Chemistry: The Central Science (13th Edition)

Published by Prentice Hall
ISBN 10: 0321910419
ISBN 13: 978-0-32191-041-7

Chapter 3 - Chemical Reactions and Reaction Stoichiometry - Exercises - Page 115: 3.39b


5.00 mg of that substance contains $ 3.08 \times 10^{-5} \space mole \space of \space allicin$.

Work Step by Step

1. Identify the conversion factors: - $mg$ to grams : $\frac{1 \space g}{1000 \space mg}$ - Grams (allicin) to moles (allicin): $\frac{1 \space mole \space (allicin)}{162.28 \space g \space (allicin)**}$ ** This was calculated in 3.39a 2. Calculate the number of moles in 5.00 mg of that substance. $5.00 \space mg \space (allicin) \times \frac{1 \space g}{1000 \space mg} \times \frac{1 \space mole \space (allicin)}{162.28 \space g \space (allicin) }=$ $ 3.08 \times 10^{-5} \space mole \space (allicin)$
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