Exam 3 Solutions

Version 048 – Exam Three: Shear – Shear – (52375) 1

This print-out should have 27 questions.Multiple-choice questions may continue onthe next column or page – find all choicesbefore answering.

DrRuth says: In all cases, select the answerthat BEST answers the question asked. Thetotal number of points on this exam is 350.Read the whole thing first and plan which toanswer first because you feel confident aboutthem. Notice the harder ones and save themup till last so you don’t use up all your timeworking on them. Unless otherwise stated,assume the gases in all questions are ideal.

Shear 52375

g = 9.8 m/s2

1 atm = 101325 Pa = 760 torr

R = 0.08206 L atm K−1mol−1

R = 0.08314 L bar K−1mol−1

R = 8.314 J K−1mol−1

Effusion rate of gas 1

Effusion rate of gas 2=

√

M2

M1

Effusion rate at T1

Effusion rate at T2

=

√

T1

T2

KE = 2/3RT

urms =

√

3RT

M

001 20.0 pointsCalculate the pressure of 4 g of nitrogen gasin a 1 L container at 20.0◦C.

1. 96.2 atm

2. 0.235 atm

3. 6.87 atm

4. 3.44 atm correct

5. 0.469 atm

Explanation:

m = 4 g V = 1 LT = 20◦C + 273.15 = 293.15 K

For the nitrogen,

n = (4 g N2) ·1 mol N2

28.0134 g N2

= 0.142789 mol N2

P V = n RT

P =n R T

V

=(0.142789 mol N2)

(

0.08206 L·atmmol·K

)

1 L× (293.15 K)

= 3.43491 atm

002 10.0 pointsA mixture of CO, CO2 and O2 is containedwithin a 275 mL flask at 0◦C. If the total pres-sure is 780 torr, the CO has a partial pressureof 330 torr and the CO2 has a partial pressureof 330 torr, what is the partial pressure of O2?

1. 330 torr

2. 120 torr correct

3. 660 torr

4. 780 torr

5. 900 torr

Explanation:

Ptotal = 780 torr PCO = 330 torrPCO2

= 330 torr

Ptotal = PCO + PCO2+ PO2

PO2= Ptotal − PCO − PCO2

= 780 torr − 330 torr − 330 torr

= 120 torr

003 20.0 pointsGiven the reaction and the solubility informa-tion below, write the net ionic equation forthe reaction


potassium phosphate(aq)+ lead(II) nitrate(aq)

General solubility guidelines:1) Most sodium, potassium and ammonium

compounds are soluble in water.2) Most nitrates, acetates and chlorates are

soluble.3) Most chlorides are soluble, except those

for silver, mercury(I) and lead. Lead(II)chloride is soluble in hot water.

4) Most sulfates are soluble, except those ofbarium, strontium and lead.

5) Most carbonates, phosphates and silicatesare insoluble, except those of sodium,potassium and ammonium.

6) Most sulfides are insoluble, except thoseof calcium, strontium, sodium, potassiumand ammonium.

1. 2 K3PO4(aq) + 3 Pb(NO3)2(aq) −→6 KNO3(aq) + Pb3(PO4)2(s)

2. 2 K3PO4(aq) + 3 Pb(NO3)2(aq) −→6 K+(aq) + 6 NO−

3 (aq) + Pb3(PO4)2(s)

3. 3 Pb2+(aq) + 2 PO3−4 (aq) −→Pb3(PO4)2(s) correct

4. None of these

5. 6 K+(aq) + 2 PO3−4 (aq)

+3 Pb2+(aq) + 6 NO−

3 (aq) −→

6 K+(aq) + 6 NO−

3 (aq) + Pb3(PO4)2(s)

Explanation:

K3PO4(aq) + Pb(NO3)2(aq)

net ionic equation = ?The ionic equation is

6 K+(aq) + 2 PO3−4 (aq)

+3 Pb2+(aq) + 6 NO−

3 (aq) −→

6 K+(aq) + 6 NO−

3 (aq) + Pb3(PO4)2(s)and the net ionic equation is

3 Pb2+(aq) + 2 PO3−4 (aq) −→ Pb3(PO4)2(s)

004 10.0 pointsConsider the two water molecules

Ob

b b

b

H H

Ob

b b

b

H H

→A

→

B

Which statement is correct?

