A 2.455 g sample of a new organic material is combusted in a bomb calorimeter. The temperature of the calorimeter and its contents increase from 23.61 ∘C to 27.04 ∘C. The heat capacity (calorimeter constant) of the calorimeter is 38.29 kJ/ ∘C, what is the heat of combustion per gram of the material?

The heat of combustion per gram of the material is 53.5 kJ

Explanation:

Let the heat released during reaction be q.

= Heat gained by calorimeter

Heat capacity of bomb calorimeter ,C = 38.29 kJ/°C  

Change in temperature = ΔT = (27.04-23.61) °C = 3.43 °C

Total heat released during reaction is equal to total heat gained by bomb calorimeter.

Thus 2.455 g of material releases 131.3 kJ of heat

1 g of material releases = of heat

Thus the heat of combustion per gram of the material is 53.5 kJ

c u c l 2 + 2 n a n o 3 → c u ( n o 3 ) 2 + 2 n a c l

  long weird answer but i hope this works

the mass of the cathode increases because ions from the solution get reduced to atoms and become part of the cathode.

explanation:

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