General Chemistry 10th Edition

Published by Cengage Learning
ISBN 10: 1-28505-137-8
ISBN 13: 978-1-28505-137-6

Chapter 6 - Thermochemistry - Questions and Problems - Page 263: 6.97


Please see the work below.

Work Step by Step

We know that $E_k=\frac{1}{2}mv^2$ This can be rearranged as $v=\frac{1}{2}\sqrt{\frac{226.433\frac{Kg.m^2}{s^2}}{\frac{1}{2}\times 1.00lb\times 0.4536\frac{Kg}{lb}}}=31.60\frac{m}{s}$
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