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Dilute $H_2SO_4$ and dilute $HCl$ in the presence of air react with

$\begin {array} {1 1} (A)\;\text{copper only} \\ (B)\;\text{both copper and silver and not gold} \\ (C)\;\text{all the three coinage metals} \\ (D)\;\text{none of the three coinage metals} \end {array}$

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Answer : copper only
$2Cu+2H_2SO_4+O_2 \rightarrow 2CuSO_4+2H_2O$
$ \qquad \quad (\text{dilute})$
$2Cu+4HCl+O_2 \rightarrow 2CuCl_2+2H_2O$
answered Jul 31, 2014 by thanvigandhi_1
 

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