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(Solved): You mix 60.0 ml of 0.200M lead (II) nitrate solution with 40.0 ml of 0.500M potassium iodide solutio ...



You mix 60.0 ml of 0.200M lead (II) nitrate solution with 40.0 ml of 0.500M potassium iodide solution, and a yellow precipitate of lead (II) iodide results. Assume 100% yield of the precipitated solid. Assume the precipitation of the solid has a negligible effect on the volume of the remaining liquid.

 

a. How many moles of lead(II) iodide are produced?

b. What concentrations of which ions remain in the solution?



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