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: 1.136

Answer

a - 7.19 x $10^{-3}$ mg b - 1.04 Å c- 1 x $10^{-3}cm$ d - 6.05 x $10^{3}$ cPa

Work Step by Step

a - 7.19 µg to milligrams 7.19 µg = 7.19 µg x $\frac{1g}{10^{6}µg}$ x $\frac{10^{3}mg}{1g}$ = 7.19 x $10^{-3}$ mg b - 104 pm to angstrom 104 pm = 104 pm x $\frac{1m}{10^{12}pm}$ x $\frac{10^{10}Å }{1m}$ = 104 x $10^{-2}$ Å 104 pm = 1.04 Å c - 0.010 mm to cm 0.010 mm = 0.010 mm x $\frac{1m}{10^{3}mm}$ x $\frac{10^{2}cm}{1m}$ = 0.010 x $10^{-1}cm$ 0.010 mm = 1 x $10^{-3}cm$ d - 0.0605 kPa to centiPaskal 0.0605 kPa = 0.0605 kPa x $\frac{1 Pa}{10^{-3}kPa}$ x $\frac{10^{2}cPa}{1Pa}$ = 0.0605 x $10^{5}$ cPa 0.0605 kPa = 6.05 x $10^{3}$ cPa
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