1)
molar heat of solution = +25.7 kJ/mol
mass of water = 55.0 g
mass of NH4NO3 = 3.0 g
moles of NH4NO3 = 3.0 / 80.04 = 0.0375 mol
delta H = - Q / n
25.7 = - Q / 0.0375
Q = - 0.963 kJ = 963 kJ
Q = m Cp dT
963 = - 55 x 4.184 x (Tf - 22.9)
Tf = 18.7 oC
final temperature = 18.7 oC
Prelab: 1. The molar heat of solution of ammonium nitrate is +25.7 kJ/mol. Assuming the specific heat of the resulting...
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