General Chemistry: Principles and Modern Applications (10th Edition)

Published by Pearson Prentice Hal
ISBN 10: 0132064529
ISBN 13: 978-0-13206-452-1

Chapter 1 - Matter: Its Properties and Measurement - Exercises - Units of Measurement: 40

Answer

d = 5.56* $10^{-3}$ mm

Work Step by Step

V = 90.0 * $10^{-12}$ $cm^{3}$ V of sphere = $\frac{4}{3} * π * r^{3}$ $\frac{4}{3} * π * r^{3}$ = 90.0 * $10^{-12}$ $π * r^{3}$ = $\frac{3}{4}$ * 90.0 * $10^{-12}$ $π * r^{3}$ = $6.75 * 10^{-11}$ $ r^{3}$ = $\frac{6.75 * 10^{-11}}{π}$ r = $\sqrt[3] \frac{6.75 * 10^{-11}}{π}$ r = 2.780041634 * $10^{-4}$ cm d = 2 * r d = 2 * 2.780041634 * $10^{-4}$ cm = 5.560083268 * $10^{-4}$ cm d = 5.560083268 * $10^{-3}$ mm Rounding to 3 significant figures gives d = 5.56* $10^{-3}$ mm
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