Chemistry 9th Edition

Published by Cengage Learning
ISBN 10: 1133611095
ISBN 13: 978-1-13361-109-7

Chapter 6 - Thermochemistry - Exercises: 61

Answer

$66.9\text{ kJ/mol}$

Work Step by Step

Moles of $AgNO_3 = 0.050*0.100=0.0050$ $q=mc\Delta T$ $q=100.0*4.18*(23.40-22.60)$ $q=334.4\text{ J per 0.0050 mol of $AgNO_3$ consumed}$ $\frac{334.4\text{ mol}}{0.0050\text{ mol $AgNO_3$}}=66.9\text{ kJ/mol}$
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