Determine the concentration of fluoride ions, [F-], when solid MgF2 (Ksp = 5.52x10^-9) is added to a 6.08x10^-
436191
Uh, the rest of the question isn't appearing. But it's an equilibrium solubility problem. Write a reaction:\n\nMgF2 --> Mg+2 + 2F-\n\nKsp = 5.52e-9\n\nFill in the values on the ICE table (I don't know them, the numbers are missing), and set an equation up.\n\nKsp = [F-]^2 * [Mg+2]\n\nSolve with the numbers from the ICE table. Remember that those concentrations in the above equation are the equilibrated concentrations, not the initial.\n\n\nHope this helped.
429534
manesiun and fluorine....magnesium forms a positively charged ion, while fluorine forms a negatively charged ion, different charges attract, so they form ionic bond to each other forming a ionic compound.\n\nnitrogen and sulphur both form negative charged ions, so there is noway to make them react unless in a prescence of a catalyst.\n\nionic compounds generally fom using, metals and nonmetals. magnesium is a metal, and fluorine nitrogen and sulphur are all non metals
508539
A) F = mg = 50 kg x 9,81 m/s2 = 0,49 kN.\nB) F = GMm/R2 = 6.67E-11 x 5.98E24 x 50 / (6.370E6)^2 N\n   = 0,49 N
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.
762286
I agree with Keith P.\n\nYou solve this by finding the number of moles of Sn and F.  You divide the sample (g) by the formula weight of the element.\n1.0g/118.71g = 0.0084\n0.640g/ 19g = 0.0336\n\nNext you divide the smaller of the two by the other to establish a sort of ratio.\n0.0336>0.0084\n\n0.0336/0.0084 = 4\nThus there is one atom of Sn for every 4 atoms of F and the empirical formula is SnF4.
788892
OF2 (~104), NF3(~107), CF4 (~109.5),  BF3 (120), BeF2(~180)
910910
Firstly, please forgive me for not having a periodic table at hand. :)\n\nSecondly, please forgive yourself as even the most advanced scientists in the world sometimes forget how to do this calculation.\n\nThirdly, it is such a simple calculation, that it is no wonder that it is so easily forgotten.\n\n1.5% solution (mass/volume) means 1.5 g of KBr in 100 ml of solution\n\nIt's that simple. :)\n\nEnjoy your science. :)\n\nSincerely,\n\nCurtis
813425
F = GmM/r^2; where G is a constant, m and M are the two masses, and r is the distance between them.\n\nm + M = 4 kg; so m = 4 - M\n\nF = G(4 - M)M/(.25)^2 = 2.5*10^-10; where G = 6.67 × 10−11 N m^2 kg^−2 \n\nUse the above to solve for M, then m = 4 - M to find m.  \n\nAnd there you have it, you do the math.
808333
You have a common ion effect problem\n\nPbCl2 <=> Pb+2 +2Cl-\n\nKsp=[Pb+2][Cl-]^2\nYou know that [Pb+2]=5.5*10^-3\nAlso from the reaction you know that the amount of Cl- coming from the PbCl2 is double the amount of Pb+2.\nSo if C is the concentration of Cl- coming from HCl, the amount of Cl- at equilibrium is [Cl-]=C+2*[Pb+2] \n\nSo Ksp= [Pb+2]* (C+2*[Pb+2])^2=>\nif we assume that C>>2[Pb+2] then C+2*[Pb+2] =C and \nKsp=[Pb+2]C^2 =>1.7*10^-5= 5.5*10^-3 C^2 =>\nC=55.6 * 10^-3 but 2*[Pb+2]=11*10^-3, so we can't do the approximation...\n\nLet's name [Pb+2]=p for simplicity\n\nKsp=p(C+2p)^2 =>\nKsp/p= C^2+4pC+4p^2 =>\nC^2+4pC+4p^2-Ksp/p=0 =>\nC^2+2.2*10^-3 C -2.97*10^-3 =0\n\nC=0.5*(-2.2*10^-3 + SQRT( (2.2*10^-3)^2 +4*2.97*10^-3) ) =>\nC= 5.34*10^-2\n\nIf you added Va L of HCl, then\nC=mole HCl/Vtotal = Ca*Va/(Va+0.12) =>\nCVa+0.12C=CaVa =>\nVa=0.12C/(Ca-C)= 0.12*0.0534/(0.12-0.0534) = 0.0962L or 96.2 ml
464163
FORMATION CONSTANT IS :\n\nkf = 1.5*10^7
504715
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.
83305
I suppose the problem was already due so I'll finish it for you now.\n\nF=m*a\na=F/m= 2*10^6N/1.2*10^6kg = 1.67m/s^2 * g/9.81m/s^2 = 0.17g
460952
Two hundred quattuordecillions\n\nCan be written using scientific notation as 2.0 x 10^47
462280
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...
400015
By newron's ii law\nMa=F-f\nf=10-10*0.6\nf=4 N\nf= -4 i (vector notation)
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.
558772
the correct answer is 1.30.  the answer would be two if the concentration was 0.01M.  you just take the negative log of the concentration of hydronium ion.