A saturated solution of barium fluoride contains 1.5E-2 M F-. Starting w/ this info calculate the Ksp of BaF2?
454203
The Solubility Product, Ksp, is defined as (for the hypothetical compound MmAn, which dissolves in water according to the equation:\n\nMmAn (s) ---->  mM(+)(aq) + nA(-)(aq)\n\nis:  Ksp = [M(+)]^m * [A(-)]^n\n\nSo, for the dissolution of Barium fluoride:\n\nBaF2 ---> Ba(+2)  + 2 F(-)\n\nthus Ksp = [Ba]*[F]^2\n\nIf [F] = 1.5E-02, then [Ba] = 1/2*[F], or 7.5E-03\n\nKsp  = [7.5E-03]*[1.5E-02]^2 = 1.7E-06\n\nThe textbook value of Ksp for barium fluoride is 1.84E-07, so if your value for the concentration of fluoride in solution is from an experiment, it is a bit high.
315574
I think this is how you do it:\nPb + F --> PbF\n\nSo Ksp = [Pb][F]\nBecause they react in a 1 to 1 ratio, their concentrations will be the same.\nSo Ksp = X^2 or [Pb]=[F]=(3.2E-8)^2\n-Not 100% sure but I think thats it
1167833
sounds like an AP question.  So you are trying to calculate Ksp for silver acetate. \n\nAgC2H3O2 <--> Ag+  + C2H3O2-\n\nSolubility will be darn low so from there set up your ice chart and go to town.
744922
HClO2 + Ba(OH)2 -> BaClO2 + H2O\n\n2(NH4)3PO4 + 3Ca(NO3)2 -> 6(NH4)(NO3) + Ca3(PO4)2\n\nFinished and balanced.
558215
F = KM1M2/r^2\n\nwhere K is the constant, M1 and M2 are the magnetic moments, and r is the distance between them.
781614
Na2SO4 (aq) + Ba(NO3)2 (aq) --> 2 NaNO3 (aq) + BaSO4 (s)\n\nThe moles of Barium = 5.00 L * 0.15 mol/L = 0.75 moles\n\nThis is also the moles of sulfate needed:\n\n0.75 mol / 0.36 mol/L = 2.08 L of Na2SO4
869137
2H3PO4 + 3Ba(OH)2 ---> 6H2O + Ba (PO4)2 +  2Ba
464163
FORMATION CONSTANT IS :\n\nkf = 1.5*10^7
217978
Barium +2 is a positive ion.  It belongs to the alkaline earth metals group together with Beryllium, Magnesium, Cesium and Francium.
649164
Francium and cesium are about 0.5.\nFluorine is highest at 3.9-4.0 (by Pauling's estimate)
1008732
Ksp = [Pb+2]*[Cl-]^2/[PbCl2]\n\nby definition, [PbCl2] = 1\n\nAlso, based on the reaction:\n\nPbCl2 =====> Pb+2  +  2 Cl-\n\n[Cl-] = 2 [Pb+2]\n\nSubstitute this into the equation, and you get:\n\nKsp = [Pb+2] * [Cl-]^2 = [Pb+2]*(2[Pb+2])^2 = 4[Pb+2]^3\n\nNow, since you know that 0.99 grams of PbCl2 is dissolved in your liter of solution, you can calculate the molarity from:\n\n(0.99 g)/ (278.1 grams/mole) = 3.56x10^-3 moles = [Pb+2]\n\nKsp = 4*(3.56x10^-3)^3 = 1.8 x 10^-7\n\nThe text value for the Ksp of lead (II) chloride is 1.7 x 10^-5\n=================\nOkay, now for the second part of your question:\n\nIf you have 0.5M lead nitrate, which is soluble, you will have 0.5M Pb+2\n\nSo, let's plug 'n chug (we'll use the text value for Ksp to show the process):\n\nKsp =  1.7x10^-5 = [Pb+2]*[Cl-]^2\n\nbut, with the lead nitrate in solution, [Pb+2] = 0.5 moles/liter (the amount you'd get from the dissolution of lead chloride will be so small as to be negligible), so:\n\nKsp = 1.7x10^-5 = [0.5]*[Cl-]^2\n\n[Cl-]^2 = (1.7x10^-5)/(0.5) = 3.4x10^-5\n\nso [Cl-] = sqrt(3.4x10^-5) = 5.83 x 10^-3  and since [Cl-] = 2 [Pb+2], we halve that value to get:\n\nsolubility = 5.83x10^-3/2 = 2.92x10^-3 moles per liter\n\n(2.92x10^-3 moles/liter) * 278.1 grams/mole = 0.811 grams/liter\n\nIf you use the value of the Ksp calculated in the first half of the question, then the solubility of PbCl2 in your solution is 3x10^-4 moles/liter, or 0.083 grams/liter
753910
CaF2 is calcium fluoride. Sorry I messed up before.
1153335
barium acetate and carbon dioxide and water
412413
I'm assuming you're asking for the reactions.  Here are a couple:\nH2 + F2 → 2HF \n\n4Fe + 3O2 → 2 Fe2O3
210085
In order to determine the molar mass of 2.00 g of para-dichlorobenzene in 50.0 g of\ncyclohexane using colligative equations, the following equations must be employed:\nΔTf = Kf m\n5.55̊C = 20.4 ̊C/m * x m\nx m = .272\n.272 m = x mol / .0500 kg\nx mol = .0136\n2.00 g / .0136 mol = 147.1 g/mol\n\nI'm sure this'll help,\nUltimeaciax
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
407618
k e =1/2  * m *v *v\n    =1/2 *3 * (3.5*10) * (3.5*10)\n    =0.18j (ans)