Version 169 – Exam 1 – mccord – (52125)
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mccord - ch301
— Exam 1 —
Dat Acid
7-9pm Sept 17, 2013
BUR 106
2pm class only - 52125
001 4.0 points
Which molecules of the following gases will
have the greatest average kinetic energy?
1. H2 at 0.5 atm and 298 K
2. CO2 at 1 atm and 298 K
3. All of the molecules have the same kinetic
energy. correct
4. He at 0.1 atm and 298 K
5. N2 at 1 atm and 298 K
Explanation:
From kinetic molecular theory, the temperature is directly proportional to mean KE.
If all of the gases are at the same temperature they have the same average KE. (Note:
The average speeds will be different since they
have different molecular weights.)
002 4.0 points
How many atoms are present in 0.35 mol of
carbon dioxide?
1. 5.81 × 10−25 atoms
2. 15 atoms
3. 2.11 × 1023 atoms
1
4. 6.32 × 1023 atoms correct
Explanation:
Each molecule of carbon dioxide contains
one carbon atom and has two oxygen atoms;
three atoms in total, so
3 mol atoms
0.35 mol CO2 ×
1 mol CO2
6.02 × 1023 atoms
×
= 6.321 × 1023 atoms
1 mol atoms
003 4.0 points
What is the root mean square speed of argon
gas at 300 K?
1. 3.96 × 102 m/s
2. 5.92 × 102 m/s
3. 4.33 × 102 m/s correct
4. 1.37 × 101 m/s
5. 4.30 × 101 m/s
Explanation:
Use the following equation for vrms and remember that the molar mass for Argon must
be in kg/mol.
r
3RT
vrms =
M
s
3(8.314)(300)
vrms =
(0.03995)
vrms = 433
004
4.0 points
For the skeletal chemical equation
BCl3(g) + H2 O(ℓ) → B(OH)3 (aq) + HCl(aq)
what is the coefficient of HCl in the balanced
equation? (Balance the equation with the
smallest possible whole number coefficients.)
1. 5
Version 169 – Exam 1 – mccord – (52125)
2
0.997 atm
80.9 g
·
L·atm
0.08206 mol·K (303.55 K) mol
= 3.23804 g/L
=
2. 18
3. 4
4. 6
5. 3 correct
006
For the reaction
4.0 points
6. 9
2 S (s) + 3 O2 (g) → 2 SO3 (g) ,
7. 2
how much SO3 can be produced from 2 g O2
and excess S?
8. 1
1. 0.08 mol SO3
9. 10
2. 0.04 mol SO3 correct
10. 8
3. 0.03 mol SO3
Explanation:
The balanced equation is
BCl3(g) + 3 H2 O(ℓ) →
B(OH)3 (aq) + 3 HCl(aq)
4. 0.13 mol SO3
005 4.0 points
A gas has a molar mass of 80.9 grams/mole, a
temperature of 30.5◦C, and a pressure of 758
torr. What is the density of this gas?
1. 32.2 g/L
5. 0.06 mol SO3
Explanation:
mO 2 = 2 g
Oxygen is the limiting reactant. Convert
mass O2 to moles SO2 using
O2 → mol O2 → mol SO3 :
mol O2
2 mol SO3
(2 g O2 ) ×
×
32 g O2
3 mol O2
= 0.04 mol SO3
2. 0.309 g/L
3
3. 2.46 × 10 g/L
4. Cannot be determined from this data.
5. 3.24 g/L correct
Explanation:
Make sure that all your known values are in
the right units:
MW = 80.9 g/mol
T = 30.4◦C + 273.15 = 303.55 K
1 atm
P = (758 torr)
= 0.997 atm
760 torr
P V = nRT
P
n
=
V
RT
007 4.0 points
The following experiment was carried out using a newly synthesized chlorofluorocarbon.
Exactly 50 mL of the gas effused through a
porous barrier in 172 s. The same volume of
argon effused in 76 s under the same conditions. Which compound is the chlorofluorocarbon?
1. C2 Cl2 F4
2. C2 Cl3 F3
3. C2 Cl4 F2 correct
4. C2 Cl5 F
5. C2 ClF5
Version 169 – Exam 1 – mccord – (52125)
3
Explanation:
2. 21.0 L correct
008 4.0 points
Consider the following reaction:
2 A (g) + 3 B (g) −→ C (g) + 2 D (ℓ)
If 27 L of gas A is mixed with 30 L of gas B at
the same pressure and temperature to give an
initial volume of 57 L. After A and B react,
the total volume changes. Assuming that the
reaction runs to completion, what is the final
volume?
