Chemistry (4th Edition)

Published by McGraw-Hill Publishing Company
ISBN 10: 0078021529
ISBN 13: 978-0-07802-152-7

Chapter 1 - Questions and Problems - Page 31: 1.25

Answer

1 $Ans) T = 35^{\circ}C$ 2 $Ans) T = -11.11^{\circ}C$ 3 $Ans) T = 38.8889^{\circ}C$ 4 $Ans) T = 1011.11^{\circ}C$ 5 $Ans) T = -459.67^{\circ}F$

Work Step by Step

Fahrenheit to celcius $$(F-32)\times \frac{5}{9} = ^{\circ}C$$ celcius to Fahrenheit $$\frac{9}{5}\times ^{\circ}C + 32 = F$$ 1)$F = 95^{\circ}F$. $$(F-32)\times \frac{5}{9} = ^{\circ}C$$ $$(95-32)\times \frac{5}{9} = ^{\circ}C$$ $Ans) T = 35^{\circ}C$ 2)$F = 12^{\circ}F$. $$(F-32)\times \frac{5}{9} = ^{\circ}C$$ $$(12-32)\times \frac{5}{9} = ^{\circ}C$$ $Ans) T = -11.11^{\circ}C$ 3)$F = 102^{\circ}F$. $$(F-32)\times \frac{5}{9} = ^{\circ}C$$ $$(102-32)\times \frac{5}{9} = ^{\circ}C$$ $Ans) T = 38.8889^{\circ}C$ 1)$F = 1852^{\circ}F$. $$(F-32)\times \frac{5}{9} = ^{\circ}C$$ $$(1852-32)\times \frac{5}{9} = ^{\circ}C$$ $Ans) T = 1011.11^{\circ}C$ 5) $T=-273.15^{\circ}C$ $$\frac{9}{5}\times ^{\circ}C + 32 = F$$ $$\frac{9}{5}\times (-273.15) + 32 = F$$ $Ans) T = -459.67^{\circ}F$
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