a 0.3511 gram sample of an unknown compound is burned in a bomb calorimeter. The

Question

a 0.3511 gram sample of an unknown compound is burned in a bomb calorimeter. The 101.5 grams of water warm from 22.14 degrees C to 26.82 degrees C. what is the heat of combustion (kJ/g) of the substance a 0.3511 gram sample of an unknown compound is burned in a bomb calorimeter. The 101.5 grams of water warm from 22.14 degrees C to 26.82 degrees C. what is the heat of combustion (kJ/g) of the substance a 0.3511 gram sample of an unknown compound is burned in a bomb calorimeter. The 101.5 grams of water warm from 22.14 degrees C to 26.82 degrees C. what is the heat of combustion (kJ/g) of the substance

Explanation / Answer

heat of reaction q = mCpdT

m = 101.5 g

Cp = 4.184 J/g.oC

dT = (26.82-22.14) = 4.68 oC

Feed values,

q = 101.5 x 4.184 x 4.68 = 1.99 kJ

So the heat of combustion of the substance = 1.99/0.3511 = 5.67 kJ/g

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