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What is the minimum amount of water (in L) needed to dissolve 8.37 g of PbCl2?...

What is the minimum amount of water (in L) needed to dissolve 8.37 g of PbCl2? The Ksp is 0.000015. Also for reference, Pb = 207.19 g/mol and Cl = 35.453 g/mol.

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Answer #1

Pb 2+ + 2W lo s 25 Rsp a kapa [5] [2502 ust O us 0.00ools 15x10 6 = 3.75 X 100 as

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