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Chapter 11 - Practice Test, Lecture notes of Law

Continental Theological SeminaryLaw

Chapter 11. Practice Test - Answers. Kinetic Molecular Theory. 1. List the five postulates of the Kinetic-Molecular Theory. Name the Gas Law.

Typology: Lecture notes

2021/2022

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Revised: 2021-03-18
Answers: 12) a. 0.295 mol, b. 0.595 g 13) 3.29 atm 14) 3760 K 15) 36.2 L 16) 5040 L, 3.47x10-4 g/L 17) 24.8 L 18) 5.00 atm
Name:
Date: Per:
Chapter 11
Practice Test - Answers
Kinetic Molecular Theory
1. List the five postulates of the Kinetic-Molecular Theory.
Name the Gas Law
2. A potato chip bag pops when taken up in the mountains.
3. A balloon put in the freezer shrinks.
4. A lighter gas moves faster than a heavier gas.
5. The pressure of two gases is the sum of their partial pressures.
Calculations, Etc.
6. Equal amounts of gas at the same temperature and pressure have the same .
7. If the temperature of a sample of gas is halved at constant volume, the pressure will be .
8. The temperature at which matter stops moving is called .
9. The values of standard temperature are .
10. The values of standard pressure (in atm & mmHg) are .
11. Derive the value of R for pressure in atmospheres. (Must show work)
How would this process be different if you were to calculate the value of R for mmHg?
12. A sample of hydrogen gas has a volume of 4.40 L at a temperature of 145 °C and a pressure of 2.30 atm.
a) How many moles are in the sample?
V = 4.40 L
PV = nRT
T = 145 ℃ (418 K)
P = 2.30 atm
n =
PV
=
(2.30 atm)(4.40 L)
= 0.2948 mol 0.295 mol H2
n = ?
RT
(0.0821 (atm•L)/(mol•K))(418 K)
R = 0.0821 (atm•L)/(mol•K)
b) What is the mass of the sample?
0.2948 mol H2
2.018 g H2
= 0.5943 g 0.594 g H2
1 mol H2
13. A sample of gas measures 5.00 liters at 2.30 atmospheres of pressure. To change the volume to 3.50 liters at constant
temperature, what pressure must be applied?
V1 = 5.00 L
P1V1 = P2V2
P1 = 2.30 atm
V2 = 3.50 L
P2 =
P1V1
=
(2.30 atm)(5.00 L)
= 3.2857 atm 3.29 atm
P2 = ?
V2
(3.50 L)
pf2

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Download Chapter 11 - Practice Test and more Lecture notes Law in PDF only on Docsity!

Revised: 2021 - 03 - 18 Answers: 1 2 ) a. 0.295 mol, b. 0.595 g 13) 3.29 atm 14 ) 3760 K 15 ) 36.2 L 16 ) 5040 L, 3.47x10-^4 g/L 17 ) 24.8 L 18 ) 5.00 atm

Name:

Date: Per:

Chapter 1 1

Practice Test - Answers

Kinetic Molecular Theory

  1. List the five postulates of the Kinetic-Molecular Theory. Name the Gas Law
  2. A potato chip bag pops when taken up in the mountains.
  3. A balloon put in the freezer shrinks.
  4. A lighter gas moves faster than a heavier gas.
  5. The pressure of two gases is the sum of their partial pressures. Calculations, Etc.
  6. Equal amounts of gas at the same temperature and pressure have the same.
  7. If the temperature of a sample of gas is halved at constant volume, the pressure will be.
  8. The temperature at which matter stops moving is called.
  9. The values of standard temperature are.
  10. The values of standard pressure (in atm & mmHg) are.
  11. Derive the value of R for pressure in atmospheres. (Must show work) How would this process be different if you were to calculate the value of R for mmHg?
  12. A sample of hydrogen gas has a volume of 4.40 L at a temperature of 145 °C and a pressure of 2.3 0 atm. a) How many moles are in the sample? V = 4.40 L T = 145 ℃ (418 K) PV = nRT P = 2.30 atm n =

PV

(2.30 atm)(4.40 L) = 0.29 48 mol  0.295 mol H 2 n =? RT (0.0821 (atm•L)/(mol•K))(418 K) R = 0.0821 (atm•L)/(mol•K) b) What is the mass of the sample? 0.29 4 8 mol H 2 2.018 g H 2 1 mol H 2 = 0.59^4 3 g^ ^ 0.594 g H^2

