Chemistry 9th Edition

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

Chapter 1 - Chemical Foundations - Exercises: 57

Answer

a) -272.778 $^{\circ}$C b) -40 $^{\circ}$C c) 20 $^{\circ}$C d) 38888871.111$^{\circ}$C

Work Step by Step

Let: Celsius -$^{\circ}$ C Fahrenheit - $^{\circ}$F Using formula: $\frac{C}{5}$= ($\frac{F - 32}{9})$ Or, $^{\circ}$C = $\frac{5(F - 32)}{9}$ a) Temperature = -459 $^{\circ}$F Thus in $^{\circ}$C = $ \frac{5(-459-32)}{9}$ = -272.778 $^{\circ}$C b) Temperature = -40 $^{\circ}$F Thus, in $^{\circ}$C = $\frac{5(- 40-32)}{9}$ = -40 $^{\circ}$C c) Temperature = 68 $^{\circ}$F Thus, in $^{\circ}$C = $\frac{5(68-32)}{9}$ = 20 $^{\circ}$C d) Temperature = 7 $\times$ $10^{7}$ $^{\circ}$F = 70000000 $^{\circ}$F Thus, in $^{\circ}$C = $\frac{5(70000000 - 32)}{9}$ = 38888871.111$^{\circ}$C
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