Asked by gentil
The heat of solution of ammonium nitrate is 26.2 kJ/mol, and a 7.315 g sample of NH4NO3 is added to 40.0 mL of water in a constant pressure calorimeter initially at 25.2°C, what is the final temperature reached by the solution? (The specific heat of water = 4.184 J/g•°C)
Answers
Answered by
DrBob222
Heat lost by KNO3 + heat gained by water = 0
q lost by NH4NO3 is
26.3 kJ x (7.315/80) = approx 2.5 kJ but you need a better answer.
Since dH is + you know the reaction is endothermic and the solution will get colder.
2.5 = mass H2O x specific heat H2O x delta T.
Solve for delta T and subtract from 25.2. Remember that 2.5 is approximate.
q lost by NH4NO3 is
26.3 kJ x (7.315/80) = approx 2.5 kJ but you need a better answer.
Since dH is + you know the reaction is endothermic and the solution will get colder.
2.5 = mass H2O x specific heat H2O x delta T.
Solve for delta T and subtract from 25.2. Remember that 2.5 is approximate.
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