  1. A sample of gas measures 5.00 liters at 2.3 0 atmospheres of pressure. To change the volume to 3.5 0 liters at constant temperature, what pressure must be applied? V 1 = 5.00 L (^) P P^1 V^1 =^ P^2 V^2 1 =^ 2.30 atm V 2 = 3.50 L P 2 =

P 1 V 1

(2.30 atm)(5.00 L) = 3.2 857 atm  3.29 atm P 2 =? V 2 (3.50 L)

Revised: 2021 - 03 - 18 Answers: 1 2 ) a. 0.295 mol, b. 0.595 g 13) 3.29 atm 14 ) 3760 K 15 ) 36.2 L 16 ) 5040 L, 3.47x10-^4 g/L 17 ) 24.8 L 18 ) 5.00 atm

Name:

Date: Per:

Chapter 1 1

Practice Test - Answers

  1. A 2.5 0 L gas container is designed to hold gases with a pressure of up to 11000. mmHg. If a gas sample that has a pressure of 740. mmHg at - 20 .0 °C is placed in the container, at what temperature will the container burst? P 2 = 11000. mmHg T 1 P 2 = T 2 P 1 The volume of the container is not important in this case as it cannot expand P^ or contract.^ The^ volume will remain constant until the moment it bursts. 1 =^ 740. mmHg T 1 = − 20.0 ℃ (253 K) T 2 =

T 1 P 2

(253 K)(11000. mmHg) = 376 0.8 K  3760 K T 2 =? P 1 (740. mmHg)

  1. A quantity of gas has a volume of 23 .0 L at - 45 .0 °C and 1000. mmHg of pressure. If the conditions are changed to STP (STP = 273 K & 760. mmHg), what will the new volume be? V 1 = 23.0 L P 1 V 1 T 2 = P 2 V 2 T 1 T 1 = −45.0 ℃ ( 228 K) P 1 = 1000. mmHg V 2 =

P 1 V 1 T 2

(1000. mmHg)(23.0 L)(273 K) = 36. 23 L  36.2 L T 2 = 273 K P 2 T 1 ( 760. mmHg)( 228 K) P 2 = 76 0. mmHg V 2 =?

  1. A quantity of gas has a volume of 650. L at 65.0 °C and 730 0. mmHg of pressure. If the gas has a mass of 1. 75 g, what is the density of the gas at STP (STP = 273 K & 760. mmHg)? V 1 = 650. L P 1 V 1 T 2 = P 2 V 2 T 1 T 1 = 6 5.0 ℃ ( 338 K) P 1 = 73 00. mmHg V 2 =

P 1 V 1 T 2

( 7300. mmHg)(650. L)(273 K) = 5042 L T 2 = 273 K P 2 T 1 ( 760. mmHg)( 338 K) P 2 = 76 0. mmHg V 2 =? m = 1.75 g D = m = 1.75 g V = 50 4 2 L V 504 2 L = 3.4^7 0 x 10−^4 g/L^ ^ 3.47 x 10−^4 g/L

  1. Given the equation, ___ CuO(s) + ___ H 2 (g) → ___ Cu(s) + ___ H 2 O(g), how many liters of hydrogen are needed to react with 88.0 g of copper (II) oxide at STP (STP means we can use 1 mol of gas = 22.4 L)? Analyze, rewrite, balance equation 1 CuO(s) + 1 H 2 (g) → 1 Cu(s) + 1 H 2 O(g) 88.0 g? L Solve for only given 88.0 g CuO 1 mol CuO 1 mol H 2 22.4 L H 2 = 24. 78 L  24 .8 L H 2 79.545 CuO 1 mol CuO 1 mol H 2
  2. Two gases are combined in a 2.00 L container. If the first gas has a pressure of 1.50 atm at a volume of 4.00 L and the second gas has a pressure of 4.00 atm at a volume of 1.00 L, what is the pressure of the combined gases? First Gas Second Gas Combined Gases

V 1 = 4.00 L V 1 = 1.00 L V 2 = 2.00 L

P 1 = 1.50 atm P 1 = 4.00 atm PTotal =? P 2 = P 1 V 1 /V 2 P 2 = P 1 V 1 /V 2 P 2 = (1.50 atm)(4.00 L) P 2 = ( 4. 0 0 atm)( 1 .00 L) (2.00 L) (2.00 L) P 2 = 3.00 L + P 2 = 2.00 L = PTotal = 5.00 L