Need help! A copper penny dropped into a solution of nitric acid produces a mixture of nitrogen oxides...?\nThe following reaction describes the formation of one of them, NO:\n3Cu + 8H^+ + 2NO3^- = 2NO + 3Cu^2+ + 4H2O\n\na) Calculate E for this reaction.\nb) Calculate E at 298 K when (H^+) = 0.100M, (NO3^-) = 0.0250M, (Cu^2+) = 0.0375M, and the partial pressure of NO = 0.00150atm. \n\nPlease show me, Not Tell Me, how to do this if the answers are:\na)Ecell = +0.62 V\nB)Ecell = +0.61 V \n\nI understand the equation and all for this I am just not geting it though.
200463
Part A\n\nReduction: 3e^- + 4H^+ + NO3^- = NO + 2H2O    Eo = \nOxidation:                 Cu  = Cu^2+ + 2e^- Eo =\n\nMultiply the reduction half equation by 2 and the oxidation half equation by 3 (so that you have 6e^- in both) then add together to get:\n\n3Cu + 8H^+ + 2NO3^- = 2NO + 3Cu^2+ + 4H2O\n\nNow Eo cell = Eo cathode - Eo anode = 0.96 - 0.34 = +0.62V\n(Note: you do NOT multiply the Eo values by the number of electrons in the half equations because Eo is a voltage, measure of electrical pressure.\n\n\nPart B\n\nThe Nernst equation can be found at:\n\nhttp://www.science.uwaterloo.ca/~cchieh/cact/c123/nernsteq.html
466646
Well, for a start your equation shows copper(II) sulphide, not sulphate, and your equation is wrong even then.  That makes a bit of a difference!\n\nLet us imagine that you've got the formula right, but the name wrong.\n\nThe equation is simply Cu + S ---> CuS\n\nThen 64g will give (64+32)g.\n\nThus 64g will give 96g.\n\nSo 64 x (200/96)g of Cu will give 200g of product.
531209
Your first conclusion is correct, but your second isn't.  Since you already have the mechanism given to you, you can see that the 2nd step is the slow step.  So, the rate law should be:\n\nrate = k (N2O2) (H2)\n\nSince N2O2 is an intermediate, it can't appear in the overall rate law, so you have to get rid of that.  From the first reaction, you see that N2O2 is formed from the collision of 2 NO.  So, the concentration of N2O2 is proportional to [NO]^2.  That gives you the overall rate law that you had, and the overall order of the reaction is 3rd order.\n\nYour second conclusion is incorrect because the slow step requires the collision of N2O2 and H2.  The slow step is just a second order reaction.  \n\nSo, you've got things pretty well in hand...Hope this helps a little...
423536
Here you are:\n\nFirst write the combustion equation and balance it:\n\nH3NNH2(l) + O2(g) ---> N2(g) + 2H2O(g).\nThe heat will be the diffrence of enthalpies of formation between the products and the reactants. Do not forget to multiply by the stoichiometric coefficient.\n\nRemember that the enthalpies of formation of O2 and N2 are 0.\n\nYou will also need to find the enthalpy of formation of H2O(g) in the table of your book.  It is -241.8 kJ/mol\n\nHf(H2NNH2) = [2*(-241.8) + 1*0] - [1*50.6 + 1*0]\n\nHf(H2NNH2) = -483.6 - 50.6 = - 534.2 kJ/mol
549229
You should have gotten elemental copper via the following equation:\n\n3CuCl2 + 2Al ---> 3Cu + 2AlCl3\n\nThe water is not part of the reaction.
459512
2Al + 3*(CuO) -> Al2O3 + 3*Cu\n\nI2 + 2*(Na2S2O3) -> 2*(NaI) + Na2S4O6\n\nI2 + 5*(Cl2) -> 2*(ICl5)
574455
Silver is being oxidized, nitrogen from the nitrate is being reduced.\n\nOxidation half reaction:\nAg --> Ag+  +  e\n\nReduction half reaction:\nNO3- +  4H+ + 3e  -->  NO + 2H2O\n\nNet balanced and reduced equation:\n3Ag +  NO3- +  4H+ --> 3Ag+ + NO + 2H2O
8101
Cr - 51.995g/mol\n49048gCr / 51.995gCr/mol = 943.32mol Cr\nStoichiometrically 2Cr to 3Cu so 943.32 x 3/2 = 1415mol Cu.\n\nPlease correct me if I'm wrong but that makes sense to me.\nOh and please award me best answer...\n\noh and part b:\n200ml x 0,75mmol/ml = 0.15mol\n0.15mol x 2/3 = 0.1molCr\n0.1molCr x 51.995g/mol =5.2g Cr.\n\nAgain correct me if I'm wrong...
