Chemistry 9th Edition

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

Chapter 1 - Chemical Foundations - Exercises: 41

Answer

a) Weight: $8\space lb\space 9.9\space oz$ Height: $20.25\space in$ b) $4.02\times 10^{4}\space km$ $4.02\times 10^{7}\space m$ c) $1.2\times 10^{-2}\space m^3$ $12\space L$ $7.3\times 10^2\space in^3$ $0.42\space ft^3$

Work Step by Step

a) $3.91 kg\times\frac{35.27\space oz}{1\space kg}=137.9\space oz$. To convert this to pounds and ounces, we subtract the largest multiple of $16\space oz$ $(128\space oz)$ from this value to get the answer. $51.4\space cm\times \frac{1\space in}{2.54\space cm}=20.2\space in$. Since we are asked to round our final answer to the nearest quarter inch, the answer is $20.25\space in$. b) $25000\space mi\times \frac{1.61\space km}{1\space mi}=4.02\times 10^{4}\space km$. To convert from kilometers to meters, simply multiply the number of kilometers by $10^3$. c) $1.0\space m\times 5.6\space cm\times\frac{1\space m}{100\space cm}\times 2.1\space dm\times\frac{1\space m}{10\space dm}=1.2\times 10^{-2}\space m^3$ $1.2\times 10^{-2}\space m^3\times\frac{1000\space L}{1\space m^3}=12\space L$ $12\space L\times\frac{61 in^3}{1\space L}=7.3\times 10^2\space in^3$ $7.3\times 10^2\space in^3\times\frac{1\space ft^3}{1728\space in^3}=0.42\space ft^3$
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