A) \[54\text{ }kcal\]
B) \[\text{98 }kcal\]
C) \[\text{196 }kcal\]
D) \[\text{108 }kcal\]
Correct Answer: B
Solution :
Equation (i) can be obtained by adding eq. (ii) and (iii). \[\Delta {{H}^{o}}=152+(-44)\] \[=152-44\] \[=98kcal\]You need to login to perform this action.
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