13.106 Hydrazine, N 2 H 4 , has been proposed as the fuel in a fuel cell in which oxygen is the oxidizing agent. The reactions are N 2 H 4 ( aq ) + 4 OH - ( aq ) → N 2 ( g ) + 4 H 2 O ( l ) + 4 e − O 2 ( g ) + 2 H 2 O ( l ) + 4 e − → 4 OH - ( aq ) (a) Which reaction occurs at the anode and which at the cathode? (b) What is the net cell reaction? (C) If the cell is to produce 0.50 A of current for 50.0 h, what mass in grams of hydrazine must be present? (d) What mass in grams of O 2 must be available to react with the mass of N 2 H 4 determined in part (c)?

Question

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Answer 1

Textbook Question

Chapter 13, Problem 13.106PAE

13.106 Hydrazine, $N 2 H 4$ , has been proposed as the fuel in a fuel cell in which oxygen is the oxidizing agent. The reactions are

$N 2 H 4 ( aq ) + 4 OH - ( aq ) → N 2 ( g ) + 4 H 2 O ( l ) + 4 e − O 2 ( g ) + 2 H 2 O ( l ) + 4 e − → 4 OH - ( aq )$

(a) Which reaction occurs at the anode and which at the cathode?

(b) What is the net cell reaction?

(C) If the cell is to produce 0.50 A of current for 50.0 h, what mass in grams of hydrazine must be present?

(d) What mass in grams of $O 2$ must be available to react with the mass of $N 2 H 4$ determined in part (c)?

(b)

Expert Solution

Interpretation Introduction

Interpretation:

To identify the net reaction.

Concept introduction:

The net reaction is the total reaction including all reactants and products.
It is obtained via addition of the half reactions.

Answer to Problem 13.106PAE

Solution:

Chemistry for Engineering Students, Chapter 13, Problem 13.106PAE , additional homework tip 1

Explanation of Solution

In order to find out the reaction, we must include the reduction-oxidation reactions. They always take place together, since one donates electrons and the other receives them.

In this case, the oxidation reaction: hydrazine oxidation to nitrogen gas. The reduction is oxygen forming hydroxide ion via the addition of electron

Adding both reactions, getting rid of repeating/same units on the left/right side of the equation:

Chemistry for Engineering Students, Chapter 13, Problem 13.106PAE , additional homework tip 2

Therefore, the reaction taking place is:

Chemistry for Engineering Students, Chapter 13, Problem 13.106PAE , additional homework tip 3

(c)

Expert Solution

Interpretation Introduction

Interpretation:

To calculate the mass of hydrazine present in given amount of time and current.

Concept introduction:

A stands for ampere, the amount of current.
Time and current can be used to calculate total charge.
The total charge relates mol of e- and Faraday constant.

Answer to Problem 13.106PAE

Solution:

$7.46 g$

Explanation of Solution

This consist of the following steps:

1^s^t Step: Calculate the total amount of charge.

$t = 50.0 h = 50 h ×3600 s1 h = 180,000 sI = 0.50 A = 0.50 C/sq = I×t = 0.50 C/s × 180,000 s = 90,000 C1 mol e- =96500 Cx mol e- = 90000 Cx = 0.9326 mol e-$

2ⁿ^d Step: Calculate the total amount of moles of hydrazine

$4 mol e- = 1 mol N2H40.9326 mol e- = y mol N2H4y =0.2332 mol N2H4$

3^r^d Step: Calculate the mass of hydrazine

$mass N2H4 = mol × Molar Mass = 0.2332 mol × 32.0 g/mol = 7.46 g$

(d)

Expert Solution

Interpretation Introduction

Interpretation:

To calculate the mass oxygen required

Concept introduction:

Use stoichiometric ratios.

Answer to Problem 13.106PAE

Solution:

$mass O2 = 7.46 g$

Explanation of Solution

Relate the moles of oxygen and hydrazine:

1^s^t Step: Calculate total moles of oxygen required

The ratio is 1:1 since 1 mol of hydrazine requires 1 mol of Oxygen gas to react.

Chemistry for Engineering Students, Chapter 13, Problem 13.106PAE , additional homework tip 4

2ⁿ^d Step: Calculate the total mass of oxygen gas

$mass O2 = mol × Molar Mass = 0.2332 mol × 32.0 g/mol = 7.46 g$

The mass is exactly the same as hydrazine since they have 1:1 ratio and similar molar masses.

