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calculate number of Ca^2+ ions in 1.97 g Ca3(PO4)2?
thanks!
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- Trevor HLv 76 years ago
Molar mass Ca3(PO4)2 = 310.18 g/mol
mol in 1.97g = 1.97/310.18 = 6.35*10^-3 mol Ca3(PO4)2
1 mol contains 6.02*10^23 molecules
molecules of Ca3(PO4)2 = (6.35*10^-3) * ( 6.02*10^23) = 3.82*10^21 molecules
Each molecule contains 3 Ca 2+ ions
No of Ca2+ ions = 3*(3.82*10^21) = 1.15*10^22 Ca2+ ions.
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