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: 65

Answer

a) $100^{\circ}C$ b) $-321.07^{\circ}F$ c) $298.15 K$ d) $310.15 K$

Work Step by Step

Using the conversion factors below, we can convert from Celsius to Fahrenheit or vice-versa. To find the Celsius temperature we can convert from Fahrenheit using the following formula $(^{\circ}F-32)\times\frac{5}{9}$ and to convert from Celsius to Fahrenheit, the following formula can be used; $(^{\circ}C\times\frac{9}{5})+32$. Additionally to convert to Kelvin from Celsius, we can do so by simply adding 273.15 and to covert from Kelvin to Celsius, we simply subtract 273.15. a) $(212^{\circ}F-32)\times\frac{5}{9} = 100^{\circ}C$ b) $77K - 273.15 = -196.15^{\circ}C$ $(-196.15^{\circ}C\times\frac{9}{5})+32 = -321.07^{\circ}F$ c) $25^{\circ}C + 273.15 = 298.15 K$ d) $(98.6^{\circ}F-32)\times\frac{5}{9} = 37^{\circ}C$ $37^{\circ}C + 273.15 = 310.15 K$
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