Answer for I2 + NaCI?
78988
I2 + 2NaCl ---> 2NaI + Cl2\n\nSodium Iodide and Chlorine
566534
There are three possible answers:\n\n1,1-diiodoethylene, which is I2C=CH2\ncis-1,2-diiodoethylene, which is IHC=CHI, with both iodines on the same side\ntrans-1,2-diiodoethylene, which is also IHC=CHI, with the iodines on opposite sides of the molecule
104563
My answer is:\n\n.217N
1029892
Na2S (aq) + 2H2O (l) = 2NaOH + H2S (g)
770494
1.) 2Ni + O2 -->  2NiO  (since it's Nickel II and Oxygen has a charge of -2, but you always have to put oxygen in groups of 2)\n\n1.) 2H2O --> 2H2 +O2  (same with hydrogen)\n2.) 2Al2O3 (because Aluminum has a charge of +3) --> 4Al + 3O2\n3.) 2NaCl --> 2Na + Cl2 (basically the rest follow the same rules)\n4.) 2NH3 --> 3H2 + 2N\n5.) 2HgO --> 2Hg + O2
606075
Cu(s) + 4HNO3 (aq)--> Cu(NO3)2(aq) + 2H2O(l) +2NO2(g)
24810
Cu + 2H2SO4 --> 2H2O + SO2 + CuSO4\nDouble the H2SO4 and the H2O.
991807
uh a pH meter will help with the hydrogen iodide because it's a strong acid.  As for the I2 I have no idea...
857898
2Al + 3*(CuO) -> Al2O3 + 3*Cu\n\nI2 + 2*(Na2S2O3) -> 2*(NaI) + Na2S4O6\n\nI2 + 5*(Cl2) -> 2*(ICl5)
391861
gh = f\no  = i\nti = sh\n\nfish
470703
reduced: 2H2O + 2e- ---> H2 + 2OH-\n\noxidized: 2I- ---> I2 +2e-
748872
Cu2+ (aq)  +  Zn (s) -->  Cu (s) + Zn2+ (aq)
872094
A . 4LiH + AlCl3 → 2LiAlH4 + 2LiCl <-- there's only 1 alluminum so it can't yield 4 in the product\n\nB. Sn + 2H2SO4 → SnSO4 + SO2 + 2H2O <-- there are 1 sn, 4 H, 2 S and 8 O ( multiply the coefficient [2] by the subscript [4])\nit matches in the product\nCORRECT\n\nC. 4BF3 + 3H2O → H3BO3 + 3HBF4< same goes here, just write out how many of each element are in the beginning of the equation and in the product and compare\nCORRECT\n\nD. 3Cl2 + 6NaOH → 5NaCl + NaClO3 + 3H2O <-- this one is hard to read, I'm assuming that I is not an L because it's followed by a subscript, so it would be iodine\nbut either way, this one is wrong because there are 5 C/Cl in the product which is wrong
871212
i2=i1R1/R2\ni1=R2i2/R1\n\nR1/R2=i2/i1\n\nR1/Rx = R2/R3\n(R1/R2)*R3=Rx\nRx=(i2/i1)*R3
692301
If the molar solubility of PbI2 is .00165 M, then we can calculate the solubility product constant, K. \n\nK = [Pb2+][I-]^2\nK = (.00165 M)(2*.00165 M)^2\nI doubled the PbI2 concentration to find the I- concentration because one mole of PbI2 is the same as two moles I-.  \nK= 1.80E-8\n\nNow, we can use the K to find the molar solubility when .100 M I- is already in solution. \n\nK = [Pb2+][I-]^2\n1.80E-8 = (x)(2x+.100 M)^2\nx=1.8E-6 M\nThe molar solubility at the same temperature in a .100 M solution of NaI is 1.8E-6 M.
971239
When the two solutions are mixed, you'll get a bright yellow precipitant.\n\nPb(NO3)2 + 2 NaI ---> PbI2 + 2 NaNO3
711424
The Equation should be:\nNaCl + H2SO4 = Na2SO4 + HCl\nSo the coefficients are (2,1,1,2)