General Chemistry 10th Edition

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

Chapter 6 - Thermochemistry - Exercises - Page 229: 6.1

Answer

$2.72\times 10^{-18}cal$

Work Step by Step

We know that $KE=\frac{1}{2}mv^2$ We plug in the known values to obtain: $KE=\frac{1}{2}(9.11\times 10^{-31})(5.0\times 10^6)^2=1.13\times 10^{-17}J$ $KE=1.13\times 10^{-17}\times \frac{1cal}{4.18J}=2.72\times 10^{-18}cal$
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