A chemical system produces 155 kJ of heat and does 22 kJ of work. What is ΔE for the surroundings? -177 kJ 177 kJ -133 kJ 133 kJ

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Answer 1

Textbook Question

Chapter 10, Problem 1SAQ

A chemical system produces 155 kJ of heat and does 22 kJ of work. What is ΔE for the surroundings?

-177 kJ
177 kJ
-133 kJ
133 kJ

Expert Solution & Answer

Interpretation Introduction

Interpretation:

To find out $ΔE$ for the surroundings of a chemical system that produces $155 kJ$ of heat and does $22 kJ$ of work.

Concept Introduction: The relationship between heat, $q$ , work, $w$ , and $ΔEsys$ is given as:

$ΔEsys=q+w$

Answer to Problem 1SAQ

Correct answer:

Therefore, option (b) is correct and options (a), (c) and (d), are incorrect.

Explanation of Solution

Reasons for the correct statement: The heat is produced by the chemical system and released to the surroundings so, $q = - 155 kJ$

The work is done by the chemical system on the surroundings so, $w = - 22 kJ$

Substituting the values:

$ΔEsys=q+w.........(1)ΔEsys=−155 kJ+(−22 kJ)=−177 kJΔEsys=−ΔEsurr(or)ΔEsurr=−ΔEsys=−(−177 kJ)=177 kJ$

Hence, option (b) is correct.

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Answer 2

Textbook Question

Chapter 10, Problem 1SAQ

A chemical system produces 155 kJ of heat and does 22 kJ of work. What is ΔE for the surroundings?

-177 kJ
177 kJ
-133 kJ
133 kJ

Expert Solution & Answer

Interpretation Introduction

Interpretation:

To find out $ΔE$ for the surroundings of a chemical system that produces $155 kJ$ of heat and does $22 kJ$ of work.

Concept Introduction: The relationship between heat, $q$ , work, $w$ , and $ΔEsys$ is given as:

$ΔEsys=q+w$

Answer to Problem 1SAQ

Correct answer:

Therefore, option (b) is correct and options (a), (c) and (d), are incorrect.

Explanation of Solution

Reasons for the correct statement: The heat is produced by the chemical system and released to the surroundings so, $q = - 155 kJ$

The work is done by the chemical system on the surroundings so, $w = - 22 kJ$

Substituting the values:

$ΔEsys=q+w.........(1)ΔEsys=−155 kJ+(−22 kJ)=−177 kJΔEsys=−ΔEsurr(or)ΔEsurr=−ΔEsys=−(−177 kJ)=177 kJ$

Hence, option (b) is correct.

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Answer 3

Textbook Question

Chapter 10, Problem 1SAQ

A chemical system produces 155 kJ of heat and does 22 kJ of work. What is ΔE for the surroundings?

-177 kJ
177 kJ
-133 kJ
133 kJ

Expert Solution & Answer

Interpretation Introduction

Interpretation:

To find out $ΔE$ for the surroundings of a chemical system that produces $155 kJ$ of heat and does $22 kJ$ of work.

Concept Introduction: The relationship between heat, $q$ , work, $w$ , and $ΔEsys$ is given as:

$ΔEsys=q+w$

Answer to Problem 1SAQ

Correct answer:

Therefore, option (b) is correct and options (a), (c) and (d), are incorrect.

Explanation of Solution

Reasons for the correct statement: The heat is produced by the chemical system and released to the surroundings so, $q = - 155 kJ$

The work is done by the chemical system on the surroundings so, $w = - 22 kJ$

Substituting the values:

$ΔEsys=q+w.........(1)ΔEsys=−155 kJ+(−22 kJ)=−177 kJΔEsys=−ΔEsurr(or)ΔEsurr=−ΔEsys=−(−177 kJ)=177 kJ$

Hence, option (b) is correct.

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Answer 4

Textbook Question

Chapter 10, Problem 1SAQ

A chemical system produces 155 kJ of heat and does 22 kJ of work. What is ΔE for the surroundings?

-177 kJ
177 kJ
-133 kJ
133 kJ

Expert Solution & Answer

Interpretation Introduction

Interpretation:

To find out $ΔE$ for the surroundings of a chemical system that produces $155 kJ$ of heat and does $22 kJ$ of work.

Concept Introduction: The relationship between heat, $q$ , work, $w$ , and $ΔEsys$ is given as:

$ΔEsys=q+w$

Answer to Problem 1SAQ

Correct answer:

Therefore, option (b) is correct and options (a), (c) and (d), are incorrect.

