How to determine the half-reaction formulas of copper(II) chloride and a (sharpened) pencil.?
485711
What are you asking here?  If there is an equation for the half-reactions of a pencil dipped in cupric chloride solution?  If so, it is easy, since there is no reaction.\n\nCuCl2 ============> Cu2+  + 2 Cl-  in solution\n\nPencil lead  = graphite and clay.  No ions in solution.\n\nThus, no reaction.\n\nOTOH, if you are referring to lead ions in solution (say, from lead nitrate) then you get:\n\nCuCl2 + Pb2+ ===========> PbCl2(s) + Cu2+\nThe half-reactions are:\n\nCuCl2 ===========> Cu2+  + 2 Cl-\nPb(NO3)2 ==========> Pb2+  + 2 NO3-\nPb2+  + 2Cl- ============> PbCl2 (s)\n\nThe copper ions and nitrate ions would remain in solution.
888667
Cu2S + 12HNO3 -> Cu(NO3)2 + CuSO4 + 10NO2 + 6H2O\n\nThere you go.
853922
These are the right ones:\n\n2K(s) + 2H2O(l) ---> 2KOH(aq) + H2(g)\nH2SO4(aq) + 2KOH(aq) ---> K2SO4(aq) + 2H2O(l)\nCa(OH)2(aq) + 2HCl(aq) ---> CaCl2(aq) + 2H2O(l)
906651
2KOH=>K(2)O+H(2)O\n\nFeO+C=>CO+Fe\n\nCa+2AlCl=>CaCl(2)+2Al\n\nP(2)O(5)+3H(2)O=>2H(3)PO(4)\n\n2KI+Pb(NO3)2=>2KNO(3)+PbI(2)
854997
CuS + H2SO4 ===> H2S + CuSO4
1030709
1 H2 +1 Cl2 ------> 2 HCl\n\n     2 - H - 2\n     2 - Cl - 2
958410
1.\nMn(s) + SnCl2(aq) --> MnCl2(aq) + Sn(s)\nCr(s) + 2HBr(aq) --> CrBr2(aq) + H2(g)\nCo(s) + CaBr2(aq) --> N.R\n2Al(s) + 3Ni(NO3)2(aq) --> 2Al(NO3)3(aq) + 3Ni(s)\n\n2.\nMg + 2AgNO3 --> Mg(NO3)2 + 2Ag\nmole of Ag = 3.580 / 108 = 0.033 mole\nmole of AgNO3 = 0.033 mole = 33 mmole\nmolarity of AgNO3 = 33 / 50 = 0.66 M\n\n3.\na.soluble\nb.soluble\nc.soluble\nd.insoluble\ne.soluble\nf.soluble\ng.insoluble in water, soluble in acid\nh.soluble\n\n4.\n3NaOH(aq) + Fe(ClO4)3(aq) --> Fe(OH)3(s) + 3NaClO4(aq)\n3Na+ + 3OH- + Fe3+ + 3ClO4- --> Fe(OH)3 + 3Na+ + 3ClO4-\nFe3+ + 3OH- --> Fe(OH)3\n(NH4)2SO4(aq) + Pb(NO3)2(aq) --> PbSO4(s) + 2NH4NO3(aq)\n2NH4+ + SO42- + Pb2+ + 2NO3- --> PbSO4 + 2NH4+ + 2NO3-\nPb2+ + SO42- --> PbSO4\n\nHope it helps\npeace\nvixklen
726370
There should be no reaction.  Provided there is enough water to dissolve the CuCl, the Compounds would separate into their Ion counterparts.\n\nSo, youd end up with Cu+, K+, and Cl- in solution
822325
a) Find some Acetic Acid (CH3COOH)\nb) Find some Copper (II) Hydroxide (Cu(OH)2)\nc) Add to form Cu(CH3CO2)2 + 2H2O\n\nWithout the volume or the number of moles, preparation of a 0.530M can be done several ways. \n\n2CH3COOH + Cu(OH)2 --> Cu(CH3CO2)2 + 2H2O\n\nUsing 1L:\n0.530 mol of Cu(CH3CO2)2 means 1.060 mol of CH3COOH\n\n1.060 / 0.5L = 2.120M\n\n0.530 mol of Cu(CH3CO2)2 means 0.530 mol of Cu(OH)2\n\n0.530 / 0.5L = 1.060M\n\nSo mix 500mL of 2.120M Acetic Acid with 500mL of 1.060M Copper (II) Hydroxide.\n\nN.B. Being a weak acid, acetic acid will not fully react with copper hydroxide, and hence the actual molarity requiredmay be higher.
747348
Hi Kitty!  Dr. J to the rescue with a full answer.\n\n1) 4Li + O2 ---> 2Li2O\n\n2) 4HF + SiO2 ---> SiF4 + 2H2O\n\n3) It is simpler if we eliminate the K and the sodium and write the equation ionic because it is a redox reaction and we need to balance.\n\nThe overall redox equation will be:\n\nCr2O7(=) + SO3(=) + H(+) ---> Cr(+3) + SO4(=).\n\nNow you balance by the half reaction method.  The two reactions are:\n\n6e(-) + Cr2O7(=) + 14H(+) ---> 2Cr(+3) + 7H2O\n3SO3(=)  + 3H2O --->  3SO4(=) + 6H(+) + 6e(-)\n\nEliminating the electrons:\n\nCr2O7(=) + 3SO3(=) + 8H(+) ---> 2Cr(+3) + 3SO4(=) + 4H2O\n\nThis one was tough!!!  Thumbs up!\n\nIf you want you could put the K and the Na back.\n\nBest answer?
