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18 Nov 2019
a solution containing 38.57 g of lead(l) acetate is allowed to react completely with a solution containing 6.256 g of sodium sulfide, how many grams of solid precipitate will be formed? Number 19.182 How many grams of the reactant in excess will remain after the reaction? Number 37.0713g Assuming complete precipitation, how many moles of each ion remain in solution? If an ion is no longer in solution, enter a zero (0) for the number of moles. Number Number Pb2+- mol C2H,01 mol Tools x 102 Number Number Na' = mol mol
a solution containing 38.57 g of lead(l) acetate is allowed to react completely with a solution containing 6.256 g of sodium sulfide, how many grams of solid precipitate will be formed? Number 19.182 How many grams of the reactant in excess will remain after the reaction? Number 37.0713g Assuming complete precipitation, how many moles of each ion remain in solution? If an ion is no longer in solution, enter a zero (0) for the number of moles. Number Number Pb2+- mol C2H,01 mol Tools x 102 Number Number Na' = mol mol
Bunny GreenfelderLv2
21 Oct 2019