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Students have asked these similar questions

9:27 AM Tue Mar 4 ← Problem 64 of 15 #63% Submit Curved arrows are used to illustrate the flow of electrons. Using the provided starting and product structures, draw the curved electron-pushing arrows for the following reaction or mechanistic step(s). Be sure to account for all bond-breaking and bond-making steps. 0:0 0:0 :0: N. :0: :O :0: H H. :0: Select to Add Arrows O :0: H O :0: 0:0. S. H Select to Add Arrows S :0: :0: H H

Order the following organic reactions by relative rate. That is, select '1' next to the reaction that will have the fastest initial rate, select '2' next to the reaction that will have the next fastest initial rate, and so on. If two reactions will have very similar initial rates, you can select the same number next to both. If a reaction will have zero or nearly zero initial rate, don't select a number and check the box in the table instead. Note: the "Nu" in these reactions means "a generic nucleophile." ملی CI :Nu 2 он 3 H Reaction Relative Rate (Choose one) ▼ Nu :CI: zero or nearly zero Nu :Nu bi (Choose one) zero or nearly zero : Nu لی Nu :H (Choose one) zero or nearly zero

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Answer 2

Textbook Question

Chapter 13, Problem 13.106PAE

13.106 Hydrazine, $N 2 H 4$ , has been proposed as the fuel in a fuel cell in which oxygen is the oxidizing agent. The reactions are

$N 2 H 4 ( aq ) + 4 OH - ( aq ) → N 2 ( g ) + 4 H 2 O ( l ) + 4 e − O 2 ( g ) + 2 H 2 O ( l ) + 4 e − → 4 OH - ( aq )$

(a) Which reaction occurs at the anode and which at the cathode?

(b) What is the net cell reaction?

(C) If the cell is to produce 0.50 A of current for 50.0 h, what mass in grams of hydrazine must be present?

(d) What mass in grams of $O 2$ must be available to react with the mass of $N 2 H 4$ determined in part (c)?

(b)

Expert Solution

Interpretation Introduction

Interpretation:

To identify the net reaction.

Concept introduction:

The net reaction is the total reaction including all reactants and products.
It is obtained via addition of the half reactions.

Answer to Problem 13.106PAE

Solution:

Chemistry for Engineering Students, Chapter 13, Problem 13.106PAE , additional homework tip 1

Explanation of Solution

In order to find out the reaction, we must include the reduction-oxidation reactions. They always take place together, since one donates electrons and the other receives them.

In this case, the oxidation reaction: hydrazine oxidation to nitrogen gas. The reduction is oxygen forming hydroxide ion via the addition of electron

Adding both reactions, getting rid of repeating/same units on the left/right side of the equation:

Chemistry for Engineering Students, Chapter 13, Problem 13.106PAE , additional homework tip 2

Therefore, the reaction taking place is:

Chemistry for Engineering Students, Chapter 13, Problem 13.106PAE , additional homework tip 3

(c)

Expert Solution

Interpretation Introduction

Interpretation:

To calculate the mass of hydrazine present in given amount of time and current.

Concept introduction:

A stands for ampere, the amount of current.
Time and current can be used to calculate total charge.
The total charge relates mol of e- and Faraday constant.

Answer to Problem 13.106PAE

Solution:

$7.46 g$

Explanation of Solution

This consist of the following steps:

1^s^t Step: Calculate the total amount of charge.

$t = 50.0 h = 50 h ×3600 s1 h = 180,000 sI = 0.50 A = 0.50 C/sq = I×t = 0.50 C/s × 180,000 s = 90,000 C1 mol e- =96500 Cx mol e- = 90000 Cx = 0.9326 mol e-$

2ⁿ^d Step: Calculate the total amount of moles of hydrazine

$4 mol e- = 1 mol N2H40.9326 mol e- = y mol N2H4y =0.2332 mol N2H4$

3^r^d Step: Calculate the mass of hydrazine

$mass N2H4 = mol × Molar Mass = 0.2332 mol × 32.0 g/mol = 7.46 g$

(d)

Expert Solution

Interpretation Introduction

Interpretation:

To calculate the mass oxygen required

Concept introduction:

Use stoichiometric ratios.

Answer to Problem 13.106PAE

Solution:

$mass O2 = 7.46 g$

Explanation of Solution

Relate the moles of oxygen and hydrazine:

1^s^t Step: Calculate total moles of oxygen required

The ratio is 1:1 since 1 mol of hydrazine requires 1 mol of Oxygen gas to react.

