# Chemistry B CFA 1 Review Quiz - Trimester 3 2014

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This is the review quiz for the Chemistry B CFA 1. Please take and review answers at the end. You may take this quiz unlimited amount of times until a passing score is achieved. You will be given a test grade for this quiz; therefore please retake until you reach a score of 90% or above.

• 1.

### Molar Mass of glucose (CO2) in grams per mole? (Hint: Use periodic table and add molecular weights of the atoms)

• A.

340.0

• B.

180.0

• C.

110.0

• D.

78.0

• E.

44.0

E. 44.0
Explanation
Glucose has the chemical formula C6H12O6. To calculate the molar mass of glucose, we need to add up the atomic masses of its constituent atoms. The atomic mass of carbon (C) is 12.01 g/mol, hydrogen (H) is 1.01 g/mol, and oxygen (O) is 16.00 g/mol. Since there are 6 carbon atoms, 12 hydrogen atoms, and 6 oxygen atoms in one molecule of glucose, the molar mass can be calculated as follows: (6 * 12.01) + (12 * 1.01) + (6 * 16.00) = 180.18 g/mol. Rounded to the nearest tenth, the molar mass of glucose is 180.0 g/mol. Therefore, the answer choice of 44.0 g/mol is incorrect.

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• 2.

### What is the molar mass of H2 in grams per mole (Hint: Use periodic table and add molecular weights of the atoms)

• A.

15.0

• B.

2.0

• C.

4.0

• D.

5.2

• E.

3.5

B. 2.0
Explanation
The molar mass of H2 in grams per mole is 2.0. This is because the molar mass of hydrogen (H) is approximately 1.0 g/mol, and since there are two hydrogen atoms in H2, the molar mass of H2 is 2.0 g/mol.

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• 3.

### What is the molar mass of NO2  in grams per mole. (Hint: Use periodic table and add molecular weights of the atoms)

• A.

16.2

• B.

46.0

• C.

44.0

• D.

13.1

• E.

None of the above

B. 46.0
Explanation
The molar mass of NO2 can be calculated by adding the atomic masses of nitrogen and oxygen. The atomic mass of nitrogen is 14.01 g/mol, and the atomic mass of oxygen is 16.00 g/mol. Therefore, the molar mass of NO2 is 14.01 + (2 * 16.00) = 46.01 g/mol.

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• 4.

### What is the molar mass of CH4. (Hint: Use periodic table and add molecular weights of the atoms)

• A.

16 grams per mole

• B.

20 grams per mole

• C.

24 grams per mole

• D.

30 grams per mole

• E.

14 grams per mole

A. 16 grams per mole
Explanation
The molar mass of CH4 can be calculated by adding the atomic masses of carbon (C) and hydrogen (H). The atomic mass of carbon is 12 grams per mole, and the atomic mass of hydrogen is 1 gram per mole. Since there is one carbon atom and four hydrogen atoms in CH4, the molar mass can be calculated as follows: (1 * 12) + (4 * 1) = 16 grams per mole.

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• 5.

### A gas in a balloon measures 2.2 L 1.1 atmospheres and 32 degrees Celsius. The environment is changed and the balloon is now 1.8 L and 1.4 atmospheres. What is the new temperature? (Hint 1: Always change temperature to Kelvin scale. To change from Celsius to Kelvin: add 273) Hint 2: Pressure volume and temperature are all changed. Use the combined gas law:

• A.

317.6 K

• B.

33.32 K

• C.

384.3 K

• D.

0.02 K

A. 317.6 K
Explanation
The correct answer is 317.6 K. The combined gas law states that the product of pressure and volume is directly proportional to the temperature. By rearranging the formula and plugging in the given values, we can solve for the new temperature. The initial temperature is given as 32 degrees Celsius, which is equivalent to 305.15 K. Using the formula, we can calculate the new temperature to be 317.6 K.

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• 6.

### A gas at a temperature of 80Celsius and a volume of 3.2 Liters is brought into an area with the temperature of 100 Celsius, what is the new volume if the pressure remains constant? (Hint : ALL GAS PROBLEMS MUST use the KELVIN temperature scale.) (To change from Celsius to Kelvin: add 273)

• A.

