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The reaction of concentrated $\text{HNO}_3$ with $\text{Cu}$ or $\text{Ag}$ liberates

$\begin {array} {1 1} (A)\;\text{NO} & \quad (B)\;\text{NO}_2 \\ (C)\;\text{N}_2\text{O}_3 & \quad (D)\;\text{N}_2\text{O}_5 \end {array}$

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Answer : $NO_2$
$ Cu+ 4HNO_3 \rightarrow Cu(NO_3)_2+2NO_2+2H_2O$
$ \qquad (\text{Cone})$
answered Jul 31, 2014 by thanvigandhi_1
 

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