## Chemistry: Molecular Approach (4th Edition)

$\underline{73{}^\circ \text{J}}$
Calculate the conversion factor between the J scale and the H scale as follows: $\begin{array}{*{35}{l}} \frac{\left( J-17 \right)}{\left( 97-17 \right)} & = & \frac{\left( H-0 \right)}{\left( 120-0 \right)} \\ \frac{\left( J-17 \right)}{80} & = & \frac{H}{120} \\ H & = & \left( \frac{120}{80} \right)\times \left( J-17 \right) \\ H & = & 1.5\left( J-17 \right) \\ \end{array}$ Substitute $84{}^\circ$ for H in the above equation as follows: \begin{align} & 84{}^\circ =1.5J-25.5 \\ & J=73{}^\circ \end{align} The boiling point on the Jekyll scale is $\underline{73{}^\circ \text{J}}$.