3.38 L

• B.

2.84 L

• C.

3.75 L

• D.

2.4 L

A. 3.38 L
Explanation
(3L)/(353K) = V2/(373K)
Math: 3 x 373 divided by 353.

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• 7.

### If pressure of a gas is 1.2 atmospheres with a volume of 12 L, how much pressure is exerted by the gas if the volume is changed to 9L? (Assume constant temperature)

• A.

.9 L

• B.

129.6 L

• C.

6.0 L

• D.

1.6 L

D. 1.6 L
Explanation
If the variables are only Pressure and Volume, (with constant temperature), this indicates Boyle's Law:
P1V1 = P2 V2 (Where 1 indicates the first set of conditions and 2 indicates the second set of conditions).
(1.2 atm) (12L) = (P2) (9L)
Math: 1.2 times 12 divided by 9.
You may also use the combined gas law and eliminate the T1 and T2

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• 8.

### 1. According to the Law of Conservation of Matter, the total mass of all the reacting substance is—

• A.

A. more than the mass of all the products.

• B.

B. more than the total mass of all the products and the reactants.

• C.

C. less than the total mass of all the products.

• D.

D. equal to the mass of all the products.

D. D. equal to the mass of all the products.
Explanation
Law of conservation of mass states that the mass of products will equal mass of reactants. Therefore mass is conserved.

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• 9.

### 2. What are the coefficients that would correctly balance this reaction? ____Ni + ____C4H8N2O2 → ____Ni(C4H8N2O2)2

• A.

F. 1, 2, 1

• B.

G. 2, 2, 1

• C.

H. 2, 1, 1

• D.

J. 2, 2, 2

A. F. 1, 2, 1
Explanation
The coefficients that would correctly balance this reaction are 1, 2, 1. This means that 1 molecule of Nickel (Ni) reacts with 2 molecules of C4H8N2O2 to form 1 molecule of Ni(C4H8N2O2)2.

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• 10.

### 3. For the reaction HCl + NaOH --> NaCl + H2O, which reactant is the limiting reactant given 100.0 g of sodium hydroxide and 100.0 g of hydrochloric acid?

• A.

A. hydrochloric acid

• B.

B. sodium hydroxide

• C.

C. sodium chloride

• D.

D. water

B. B. sodium hydroxide
Explanation
NaOH has the higher molar mass; therefore if all things are equal NaOH will be the limiting reagent.

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• 11.

### 4. In the reaction 2RbNO3 → 2RbNO2 + O2, how many moles of O2 are produced when 5.0 mol of RbNO3 decompose?

• A.

F. 1.0 mol

• B.

G. 2.5 mol

• C.

H. 3.0 mol

• D.

J. 7.5 mol

B. G. 2.5 mol
Explanation
In the given reaction, it is stated that 2 moles of RbNO3 decompose to produce 1 mole of O2. Therefore, if 5.0 moles of RbNO3 decompose, it will produce half of that amount of O2, which is 2.5 moles. Hence, the correct answer is G. 2.5 mol.

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• 12.

### 5. In the reaction 2 H2 + O2 → 2 H2O, how many moles of oxygen are required to fully react with 6.0 mol of hydrogen?

• A.

A. 3.0 mol

• B.

B. 6.0 mol

• C.

C. 12.0 mol

• D.

D. 18.0 mol

A. A. 3.0 mol
Explanation
In the balanced chemical equation, it shows that 2 moles of hydrogen react with 1 mole of oxygen to form 2 moles of water. Therefore, to fully react with 6.0 moles of hydrogen, we would need half the amount of oxygen, which is 3.0 moles.

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• 13.

### 6. In the reaction 2 H2 + O2 → 2 H2O, what is the mole ratio of hydrogen to water?

• A.

A. 2 : 2

• B.

B. 2 : 1

• C.

C. 1 : 2

• D.

D. 4 : 4

A. A. 2 : 2
Explanation
The mole ratio of hydrogen to water in the given reaction is 2:2. This means that for every 2 moles of hydrogen, 2 moles of water are produced.

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• 14.

### 7. In the reaction N2 + 3 H2 → 2 NH3, what is the mole ratio of hydrogen to nitrogen?

• A.

F. 1 : 1

• B.

