chemistry - mccord - exam 1

Version 078 – Exam 1 – mccord – (51315) 1

This print-out should have 26 questions.Multiple-choice questions may continue onthe next column or page – find all choicesbefore answering.

mccord - ch30112noon class only - 51315

7-9pm Sept 18, 2012

in WEL 2.224

R = 0.08206 L·atm/mol·K

R = 62.36 L·torr/mol·K

R = 8.314 J/mol·K

NA = 6.022× 1023 mol−1

1 atm = 101325 Pa

1 bar = 100000 Pa

1 atm = 14.7 psi

001 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 pressures in the two bulbs are thesame.

2. The bulb containing O2 contains moremolecules of gas.

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

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

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.

002 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; 10.7 g left over



4. H2O; 7.20 g left over correct

5. H2O; 10.7 g left over

6. 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

003 10.0 points

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

1. 56.0 g ·mol−1

2. 44.8 g ·mol−1

3. 28.0 g ·mol−1

4. 8.00 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

004 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. −154◦C correct

3. −45.1◦C

4. −261◦C

5. 347◦C

6. 45.1◦C

7. −347◦C

8. 154◦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 .


005 10.0 points

Balance the equation

C2H6 +O2 → CO2 +H2O

using the smallest possible integers. The sumof the coefficients is

1. 9.

2. 10.

3. 19. correct

4. 4.

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

006 10.0 points

A gas sample occupies 9.42 L at 2.0◦C. Whatis the pressure given that there are 1.35 molof gas in the sample?1. 8.974712. 3.355773. 6.329574. 2.936495. 3.235636. 10.24347. 6.689458. 5.896519. 4.6237210. 8.55431

Correct answer: 3.23563 atm.

Explanation:

n = 1.35 mol

T = 2.0◦C+ 273 = 275 K

V = 9.42 L

P = ?

P =nRT

V

=(1.35 mol)

(

0.0821 L·atm

mol·K

)

(275 K)

9.42 L= 3.23563 atm

007 10.0 points

The density of the vapor of allicin, a compo-nent of garlic, is 1.14 g · L−1 at 125◦C and 175Torr. What is the molar mass of allicin?

1. 869 g ·mol−1

2. 21.6 g ·mol−1

3. 273 g ·mol−1

4. 162 g ·mol−1 correct

5. 50.8 g ·mol−1

Explanation:

T = 125◦C + 273.15 K = 398.15 K

P = (175 Torr)1 atm

760 Torr= 0.230263 atm

ρ = 1.14 g/LThe 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

=(1.14 g/L)

(

0.08206 L·atm

mol·K

)

0.230263 atm× (398.15 K)

= 161.755 g/mol


008 10.0 points

Absolute zero (0K) is the temperature atwhich

1. gaseous helium liquefies.

2. a graph of V versus1

Pintersects the

1

Paxis.

3. None of these

4. a graph of P versus1

Vintersects the

1

Vaxis.

5. the straight line graph of V versus Tintersects the T axis. correct

Explanation:

A graph of V vs. T is a straight line. Thisline intersects the temperature axis at −273◦

C, at which is 0K or absolute zero.

009 10.0 points

Ludwig Boltzmann performed a simple, butpowerful 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 temperature of the gas, not the molarmass of the gas.

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

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

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. must be the same. correct

3. must be different.

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

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

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 moles of gas; molar volume ofthe particles. correct

2. number of molecules of gas; molar volumeof the particles.

3. number of electrons within the gas; molec-ular radius of the particles.

4. number of moles of gas; molecular radiusof the particles.

5. number of moles of gas; volume of thecontainer itself.


Explanation:

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

gas molecules (aka: the particles) really dotake up a finite amount of space themselves. brepresents the space that the particles cannotmove into because it is already occupied.

012 10.0 points

How many molecules are contained in1204 mL of O2 gas at 125

◦C and 837 torr?1. 4.50218e+222. 3.03123e+223. 4.01328e+224. 2.15205e+225. 2.04312e+226. 6.77392e+227. 4.18495e+228. 2.60369e+229. 2.44492e+2210. 1.51595e+22

Correct answer: 2.44492× 1022 molec.

Explanation:

V = 1.204 L T = 125◦C+ 273 = 398 K

P = 837 torr ·1 atm

760 torr= 1.10132 atm

Applying the ideal gas law equation,

P V = nRT

n =P V

RT

=(1.204 L) (1.10132 atm)(

0.08206 L·atm

mol·K

)

(398 K)

·6.022× 1023 molec

mol= 2.44492× 1022 molec

013 10.0 points

The pressure on a gas at −44◦C is doubled,but its volume is held constant. What will thefinal temperature be in degrees Celsius?1. 177.02. 143.03. 131.04. 111.0

5. 149.06. 183.07. 99.08. 185.09. 137.010. 151.0

Correct answer: 185◦C.

Explanation:

P2 = 2P1 T1 = −44◦C+ 273 = 229 KT2 = ?Applying the Gay-Lussac law,

P1

T1=

P2

T2

T2 =P2 T1P1

=2P1 (229 K)

P1

= 458 K = 185◦C

014 10.0 points

A 6.00 L sample of C2H4(g) at 2.00 atm and293 K is burned in 6.00 L of oxygen gas atthe same temperature and pressure to formcarbon dioxide gas and liquid water. If thereaction goes to completion, what is the finalvolume of all gases at 2.00 atm and 293 K?

