Chemistry: The Molecular Science (5th Edition)

Published by Cengage Learning
ISBN 10: 1285199049
ISBN 13: 978-1-28519-904-7

Chapter 2 - Chemical Compounds - Questions for Review and Thought - Topical Questions - Page 90d: 78b


$92.8 \ mmol$

Work Step by Step

We use unit conversions to obtain: $$ \frac{2.458 g}{1} \frac{.0240 \ g \ copper}{1 \ g \ penny} \frac {1 mol}{63.55 g}= 92.8 \times 10^{-3} mol = 92.8\ mmol$$
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