Hey thanks so much for the reply here are my 2 choices for white precipitate   K2C2O4  or 2NAI?
117753
Okay you have a choice between Potassium oxalate (K2C2O4 ) and Sodium Iodide (NaI). \n\nPotassium oxalate is highly soluble (30g/100g) and sodium iodide is even more soluble(184 g/100 ml).\n\nWithout knowing the reactants it is almost impossible to predict a product, especially with two choices. Given those two products, I would expect it to be K2C2O4, but that is just a guess. What was in the solution?
1088014
你很漂亮. ni3 hen3 piao4 liang
363162
When you ask about the formula that is a pretty broad question.  The only formula I know is the balanced compound formula which is Ga2(SO4)3, direct combination formula which is 2Ga + 3S04 --> Ga2(SO4)3, or the decomposition formula which is, Ga2(SO4)3 --> 2Ga + 3S04.  If this isn't what you wanted I'm sorry, if you want you can email me and be more specific.
672061
Here is their products page.\n\nhttp://www.cpchem.com/enu/products.asp
893375
I think the reactants that you've listed my be wrong or balanced correctly to give a possible vanadium oxide.  Please check out: http://www.webelements.com/webelements/elements/text/V/key.html\n\nThis site lists all 5 possible vanadium oxides (right side half way down the page)\n\nThe most is likely is vanadium (III) oxide (aka vanadium oxide or\ndivanadium trioxide) but to produce two moles the reaction would be:\n\n4V + 3O2 --> 2 V2O3
979040
1) ZnNO2\n2) Co(HSO3)3\n3) HgF\n4) Al2(CO3)3\n5) CeC2O4\n6) BaClO\n7) SnBr2\n8) Mg(HSO3)2
746627
Cl2 + 2KI --> 2KCl + I2\n\nsingle replacement
187439
the first answer is wrong, all nitrates are soluble, you learn that in first year chemistry. Therefore the precipitate cannot be ammonium nitrate. Althought i have no answer yet myself for the question, i do know the 1st answer cannot be correct.
129663
Please do your own home work.\n\nI suspect you are missing a symbol, it probably should be,\nN2O4  <==>  2NO2\n\nThink about how each of the changes will effect the equilibrium of the reaction.
15061
K2Cr2O7 + 2Pb(C2H3O2)2 + H2O --> 2PbCrO4 + 2HC2H3O2 + 2KC2H3O2
712912
BaI2(aq) + MgSO4(aq)  --> BaSO4(s) + MgI2(aq)\n\nSo BaSO4 will precipitate.
693004
(1) The color will change from blue to orange to clear with a muddy brown precipitate\n\n(2) Pale brown to colorless solution\n\n(3) Pale brown to colorless solution\n\n(4) No color changes\n\n(5) Blue to colorless\n\nI read this inorganic chemistry long baack... to my knowledge they are correct. I hope this helps\nGOOD LUCK
12446
Both group 1 and 2 nitrates give oxides on heating.\ngroup 1 carbonates are thermally very stable but group 2 carbonates are thermally unstable, the unstability increasing with cationic size.
564401
Silver Carbonate  - AgCO3 (3 being the small number on teh O)\nSilver Nitrate - AgNO3\nSilver Sulphate - Ag2SO4
771965
I can't tell you the name because i can't remember that stuff but the product will be K(NO3) PbI2\nif you're writing it in the long form it will be\n2KI + Pb(NO3)2 yeilds (arrow) 2K(NO3) + PbI2
72603
Wow, I hope these are right.  I wish i had a solubility chart or something (Which you probably have)  \n\nCuSO4(aq) + Na2S(aq) ---> CuS (s) + NaSO4 (aq)\nNet ionic - Cu2+ (aq) + S2- (aq)  ---> CuS (s)\n\nNiCl2 + 2KNO3 --> Ni(NO3)2 (aq) + 2KCl (aq)\nnet ionic - none (if NiCl2 isn't aq, then this is wrong)\n\nCa(OH)2 + 2HC2H3O2 ---> Ca(C2H3O2)2 (aq) + 2H2O (l) \nnet ionic - H+ (aq) + OH-(aq) ---> H2O (l)\nactually, since the acetic acid is a weak acid, this might be a little different.  If you learned it this way, then your net ionic is:\nHC2H3O2 (aq) + OH- (aq) ---> H2O (l) + C2H3O2- (aq)\n\nHope this helped
215729
K2IC4H6 + HNO3 ---  K2HI + NO3C4H6