Chemistry 9th Edition

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

Chapter 6 - Thermochemistry - Exercises - Page 288: 63


$-170 \text{ J/g}$ $-20.\text{ kJ/mol}$

Work Step by Step

$\Delta H = q$ (because pressure is constant) $\Delta H = mc\Delta T$ $\Delta H = (125+10.5)*4.18*(21.1-24.2)=-1.8$ kJ $\frac{-1.8 kJ}{10.5\text{ grams}}=-170 \text{ J/g}$ $\frac{-1.8 kJ}{10.5\text{ grams}}*\frac{119 \text{ grams}}{1\text{ mol }}=-20.\text{ kJ/mol}$
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