Use the calorimetry equation to find the heat involved

$\overline{){\mathbf{Q}}{\mathbf{}}{\mathbf{=}}{\mathbf{}}{\mathbf{mC}}{\mathbf{\u2206}}{\mathbf{T}}}\phantom{\rule{0ex}{0ex}}\mathbf{Q}\mathbf{}\mathbf{=}\mathbf{}\mathbf{(}\mathbf{60}\mathbf{}\overline{)\mathbf{g}}\mathbf{)}\mathbf{(}\mathbf{4}\mathbf{.}\mathbf{18}\mathbf{}\frac{\mathbf{J}}{\overline{)\mathbf{g}\mathbf{\xb0}\mathbf{C}}}\mathbf{)}\mathbf{\left(}\mathbf{22}\overline{)\mathbf{\xb0}\mathbf{C}\mathbf{-}\mathbf{16}\mathbf{.}\mathbf{9}\overline{)\mathbf{\xb0}\mathbf{C}}}\mathbf{\right)}\phantom{\rule{0ex}{0ex}}\mathbf{Q}\mathbf{}\mathbf{=}\mathbf{}\mathbf{1279}\mathbf{}\mathbf{J}$

Calculating the change in enthalpy for 1 mole ( MM of NH_{4}NO_{3 }is 80 g/mol)

When a 4.25-g sample of solid ammonium nitrate dissolves in 60.0 g of water in a coffee-cup calorimeter, the temperature drops from 22.0 ^{o}C to 16.9 ^{o}C.

You may want to reference (Pages 179 - 183) Section 5.5 while completing this problem.

Assume that the specific heat of the solution is the same as that of pure water.

Calculate delta(H) (in kJ/mol NH_{4}NO_{3}) for the solution proces.

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