calculate the E cell for this reaction 3Cu(s) + 8H+(aq) + 2NO3-(aq) ---> 2NO(g) + 3Cu2+(aq) + 4H20(l)?
343278
Half reactions are:\nOxidation: Cu ----> Cu2+ + 2e    Eo = -0.34V\nReduction: NO3- + 4H+ + 3e --->  NO + 2H2O   Eo = +0.96V\n\nAdd the two voltages (do not multiply this number by the coefficients you use to cancel the number of electrons out) and you get the +0.62V for the Eo of the cell.\n\nThe Nernst equation will take this form:\nE = Eo - 0.0592/n * log {(PNO)^2 * [Cu2+]^3 / [NO3-]^2 * [H+]^8}\n\nn will equal 6 here, not 8, as that will be the number of electrons that will be cancelled out when you add the oxidation and reduction reactions together (multiplied by the correct number to cancel those electrons out).\n\nThe concentration's portion simplifies, once you plug all of the numbers in, to about 18.94, whose log is 1.278.  Multiply that by 0.0592 and divide by 6 and you get 0.01, so only 0.01 V will be lost from the standard potential for that reaction at those concentrations.
890894
C12H22O11+12O2-------12CO2+11H2O+E
893153
2Fe (s) + 3F2 (g) ---> 2FeF3(s)
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)
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)
978323
*your equation is written wrong it should be:\n3Cu(NO3)2 + 2Na3(PO4) -> Cu3(PO4)2 + 6Na(NO3)\n\nyou have to find your limiter and your excess.\n 1.) change grams to moles using moler mass\n2.2 g Cu(NO3)2 times 1mol/188 g = .0117mol\n1.29 Na3(PO4) times 1mol/164 g = .00787 mol\n\n2.) divide the two by the # of moles in the balanced equation:\nCu(NO3)2: .0117 mol/3 = .00390 mol\n Na3(PO4): .00787 mol/2=.00394 mol\n\n3.) So now you know you don't really have a limiter, but we'll go with Cu(NO3)2 b/c there's a little less. \n\nthere are 3 moles of Cu(NO3)2 used for every 1 mol of Cu3(PO4) produced.\n\n.0117 mol Cu(NO3)2 times (1 mol Cu3(PO4) / 3 mol Cu(NO3)2) = .00390 mol Cu3(PO4) produced\n\nconvert that to grams\n\n.00390 mol Cu3(PO4) times 286 g/mol = 1.12 g \n\nthat's your actual\n\nso % yield = .89/1.12 x 100 = 79.5%\n\noh yeah you find moler mass by adding the molecular weights of the atoms in the compound (times the number of the atoms in that compound) for ex. Cu3(PO4) 63.6 x 3 + 31.0 + 16.0 x 4 = 286\n\nDon't forget significant figures!!\n\nhope that helped!
262916
4 OH-(aq) --->  O2(g) + 2H2O(l) + 4e-
844624
FeCl3   +    3NaOH  ----->    Fe(OH)3   +   3NaCl\n\nthats it
863022
Here you are:\n\nFe+2 + 1/2O2 (g) + H2O (l)  ---> Fe (OH)2 (s)\n\nFe (OH)2 (s) + 1/2O2 (g) ---> Fe2O3 (s) + 2H2O (l)
1097630
Na2CO3 + 2HCl-> NaCl + H2O + CO2
788070
Well its easy dear.\n\n2Li (s) + Cl2 (g) ----> 2 LiCl (s)
770691
CoCl2(aq) + FeSO4(aq) = CoSO4 (aq) + FeCl2(aq)
892172
44 A) 2AgO --> 4Ag + O2\n     B)NiCO3 --> NiO + CO2\n     C)NH4NO3 --> N2O + 2H2O\n48)A)2HCl + Ca(OH)2 --> CaCl2 + 2H2O\n     B)3Ag2SO4 + 2AlCl3--> 6AgCl + Al2(SO4)3\n     C)&D) I've no idea, sry\n50)A)C4H8 + 6O2 --> 4CO2 + 4H2O\n     B)2C8H18 + 25O2 --> 16CO2 + 18H2O\n     C)2C3H8O3 + 7O2 --> 6CO2 + 8H2O\n     D)2C3H6O + 5O2 --> 6CO2 + 6H2O\n52)A)2Al + 3H2SO4 --> Al2(SO4)3 + 3H2\n     B)2HCl + Ba(OH)2--> BaCl2 + 2H2O\n     C)Au + HCl--> No Reaction(gold is below hydrogen in the metal reactivity series)\n54)A)2Cl2 + 2NaI --> 2NaCl2 + I2\n     B)2NH3--> N2 + 3H2\n     C)4Na + O2--> 2Na2O\n     D)Mg + H2SO4--> MgSO4 + H2\n     E)MgCl2 + Ca(OH)2--> Mg(OH)2 + CaCl2\n     F)H2 + Cl2--> 2HCl
739042
2Al + 6HNO3 ----> 2Al(NO3)3(s) + 3H2(g)
136146
The 1st answer is wrong. The reaction is\n\nNa2CO3(aq) + SrCl2 (aq)--> 2NaCl (aq) +SrCO3 (s)
945429
Equation becomes CuCl2 + 2KI --> CuI2 + 2KCI\n\nProducts are Cuprous Iodide and Potassium Chloride
701763
First, notice you have two S atoms on right hand side.\nTo get two S atoms on left, you must have two H2SO4 molecules.\nYour equation is now Cu + 2H2SO4 ---> CuSO4 + SO2 + ??\nIf you subtract right from left, you are left with 4H + 2O atoms, for 2H2O\nSo final eq is Cu + 2H2SO4 ---> CuSO4 + SO2 + 2H2O\n\nHope this helps you to approach these equation-balancing problems methodically.