Chemistry: Principles and Practice (3rd Edition)

Published by Cengage Learning
ISBN 10: 0534420125
ISBN 13: 978-0-53442-012-3

Chapter 1 - Introduction to Chemistry - Questions and Exercises - Exercises - Page 37: 1.83

Answer

a) $-268.94^{\circ}C$, $-452.09^{\circ}F$ b) $204^{\circ}C$

Work Step by Step

a) Converting temperature from Kelvin to Celsius, we use the formula $$T(^{\circ}C)=T(K)-273.15$$ Substituting, $$T(^{\circ}C)=4.21-273.15$$ $$T(^{\circ}C)=-268.94^{\circ}C$$ In converting from Celsius to Fahrenheit, the formula is $$T(^{\circ}F)=\frac{9}{5}\times T(^{\circ}C)+32$$ Substituting, $$T(^{\circ}F)=\frac{9}{5}\times (-268.94)+32$$ $$T(^{\circ}F)=-484.092+32$$ $$T(^{\circ}F)=-452.09^{\circ}F$$ b) In converting from Fahrenheit to Celsius, the formula is $$T(^{\circ}C)=\frac{5}{9}\times(T(^{\circ}F)-32)$$ Substituting, $$T(^{\circ}C)=\frac{5}{9}\times(400-32)$$ $$T(^{\circ}C)=\frac{5}{9}\times(368)$$ $$T(^{\circ}C)=204^{\circ}C$$
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