Chemistry for Engineering Students, Chapter 13, Problem 13.106PAE , additional homework tip 4

2ⁿ^d Step: Calculate the total mass of oxygen gas

$mass O2 = mol × Molar Mass = 0.2332 mol × 32.0 g/mol = 7.46 g$

The mass is exactly the same as hydrazine since they have 1:1 ratio and similar molar masses.

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Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 3

Textbook Question

Chapter 13, Problem 13.106PAE

13.106 Hydrazine, $N 2 H 4$ , has been proposed as the fuel in a fuel cell in which oxygen is the oxidizing agent. The reactions are

$N 2 H 4 ( aq ) + 4 OH - ( aq ) → N 2 ( g ) + 4 H 2 O ( l ) + 4 e − O 2 ( g ) + 2 H 2 O ( l ) + 4 e − → 4 OH - ( aq )$

(a) Which reaction occurs at the anode and which at the cathode?

(b) What is the net cell reaction?

(C) If the cell is to produce 0.50 A of current for 50.0 h, what mass in grams of hydrazine must be present?

(d) What mass in grams of $O 2$ must be available to react with the mass of $N 2 H 4$ determined in part (c)?

(b)

Expert Solution

Interpretation Introduction

Interpretation:

To identify the net reaction.

Concept introduction:

The net reaction is the total reaction including all reactants and products.
It is obtained via addition of the half reactions.

Answer to Problem 13.106PAE

Solution:

Chemistry for Engineering Students, Chapter 13, Problem 13.106PAE , additional homework tip 1

Explanation of Solution

In order to find out the reaction, we must include the reduction-oxidation reactions. They always take place together, since one donates electrons and the other receives them.

In this case, the oxidation reaction: hydrazine oxidation to nitrogen gas. The reduction is oxygen forming hydroxide ion via the addition of electron

Adding both reactions, getting rid of repeating/same units on the left/right side of the equation:

Chemistry for Engineering Students, Chapter 13, Problem 13.106PAE , additional homework tip 2

Therefore, the reaction taking place is:

Chemistry for Engineering Students, Chapter 13, Problem 13.106PAE , additional homework tip 3

(c)

Expert Solution

Interpretation Introduction

Interpretation:

To calculate the mass of hydrazine present in given amount of time and current.

Concept introduction:

A stands for ampere, the amount of current.
Time and current can be used to calculate total charge.
The total charge relates mol of e- and Faraday constant.

Answer to Problem 13.106PAE

Solution:

$7.46 g$

Explanation of Solution

This consist of the following steps:

1^s^t Step: Calculate the total amount of charge.

$t = 50.0 h = 50 h ×3600 s1 h = 180,000 sI = 0.50 A = 0.50 C/sq = I×t = 0.50 C/s × 180,000 s = 90,000 C1 mol e- =96500 Cx mol e- = 90000 Cx = 0.9326 mol e-$

2ⁿ^d Step: Calculate the total amount of moles of hydrazine

$4 mol e- = 1 mol N2H40.9326 mol e- = y mol N2H4y =0.2332 mol N2H4$

3^r^d Step: Calculate the mass of hydrazine

$mass N2H4 = mol × Molar Mass = 0.2332 mol × 32.0 g/mol = 7.46 g$

(d)

Expert Solution

Interpretation Introduction

Interpretation:

To calculate the mass oxygen required

Concept introduction:

Use stoichiometric ratios.

Answer to Problem 13.106PAE

Solution:

$mass O2 = 7.46 g$

Explanation of Solution

Relate the moles of oxygen and hydrazine:

1^s^t Step: Calculate total moles of oxygen required

The ratio is 1:1 since 1 mol of hydrazine requires 1 mol of Oxygen gas to react.

Chemistry for Engineering Students, Chapter 13, Problem 13.106PAE , additional homework tip 4

2ⁿ^d Step: Calculate the total mass of oxygen gas

$mass O2 = mol × Molar Mass = 0.2332 mol × 32.0 g/mol = 7.46 g$

The mass is exactly the same as hydrazine since they have 1:1 ratio and similar molar masses.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 4

Textbook Question

Chapter 13, Problem 13.106PAE

13.106 Hydrazine, $N 2 H 4$ , has been proposed as the fuel in a fuel cell in which oxygen is the oxidizing agent. The reactions are

$N 2 H 4 ( aq ) + 4 OH - ( aq ) → N 2 ( g ) + 4 H 2 O ( l ) + 4 e − O 2 ( g ) + 2 H 2 O ( l ) + 4 e − → 4 OH - ( aq )$

(a) Which reaction occurs at the anode and which at the cathode?

