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Question 3 of 20 When 0.0701 mol of an unknown hydrocarbon is burned in a bomb calorimeter, the calorimeter increases in temp

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Answer #1

Heat (q) = C.∆T

q = 1.229×2.19

= 2.69 kJ

∆H = - q/n

= - 2.69/0.0701

= -38.4 kJ

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