Chemistry 9th Edition

Published by Cengage Learning
ISBN 10: 1133611095
ISBN 13: 978-1-13361-109-7

Chapter 1 - Chemical Foundations - Exercises - Page 35: 37

Answer

a) $188.1$ b) $12$ c) $4\times 10^{-7}$ d) $6.3\times 10^{-26}$ e) $4.9$ f) $0.22$

Work Step by Step

a) $\frac{2.526}{3.1}+\frac{0.470}{0.623}+\frac{80.705}{0.4326}\approx 188.13$. After reducing the fractions, we get $0.81 + 0.754 + 186.6$ (note: terms are rounded here only to show sig figs, and these rounded terms will not be used in the calculation of the final answer). Because these terms are added together and uncertainty appears in the first decimal place of the third term, the final answer will have $1$ decimal place. b) $(6.404\times 2.91)/(18.7-17.1)\approx 11.647$. After simplifying the parentheses, we get $18.6/1.6$ (note: terms are rounded here only to show sig figs, and these rounded terms will not be used in the calculation of the final answer). Because the divisor only has $2$ significant figures, the answer will also have $2$ significant figures. c) $6.071\times 10^{-5}-8.2\times 10^{-6}-0.521\times 10^{-4}\approx 4.1\times 10^{-7}$. Subtracting the terms gives an uncertainty in the first place value to the left, so the final answer will have no decimal places. d) $(3.8\times 10^{-12}+4.0\times 10^{-13})/(4\times 10^{12}+6.3\times 10^{13})\approx 6.2687\times 10^{-26}$. Simplifying the parentheses gives $(4.2\times 10^{-12})/(6.7\times 10^{13})$. Since both terms have $2$ significant figures, the answer will also have $2$ significant figures. e) $\frac{9.5+4.1+2.8+3.175}{4}\approx 4.8938$. Since we are asked to find an average and three of the terms have $2$ significant figures, the answer will have $2$ significant figures. f) $\frac{8.925-8.905}{8.925}\times 100\approx 0.22409$. Simplifying the fraction gives a term with $2$ significant figures. Since we are asked to consider the $100$ as an exact number, the answer will also have $2$ significant figures.
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