Chemistry: Molecular Approach (4th Edition)

Published by Pearson
ISBN 10: 0134112830
ISBN 13: 978-0-13411-283-1

Chapter 1 - Exercises - Page 40: 110


\[\underline{73{}^\circ \text{J}}\]

Work Step by Step

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 \[\underline{73{}^\circ \text{J}}\].
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