Ksp for BaF2 = 2x10^-6  \n\ni have no idea how to solve this
127814
The reaction is\n\nBa+2 + 2F- <==> BaF2\n\nLet x = [Ba+2] and [F-] = .1 M\n\nQ = [Ba+2][F-]^2 = .01x\n\nA precipiate forms when Q > Ksp.\n\nThis happens when .01x > 2*10^-6\nx = [Ba+2] = 2*10^-4\n\nYeah, sorry, I got confused for a second.  There's a few listings here if you're interested: http://www.towson.edu/~ladon/solprod.html
547088
NaNO3 is very soluble\nBaF2 has Ksp=1.84*10^-7\n\nBaF2 <=> Ba(+2) + 2F(-)\n\nso Ksp=[Ba(+2)][F(-)]^2 This is for the concentrations at equilibrium.\nIf the product P=[Ba(+2)][F(-)]^2 for the initial concentrations (and not the concentrations at equilibrium) is P>Ksp then you have a precipitate, if P=Ksp then you have a saturated solution and if P<Ksp you have a non saturated solution.\n\n[Ba+2]=mole/V= 5*10^-4/1=  5*10^-4\n\nNaF -> Na(+) + F(-)\nTherefore [F(-)]=[NaF]=6*10^-3\n\nP=[Ba(+2)][F(-)]^2 = (5*10^-4)*(6*10^-3)^2 =1.80*10^-8 <Ksp so no precipitate
304365
Set h(t) = 26 and solve,\n\n26 = 11 +34t -16t^2\n\n16t^2 - 34t +15 = 0\n\nnow, factoring by parts:\n\n(16t^2 - 24t - 10t +15)\n\n8t(2t - 3) - 5(2t - 3)\n\n(8t - 5)(2t - 3) = 0\n\nSo, it passes the window at t= 5/8 and t = 3/2.
407182
Keq = [HF]^2/[H2][F2] = (0.4)^2/0.5)(0.1) = 0.16/0.05 = 3.2\nNow add 0.150 mole F2.
484301
Get the unbalanced force from the first 2 #s in your ? :  2.8-2.6 = .2\n\nb)  F = ma---> m = F/a = 0.2/.11 = 1.818 kg
392163
delta S = delta H vap in J / Temp in K\n\nPLEASE READ WAD WE HAVE WROTE IN YOUR EARLIER QNS
111128
I think I can do this one...I'll get right back to you when I figure it out... *grabs a pen and paper...and also tries to figure out why people are answering, saying simply that they can't or won't answer your question*\n\nEDIT...k, here's what I got...not sure if I understood what your question was asking, so I may be mistaken...\n\nm=h(x)-2x\nk=(2-3h(x))/6x
355788
Ka = [H+][A-]/[HA]\nin\n(a), [H+]=[A-]=x, [HA] = 0.1 -x\nso x^2/(0.1 - x) = 1.37 x 10^-4\nx = 0.0037 M\n% diss = (0.0037)/(0.10) = 3.7%\n\n(b), [H+]=[A-]=x, [HA] = 1 -x\nso x^2/(1 - x) = 1.37 x 10^-4\nx = 0.0117 M\n% diss = (0.0117)/(1) = 1.2%
630083
Jennifer, I'm a little bit puzzled about your question because you're asking for what the funtion is and you have it written there.  The funtion is s = -16t2 + v0t + s0\n     What I really think you want is the solution.  Correct?  Well, here's how you get that.  You put in 0 for s0, and 64 for v0.  ALSO, s, on the left side of the equation, is also zero because it will land back on the ground when it's done.  When you solve using these numbers, the only variable left is t.  And when you do the algebra, \n  0  = -16t2  +  64t  + 0\n \n16t2  =  64t\n\n16t  =  64\n\nt  = 4      seconds, that is.\n\nI hope this works.  I'm a physics teacher.  If you need further clarification for this problem, or if you need help with any others, feel free to write me at fortitudinousskeptic@yahoo.com\nI'll be glad to help!
519078
T=m1*-9.8+m1*a\nT=m2*a--m2*-9.8\n\nset them equal and solve for a\n\ntrust me, i had a test on this today
219453
AgBr-------->Ag+ + Br-\n(a) Ksp=[Ag][Br] solubility=s=[Ag]=[Br]=Ksp^0.5\ns=(6*10^-7)^0.5\n(b) [Br]=0.01 (because all of kBr was dissuciated)\ns=Ksp/[Br]=(6*10^-7)/0.01 mol/lit
404410
F=-kx and F=ma\n\nfind mass from period     T = 2pi(m/k)^(1/2)\n\nhmmmm,  you seem to be missing k\n\nfor fixed ratio m/k the period of 0.51s and amplitude 4 cm is not enough to determine the mass (and therefore acceleration).
377523
2C12H22O11 +24O2 ---> 24CO2 +22H2O\na)F\nb)F\nc)T\nd)F\ne)T\nf)T
533662
k = .7/.25 = 2.8 N/m\n\nT = 2π√(m/k) = 2π√(.25/2.8) = 1.877 sec\n\nergo, t = .71*1.877 = 1.33267 sec\n\nYou haven't told us where the glider is at t = 0\nIf it's at its deflected position, then x = .25cos(2πt/T) = -.062\nIf it's at x = 0 moving left, then x = -.25sin(2πt/T) = +.218
173450
Write the equation for K.  You know K, [NO2}, so solve for the only variable you don't know [N2O4].  By the way, these are gasses so it is not a solubility product.  That involves a solid.
665249
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.
176646
By newron's ii law\nMa=F-f\nf=10-10*0.6\nf=4 N\nf= -4 i (vector notation)