Bundle: Introductory Chemistry: A Foundation, Loose-leaf Version, 9th + OWLv2 with MindTap Reader, 1 term (6 months) Printed Access Card
Bundle: Introductory Chemistry: A Foundation, Loose-leaf Version, 9th + OWLv2 with MindTap Reader, 1 term (6 months) Printed Access Card
9th Edition
ISBN: 9780357000922
Author: Steven S. Zumdahl, Donald J. DeCoste
Publisher: Cengage Learning
Question
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Chapter 10, Problem 17QAP
Interpretation Introduction

(a)

Interpretation:

ΔE should be calculated when q = + 65 kJ and w = -22 kJ

Concept Introduction:

The change in internal energy is equal to sum of heat and work done.

The mathematical expression is given as:

ΔE =q+w

Where, q = heat and w = work done.

If heat is released from the system, q is negative and if heat is absorbed by the system, q is positive.

If work is done on the system, w is positive and if work is done by the system, w is negative.

Interpretation Introduction

(b)

Interpretation:

ΔE should be calculated when q = + 200. kJ and w = -73 kJ

Concept Introduction:

The change in internal energy is equal to sum of heat and work done.

The mathematical expression is given as:

ΔE =q+w

Where, q = heat and w = work done.

If heat is released from the system, q is negative and if heat is absorbed by the system, q is positive.

If work is done on the system, w is positive and if work is done by the system, w is negative.

Interpretation Introduction

(c)

Interpretation:

ΔE should be calculated when q = -18 kJ and w = -40. kJ

Concept Introduction:

The change in internal energy is equal to sum of heat and work done.

The mathematical expression is given as:

ΔE =q+w

Where, q = heat and w = work done.

If heat is released from the system, q is negative and if heat is absorbed by the system, q is positive.

If work is done on the system, w is positive and if work is done by the system, w is negative.

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Chapter 10 Solutions

Bundle: Introductory Chemistry: A Foundation, Loose-leaf Version, 9th + OWLv2 with MindTap Reader, 1 term (6 months) Printed Access Card

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