Please explain in details why if I give the following cell potentials:\n\nK+ + e- ---> K(s)            E = -2.924\nI2 + 2e- ---> 2I-            E = +0.535\n2H2O + 2e- ---> H2 + 2OH-    E = -0.828\n2I- ---> I2 +2e-             E = -.0535\n2H2O --> O2 +4H+ + 4e-       E = -1.229
213351
reduced: 2H2O + 2e- ---> H2 + 2OH-\n\noxidized: 2I- ---> I2 +2e-
478861
Well, in pure water, if you let x = [Mg2+] the [OH-] = 2x.  Per the equilibrium expression 5.61 x 10E-11 = 4xE3 so x=2.4 X 10E-4 M\n\nIn the NaOH solution, the [OH-] is 0.180 M (the contribution from the Mg(OH)2 is negligible).  So in this case\n                5.61 x 10E-11 = [Mg](.180)E2  \nSolving for [Mg], I get 1.7 X 10E-9 M\n\nCheck my arithmetic, I often get careless punching the calculator.
450233
Use the Nernst equation:\ndE = dE0 - 0.0592 V/n ln [C]^c [D]^d /([A]^a [B]^b)\n\nn is the number of electrons involved, 2 in this case.\nTherefore we have in all cases:\n dE = dE0 - 0.0296 V ln [Sn2+] /([Fe2+]\nFurthermore we use the standard electrode potentials to calculate  dE0 \nSn2+(aq) + 2e- -> Sn(s) -0.14 V\nFe2+(aq) + 2e- -> Fe(s) -0.41 V\nTherefore dE0 = 0.41 - 0.14 = 0.27 V\na) The concentration are 1 M/L and we have dE = 0.27 V\nb) dE = 0.27 - 0.0296 ln 0.01/0.2 = 0.27 + 0.09 = 0.36 V\nc) dE = 0.27 - 0.0296 ln 0.4/0.005 = 0.27 - 0.13 = 0.14 V\n\nOnly in the last case the reaction moves to the left: the concentration of Sn2+ is the highest.\n\nYou can take a look here for explanations and examples:\n\nhttp://www.science.uwaterloo.ca/~cchieh/cact/c123/nernsteq.html
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
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)
394654
1.) 2Ni + O2 -->  2NiO  (since it's Nickel II and Oxygen has a charge of -2, but you always have to put oxygen in groups of 2)\n\n1.) 2H2O --> 2H2 +O2  (same with hydrogen)\n2.) 2Al2O3 (because Aluminum has a charge of +3) --> 4Al + 3O2\n3.) 2NaCl --> 2Na + Cl2 (basically the rest follow the same rules)\n4.) 2NH3 --> 3H2 + 2N\n5.) 2HgO --> 2Hg + O2
339698
1)\nM = Mo e^(-kt)\ne^(-kt) = (M/Mo)\n-kt = ln(M/Mo)\nk = (-1/t)·ln(M/Mo)\n= (-1/18.8)·ln(57.9/93) = 0.025 yr^-1\n\n2)\nM = Mo/2 = Mo e^(-kt) ──► ½ = e^(-kt)\n-kt = ln(½)\nt = (-1/k)·ln(½)\n= (-1/0.025)·ln(½) = 27.5 yr
462271
Li2O + CO2 ---> LiOH\nNa2O2 + CO2 ----> Na2CO3 + O2\nKO2 + CO2 ---->  K2CO3  + O2
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!
180554
k e =1/2  * m *v *v\n    =1/2 *3 * (3.5*10) * (3.5*10)\n    =0.18j (ans)
28854
the earlier equation is not balanced.\nCaO + 2HCl --------> CaCl2 + H2O
374337
Cl2 + 2 KI → I2 + 2 KCl
326793
MnO4- + 5e- + 8H+ -> Mn2+ + 4H2O\n\nVO+2  + H2O ->  VO2+ + e- + 2H+\n\nMultiplying eqn. below by 5\n\nMnO4- + 5VO+2 + H2O -> Mn2+ + 5VO2+ + 2H+
293029
The previous responses are right but ko is not a potential. It's the (average) momentum for a particle in free space. With psi defined by:\n\npsi(x) = A exp(ikx - x^2/(2sigma))\n==>\npsi*(x)psi(x) = |A|^2 exp(-x^2/sigma)\n\nYou need the integrals:\n\n&#92;int_(-oo)^(+oo) exp(-x^2/sigma) dx = sqrt(pi*sigma)\n&#92;int_(-oo)^(+oo) exp(-x^2/sigma) x^2 dx = (sqrt(pi)/2) sigma^(3/2)\n\n<x^2> is the second expression divided by the first:\n\n<x^2> = 1/(2 sigma)\n\nSince H = p^2/2m, \n\nH psi(x)\n= -(1/2m)(d^2/dx^2) psi(x)\n= -(1/2m) [(ik-x/sigma)^2 - 1/sigma] psi(x)\n= (1/2m) [k^2-x^2/sigma^2 + 2ikx/sigma + 1/sigma] psi(x)\n\n==>\n\n<H>\n=(1/2m)[k^2-<x^2>/sigma^2 + 2ik<x>/sigma + 1/sigma]\n=(1/2m)[(k^2+1/sigma-(1/2sigma)]\n=(1/2m)[(k^2+1/(2sigma)]
479190
For reaction aA <--> bB + cC, the Eq.constant K is defined as\n\nK = (B^b * C^c) / A^a\n\nSince in your equation a=b=c=1, and concentration for B=C, the equation simplifies to\n\nK = B^2 / A\n\nwith B=0.046 and A=0.454    (0.454=0.500-0.046)\n\nso K = 0.046^2 / 0.454 = 0.0047
348952
You have to start with a balanced equation.  The subscripts will give you the correct charge and a prefix will give you the right amount of atoms on each side of the equation.  Some elements, like O, F, and I are diatomic and come in pairs.
505326
its actually pretty simple. U multiply by -1 when u flip over an equation...so the reactants become the products and the products the reactants. that way instead of having a negative number u will have a positive number for the kJ. as in this problem. U have to flip #1 so u can cancel a few things with #2. so it becomes..\n1)3 H2O(l) + 2 CO2(g) AH---->CH3CH2OH(l) + 3 O2(g) (multiply by -1 gives u a postive +1367kJ. part two stays the same.\n2) C2H4(g) + 3 O2 ---> 2 H2O(l) + 2 CO2 (g) AH= -1411 kJ\n\nNow the 3 O2 from #1 and #2 cancel since they are on opposite sides. Now the 2 H20 cancel from #2 and you are left with only one in #1 since they are also on opposite sides...which after canceling..leaves u with your original equation.\n\nC2H4(g) + H2O(l) ------> CH3CH2OH(l)\n\nnow since u flipped over #1.....and u have a positive +1367kJ and\n#2 is left the same...its still at negative -1411kJ. u add them together and u get an enthalpy change of -44kJ. sorry its pretty confusing to explain..if anything message me and i willl try to help.