The correct answer is -2.09x10 to the power of 4 kJ/mol.
First, calculate the amount of heat released during the reaction:
q = mcΔT
q = (1071 kJ/°C)(3.44 °C)(0.0150 kg) (convert grams to kg)
q = - 52.99 kJ
Note that the "minus" sign in the answer indicates that heat is released during the reaction.
Next, calculate the moles of NaNO3:
moles NaNO3 = mass/molar mass = 15.0 g/(22.990 g/mol + 14.007 g/mol + 3(15.999 g/mol)) = 0.0881 mol
Finally, calculate the enthalpy change per mole of NaNO3:
ΔH = q/moles of NaNO3 = -52.99 kJ/0.0881 mol = -2.09x10 to the power of 4 kJ/mol
When 15.0 g of sodium nitrate (NaNO3) was dissolved in water in a bomb calorimeter, the temperature reduced from 25.00 °C to 21.56 °C. If the heat capacity of the solution and the calorimeter is 1071kJ/°C, what is the enthalpy change for the process per mol of sodium nitrate that dissolves in water? (Atomic masses of N-14.007 g/mol; Na-22.990 g/mol;O-15.999 g/mol)
-13.03x10 to the power of 4 kJ/mol
+13684.24 x10 to the power of 4 kJ/mol
-52.99x10 to the power of 4 kJ/mol
-2.09x10 to the power of 4 kJ/mol
1 answer