Formulas for compounds...??
966082
1 - NH4OH\n2 - PbO\n3 - KBr\n4 - PCl5 (five)\n5 - Al(NO3)3\n6 - Ba3(PO4)2  (not sure)\n7 - C4H10\n8 - Sn(OH)4\n9 - NH4NO3\n10 - Li3P\n11 - Ba3ClO2   (not sure)\n12 - MgSO3
839940
1. 2C6H6 + 15O2 --> 12CO2 + 6HOH\n\n2. C12H22O11 + 12O2 --> 12CO2 + 11HOH\n\n3. CH3COCH3 + 4O2 --> 3CO2 + 3HOH\n\n4. C2H5OC2H5 + 6O2 --> 4CO2 + 5HOH\n\n5. C4H9OH + 6O2 --> 4CO2 + 5HOH
882098
I'm not sure what you mean by relative formula as you seem to know the formula already. I presume that this compound is potassium ferro cyanide.
906651
2KOH=>K(2)O+H(2)O\n\nFeO+C=>CO+Fe\n\nCa+2AlCl=>CaCl(2)+2Al\n\nP(2)O(5)+3H(2)O=>2H(3)PO(4)\n\n2KI+Pb(NO3)2=>2KNO(3)+PbI(2)
890894
C12H22O11+12O2-------12CO2+11H2O+E
574078
This might help :)\nhttp://www.chemicalforums.com/index.php?board=8;action=display;threadid=3263
813421
Na2CO3(aq) + MgCl2(aq) --> 2NaCl(aq) + MgCO3(aq)
935686
1. Look at the periodic table and figure out the formula mass of your compounds.  .005*formula mass of  Pb(NO3)2 and ditto for the 2nd one...\n\n2. 1/2 of 454g...
727014
start with the O... they are always -2\nsince there are 2, and the compound must balance...\nPb must be +4\n\nso you get +4 -2
910345
The formula would be Al2(CO3)3; the numbers are subscripts. \n\nLearn your charges for the polyatomic ions and for the other elements, it will really help.
895751
Al2S3 + 6H2O -> 2Al(OH)3 + 3H2S\n\nP4O10 + 6H2O -> 4H3PO4\n\n\nShould be right.
422969
AgNO3(aq)+2Na2S2O3(aq)--> [Ag(S2O3)2]3-(aq) + 4Na+(aq)+ NO3-(aq)
921488
2ClO4(-) --->  2ClO4(-)\n2H(+) + 2OH(-) ---> 2H2O\n\nwhere sign in brackets are superscripts.
853922
These are the right ones:\n\n2K(s) + 2H2O(l) ---> 2KOH(aq) + H2(g)\nH2SO4(aq) + 2KOH(aq) ---> K2SO4(aq) + 2H2O(l)\nCa(OH)2(aq) + 2HCl(aq) ---> CaCl2(aq) + 2H2O(l)
832783
e.g. of formula: Magnesium+oxygen=magnesium oxide\ne.g. of Mixture: Air\ne.g. of Compound: Magnesium oxide
888667
Cu2S + 12HNO3 -> Cu(NO3)2 + CuSO4 + 10NO2 + 6H2O\n\nThere you go.
974325
You should probably need your periodic table.\nGenerally chemical formulas are for ionic compounds, hence you need to know the charges of the ions formed.\nAll Group I metals form ions of +1 charge, Group II metals forms +2 charged ions, Group III (like Al) forms +3 charge, Group IV prefers to share, seldom will they form ions, hence we shall keep them out of the picture.\nGroup V non-metals form -3 charge, Group VI -2 and Group VII -1. Group O elements have full outermost shell, hence they seldom participate in bonding.\nSo if you are looking at magnesium oxide, the ionic compound is formed between magnesium ion (Group II, hence Mg2+) and the oxide ion (from oxygen, Group VI, hence O2-).\nYou will realise that to balance the charges, you need exactly 1 Mg2+ and 1 O2-. Hence the chemical formula of magnesium oxide is MgO.\nAnother example, for sodium chloride. Sodium ion (Na+) and chloride ion (Cl-). Hence 1 of each is needed to balance the charges and the formula is NaCl.\nAnother example, lead (IV) oxide. Lead is a transition metal and transition metals form ions of variable valencies. Lead (IV) means the charge of the lead ion here is Pb4+. Thus you need 1 Pb4+ and 2 O2- (so that the charges are cancelled) for lead (IV) oxide, PbO2.\nYou have also polyatomic ions, like hydroxide (OH-), sulphate [(SO4)2-], nitrate [(NO3)-] and carbonate [(CO3)2-]. Similarly, ionic compounds formed will require the balance of charges.\nExample, with sodium sulphate.\nYou have Na+ and sulphate [(SO4)2-].\nYou will need 2 sodium ions to balance the -2 charge of the sulphate ion. Hence the chemical formula is Na2SO4.\nHope this helps. :)