A) \[57.0\,\,kJ\]
B) \[11.4\,\,kJ\]
C) \[28.5\,\,kJ\]
D) \[34.9\,\,kJ\]
Correct Answer: B
Solution :
\[0.2\] mole of \[KOH\] will neutralise \[0.2\] mole of\[HN{{O}_{3}}\]. Thus, heat evolved\[=57\times 0.2=11.4\,\,kJ\]You need to login to perform this action.
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