How do I determine the charge of the iron ion in FeCl3?
816320
Since chloride is an halogen, it has a charge of -1.\nBecause a molecule always have a charge of zero, the iron must be a positive charge.\nThere are three Cl, for a total of -3, and only one Fe, that has to have the whole +3 all to itself to balance out the -3 of the 3 Cl.
643732
The six cyanide (CN) radicals makes minus six charge. Fe can either have +2 or +3 charge. Even considering +3 for Fe, the net charge on [Fe(CN)6] is -6+3=-3 which can not be balanced by Zn as it can have only +2 charge. Because -3+2=-1 still remains i.e. charges are not balanced, the compound can not exist.
775039
For the groups on the table, the ion charges are\n\n          1       +1\n          2       +2\n         13      +3\n         14    +4/-4\n         15       -3\n         16       -2\n         17       -1\n\nThe ion charge equals the number of electrons gained (-) or \nlost (+) by the atom to form the ion.
175051
Iron (III) oxalate ---> Fe2(C2O4)3\n{valency of oxalate i.e. C204 is -2 and of Fe is +3}\n\nLead (II) carbonate---> PbCO3\n{valency of Pb is +2 and of CO3 is -2}
819316
CuSO4+Fe = FeSO4+Cu\nCompute the number of atoms of CuSO4 in 7 gms of it, the number of atoms in 2gm of Fe. Choose the smaller number of the two. That is the number of atoms of Cu formed. Compute its weight.
607593
K2PtCl6 only reacts with Ferric ions of Iron i.e. Fe3+ this is rate determining reaction
616395
This compund has an octahedral geometry and is coordinated with 3 bidentate oxalate (C2O4) ligands.\nReaction of potassium oxalate hydrate and ferric chloride\n\n3K2C2O4ˇH2O + FeCl3 --> K3[Fe(C2O4)3]ˇ3H2O + 3KCl\n\nWater molekules do not act as ligands and potassium ions act as counter-ions
918374
Fe2O3 -- Fe+O2\n\nthis is the balanced form\n\n2(Fe2O3)--- 4Fe+3O2\nwithout the parentheses
414002
one cannot measure electronegativity.it has been given on the Pauling scale given by Pauling .and charge can be determined by the valency of the element.
727004
(NH4)2Cr2O7 -> Cr2O3 + 4*(H2O) + N2\n\n(NH4)2SO4 + 2*(NaOH) -> 2*(NH3) + 2*(H2O) + Na2SO4\n\nFe2(SO4)3 + 6*(NH3) + 6*(H2O) -> 2*(Fe(OH)3) + 3*((NH4)2SO4)
1114092
All you really need to know is what column the element is in.  The first column from the left is +1 and the second column from the left is +2.  Here you have to remember to skip over to column 13, which is +3.  14 is + or -4, 15 is -3, 16 is -2, 17 is -1, and 18 is 0.  Columns 3-12 are metals, and there are usually several small numbers above the symbol on a good table.  They will tell you the charge of the element.  If there are more than one for the metal, the bold one is the most common.\n\nHope this helps and isn't too confusing!
426422
The official English name for the element in question is Iron.\n\nThe official symbol for the element is Fe, which comes from the Latin name for iron: "Ferrum."\n\nThe latest international convention of naming iron compounds (using English) is to  call the compound Iron and use Roman numerals to indicate the oxidation state of the iron.  Therefore, FeCl3 is Iron (III) Chloride, while FeS is Iron (II) Sulfide.\n\nOlder methods called Iron (III) compounds "Ferric" and Iron (II) compounds "Ferrous," so that the examples above would be called Ferric Chloride and Ferrous Sulfide.
706351
KCl(aq) + AgNO3(aq)  --> KNO3(aq) + AgCl(s)\n\nNet ionic: Cl-(aq) + Ag+(aq)  --> AgCl(s)\n\nNa2SO4(aq) + Ba(NO3)2  --> 2NaNO3(aq) + BaSO4(s)\n\nnet ionic: SO4(-2)(aq) + Ba(+2)(aq)  --> BaSO4(s)
834501
First let's find how many moles of each one we have\n\nmole= M*V, so for \nFeCl3 =0.234*25*10^-3= 5.85*10^-3\nNaOH =0.453*42.5*10^-3 =19.25*10^-3\n\nFeCl3 + 3NaOH -> Fe(OH)3 + 3NaCl\n\nso\n\n1mole FeCl3 reacts with 3mole NaOH\n5.85*10^-3 mole react with x\n\nThus x =3*5.85*10^-3=17.55*10^-3 mole NaOH but we have 19.25*10^-3 mole NaOH, therefore NaOH is in excess\nTherefore the amount of Fe(OH)3 will be calculated according to the limiting reageant FeCl3.\n\nso from the reaction\n\n1 mole FeCl3 gives 1 mole Fe(OH)3\nor better yet, since the molecular weight of Fe(OH)3 is 106.85 \n1 mole FeCl3 gives 106.85 g Fe(OH)3\n5.85*10^-3 mole give x g\nx= 106.85*5.85*10^-3= 0.625 g\n\nWe calculated that 17.55*10^-3 mole NaOH reacted.\nSo we have remaining (19.25-17.55) *10^-3= 1.7*10^-3 mole NaOH\n\nThese are in a total volume of 25.0+42.5=67.5 ml= 67.5*10^-3L.\n\nSo the concentration of the excess NaOH is\nC=mole/V= 1.7*10^-3/67.5*10^-3= 0.025 M
971905
First, look at the oxygen.  Oxygen has a charge of 2-, and there are 4, so in the entire molecule, oxygen gives a charge of 8-. The total charge of the molecule is 1-, therefore to find Cl, create the equation -8+x = -1\nx=7\nTherefore chlorine has a charge of 7+\nThe total charge of the molecule = the sum of the charges of each atom
87280
MBrO3-9H2O where M can be Na,K,Li, other +1 cations\n\nIf M is other than these you have to adjust subscripts
165799
You want us to calculate acitivites?\n\nYou need to determine the ionic strength and atomic radius in order to do that.