Chemistry: A Molecular Approach (3rd Edition)

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

Chapter 1 - Sections 1.1-1.8 - Exercises - Challenge Problems - Page 42: 137

Answer

$$ 86.4 \space mg$$

Work Step by Step

1. Volume of air: $$8 \space h \times \frac{60 \space min}{1 \space h} \times \frac{20 \space breaths}{1 \space min} \times \frac{0.50 \space L \space air}{1 \space breath} = 4.8 \times 10^3 \space L \space air$$(1 significant figure) 2. Volume of CO: $$4.8 \times 10^3 \space L \space air \times \frac{15.0 \space L \space CO}{10^6 \space L \space air} = 0.072 \space L \space CO$$ 3. Mass of CO: $$0.072 \space L \space CO \times \frac{1.2 \space g}{1 \space L} = 0.0864 \space g \times \frac{1000 \space mg}{1 \space g} = 86.4 \space mg$$
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