Explanation of Solution

Reasons for the correct statement: The heat is produced by the chemical system and released to the surroundings so, $q = - 155 kJ$

The work is done by the chemical system on the surroundings so, $w = - 22 kJ$

Substituting the values:

$ΔEsys=q+w.........(1)ΔEsys=−155 kJ+(−22 kJ)=−177 kJΔEsys=−ΔEsurr(or)ΔEsurr=−ΔEsys=−(−177 kJ)=177 kJ$

Hence, option (b) is correct.

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Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

Expert Solution & Answer

Interpretation Introduction

Interpretation:

To find out $ΔE$ for the surroundings of a chemical system that produces $155 kJ$ of heat and does $22 kJ$ of work.

Concept Introduction: The relationship between heat, $q$ , work, $w$ , and $ΔEsys$ is given as:

$ΔEsys=q+w$

Answer to Problem 1SAQ

Correct answer:

Therefore, option (b) is correct and options (a), (c) and (d), are incorrect.

Explanation of Solution

Reasons for the correct statement: The heat is produced by the chemical system and released to the surroundings so, $q = - 155 kJ$

The work is done by the chemical system on the surroundings so, $w = - 22 kJ$

Substituting the values:

$ΔEsys=q+w.........(1)ΔEsys=−155 kJ+(−22 kJ)=−177 kJΔEsys=−ΔEsurr(or)ΔEsurr=−ΔEsys=−(−177 kJ)=177 kJ$

Hence, option (b) is correct.

Answer 8

Expert Solution & Answer

Interpretation Introduction

Interpretation:

To find out $ΔE$ for the surroundings of a chemical system that produces $155 kJ$ of heat and does $22 kJ$ of work.

Concept Introduction: The relationship between heat, $q$ , work, $w$ , and $ΔEsys$ is given as:

$ΔEsys=q+w$

Answer to Problem 1SAQ

Correct answer:

Therefore, option (b) is correct and options (a), (c) and (d), are incorrect.

Explanation of Solution

Reasons for the correct statement: The heat is produced by the chemical system and released to the surroundings so, $q = - 155 kJ$

The work is done by the chemical system on the surroundings so, $w = - 22 kJ$

Substituting the values:

$ΔEsys=q+w.........(1)ΔEsys=−155 kJ+(−22 kJ)=−177 kJΔEsys=−ΔEsurr(or)ΔEsurr=−ΔEsys=−(−177 kJ)=177 kJ$

Hence, option (b) is correct.

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Interpretation Introduction

Interpretation:

To find out $ΔE$ for the surroundings of a chemical system that produces $155 kJ$ of heat and does $22 kJ$ of work.

Concept Introduction: The relationship between heat, $q$ , work, $w$ , and $ΔEsys$ is given as:

$ΔEsys=q+w$

Answer 12

Answer to Problem 1SAQ

Correct answer:

Therefore, option (b) is correct and options (a), (c) and (d), are incorrect.

Answer 13

Explanation of Solution

Reasons for the correct statement: The heat is produced by the chemical system and released to the surroundings so, $q = - 155 kJ$

The work is done by the chemical system on the surroundings so, $w = - 22 kJ$

Substituting the values:

$ΔEsys=q+w.........(1)ΔEsys=−155 kJ+(−22 kJ)=−177 kJΔEsys=−ΔEsurr(or)ΔEsurr=−ΔEsys=−(−177 kJ)=177 kJ$

Hence, option (b) is correct.

Answer 14

Ch. 10 - A chemical system produces 155 kJ of heat and does...Ch. 10 - Which sample is most likely to undergo the...Ch. 10 - Prob. 3SAQ Ch. 10 - A 12.5-g sample of granite initially at 82.0 C is...Ch. 10 - A cylinder with a moving piston expands from an...Ch. 10 - When a 3.80-g sample of liquid octane (C8H18)...Ch. 10 - Hydrogen gas reacts with oxygen to form water....Ch. 10 - Manganese reacts with hydrochloric acid to produce...Ch. 10 - Consider the reactions: A2BH1A3CH2 What is H for...Ch. 10 - Use standard enthalpies of formation to determine...

A chemical system produces 155 kJ of heat and does 22 kJ of work. What is ΔE for the surroundings? -177 kJ 177 kJ -133 kJ 133 kJ

Solution Summary: The author explains the relationship between heat, q, work, and Delta 'E_sys'.

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Answer to Problem 1SAQ

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