1. 29.0
2. 25.0
3. 17.0
4. 23.0
5. 19.0
6. 15.0
7. 31.0
8. 13.0
9. 27.0
10. 21.0
Correct answer: 17 L.
Explanation:
Because the pressure and temperature are
constant, the problem can be worked in liters
(volume units) instead of moles.
2 L of A
= 20 L A reacts
3 L of B
This will leave behind (excess) A = 27 - 20 =
7 L excess A
30 L of B
3. 84.0 L
4. 42.0 L
5. 32.0 L
6. 672 L
7. 2.02 L
8. 22.4 L
Explanation:
At STP, the reaction can be shown to relate
volumes. It may thus be stated that
2 L hydrogen gas reacts with 1 L oxygen
gas to form 2 L water.
Therefore 21.0 L of oxygen is needed to
react with 42.0 L hydrogen.
010 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?
1. 0.56
2. 0.61 correct
3. 0.44
amount of C made = 30 L B (1/3) = 10 L of
C made.
4. 0.73
So the total volume after the reaction is the C
made PLUS the excess of A which is 10 + 7 =
17 L for the system.
5. 0.27
009 4.0 points
A fuel cell car is powered by electrons harvested from the flameless, low-temperature
reaction
2 H2 + O2 → 2 H2O
7. 0.39
How many liters of oxygen is needed to react
with 42.0 L of hydrogen at STP?
1. 63.0 L
6. 0.50
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 CO2 .
44/28 = 1.57 mol of N2 gas
XN2 = 1.57/(1+1.57) = 0.61
011
4.0 points
Version 169 – Exam 1 – mccord – (52125)
The density of a gas is 1.96 g/L at STP. What
would the mass of 2.00 moles of the gas be at
STP?
3. balloon B correct
4. they are identical assuming they are both
behaving ideally
1. 0.510 g
2. 43.9 g
3. 44.8 g
4. 0.0875 g
5. There is not enough information to
solve.
6. 87.8 g correct
Explanation:
d = 1.96 g/L
T = 0◦ C = 273 K
m
d = , so
V
P = 1 atm
L · atm
R = 0.0821
mol · K
m R T
mRT
dRT
=
=
PV
V
P
P
L·atm
(1.96 g/L) 0.0821 mol·K (273 K)
=
1 atm
= 43.9301 g/mol
M=
Two moles of the gas will weigh
(2 mol)
4
43.9301 g
= 87.8601 g .
mol
012 4.0 points
You have two balloons with the same volume and temperature. One balloon is filled
with some number of moles of gas A. The
other balloon contains an identical number of
moles of gas B. The density of balloon A is
5 times larger than that of balloon B. The
gas in each balloon is slowly leaking out as a
result of effusion through small pores in the
balloon surface. Which balloon will deflate
the fastest?
1. balloon A
2. it depends on the pressure
Explanation:
Gas A and gas B have the same n, V, T,
and P. The only difference is in the mass of the
gas particles. Gas A is 5 times more massive
than gas B since its density is 5 times larger.
The more massive gas will have
√ a smaller
rms velocity (by a factor of 5) and thus
an effusion rate that is slower as well. Thus
balloon B, with the less massive particles, will
deflate first.
013 4.0 points
Consider two equal-sized containers, one filled
with H2 gas and one with O2 gas at the same
temperature and pressure. The mass of the
H2 gas is (greater than, equal to, less than)
that of the O2 gas.
1. less than correct
2. greater than
3. equal to
Explanation:
Avogadro’s Law states that at the same
temperature and pressure, equal volumes
of all gases contain the same number of
molecules. The standard molar volume of
an ideal gas is taken to be 22.414 Liters per
mol at STP.
1 mol of H2 weighs less than 1 mole of O2
because their molecular weights are different.
014 4.0 points
A lungful of air (393 cm3 ) is exhaled into a
machine that measures lung capacity. If the
air is exhaled from the lungs at a pressure
of 1.76 atm at 39.1 ◦ C but the machine is at
ambient conditions of 0.958 atm and 23 ◦ C,
what is the volume of air measured by the
machine?
1. 754.12
2. 477.836
Version 169 – Exam 1 – mccord – (52125)
3. 440.952
4. 684.759
5. 551.798
6. 951.409
7. 744.022
8. 233.089
9. 302.37
10. 413.84
Correct answer: 684.759 cm3 .
Explanation:
T1 = 39.1◦ C + 273 = 312.1 K P1 = 1.76 atm
T2 = 23◦ C + 273 = 296 K
P2 = 0.958 atm
3
V1 = 393 cm
Because P , V , and T change,
P1 · V1
P2 · V2
=
T1
T2
P1 · V1 · T2
V2 =
P2 · T1
(1.76 atm) (393 cm3 ) (296 K)
=
(0.958 atm) (312.1 K)
= 684.759 cm3 .
