For the reaction   BaI2   +   MgSO4 (aq)       BaSO4   precipitates out.?
918301
.06 bal2 + .0501 mgso4 = x BaSO4\n2.150 L x (square root of .06) = .5266401\n1.980 L X  .0501 = .099198\n.099198 / .5266401=.1883601\nTHEREFORE YOUR % PERCENTAGE YEILD IS 18.83%
782238
Here's what you do:\n\n1) Write the balanced equation for the reaction.  How many moles of BaSO4 does each mole of BaX2 create? \n2) Determine the F.W. for each BaX2, and for BaSO4 (by the way, the unknown won't be BaF2. It's insoluble: http://www.ktf-split.hr/periodni/en/abc/kpt.html ) Determine the number of moles there are in 0.0944 g of BaSO4, and in 0.158 g of each BaX2.\n3) For which BaX2 does 0.158 g represent the the correct number of moles?  Note: I worked this out, and got a number that was very close, but not an exact match.  That's realistic, since the BaX2 would not be absolutely pure, SOME minute amount of BaSO4 will dissolve, etc.
782516
a) 3SiO2 +2 Al --->3 Si + 2Al2O3\n\nb)AgNO3 +HCl---> AgCl +HNO3\n\nc) Mg + CuSO4 ---> MgSO4 + Cu\n\nd) Pb(NO3)4 +4 NaI --->4 NaNO3 + PbI4 (s)
861717
2 HNO3(aq) + Sr(OH)2(aq) --> Sr(NO3)2 (aq) + 2 H2O\n\n2 H3PO4(aq) + 3 Ba(OH)2(aq) --> Ba3(PO4)2 (cr)* + 6 H2O\n\n* Barium Phosphate will precipiate out of an aqueous solution in this reaction
865869
Your question is hard to interpret.  Are you saying you have a solution of sodium nitrate (NaNO3) and lead (II) sulfate?  The combination looks like the result of an ionic recombination (double replacement) reaction between sodium sulfate and lead nitrate.  The lead sulfate is a solid (low solubility in water) and precipitates out.  The sodium and nitrate are spectator ions.
813911
HBr ---> H(+) + Br(-)\nH2SO3 ---> H(+) + HSO3(-)\nCH3-COOH ---> H(+) + CH3-COO(-)
714201
NaOH + NaH2PO4 ===> Na2HPO4 + H2O\n\nH2S + 2KOH ===> K2S + 2H2O\n\n(NH4)2SO4 + 2KOH ===> 2NH3 + K2SO4 + 2H2O\n\nSO2 + Ca(OH)2 ===> CaSO3 + H2O\n\nIn the above equation, further oxidation in air could occur by\n\n2CaSO3 + O2 ===> 2CaSO4
751720
chromium (II) sulfate fo sho
216402
It's:\n\nBaSO4(s)---water <--> Ba2+(aq)  +  SO4 2-(aq)\n\nthe solubility for this compound is very low  per  g/L\n\n\n about 0.0023g/L
595177
Actually, this is pretty simple.  You have a saturated  solution of lead sulfate (PbSO4) and strontium sulfate (SrSO4).  Probably the biggest hurdle here is to realize that the concentration of sulfate ion will be the same for both salts!  Therefore, the concentration of sulfate ion will be the result of the most soluble of the two salts:\n\nKsp for PbSO4 = [Pb+2]*[SO4-2] = 1.6x10^-8\nKsp for SrSO4 = [Sr+2]*[SO4-2] = 2.8x10^-7\n\nStrontium sulfate is the more soluble of the two salts, so it will control the concentration of sulfate:\n\nSrSO4 ===> Sr+2  + SO4-2, thus [Sr+2] = [SO4-2]\n\nKsp for SrSO4 = 2.8x10^-7 = [SO4-2]^2 , so \n\n[SO4-2] = sqrt(2.8x10^-7) = 5.29x10^-4 = [Sr+2]\n\nNow, to find out the concentration of Pb+2, we use the value we have just calculated for [SO4-2] and the Ksp:\n\nKsp for PbSO4 = [Pb+2] * [SO4-2] = 1.6x10^-8\n\n[Pb+2] = 1.6x10^-8/[SO4-2] = (1.6x10^-8)/(5.29x10^-4)\n[Pb+2] = 3.02x10^-5\n\nand there we have your answers:\n[Sr+2] = [SO4-2] = 5.29x10^-4 moles/liter\n[Pb+2] = 3.02x10^-5 moles/liter
298301
Hmm. This is exactly what I've been doing for the past few weeks. I'm assuming these are aqueous solutions of the three, right?\n\nWell, both magnesium and strontium form precipitates with hydroxides, so you could add some sodium hydroxide and centrifuge it down. To the liquid remaining, add some kind of sulfate, like potassium sulfate, because barium precipitates with sulfates, forming BaSO4.\n\nTo the solid containing strontium and magnesium, add ammonia to make the solution basic, then add sodium carbonate. The magnesium should complex with the carbonate and fall out, forming Mg4(CO3)3(OH)2. Then, whatever's left is strontium.
