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LaBrake CH301 Exam 1 Fall 2016
49945 / 49950
Remember that the bubble sheet has many conversion factors and constants on the back.
R = 0.08206 L atm/mol K
R = 62.36 L torr/mol K
R = 0.08314 L bar/mol K
R = 8.314 J/mol K
NA = 6. 022 × 1023 mol−^1
1 atm = 1.01325 × 105 Pa
1 atm = 760 torr
1 atm = 14.7 psi
1 bar = 10^5 Pa
1 in = 2.54 cm
1 lb = 453.6 g
1 gal = 3.785 L
ρwater = 1.00 g/mL
ρmercury = 13.6 g/mL
NOTE: Please keep your Exam copy intact (all pages still stapled). You must turn in your exam
copy plus your bubble sheet, and scratch paper.
This print-out should have 18 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering.
001 4.0 points Determine the molecular weight of a com- pound given the gas phase data: mass of 27 .64 g, temperature of 18.4◦C, pressure of 956 torr, volume of 14.32 L.
- 0.046 g/mol
- 191 g/mol
- 279 g/mol
- 36.7 g/mol correct
- 24.9 g/mol
Explanation: m = 27.64 g V = 14.32 L T = 18. 4 ◦C + 273.15 = 291.55 K P = 956 torr R = 0. 08206 L mol·atm·K Using the ideal gas law,
P V = n R T n =
P V
R T
(956 torr) (14.32 L) (0. 08206 L mol·atm·K ) (291.55 K)
×
1 atm 760 torr = 0.752909 mol gas.
If 0.752909 moles of the gas weigh 27.64 g, the molecular weight is
MW =
27 .64 g 0 .752909 mol
= 36.7109 g/mol.
002 4.0 points Consider two separate 1 L gas samples, both at the same temperature and pressure. The two gases have different molar masses. Which is true?
- The particles in both gas samples have the same average kinetic energies. correct
- You would need more information to be able to compare the average kinetic energies of these two gas samples.
- The particles in both gas samples have different average kinetic energies. Explanation: Average kinetic energies of gases are deter- mined by temperature.
003 4.0 points
a b (L^2 bar/mol^2 ) (L/mol) Acetonitrile 17.81 0. Butane 14.66 0. Freon 10.78 0.
Which gas molecule do you expect to be the largest? (a and b are Van der Waals constants.)
- Freon
- Butane correct
- Acetonitrile Explanation: The Van der Waals constant a represents the role of attractions, so it is relatively large for molecules that attract each other strongly. The Van der Waals constant b represents the role of repulsions and can be thought of as representing the volume per mole of molecules. The larger the b value, the larger the molecules in the gas.
004 4.0 points Calculate the mass of NH 3 that can be pro- duced from 30.0 g N 2 and 5.0 g H 2 in the following reaction.
N 2 (g) + 3 H 2 (g) → 2 NH 3 (g)
- 36.4 g
EffCO 2 Effunknown
MWunknown MWCO 2
MWunknown = MWCO 2
EffCO 2 Effunknown
= 44.01 g/mol
1 .77 Effunknown Effunknown
= 137.879 g/mol.
007 4.0 points A 6.35 L sample of carbon monoxide is col- lected at 55.0◦C and 0.892 atm. What volume will the gas occupy at 1.05 atm and 59.0◦C?
- 6.10 L
- 6.68 L
- 4.82 L
- 1.96 L
- 5.46 L correct Explanation: V 1 = 6.35 L P 1 = 0.892 atm T 1 = 55◦C = 328 K P 2 = 1.05 atm T 2 = 59◦C = 332 K We can use the combined gas law and solve for V 2 :
P 1 V 1 T 1
P 2 V 2
T 2
V 2 =
P 1 V 1 T 2
T 1 P 2
(6.35 L) (0.892 atm) (332 K) (328 K) (1.05 atm) = 5.46 L
008 4.0 points Consider
2 Al(s) + 6 HCl(ℓ) → 2 AlCl 3 (s) + 3 H 2 (g) ,
the reaction of 4.5 mol Al with excess HCl to produce hydrogen gas. What is the pressure
of H 2 (g) if the hydrogen gas collected occupies 14L at 300K?
- 5.28 atm
- 1.07 atm
- 0.0763 atm
- 11.9 atm correct
- 0.233 atm
- 7.9 atm Explanation: V = 14 L T = 300 K nH 2 = 4.5 mol First calculate the number of moles of H 2 produced:
? mol H 2 = 4.5 mol Al ×
3 mol H 2 2 mol Al = 6.75 mol H 2
P V = n R T
PH 2 =
nH 2 R T V
=
(6.75 mol)
- 0821 L mol·atm·K
(300 K)
14 L
= 11.8752 atm.
