Introductory Chemistry (5th Edition)

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

Chapter 3 - Matter and Energy - Exercises - Problems - Page 89: 73

Answer

a) $-459.4^{\circ}F$ is the missing blank. b) $28.06^{\circ}C$ is the missing Celsius blank and $301.21 K$ is the missing Kelvin blank. c) $47.3^{\circ}F$ is the missing blank for Fahrenheit and $281.65 K$ is the missing blank for the Kelvin temperature.

Work Step by Step

a) For part a) to fill in one of the three blanks we only need to use one of the conversion factors as the other one will give the same answer. So to find the missing Fahrenheit temperature we can convert from Celsius. using the following formula $(^{\circ}C\times\frac{9}{5})+32$, therefore $(-273^{\circ}C\times\frac{9}{5})+32 = -459.4^{\circ}F$ b) To find the missing Celsius temperature we can convert from Fahrenheit using the following formula $(^{\circ}F-32)\times\frac{5}{9}$; therefore $(82.5^{\circ}F-32)\times\frac{5}{9} = 28.06^{\circ}C$ and to find the other blank for Kelvin, we can use our celcius temperature and simply add $273.15$. Therefore, $28.06^{\circ}C + 273.15 = 301.21 K$ c) To find the missing Fahrenheit temperature we can convert from Celsius using the following formula $(^{\circ}C\times\frac{9}{5})+32$. Therefore, $(8.5^{\circ}C\times\frac{9}{5})+32 = 47.3^{\circ}F$, and to find the other blank for Kelvin, we can use our celcius temperature and simply add $273.15$. Therefore, $8.5^{\circ}C + 273.15 = 281.65 K$
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