Chemistry: The Molecular Nature of Matter and Change 7th Edition

Published by McGraw-Hill Education
ISBN 10: 007351117X
ISBN 13: 978-0-07351-117-7

Chapter 1 - Problems: 1.59

Answer

a) 3.8g b) 1.03mL c) 564cm^3

Work Step by Step

a) $18.02g\div4.8g$=3.7542g, which is rounded to 2 sig figs because the least amount of sig figs in the initial calculation is 2. b) $7.87mL\div7.66mL$= 1.0274mL, which is rounded to 3 sig figs because the least amount of sig figs in the initial calculation is 3. c) (pi is assumed to be 3.14) $V=3.14\times6.23cm^{2}\times4.630cm$ is the calculation with all of the variables plugged in. V=564.230, which is rounded to 3 sig figs because the least amount of sig figs in the initial calculation is 3.
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