524190
You are using the wrong number of moles...\nHNO3 mol = (40.5ml/1000 ml/L)(0.620 mol/L) = 0.02511\n\nfrom chemical eqn...\n\n(3 mol H2/6 mol HNO3) 0.02511 mol HNO3 = 0.012555 mol H2\n\nNow...\n\nPV = nRT\n\nT = 27.5 + 273.15 = 300.65K\nP = 0.755 atm\nn = 0.012555 mol \nR = 0.08206 L atm/mol K\n\nPV = nRT\nV = nRT/P = (0.012555 x 0.08206 x 300.65)/0.755 atm = 0.41 L
432576
Copper is more reactive than silver, hence displacement of silver occurs.\nCu (s) + 2AgNO3 (aq) --> 2Ag (s) + Cu(NO3)2 (aq)\n\nSum of the coefficients? 6...\n\nHope this helps :)
483757
Al + Fe(NO3)3 ---> Al(NO3)3 + Fe.\n\n2K + 2H2O ---> 2KOH + H2\n\n2KOH + CuSO4 ---> Cu(OH)2 + K2SO4\n\n3NaOH + H3PO4 ---> 3H2O + Na3PO4\n\nBest?
51199
Please do your own home work.\n\nI suspect you are missing a symbol, it probably should be,\nN2O4  <==>  2NO2\n\nThink about how each of the changes will effect the equilibrium of the reaction.
355448
Concept: An element and a compound can react to form another element and another compound. \n\nMaterials: 200 mL of 0.1M copper(II) chloride (CuC12), 5 cm. square piece of aluminum foil (Al), 400-mL beaker, safety goggles. \n\nProcedure: 1. Place the beaker of copper(II) chloride in front of the class. 2. Crumple the aluminum foil and throw it in the beaker. 3. Soon the solution starts to bubble as hydrogen gas is produced. 4. A red brown precipitate of copper falls to the bottom of the beaker, and the aluminum foil dissolves completely. 5. Aluminum would react spontaneously with water without its protective oxide coating. The aluminum-copper(II) chloride reaction removes this protection. Caution: Copper(II) chloride is an irritant. \n3CuCl2 + 2Al --> 2AlCl3 + 3Cu\n\nInput = 3CuCl2 + 2Al \nOutput = 2AlCl3 + 3Cu
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
534322
1.\nMn(s) + SnCl2(aq) --> MnCl2(aq) + Sn(s)\nCr(s) + 2HBr(aq) --> CrBr2(aq) + H2(g)\nCo(s) + CaBr2(aq) --> N.R\n2Al(s) + 3Ni(NO3)2(aq) --> 2Al(NO3)3(aq) + 3Ni(s)\n\n2.\nMg + 2AgNO3 --> Mg(NO3)2 + 2Ag\nmole of Ag = 3.580 / 108 = 0.033 mole\nmole of AgNO3 = 0.033 mole = 33 mmole\nmolarity of AgNO3 = 33 / 50 = 0.66 M\n\n3.\na.soluble\nb.soluble\nc.soluble\nd.insoluble\ne.soluble\nf.soluble\ng.insoluble in water, soluble in acid\nh.soluble\n\n4.\n3NaOH(aq) + Fe(ClO4)3(aq) --> Fe(OH)3(s) + 3NaClO4(aq)\n3Na+ + 3OH- + Fe3+ + 3ClO4- --> Fe(OH)3 + 3Na+ + 3ClO4-\nFe3+ + 3OH- --> Fe(OH)3\n(NH4)2SO4(aq) + Pb(NO3)2(aq) --> PbSO4(s) + 2NH4NO3(aq)\n2NH4+ + SO42- + Pb2+ + 2NO3- --> PbSO4 + 2NH4+ + 2NO3-\nPb2+ + SO42- --> PbSO4\n\nHope it helps\npeace\nvixklen
379368
NaCl2 + NaCO3 ==> NaCO3 + NaCl2 NOTE: these two reactants will not react because the reactants and the products are identical\n\nCu + O2 ==> CuO2   This is already balenced\nCuSO3 + Fe ==> FeSO3 + Cu   The charge is not given so procuts could be different\n\n2HCl + 2Zn ==> 2ZnCl + H2\nMnO2 + H2O2 ==> No reaction: similar to NaCl2+NaCo3\n\nGood luck
63081
[Ca(NO3)2]=n/V  ---> n=[]*V= 0.390M * 0.044L=0.01716 mol\nn=m/M ---> m=n*M= 0.01716mol * 164 g/mol=2.81424 g