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: 1.134

Answer

a- 127 Å = 1.27 x $10^{-10}$ cm b - 21.0 kg = 2.1 x $10^{7}$ mg c- 1.09 x $10^{1}$ mm d- 4.6ns = 4.6 x $10^{-3}$ ms

Work Step by Step

a - 127 Angstrom to centimeter 127 Å = 127 x $\frac{10^{-10}(m) }{1Å}$ x $\frac{10^{-2}(cm)}{1m}$ = 127 x $10^{-12}$ cm or 127 Å = 1.27 x $10^{-10}$ cm b - 21.0 kg to milligrams 21.0 kg = 21.0 kg x $\frac{10^{3}g}{1kg}$x $\frac{10^{3}mg}{1g}$ = 21 x $10^{6}$ mg or 21.0 kg = 2.1 x $10^{7}$ mg c - 1.09 cm to millimeters 1.09 cm = 1.09 cm x $\frac{1m}{10^{2}cm}$ x $\frac{10^{3}mm}{1m}$ = 1.09 x $10^{1}$ mm d - 4.6ns to microseconds 4.6 ns = 4.6 ns x $\frac{1s}{10^{9}ns}$ x $\frac{10^{6}µs}{1s}$ 4.6ns = 4.6 x $10^{-3}$ ms
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