Calorimetry Problem (chemistry) Worked it out and keep getting the wrong answer. Help?
A calorimeter contains 16.0mL of water at 11.5 Celsius. When 2.10g of (a substance with a molar mass of 79.0g/mol) is added, it dissolves via the reaction X(s) + H2O(l) ------> X(aq)
and the temperature of the solution increases to 29.0 Celcius .
Calculate the enthalpy change, delta H , for this reaction per mole of .
Assume that the specific heat and density of the resulting solution are equal to those of water [4.18J/(gxC) and 1.00g/ml ] and that no heat is lost to the calorimeter itself, nor to the surroundings.
Express the change in enthalpy in kilojoules per mole to three significant figures.
Worked it out and got 44.0kJ/mol but it keeps telling me I am wrong. So no I am not asking you to do my homework for me, I really just don't understand how I am wrong.
A calorimeter contains 16.0mL of water at 11.5 Celsius. When 2.10g of (a substance with a molar mass of 79.0g/mol) is added, it dissolves via the reaction X(s) + H2O(l) ------> X(aq)
and the temperature of the solution increases to 29.0 Celcius .
Calculate the enthalpy change, delta H , for this reaction per mole of .
Assume that the specific heat and density of the resulting solution are equal to those of water [4.18J/(gxC) and 1.00g/ml ] and that no heat is lost to the calorimeter itself, nor to the surroundings.
Express the change in enthalpy in kilojoules per mole to three significant figures.
Worked it out and got 44.0kJ/mol but it keeps telling me I am wrong. So no I am not asking you to do my homework for me, I really just don't understand how I am wrong.
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(4.18 J/g·°C) x (16.0 g) x (29.0 - 11.5)°C = 1170.4 J gained by the water
(1170.4 J) / ((2.10 g) / (79.0 g/mol)) = 44029 J/mol = 44.0 kJ
I agree with you. I don't see anything wrong. Are you sure you copied all the amounts correctly?
(1170.4 J) / ((2.10 g) / (79.0 g/mol)) = 44029 J/mol = 44.0 kJ
I agree with you. I don't see anything wrong. Are you sure you copied all the amounts correctly?