redox oxidation state?
959805
HBr, -1\nHOBr, +1\nBr2, 0\nHBrO4,  +7\nBrF3  +3\n\nThere is nothing to explain.  There are rules, and you need to learn them and follow them.\n\nHere is a site with a good set of rules\nhttp://www.science.uwaterloo.ca/~cchieh/cact/c123/oxidstat.html
819193
If you look at how the -CN is bonded [C≡N], you can see that the oxidation number of N is -3.  From that it will follow, as you note, that C is +2.
795154
2CO(g) + O2(g) --> 2CO2(g)
871537
It is not an redox or acid-base reaction. so you can't employ oxidation number method or ion-electron method. So the options you have left are the inspection method & algebraic method. Among them, I think this equation is way too simple to employ algebraic method.Why don't you employ the simple inspection method.\n\nC2H5OH-----------> Oh! BY THE WAY! WHAT IS HCO2. THERE IS NO SUCH COMPOUND AS THIS.\n\nHowever the possible products are:-\nC2H5OH----------------->CH3CHO + H2(Cu,300deg C)\nC2H5OH----------------->H2CCH2 + H2O(H2SO4,170 deg. C)\nC2H5OH----------------->C2H5OC2H5(H2SO4,140 deg. C)\nC2H5OH----------------->H2CCH2 + H2O(Al2O3,300 deg. C)\nC2H5OH----------------->H2CCH2 + H2O(H3PO4,250 deg. C)
825248
Basically it is\n\n2 H2O2 -> O2 +2 H2O,\n\nwith catalase written above the reaction arrow.\n\nA more detailed formula would show intermediary reactions of H2O2 with catalase, and then how catalase returns to the original state giving up O2 and H2O.
116069
The balance is :\n\n     H2O2(aq) -> O2(g) + 2H+(ag) + 2e-\n MnO2(s) + 4H+(aq)  +  2e-  ->  Mn2+(aq)  + 2H2O\n \n Combine the two half-reactions and cancel on both sides yield the     complete balance equation.\n\n MnO2(s) + H2O2(aq) + 2H+(aq) -> Mn2+(aq) + O2(g) + 2H2O\n\n this is an oxidation-reduction reaction\ntype  decomposition reaction.
606075
Cu(s) + 4HNO3 (aq)--> Cu(NO3)2(aq) + 2H2O(l) +2NO2(g)
5016
O\n                       ^\n                       |\n                      xx\n                  H-O-Br:->O\n                      xx\n                       |\n                       V\n                       O\n\nPlace the xx above and below Br. Br donates electron pairs to three oxygen atoms. There is also one Br-O-H bond in the molecule.
1029753
Within each half-reaction, the charge on each side must be balanced, as well as the number of each element.\n\nFor the overall oxidation-reduction reaction, the charge on each side must be the same, as well as the number of each element, and no free electrons should be in the equation.
942551
Well Im not sure what youre exactly looking for but the basics of it is oxidation passes the electron and reduction accepts the electron.  There are some basic rules for assigning oxidation number and its best to memorize them for a test or something
918374
Fe2O3 -- Fe+O2\n\nthis is the balanced form\n\n2(Fe2O3)--- 4Fe+3O2\nwithout the parentheses
603812
The H's are +1, and the O's are -2, and the whole thing is 0, so...\n4(+1) + 2x + 7(-2) = 0\nx=5, so the oxidation number of phosphorus is +5.
601089
Actually, H2O2 is a very strong oxidizing agent. \nNaClO + H2O2 ---> O2 + H2O + NaCl\nNet ionic equation:\nClO- + H2O2 ---> O2 + H2) + Cl-
242316
Add up the charges in the molecule. Like, in ClO4-,\n\nOxygen has an oxidation state of -2 by default. Therefore, Cl must have an oxidation state of +7 to balance out the oxygens. Notice that 7 + (-2)*4 = -1, which is the charge of the ion.\n\nThe oxidation state of Pb in PbSO4 is +2, because the sulfate ion has an charge of -2. To balance this out, the Pb must be +2, as there is no charge on the molecule.\n\nHope this helped.
230838
Assuming that you mean the triplet state of oxygen, it's zero. The heat of formation of the naturally occuring ground state of an element is zero.
244576
Ru has oxidation states 2,3,4,6,8\nwith three and four most stable
832005
mn : what a great resource!\n\nhmmm... is it a redox question?