0.24J/g*degC * 4.37g * 2.5degC = 2.622J
The 2.5 degC is the difference between 25 and 27.5 deg C.
Neutralization reaction is the reaction between acid and base.
There the neutralization reaction is:
HBr + LiOH ----------> LiBr + H2O.
Hope this helps!
Answer: The molar enthalpy change is 73.04 kJ/mol
Explanation:
moles of HCl=
As NaOH is in excess 0.0415 moles of HCl reacts with 0.0415 moles of NaOH.
volume of water = 100.0 ml + 50.0 ml = 150.0 ml
density of water = 1.0 g/ml
mass of water =
q = heat released
m = mass = 150.0 g
c = specific heat =
= change in temperature =
Thus 0.0415 mol of HCl produces heat = 3031.3 J
1 mol of HCL produces heat =
Thus molar enthalpy change is 73.04 kJ/mol