Complete and balance the following redox equation. (check my answer)?
823470
First write the two half reactions:\n\n3 e(-) + Bi(OH)3 ---> Bi\nSnO2(=) ---> snO3(=) + 2 e(-)\n\nMultiply the first equation by 2 and the second by 3 and sum to get:\n\n3SnO2(=) + 2Bi(OH)3 ---> 3SnO3(=) + 2Bi\n\n12 O in the left, 9 on the right, so add 3H2O\n\n3SnO2(=) + 2Bi(OH)3 ---> 3SnO3(=) + 2Bi + 3H2O\n\nCheck that H is OK.  Yes. Done\n\nSo the sum is 13. (d) is the answer.
971763
1) Fe(s) +Cu^2+(aq) --> Fe^2+(aq) + Cu(s)  Single replacement, redox reaction where iron is oxidized and copper is reduced\n\n2) 2Al(s)+(3/2)O2(g)-->Al2O3(s)\nredox reaction, oxygen is reduced (neutral to -2) and aluminum is oxidized (neutral to +3)
73319
the earlier equation is not balanced.\nCaO + 2HCl --------> CaCl2 + H2O
691513
Check that both sides of equation match. Here is an example:\n\n3 CH3CH2OH + 2 Cr2O7 2- + 16 H+ --> 3 CH3COOH + 4 Cr3+ + 11 H2O\n\nThere should be an equal amount of H, C, O, Cr in both sides of the equation. \nIt is sometimes difficult to know where to start balancing an equation. In such case you should try to find an element/molecule that does not appear in every molecule on the other side of the equation. Eg. when balancing this equation above, I first balance the number of Cr because it is on its own on the right side. Sometimes you just need to try and play with numbers for a while to find the right answer. Hope this helped...
1018313
Silver is being oxidized, nitrogen from the nitrate is being reduced.\n\nOxidation half reaction:\nAg --> Ag+  +  e\n\nReduction half reaction:\nNO3- +  4H+ + 3e  -->  NO + 2H2O\n\nNet balanced and reduced equation:\n3Ag +  NO3- +  4H+ --> 3Ag+ + NO + 2H2O
157730
I think FePO4 +O2 >>>>> Fe2O3 + PO2. balance it and solve it . n its simple stoichiometry.
455269
H2SO4 + Mg(OH)2 -> 2H2O + MgSO4
832296
A) Ag2O (s) >>> 2 Ag(s) + 1/2 O2 (g)\nB) Nickel (II) Carbonate >>>>>Nickel (II) Oxide + [Carbon Dioxide], NiCO3 -> NiO + CO2\nC) Ammonium Nitrate >>>>> Dinitrogen monoxide + Water,  \nC does not balance as is... but this does:\nNH3NO3 -> N2O +  1.5 H2O + 0.5 O2\n\n48) A) 2 HCl (aq) + Ca (OH)2 (aq) >>>> 2 H2O + CaCl2 (aq)\nB) 3 Ag2 SO4 (aq) + 2 AlCl3 (aq) >>>> 6 AgCl (s) + Al2(SO4)3 (aq) (silver chloride is a precipitate)\nC) H2C2O4 (aq) + 2 KOH (aq)>>>> 2 H2O + K2C2O4 (aq)\nD) CdBr2 (aq) + Na2S (Aq)>>> CdS(s) + 2 NaBr (aq) (Cadium sulfide is a precipitate)\n50 Write a Balanced equation for the complete combustion of each compound:\nA) Butene (C4H8): C4H8 + 6 O2 -> 4 CO2 + 4 H2O\nB) Octane (C8H18): C8H18 + 12.5 O2 -> 8 CO2 + 9 H2O\nC) glycerol (C3H8O3): C3H8O3 + 3.5 O2 -> 3 CO2 + 4 H2O\nD) Acetone (C3H6O): C3H6O + 4 O2 -> 3 CO2 + 3 H2O\n52: Complete these:\nA) Al (s) + H2SO4 (aq)>>>> AlSO4(aq) + H2(g)\nB) 2 HCl (aq) + Ba (OH)2 (aq)>>>> BaCl2 (aq) + 2 H2O\nC) Au (s) + 3 HCl (aq) >>>>> AuCl3(aq) + 1.5 H2 (g)\n54) EACH IS BALANCED BUT "INCORRECT", so correct these:\nA) Cl-2 + NaI(aq) >>> NaCl2(aq) + I-\nB) NH3(aq) >>> N- + H+3\nC) Na (s) + O2(g) >>NaO2(s)\nD) 2Mg (s) + H2SO4(aq) >>Mg2SO4(aq) + H2(g)\nE) MgCl2(aq) + Ca(OH)2(s) >>> Mg(OH)2(s) + CaCl2(aq)\nF) H2 (g) + Cl2 (g) >> H2Cl2 (g)
781203
2H3PO4(aq) + 3Mn(OH)2(s)  --> Mn3(PO4)2(s) + 6H2O(l)
764101
In all reactions one uses that 2 OH -> H2O + O, the split of oxygen goes to create the ion. Finally balance the equation:\n \nSO3 + 2 NaOH -> Na2SO4 + H2O\n\nCO2 + 2 KOH -> K2CO3 + H2O \n\nSO2 + Ba(OH)2 -> BaSO3 + H2O
844624
FeCl3   +    3NaOH  ----->    Fe(OH)3   +   3NaCl\n\nthats it
98695
My answer is,\n \nNH4NO3(ag) + KOH(aq) -> NH4OH(aq) + KNO3(aq)
222097
OH-(aq) + HC2H3O2(aq)---> H2O(l) + C2H3O2-(aq)
888667
Cu2S + 12HNO3 -> Cu(NO3)2 + CuSO4 + 10NO2 + 6H2O\n\nThere you go.
904934
2HNO3 + Mg(OH)2 -->Mg(NO3)2 +2H20\n\n2H+(aq) + 2NO3-(aq) + Mg+(aq) + 2(OH)-(aq) -->Mg+(aq) + 2(NO3)-(aq) +2H20(l)\n\n2+(aq) + 2(OH)-(aq) --> 2H20(l)
608403
NaOH (aq) + HCl (aq) --> NaCl (aq) + H2O (l)
148175
The correct balanced equation is:\n2ZnS + 3O2 --> 2ZnO + 2SO2\nThis is BALANCED\nMr chemistry teacher, you didnt balance your equation,...and i'm in the ninth grade