See my comments to your other post regarding this being a made up problem.
I don't like this answer either based on the same rationale as I posted earlier; however,
q = qH2O + qCal
q = [mass H2O x specific heat H2O x dT) + (CCal*dT)
q = (100.0 x 4.18 x 9.80)+(6.50*9.80) = 4160.1 J
q/mol = -4160.1 J/0.050 mol = -83.2 J/mol which is -0.0832 kJ/mol. This is a ridiculous answer based on literature values of 57.1 kJ/mol.
Again, please let me know how this turns out. If I'm making an error I need to know it.
Calculate the Molar Enthalpy of Neutralization in kJ/mol of the reaction between a monoprotic acid and a monoprotic base, given the following information: *Make sure you report your answer using the correct sign*
The temperature change equals 9.80 degrees C, 50.0mL of 1.00 M concentration of Acid
50.0mL of 1.00 M concentration of Base, heat capacity of the calorimeter is 6.50 J/degree C. The specific heat of
water is 4.180 J/g degree C.
1 answer