(b) What is the net cell reaction?

(C) If the cell is to produce 0.50 A of current for 50.0 h, what mass in grams of hydrazine must be present?

(d) What mass in grams of $O 2$ must be available to react with the mass of $N 2 H 4$ determined in part (c)?

(b)

Expert Solution

Interpretation Introduction

Interpretation:

To identify the net reaction.

Concept introduction:

The net reaction is the total reaction including all reactants and products.
It is obtained via addition of the half reactions.

Answer to Problem 13.106PAE

Solution:

Chemistry for Engineering Students, Chapter 13, Problem 13.106PAE , additional homework tip 1

Explanation of Solution

In order to find out the reaction, we must include the reduction-oxidation reactions. They always take place together, since one donates electrons and the other receives them.

In this case, the oxidation reaction: hydrazine oxidation to nitrogen gas. The reduction is oxygen forming hydroxide ion via the addition of electron

Adding both reactions, getting rid of repeating/same units on the left/right side of the equation:

Chemistry for Engineering Students, Chapter 13, Problem 13.106PAE , additional homework tip 2

Therefore, the reaction taking place is:

Chemistry for Engineering Students, Chapter 13, Problem 13.106PAE , additional homework tip 3

(c)

Expert Solution

Interpretation Introduction

Interpretation:

To calculate the mass of hydrazine present in given amount of time and current.

Concept introduction:

A stands for ampere, the amount of current.
Time and current can be used to calculate total charge.
The total charge relates mol of e- and Faraday constant.

Answer to Problem 13.106PAE

Solution:

$7.46 g$

Explanation of Solution

This consist of the following steps:

1^s^t Step: Calculate the total amount of charge.

$t = 50.0 h = 50 h ×3600 s1 h = 180,000 sI = 0.50 A = 0.50 C/sq = I×t = 0.50 C/s × 180,000 s = 90,000 C1 mol e- =96500 Cx mol e- = 90000 Cx = 0.9326 mol e-$

2ⁿ^d Step: Calculate the total amount of moles of hydrazine

$4 mol e- = 1 mol N2H40.9326 mol e- = y mol N2H4y =0.2332 mol N2H4$

3^r^d Step: Calculate the mass of hydrazine

$mass N2H4 = mol × Molar Mass = 0.2332 mol × 32.0 g/mol = 7.46 g$

(d)

Expert Solution

Interpretation Introduction

Interpretation:

To calculate the mass oxygen required

Concept introduction:

Use stoichiometric ratios.

Answer to Problem 13.106PAE

Solution:

$mass O2 = 7.46 g$

Explanation of Solution

Relate the moles of oxygen and hydrazine:

1^s^t Step: Calculate total moles of oxygen required

The ratio is 1:1 since 1 mol of hydrazine requires 1 mol of Oxygen gas to react.

Chemistry for Engineering Students, Chapter 13, Problem 13.106PAE , additional homework tip 4

2ⁿ^d Step: Calculate the total mass of oxygen gas

$mass O2 = mol × Molar Mass = 0.2332 mol × 32.0 g/mol = 7.46 g$

The mass is exactly the same as hydrazine since they have 1:1 ratio and similar molar masses.

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Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

(b)

Expert Solution

Interpretation Introduction

Interpretation:

To identify the net reaction.

Concept introduction:

The net reaction is the total reaction including all reactants and products.
It is obtained via addition of the half reactions.

Answer to Problem 13.106PAE

Solution:

Chemistry for Engineering Students, Chapter 13, Problem 13.106PAE , additional homework tip 1

Explanation of Solution

In order to find out the reaction, we must include the reduction-oxidation reactions. They always take place together, since one donates electrons and the other receives them.

In this case, the oxidation reaction: hydrazine oxidation to nitrogen gas. The reduction is oxygen forming hydroxide ion via the addition of electron

Adding both reactions, getting rid of repeating/same units on the left/right side of the equation:

Chemistry for Engineering Students, Chapter 13, Problem 13.106PAE , additional homework tip 2

Therefore, the reaction taking place is:

Chemistry for Engineering Students, Chapter 13, Problem 13.106PAE , additional homework tip 3

(c)

Expert Solution

Interpretation Introduction

Interpretation:

To calculate the mass of hydrazine present in given amount of time and current.