015 4.0 points
Given a percent yield of 45.0% for the reaction:
H2 + Cl2 → 2 HCl(g)
How many moles of HCl gas are produced
if 15.5 L of Cl2 at STP and excess H2 are
reacted?
1. 3.08 mol
2. 1.38 mol
3. 0.622 mol correct
4. 0.769 mol
5. 0.346 mol
6. 0.156 mol
Explanation:
Use P V = nRT to calculate the number of
moles of Cl2 .
n = P V /(RT ) = 1(15.5)/(0.08206)273.15
5
n = 0.6915 mol Cl2
Now adjust for 2:1 ratio of HCl to Cl2 .
0.6915(2/1) = 1.383 mol HCl (100% yield)
Now adjust for 45% yield...
1.383(45%) = 0.622 moles HCl
016 4.0 points
The following shows three plots of the pressure versus volume for a fixed amount of
an ideal gas at three different temperatures.
Which line is the P-V plot of the gas at the
highest temperature?
P
V
1. the solid line
2. there is no way to tell without knowing
the units
3. the dashed line
4. the dotted line correct
Explanation:
The plot shows that pressure is inversely related to volume. This is because the product
of PV = constant (nRT). The highest temperature will have the highest constant. The plot
with the largest combination of PV values is
the dotted plot.
017 4.0 points
A sample of air has a volume of 146 mL at
68 ◦ C. At what temperature will its volume
be 55 mL at constant pressure?
1. -172.324
2. -144.541
3. -152.767
4. -139.37
5. -169.437
6. -122.197
Version 169 – Exam 1 – mccord – (52125)
7. -156.72
8. -108.0
9. -124.457
10. -142.6
2. NO, leave my score alone, I prefer the
lower score
Correct answer: −144.541 C.
◦
Explanation:
V1 = 146 mL
T1 = 68 ◦ C + 273 = 341 K
Applying Charles’ Law,
V2 = 55 mL
T2 = ?
V1
V2
=
T1
T2
(55 mL)(341 K)
T1 V2
=
T2 =
V1
146 mL
= 128.459 K
◦
C = K − 273 = −144.541 ◦ C
018 4.0 points
Which of the gases given will have the greatest
value for the van der Waal’s coefficient b ?
1.
6
correct
2. All 4 will have the same b value because
they are all spherical.
3.
4.
5.
Explanation:
The van der Waal’s coefficient b correlates
to the volume (size) of the gas molecule. So
the largest circle is the largest molecule and
therefore will have the largest value for b.
019 0.0 points
This question starts out at zero points but
could very well increase after the grading.
Now, if more points are awarded (the curve)
on this assignment, would you like them
added to your score?
1. YES, I would like the points and the
higher score. correct
Explanation:
This should be a no-brainer. Most students
want higher scores. If you picked yes, you got
credit for the question and you got the extra
points you asked for (if they were granted
by your instructor). If you answered NO,
you also got what you wanted... no points
awarded.
020 4.0 points
You have two balloons of equal volume at
room temperature and 1 atm pressure. One
balloon is filled with He and the other is filled
with H2 . Which statement is true about the
relative number of gas particles in the two
balloons?
1. Both balloons have approximately the
same number of gas particles. correct
2. The He balloon has approximately half
the number of particles, since each particle is
only composed of 1 atom.
3. The H2 balloon has approximately half
the number of particles since each particle
has twice as many atoms.
4. The He balloon has approximately half
the number of particles since each particle
has about twice the mass of H2
Explanation:
One mole of any ideal gas takes up approximately 22.4L of space at STP.
021 4.0 points
Three of the primary components of air are
carbon dioxide, nitrogen, and oxygen. In a
sample containing a mixture of only these
gases at exactly one atmosphere pressure, the
partial pressures of carbon dioxide and nitrogen are given as PCO2 = 0.285 torr and
PN2 = 640.554 torr. What is the partial pressure of oxygen?
1. 148.118
Version 169 – Exam 1 – mccord – (52125)
7
× (293.15 K)
= 3.43491 atm
2. 141.698
3. 131.328
4. 126.614
5. 175.298
6. 206.566
7. 165.715
8. 119.161
9. 138.817
10. 170.845
023 4.0 points
A 4.0 L flask containing N2 at 15 atm is
connected to a 4.0 L flask containing H2 at
7.0 atm and the gases are allowed to mix.
What is the mole fraction of N2 ?
