... Data: Trial #1: HCl (aq) + NaOH (aq)--> H20 (l) + NaCl (aq) Trial #2: NaOH (aq) + NH4Cl (aq)--> NH3 (aq) + NaCl (aq) + H20 (l) Trial #3: HCl (aq) + NH4OH (aq) --> NH4Cl (aq) + H20 (l) Trial #1 Trial #2 Trial #3 Initial Temperature C ? 21.93 22.34 21.62 Final Temperature C ? 35.22 23.24 34.38 Data Analysis: * Calculating the amount of heat energy: (q) = Cp x m x ?T Trial #1: q = (4.18 J/g ?C) x (1.03 g/mL x 50mL) (35.22-21.93) = 2860.94 J Trial #2: q = (4.18 J/g ?C) x (1.03 g/mL x 50mL) (23.24-22.34) = 193.74 J Trial #3: q = (4.18 J/g ?C) X (1.03 g/mL X 50mL) (34.38-21.62) = 2746.85 J * Calculating the Enthalpy Change: Indirect Method Calculating the net ionic equation: ?H = ?H (products) - ?H (reactants) Trial #1: HCl (aq) + NaOH (aq)--> H20 (l) + NaCl (aq) --> overall H+ (aq) + OH- (aq)--> H20 (l) --> net ionic ...

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