Version 001 – HW02-gas laws – vandenbout – …

Version 001 ? HW02-gas laws ? vandenbout ? (52130)

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This print-out should have 24 questions. Multiple-choice questions may continue on the next column or page ? find all choices before answering.

Mlib 76 2075 001 10.0 points A gas is enclosed in a 10.0 L tank at 1200 mm Hg pressure. Which of the following is a reasonable value for the pressure when the gas is pumped into a 5.00 L vessel?

1. 2400 mm Hg correct

2. 24 mm Hg

3. 0.042 mm Hg

4. 600 mm Hg

Explanation: V1 = 10.0 L P1 = 1200 mm Hg

V2 = 5.0 L

Boyle's law relates the volume and pressure of a sample of gas:

P1 V1 = P2 V2

P2

=

P1 V1 V2

=

(1200

mm Hg)(10.0 5L

L)

= 2400 mm Hg

Holt da 10 rev 22 003 10.0 points A flask containing 130 cm3 of hydrogen was collected under a pressure of 27.4 kPa. What pressure would have been required for the volume of the gas to have been 93 cm3, assuming the same temperature?

Correct answer: 38.3011 kPa.

Explanation: V1 = 130 cm3 P1 = 27.4 kPa

Applying Boyle's law,

V2 = 93 cm3 P2 = ?

P1 V1 = P2 V2

P2

=

P1 V1 V2

=

(27.4

kPa)(130 93 cm3

cm3)

= 38.3011 kPa

Gas Vol

004 10.0 points A 5.00-L sample of a gas exerts a pressure of 1040 torr at 50.0C. In what volume would the same sample exert a pressure of 1.00 atm at 50.0C?

1. 6.84 L correct

2. 1.95 L

Holt da 10 4 sample 3 002 10.0 points

A sample of gas in a closed container at a temperature of 75C and a pressure of 5 atm is heated to 293C. What pressure does the gas exert at the higher temperature?

Correct answer: 8.13218 atm.

Explanation:

T1 = 75C + 273 = 348 K

P1 = 5 atm

T2 = 293C + 273 = 566 K

P2 = ?

Applying the Gay-Lussac law,

P1 T1

=

P2 T2

P2

=

P1 T2 T1

=

(5

atm) (566 348 K

K)

= 8.13218 atm

3. 0.581 L

4. 3.33 L

5. 10.5 L

Explanation:

P1 = (1040 torr)

1 atm 760 torr

= 1.36842 atm

P2 = 1 atm

V1 = 5 L

Since the number of moles and the temper-

ature of the gas sample are held constant, the

change in the volume is determined by Boyle's

Law:

P1 V1 = P2 V2

V2

=

P1 V1 P2

=

(1.36842 atm)(5 1 atm

L)

= 6.84211 L

Version 001 ? HW02-gas laws ? vandenbout ? (52130)

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Brodbelt 430abc 005 10.0 points Consider the following reaction of Al with HCl to produce hydrogen gas.

A sample of nitrogen gas is contained in a piston with a freely moving cylinder. At 0 C, the volume of the gas is 317 mL. To what temperature must the gas be heated to occupy a volume of 501 mL?

2 Al + 6 HCl 2 AlCl3 + 3 H2

This reaction has a yield of 82.5 percent. How many moles of HCl are needed to produce 14.0 L of H2 at 351 K and 1.11 atm?

1. 1.08 mol

2. 0.446 mol

3. 0.655 mol

4. 0.540 mol

5. 0.890 mol

6. 0.223 mol

7. 1.31 mol correct

Explanation: V = 14 L P = 1.11 atm

T = 351 K p = 82.5%

P V = nRT

n

=

PV RT

=

(1.11 (0.0821

atm)(14 L)

L?atm mol?K

)(351

K)

= 0.539263 mol H2

For 82.5% yield

nH2

=

0.539263 mol 0.825

H2

=

0.653652

mol

H2

nH2

=

(0.653652

mol

H2)

6 mol HCl 3 mol H2

= 1.3073 mol HCl

Correct answer: 158.461C.

Explanation: V1 = 317 mL T1 = 0 C + 273 = 273 K

Applying Charles' Law,

V2 = 501 mL T2 = ?

V1 T1

=

V2 T2

T2

=

T1 V2 V1

=

(273 K)(501 mL) 317 mL

= 431.461 K

C = K - 273 = 158.461C

Mlib 04 1093 007 10.0 points If I have 44.8 liters of nitrogen gas at standard temperature and pressure, how much will it weigh?

1. 56 g correct

2. 28 kg

3. 28 g

4. 44.8 g

5. 56 u

Explanation: P = 1 atm T = 273 K

V = 44.8 L

P V = nRT

n

=

PV RT

n=

(1 atm)(44.8 L)

28 g

0.08206

L?atm K?mol

? (273 K) 1 mol

= 55.9941 g

Holt da 10 3 sample 2 006 10.0 points

ChemPrin3e T04 33 008 10.0 points

Version 001 ? HW02-gas laws ? vandenbout ? (52130)

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At 80.0C and 12.0 Torr, the density of camphor vapor is 0.0829 g ? L-1. What is the molar mass of camphor?

1. 243 g ? mol-1

2. 3490 g ? mol-1

3. 152 g ? mol-1 correct

4. 20.3 g ? mol-1

5. 34.5 g ? mol-1

Explanation:

T = 80C + 273.15 K = 353.15 K

P

=

(12

Torr)

1 atm 760 Torr

=

0.0157895

atm

= 0.0829 g/L

The ideal gas law is

P V = nRT

n 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

=

,

with units of g/L.