Concept introduction:

A stands for ampere, the amount of current.
Time and current can be used to calculate total charge.
The total charge relates mol of e- and Faraday constant.

Answer to Problem 13.106PAE

Solution:

$7.46 g$

Explanation of Solution

This consist of the following steps:

1^s^t Step: Calculate the total amount of charge.

$t = 50.0 h = 50 h ×3600 s1 h = 180,000 sI = 0.50 A = 0.50 C/sq = I×t = 0.50 C/s × 180,000 s = 90,000 C1 mol e- =96500 Cx mol e- = 90000 Cx = 0.9326 mol e-$

2ⁿ^d Step: Calculate the total amount of moles of hydrazine

$4 mol e- = 1 mol N2H40.9326 mol e- = y mol N2H4y =0.2332 mol N2H4$

3^r^d Step: Calculate the mass of hydrazine

$mass N2H4 = mol × Molar Mass = 0.2332 mol × 32.0 g/mol = 7.46 g$

(d)

Expert Solution

Interpretation Introduction

Interpretation:

To calculate the mass oxygen required

Concept introduction:

Use stoichiometric ratios.

Answer to Problem 13.106PAE

Solution:

$mass O2 = 7.46 g$

Explanation of Solution

Relate the moles of oxygen and hydrazine:

1^s^t Step: Calculate total moles of oxygen required

The ratio is 1:1 since 1 mol of hydrazine requires 1 mol of Oxygen gas to react.

Chemistry for Engineering Students, Chapter 13, Problem 13.106PAE , additional homework tip 4

2ⁿ^d Step: Calculate the total mass of oxygen gas

$mass O2 = mol × Molar Mass = 0.2332 mol × 32.0 g/mol = 7.46 g$

The mass is exactly the same as hydrazine since they have 1:1 ratio and similar molar masses.

Answer 8

(b)

Expert Solution

Interpretation Introduction

Interpretation:

To identify the net reaction.

Concept introduction:

The net reaction is the total reaction including all reactants and products.
It is obtained via addition of the half reactions.

Answer to Problem 13.106PAE

Solution:

Chemistry for Engineering Students, Chapter 13, Problem 13.106PAE , additional homework tip 1

Explanation of Solution

In order to find out the reaction, we must include the reduction-oxidation reactions. They always take place together, since one donates electrons and the other receives them.

In this case, the oxidation reaction: hydrazine oxidation to nitrogen gas. The reduction is oxygen forming hydroxide ion via the addition of electron

Adding both reactions, getting rid of repeating/same units on the left/right side of the equation:

Chemistry for Engineering Students, Chapter 13, Problem 13.106PAE , additional homework tip 2

Therefore, the reaction taking place is:

Chemistry for Engineering Students, Chapter 13, Problem 13.106PAE , additional homework tip 3

Answer 9

(b)

Expert Solution

Answer 10

(b)

Expert Solution

Answer 11

Expert Solution

Answer 12

Interpretation Introduction

Interpretation:

To identify the net reaction.

Concept introduction:

The net reaction is the total reaction including all reactants and products.
It is obtained via addition of the half reactions.

Answer 13

Answer to Problem 13.106PAE

Solution:

Chemistry for Engineering Students, Chapter 13, Problem 13.106PAE , additional homework tip 1

Answer 14

Explanation of Solution

In order to find out the reaction, we must include the reduction-oxidation reactions. They always take place together, since one donates electrons and the other receives them.

In this case, the oxidation reaction: hydrazine oxidation to nitrogen gas. The reduction is oxygen forming hydroxide ion via the addition of electron

Adding both reactions, getting rid of repeating/same units on the left/right side of the equation:

Chemistry for Engineering Students, Chapter 13, Problem 13.106PAE , additional homework tip 2

Therefore, the reaction taking place is:

Chemistry for Engineering Students, Chapter 13, Problem 13.106PAE , additional homework tip 3

Answer 15

(c)

Expert Solution

Interpretation Introduction

Interpretation:

To calculate the mass of hydrazine present in given amount of time and current.

Concept introduction:

A stands for ampere, the amount of current.
Time and current can be used to calculate total charge.
The total charge relates mol of e- and Faraday constant.

