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


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

Work Step by Step

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

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.