Chemistry help

Question

In a bomb calorimeter (Ccal = 20.00 kJ•K–1), 1.211 g of a hydrocarbon (FW = 242.21 amu) is combusted. The termpeature of the calorimetry increased from 20.00 °C to 20.80 °C. What is the LaTeX: \Delta Δ Ecomb in kJ/mole?

Answer 1

∆T = change in temperature = 20.80 deg - 20.00 deg = 0.80 degrees

20.00 kJ/deg x 0.80 deg = 16.00 kJ

moles combusted = 1.211 g x 1 mole/242.21 g = 0.0049998 moles

∆H = 16.00 kJ/0.0049998 moles = 3200.13 kJ = 3.200x10³ kJ/mole