UT CH 301 UNIT 1 EXAM-Solutions

Version 398 – UNIT 1 EXAM – vandenbout – (51335) 1

This print-out should have 21 questions.Multiple-choice questions may continue onthe next column or page – find all choicesbefore answering.

001 10.0 points

The pressure on a gas at −68◦C is doubled,but its volume is held constant. What will thefinal temperature be in degrees Celsius?1. 151.02. 123.03. 117.04. 161.05. 137.06. 185.07. 187.08. 127.09. 119.010. 177.0

Correct answer: 137◦C.

Explanation:

P2 = 2P1 T1 = −68◦C+ 273 = 205 KT2 = ?Applying the Gay-Lussac law,

P1

T1=

P2

T2

T2 =P2 T1P1

=2P1 (205 K)

P1

= 410 K = 137◦C

002 10.0 points

The same number of grams of NH3 and O2

are placed in separate bulbs of equal vol-ume and temperature under conditions whenboth gases behave ideally. Which statementis true?

1. The pressure in the O2 bulb is greaterthan the pressure in the NH3 bulb.

2. Both bulbs contain the same number ofmoles of gas.

3. The bulb containing O2 contains moremolecules of gas.

4. The pressures in the two bulbs are thesame.

5. The pressure in the NH3 bulb is greaterthan the pressure in the O2 bulb. correct

Explanation:

The molecular weight of NH3 is less thanthat of O2, so in equal masses there are moremoles of NH3 than of O2. At the same volumeand temperature, the larger number of molesof NH3 would exert a higher pressure.

003 10.0 points

Balance the equation

C2H6 +O2 → CO2 +H2O

using the smallest possible integers. The sumof the coefficients is

1. 19. correct

2. 4.

3. 9.

4. 10.

Explanation:

A balanced equation must have the samenumber of each kind of atom on both sidesof the equation. We find the number of eachkind of atom using equation coefficients andcomposition stoichiometry. For example, wefind there are 4 C atoms on the reactant side:

?C atoms = 2C2H6 ×2 C

1 C2H6

= 4C .

The balanced equation is

2C2H6 + 7O2 → 4CO2 + 6H2O

and has 4 C, 12 H, and 14 O atoms on eachside.? sum coefficients = 2 + 7 + 4 + 6 = 19

004 10.0 points

In the chemical reaction

2H2O+ energy → 2H2 +O2


1. twice as much O2 as H2 will be pro-duced.

2. there is no way to know how much of eachwill be produced.

3. equal amounts of H2 and O2 will be pro-duced.

4. twice as much H2 as O2 will be produced.correct

Explanation:

005 10.0 points

Consider the following reaction:

CaCN2 + 3H2O → CaCO3 + 2NH3

105.0 g CaCN2 and 78.0 g H2O are reacted.Assuming 100% efficiency, which reactant isin excess and how much is leftover? Themolar mass of CaCN2 is 80.11 g/mol. Themolar mass of CaCO3 is 100.09 g/mol.

1. CaCN2; 7.20 g left over

2. H2O; 10.7 g left over

3. H2O; 7.20 g left over correct

4. H2O; 70.8 g left over



Explanation:

CaCN2 + 3H2O → CaCO3 + 2NH3

According to the stoichiometry 1 mol ofCaCN2 reacts with 3 moles of H2O. First wecalculate the moles of CaCN2 and of H2O asshown below.

? mol CaCN2 = 105.0 g NaCN2

×1 mol CaCN2

80.11 g CaCN2

= 1.31 mol CaCN2

? molH2O = 78.0 g H2O×1 mol H2O

18.0152 g H2O= 4.33 mol H2O

Since the stoichiometric ratio CaCN2 toH2O is 1 to 3, to react completely 1.31 molCaCN2 we will require to use:

3× 1.31 = 3.93 mol H2O

We calculated that we have 4.33 mol H2O.Therefore water is in excess:

4.33− 3.93 = 0.40 mol H2O

? excess H2O = 0.40 mol H2O

×18.0152 g H2O

1 mol H2O= 7.20 g H2O

006 10.0 points

This question is merely a placeholder forthe points in the hand-graded portion of theexam.

