What is the equilibrium constant (numarical value) of Fe3+ +SCN- <=>FeSCN 2+?
150280
The equilibrium constant is determined experimentally.  Where you give any experimental data in your homework problem?
770269
For aA+bB -> cC+dD\n\nE=E0-(RT/nF)* ln([C]^c*[D]^d / ([A]^a*[B]^b))\n\nwhere n=number of electrons taking part in the reaction\n\nSolids are not considered in the equation, so in your case\n\nE=E0 -(RT/nF)*ln([Fe2+]/[Cd+2])\n\nYou can find the standard reduction potentials for the Fe+2 and Cd+2 at \nhttp://www.northland.cc.mn.us/Chemistry/standard_reduction_potentials.htm\nThen E0= E0(Cd2+)-E0(Fe2+)\nR=8.314\nn=2\nF=96500\n\nYou have to specify T ...
138848
Here is one example ,\n\nAgNO3(aq) + NaCl(aq)-> AgCl(s)white precipitate + NaNO3(aq)
375945
Nickel (II) oxide, sulphur dichloride, and chlorine\n\n2NiCl2 + S02 ---> SCl2 + 2NiO + Cl2
661226
It doesn't matter how much iron you happen to dissolve as long as you have the exact mass you dissolved for the later calculations. Use just enough sulphuric acid to dissolve all of the iron, then transfer the solution into a measuring bulb and fill up to the mark - the theoretical concentration of iron is the mass you used divided by the volume. \n\nTheoretically you need 1 mol of acid for each mol of iron, but you will want some excess acid to keep the ions in solution. Make sure that oxygen has no chance to react with your solution during dissolving and titration.\n\nTitration with potassium permanganate in acid solution will require 1 mol permanganate for each 5 moles of iron. Usual lab concentrations are 0.02 mol/l (1 ml of that solution detects 0.1 mmol iron). You will want to use roughly 20 ml permanganate solution per titration for a volume error <1%.\n\nNow we start calculating (using SI units). Iron has a molar mass of 55.8... g/mol. If you want to use about 20 ml, you will need about 0.002 mol of iron. \n\nYour iron solution will have a concentration of \n\nC = m / (M * V)\n\nm : mass of (impure) iron you dissolved, in grams\nM : molar mass of iron, in gram/mol\nV : volume of the volumetric bottle you filled your solution up with, in ml (for convenience)\n\nC : concentration of impure iron, in mol/ml\n\nDivide 0.002 by C to see (roughly) how many ml of your iron solution you want for your first titration. Take an appropriate full pipette which fits the order of magniture, or else dilute your original solution in a volumetric bottle so that you can use a full pipette you have at hand.\n\n\nNow to get percentage purity.\n\nI'll assume you can calculate a concentration from titration.\n\nWith your titration you will determine the measured concentration of iron (in small numbers of mol/ml). Divide that result by C (calculated above), and multiply by 100 to get percentiles.\n\nHave fun!
945366
Bromine is in it's ionic form for both reactions (solubility rules)\nSo you have the equation:\nFe+3 + Na(s) > Fe(s) + Na+\n\nHalf Reactions:\nFe+3 + 3e- > Fe(s)\nNa > Na+ + 1e-\n\nYou'll need 3 sodiums to take in the 3 electrons given off from the Iron.\n\nMaking it:\n\nNet-Ionic: Fe+3 + 3Na(s) > Fe(s) + 3Na+\nStandard: FeBr3 + 3Na(s) > Fe(s) + 3NaBr
253611
Li+\nBe2+\nB3+\nN3-\nO2-\nF-
908510
It should be Fe2+ + 2OH- --> Fe(OH)2 since you not only need to balance the elements but you also need to balance the charges.
825401
First of all I assume  Na2Co3 should be Na2CO3.\n\nFe(CO)5 + 4NaOH = Na2Fe(CO)4 + Na2CO3+ 2H2O\n\nFirst, balance out the metals (Na and Fe) then C, which was already balanced. Then I chose to balance out H and which automatically balanced out O.
857898
2Al + 3*(CuO) -> Al2O3 + 3*Cu\n\nI2 + 2*(Na2S2O3) -> 2*(NaI) + Na2S4O6\n\nI2 + 5*(Cl2) -> 2*(ICl5)
103867
NaClO + 2FeSO4 + H2O -> NaCl + 2 Fe(3+) + 2 (SO4)(2-) + 2 OH(1-)
780875
a)4 Fe + 3O2--->2 Fe2O3\n\nCH4 + 2O2 ---> CO2 +2 H2O
183200
It is very obvious that the answer is Fe(CO2)2 or (OOCCOO)Fe.
374219
For the cell reaction: FeO(s) + H2(g) = Fe(s) + H2O(l) \n\nDG0(298 K) = DGf0 Fe(s) + DGf0 H2O(l) – DGf0 FeO(s) – DGf0 H2(g)\n\nFrom any table of Thermodynamic Properties:\nDGf0 Fe(s) = 0 KJ/mol\nDGf0 H2O(l) = -237.192 KJ/mol\nDGf0 FeO(s) = -251.4 KJ/mol\nDGf0 H2(g) = 0 KJ/mol\n\nThen: \nDG0(298 K) = 14.204 KJ/mol\n\nFurthermore: E0 = -DG0/(nF) = -14204/(2 x 96500) = -0.074 V vs NHE (normal hydrogen electrode), where n: number of electrons in each half reaction (FeO/Fe and 2H+/H2, n=2) and F is the Faraday constant (96500 C/mol).
375272
My guess is this:\n\nFeCl3 is a Lewis Acid. KSCN is a lewis base. AgNO3 is neither and it's a spectator.\n\nFeCl3 + KSCN ---> Fe(SCN)3 + KCl\n\nFe(SCN)3 may well be an orange color, depending on the field strength of SCN. The D orbitals in the iron are split in a 2X3 arrangement. The bottom 3 will each have one electron, according to Hund's rule. The top orbital can be reached, if SCN is a weak field ligand (I don't feel like looking up the spectrochemical series to find out). What's nice is that the total number of unpaired electrons will not change, so you don't have to worry about any kind of Spin-forbidden transition, that would render the compound colorless.
742926
Cu(s) + 2HNO3(aq) = Cu(NO3)2(aq) + H2(g)\n\nCu + 2H+ + 2NO3- = Cu+2 + 2NO3- + H2 (leaving off the s,aq,g)\n\nCu + 2H+ = Cu+2 + H2 (net ionic equation, the NO3-'s just cancel out, they don't change in the reaction).
429906
Never trust materials science majors ...\n\n1. NaCl (s) --> Na+ (aq) + Cl- (aq)\n2. Ca (s) + 2H2O (l) --> Ca(OH)2 (s) + H2 (g)\n3. Ca(OH)2 (s) --> H2O (g) + CaO (s)