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 35: 1.77


The mass of 5.9 $ cm^{3}$ platinum is 1.3 x $10^{2}$ g .

Work Step by Step

1- We know the density of platinum and the volume of the piece we want to find the mass. Using the formula for density we can find mass: d = $\frac{m}{v}$ multiplying both sides with v we will get: d x v =$\frac{m}{v}$ x v So m = d x v 2- Replace density of platinum 21.4 $\frac{g}{cm^{3}}$ and the volume 5.9 $cm^{3}$ in the formula of mass to find the mass: m = 21.4 $\frac{g}{cm^{3}}$ x 5.9 $cm^{3}$ m = 126 g (Note: the real number would be 126.26 g , but considering the significant number we wrote 126 g.) Further considering significant number would bring the mass of platinum to 1.3 x $10^{2}$ g. ( 126 g $\approx$ 130 g = 1.3 x $10^{2}$ g)
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