1. The covalent bond B is stronger than thehydrogen bond A.

2. The hydrogen bond A is stronger thanthe covalent bond B.

3. The hydrogen bond B is stronger thanthe covalent bond A.

4. The covalent bond A is stronger than thehydrogen bond B. correct

Explanation:

Ob

b b

b

H Hδ−δ−

δ+

Ob

b b

b

H Hδ−δ−

δ+

→A

→B

A is a covalent bond while B is a hydrogenbond. Covalent bonds are much stronger thanH-bonds.

005 10.0 pointsWhich is the weakest type of attractive forcebetween particles?

1. covalent bond

2. ionic bond

3. dispersion forces correct

4. hydrogen bond

Explanation:

London forces, dispersion forces, or induceddipoles all describe the same intermolecularforce. London forces are induced, short-lived,and very weak. Molecules and atoms canexperience London forces because they have


electron clouds. London forces result from thedistortion of the electron cloud of an atom ormolecule by the presence of nearby atoms ormolecules.

Permanent dipole-dipole interactions arestronger than London forces and occur be-tween polar covalent molecules due to chargeseparation.

H-bonds are a special case of very strongdipole-dipole interactions. They only occurwhen H is bonded to small, highly electro-magnetic atoms – F, O or N only.

Ion-ion interactions are the strongest due toextreme charge separation and occur betweenionic molecules. They can be thought of asboth inter- and intramolecular bonding.

006 10.0 pointsSally has an inflatable doll filled with heliumat the circus. Jane has an identical doll filledwith argon at the circus. They sit next toeach other and discuss chemistry. They argueover whose balloon has gas molecules zippingaround at greater average velocities. Resolvethe argument.

1. The velocities are the same because theballoons are at the same temperature.

2. It depends on the volume of the identicalballoons.

3. Helium has the greater velocity. correct

4. The velocities cannot be discussed be-cause gases don’t have velocities – they havepressures.

5. Argon has the greater velocity.

Explanation:

By the kinetic molecular theory,Average molecular kinetic energyKE ∝ T .

Average molecular speed U ∝

√

T

MW

UH2

UAr

=

√

T

MWH2√

T

MWAr

=

√

MWAr

MWH2

=

√

40

1= 6.3

UH2= 6.3 UAr

Therefore, the H2 molecules are moving anaverage of 6.3 times faster than the Armolecules. (Smaller molecules are faster.)

007 10.0 pointsWhich of the molecules

H2, HF, CH4, CH3OHcan be involved in hydrogen bonding?

1. HF, CH4

2. CH3OH, H2

3. CH3OH, CH4

4. HF, CH3OH correct

Explanation:

For a molecule to be involved in hydrogenbonding, the molecule must have a hydrogenatom and a highly electronegative atom suchas F, O, N, or Cl. The only two moleculesthat satisfy these requirements are HF andCH3OH.

008 10.0 pointsTo promote ideal behavior of a gas, one should(raise/lower) the pressure and (raise/lower)the temperature of the gas.

1. raise, lower

2. raise, raise

3. lower, lower

4. lower, raise correct

Explanation:

Low pressure and high temperature bothfavor ideal behavior. Low pressure helps toensure that the likelihood of molecules inter-acting with each other is small, thus helpingto satisfy one of the assumptions of kineticmolecular theory, namely that gas molecules


are not attracted to each other. High temper-ature gives the molecules a greater kinetic en-ergy, helping to ensure that collisions will beas elastic as possible, again helping to satisfyone of the assumptions of kinetic moleculartheory.