MM

=

RT P

=

(0.0829 g/L) 0.08206 0.0157895 atm

L?atm mol?K

? (353.15 K)

= 152.152 g/mol

5. 4.00 g/L Explanation:

Lyon end quiz

010 (part 1 of 3) 10.0 points A chemist has synthesized a greenish-yellow gaseous compound that contains only chlorine and oxygen and has a density of 7.71 g/L at 36.0 degrees Celsius and 2188.8 mm Hg. What is the molar mass of the compound?

1. 25.8 g/mol

2. 86.9 g/mol

3. 51.5 g/mol

4. 67.9 g/mol correct

5. 109 g/mol

Explanation:

T = 36C + 273 = 309 K

P

=

2188.8

mm

Hg

atm 760 mm

Hg

=

2.88

atm

P V = nRT

n V

=

P RT

=

(2.88 atm)

(309

K)(0.08206

L?atm molK

)

=

0.1136

mol L

MW =

7.71 g/L

0.1136

mol L

= 67.87 g/mol

Lyon e5 q4a 009 10.0 points What is the density of nitrogen gas at STP?

011 (part 2 of 3) 10.0 points What is the molecular formula of the compound?

1. 2.50 g/L 2. 1.25 g/L correct 3. 0.500 g/L

1. ClO4 2. ClO3 3. ClO

4. 0.625 g/L

4. ClO2 correct

Version 001 ? HW02-gas laws ? vandenbout ? (52130)

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5. ClO5 Explanation:

MWClO2 = (35.5) + 2(16) = 67.5 g/mol

012 (part 3 of 3) 10.0 points How many moles of this gaseous compound are there in 15 L at STP?

VCH4 = 10 L

VO2 = 20 L

Avogadro's Principle tells us that because

P and T are the same, V n, so we can

work with the volume instead of the number

of moles. From the equation above, we need 2

times more L of O2 than that of CH4, and we

DO have this! Both reagents are present in

stoichiometric amounts. We will have no left

over reagents.

Find out how much CO2 is made based on

the L of O2:

1. 1.0 moles

2. 5.0 moles

3. 0.67 moles correct

4. 0.50 moles

5. 3.0 moles

Explanation:

At STP, 1 mole of an ideal gas has a volume

of

22.4

L;

i.e.,

22.4 L mol

.

(15

L)

mol 22.4 L

=

0.67

mol

Mlib 04 1147 013 10.0 points What is the final volume if 10 L methane reacts completely with 20 L oxygen

V = (20 L O2) ?

1 L CO2 2 L O2

= 10 L CO2

Note we could have just as easily done this calculation with the volume of CH4, because both reagents are fully used up in the reaction. We also assume the volume of the liquid water formed is negligible.

Msci 02 1216 014 10.0 points Calculate the volume of methane (CH4) produced by the bacterial breakdown of 0.65 kg of sugar (C6H12O6)

C6H12O6 3 CH4 + 3 CO2

at 295 K and 800 torr.

1. 455 L

2. 149 L

CH4(g) + 2 O2(g) CO2(g) + 2 H2O() at 100C and 2 atmospheres?

1. 30 L 2. 20 L 3. 15 L 4. 10 L correct 5. Cannot be determined from the information given. Explanation:

3. 252 L

4. 3036 L

5. 249 L correct

Explanation:

n

=

0.65

kg

C6H12O6

?

1000 g 1 kg

mol C6H12O6

3 mol CH4

?

?

180 g C6H12O6 1 mol C6H12O6

= 10.8333 mol CH4

T = 295 K

P

=

800

torr

?

1 atm 760 torr

=

1.05263

atm

Version 001 ? HW02-gas laws ? vandenbout ? (52130)

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Applying the ideal gas law equation,

P V = nRT

V

=

nRT P

V = (10.8333 mol)

0.08206

L?atm mol?K

(295 K)

1.05263 atm

= 249.138 L

1. 4.48 L 2. 2.24 L correct 3. 3.36 L 4. 6.72 L

Msci 12 1505 015 10.0 points If the reaction

N2(g) + 3 H2(g) 2 NH3(g)

is carried out at constant temperature and pressure, how much H2 is required to react with 6.1 L liters of N2?

1. 9.15 L

2. 18.3 L correct

3. 24.4 L

5. 1.12 L

Explanation:

nO2

=

0.1

mol

Ag2O

?

1 mol O2 2 mol Ag2O

= 0.05 mol O2

P = 1 atm

T = 546 K

PV = nRT

V

=

nRT P

=

(0.05)

(0.08206) 1

(546

K)

= 2.24024 L

4. Cannot be determined.

5. 12.2 L

Explanation: VN2 = 6.1 L nN2 = 1 mol

P V = n R T , so n V , and

nH2 = 3 mol

nCH4 nCO2

=

VCH4 VCO2

VCO2

=

VCH4 nCO2 nCH4

=

(6.1

L N2)(3 mol 1 mol N2

H2)

= 18.3 L H2

Msci 12 1514 016 10.0 points What volume of pure oxygen gas (O2) measured at 546 K and 1.00 atm is formed by complete dissociation of 0.1 mol of Ag2O?

2 Ag2O(s) 4 Ag(s) + O2(g)

LDE Ideal Gas Calculation 001 017 10.0 points

If we increase the volume of a gaseous system by a factor of 3 and raise the temperature by a factor of 6, then the pressure of the system will (increase/decrease) by a factor of (2/18):

1. decrease, 18

2. increase, 18

3. increase, 2 correct

4. decrease, 2

Explanation: Tripling the volume will decrease the pres-

sure by a factor of 3 and sextupling the temperature will increase the pressure by a factor of 6, resulting a double the original pressure.

DVB Avagadro's Gas Law 018 10.0 points

You have a sample of H2 gas and Ar gas at

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