helium can be calculated using the pressure of helium in the first flask and the volume of the two flasks:
A derivation of this is given when Part A is completed.
Part C
If we connect the two flasks through a stopcock and we open the stopcock, what is the total pressure?
Express the total pressure in to three significant figures.
ANSWER:
Correct
The total pressure is the sum of the partial pressures of the two gases after the stopcock is opened. The total pressure can be calculated by
summing up the two partial pressures calculated in Parts A and B.
Exercise 6.70 - Enhanced - with Feedback and Hints
=
281
=
458
=
739
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MISSED THIS?
Read Section 6.6 (Page) ; Watch KCV 6.6
, IWE 6.11
.
A heliox deep-sea diving mixture contains 2.0 of oxygen to every 98.0 of helium.
Part A
What is the partial pressure of oxygen when this mixture is delivered at a total pressure of 9.5 ?
Express your answer in atmospheres using two significant figures.
Hint 1. Calculate the amount of oxygen in a 2.0 \rm g sample
To determine the partial pressure of oxygen, the amounts of all gases in a mixture should be calculated. These will then be used to calculate mole
fractions for each component.
Determine the amount of oxygen in a 2.0 sample.
Express your answers to two significant figures using the appropriate units.
ANSWER:
Hint 2. Calculate the amount of helium in a 98.0 \rm g sample
To determine the partial pressure of oxygen, the amounts of all gases in a mixture should be calculated. These will then be used to calculate mole
fractions for each component.
Determine the amount of helium in a 98.0 sample.
Express your answers to three significant figures using the appropriate units.
ANSWER:
ANSWER:
Correct
To calculate the partial pressure of oxygen in a mixture at a total pressure of 9.5 , first, find its mole fraction. Mole fraction is the
ratio between the number of moles of a component in a mixture and the total number of moles in it. You will need to convert the mass of each gas
to moles, which is summarized in the hints for this part, resulting in oxygen and 24.5 helium.
The mole fraction of is then calculated by
Finally, calculate the partial pressure of :
Exercise 6.75 - Enhanced - with Feedback and Hints
MISSED THIS?
Read Section 6.7 (Page) ; Watch IWE 6.12
.
amount of oxygen =
6.3×10
−2
amount of helium =
24.5 =
2.4×10
−2
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Automobile air bags inflate following a serious impact. The impact triggers the following chemical reaction:
Part A
If an automobile air bag has a volume of 11.5 , what mass of (in ) is required to fully inflate the air bag upon impact? Assume STP conditions. Express your answer in grams to three significant figures.
Hint 1. Calculate the amount of in an 11.5 sample at STP conditions
Determine the amount of in an automobile air bag that has a volume of 11.5 at STP conditions.
Express your answer in moles to three significant figures.
ANSWER:
ANSWER:
Correct
Since the reaction occurs under standard temperature and pressure, the volume of can be converted directly to the amount in moles using
the molar volume, 22.4 , to create a conversion factor:
Then use the stoichiometric relationship from the balanced equation to find the number of moles of involved in the reaction:
Finally, use the molar mass of to obtain the mass that is required to fully inflate the air bag upon impact:
Exercise 6.80 - Enhanced - with Feedback and Hints
MISSED THIS?
Read Section 6.7 (Page) ; Watch IWE 6.12
.
Carbon monoxide gas reacts with hydrogen gas to form methanol via the following reaction:
A 1.05 reaction vessel, initially at 305 , contains carbon monoxide gas at a partial pressure of 232 and hydrogen gas at a partial pressure of 388 .
Part A
Identify the limiting reactant and determine the theoretical yield of methanol in grams. Express your answer with the appropriate units.
Hint 1. Calculate the amount of carbon monoxide gas and hydrogen gas in the reaction vessel
=
0.513
=
22.3
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Calculate the amount of carbon monoxide gas in the 1.05 sample at a temperature of 305 , a pressure of 232 , and the amount of
hydrogen gas in the same sample at a pressure of 388 . Express your answers in moles separated by a comma.
ANSWER:
ANSWER:
Correct
Use the ideal gas law to find the moles of each reactant present, , where is the number of moles, making certain to convert to . Then, determine which reactant is limiting by calculating which reactant forms the least
amount of product. Finally, convert the
amount of product to the mass of product to obtain the theoretical yield. This series of calculations is summarized as follows:
Exercise 6.82 - Enhanced - with Feedback
MISSED THIS?
Read Section 6.8 (Page) ; Watch IWE 6.8 .
Part A
A flask at room temperature contains exactly equal amounts (in moles) of nitrogen and xenon. Sort the conditions based on the gas described.
