Step 1

**Total volume = 50.0 mL + 50.0 mL = 100.0 mL**

**$\mathbf{m}\mathbf{}\mathbf{=}\mathbf{}\mathbf{100}\mathbf{}\overline{)\mathbf{mL}}\mathbf{}\mathbf{\times}\frac{\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{g}}{\mathbf{1}\mathbf{}\overline{)\mathbf{mL}}}$ = 100 g**

**$\mathbf{q}\mathbf{}\mathbf{=}\mathbf{}\mathbf{mc}\mathbf{\u2206}\mathbf{T}\phantom{\rule{0ex}{0ex}}\mathbf{q}\mathbf{}\mathbf{=}\mathbf{}\mathbf{(}\mathbf{100}\mathbf{}\mathbf{g}\mathbf{)}\mathbf{(}\mathbf{4}\mathbf{.}\mathbf{18}\mathbf{}\frac{\mathbf{J}}{\mathbf{g}\mathbf{\xb7}\mathbf{\xb0}\mathbf{C}}\mathbf{)}\mathbf{(}\mathbf{24}\mathbf{.}\mathbf{21}\mathbf{\xb0}\mathbf{C}\mathbf{}\mathbf{-}\mathbf{}\mathbf{23}\mathbf{.}\mathbf{40}\mathbf{\xb0}\mathbf{C}\mathbf{)}$**

**q = 338.58 J**

The addition of hydrochloric acid to a silver nitrate solution precipitates silver chloride according to the reaction:

AgNO_{3}(aq) + HCl(aq) → AgCl(s) + HNO_{3}(aq)

When 50.0 mL of 0.100 M AgNO_{3} is combined with 50.0 mL of 0.100 M HCl in a coffee-cup calorimeter, the temperature changes from 23.40˚C to 24.21˚C. Calculate ΔH_{rxn} for the reaction as written. Use 1.00 g/mL as the density of the solution and C_{s} = 4.18 J/(g • ˚C) as the specific heat capacity of the solution.

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