G. 1 : 2

• C.

H. 3 : 1

• D.

J. 3 : 2

C. H. 3 : 1
Explanation
The mole ratio of hydrogen to nitrogen in the reaction N2 + 3 H2 -> 2 NH3 is 3:1. This means that for every 3 moles of hydrogen, there is 1 mole of nitrogen involved in the reaction.

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• 15.

### 8. A student burns 1.50 mol C3H8 according to the following reaction:          C3H8 + 5O2 → 3CO2 + 4H2O How many grams of carbon dioxide are produced?

• A.

A. 4.40 × 101 g

• B.

B. 6.61 × 101 g

• C.

C. 1.32 × 102 g

• D.

D. 1.98 × 102 g

D. D. 1.98 × 102 g
Explanation
When 1.50 mol of C3H8 is burned, according to the balanced equation, 3 mol of CO2 is produced. To find the mass of CO2 produced, we need to use the molar mass of CO2, which is 44.01 g/mol.

First, we calculate the number of moles of CO2 produced:
1.50 mol C3H8 * (3 mol CO2 / 1 mol C3H8) = 4.50 mol CO2

Then, we calculate the mass of CO2 produced:
4.50 mol CO2 * 44.01 g/mol = 198.045 g

Rounding to the appropriate number of significant figures, the answer is 1.98 × 102 g.

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• 16.

### 9. A three (3.00) mole sample of K2S would have how many atoms of potassium?

• A.

F. 2.01 x 1023

• B.

G. 1.21 x 1024

• C.

H. 1.81 x 1024

• D.

J. 3.62 x 1024

D. J. 3.62 x 1024
Explanation
A mole is a unit of measurement in chemistry that represents a certain number of particles, which is approximately 6.022 x 10^23 particles. In this case, we have a three mole sample of K2S, which means we have 3 times the Avogadro's number of K2S particles. Each K2S molecule contains 2 potassium atoms, so we can multiply the number of K2S particles by 2 to find the number of potassium atoms. Therefore, the correct answer is 3.62 x 10^24 atoms of potassium.

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• 17.

### 11. What is the percent of hydrogen in methane, CH4?

• A.

F. 16.0

• B.

G. 25.1

• C.

H. 74.9

• D.

J. 100.0

B. G. 25.1
Explanation
Methane, CH4, consists of one carbon atom and four hydrogen atoms. To find the percent of hydrogen in methane, we need to calculate the mass of hydrogen and divide it by the total mass of methane. The molar mass of hydrogen is 1 g/mol and the molar mass of methane is 16 g/mol. Therefore, the mass of hydrogen in methane is 4 g/mol. To find the percent, we divide the mass of hydrogen by the total mass of methane and multiply by 100: (4 g/mol / 16 g/mol) * 100 = 25%. Therefore, the correct answer is 25.1%.

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• 18.

### 12. Which of the following is the mass in grams of 4.25 × 10³ mol of N2?

• A.

A. 2.35 × 10-4 g

• B.

B. 1.52 × 102 g

• C.

C. 5.95× 104 g

• D.

D. 1.19 × 105 g

C. C. 5.95× 104 g
Explanation
This question is asking for the mass in grams of 4.25 x 10^3 mol of N2. To find the mass, we can use the molar mass of N2, which is 28.02 g/mol. We can then multiply the molar mass by the number of moles to get the mass in grams. Therefore, the correct answer is C. 5.95 x 10^4 g.

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• 19.

### 13. What is the new pressure of 150 mL of a gas that is compressed to 50 mL when the original pressure was 3.0 atm and the temperature is held constant?

• A.

F. 100 atm

• B.

G. 0.90 atm

• C.

H. 1.0 atm

• D.

J. 9.0 atm

D. J. 9.0 atm
Explanation
When a gas is compressed, its volume decreases while the pressure increases. According to Boyle's Law, the product of pressure and volume is constant when the temperature is held constant. In this case, the initial pressure is 3.0 atm and the initial volume is 150 mL. The final volume is 50 mL. Using the equation P1V1 = P2V2, we can calculate the final pressure. P1 = 3.0 atm, V1 = 150 mL, V2 = 50 mL. P2 = (P1 * V1) / V2 = (3.0 atm * 150 mL) / 50 mL = 9.0 atm. Therefore, the new pressure is 9.0 atm.

