A) \[N{{O}_{2}}\]
B) \[{{N}_{2}}\]
C) NO
D) \[{{N}_{2}}O\]
Correct Answer: A
Solution :
\[Cu+4HN{{O}_{3(conc.)}}\to \underset{(A)}{\mathop{Cu{{\left( N{{O}_{3}} \right)}_{2}}}}\,+\underset{(B)}{\mathop{2N{{O}_{2}}}}\,+2{{H}_{2}}O\]You need to login to perform this action.
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