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$H_{2} S$ can be distinguished from ${SO}_{2}$ by its action on

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detailed solution

Correct option is D

Lead acetate and $H_{2} S$ combine to form black $PbS$ .

$Pb {({CH}_{3} COO)}_{2} + H_{2} S \to PbS + {CH}_{3} COOH$ .

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H2S can be distinguished from SO2 by its action on