10 g of water are placed in a coffee-cup calorimeter. 2.00 g of NaNO3 is poured into the coffee-cup calorimeter. T(final-initial) is -3.88 degrees Celsius. The Heat of Formation for NaNO3 is -466.68 kJ/mol.
How do I find qsytem and deltaH(soln)?
dHsolution:
q = mass water x specific heat H2O x (Tfinal - Tinitial)
q = 10 g x 4.184 J/g*C x -3.88 = -162.3 J
So 2 g NaNO3 (2/85 = 0.0235 moles) caused a decrease in T of 3.88 C.
162.3 J/2 g x (85 g/mol) = ? J/mol is the delta H solution.
I don't know what you mean by qsystem.