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Cr+2 P-3 ----> these are the ions and their charges. You know chromium's charge because of the roman numeralCr+2 Cr+2 Cr+2 P-3 P-3 ------> the charges have to add up to zero, so three +2 Cr ions plus two -3 P ions cancel each other outCr3P2 -----> simplify
chromium(III) oxalate [Cr(ox)3]3- this is the actual formula. Chromuim (III) oxalate is the name of the complex. The (ox) being the ligands attached to the metal centre.
V+4 PO4-3 <--- these are the ions and their charges V+4 V+4 V+4 PO4-3 PO4-3 PO4-3 PO4-3 <--- the charges have to add up to zero, so three +4 vanadium ions cancels out four -3 phosphate ions V3(PO4)4 <--- simplify
Al+3+ PO4^-3 AlPO4
a phosphate ion is PO43-
Cr+2 P-3 ----> these are the ions and their charges. You know chromium's charge because of the roman numeralCr+2 Cr+2 Cr+2 P-3 P-3 ------> the charges have to add up to zero, so three +2 Cr ions plus two -3 P ions cancel each other outCr3P2 -----> simplify
PO4
Cr3(N)3 ===>CrN
(Hg2)3(PO4)2
The chemical formula of barium phosphate is Ba3(PO4)2.
Mo3(PO4)4
chromium(III) oxalate [Cr(ox)3]3- this is the actual formula. Chromuim (III) oxalate is the name of the complex. The (ox) being the ligands attached to the metal centre.
formula for Copper (I) is Cu- and the formula for Phosphate is (PO4)3- the two have to have a net charge of zero, but (Cu)-(PO4)3- is uneven +1+-3=-2 so adding 2 more Copper (I) to the compound the formula Cu3PO4 ends up as leaving the net charge to be 0 = +3 + -3
V+4 PO4-3 <--- these are the ions and their charges V+4 V+4 V+4 PO4-3 PO4-3 PO4-3 PO4-3 <--- the charges have to add up to zero, so three +4 vanadium ions cancels out four -3 phosphate ions V3(PO4)4 <--- simplify
Formula: (Hg2)3(PO4)2
Cu3PO4
T he correct formula for iron(III) phosphate is Fe3(PO4)2.