1. yes CORRECT2. do I have a choice?3. without a doubt4. absolutely5. sure

Explanation:

This question is merely a placeholder forthe points in the hand-graded portion of theexam.

007 10.0 points

Two gases are contained in gas bulbs con-nected by a valve. Gas A is present in a 36 Lbulb at a pressure of 273 torr. Gas B exertsa pressure of 716 torr in a 84 L bulb. Whatis the partial pressure of gas B after theyequilibrate once the valve has been opened?


1. 728.02. 402.03. 566.04. 642.05. 558.06. 578.07. 550.08. 476.09. 847.010. 501.0

Correct answer: 501 torr.

Explanation:

VA = 36 L PA = 273 torrVB = 84 L PB = 716 torrVtotal = VA + VB = 120 L

PB VB = Pt Vt

Pt =PB VBVt

=(716 torr) (84 L)

120 L= 501 torr

008 10.0 points

In an improved version of the gas law, V isreplaced by (V − n b). The two quantities nand b in this equation represent, respectively,the

1. number of molecules; excluded molecularvolume.

2. number of moles; excluded molecular vol-ume. correct

3. number of moles; molecular radius.

4. number of electrons; molecular radius.

5. number of moles; the container size.

Explanation:

n represents moles (as it does in P V =nRT ).b is the correction factor for volume because

molecules really do take up space.

009 10.0 points

Ludwig Boltzmann performed a simple, but

powerful experiment to gather evidence con-cerning the velocity distribution of a sampleof gas particles. His experiment revealed thatthe velocities of gases:

1.Are distributed in a characteristic manneracross a range of temperatures that dependson the molar mass of the gas, but not thetemperature of the gas.

2.Are distributed in a characteristic manneracross a range of temperatures that dependson the temperature of the gas, not the molarmass of the gas.

3.Are distributed in a characteristic manneracross a range of temperatures that dependson the molar mass of the gas and the temper-ature of the gas. correct

4. Are distributed in the same characteristicmanner for all gases, regardless of the tem-perature or molar mass, as long as the gas isbehaving ideally.

Explanation:

.

010 10.0 points

Consider two balloons filled with gas and ar-ranged so that P , V , T are the same in both.The number of molecules in each balloon

1. could be different if the filling gases aredifferent.

2. would be the same only if the filling gasesare the same.

3. must be different.

4. must be the same. correct

Explanation:

P1 = P2 V1 = V2 T1 = T2

P V = nRT, so R =P V

nTThus

P2 V2n2 T2

=P1 V1n1 T1


n1

n2

=P1 V1 T2P2 V2 T1

= 1

n1 = n2

011 10.0 points

A steel tank containing helium is cooled to15◦C. If you could look into the tank and seethe gas molecules, what would you observe?

1. The molecules would move to the centerof the tank because their velocities would belower, thus giving them less pressure.

2. The gas molecules would become uni-formly distributed near the entire wall of thetank because the molecules would try to es-cape the container due to their kinetic ener-gies.

3. The molecules would sink to the bottomof the tank because of the loss of pressure.

4. The gas molecules would still be uni-formly distributed around the tank becausegases expand to fill up the whole volume dueto their constant molecular motion. correct

5. The gas molecules would have higher ki-netic energies and lower velocities, thus creat-ing no net change.

Explanation:

The average kinetic molecular energy woulddecrease, but since it is still in gas phase, themolecules would still expand uniformly to fillthe tank.

012 10.0 points

A gas sample occupies 3.59 L at 8.0◦C. Whatis the pressure given that there are 1.63 molof gas in the sample?1. 10.64482. 6.056243. 6.778344. 4.104065. 4.428266. 3.726757. 8.833268. 5.303039. 4.9661

10. 10.4747

Correct answer: 10.4747 atm.

Explanation:n = 1.63 mol

T = 8.0◦C+ 273 = 281 K

V = 3.59 L

P = ?

