Chemistry: A Molecular Approach (3rd Edition)

Published by Prentice Hall
ISBN 10: 0321809246
ISBN 13: 978-0-32180-924-7

Chapter 2 - Sections 2.1-2.9 - Exercises - Problems by Topic - Page 82: 89a

Answer

$36\ g\ Au $

Work Step by Step

1. Use Avogadro's number and the molar mass of gold as conversion factors in order to determine the number of grams of gold. 2. $1.1\times10^{23}\ atoms\ Au\times\frac{1\ mol\ Au}{6.022\times 10^{23}\ atoms\ Au}\times\frac{196.96\ g \ Au}{1\ mol\ Au} = 36\ g\ Au $
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