Drag the appropriate items to their respective bins.
ANSWER:
amount of , amount of =
1.28×10
−2
,2.14×10
−2
0.343 Reset
Help
Gas for which the molecules or atoms
have the greatest average velocity.
Gas that would have a higher rate of
effusion through a small hole opened
in the flask.
Gas for which the molecules or atoms
have the greatest average kinetic
energy.
Gas that exerts the greater partial
pressure.
Nitrogen has the greater value
Xenon has the greater value
Both gases have the same value
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Correct
The partial pressure and the average kinetic energy of a gas are not related to the mass of the gas particles. The rate of effusion and the average
velocity of gas particles decrease as the mass of the particles increases.
Exercise 6.88 - Enhanced - with Feedback
MISSED THIS?
Read Section 6.9 (Page) ; Watch IWE 6.15
.
A sample of effuses from a container in 44 .
Part A
How long would it take the same amount of gaseous to effuse from the same container under identical conditions?
Express your answer using two significant figures.
ANSWER:
Correct
The ratio of effusion rates of two different gases is given by Graham’s law of effusion:
where and are the molar masses of each gas. The time of effusion is inversely proportional to the rate of effusion, as follows:
The final equation is then
so
Exercise 6.89 - Enhanced - with Feedback
MISSED THIS?
Read Section 6.8 (Page) ; Watch KCV 6.8 .
The following graph shows the distribution of molecular velocities for two different molecules (A and B) at the same temperature.
Part A
Which molecule has the higher molar mass? Which molecule would have the higher rate of effusion?
Match the words in the left column to the appropriate blanks in the sentences on the right.
= 110
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ANSWER:
Correct
The average velocity of a molecule is inversely proportional to its molar mass; therefore, gas A has a higher molar mass because it has a lower
average velocity. The average velocity of a molecule is directly proportional to its effusion rate; therefore, gas B will have a higher effusion rate
because it has a higher average velocity.
Chapter 6 Algorithmic Question 31
Part A
Determine the volume of 15.0 g of argon gas at 157°C and 2.50 kPa pressure.
ANSWER:
Correct
Chapter 6 Algorithmic Question 30
Part A
A 70.0-L steel tank at 20.0°C contains acetylene gas, C
2
H
2
, at a pressure of 1.39 atm. Assuming ideal behavior, how many grams of acetylene are in the
tank?
ANSWER:
Reset
Help
Gas A
higher
Gas B
lower
Gas A
has the higher molar mass because it has the lower
average velocity.
Gas B
will have the higher effusion rate because it has the higher
average velocity.
1.94 L
196 L
33.5 L
5.30 L
537 L
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Correct
Chapter 6 Multiple Choice Question 21
Part A
What mass of NO
2
is contained in a 13.0 L tank at 4.58 atm and 385 K?
ANSWER:
Correct
Exercise 6.52 - Enhanced - with Feedback
MISSED THIS?
Read Section 6.4 (Page) .
A sample of nitrogen gas in a 1.76-
container exerts a pressure of 1.46 at 24 .
Part A
What is the pressure if the volume of the container is maintained constant and the temperature is raised to 331 ?
Express the pressure in atmospheres to three significant figures.
ANSWER:
Correct
The general gas law can be used to solve for the final pressure:
Since the volume remains constant, it can be canceled from the equation, giving
Rearranging to solve for the final pressure then gives
1540 g
105 g
6.01 g
4.04 g
69.2 g
24.4 g
18.8 g
53.1 g
86.7 g
= 2.97
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Exercise 6.110
The emission of by fossil fuel combustion can be prevented by injecting gaseous urea into the combustion mixture. The urea reduces (which
oxidizes in air to form ) according to the following reaction: Suppose that the exhaust stream of an automobile has a flow rate of 2.32 at 643 and contains a partial pressure of of 12.6 .
Part A
What total mass of urea is necessary to react completely with the formed during 8.3 hours of driving?
Express your answer using two significant figures.
ANSWER:
Correct
Chapter 6 Algorithmic Question 41
Part A
Give the temperature and pressure at STP.
ANSWER:
Correct
Chapter 6 Algorithmic Question 45
Part A
What volume will 0.800 moles of Ar occupy at STP?
ANSWER:
=
650
0°C and 760 torr
32 K and 760 torr Hg
0 K and 760 torr
0°C and 1 mm Hg
25°C and 30.00 in Hg
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Chapter 6 Algorithmic Question 53
Part A
What is the density of oxygen gas at STP?
