General Chemistry 10th Edition

Published by Cengage Learning
ISBN 10: 1-28505-137-8
ISBN 13: 978-1-28505-137-6

Chapter 1 - Chemistry and Measurement - Questions and Problems - Page 38: 1.135

Answer

a - 5.91 x $10^{6}$ mg b - 7.53 x $10^{5}$ µg c- 9.01 x $10^{4}$ kHz d - 4.98 x $10^{-4}$ kJ

Work Step by Step

a - 5.91 kg to milligrams 5.91 kg = 5.91 kg x $\frac{10^{3}g}{1kg}$x $\frac{10^{3}mg}{1g}$ = 5.91 x $10^{6}$ mg b - 753 mg to micrograms 753 mg = 753 mg x $\frac{1g}{10^{3}mg}$ x $\frac{10^{6}µg}{1g}$ = 753 x $10^{3}$ µg 753 mg = 7.53 x $10^{5}$ µg c - 90.1 MHz to kilohertz 90.1 MHz = 90.1 MHz x $\frac{1Hz}{10^{-6}MHz}$ x $\frac{10^{-3}kHz}{1Hz}$ = 90.1 x $10^{3}$ kHz 90.1 MHz = 9.01 x $10^{4}$ kHz d - 498 mJ to kilojouls 498 mJ = 498 mJ x $\frac{1J}{10^{3}mJ}$ x $\frac{10^{-3}kJ}{1J}$ = 498 x $10^{-6}$ kJ 498 mJ = 4.98 x $10^{-4}$ kJ
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