In a coffee-cup calorimeter, 1 mol NaOH and 1 mol HBr initially at 23.5 oC (Celsius) are mixed in 100g of water to yield the following reaction:
NaOH + HBr → Na+(aq) + Br-(aq) + H2O(l)
After mixing the temperature rises to 84 oC. Calculate the change in enthalpy of this reaction.
Specific heat of the solution = 4.184 J/(g oC)
State your answer in kJ with 3 significant figures. Don't forget to enter the unit behind the numerical answer.
The molecular weight of NaOH is 40.0 g/mol, and the molecular weight of HBr is 80.9 g/mol.
ΔH = ?
I keep doing it as such
ΔH=-(c)(m)(ΔT)
So,
ΔH=-(4.184)(100+80.9+40.0)(84-23.5)
Where am I going wrong?