Chemistry: The Central Science (13th Edition)

Published by Prentice Hall
ISBN 10: 0321910419
ISBN 13: 978-0-32191-041-7

Chapter 5 - Thermochemistry - Exercises - Page 206: 5.45a


-29.475 kJ

Work Step by Step

Given Ag+ + Cl- --> AgCl $\Delta H = -65.5 kJ $, it means that for each mole of AgCl formed, 65.5 kJ of heat are released. To find $\Delta H$ for the production of 0.450 mol, it would be 0.450 * -65.5 kJ = -29.475 kJ
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.