Finding the Oxidation numbers of CN-?
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.
904652
CH3CH2C(CH3)3.\nI guess you forgot to type the C before the parenthesis. If you didn't forget, then none of the answers is correct (you would be missing a carbon atom).\n\nChem-man I would have contacted you in private about this but you do not give us the option to e-mail you. I have seen several of your answers prompting askers to verify your answer through echemistryhelp. I checked the website and it is not a free one; you post a question and pay for the answer. I'd say that this is a form of indirect advertisement  which is against the community guidelines. Please refrain from doing that or I will have to report you.
306342
in H3P04 u need the oxidation number of what? all these elements have different oxidation numbers in different compounds. the procedure for u to find out the oxidation number is .... if u want the oxidation number of P then write there as x, for H as u know it is +1, for O it is -2.\n then, (+1*3)+(x*1)+(-2*4)=0; 3+x-8=0; x=5 thus, the oxidation number of P in H3P04 is 5.\nin the same way, in H2O2; O is -1\nSi(OH)4 ; Si is x+(4*-2)+4=0; x=4\nin KH2PO4, P is  +1+(+1*2)+x+(-2*4)=0; x=5\nin Fe(HCO3)2 if Fe is +2, H is +1, O is -2 then C is \n+2+2+2x-12=0; x=4\nCu2CrO4, Cu is +2, O is -2 and Cr is 4+x-8=0; x=4\nKMnO4, K is +2, O is -2 and Mn is +2+x-8=0 x=6\nN03, Ois -2, x+(-2*3)=0, x=6\nin HN03, H is +1, O is -2,then N is +1+x-6=0, x=5\nBeH2, H is +2\n\nany doubts u can ask again.
383259
go to\nwww.chemistry.com
957244
2NO + O2 --> 2NO2\nMr of NO = 14 + 16 = 30\nMoles of NO = (29.03/30) = 0.9677 (to 4 s.f.)\nMr of O2 = 32\nMoles of O2 = (19.35/32) = 0.6047 (to 4 s.f.)\n\nComparing the moles of NO and O2 given, you should notice that NO is the limiting reagent.\nHence moles of NO2 produced = moles of NO used = 0.9677\n(since the ratio of NO2 to NO is 2:2)\nMass of NO2 produced = 0.9677 x (Mr of NO2) = 0.9677 x 46 = 44.5 g (to 3 s.f.)\n\nHope this helps :)
77087
single replacement, for example: CO2+N3=CN3+O2
205870
CaCO3(s) + 2HC2H3O2(aq)-> Ca(C2H3O2)2(aq) + H2O(l) + CO2(g)
850606
Mg       2+\nPb       4+\nCrO4  2-\nNO2    1-\nI           1-\nClO3   1-\nOH      1-\nFe       2+\nNH4    1+\nCu      1+
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)
887797
It looks like you got it. This is how it came out for me:\n\nThis is the reaction:\nC10H8(s) + 12 O2(g) ® 10 CO2(g) + 4 H2O(l)\n\nand Hc=(Summation of Hf of products)-(Summation of Hf of reactants)\nSo...\n\nHc (C10H8) = 10 Hf(CO2) + 4 Hf(H2O) – Hf(C10H8)\n-5156.8kJ/mol=(10)(-393.5 kJ/mol)+(4)(-285.9kJ/mol0-Hf (C10H8)\n-5156.8kJ/mol=(-3935 kJ/mol)+(-1143.6 kJ/mol)- Hf (C10H8)\n-5156.8kJ/mol=(-5078.6)- Hf(C10H8)\n-78.2kJ/mol=-Hf (C10H8)\n78.2kJ/mol=Hf (C10H8)\n\nI believe that's how it's done, but it's been a while since I've done that kind of problem, so it's probably not a terrible idea to double check with someone else.
888667
Cu2S + 12HNO3 -> Cu(NO3)2 + CuSO4 + 10NO2 + 6H2O\n\nThere you go.
961793
Call it (CxHyOz)\nMass of C = 12x\nMass of H = 2y\nMass of O = 16z\n-----------------------\nIts molecular weight M = 12x + 2y + 16z = 90\n\nPercentage of C: (12x / M) *100% = 26.7% <-> 12x = 90 * 26.7 / 100 <-> 12x = 24 <-> x = 2\n\nYou can do the same with H and O to find y = 2 and z = 4.\n\nThe formula is, hence, C2H2O4.
755565
1.) Cr(NO2)2+(NH4)2SO4\n---->CrSO4+(2)NH4NO2\n2.) (2)KNO3+H2CO3\n----->K2CO3+(2)HNO3\n3.) (3)KBr+Fe(OH)3\n----->(3)KOH+FeBr3\n4.) B2Br6+(6)HNO3\n----->(2)B(NO3)3+(6)HBr\n\nHope this helps=)\nTip: Start by balancing those more isolated atoms first, especially when you are balancing more complex equations (they may have atoms that appear in more than 1 products).\nAlso, feel free to leave numbers in fractions while you balance ur equation. You can always multiply all the numbers later to make them all whole numbers. You will find this useful later when you have to balance long equations. Itd definitely more tedious to alter every number in ur mind and writing them all out at one go. Do it progressively=)
391116
2Ca(s) + O2(g) -> 2CaO(s)
926891
just count how many of each element you have, and make sure they add up\n\n\nC2H6 + 5O2 ---> 2CO2 + 3H2O\n2 carbon 6 hydrogen 2 oxygen----> 1 carbon 2 oxygen 2 hydrogen 1 oxygen\n\n\nso to even it out it would be...\n\nC2H6 + 7O2--->2CO2 + 3H2O\n2 carbon 6 hydrogen + 7oxygen--->2 carbon 4 oxygen + 6 hydrogen 3 oxygen\n\niff you add up each element they are the same on each side
985592
Carbon:  12.01 g\nNeon:  20.18 g\nIron:  55.85 g\nUranium:  238.03 g
754704
You should write NH4(+) otherwise you will not get the oxidation number right.  H is 1 so H4 is 4.\n\nIn order to be left with 1(+) you need 4 - 3 = 1.  So N will be -3.\n\nCheck -3 + 4*1 = +1 OK.