q = [
mass H2O x specific heat H2O x (Tfinal-Tinitial)] + [Qcal*(Tfinal-Tinitial)]
This gives J/0.05 mol. Convert to kJ/mol
Calculate the molar heat of neutralization in kJ/mol of the reaction between HA and BOH given the following information:
The temperature change equals 8C,
50 mL of 1 M concentration of Acid
50 mL of 1 M concentration of Base
Heat capacity of the calorimeter is 6.5 J/C.
Remember that the specific heat of water is 4.18 J/gC
1 answer
1 answer