Introductory Chemistry (5th Edition)

Published by Pearson
ISBN 10: 032191029X
ISBN 13: 978-0-32191-029-5

Chapter 2 - Measurement and Problem Solving - Exercises - Problems - Page 50: 54

Answer

a) $1.2\times10^{3}$ is the correct answer to 2 significant figures. b) $40$ is the correct answer to 2 significant figures. c) $56$ is the correct answer to 2 significant figures. d) $0.0100 $ is the correct answer to 2 significant figures.

Work Step by Step

a) $1.249\times10^{3}$ should be rounded down to $1.2\times10^{3}$ instead of $1.3\times10^{3}$ because after the second digit the value of the digit is 4 which is less than 5 and thus the digit prior should be rounded down to 2. b) $3.999\times10^{2}$ correctly rounds up to 40 as the next digit is 9 which rounds up and since the digit before is also a 9, the next nearest whole number is correct with no decimal places. c) $ 56.21$ correctly rounds down to 56 as the hundredth decimal is less than 50 (i.e. its 21) so the correct rounding to two significant figures would be the closest whole number which is 56. d) $0.009964$ rounds to $0.0100$ because $0.010$ has only one significant figure and not two so an extra zero is needed.
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