What is the equilibrium constant (numarical value) of Fe3+ +SCN- <=>FeSCN 2+?
138065
The equilibrium constant of any equation is [C]^c[D]^d/[A]^a[B]^b.\n\nthe equation is already balanced. But you have to know how many moles of each element there are in order to compute the problem
51552
AgNO3 + NaCl -----> AgCl + NaNO3\n\ntotal ionic equation\n\nAg+  +  NO3-  +  Na+  +  Cl-\n\n                      ----->\n\n                            AgCl  +  Na+  +  NO3-\n\nnet ionic equation\n\n    Ag+  +  Cl-  ----->  AgCl
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.
729557
_NaOH(aq) + _Fe(ClO4)3(aq) => _Fe(OH)3(s) + _NaClO4(aq)\n\n3  NaOH(aq) + 1  Fe(ClO4)3(aq) => 1  Fe(OH)3(s) + 3  NaClO4(aq)\nThe coefficients of one could be excluded but I put one which is also equivalent.\n\n_((NH4)2)SO4(aq) + _Pb(NO3)2(aq) => _PbSO4(s) + _NH4NO3(aq)\n\n1  ((NH4)2)SO4(aq) + 1  Pb(NO3)2(aq) => 1  PbSO4(s) + 2  NH4NO3(aq)\n\nThese are only the balanced equations.\n\nThe following is the ionic equations:\n(Na)+ (aq) + (OH)- (aq) + (Fe)3+ (aq) + (ClO4)- (aq) => Fe(OH)3(s) + (Na)+ (aq) + (ClO4)- (aq)\nand\n(NH4)+ (aq) + (SO4)2- (aq) + (Pb)2+ (aq) + (NO3)- (aq) => PbSO4(s) + (NH4)+ (aq) + (NO3)- (aq)\n\nThe ionic equations are already balanced.  It is simply the breaking down of the equations into ions.  The solid remains intact while the compounds in aqueous solution become ions.\n\nThe + and - signs are the charge of the ions.  I can't find any other way to write them so I separated the compound with parentheses.  Sorry if it's confusing.\n\nThe net ionic equations are "simplified" ionic equations.  Basically, the things that are repeated on both sides get crossed off.  What is remaining is simply the net ionic equation.\n\n(OH)- (aq) + (Fe)3+ (aq) => Fe(OH)3(s)\nand\n(SO4)2- (aq) + (Pb)2+ (aq) => PbSO4(s)
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!
893153
2Fe (s) + 3F2 (g) ---> 2FeF3(s)
777428
well actually me and you are in the same situation. im trying to find out how to write wrod equations into a reaction. lol but neway i just learned how to balance equations its very simple.\nFe2O3+CO--Fe+CO2\n:draw a box around each set of elements:Fe2O3 would b in a box CO would be in a box and so forth.\n:then write out a diagram of how many atoms each element has:\nEx: Element/ Before/After\n      Fe          2         1\n      O            4         1\n      C            1          1\n:Then Write numbers in front of each of the boxes until the inventory for each element is the same both before and after the reaction. Now, what happens when we put a number in front of a formula? Basically, anything in that box is multiplied by that number, because we're saying that we have that many of that kind of molecule. \n:Now do this until the diagram is equal on both sides\nAND YOU HAVE YOUR BALANCED EQUATION
283215
[Cu(NH3)4]SO4*H2O (s) + 2HCl(aq) --->\n\nCu(NH3)4 +2 (aq) + H2SO4(aq) + H2O(l) + 2Cl-(aq)
677826
1) is already balanced... to be more precise\n2C2H6 + 7O2 --> 4 C02 + 6H2O\n\n2) Fe2O3 + (3)H2SO4 --> Fe2(SO4)3 + (3)H2O
861343
Ferrous means fe (II)\n1) 3Fe2+ + MnO4- + 4H+ to give  3Fe3+ + Mn2+ + 2H2O \n\nThere are two half equations\nthe first is pretty simple: 3 Fe2+ to give 3 Fe3+. That is the oxidation reaction\n\nthe other is the reduction. \nMn +7 is reduced to Mn+2\nthe secret to balancing this is to add water to the side with no oxygen, and H+ to the other side to balance it out. Check that the charges are balanced\n\n2) Fe3+ + Sn2+ to give Fe2+ + Sn3+\n \n3) cant help u here u need a textbook!
382458
H2 + Cl2 -> 2HCl\n2Ag + H2S -> Ag2S + H2 \nCH4 + 2O2 -> CO2 + 2H2O\n2HgO -> 2Hg + O2\n3Fe + 4H2O -> Fe3O4 + 4H2
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}
375945
Nickel (II) oxide, sulphur dichloride, and chlorine\n\n2NiCl2 + S02 ---> SCl2 + 2NiO + Cl2
170129
CuSO4 (s)  --H2O-->  Cu+2 (aq) + SO4-2 (aq)
888667
Cu2S + 12HNO3 -> Cu(NO3)2 + CuSO4 + 10NO2 + 6H2O\n\nThere you go.
612107
a. 2k+ = .20moles , SO42-=.10moles\n\nb.Fe3+=.10moles,3NO3-= .30moles\n\nc.2Al3+=.20moles,3SO42-=.30moles\n\nd. Ni2+=.10moles,SO42-=.10moles
350981
This isn't too bad.\n\nFirst, find the enthalpy of formation for each of the reactants and products.\n\nFe2O3(cr) = -272 kJ/mol\nCO(g) = -110.525 kJ/mol\nCO2(g) = -393.509 kJ/mol\nFe = 0 kJ/mol\n\nThen add the enthalpies of formation for the reactants and subtract the enthalpies of formation for the products:\n\nFe2O3 + 3CO - (3CO2 + 2Fe)\n\n-272 kJ/mol + 3(-110.525 kJ/mol) - 3(-393.509 kJ/mol) - 2(0) = -603.575 kJ/mol - -1180.53 = -576.952 kJ/mol \n\nThat should do it.