Question

I've tried solving this problem 100x and i keep getting it wrong. I don't know where i am going wrong. Wouldnt it be:

(100ml * 4.180 * 6.49) + (6.50*6.49)
2713+42.2=2755.2=
2755.2/0.05=55104

Then convert to kj so
55104/1000=55.104 which would be 55.1 for correct sig digits????




Calculate the Molar Enthalpy of Neutralization (ΔHn) in kJ/mol of the exothermic reaction between a monoprotic acid and a monoprotic base, given the following information:

The temperature change equals 6.49°C,
50.0 mL of 1.00 M concentration of Acid
50.0 mL of 1.00 M concentration of Base
Heat capacity of the calorimeter is 6.50 J/°C.
The specific heat of water is 4.180 J/g°C

Answer 1

Have you tried making that -55.1 kJ/mol. It must be negative since it is exothermic. It's time for me to go to bed. Let me know quickly if that is right.