Can someone help me compare and contrast Q and Ksp if both are calculated using the same equations of ion concentrations.  I figured someone would know what I was talking about because the last question I asked was answered really well.  I kind of have an idea about this but I am not sure.  I have a test on Wednesday and it is an essay on this and my teacher is extremely picky and it is not in our notes.
199851
Q which is the reaction quotient is the ratio of concenctration of reactants to concention of products each concentrations raised to their respective stiochiometry (the number infront of the products or reactants when u balance the equation) at any given time other than equilibrium. Unlike Ksp which is solubility product is the same except it's at equilibrium.\nHowever, Q can help to to find out whether or not there is a precipitate will be formed so if Qsp is < Ksp no precipitate but if Qsp > Ksp a precipitate is formed  and will continue until the concentration of ions remaining in solution satisfy Ksp\nif u don't understand maybe u can send a e-mail and i can explain what u don't understand
227602
H2S---> 2 H^+ + S^2-\nK(eq) = \n[2H^+]^2 [S^2-]\n ---------------\n     [H2S]\n\nNH4+----> NH3  + H^+\nK(eq) =\n [H^+][NH3]\n       ------------\n         [NH4+]\n\nC6H5COOH----> C6H5COO-  + H+\n\nK(eq) = \n[H^+][C6H5COO-]\n        ---------------\n         [C6H5COOH]
46865
Kerplode is right - but that's kind of complicated.  Basically you solve both in exactly the same way.  The only difference is that instead of an = you have an > or <.  \n\nIt just looks different, there really is no difference in how you solve them.
177137
first, you find out the equation of Keq and delta G. I think it should be Keq = e^ [-deltaG/(RT)].\nThis equation and constant R can be found in any general chemistry textbook. And T is usu. 25+273=298K\nthen you can find out deltaG for the equation.\n\nwe also have deltaG= [- delta G for Ag(NH3)2^+1(aq) + (-32.0)] - [-26.2 - 3.8]\n\nand there comes the answer!!! ;)
28552
Wow, I hope these are right.  I wish i had a solubility chart or something (Which you probably have)  \n\nCuSO4(aq) + Na2S(aq) ---> CuS (s) + NaSO4 (aq)\nNet ionic - Cu2+ (aq) + S2- (aq)  ---> CuS (s)\n\nNiCl2 + 2KNO3 --> Ni(NO3)2 (aq) + 2KCl (aq)\nnet ionic - none (if NiCl2 isn't aq, then this is wrong)\n\nCa(OH)2 + 2HC2H3O2 ---> Ca(C2H3O2)2 (aq) + 2H2O (l) \nnet ionic - H+ (aq) + OH-(aq) ---> H2O (l)\nactually, since the acetic acid is a weak acid, this might be a little different.  If you learned it this way, then your net ionic is:\nHC2H3O2 (aq) + OH- (aq) ---> H2O (l) + C2H3O2- (aq)\n\nHope this helped
253572
1.) Q = (nh3)^2/(n2*h2^3) = 0.16\n\nAs Q>Kp, reaction will move backward.\n\n2.) Kc = Kp*(RT)^(1+3-2) = 3.1*10^-4*(8.314*700)^2 = 3.55*10^11\n\n3.)\n\nGiven reaction is 2*(ii)+(i)\n\nSo, Kp = (8.3*10^-3)^2*3.1*10^-4=2.13*10^-8\n\nLet original P of HCN = 1\n\nso, partial pressure of C2N2 and H2 = p (assuming 2p of HCN decomposes at equilibrium)\n\nEqn is\n\np*p/(1-2p)^2 = 4*10^-4\n\nSo, p/(1-2p) = 2*10^-2\n\nSo, p = 0.019 atm.\n\nEnjoy.............