009 15.0 pointsAs we increase the temperature of a liquid,its properties change. Which of the follow-ing would NOT be an expected change in theproperties of a liquid as we increase its tem-perature?

1. decrease in density

2. increase in tendency to evaporate

3. increase in surface tension correct

4. increase in vapor pressure

5. decrease in viscosity

Explanation:

Surface tension typically decreases with in-creasing temperature.

010 10.0 pointsThe vapor pressure of all liquids

1. increases with temperature. correct

2. is the same at their freezing points.

3. decreases with the increasing volume ofthe container.

4. increases with volume of liquid present.

5. is the same at 100◦C.

Explanation:

As temperature (kinetic energy) increases,rate of evaporization increases and rate of con-densation decreases; therefore, vapor pressurewill increase with increasing temperature.

011 20.0 pointsAt STP a gas occupies 121 mL. How many

milliliters will this gas occupy at −52◦C and1.18 atm?

1. 115 mL

2. 83.0 mL correct

3. −19.5 mL

4. 127 mL

Explanation:

P1 = 1 atm P2 = 1.18 atmT1 = 273.15 K V1 = 121 mLT2 = −52◦C + 273.15 = 221.15 K

P1 V1

T1

=P2 V2

T2

V2 =P1 V1 T2

T1 P2

=(1 atm) (121 mL) (221.15 K)

(273.15 K) (1.18 atm)

= 83.0212 mL

012 10.0 pointsWhich molecule would you expect to have thelargest value for the van der Waals constant“b”?

1. NH3

2. Ne

3. He

4. PH3 correct

Explanation:

013 20.0 pointsA 250 milliliter sample of hydrogen was col-lected over water at 20◦C on a day when theatmospheric pressure was 756 torr. The va-por pressure of water under these conditionsis 17.54 torr. What volume will the dry hy-drogen (no water vapor present) occupy atSTP?

1. 20.0 liters


2. 0.226 liters correct

3. 0.105 liters

4. 0.500 liters

5. 0.385 liters

Explanation:

V = 250 mL = 0.25 L Patm = 756 torrT = 20◦C = 293.15 K PH2O = 17.54 torr

Patm = PH2+ PH2O

PH2= Patm − PH2O

= 756 torr − 17.54 torr

= 738.46 torr ·1 atm

760 torr= 0.971658 atm

Applying the ideal gas law equation,

P V = n R T

nH2=

PH2V

R T

=(0.971658 atm) (0.25 L)

(


)

(293.15 K)

= 0.0100979 mol

Dry H2 at STP:nH2

= 0.0100979 mol T = 0◦C = 273.15 KP = 1 atm

Applying the ideal gas law equation,

P V = n R T

V =n R T

P

=(0.0100979 mol) (0.08206 L·atm

mol·K)

1 atm× 273.15 K

= 0.226342 L

014 10.0 pointsDrRuth says: the answers are talking aboutwhat would you observe about the state of thegas now, compared to before it was heated.

A steel tank containing helium is heated to555◦C. If you could look into the tank and seethe gas molecules, what would you observe?

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

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

3. The gas molecules would be uniformlydistributed near the entire wall of the tankbecause the molecules would try to escape thecontainer due to their kinetic energies.

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

5. The gas molecules would have higher ve-locities and higher kinetic energies and movein straight lines until hitting the walls of thetank. correct

Explanation:

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

015 10.0 pointsA 6.5 L sample of nitrogen at 25◦C and1.5 atm is allowed to expand to 13.0 L. Thetemperature remains constant. What is thefinal pressure?

1. 0.38 atm

2. 3.0 atm

3. 0.063 atm

4. 0.75 atm correct

5. 0.12 atm

Explanation:

V1 = 6.5 L V2 = 13.0 LP1 = 1.5 atm

Boyle’s law relates the volume and pressureof a sample of gas:

P1 V1 = P2 V2


P2 =P1 V1

V2

=(1.5 atm) (6.5 L)

13.0 L= 0.75 atm

016 10.0 pointsSublimation describes which of the followinginstances?

