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126 mL of 0.022 M Pb(NO3)2 solution is mixed with 52.0 mL of 0.103 M NaCl solution. A precipitate is observed to form. Assuming the

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126 mL of 0.022 M Pb(NO3)2 solution is mixed with 52.0 mL of 0.103 M NaCl solution. A precipitate is observed to form. Assuming the reaction goes to completion (as you will find out in Chem 1220, this is a lie, but for now we'll live with it), how many grams of precipitate do you expect to form? (Answers given in g.) O 0.67

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