3. A 3.51 g sample of benzene, C 6 H 6, was burned in a bomb calorimeter in an excess of oxygen. The initial temperature was 25.00 o C and rose to 37.18 o C. The heat capacity of the calorimeter and contents was 12.05 kJ/ o C. Find q (heat given off) for the combustion of 1 mol of benzene.

4. A 3.51 g sample of benzene, C 6 H 6, was burned in a bomb calorimeter in an excess of oxygen. The initial temperature was 25.00 o C and rose to 37.18 o C. the heat capacity of the calorimeter and contents was 12.05 kJ/ o C. Find q for the combustion of 1 mol of benzene. C 6 H 6 (l) + 15/2 O 2 (g)  6CO 2 (g) + 3H 2 O(l) 3.51 g  T = 12.18 o C What units do we desire?

5. A 3.51 g sample of benzene, C 6 H 6, was burned in a bomb calorimeter in an excess of oxygen. The initial temperature was 25.00 o C and rose to 37.18 o C. the heat capacity of the calorimeter and contents was 12.05 kJ/ o C. Find q for the combustion of 1 mol of benzene. C 6 H 6 (l) + 15/2 O 2 (g)  6CO 2 (g) + 3H 2 O(l) 3.51 g  T = 12.18 o C

6. If we could find the number of kJ and the number of moles of benzene, we could divide the kJ by the moles to get our answer!

7. A 3.51 g sample of benzene, C 6 H 6, was burned in a bomb calorimeter in an excess of oxygen. The initial temperature was 25.00 o C and rose to 37.18 o C. the heat capacity of the calorimeter and contents was 12.05 kJ/ o C. Find q for 1 mol of benzene. C 6 H 6 (l) + 15/2 O 2 (g)  6CO 2 (g) + 3H 2 O(l) 3.51 g  T = 12.18 o C = kJ 12.18 o C 146.8 = mol BZ 3.51 g 0.04497

8. Find q for 1 mol of benzene. C 6 H 6 (l) + 15/2 O 2 (g)  6CO 2 (g) + 3H 2 O(l) 3.51 g  T = 12.18 o C = kJ 12.18 o C 146.8 = mol BZ 3.51 g 0.04497 And now?

9. Find q for 1 mol of benzene. C 6 H 6 (l) + 15/2 O 2 (g)  6CO 2 (g) + 3H 2 O(l) 3.51 g  T = 12.18 o C = kJ 12.18 o C 146.8 = mol BZ 3.51 g 0.04497 146.8 kJ 0.04497 mol Bz 3264

10. Find q for 1 mol of benzene. C 6 H 6 (l) + 15/2 O 2 (g)  6CO 2 (g) + 3H 2 O(l) 3.51 g  T = 12.18 o C = 12.18 o C 3264 3.51 g