Chemistry: A Molecular Approach (3rd Edition)

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

Chapter 3 - Sections 3.1-3.12 - Exercises - Problems by Topic: 64c

Answer

$$1.039\times10^{-3}molsC_{8}H_{18}$$

Work Step by Step

Use molar mass to convert from grams to moles. $0.1187 g$ $C_{8}H_{18}$$\times\frac{1 mol C_{8}H_{18}}{114.2278 g C_{8}H_{18}} =$ $1.039\times10^{-3}$ $mols$ $C_{8}H_{18}$
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