A saturated solution of barium fluoride contains 1.5E-2 M F-.  Starting w/ this info calculate the Ksp of BaF2
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Themes > Science > Chemistry > Miscellenous > Help file Index > Precipitation Equilibrium > Ksp and Ion Concentration \n\nThe solubility product constant can be used to compute the equilibrium concentration of ions in solution in the same way that you use other equilibrium constant expressions to compute equilibrium concentrations. \n\nUse the Initial, Change, Equilibrium method to compute the concentrations: write the equilibrium expression, the initial amounts of each thing in the expression, how much they change by, then the equilibrium amounts. From that, you can compute the concentrations of all species. \n\nExample: BaF2 has a Ksp of 1.8*10-7. What is the concentration of barium in a saturated solution of BaF2? \n\nSolution:. First, write the balanced equation and the solubility constant expression: \n\nBaF2(s) < = > Ba+2(aq) + 2F-(aq) \nKsp = [Ba+2][F-]2\nWe start with no barium or fluoride ions in solution. If we set the amount of barium ions formed to x, then the number of flouride ions formed is just 2x \n\n [Ba+2] (M) [F-] (M) \nInitial 0 0 \nChange +x +2x \nEquilibrium x 2x \n\n\nNow plug in the equilibrium values shown above and solve for x \nKsp = [Ba+2][F-]2 \nKsp = (x)(2x)2 \nx = 3.6*10-3\nSince we defined the concentration of Ba+2 to be x, we've solved the problem \n[Ba+2] = 3.6*10-3 \n\nInformation provided by: http://learn.chem.vt.edu\n \n**try to visit the site below..** You can calculate the Ksp based on the above example..I'm sure the answer is in your books and notes..have fun learning..
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according to the common ion effect, placing CaF2 in either solution will decrease its solubility.  However when having to choose between the two, it will be most soluble in Ca(NO3)2.  This is because Ksp=[Ca+][F-]^2  The fluorine component is squared making it a bigger factor in lowering the solubility.
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Okay. This is a multi-step pH problem. First, find the Kb of the fluoride ion (F-). To do this, follow this formula:\n\nKw/Ka = Kb\n\nThis formula can be rearranged to be Kw/Kb = Ka, by the way. Kw is the K of dissociation of water, which is 1.0e-14.\n\nPlug in the numbers and you get 1.47e-11 as a Kb for fluoride. Now, write an equilibrium ICE table, filling in the values you know.\n\n_F- + H2O --> HF + OH-\nI\nC\nE\n\nSolve the equilbrium table and you get the following equation:\n\nx^2/(0.00339-x) = 1.47e-11\n\nx is the concentration of both HF and OH-.\n\nSolve the equation and you get 2.23e-7 as the concentration of OH-. The easiest way to finish the problem now is to find the pOH. Take the -log of the concentration of OH-. The pOH is 6.651. Now, follow this equation: pH + pOH = 14. The pH of the solution is 7.34876. To the right amount of significant figures, the answer is 7.349 (the numbers to the left of the decimal place doesn't count toward the sig figs in pH's).\n\nHope this helped.
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Ba(NO3)2 + Na2CO3-->BaCO3 +2NaNO3\n\nto do a double displacement follow this formula\n\nAB + CD --> AD + CB
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equation is F= (Gm1m2)/(d^2)\n\nG is 6.67e-11Nm2/kg2.\n\nso...F= (6.67E-11)(100)(5.95E24)/((6.37E6)^2)\n\nuse a calculator and type it in exactly as you see it above, and you will get the right answer.  just be sure to put N at the end.
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S(I)+3f (g)--> SF (g)\n..........2...........6
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Here it is:\n\n(98,42)Mo + (1,0)n ---> (99,42)Mo
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5H2O2(aq)+2KMnO4(aq)+\n3H2SO4(aq)-->\n2MnSO4(aq)+K2SO4(aq)+\n8H2O(l)+5O2(g)
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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
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OF2 (~104), NF3(~107), CF4 (~109.5),  BF3 (120), BeF2(~180)
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You need the formula:\ndelta Tf=iKfXmolality\nyour delta Tf is 1.96\nyour i is 1 (because it does not dissociate)\nyour Kf is 5.12\nmolality is moles of solute divided by kilogram solvent, which is .0250 kg\nYour final equation should read:\n1.96=(1)(5.12)(x/.0250)\nThen use algebra to find x.
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6Rb(s)+N2(g)--->2Rb3N(s)
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would you like to supply me with some formulas? im out of grade 12 physics and i forgot most of the formulas. trust me from experience, just plug numbers into a formula and things will work\n\nF=mg = ma\nFk=μkN\na= vf^2 - vi^2 / 2d
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E = Fd = (2E-9 N) (200E-12 m) = x J\nx J (1 kJ / 1000 J) = xE-3 kJ\n\nI think this is already per mol...
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Ca2+=[Ar]\nMg2+=[Ne]\nI-=[Xe]\nF-=[Ne]
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2Fe (s) + 3F2 (g) ---> 2FeF3(s)
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bcd\n\n\n\nSolution:\nCaCN2 + 3H2O ---> 2NH3 + CaCO3