Answer
Please see the work below.
Work Step by Step
We know that
$E=mc^2$
Now $\Delta m=\frac{E}{c^2}$
We plug in the known values to obtain:
$\Delta m=\frac{1.715\times 10^6J}{3.00\times 10^8\frac{m}{s}}=0.00572Kg=5.72g$
Chemistry 12th Edition
by Chang, Raymond; Goldsby, Kenneth
Published by
McGraw-Hill Education
ISBN 10:
0078021510
ISBN 13:
978-0-07802-151-0
Chapter 2 - Atoms, Molecules, and Ions - Questions & Problems - Page 73: 2.104
Update this answer!
You can help us out by revising, improving and updating this answer.
Update this answerAfter 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.
