q(NH4NO3) + q(H2O) + q(cal) = 0
[(7.65/80.04)*36,000]+[(415*4.184*1.92)]+(C*1.82)=0
Solve for C.
I have done everything to solve this problem and I can't get the right answer (153.8J/degrees C).
Ammonium nitrate, a component of fertilizer, decomposes: NH4NO3(s) > N2O(g) + 2H2) (g)
the delta H=-36.0kJ/mol.
7.65g of ammonium nitrate was allowed to decompose in a bomb calorimeter containing 415g of water. The heat produced resulted in a temperature increase of 1.82 degrees C. What is the heat capacity of the calorimeter, in J/K? The specific heat capacity of water=4.184 J/K/g
Thanks!
3 answers
I still don't get 153.8 and I am unsure as to where the 1.92 came from?
The 1.92 is my typo. It should be 1.82 from the problem. It doesn't give 153.8 but it's close. I think that's the way to do the problem.
1 answer