Yahoo Answers: Answers and Comments for What number of atoms of phosphorus are present in 1.01 g of Ca3(PO4)2 and Na2HPO4? [Chemistry]
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Fri, 24 May 2019 00:11:45 +0000
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Yahoo Answers: Answers and Comments for What number of atoms of phosphorus are present in 1.01 g of Ca3(PO4)2 and Na2HPO4? [Chemistry]
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From az_lender: Molar mass of Ca3(PO4)2 is 310.18 g/mol.
Molar...
https://answers.yahoo.com/question/index?qid=20190524001145AAlB0PQ
https://answers.yahoo.com/question/index?qid=20190524001145AAlB0PQ
Fri, 24 May 2019 00:47:46 +0000
Molar mass of Ca3(PO4)2 is 310.18 g/mol.
Molar mass of Na2HPO4 is 141.96 g/mol.
So 1.01 gram of Ca3(PO4)2 would be 3.256 millimoles,
i.e. 1.961 x 10^21 formula units. The number of P atoms would be
3.922 x 10^21.
And 1.01 gram of Na2HPO4 would be 7.115 millimoles,
i.e. 4.284 x 10^21 formula units. The number of P atoms would be
4.284 x 10^21.