## General Chemistry 10th Edition

Published by Cengage Learning

# Chapter 1 - Chemistry and Measurement - Questions and Problems - Page 35: 1.70

#### Answer

a. $t_{C}$ = $11^{0}$C b. $t_{C}$ = $-24^{0}$C c. $t_{F}$ = $-42^{0}$F d. $t_{F}$ = $72^{0}$F

#### Work Step by Step

a. We can convert temperature in Fahrenheit into Celsius, by applying the formula: $t_{C}$ = $\frac{5^{0}C}{9^{0}F}$ x ($t_{F}$-$32^{0}$F) We know $t_{F}$ = $51^{0}$F , so we can write: $t_{C}$ = $\frac{5^{0}C}{9^{0}F}$ x ($51^{0}$F-$32^{0}$F) $t_{C}$ = $\frac{5^{0}C}{9^{0}F}$ x$19^{0}$F $t_{C}$ = $10.6^{0}$C $t_{C}$ = $11^{0}$C b. Apply the formula: $t_{C}$ = $\frac{5^{0}C}{9^{0}F}$ x ($t_{F}$-$32^{0}$F) We know $t_{F}$ = $-11^{0}$F , so we can write: $t_{C}$ = $\frac{5^{0}C}{9^{0}F}$ x ($-11^{0}$F-$32^{0}$F) $t_{C}$ = $\frac{5^{0}C}{9^{0}F}$ x($-43^{0}$)F $t_{C}$ = $-23.9^{0}$C $t_{C}$ = $-24^{0}$C c. Apply the formula to convert Ceslius to Fahrenheit: $t_{F}$ =($t_{C}$ x$\frac{9^{0}F}{5^{0}C}$) +$32^{0}$F We know $t_{C}$ = $-41^{0}$C, so : $t_{F}$ =($-41^{0}$C x$\frac{9^{0}F}{5^{0}C}$) +$32^{0}$F $t_{F}$ = $-73.8^{0}$F+$32^{0}$F $t_{F}$ = $-41.8^{0}$F $t_{F}$ = $-42^{0}$F d. Apply the formula to convert Ceslius to Fahrenheit: $t_{F}$ =($t_{C}$ x$\frac{9^{0}F}{5^{0}C}$) +$32^{0}$F We know $t_{C}$ = $22^{0}$C, so : $t_{F}$ =($22^{0}$C x$\frac{9^{0}F}{5^{0}C}$) +$32^{0}$F $t_{F}$ = $39.6^{0}$F+$32^{0}$F $t_{F}$ = $71.6^{0}$F $t_{F}$ = $72^{0}$F

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