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.

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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 o C and rose to o C. The heat capacity of the calorimeter and contents was kJ/ o C. Find q (heat given off) for the combustion of 1 mol of benzene.

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 o C and rose to o C. the heat capacity of the calorimeter and contents was 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 = o C What units do we desire?

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 o C and rose to o C. the heat capacity of the calorimeter and contents was 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 = o C

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!

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 o C and rose to o C. the heat capacity of the calorimeter and contents was 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 = o C = kJ o C = mol BZ 3.51 g

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 = o C = kJ o C = mol BZ 3.51 g And now?

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 = o C = kJ o C = mol BZ 3.51 g kJ mol Bz 3264

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 = o C = o C g