orin6lenPolsousing
orin6lenPolsousing orin6lenPolsousing
  • 20-12-2016
  • Chemistry
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a 10.00 gram sample of a soluble barium salt is treated with an excess of sodium sulfate to precipitate 11.21 grams BaSO4, (M= 233.4). which barium salt is it?

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Аноним Аноним
  • 20-12-2016
Hope it helps. All the best!
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MissPhiladelphia
MissPhiladelphia MissPhiladelphia
  • 22-12-2016
For the answer to the question above, first divide the 
BaSo4 and 233.4 and it will look like this.

(11.21 g BaSO4) / (233.4 g/mol BaSO4) = 0.0480 mol BaSO4 and original barium salt 
Then,
(10.0 g) / (0.0480 mol) = 208.3 g/mol 
So the answer would be BaCl2.
I hope my answer helped you.
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