ANSWER:
Correct
Chapter 6 Algorithmic Question 81
Part A
How many liters of oxygen are needed to exactly react with 13.8 g of methane at STP? CH
4
(
g
) + 2 O
2
(
g
) → CO
2
(
g
) + 2 H
2
O(
l
)
ANSWER:
Correct
17.9 L
16.0 L
22.4 L
45.9 atm
75.0 L
0.0450 g/L
1.30 g/L
0.700 g/L
1.43 g/L
9.7 L
19.4 L
97.4 L
41.9 L
38.6 L
10/4/22, 12:46 PM
Gases Workshop Activity - Oct 3rd/4th (Workshop 6)
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Chapter 6 Algorithmic Question 82
Part A
When 8.0 g of zinc metal reacts with excess HCl, how many liters of H
2
gas are produced at STP?
ANSWER:
Correct
Chapter 6 Algorithmic Question 86
Part A
How many grams of zinc metal are required to produce 2.5 liters of hydrogen gas at STP according to the chemical equation shown below?
Zn(
s
) + 2 HCl(
aq
) → ZnCl
2
(
aq
) + H
2
(
g
)
ANSWER:
Correct
Chapter 6 Algorithmic Question 92
Part A
How many mL of O
2
gas at 25°C and 755 mm Hg pressure can be produced from the thermal decomposition of 0.600 grams of KClO
3
(
s
) according to the
chemical equation shown below?
2.74 L
5.49 L
0.458 L
0.265 L
0.137 g
14.6 g
7.3 g
31 g
163 g
10/4/22, 12:46 PM
Gases Workshop Activity - Oct 3rd/4th (Workshop 6)
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2 KClO
3
(
s
) → 2 KCl(
s
) + 3 O
2
(
g
)
ANSWER:
Correct
Chapter 6 Algorithmic Question 98
Part A
Identify the gas particle that travels the fastest
.
ANSWER:
Correct
Chapter 6 Algorithmic Question 100
Part A
What is the average speed (actually the root-mean-square speed) of a argon atom at 27°C?
ANSWER:
Correct
181 mL
362 mL
60.2 mL
80.4 mL
N
2
Ne
Kr
Cl
2
H
2
130 m/s
4.11 m/s
433 m/s
13.7 m/s
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Gases Workshop Activity - Oct 3rd/4th (Workshop 6)
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Chapter 6 Algorithmic Question 107
Part A
If CO
2
and NH
3
are allowed to effuse through a porous membrane under identical conditions, the rate of effusion for NH
3
will be ________ times that of
CO
2
.
ANSWER:
Correct
Chapter 6 Algorithmic Question 108
Part A
Rank the following in order of decreasing rate of effusion.
F
2
SF
6
He
Ar ANSWER:
Correct
Chapter 6 Algorithmic Question 112
Part A
Identify the element or molecule that has the highest effusion rate.
2.6
0.39
0.62
1.6
Ar > He > SF
6
> F
2
He > F
2
> SF
6
> Ar
He > F
2
> Ar > SF
6
SF
6 > Ar > F
2
> He
F
2
> Ar > He > SF
6
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Gases Workshop Activity - Oct 3rd/4th (Workshop 6)
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ANSWER:
Correct
Chapter 6 Algorithmic Question 114
Part A
Identify the element or molecule that has the highest diffusion rate.
ANSWER:
Correct
Chapter 6 Multiple Choice Question 56
Part A
Define effusion
.
ANSWER:
Correct
Ne
SO
2
Xe
F
2
N
2
Ar
CO
Ne
SO
2
Kr
Effusion is the process by which gas escapes from a container into a vacuum through a small hole.
Effusion is the process by which solid molecules spread out in response to a concentration gradient.
Effusion is the process by which gas molecules spread out in response to a concentration gradient.
Effusion is the process by which liquid escapes from a container into a vacuum through a small hole.
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Chapter 6 Algorithmic Question 117
Part A
Which of the following compounds will behave least like an ideal gas at low temperatures?
ANSWER:
Correct
Exercise 6.102 - Enhanced - with Feedback and Hints
MISSED THIS?
Read Section 6.10. You can click on the Review link to access the section in your eText.
Part A
Use the van der Waals equation to calculate the pressure exerted by 1.490 of in a volume of 5.255 at a temperature of 286.0 .
Express your answer in atmospheres.
You did not open hints for this part.
ANSWER:
Incorrect; Try Again; 4 attempts remaining
Part B
Use the ideal gas equation to calculate the pressure exerted by 1.490 of in a volume of 5.255 at a temperature of 286.0 .
Express your answer in atmospheres.