P =nRT

V

=(1.63 mol)

(

0.0821 L·atm

mol·K

)

(281 K)

3.59 L= 10.4747 atm

013 10.0 points

The root mean square speed of nitrogenmolecules in air at 20◦C is 511 m/s in a cer-tain container. If the gas is allowed to expandto twice its original volume, the root meansquare velocity of nitrogen molecules drops to325 m/s. Calculate the temperature after thegas has expanded.

1. −261◦C

2. −347◦C

3. 347◦C

4. 154◦C

5. 261◦C

6. 45.1◦C

7. −154◦C correct

8. −45.1◦C

Explanation:

T1 = 20◦C+ 273.15 = 293.15 KvT1

= 511 m/s vT2= 325 m/s

From kinetic molecular theory, tempera-ture is directly proportional to mean KE.

KEmean =1

2(MW)(average molecular speed)2

and knowing MW is constant (it’s the samegas), T ∝ v2rms and

vT1

vT2

=

√T1√T2


T2 =

(

vT2

vT2

)2

T1 =

(

325 m/s

511 m/s

)2

(293.15 K)

= 118.581 K ,

so the final temperature is

118.581 K− 273.15 = −154.569◦C .

014 10.0 points

If 250 mL of a gas at STP weighs 2 g, what isthe molar mass of the gas?

1. 28.0 g ·mol−1

2. 8.00 g ·mol−1

3. 56.0 g ·mol−1

4. 44.8 g ·mol−1

5. 179 g ·mol−1 correct

Explanation:

V = 250 mL P = 1 atmT = 0◦C = 273.15 K m = 2 gThe density of the sample is

ρ =m

V=

2 g

0.25 L= 8 g/L

The ideal gas law is

P V = nRTn

V=

P

RT

with unit of measure mol/L on each side.Multiplying each by molar mass (MM) gives

n

V·MM =

P

RT·MM = ρ ,

with units of g/L.

MM =ρRT

P

=(8 g/L)(0.08206 L · atm/mol/K)

1 atm× (273.15 K)

= 179.318 g/mol

015 10.0 points

Real gases behave most nearly like ideal gasesat

1. low temperatures and low pressures.

2. high temperatures and low pressures.correct

3. high pressures and low molar masses.

4. high temperatures and high pressures.

5. low temperatures and high pressures.

Explanation:

At high temperatures the gas molecules aremoving more rapidly and the effects of theattractive forces are less significant. At lowpressures the molecules are on average muchfurther apart and the effects of the attractiveforces are less significant because there arefewer ‘close encounters’.

016 10.0 points

Calculate the number of carbon atoms in 4.56grams of ethanol (CH3CH2OH).

1. 5.97× 1022 atoms

2. 5.49× 1024 atoms

3. 1.19× 1023 atoms correct

4. 2.53× 1026 atoms

5. 1.79× 1023 atoms

Explanation:

mCH3CH2OH = 4.56 gEach CH3CH2OH molecule contains two

carbon atoms. There are Avogadro’s numberof ethanol molecules in one mole of ethanol.We need the molecular mass of ethanol so wecan convert grams of ethanol to moles ethanol:Molecular mass of CH3CH2OH

= 2(12.01 g/mol) + 6(1.01 g/mol)

+1(16.00 g/mol)

= 46.08 g/mol


We can use this molecular mass to convertg ethanol to mol ethanol:

? mol ethanol = 4.56 g CH3CH2OH

×1 mol CH3CH2OH

46.08 g CH3CH2OH

= 0.09896 mol CH3CH2OH

We can now use Avogadro’s number and theratio of C atoms to CH3CH2OH molecules tofind the number of carbon atoms:? atoms C

= 0.09896 mol CH3CH2OH

×6.022× 1023 molec CH3CH2OH

1 mol CH3CH2OH

×2 atoms C

1 molec CH3CH2OH

= 1.192× 1023 atoms C

017 10.0 points

For the reaction

2 NH3 +CH3OH → products ,

how much CH3OH is needed to react with93.5 g of NH3?

