Chemistry (7th Edition)

Published by Pearson
ISBN 10: 0321943171
ISBN 13: 978-0-32194-317-0

Chapter 1 - Chemical Tools: Experimentation and Measurement - Section Problems - Page 29: 52

Answer

(a) $5pm = 5 \times 10^{-10} cm = 5 \times 10^{-3} nm$ (b) $8.5 cm^3 = 8.5 \times 10^{-6}m^3 = 8.5 \times 10^3mm^3$ (c) $65.2 mg$ $ = 6.52 \times 10^{-2}g = 6.52 \times 10^{10}pg$

Work Step by Step

We use unit conversions to find: (a) $5pm \times \frac{10^{-12}m}{1pm} = 5 \times 10^{-12}m$ $5 \times 10^{-12}m \times \frac{1cm}{10^{-2}m} = 5 \times 10^{-10}cm$ $5 \times 10^{-12}m \times \frac{1nm}{10^{-9}m} =5 \times 10^{-3}nm$ (b) $8.5 cm^3 \times \frac{10^{-6}m^3}{1cm^3} = 8.5 \times 10^{-6}m^3$ $8.5 \times 10^{-6}m^3 \times \frac{1mm^3}{10^{-9}m^3} = 8.5 \times 10^3mm^3$ (c) 65.2 mg $\times \frac{10^{-3}g}{1mg} = 6.52 \times 10^{-2}g$ $6.52 \times 10^{-2}g \times \frac{1pg}{10^{-12}g} = 6.52 \times 10^{10}pg$
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