247376
I don't know if this is right ... but check it and tell me\n\nV*C KOH = V*C HBr\n\nV KOH = 18.6ml C KOH = 0.752M\nV HBr = 22.1ml C HBr = ?\n\n18.6*0.752 = 22.1*C HBr\n\nC HBr = 0.6329M\nIts seems reasonable cause the difference between concentrations isn't significant
419805
Neutralization Reaction:\nHC2H3O2(aq) + KOH(aq) -->H2O(l) + KC2H3O2(aq)\nThe equation is already balanced.\n\na. False KC2H3O2(aq) and H2O are products.\nb. True\nc. False, some ions will cancel out but some will remain.\n\nThe ionic equation is as follows:\n(H^+)(aq) + (C2H3O2^-)(aq) + (K^+)(aq) + (OH^-)(aq) -->H2O(l) + (K^+)(aq) + (C2H3O2^-)(aq)\nRepeated ions are crossed off from both sides and the net ionic equation is:\n(H^+)(aq) + (OH^-)(aq) -->H2O(l)\n\nd. True\ne. False\n\nNote: the + and - signs were written with ^ to indicate that they were raised up (like a power).  Sorry if this notation is confusing.  Good luck and I hope that this helps!
224245
Homework, huh?\n\nWell, it's because K+ contains more electrons than the Na+. Which means that K+ has 3 layers of electrons (filled in the 1s2 2s2 2p6 3s2 3p6) while while Na+ has only 2 layers of Electrons (1s2 2s2 2p6)
330317
Jaleesa, I think that you can use the same principle that I explained in the earlier question. The Sodium ion has fewer electrons and fewer shells, and is thus smaller than the potassium ion. But I think you should do the rest of your homework yourself.
439197
If C is your initial concentration (C=0.114 M) then\n\n.. .. .. .. .. .. .. .. HC3H5O2 <=>  C3H5O2- + H+\nIntial .. .. .. .. .. .. .. C\nDissociate .. .. .. .. x\nProduce .. .. .. .. .. .. .. .. .. .. .. .. .. .. x .. .. .. .. x\nAt Equilbrium .. .. C-x .. .. .. .. .. .. .. x. .. .. .. .. x\n\nKa=[C3H5O2-][H+] / [HC3H5O2]= x^2/(C-x)\n\nYou find x either by solving the quadratic equation or by assuming that C >> x and then C-x=C so Ka=x^2/C. You calculate x and if it is truly x << C the assumption is valid (otherwise you have to solve the quadratic)\n\nOnce you have the value of x,\n[H+]=[C3H5O2-]=x\n[HC3H5O2]=C-x= 0.114-x\npH=-logx\n[OH-]= Kw/[H+]= 10^-14/x, or if you prefer pOH=14-pH and [OH=]=10^-pOH\n\npercent dissociation is (x/C)*100%
548258
You need a variable in any experiment to compare results. Perhaps you could use cold (ice) water in one container, warm in another and hot or boiling in another. You can then make your prediction on which will dissolve first and have an experiment that will verify whether you have made an accurate prediction.
223221
I think they're both 1.ooM.  \n\n          HCL -> H+ + Cl-\n         1mole   1mole  1mole\nUsing mole ration, the hydronium and chloride ions will both be 1 M concentrations. :)\n\nI think that's right.
647359
Look up stuff like Arrhenius equations. I don't have my phys chem text book to hand.\n\nFor part two, I think it's D, as I'm fairly certain it's not any of the others.
522266
I have two problems with this: First, this is not an avenue for high school kids to get their homework done for them and second, the thought that someone would take yahoo answers 'points' as payment for doing a highschool kids homwork makes me feel ill. \n\n(i)All that aside, the reducing agent is the compound/element that looses electrons. \n\n(ii) First balance the equation, then find then number of moles of potassium permagenate used in the titration, then convert to the equivalent number of mols of the Be complex. \n\n(iii) multiply answer for (ii) by 50\n\n(iv) use the difference between your asnwer for (iii) and the given sample mass to find mass percent.
22824
OK. You've posted this (and several other) questions(with exactly the same words and numbers) for the fourth day in a row, and everytime you've asked, one person has responded and we've all got different answers. \n\nI can't tell if you're a student that's gonna keep posting until several people give you the same answer, of if you're just someone that likes making the nerdy people solve chem problems. If you're the latter, you should really find a better use for your time. \n\nIf you're the former, try working the problem yourself and then compare your answers to ours, or compare our approaches, your notes, and textbook and see where they differ. Our different answers can probably be attributed to some math or conversion error. \n\nRemember, when all else fails follow the units and you'll be right more times than not.
352207
Here's a nice explanation:\nhttp://www.4to40.com/QA/index.asp?counter=382&category=science