Chemistry 12th Edition

Published by McGraw-Hill Education
ISBN 10: 0078021510
ISBN 13: 978-0-07802-151-0

Chapter 2 - Atoms, Molecules, and Ions - Questions & Problems: 2.104

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: $\Delta m=\frac{1.715\times 10^6J}{3.00\times 10^8\frac{m}{s}}=0.00572Kg=5.72g$
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