009 4.0 points A 50/50 mix (by mass) of nitrogen gas and carbon dioxide is made. What is the mole fraction of nitrogen in this mixture?
- 0.61 correct
Explanation: 50/50 mix means the same mass for both gases. So let that mass be 44 g each. 44g is exactly 1 mole of the CO 2. 44/28 = 1.57 mol of N 2 gas XN 2 = 1.57/(1+1.57) = 0.
010 4.0 points Consider the Maxwell-Boltzmann distribu- tion plots for a series of gases with differ- ent molar masses but with identical temper- atures. Which of the statements below is a correct description of the shape and position of the plot as it pertains to the relative mass of the gas.
- Heavier molecules tend to have much broader ranges of velocities and larger average velocities.
- Lighter molecules tend to have much broader ranges of velocities and smaller av- erage velocities.
- Heavier molecules tend to have much nar- rower ranges of velocities and smaller average velocities. correct
- Lighter molecules tend to have much nar- rower ranges of velocities and larger average velocities.
Explanation: more massive particles carry the same ki- netic energy as smaller ones at much lower velocities because m is much bigger in the equation:
Ek = 1/ 2 mv^2
Same temperature means the same Ek so the velocities are relatively less than lighter molecules. In addition, the ranges is much more narrow (less spread) in the distribution.
011 4.0 points Under what conditions is a gas most likely to deviate from ideal behavior?
- low density
- high pressure correct
- when considering noble gases
- high temperatures Explanation: Under high pressure, there are many more collisions between molecules. This increases the likelihood that a non-ideal interaction will occur.
012 4.0 points A 6.00 L sample of C 2 H 4 (g) at 2.00 atm and 293 K is burned in 6.00 L of oxygen gas at the same temperature and pressure to form carbon dioxide gas and liquid water. If the reaction goes to completion, what is the final volume of all gases at 2.00 atm and 293 K? The balanced equation is:
C 2 H 4 (g) + 3 O 2 (g) → 2 H 2 O(ℓ) + 2 CO 2 (g)
1. 4.00 L
- 8.00 L correct
- 12.00 L
- 2.00 L
- 6.00 L
- 2.66 L Explanation: The balanced equation is
C 2 H 4 (g) + 3 O 2 (g) → 2 H 2 O(ℓ) + 2 CO 2 (g) Avogadro’s Principle tells us that when P and T are constant (which they are in this problem), V ∝ n, so we can work with the volume of the gases instead of the number of moles. From the equation above, we need 3 times more L of O 2 than of C 2 H 4 (18 L, not the given 6 L!) so O 2 is the limiting reagent. Find out how much CO 2 is made based on the L of O 2 :
LCO 2 = 6 L O 2 ×
2 L CO 2
3 L O 2
= 4 L CO 2
- twelve correct
- ten
Explanation: The balanced chemical equation is 1 Al 2 (SO 4 ) 3 + 6 NaOH → 2 Al(OH) 3 + 3 Na 2 SO 4 which gives 2 Al, 3 SO 4 , 6 Na, and 6 OH− on both sides of the equation. The sum of coefficients is 1 + 6 + 2 + 3 = 12.
017 4.0 points A chemist prepares a sample of helium gas at a certain pressure, temperature and volume and then removes all but a fourth of the gas molecules (only a fourth remain). How must the temperature be changed (as a multiple of T 1 ) to keep the pressure and the volume the same?
- T 2 = 12 T 1
- T 2 = 161 T 1
- T 2 = 14 T 1
- None of these
- T 2 = 4 T 1 correct
- T 2 = 16 T 1
- T 2 = 2 T 1
Explanation: n 2 = n 41 Because we want P and V to be constant, we can use the relationship
n 1 · T 1 = n 2 · T 2.
The fraction of the moles is the same as the fraction of the molecules, so
n 1 · T 1 =
(n 1 4
T 2
T 2 = 4 T 1.
For V and P to remain constant, the temper- ature must be four times.
018 4.0 points How many grams of NH 3 can be produced from the reaction of 17.8 moles of H 2 and a sufficient supply of N 2?
N 2 + 3 H 2 → 2 NH 3
- 202 grams correct
- 225 grams
- 408 grams
- 11.9 grams
- 455 grams
- 100.grams Explanation:
? g NH 3 = 17.8 moles H 2 ×
2 moles NH 3 3 moles H 2
×
17 .034 g NH 3 1 mole NH 3 = 202.137 grams