1. solid → liquid

2. solid → gas correct

3. liquid → gas

4. gas → solid

Explanation:

Sublimation is the direct vaporization of asolid by heating without passing through theliquid state.

017 15.0 pointsAt what temperature will 2.50 moles of idealgas produce a pressure of 25.0 atm in a 10.0 Lcontainer?

1. 12.0◦C

2. 1,218 K correct

3. 760 K

4. 1,490 K

Explanation:n = 2.5 mol

P = 25 atm

V = 10 L

Applying the ideal law equation,

P V = n RT

T =P V

n R

=(25 atm) (10 L)

(


)

(2.5 mol)

= 1218.62 K

018 10.0 pointsWhat types of intermolecular interactionsdoes chloroform (CH3Cl) exhibit?

I) London dispersionII) dipole-dipole

III) hydrogen bondingIV) covalent bonding

Recall that chloroform is tetrahedralelctronic geometry with C as the central atom.

1. I and II only correct

2. II and IV only

3. II only

4. I, II, and III only

5. II and III only

Explanation:

All molecules, because they have elec-tron clouds, experience instantaneous dipoles.CH3Cl is polar and thus experiences dipole-dipole interactions. CH3Cl does not containany O H, N H, or F H bonds andtherefore does not experience any hydrogenbonding interactions. Covalent bonding is anintramolecular interaction.

019 10.0 pointsWhich of the following is NOT true aboutgases?

1. The density of a gas can be increased byapplying increased pressure.

2. The gas is at STP if it is at 273 K and 1atm.

3. The volume a gas occupies is directlyproportional to its molecular weight. correct

4. Gases can expand without limit.

5. Gases exert pressure on their surround-ings.

Explanation:

Volume is directly proportional to numberof moles of a gas present, not the molecularweight:


P V = n RT

V ∝ n

020 10.0 pointsWhich of the following statements about dis-persion forces is NOT correct? Dispersionforces

1. decrease in strength with increasingmolecular size. correct

2. are the only forces between nonpolarmolecules.

3. are also called London forces.

4. are temporary rather than permanentdipole-dipole interactions.

Explanation:

Polarizability increases with increasingsizes of molecules and therefore with increas-ing numbers of electrons. Therefore, Lon-don forces are generally stronger for moleculesthat are larger or have more electrons.

021 20.0 pointsThe 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. 21.6 g · mol−1

2. 50.8 g · mol−1

3. 273 g · mol−1

4. 162 g · mol−1 correct

5. 869 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 = n RTn

V=

P

R T

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

n

V· MM =

P

R T· MM = ρ ,

with units of g/L.

MM =ρ RT

P

=(1.14 g/L)

(


)

0.230263 atm× (398.15 K)

= 161.755 g/mol

022 10.0 pointsIf it takes 15 s for a certain sample of neonto effuse through a porous barrier, how muchtime will it take for the same amount of ni-trogen gas to effuse through the barrier underthe same conditions?

1. greater than 15 s correct

2. less than 15 s

3. 15 s

Explanation:

Rate of effusion ∝1

√MW

MWs for each gas:Ne : 20.179 g/mol N2 : 28.0134 g/mol

The heaviest gas will have particles movingat the lowest average speed so it will effusemore slowly.

023 15.0 pointsRank the molecules

CH3F, C2H6, H2O, H2, He

in terms of increasing viscosity.