1. 46.8 mol

2. 2.75 mol correct

3. 5.50 mol

4. 11.3 mol

5. 88.1 mol

6. 1.31 mol

7. 3.32 mol

Explanation:

mNH3= 93.5 g

? mol CH3OH = 93.5 g NH3 ×1 mol NH3

17 g NH3

×1mol CH3OH

2molNH3

= 2.75 mol CH3OH

018 10.0 points

If the average speed of a carbon dioxidemolecule is 410 m · s−1 at 25◦C, what is theaverage speed of a molecule of methane at thesame temperature?

1. 679 m · s−1 correct

2. 410 m · s−1

3. 247 m · s−1

4. 1130 m · s−1

5. 1000 m · s−1

Explanation:

From kinetic molecular theory, the temper-ature is directly proportional to mean KE.

KEmean=1

2(MW)(average molecular speed)2

and knowing T is constant,

vCO2

vCH4

=

√

MWCH4

MWCO2

vCO2= vCH4

√

MWCO2

MWCH4

= (410 m/s)

√

44.0098 g/mol

16.0426 g/mol

= 679.08 m/s

019 10.0 points

Isoamyl acetate (C7H14O2) is a chemicalcompound that has a strong odor similar tobananas. At room temperature, the rms ve-locity of a isoamyl acetate molecule in the gasphase is approximately 240 m s−1.

About how long will it take an isoamylacetate molecule to diffuse 24 m across a roomfilled with air at SATP?

1. About 0.1 second, since it is traveling240 m s−1.2. The molecule will never diffuse across

the room because it is too large.


3. Faster than 0.1 second, since 240 m s−1

is the rms velocity which is not the same asthe average velocity.4. Much longer than 0.1 second, because

the molecule will have many collisions withother gas molecules. CORRECT

Explanation:

Because the isoamyl acetate will undergomany (too many too count) collisions it willtake much much much longer than 0.1 secondto diffuse 24 m. The answer could only be 0.1seconds if the molecule had no collisions withother gas molecules. It is true that the rmsvelocity is not the average. It is also true thatthere is a distribution of velocities and somemolecules will be moving faster. However,they will also undergo many many collisionscausing them to take much more than 0.1 s tomove 24 m.

020 10.0 points

Use van der Waals’ equation to calculatethe pressure exerted by 1.25 mol of ammo-nia at −0.1 ◦C in a 1.54 L container. Thevan der Waals’ constants for ammonia area = 4.00 L2·atm/mol2 and b = 0.0400 L/mol.(The values for a and b have been rounded offto simplify the arithmetic.)

1. 8.08102 atm

2. 32.3241 atm

3. 21.5494 atm

4. 16.162 atm correct

5. 12.1215 atm

Explanation:

n = 1.25 mol T = −0.1◦C = 272.9 K

V = 1.54 L a = 4.0 L2 · atm/mol2

b = 0.04 L/mol(

P +n2 a

V 2

)

(V − n b) = nRT

P =nRT

V − n b−

n2 a

V 2

=

(1.25 mol)

(

0.08206L · atm

mol ·K

)

(272.9 K)

1.54 L− (1.25 mol)(0.04 L/mol)

−(1.25 mol)2(4.0 L2 · atm/mol2)

(1.54 L)2

= 16.162 atm

021 10.0 points

A mixture of three gases, A, B, and C, is at atotal pressure of 6.2 atm. The partial pressureof gas A is 1.63 atm; that of gas B is 3.78 atm.What is the partial pressure of gas C?1. 1.432. 2.243. 2.094. 1.45. 1.916. 0.797. 1.978. 0.859. 2.5910. 1.17

Correct answer: 0.79 atm.

Explanation:

PT = 6.2 atm PB = 3.78 atmPA = 1.63 atm PC = ?

PT = PA + PB + PC

PC = PT − (PA + PB)

= 6.2 atm− (1.63 atm+ 3.78 atm)

= 0.79 atm

UT CH 301 UNIT 1 EXAM-Solutions

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