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• 20.

### 13. What is the new pressure of 150 mL of a gas that is compressed to 50 mL when the original pressure was 3.0 atm and the temperature is held constant?

• A.

F. 100 atm

• B.

G. 0.90 atm

• C.

H. 1.0 atm

• D.

J. 9.0 atm

D. J. 9.0 atm
Explanation
When a gas is compressed, its volume decreases while the pressure increases. According to Boyle's Law, the product of the initial pressure and volume is equal to the product of the final pressure and volume when the temperature is held constant. In this case, the initial pressure is 3.0 atm and the initial volume is 150 mL. The final volume is 50 mL. Using the formula P1V1 = P2V2, we can calculate the final pressure. P1 = 3.0 atm, V1 = 150 mL, V2 = 50 mL. Solving for P2, we get P2 = (P1 x V1) / V2 = (3.0 atm x 150 mL) / 50 mL = 9.0 atm. Therefore, the new pressure is 9.0 atm.

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• 21.

### 14. If the temperature of a gas changes, but the number of moles and volume remain constant, which of the following values must also change?

• A.

A. density

• B.

B. pressure

• C.

C. number of atoms

• D.

D. mass

B. B. pressure
Explanation
When the temperature of a gas changes, but the number of moles and volume remain constant, the pressure of the gas must also change. This is because according to the ideal gas law, pressure is directly proportional to temperature. As the temperature increases, the kinetic energy of the gas particles increases, causing them to collide more frequently and with greater force against the walls of the container, resulting in an increase in pressure. Conversely, if the temperature decreases, the kinetic energy and frequency of collisions decrease, leading to a decrease in pressure.

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• 22.

### 15. A sample of a gas in a rigid container at 30.0°C and 2.00 atm has its temperature increased to 40.0°C. What will be the new pressure?

• A.

F. 1.03 atm

• B.

G. 1.94 atm

• C.

H. 2.07 atm

• D.

J. 6.04 atm

C. H. 2.07 atm
Explanation
When the temperature of a gas increases, its pressure also increases, assuming the volume and amount of gas remain constant. This relationship is known as Charles's Law. In this question, the initial pressure is 2.00 atm and the initial temperature is 30.0°C. The final temperature is 40.0°C. Using Charles's Law, we can calculate the new pressure by using the formula: P1/T1 = P2/T2. Plugging in the values, we get (2.00 atm)/(30.0°C) = P2/(40.0°C). Solving for P2, we find that the new pressure is approximately 2.07 atm.

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• 23.

### 12. Calculate the approximate volume of a 1.50 mol sample of gas at 15.0oC and a pressure of 3.75 atm.

• A.

A. 1.73 L

• B.

B. 5.93 L

• C.

C. 8.65 L

• D.

D. 9.46 L

D. D. 9.46 L
Explanation
The volume of a gas can be calculated using the ideal gas law equation: PV = nRT. In this equation, P represents the pressure, V represents the volume, n represents the number of moles of gas, R is the ideal gas constant, and T represents the temperature in Kelvin. To solve for V, we rearrange the equation to V = (nRT) / P. Plugging in the given values of n = 1.50 mol, P = 3.75 atm, and T = 15.0oC + 273.15 (to convert to Kelvin), we can calculate the approximate volume to be 9.46 L. Therefore, the correct answer is D. 9.46 L.

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• 24.

### 13. The volume of a gas is 50.0 mL at 20.0 K. What will be the new temperature if the gas is compressed to 10.0 mL under constant pressure?

• A.

F. 2.50 K

• B.

G. 3.00 K

• C.

H. 4.00 K

• D.

J. 5.00 K

C. H. 4.00 K
Explanation
According to Charles's Law, the volume of a gas is directly proportional to its temperature when pressure is held constant. In this case, the volume is halved from 50.0 mL to 10.0 mL, so the temperature should also be halved from 20.0 K to 10.0 K. However, since the options only include whole numbers, the closest option is 4.00 K.

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• Current Version
• Mar 20, 2023
Quiz Edited by
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• Apr 02, 2014
Quiz Created by
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