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The image on an exposed and developed photographic film is due toA. `AgBr`B. `[Ag(C_(2)O_(3))_(2)]^(3+)`C. `Ag`D. `Ag_(2)O`. |
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Answer» Correct Answer - C Quinol developer (a reducing agent) reduces AgBr to Ag. `2AgBr^(+)(s)+2OH^(-)(aq)+C_(6)H_(5)(OH)_(2)(aq)to2Ag(s)+2H_(2)O+C_(6)H_(4)O_(2)(aq)+2Br^(-)(aq)` Where `AgBr^(+)` represents a molecules of AgBr exposed to light. |
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