`Xg` of ethanal was subjected to combustion in a bomb calorimeter and the heat produced is `Y J`. ThenA. `Delta E_(("combustion"))=-XJ`B. `Delta E_(("combustion"))= -YJ`C. `Delta E_(("combustion"))= -(44Y)/(X)"J mol"^(-1)`D. `Delta H_(("combustion"))= -(44Y)/(X)"J mol"^(-1)`

`Xg` of ethanal was subjected to combustion in a bomb calorimeter and

1.	`Xg` of ethanal was subjected to combustion in a bomb calorimeter and the heat produced is `Y J`. ThenA. `Delta E_(("combustion"))=-XJ`B. `Delta E_(("combustion"))= -YJ`C. `Delta E_(("combustion"))= -(44Y)/(X)"J mol"^(-1)`D. `Delta H_(("combustion"))= -(44Y)/(X)"J mol"^(-1)`
Answer» Correct Answer - C Heat of combustion is determined for 1 mole substane. In Bomb calorimeter heat of reaction `= Delta E` `Xg CH_(3)CHO` produces Y Joule heat `therefore` 1 mole `=44 g CH_(3)CHO` will produce `(Y)/(X)xx44` Joule heat `Delta E=(-44Y)/(X)J mol^(-1)`

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`Xg` of ethanal was subjected to combustion in a bomb calorimeter and the heat produced is `Y J`. ThenA. `Delta E_(("combustion"))=-XJ`B. `Delta E_(("combustion"))= -YJ`C. `Delta E_(("combustion"))= -(44Y)/(X)"J mol"^(-1)`D. `Delta H_(("combustion"))= -(44Y)/(X)"J mol"^(-1)`

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