Chemistry 9th Edition

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

Chapter 6 - Thermochemistry - Exercises: 48

Answer

$-3.55*10^{4}\text{ kJ}$

Work Step by Step

a. $1.00 \text{ gram $CH_4$}* \frac{1\text{ mol $CH_4$}}{16.04\text{ grams $CH_4$}}*\frac{-891\text{ kJ}}{1\text{ mol $CH_4$}}=-55.5\text{ kJ}$ b. $PV=nRT$ $740*1.00*10^{3}=n*62.364*298$ $n=39.8\text{ mol}$ $39.8\text{ mol}*\frac{-891\text{ kJ}}{1\text{ mol}}=-3.55*10^{4}\text{ kJ}$
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