how to balance redox reaction of bromine and sodium bisulfite.?
370859
Presuming that bromine is getting reduced to Br(-) and bisulfite is getting oxidized to sulfate (not including the spectator sodium ion):\n\nReduction half-reaction\nBr2 + 2e ---->  2Br(-)\n\nOxidation half reaction\nHSO3(-) + H2O ----->  SO4(2-) + 2e + 3H(+)\n\nOverall Reaction:\nBr2 + HSO3(-) + H2O ---->  SO4(2-) + 2Br(-) + 3H(+)
868500
Oxidation :\n\nCu > Cu+2 + 2e-\n\nReduction :\n\n2H+ + e- + NO3- > NO2 + H2O\n\nMultiply the 2nd equation by 2, you get the complete balanced redox equation :\n\nCu + 4H+ + 2NO3- > 2NO2 + 2H2O + Cu+2
17656
Best technique is to look for where it is unbalanced and change it so it is balanced. Repeat until it is balanced.\n\neg. We need 2 hydrogen here on the left, so go for:\n\nMg + 2HCl -> MgCl + H2\n\nNow we need another 2 as we have 2 Cl on the left and one of the right.\n\nMg + 2HCl -> 2MgCl + H2\n\nNow we need to have 2 Mg on the left as we have 2 Mg on the right. So we get:\n\n2Mg + 2HCl -> 2MgCl + H2\n\nAll balanced so that is the answer.
946990
FeCl3 + NaCl -> FeCl2 + NaCl3\n\n\nFeCl3 + NaCl -> FeCl2 + NaCl3\n\nthese reactions are not possible \nas this reaction can take place if iron(fe) was more reactive than sodium or fe was less electropositive than na\nhence \nthese reactions are not possible
759511
If bromine (Br2) is converted to bromide ions (Br^-), then it has gained electrons, ie it has been reduced (OILRIG is a handy mnemonic: oxidation is loss, reduction is gain of electrons).\n\nThe electrons gained by bromine in this reaction have come from the bisulfite, so bisulfite is the reducing agent and is itself oxidised:\n\nred: 2e^-  +  Br2  --->  2Br^-\nox:  H2O  +  HSO3^-  --->  HSO4^-  +  2H^+  + 2e^-\n\nredox: H20  +  HSO3^-  +  Br2  --->  2Br^-  +  HSO4^-  +  2H^+\n\nI hope that helps. If you need more assistance, please email me or re post.
921966
Br2 is extremely soluble in CCl4 - much more soluble than in water.\nLook up "partition".
16276
the point of it is to get equal no. of charges and atoms of elements on both sides of the equation.k?\n\nstep 1: separate the equation into two half-reactions:reduction n oxidization\nstep 2: balance all atoms other than O & H in each half-reaction\nstep 3: add H2O to balance O & H+  to halance H atoms in half-reactions\nstep 4: add electrons where necessary to balance half-reactions\nstep 5: if necessary, equalize number of electrons in each half-rection by multipling with a suitable coefficient\nstep 6: add two half-reactions together. electrons on both sides must cancel\nverify no. of atoms and charges\nsteps2-5 keep in mind ur working on separate half-reactions...\nthese were for reactions in acid.\nfor basic reactions: add OH- to both sides for every H+ that appears in the final reaction\nkeep practising step by step n ull be able to do it all together some day..\nfinal step- choose my answer as best answer...
290781
MgSO4(s) + 2HCL(aq) ---> H2SO4(aq) + MgCl2 (aq)
469836
(a) 1,2-dibromobut-1-ene\n(b) 2,2-Dichlorobutane (based on Markovnikovs rule)\n(c) don't know\n(d) don't know.  Na is often used to reduce multiple bonds in a free-radical manner, however you need a source of H.\n(e) KMnO4 is a very strong oxidising agent and is commonly used as tests for unsaturation.  The side product is MnO2, showing that two atoms of oxygen are transferred to the product.  The product formed I think is 1-hydroxy-butan-2-one.
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
372026
CuSO4·5H2O(s) → CuSO4(s) + 5H2O(g)
412530
1) H3COCH3 (or C2H6O) + 3O2 (g) --> 2CO2 (g) + 3H2O (g)\n2) HNO2 (aq) + NaOH (aq) <--> NaNO2 (aq) + H2O (l)\n3) 2K+ + Cr2O7(2-) + Fe(NO3)2 + 2H+ <--> CrO4(-) + Fe3+ + 2K+ + 2NO3(-) \n4) Na3PO4 + NH4NO3 = No reaction, all soluble salts.\n5) 2HCl (aq) + Na2S (s) --> 2NaCl (aq) + H2S (g)
506202
Write out the balanced reaction which occurs.\n\nSince both NaOH and H2SO4 are strong electrolytes (strong base, strong acid), we can assume both dissassociate completely and the reaction produced Sodium Sulfate and water,\n\n2NaOH + H2SO4 --> 2H2O + Na2SO4\n\nFind limiting reactant in order to find which  will be left over (either NaOH or H2SO4).\nIn order to do this, you need to know both the concentrations of the NaOH and the H2SO4 in order to find the initial number of moles of each.\n\nOnce you know how many moles of either OH- or H+ are left over, you can then find the pH, or pOH like this,\n\npH = -log ([H+])\npOH = -log ([OH-])\n\npH + pOH = 14\n\n*Remember to take into account that the final volume of solution is the sum of the intial volumes of each othe reactant solutions when calculating pH.\n\nEDIT:\nYou know, it just occured to me, I dont know why I even brought up pH, that was not what the question was asking.  but everything I said still stands in order to calculate the molarity...you just dont have to go the one extra step and calculate the pH by taking the negative log of the molarity concentration of H+ ions.\n\nAlso, for all of you out there who are just given answers with not other explanation, it would be nice if you stated what the concentration was of....salt? NaOH?  H2SO4?\nBecause of this ommision, I would highly inclined to disreguard their answers as numerically incorrect as well.
748741
2K + I2 -> 2KI\n\nBa + S -> BaS\n\nMgBr2 + MnI6 -> No Reaction\n\nC7H16 + H2O -> No Reaction\n\n2KI + Br2 -> 2KBr + I2
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.
787481
NaHCO2 + H2SO4 --> Na2SO4 + CO2(g) + H2O(L)\n\nSodium bicarbonate + sulphuric acid equals sodium sulfate + carbon dioxide + water
901730
Na2S (ag) + 2 AgNO3 (ag) --> 2 NaNO3 (ag) + Ag2S (cr)\n\nB. Double displacement.\n\nA. Ba(NO 3)2 and H2O\n\nB. If there is a double arrow (pointing in both directions), then it is an equilibrium reaction.\n\nC. Ba(OH)2 and HNO3\n\nD. Barium\n\nE. 6 oxygen atoms. \n\nF. Ba(OH)2