## Chemistry: Molecular Approach (4th Edition)

Published by Pearson

# Chapter 1 - Exercises - Page 40: 109

#### Answer

$\underline{34{}^\circ \text{F}}$

#### Work Step by Step

The boiling point of water is $130{}^\circ \text{X}$. The freezing point of water is $10{}^\circ \text{X}$. The boiling point of water on the Fahrenheit scale is $\text{212}{}^\circ \text{F}$. The freezing point of water on the Fahrenheit scale is $\text{32}{}^\circ \text{F}$. \begin{align} & \frac{\left( F-32 \right)}{\left( 212-32 \right)}=\frac{\left( X-10 \right)}{\left( 130-10 \right)} \\ & \frac{\left( F-32 \right)}{180}=\frac{\left( X-10 \right)}{120} \\ & 1.5X+17=F \end{align} When the reading on the Fahrenheit scale will be the same as that of the X scale, the above equation will become \begin{align} & 1.5X+17=X \\ & X=34{}^\circ \end{align} The temperature at which the Fahrenheit and X readings are the same is $\underline{34{}^\circ \text{F}}$.

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