**Part A**

Step 1

$\mathbf{q}\mathbf{}\mathbf{=}\mathbf{}\mathbf{0}\mathbf{.}\mathbf{1584}\mathbf{}\overline{)\mathbf{g}}\mathbf{}\mathbf{\times}\mathbf{}\mathbf{26}\mathbf{.}\mathbf{42}\mathbf{}\frac{\mathbf{kJ}}{\overline{)\mathbf{g}}}$

**q = 4.1849 J**

Step 2

$\overline{){\mathbf{q}}{\mathbf{}}{\mathbf{=}}{\mathbf{}}{{\mathbf{C}}}_{{\mathbf{cal}}}{\mathbf{\u2206}}{\mathbf{T}}}\phantom{\rule{0ex}{0ex}}{\mathbf{C}}_{\mathbf{cal}}\mathbf{}\mathbf{=}\frac{\mathbf{q}}{\mathbf{\u2206}\mathbf{T}}\phantom{\rule{0ex}{0ex}}{\mathbf{C}}_{\mathbf{cal}}\mathbf{}\mathbf{=}\frac{\mathbf{4}\mathbf{.}\mathbf{1849}\mathbf{}\mathbf{kJ}}{\mathbf{2}\mathbf{.}\mathbf{54}\mathbf{}\mathbf{\xb0}\mathbf{C}}$

The combustion of 0.1584 g benzoic acid increases the temperature of a bomb calorimeter by 2.54°C. Calculate the heat capacity of this calorimeter. (The energy released by combustion of benzoic acid is 26.42 kJ/g.) A 0.2130-g sample of vanillin (C_{8}H_{8}O_{3}) is then burned in the same calorimeter, and the temperature increases by 3.25°C. What is the energy of combustion per gram of vanillin? Per mole of vanillin?

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Our data indicates that this problem or a close variation was asked in Chemistry: An Atoms First Approach - Zumdahl Atoms 1st 2nd Edition. You can also practice Chemistry: An Atoms First Approach - Zumdahl Atoms 1st 2nd Edition practice problems.