Chemistry (7th Edition)

Published by Pearson
ISBN 10: 0321943171
ISBN 13: 978-0-32194-317-0

Chapter 1 - Chemical Tools: Experimentation and Measurement - Section Problems - Page 30: 57

Answer

(1) $7.926\times 10^3$ $mi$ (2) $7.9 \times 10^3$ $mi$

Work Step by Step

(1) Four significant figures: In 7926.381, the fourth significant figure is 6, so: 7926 is the rounded number. Now, we have to check the next number, "3"; since it is less than 5, we leave the number as 7926. Scientific notation: 1. Move the decimal point to put the number between 1 and 10 7.926 2. Now, multiply the number by 10 to the power of the number of times you moved the decimal point. If you moved to the right, the number is negative. Since we moved the decimal point by 3 times to the left: $7.926 \times 10^3$ $mi$ --------------- (2) Two significant figures: In 7926.381, the second significant figure is 9, so: 7900 is the rounded number. Now, we have to check the next number: "2". Since it is less than 5, we leave the number as 7900. Scientific notation: 1. Move the decimal point to put the number between 1 and 10 7.9 2. Now, multiply the number by 10 to the power of the number of times you moved the decimal point. If you moved to the right, the number is negative. Since we moved the decimal point by 3 times to the left: $7.9 \times 10^3$ $mi$
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