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Индира347
11.12.2021 12:19
Химия
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How many ml of water should be added to 5000 ml of sulfuric acid solution with a titer of 0.009001 so that 1ml of the resulting solution corresponds to 0.01 g of barium sulfate?

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ekatsinrom
ekatsinrom
4,6(33 оценок)

BaSO4->H2SO4, so 0.01 g of baso4 corresponds to 98*(0.01/233) = 4.291845*10^-5 g of H2SO4 - that's the resulting solution titer (i. e. 4.291845*10^-5 g/ml),

One must dilute original solution as much as 0.009001/4.291845*10^-5 = 209.723 times higher,

The resulting solution volume is 5000*209.723 = 1048615 ml,

Water to be added (ml) = 1048615-5000 = 1043615.

olga19852
olga19852
4,8(89 оценок)

а) 3   feso4+2 na3po4 = fe3(po4)2+3 na2so4

3 fe(2+)+3 so4(2-) + 6 na(+)+2 po4(3-) = fe3(po4)2+6 na(+)+3 so4(2-)

3 fe(2+)+2 po4(3-) = fe3(po4)2

б) ca(oh)2+cu(no3)2 = ca(no3)2+cu(oh)2

ca(2+)+2 oh(-) + cu(2+)+2 no3(-) = ca(2+)+2 no3(-)(+cu(oh)2

cu(2+)+2 oh(-) = cu(oh)2

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