Correct answer: 119.161 torr.
1. 0.68 correct
Explanation:
PCO2 = 0.285 torr
PN2 = 640.554 torr
2. Cannot be determined
PT
PO 2
PO 2
PO 2
PT = 1 atm = 760 torr
PO 2 = ?
= PCO2 + PN2 + PO2
= PT − (PCO2 + PN2 )
= 760 torr − (0.285 torr + 640.554 torr)
= 119.161 torr
022 4.0 points
Calculate the pressure of 4 g of nitrogen gas
in a 1 L container at 20.0◦C.
3. 0.75
4. 0.93
5. 0.26
6. 0.32
Explanation:
Applying the ideal gas law
P V = nRT
PV
=
RT
1. 0.235 atm
2. 0.469 atm
(15 atm) (4 L)
RT
(7 atm) (4 L)
=
RT
3. 6.87 atm
nN 2 =
4. 3.44 atm correct
nH 2
5. 96.2 atm
Explanation:
m=4g
T = 20◦ C + 273.15 = 293.15 K
For the nitrogen,
V =1L
nN 2
nN 2
=
ntotal
nN 2 + nO 2
(15 atm) (4 L)
RT
=
(15 atm) (4 L) (7 atm) (4 L)
+
RT
RT
(15 atm) (4 L)
=
(15 atm) (4 L) + (7 atm) (4 L)
= 0.681818
χN2 =
1 mol N2
28.0134 g N2
= 0.142789 mol N2
n = (4 g N2 ) ·
P V = nRT
n RT
P =
V
(0.142789 mol N2 ) 0.08206
=
1L
Assume the temperature of the two gases
remains the same before and after the mixing
occurs. The mole fraction of N2 is
L·atm
mol·K
024
4.0 points
Version 169 – Exam 1 – mccord – (52125)
An analysis of a volatile liquid shows that
it contains 62.04% carbon, 10.41% hydrogen,
and 27.54% oxygen by mass. At 150 ◦ C and
1 atm, 500 mL of the vapor has a mass of
1.673 g. What is the molecular weight of the
compound?
6. 178.1 g
7. 95.64 g
8. 253.1 g
1. 58 g/mol
9. 167.4 g
2. 288 g/mol
10. 121.7 g
Explanation:
mC6 H6 = 39.6 g
The balanced equation is
3. 144 g/mol
4. 232 g/mol
5. 116 g/mol correct
Explanation:
T = 150◦ C + 273 = 423 K
P = 1 atm
V = 500 mL = 0.5 L
Applying the ideal gas law equation,
P V = nRT
PV
n=
RT
(1 atm) (0.5 L)
(0.08206 L · atm) (423 K)
= 0.0144045 mol
=
MW =
1.673 g
= 116.144 g/mol
0.0144045 mol
025
For the reaction
4.0 points
? C6 H6 + ? O2 → ? CO2 + ? H2 O
39.6 grams of C6 H6 are mixed with 133.8
grams of O2 and allowed to react. How much
CO2 could be produced by this reaction?
1. 133.9 g correct
8
mO2 = 133.8 g
2 C6 H6 + 15 O2 → 12 CO2 + 6 H2 O .
The molecular weight of C6 H6 is 78.1118
g/mol, giving 0.506965 mol C6 H6 .
The molecular weight of O2 is 31.9988
g/mol, giving 4.18141 mol O2 .
15 mol O2
0.506965 mol C6 H6 ×
2 mol C6 H6
= 3.80224 mol O2 ,
which is less than what is actually present.
Therefore the limiting reactant must be C6 H6 .
The molecular weight of CO2 is 44.0095
g/mol.
The limiting reactant (C6 H6 ) will yield
3.04179 mol CO2 which is equal to 133.9 g
of CO2 .
12 mol CO2
0.506965 mol C6 H6 ×
2 mol C6 H6
44.0095 g CO2
= 133.9 g CO2
×
1 mol CO2
026 4.0 points
Gases at low pressures are modeled well by the
ideal gas law, while gases at high pressures are
not. This is because the gas particles at high
pressure
1. have average velocities that are not determined by the temperature.
2. 107.1 g
3. 207.5 g
2. have average kinetic energies that are not
determined by the temperature.
4. 281.2 g
3. are closer together. correct
5. 147.3 g
4. do not interact with each other.
Version 169 – Exam 1 – mccord – (52125)
5. no longer have elastic collisions with the
walls of their container.
Explanation:
At high pressures the distances between
molecules become smaller. This leads to interactions between the gas particles (both attractions and repulsions) that cause deviations from ideal behavior (no interactions).
9