882018
1) 2Al + Fe2O3=Al2O3 + 2Fe\n2) 8.53g BaCl2*(1mole BaCl2/ 208.3G BaCl2)=.0410mole BaCl2\n.0410mole BaCl2*(2moles KCl/1moleBaCl2)=.0820 mole KCl2\n.0820mole KCl2*(110.1gKCl2/1mole KCl2)=9.03gKCl2\n3)i don't know this one\n4) 2NaOH + H2 is reduced\n5) i don't know this one\n6) 5gNa*(1moleNa/23gNa)=.217mole Na\n.217mole Na*(6.02x10^23/1mole Na)=1.31x10^23 atoms Na
231663
_KCl3 _KCL + _o2? K2Cl4o2 thats the answer i think.:)
770385
FeCl3+3NaOH>>Fe(OH)3+3NaCl
951188
4. How many grams of zinc chloride can be formed from the reaction of 7.96 grams of zinc with excess HCL?\n\nZn + 2HCl → ZnCl2 + H2\n\n1 mole Zn produces 1 mole ZnCl2\nTherefore 65.7g Zn produces (65.7 + 2 x 35.5)g ZnCl2\nie ~65.7g Zn produces ~136.7g ZnCl2\nSo 7.96g Zn produces ~7. 96 x 136.7/65.7g ZnCl2\nie ~ 16.6g ZnCl2\n\n5. How many kilograms of copper sulfide could be formed from the reaction of 1.80 mol of copper with excess sulfur.\n\nCu + S → CuS\n\nSop 1 mole Cu produces 1 mole CuS\nTherefore 1.8 mole Cu produces 1.8 mole CuS\n1 mol CuS weighs ~(63.5 + 32.1)g = 95.6g\nSo 1.8 mol CuS weighs ~ 1.8 x 95.6g ≈ 172.1g = 0.1721kg\n\n6. If Copper(one) sulfide is partially roasted in air (reaction with 02), copper(one) sulfite is formed. Subsequently upon heating, the copper sulfite thermally decomposes to copper (one) oxide and sulfur dioxide. Write balanced chemical equations for these 2 reactions\n\nCu2S + 3O2 → Cu2SO3\nCu2SO3 → Cu2O + SO2
433211
This is simple-\n3Mg(S) + N2(G) (2 is subscript) Yields (arrow) Mg3N2(S)
1008732
Ksp = [Pb+2]*[Cl-]^2/[PbCl2]\n\nby definition, [PbCl2] = 1\n\nAlso, based on the reaction:\n\nPbCl2 =====> Pb+2  +  2 Cl-\n\n[Cl-] = 2 [Pb+2]\n\nSubstitute this into the equation, and you get:\n\nKsp = [Pb+2] * [Cl-]^2 = [Pb+2]*(2[Pb+2])^2 = 4[Pb+2]^3\n\nNow, since you know that 0.99 grams of PbCl2 is dissolved in your liter of solution, you can calculate the molarity from:\n\n(0.99 g)/ (278.1 grams/mole) = 3.56x10^-3 moles = [Pb+2]\n\nKsp = 4*(3.56x10^-3)^3 = 1.8 x 10^-7\n\nThe text value for the Ksp of lead (II) chloride is 1.7 x 10^-5\n=================\nOkay, now for the second part of your question:\n\nIf you have 0.5M lead nitrate, which is soluble, you will have 0.5M Pb+2\n\nSo, let's plug 'n chug (we'll use the text value for Ksp to show the process):\n\nKsp =  1.7x10^-5 = [Pb+2]*[Cl-]^2\n\nbut, with the lead nitrate in solution, [Pb+2] = 0.5 moles/liter (the amount you'd get from the dissolution of lead chloride will be so small as to be negligible), so:\n\nKsp = 1.7x10^-5 = [0.5]*[Cl-]^2\n\n[Cl-]^2 = (1.7x10^-5)/(0.5) = 3.4x10^-5\n\nso [Cl-] = sqrt(3.4x10^-5) = 5.83 x 10^-3  and since [Cl-] = 2 [Pb+2], we halve that value to get:\n\nsolubility = 5.83x10^-3/2 = 2.92x10^-3 moles per liter\n\n(2.92x10^-3 moles/liter) * 278.1 grams/mole = 0.811 grams/liter\n\nIf you use the value of the Ksp calculated in the first half of the question, then the solubility of PbCl2 in your solution is 3x10^-4 moles/liter, or 0.083 grams/liter
799504
FeS + 2HBr --> FeBr2 + H2S\nFe2+ + S2- + 2H+ + 2Br- --> Fe2+ + 2Br- + H2S\n2H+ + S2- --> H2S\n\nK2CO3 + CuCl2 --> 2KCl + CuCO3\n2K+ + CO32- + Cu2+ + 2Cl- --> 2K+ + 2Cl- + CuCO3\nCu2+ + CO32- --> CuCO3\n\npeace\nvixklen