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B
NaOH(aq)+BaCl(aq)→BaOH(s)+NaCl(aq)
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C
Pb(NO3)2(aq)+2KI(aq)→2KNO3(aq)+PbI2(s)
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D
CuO(s)+Mg(s)→Cu(s)+MgO(s)
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E
4Fe+3O2→2Fe2O3
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Solution
The correct option is BPb(NO3)2(aq)+2KI(aq)→2KNO3(aq)+PbI2(s) The yellow precipitate of lead di iodide is produced in : Pb(NO3)2(aq)+2KI(aq)→2KNO3(aq)+PbI2(s)↓