constant volume caliometry

Question

What is ΔE for the reaction, in units of kJ/mol?

1.11 × 10^-3 mol sucrose

ΔT = 2.14 °C

q_water = 5.37 kJ

q_cal = 0.898 kJ

Answer 1

5.37 kJ + 0.898 kJ = 6.27 kJ

6.27 kJ / 1.11x10^-3 moles = 5.65x10³ kJ/mol