A0.233 g sample of a hydrocarbon (MM-114.23 g/mol) is burned in a bomb calorimeter that has...

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Answer 1

Answer #1

Heat liberated(q) = C*DT

C = Heat capacity of calorimeter = 2.3 kj/c

DT = 28.5-25.0 = 3.5

q = 2.3*3.5 = 8.05 kjoule

q = 8.05 kj

n = No of mole of hydrocarbon burned = 0.233/114.23 = 0.00204 mole

DHrxn = - q/n

= - 8.05/0.00204

= -3947 kj/mol

answer: -3950

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Answer 2

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