Hint 1. Rearrange the ideal gas law
The ideal gas law is used to solve for the ideal pressure exerted by the gas. However, you must first rearrange it to solve for the unknown
variable. In the ideal gas law below, is pressure, is volume, is the number of moles, is the gas constant, and is temperature. Rearrange the equation for the ideal gas law to solve for pressure, .
Ne
CCl
4
H
2
F
2
Xe
=
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Express your answer in terms of , , , and .
ANSWER:
ANSWER:
Correct
The ideal gas law is . Solving for gives .
Part C
Which of the following best explains why the two values in Parts A and B are different?
ANSWER:
Correct
Gases behave nonideally at high pressures and low temperatures. Larger molecules also tend to increase the volume of the gas beyond what is
predicted by the ideal gas equation.
Score Summary:
Your score on this assignment is 96.8%.
You received 37.75 out of a possible total of 39 points.
=
=
6.65
The perceived forces between chlorine gas molecules that contribute to nonideal behavior are most strongly influenced by the polar chlorine molecules and the low system temperature.
the high system temperature and relatively large chlorine molecule size.
the high system pressure and relatively large chlorine molecule size.
the high system pressure and high system temperature.
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Chemical Reactions and Equations
3. Observation of burning match or splint.
What caused the change in the burning match or splint?
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» CO, (g)+
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5. Type of reaction:
F. Hydrogen Peroxide
Reactants
1. Appearance of
Reactants
2. Evidence of a Chemical
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H,O2(aq)
3.
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H2O(1) +.
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Q1 What evidence of a chemical reaction might you see in the following cases? Refer
to Table 1.
a. dropping an Alka-Seltzer tablet into a glass of water
b. bleaching a stain
c. burning a match
d. rusting of an iron nail
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a. Mg(s) + HCl(aq)
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b.
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PAP Chemistry-2903012-42100P-1 / Gases / Lesson 148
Aguilar
All changes saved
25. The kinetic molecular theory says that collisions between particles are elastic, but real-world collisions do not follow this assumption. What is some of the kinetic energy
changed into after a collision?
O heat
O light
O potential energy
O chemical energy
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A reaction at 31.0 °C evolves 228. mmol of dinitrogen monoxide gas.
Calculate the volume of dinitrogen monoxide gas that is collected. You can assume the pressure in the room is exactly 1 atm. Round your answer to 3
significant digits.
volume: ||
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Calcium hydride reacts with water to produce calcium hydroxide and hydrogen gas.
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If 7.003 g CaH₂ is combined with 61.000 g H₂O, the reaction proceeds until all the CaH2 is consumed. The hydrogen gas escapes to the environment, and the final weight of the flask is 67.332 g.
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What is true for this chemical reaction: 3CO(g) + Fe₂O3(s) = 3CO₂(g) + 2Fe(s)?
Select an answer and submit. For keyboard navigation, use the up/down arrow keys to select an answer.
a
b
C
AHxn> 0; ASxn > 0
AHxn 0
AHxn 0; ASxn <0
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[References]
Consider the combustion of
propane:
СзHs (9) + 502(9) — ЗСО%(9) + 4H20(1) ДН %— — 2221 kJ
A balloon is being inflated to its full extent by heating the air inside it. In the final stages of this process, the volume of the balloon changes
from 4.00 x 10° L to 4.50 x 10° L by the addition of 1.6 x 10° J energy as heat. Assume that all the heat comes from the combustion of
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Mass =
g
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I Review | Constants | Periodic Table
Open the activity depicting Hess's Law.
Part B
N: (g) + 0: (g)
2NOg)
AH is an Extensive Property
+ N: ) +0:() → NO(g)
AH Changes sign when a
Click on the process is reversed
screen when you started and its derivative demonstrate that the reaction enthalpy, AH, changes sign when a process is reversed.
button within the activity and analyze the relationship between the two reactions that are displayed. The reaction that was on the
AH Canges sign when a
process is reversed
2NO(g) → N:(gì + 0: (g)
Consider the reaction
Hess Law of Constant
icat Sumnation
+ N: (g) +0: (g) – d
NQ +0:()
B's) + 302(g)—B>0ҙ(s), дн — -1270KJ
When two reactions are
added together. the
individual reaction
enthalpies are summed.
Notice that equal quantities
of the same species, on both
sides of the reaction arrow,
cancel.
→ NO: g
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+ N: (g) + O: (g → NO: (g)
Express your answer with appropriate units.
• View Available Hint(s)
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Co(g) reacts with the 02(g) according to the following reaction.
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Which of the following diagrams correctly illustrates the reaction mixture after the reaction goes to
completion?
000
00
00
000
(A)
(C)
00
000
000 000
000
000
000
00
00
000
000 000
000
(D)
000 000
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