1 ) What volume of 0.150 M hydrochloric acid reacts with excess lead (II) nitrate solution...

1 ) What volume of 0.150 M hydrochloric acid reacts with excess lead (II) nitrate solution in order to yield 1.88 g lead ( II) chloride precipitate?

2) Calculate the boiling point and the freezing point of 50.0 g of AlCl3 dissolved in 400g of water, note that AlCl3 produces four particles

3) If 45.5 mL of 0.150 M sodium sulfate solution reacts with 50.0 mL of 0.175 M aqueous barium nitrate, what is the mass of Ba SO4 precipitate

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queen_honey_blossom answered 3 months ago

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