barium nitrate and sodium sulfate in a balanced equation?
141058
sodium sulfte is soluble, but barium sulfate would form as a precipitate. Thus\n\nBaNO3 + Na2SO4 -> BaSO4 + 2 NaNO3, or simply\n\nBa(2+) + SO4(2-) -> BaSO4
608403
NaOH (aq) + HCl (aq) --> NaCl (aq) + H2O (l)
775066
You have the products, but now you have to balance the equation. The unbalanced equation is:\n\nLi2CO3 + Cr2(SO4)3 --> Li2SO4 + Cr2(CO3)3\n\nBalanced:\n\n3 Li2CO3 + Cr2(SO4)3 --> 3 Li2SO4 + Cr2(CO3)3
708721
Here is the right equation for nitric acid and sodium bicarbonate\n\n\nHNO3 + NaHCO3 ---> NaNO3 + H2O + CO2\n\nIf you do not like the sodium:\n\nHNO3 + HCO3(-) --> NO3(-) + H2O + CO2
768332
D) sodium sulfide, NaS is a binary compound. The rest have polyatomic ions.
911120
2BF3 + 3Li2SO3 → B2(SO3)3 + 6LiF\n3CaCO3 + 2H3PO4 → Ca3(PO4)2 + 3H2CO3\n8Ag2S → 16Ag+S8
712912
BaI2(aq) + MgSO4(aq)  --> BaSO4(s) + MgI2(aq)\n\nSo BaSO4 will precipitate.
564798
Ba(NO3)2 + Na2CO3-->BaCO3 +2NaNO3\n\nto do a double displacement follow this formula\n\nAB + CD --> AD + CB
294143
add:\n\nH2S2O8(ag)+ NH3(aq) + ? -->(NH4)2S208(aq)  + ?\n\nAmmonia can be prepared in the laboratory by treating ammonium chloride with sodium hydroxide:\n\nthe net ionic equation to this reaction is:\n\n NH4+(aq) +  OH-(aq) ---> NH3(g) + H2O(l) in system 1\n \nand\n\n NH3(g)   = NH3(aq) in solution in system 2\n\nyou finish it ,if you need more help just e-mail me
813911
HBr ---> H(+) + Br(-)\nH2SO3 ---> H(+) + HSO3(-)\nCH3-COOH ---> H(+) + CH3-COO(-)
170895
Sodion Hypochlorite and Sodium Thiosulfate?  Ive never heard of Na2(s2o3)2?!  Anyways heres the balanced equation:\n\n2NaOCl + Na2S2O3 ---> Na2S2O3 + 2NaOCl\n\nPretty straight forward I guess.  Double Replacement reaction.
986618
The equation is already balanced. Count the Ag's on each side = 1. Count the Na's on each side = 1. Count the Cl's on each side = 1. Count the NO3's on each side = 1. One can count the N's on each side = 1, or the O's on each side = 3.\n\nBut one could also consider the question another way:\nAg+ + NO3- + Na+ + Cl- ===> AgCl + Na+ + NO3-
899322
First find how many moles of each reactant you have\n\nmoles BaI2 = M*V= 1.35*0.055\nmoles MgSO4 = M*V= 1.2*0.0476\n\nThe reaction is \nBaI2 + MgSO4 ->BaSO4 + MgI2\n\nthe stoichiometry is 1:1, so you should have equal moles of the two reactants. In your case however one of them is in excess. So you will have to calculate the yield based on the limiting reagent.\n\nSince the stoichiometry is also 1:1 for BaSO4 and the limiting reagent you should have moles of BaSO4 = moles of the limiting reactant. But moles=mass/MW => mass= moles*MW\nYou have calculated the moles, find the MW of BaSO4 and you'll get the theoretical mass of BaSO4 that would come from 100% reaction.\nThe percent yield is (actual mass/theoretical mass)*100%
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.
87280
MBrO3-9H2O where M can be Na,K,Li, other +1 cations\n\nIf M is other than these you have to adjust subscripts
577084
Acid + base = salt + water.\n\nUnbalanced eqn is: Al(OH)3 + H2CO3 --> Al2(CO3)3 + H2O
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)