Chemistry, asked by Monya8171, 1 year ago

A 10.0 ml of 0.600m sodium chloride is added to a 50.0 ml of 0.100m lead (ii) nitrate solution, and a precipitate is formed. The % yield of the reaction was 33.3% (% yield= actual yield/theoretical yield x 100). Calculate the actual yield of the precipitate in grams.

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Answered by airteluser

1

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