1. 12.00 L

2. 2.66 L

3. 2.00 L

4. 8.00 L correct

5. 4.00 L

6. 6.00 L

Explanation:

The balanced equation is

C2H4(g) + 3O2(g) → 2H2O(ℓ) + 2CO2(g)

Avogadro’s Principle tells us that when Pand T are constant (which they are in thisproblem), V ∝ n, so we can work with thevolume of the gases instead of the number ofmoles. From the equation above, we need 3


times more L of O2 than of C2H4 (18 L, notthe given 6 L!) so O2 is the limiting reagent.Find out how much CO2 is made based on

the L of O2:

LCO2= 6 L O2 ×

(

2 L CO2

3 L O2

)

= 4 L CO2

We now find out how much C2H4 was usedreacting with all the L of O2:

LC2H4= 6 L O2 ×

(

1 L C2H4

3 L O2

)

= 2 L C2H4

This means 6 L− 2 L = 4 L C2H4 are unusedand still present.The final mixture is

4 L unreacted C2H4 + 4 L CO2 = 8 L of gas .

We assume the volume of the water is insignif-icant.

015 10.0 points

If the temperature of a fixed amount of gas isdecreased at constant pressure its volume will

1. Insufficient data to answer this question.

2. decrease. correct

3. remain the same.

4. increase.

Explanation:

The volume of a gas is directly proportionalto its absolute temperature (Charles’ Law).This means that as the absolute temperatureof a gas decreases, its volume decreases pro-portionately.

016 10.0 points

A 22.4 L vessel contains 0.02 mol H2 gas,0.02 mol N gas, and 0.1 mol NH3 gas. Thetotal pressure is 700 torr. What is the partialpressure of the H2 gas?

1. 100 torr correct

2. None of these

3. 14 torr

4. 7 torr

5. 28 torr

Explanation:

ntotal = 0.14 mol Ptotal = 700 torrnH2

= 0.02 mol

XH2=

nH2

ntotal

=0.02 mol

0.14 mol= 0.142857

PH2= XH2

Ptotal

= (0.142857) (700 torr)

= 100 torr

017 10.0 points

For the reaction

2 NH3 + CH3OH → products ,

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

1. 2.75 mol correct

2. 88.1 mol

3. 1.31 mol

4. 5.50 mol

5. 46.8 mol

6. 11.3 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


A mixture of three gases, A, B, and C, isat a total pressure of 7.31 atm. The partialpressure of gas A is 2 atm; that of gas B is3.78 atm. What is the partial pressure of gasC?1. 0.742. 1.623. 0.724. 2.05. 2.276. 0.957. 1.878. 1.719. 1.5310. 1.78

Correct answer: 1.53 atm.

Explanation:

PT = 7.31 atm PB = 3.78 atmPA = 2 atm PC = ?

PT = PA + PB + PC

PC = PT − (PA + PB)

= 7.31 atm− (2 atm+ 3.78 atm)

= 1.53 atm

019 10.0 points

Two gases are contained in gas bulbs con-nected by a valve. Gas A is present in a 21 Lbulb at a pressure of 388 torr. Gas B exertsa pressure of 803 torr in a 86 L bulb. Whatis the partial pressure of gas B after theyequilibrate once the valve has been opened?1. 558.02. 545.03. 476.04. 715.05. 623.06. 773.07. 645.08. 463.09. 731.010. 523.0

Correct answer: 645 torr.

Explanation:

VA = 21 L PA = 388 torr

VB = 86 L PB = 803 torrVtotal = VA + VB = 107 L

PB VB = Pt Vt

Pt =PB VBVt

=(803 torr) (86 L)

107 L= 645 torr

020 10.0 points

Real gases behave most nearly like ideal gasesat

1. high temperatures and low pressures.correct

2. low temperatures and low pressures.

3. low temperatures and high pressures.

4. high pressures and low molar masses.

5. high 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’.

021 0.0 points

This question starts out at zero points butcould very well increase after the grading.Now, if more points are awarded (the curve)on this assignment, would you like themadded to your score?

1. NO, leave my score alone, I prefer thelower score

2. YES, I would like the points and thehigher score. correct

Explanation:

This should be a no-brainer. Most studentswant higher scores. If you picked yes, you gotcredit for the question and you got the extrapoints you asked for (if they were granted


by your instructor). If you answered NO,you also got what you wanted... no pointsawarded.

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

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 move to the centerof the tank because their velocities would belower, thus giving them less 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 molecules would sink to the bottomof the tank because of the loss of pressure.

Explanation:

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

023 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. 1130 m · s−1

2. 410 m · s−1

3. 1000 m · s−1

4. 679 m · s−1 correct

5. 247 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

024 10.0 points

Use van der Waals’ equation to calculatethe pressure exerted by 2.6 mol of ammo-nia at −4 ◦C in a 4.71 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. 8.43637 atm

2. 5.62425 atm

3. 14.998 atm

4. 22.497 atm

5. 11.2485 atm correct

Explanation:

n = 2.6 mol T = −4◦C = 269 K

V = 4.71 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


=

(2.6 mol)

(

0.08206L · atm

mol ·K

)

(269 K)

4.71 L− (2.6 mol)(0.04 L/mol)

−(2.6 mol)2(4.0 L2 · atm/mol2)

(4.71 L)2

= 11.2485 atm

025 10.0 points

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

1. 1.19× 1023 atoms correct

2. 1.79× 1023 atoms

3. 2.53× 1026 atoms

4. 5.97× 1022 atoms

5. 5.49× 1024 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/molWe can use this molecular mass to convert

g 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

026 10.0 points

In the chemical reaction

2H2O+ energy → 2H2 +O2

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

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

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

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

Explanation:

chemistry - mccord - exam 1

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