1. None of the answers is correct.

2. He, H2, C2H6, CH3F, H2O correct


3. C2H6, H2O, CH3F, He, H2

4. H2O, CH3F, C2H6, H2, He

5. H2, He, CH3F, H2O, C2H6

Explanation:

The intermolecular attractions, rankedfrom weakest to strongest, are London forces(also called van der Waals forces or induceddipoles), dipole-dipole interactions, hydro-gen bonding interactions, and ion-ion interac-tions. In general, the larger the molecule, thegreater the total intermolecular forces. Strongintermolecular interactions cause molecules to“stick” to one another, and are thus moreviscous than molecules that experience weakintermolecular attractions.

H2O has two O H bonds which canhydrogen bond with neighboring watermolecules.

CH3F is polar but contains no O H ,N H , or F H bonds to participate inhydrogen bonding. Therefore the strongestintermolecular forces that CH3F experiencesare dipole-dipole interactions.

He, H2, and C2H6 are nonpolar and thusexperience only the weakest intermolecularforces, London forces.

024 20.0 pointsDrRuth says: First balance the reaction equa-tion.

For the reaction

? C6H6 + ? O2 → ? CO2 + ? H2O

41.4 grams of C6H6 are mixed with 142.5grams of O2 and allowed to react. How muchCO2 could be produced by this reaction?

1. 217 g

2. 140 g correct

3. 156.8 g

4. 125 g

5. 112 g

6. 100 g

7. 175 g

8. 294 g

9. 264.6 g

10. 186.2 g

Explanation:

mC6H6= 41.4 g mO2

= 142.5 gThe balanced equation is

2 C6H6 + 15 O2 → 12 CO2 + 6 H2O .

The molecular weight of C6H6 is 78.1118g/mol, giving 0.530009 mol C6H6.

The molecular weight of O2 is 31.9988g/mol, giving 4.45329 mol O2.

0.530009 mol C6H6 ×15 mol O2

2 mol C6H6

= 3.97507 mol O2,which is less than what is actually present.Therefore the limiting reactant must be C6H6.

The molecular weight of CO2 is 44.0095g/mol.

The limiting reactant (C6H6) will yield3.18006 mol CO2 which is equal to 140 gof CO2.

0.530009 mol C6H6 ×12 mol CO2

2 mol C6H6

×44.0095 g CO2

1 mol CO2

= 140 g CO2

025 15.0 pointsDrRuth is nice enough to tell you that sulfurdioxide is SO2 and sulfur trioxide is SO3, butthinks you should really be able to figure thatout for yourself.

Sulfur dioxide reacts with oxygen gas toproduce sulfur trioxide. How much oxygengas will react with 15.0 L of sulfur dioxide ifboth gases are at 101.3 kPa and 125◦C?

1. 0.230 mol correct

2. 0.459 mol

3. 0.919 mol


4. 0.731 mol

5. 0.115 mol

Explanation:

V = 15 L

P = (101.3 kPa)1 atm

101.325 kPa=

0.999753 atmT = 125◦C + 273.15 = 398.15 K

For the SO2, apply the ideal gas law:

P V = n RT

n =P V

RT

=(0.999753 atm)(15 L)

(


)

(398.15 K)

= 0.458993 mol SO2

The balanced equation is

2 SO2(g) + O2(g) → 2 SO3(g)

n = (0.458993 mol SO2)1 mol O2

2 mol SO2

= 0.229496 mol O2

026 10.0 pointsIf the temperature of a fixed amount of gas isincreased at constant pressure its volume will

1. Insufficient data to answer this question

2. remain the same.

3. increase. correct

4. decrease.

Explanation:

The volume of a gas is directly proportionalto its absolute temperature (Charles’s Law).This means that as the absolute temperatureof a gas is increased, its volume increasesproportionately.

027 10.0 pointsGas X has a larger value than Gas Y for the

van der Waals constant “a”. This indicatesthat

1. the molecules of X have stronger inter-molecular attractions for each other than themolecules of Y have for each other. correct

2. the molecules of gas X have a highervelocity than do the molecules of gas Y.

3. the molecules of X are larger than themolecules of Y.

4. the molecules of gas X repel other Xmolecules.

Explanation:

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