Chemistry: A Molecular Approach (3rd Edition)

Published by Prentice Hall
ISBN 10: 0321809246
ISBN 13: 978-0-32180-924-7

Chapter 1 - Sections 1.1-1.8 - Exercises - Problems by Topic: 51b


$$77 K = -321^{o}F$$

Work Step by Step

We know the temperature of liquid nitrogen is $-321^{o}F$ so 77 K must be equal to $-321^{o}F$. To solve mathematically, convert Kelvin to Celsius and then Celsius to Fahrenheit: $^{o}C = K - 273.15 $ is our first conversion factor. $^{o}C = 77K - 273.15 = -196.15^{o}C$ $^{o}F = 1.8(^{o}C) + 32 $ is our second conversion factor. $^{o}F = 1.8(-196.15^{o}C) + 32 = -321^{o}F$
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