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 - Applying Concepts - Page 90f: 123f


159.8 g of $Br_2$ has the greater number of atoms.

Work Step by Step

Molar mass $Br_2$ = $79.904 \times 2 = 159.808\ g$. Therefore: 159.8 g of $Br_2$ has about 1 mol of $Br_2$ molecules. $6.022 \times 10^{23}$ molecules of $Br_2$. $6.022 \times 10^{23} Br_2 > 1\ Br_2$ Since we are comparing the same type of molecule ($Br_2$), the number of atoms will have the same result.
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