Some chemistry wiz help me! Chemistry 1 question!?
829125
I dont understand what you mean by "correcting certain formulas", but it appears Fe bears +1 oxidation state, if oxygen is -2. However, Fe does not bear such oxidation state, normally; it can bear 0, +2, +3, +4 and +6. With oxygen, what you will get, most likely, are FeO and Fe2O3. So, I would expect the the compound was a mixture of iron metal and one/several of these oxides. It appears that your compound could be 1:1 mixture of iron and iron(II) oxide.\n\nIf the question states that the compound was pure, check your calculation again; if you have done nothing wrong, the compound could be iron(II) peroxide [iron bears +2, oxygen bears -1]; it's not stable [I'm not sure whether it exists or not], or the strange iron(I) oxide [again, never seen this compound before, not sure it actually exists or not].
574078
This might help :)\nhttp://www.chemicalforums.com/index.php?board=8;action=display;threadid=3263
788506
b. electrons\n\nI really hope you are trying to answer these first before you put them on here because although we are all certainly willing to help you, you are not going to gain anything from just having the answers given to you.  You really need to try and learn this stuff.
150925
Reaction One (CO is Co I'm assuming ... they are NOT equivalent) is a double replacement. The products will be aqeuous KNO3 and insoluble Co(OH)2\n\nReaction 2 is a redox. The product will probably be Mg2N3 (Magnesium Nitride) but I'm not entirely sure if that chemical formula is correct. The magnesium is oxidized and the nitrogen is reduced.\n\nAg + H2SO4 ---> Maybe Ag2SO4 as a redox.\n\nF2 + KBr ---> KF + BR2 Single Replacement
496291
You need to take more care in writing equations when you post your questions. \n\nPart a) is difficult to decipher - you will need to re post it.\n\nPart b) is presumably:\n\nCH3COOH(l) + 2O2(g) --> 2CO2(g) + 2H2O(l)\n\nPart c) is presumably:\n\nFe2O3(s) + 3CO(g) --> 2Fe(s) + 3CO2(g)\n\nI would next use heats of formation to calculate the enthalpy change in the above reactions using the formula:\n\ndelta Ho = sum(heats of formation of products) - sum (heats of formation of reactants).\n\nIf you have a data book or data sheet, these values can be found there (Note: heats of formation of elements are zero.) \n\nIf you want web sites that might help you, try googling "enthalpy change problems" +tutorial.\n\nI hope this helps you.
840582
1.H2O=18g\n\n2.CH3Cl=52g\n\n3.SO4=96 g\n\n4.NH3=17 g\n\n5.C3H7O=59 g\n\n6.NO2=46g
917667
Part A:\nThe answer is 12 L H2.  Here is the solution:\n(4 L N2)(1 mol N2 / 22.4 L N2)(3 mol H2 / 1 mol N2)(22.4 L H2 / 1 mol H2)\nso (4*1*3*22.4) / (22.4*1*1) = 12 L H2\n**note that in in ideal gas there are 22.4 liters/mol**\n\nPart B:\nThe answer is 8 L NH3.  Here is the solution:\n(4 L N2)(1 mol N2 / 22.4 L N2)(2 mol NH2 / 1 mol N2)(22.4 L NH3 / 1 mol NH3)\nso (4*1*2*22.4) / (22.4*1*1) = 8 L NH3\n\nHope this helps!
793636
shawty's NOT good at stoichiometry...has made a fundamental error in the final line so end result is rubbish...\nhow can you end up with MORE Cu than you started with when it's 2 moles : 2moles? \n\n(8/159)x 127 = 6.4 g
608403
NaOH (aq) + HCl (aq) --> NaCl (aq) + H2O (l)
675016
This is a calorimetry problem! There is an equation for it:       \nq = mc(change in temperature)\n\nq = the heat required \nm= mass\nc= specific heat \n\nAlso, write out the reaction and products.
727458
2. 2 Na + 2 H2O --> H2 + 2 NaOH\n3. 2 HNO3 + Ca(OH)2 --> Ca(NO3)2 + H2O\n4. C5H12 +  8 O2 --> 6 H2O + 5 CO2
843441
The answer is c.\nC12H22O11+12O2-->12CO2+11H2O
791178
I HAVE A CHEMISTRY EXAM tomorrow. Actually a small test.lol\nwant answers go to the site below!
781038
I agree with birchardv here...........we need the rest of the problem, and you would learn a lot more if you did the homework yourself.  But do whatever floats your boat.  Add the rest of the problem and I'll get back to you.\n\nOkay Chemist, this problem is really long; if you want a good answer with plenty of steps, I'll need more room.  Email me and I'll send you the solution.
734957
Na+ Cl-\nC +4 O-2 each\nK+ Br-\nN-3 H+1 each\nMg+2 O-2
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)
1074656
[V]=J/C=W/A=Kg.m^2/(s^3.A)\n\nFor experimental reasons A is the base unit and not C, you can not count electrons one by one, not yet probably soon