Answer to Problem 13.106PAE

Solution:

$7.46 g$

Explanation of Solution

This consist of the following steps:

1^s^t Step: Calculate the total amount of charge.

$t = 50.0 h = 50 h ×3600 s1 h = 180,000 sI = 0.50 A = 0.50 C/sq = I×t = 0.50 C/s × 180,000 s = 90,000 C1 mol e- =96500 Cx mol e- = 90000 Cx = 0.9326 mol e-$

2ⁿ^d Step: Calculate the total amount of moles of hydrazine

$4 mol e- = 1 mol N2H40.9326 mol e- = y mol N2H4y =0.2332 mol N2H4$

3^r^d Step: Calculate the mass of hydrazine

$mass N2H4 = mol × Molar Mass = 0.2332 mol × 32.0 g/mol = 7.46 g$

Answer 16

(c)

Expert Solution

Answer 17

(c)

Expert Solution

Answer 18

Expert Solution

Answer 19

Interpretation Introduction

Interpretation:

To calculate the mass of hydrazine present in given amount of time and current.

Concept introduction:

A stands for ampere, the amount of current.
Time and current can be used to calculate total charge.
The total charge relates mol of e- and Faraday constant.

Answer 20

Answer to Problem 13.106PAE

Solution:

$7.46 g$

Answer 21

Explanation of Solution

This consist of the following steps:

1^s^t Step: Calculate the total amount of charge.

$t = 50.0 h = 50 h ×3600 s1 h = 180,000 sI = 0.50 A = 0.50 C/sq = I×t = 0.50 C/s × 180,000 s = 90,000 C1 mol e- =96500 Cx mol e- = 90000 Cx = 0.9326 mol e-$

2ⁿ^d Step: Calculate the total amount of moles of hydrazine

$4 mol e- = 1 mol N2H40.9326 mol e- = y mol N2H4y =0.2332 mol N2H4$

3^r^d Step: Calculate the mass of hydrazine

$mass N2H4 = mol × Molar Mass = 0.2332 mol × 32.0 g/mol = 7.46 g$

Answer 22

(d)

Expert Solution

Interpretation Introduction

Interpretation:

To calculate the mass oxygen required

Concept introduction:

Use stoichiometric ratios.

Answer to Problem 13.106PAE

Solution:

$mass O2 = 7.46 g$

Explanation of Solution

Relate the moles of oxygen and hydrazine:

1^s^t Step: Calculate total moles of oxygen required

The ratio is 1:1 since 1 mol of hydrazine requires 1 mol of Oxygen gas to react.

Chemistry for Engineering Students, Chapter 13, Problem 13.106PAE , additional homework tip 4

2ⁿ^d Step: Calculate the total mass of oxygen gas

$mass O2 = mol × Molar Mass = 0.2332 mol × 32.0 g/mol = 7.46 g$

The mass is exactly the same as hydrazine since they have 1:1 ratio and similar molar masses.

Answer 23

(d)

Expert Solution

Answer 24

(d)

Expert Solution

Answer 25

Expert Solution

Answer 26

Interpretation Introduction

Interpretation:

To calculate the mass oxygen required

Concept introduction:

Use stoichiometric ratios.

Answer 27

Answer to Problem 13.106PAE

Solution:

$mass O2 = 7.46 g$

Answer 28

Explanation of Solution

Relate the moles of oxygen and hydrazine:

1^s^t Step: Calculate total moles of oxygen required

The ratio is 1:1 since 1 mol of hydrazine requires 1 mol of Oxygen gas to react.

Chemistry for Engineering Students, Chapter 13, Problem 13.106PAE , additional homework tip 4

2ⁿ^d Step: Calculate the total mass of oxygen gas

$mass O2 = mol × Molar Mass = 0.2332 mol × 32.0 g/mol = 7.46 g$

The mass is exactly the same as hydrazine since they have 1:1 ratio and similar molar masses.

Answer 29

Ch. 13 - Prob. 1CO Ch. 13 - Prob. 2CO Ch. 13 - • write and balance half-reactions for simple...Ch. 13 - Prob. 4CO Ch. 13 - Prob. 5CO Ch. 13 - • use standard reduction potentials to predict the...Ch. 13 - • calculate the amount of metal plated, the amount...Ch. 13 - Prob. 8CO Ch. 13 - Prob. 9CO Ch. 13 - Prob. 10CO

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Answer to Problem 13.106PAE

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Answer to Problem 13.106PAE

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