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Pour a solution of Sodium(or Potassium) Iodide over Lead nitrate solution. The Lead iodide will be precipitated out as a yellow solid
A precipitate of Lead iodide and Potassium nitrate are formed
Lead iodide.
This is a double displacement reaction. 2KI + Pb(NO3)2 --> 2KNO3 + PbI2 Potassium iodide + Lead(II) nitrate --> Potassium nitrate + Lead(II) iodide A bright yellow precipitate will form when these two react.
just a clorine hydroxyde :))
Pour a solution of Sodium(or Potassium) Iodide over Lead nitrate solution. The Lead iodide will be precipitated out as a yellow solid
A precipitate of Lead iodide and Potassium nitrate are formed
Aqueous lead nitrate plus aqueous sodium iodide produce solid lead iodide and aqueous sodium nitrate.
Lead iodide (Pb2I) precipitates as a yellow solid, leaving a solution of potassium and nitrate ions.
Lead iodide.
it doesn't
This is a double displacement reaction. 2KI + Pb(NO3)2 --> 2KNO3 + PbI2 Potassium iodide + Lead(II) nitrate --> Potassium nitrate + Lead(II) iodide A bright yellow precipitate will form when these two react.
Take a few drops of both samples and add some lead nitrate. A yellow precipitate indicates lead iodide and it gives the inference that it contains iodide ions, hence the solution of sodium iodide.
2KI+Pb(NO(3))(2) yields 2KNO(3)+PbI(2). You basically get potassium nitrate and lead (II) iodide when you react potassium iodide and lead nitrate dissolved in solution.
Yes. The equation is Pb(NO3)2(aq) + 2KI(aq) ==> PbI2(s) + 2 KNO3(aq)
Since lead iodide is insoluble in an aqueous solution, you could filter it out using a funnel and filter paper.