AP Chemistry problem not covered in my book...?
543314
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.............
1040448
ask your chem prof or tutor ..it's hard to explain over the internet!
975766
I think you've got the wrong exam number. I've probably got the answers (with 100% accuracy) if it's Chemistry, but, i need the correct exam number. Pick a question and then type it in here so I'll know if I have it. Let me know and I'd be glad to help.  \n\n:)
1142668
Problem \ntake D as Delta(change)\nEstimate the change in enthalpy, DH, for the following reaction:\n\nH2(g) + Cl2(g) --> 2 HCl(g) \n\nSolution \n\nTo work this problem, think of the reaction in terms of simple steps:\n\n\nFirst, the reactant molecules, H2 and Cl2, break down into their atoms\nH2(g) --> 2 H(g)\nCl2(g) --> 2 Cl(g)\n\n\nNext, these atoms combine to form HCl molecules\n2 H(g) + 2 Cl(g) --> 2 HCl(g) \n\nIn the first step, the H-H and Cl-Cl bonds are broken. In both cases, one mole of bonds is broken. When we look up the single bond energies for the H-H and Cl-Cl bonds, we find them to be +436 kJ/mol and + 243 kJ/mol, therefore for the first step of the reaction:\n\nDH1 = +(436 kJ + 243 kJ) = +679 kJ\n\nBond breaking requires energy, so we expect the value for DH to be positive for this step.\n\nIn the second step of the reaction, two moles are H-Cl bonds are formed. Bond breaking liberates energy, so we expect the DH for this portion of the reaction to have a negative value. Using the table, the single bond energy for one mole of H-Cl bonds is found to be 431 kJ:\n\nDH2 = -2(431 kJ) = -862 kJ\n\nBy applying Hess's Law, DH = DH1 + DH2\n\nDH = +679 kJ - 862 kJ\n\nDH = -183 kJ\n\n\nAnswer\n\nDH = -183 kJ
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.
844238
Li2O + H2O --> 2 LiOH\n\nIf you include the H2 it will never ballance.  You will not end up with H2 as a product anyway since OH will react with Li and the remaining H will react with the LiO hence 2 of LiOH.  Now on both sides of the equation you have 2 - Li, 2-O and 2-H.  If you did want to include a water molecule on the produst side then the equation would read as follcws:\n\nLi2O + 2H2O --> 2LiOH + H2O\n\nthis equation ballances with 2-Li, 3-O and 4-H on each side\n\nThe first equation shaow the ideal combination of the two compuonds (what most teachers are looking for) and the seccond equation shows a instance where there is not enough Li2OH to completely react with the all of the water
1138290
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.
791178
I HAVE A CHEMISTRY EXAM tomorrow. Actually a small test.lol\nwant answers go to the site below!
887064
The main thing you need to know to figure out the ionic equation is whether the product is soluble or not.  In this case, PbSO4 and Mg(NO3)2 are your products, and lead salts are insoluble.  Thus, PbSO4 and any ion that is used to make this molecule is removed from the ionic equation:\n\nPb(+2) + 2(NO3)(-1) + Mg(+2) + (SO4)(-2) -> PbSO4 + Mg(NO3)2 (aq)\n\nbecomes:\n\n2(NO3)(-1) + Mg(+2) -> Mg(NO3)2 (aq)
482555
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
863469
I think it's math! I've had trouble with it =) maybe chemistry also
880546
you should learn how to do your own homework hun you'll never be a chemist with someone else doing your work.\n\n\nuse the molar mass from the periodic table (units = g/1mol)\n\nCl = 35.5\nH = 1.00\n\nMolar mass, formula weight, molecular weight, formula mass = 36.5 grams HCl/1 mol HCl\n\n\nThis is a simple problem is you can't do this problem maybe you should try changing majors.  I hear history or english is nice.
122160
Are you sure it's not in your textbook? \n\nPlease do your homework. \n\nSee below...
606591
where were you when i was taking AP Chem in high school? lol
383531
========\n\nApparently purity, calculation or measurement errors have been there.\n\nYou repeated the thrice from the same solution to get an average.\n\nTry another solution.
660043
1.) do your own homework\n2.) this is the computer forum, not high-school chemistry.  Next time post to the right furom.
105211
Pressure * volume = number of mol * temperature (in ^oK) * 8.314\n\nYou have all the information you need right there. \nFirst solve for n for the initial state. \nSince you haven't added any more material you can then use the same value of n for the second state and solve for P.\n\nThis is sixth grade algebra. Anyone being asked to solve gas equations had better be able to do it standing on their head or they are in real trouble.