a) LiCl b) CaF_2 c) Al(NO_3)_3\nHow?
138973
The previous answer is correct if an equal number of moles is dissolved.  If you were looking at a saturated solution, it would be LiCl, which can dissolve in greater quantities than any other salt, even greater than NaCl. (36 g of NaCL will dissolve in 100 ml water compared to 55 grams LiCl per 100 ml water). The Li+ ion is very very small-it's only larger than hydrogen and helium. For an equal weight of LiCl and Al(NO3)3 dissolved in water, LiCl would also have a greater effect, because of the greater number of ions per gram.  Thus, in either a saturated solution or a weight to volume solution, LiCl would produce the greater number of ions, whereas it would not in a equamolar solution.  LiCl is used a great deal in density experiments, because you can make a very dense aqueous solution of it, and in reducing freezing point, since you can get the maximum number of ions of any aqueous solution, so the questioner is probably looking for LiCl.
515606
electron affinity\n\nas you are changing Cl2 ->2Cl-
341129
NaOH\nSn(NO3)2\nCo2O3\nFeF2\nCuS2\nFe(OH)3\nAgBr
91053
Start by balancing Al, then Cl. Al is already  balanced, so when you balance Cl everything is easy from their. \nYou final amswer will be:\n3HCl + Al(OH)3 ----> 3H2O + AlCl3
394818
CoCl2(aq) + FeSO4(aq) = CoSO4 (aq) + FeCl2(aq)
469919
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
477308
1) 2Al + Fe2O3=Al2O3 + 2Fe\n2) 8.53g BaCl2*(1mole BaCl2/ 208.3G BaCl2)=.0410mole BaCl2\n.0410mole BaCl2*(2moles KCl/1moleBaCl2)=.0820 mole KCl2\n.0820mole KCl2*(110.1gKCl2/1mole KCl2)=9.03gKCl2\n3)i don't know this one\n4) 2NaOH + H2 is reduced\n5) i don't know this one\n6) 5gNa*(1moleNa/23gNa)=.217mole Na\n.217mole Na*(6.02x10^23/1mole Na)=1.31x10^23 atoms Na
38717
Some of the matals have multiple oxidation states and these must be indicated in the names:\nAmericurium (III) fluoride\nIndium (III) fluoride\nBarium permanganate\nNeodymium (III) iodide\nCadmium sulfite\nSilicon tetrathiocyanate\nVanadium (III) fluoride\nIron (III) dichromate\nGold (I) cyanide\nBarium bromate\nMagnesium molybdate\nPlutonium (III) bromide\nSamarium (III) sulfate\nTungsten (II) iodide\nPotassium borate
407062
1. AgCl(s) + NaNO3\n\n2. AgOH(s) + NaNO3\n\n3. They would not react as both are nitrates already
18765
CuI2 Copper (II) iodide\nSF6 sulfur hexafluoride\nNaOCl Sodium hypochlorite\nPb3(PO4)2 lead (II) phosphate\nHClO Hypochlorous acid\n\nChlorine triflouride  ClF3\nChromiumIII carbonate  Cr2(CO3)3\nPotassium Chlorate KCLO3\nSodium peroxide Na2O2 \nMercuryI chloride HgCl\n\nThanks You're welcome
459512
2Al + 3*(CuO) -> Al2O3 + 3*Cu\n\nI2 + 2*(Na2S2O3) -> 2*(NaI) + Na2S4O6\n\nI2 + 5*(Cl2) -> 2*(ICl5)
192262
The second one.  The other three contains salts of calcium.
287166
a) Ru - ruthenium\nb) S - sulfur\nc) Ir - iridium; with an incomplete d-subshell, it must be a transition metal\nd) Sb - antimony
400379
a) 2NaCl ---> 2Na + Cl2\n\nb) pretty sure its 9.2793 g of Na
121334
1. electron\n2. ionic\n3. molecules of compounds\n4. decomposition (not sure)
299966
Aluminium has a valanly of 3 Oxgygen has 2\nSo\n4Al + 3O2   -> 2Al2O3
157500
Copper, Iron, Chromium, Aluminium