Chemistry 9th Edition

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

Chapter 1 - Chemical Foundations - Integrative Problems - Page 41: 121

Answer

a. 366.48 K b. 199.82 K c. 166.66 K d. The "300 Club" name only works for the Fahrenheit scale

Work Step by Step

At 200.0°F: Tc = 5/9 (200.0°F − 32°F) = 93.33°C; Tk = 93.33 + 273.15 = 366.48 K At −100.0°F: Tc = 5/9 (−100.0°F − 32°F) = −73.33°C; TK = −73.33°C + 273.15 = 199.82 K ∆T(°C) = [93.33°C − (−73.33°C)] = 166.66°C; ∆T(K) = (366.48 K −199.82 K) = 166.66 K The “300 Club” name only works for the Fahrenheit scale; it does not hold for the Celsius and Kelvin scales.
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