832296
A) Ag2O (s) >>> 2 Ag(s) + 1/2 O2 (g)\nB) Nickel (II) Carbonate >>>>>Nickel (II) Oxide + [Carbon Dioxide], NiCO3 -> NiO + CO2\nC) Ammonium Nitrate >>>>> Dinitrogen monoxide + Water,  \nC does not balance as is... but this does:\nNH3NO3 -> N2O +  1.5 H2O + 0.5 O2\n\n48) A) 2 HCl (aq) + Ca (OH)2 (aq) >>>> 2 H2O + CaCl2 (aq)\nB) 3 Ag2 SO4 (aq) + 2 AlCl3 (aq) >>>> 6 AgCl (s) + Al2(SO4)3 (aq) (silver chloride is a precipitate)\nC) H2C2O4 (aq) + 2 KOH (aq)>>>> 2 H2O + K2C2O4 (aq)\nD) CdBr2 (aq) + Na2S (Aq)>>> CdS(s) + 2 NaBr (aq) (Cadium sulfide is a precipitate)\n50 Write a Balanced equation for the complete combustion of each compound:\nA) Butene (C4H8): C4H8 + 6 O2 -> 4 CO2 + 4 H2O\nB) Octane (C8H18): C8H18 + 12.5 O2 -> 8 CO2 + 9 H2O\nC) glycerol (C3H8O3): C3H8O3 + 3.5 O2 -> 3 CO2 + 4 H2O\nD) Acetone (C3H6O): C3H6O + 4 O2 -> 3 CO2 + 3 H2O\n52: Complete these:\nA) Al (s) + H2SO4 (aq)>>>> AlSO4(aq) + H2(g)\nB) 2 HCl (aq) + Ba (OH)2 (aq)>>>> BaCl2 (aq) + 2 H2O\nC) Au (s) + 3 HCl (aq) >>>>> AuCl3(aq) + 1.5 H2 (g)\n54) EACH IS BALANCED BUT "INCORRECT", so correct these:\nA) Cl-2 + NaI(aq) >>> NaCl2(aq) + I-\nB) NH3(aq) >>> N- + H+3\nC) Na (s) + O2(g) >>NaO2(s)\nD) 2Mg (s) + H2SO4(aq) >>Mg2SO4(aq) + H2(g)\nE) MgCl2(aq) + Ca(OH)2(s) >>> Mg(OH)2(s) + CaCl2(aq)\nF) H2 (g) + Cl2 (g) >> H2Cl2 (g)
178817
manganese(II) sulfate (MnSO4)
290781
MgSO4(s) + 2HCL(aq) ---> H2SO4(aq) + MgCl2 (aq)
899611
Ba(NO3)2 and sodium phosphate\n\n1.  Ba(NO3)2 + Na3PO4 --> NaNO3 + Ba3(PO4)2\n\n2.  Ba(+2) + NO3(-1) + Na(+1) + PO4(-3) --> Na(+1) + NO3(-1) + Ba(+2) + PO4(-3)\n\n3.  No reaction (All spectator ions cancel out)\n\nBa(NO3)2 and sodium carbonate\n\n1.  Ba(NO3)2 + Na2CO3 --> BaCO3 + NaNO3\n\n2.  Ba(+2) + NO3(-1) + Na(+1) + CO3(-2) --> BaCO3 + Na(+1) + NO3(-1)\n\n3.  Forms BaCO3 (Barium Carbonate)\n\n\nBa(NO3)2 and pottasium iodide\n\n1.  Ba(NO3)2 + KI --> BaI2 + KNO3 \n\n2.  Ba(+2) + NO3(-1) + K(+1) + I(-1) --> Ba(+2) + I(-1) + K(+1) + NO3(-1)\n\n3.  No reaction (All spectator ions cancel out)\n\nBa(NO3)2 and sodium sulfate\n\n1.  Ba(NO3)2 + Na2SO4 --> NaNO3 + BaSO4\n\n2.  Ba(+2) + NO3(-1) + Na(+1) + SO4(-2) --> Na(+1) + NO3(-1) + BaSO4\n\n3.  Forms BaSO4 (Barium Sulfate)\n\nHopes this helps you!
787481
NaHCO2 + H2SO4 --> Na2SO4 + CO2(g) + H2O(L)\n\nSodium bicarbonate + sulphuric acid equals sodium sulfate + carbon dioxide + water
326922
Another easy reaction:\n\nMg(s) + HNO3 ===========> Mg(NO3)2 + H2 (g)\n\nMagnesium reacts rapidly with nitric acid to make magnesium nitrate and hydrogen gas