If it takes 100. minutes to recharge one battery, how much Ni(OH)2 (in grams) is oxidized to NiO(OH)?
457090
First you can calculate the quantity of electricity used to charge the battery using the equation:\n\nQ = It\n\nwhere\nQ = charge in Coulombs (C)\nI = current in Amps (A)\nt = time in seconds (s)\n\nIn the above question then\n\nQ = 0.5 x 100 x60 = 3,000 C\n\nNow the oxidation state of the Ni in Ni(OH)2 is 2+, while the oxidation state of Ni in NiO(OH)2 is 4+, so the change is 2+.\n\nFaraday's constant = 96,500 Cmol-1 so \nthe moles of electrons available = 3000/96500 = 0.03109 \n\nHowever, 2 moles of electrons are required to oxidise 1 mol of Ni(OH)2 therefore moles of Ni(OH)2 that can be oxidised is 0.03109/2 = 0.015545. \n\nLastly, we can now convert the moles of Ni(OH)2 to grams by multiplying by its formula mass, which is 92.708 gmol-1, ie\n0.015545 x 92.708 = 1.44 g.
900316
This is a common redox question easily balanced by the half-reaction method.   The half-reaction method is taught using a few different styles, so you'd best learn the style that your teacher or professor expects you to learn.  \n\nFirst, identify your half reactions and balance them:\n\n(Cr2O7)2- contains 2 Cr6+ ions being reduced to 2Cr3+, gaining  6 electrons:\n\n6e-  +  2Cr6+  --->  2Cr3+    (net charge now adds up to +6 on both sides)\n\nMn2+ is oxidized to Mn4+ (as MnO2) losing 2 electrons, or\n\nMn2+  ---->  MnO2  + 2e-\n\nNet electron transfer must equal zero, so we must triple the oxidizing reaction to yield 6 e- to balance the 6 e- gained, so e- gained now equals e- lost (that is, 6 e' in, 6 out)\n\nThat gives us the clue to multiply the manganese term by 3:\n\n(Cr2O7)2- + H+ + 3Mn2+ --->  2Cr3+ + 3MnO2  + H2O\n\nInspection shows the Cr, Mn, and O terms to be balanced but the left side of the equation to be down by 1 H+, so let's put a 2 by the H+....\n\n(Cr2O7)2- + 2H+ + 3Mn2+ --->  2Cr3+ + 3MnO2  + H2O\n\nWe think we're finished here, so let's make the important two final checks:\n1.  Mass Balance and 2.  Charge Balance\n\nEach side of the reaction has 2Cr, 3Mn, 7O, and 2H\n(Mass balance OK)\n\nNow add up the net charges on both sides:\n-2  +2  +6 = +6 on the left side...\n+6 + 0 + 0 = +6 on the right.  (Charge Balance OK)\n\nReaction is balanced.\n\nAgain, WARNING:  Some methods teach you to balance in ACID or in BASE.  This problem, by virtue of the H+ presence is in acid.  Balancing a redox equation in base has a couple more steps attached.  Again, consult your text or your instructor as to their preferred style.
836906
I had to read about this in order to answer your question at this web site:\n\nhttp://home.um.edu.mt/chemistry/CH230/2.Analytical/CP210a.PDF\n\nSo, two moles of DMG combines with each Ni++ ion.\n\n(0.2145 g Ni)* (1 mole Ni/58.71 g Ni) = 3.65*10^-3 mole Ni\n\nIf what you tell me is correct that DMG is 116.1194 then the Ni complex which loses 2 H+ for the Ni++ has a molecular wt of\n172.8134 (please double check)\n\n(3.65*10^-3 mole Ni)* (172.8134 g/mole) = 0.6314 g\n\nplease check for errors
798267
This is a Hess' Law problem.\n\nThe strategy I like to teach for these is to get the biggest thing you need first then get rid of the biggest thing you don't need next.  So we want to get the 2 N2O5 first then get rid of the resulting 4 HNO3.\n\nUse -2 times the first eq to get your 2 N2O5, -4 times the third to get rid of your 4 HNO3, finally -2 times the second to get rid of the water:\n\n2 N2O5 + 2H2O > 4 HNO3   -152.4 kJ\n4 HNO3              >  2 N2 + 6 O2 + 2 H2  +696.0 kJ\nO2 + 2H2           >   2 H2O  -572 kJ\n\nEverything cancels properly to get the desired equation, and delta H is -28.4 kJ.
343278
Half reactions are:\nOxidation: Cu ----> Cu2+ + 2e    Eo = -0.34V\nReduction: NO3- + 4H+ + 3e --->  NO + 2H2O   Eo = +0.96V\n\nAdd the two voltages (do not multiply this number by the coefficients you use to cancel the number of electrons out) and you get the +0.62V for the Eo of the cell.\n\nThe Nernst equation will take this form:\nE = Eo - 0.0592/n * log {(PNO)^2 * [Cu2+]^3 / [NO3-]^2 * [H+]^8}\n\nn will equal 6 here, not 8, as that will be the number of electrons that will be cancelled out when you add the oxidation and reduction reactions together (multiplied by the correct number to cancel those electrons out).\n\nThe concentration's portion simplifies, once you plug all of the numbers in, to about 18.94, whose log is 1.278.  Multiply that by 0.0592 and divide by 6 and you get 0.01, so only 0.01 V will be lost from the standard potential for that reaction at those concentrations.
538915
A - battery has 3 major componets. Acid, a salt bridge, and some type of metal (depending on the battery type... a nicad battery has nickel and cadmium for example).  Acid 'donates' a proton (H+) along the salt bridge, which in turn reacts with the metal.  This reaction is what produces energy.  The metal is oxidized.  Its been a while since I studied this, but overall, it is pretty close.
729087
1) this one is already balanced for you!!\n\n2)Count the number of Mg and O on the right side, and count the number of Mg and O on the right side.  On the left, you have 2 O and 1 Mg, and on the right you have 1 Mg and 1 O.  To make the oxygen on both sides equal, you need to double the MgO.  Now on the left, you have 2 O and 1 Mg, and on the right, you have 2Mg and 2 O.  Now you need to multiply the Mg by two so the Mg on both sides is equal.  and you're done!!!! \n2Mg + 02 --> 2MgO\n\n3) On the left, you have 1 Zn, 1H, and 1 Cl.  on the right you have 1 Zn, 2Cl and 2H.  the Zn are balanced and you just need to multply the HCl on the left to balance the whole equ.\nZn + 2HCl --->ZnCl2 + H2\n\n4) On the left, you have 1 P and 2O, and on the left you have 4 P and 10 O. If you multiply the P on the right by 4, and the O on the right by 10, you have 4P and 10 O on both sides\n4P + 5O2 ---->P4O10\n\n5)  On the left, you have 1 Fe 2 Cl, and pn the left, you have 1 Fe and 3 Cl.  The easiest way to balance this is multiply the Cl on the right by the amount of the left one and vice versa. when you do that, you have 1 Fe and 6Cl on the right, and you have 2Fe and 6Cl on the left. To finish the equation, multiply the Fe by two.\n2Fe + 3Cl2---->2FeCl3\n\nVoila!  You have balanced 5 equations!\n\ni hope this was helpful!
747348
Hi Kitty!  Dr. J to the rescue with a full answer.\n\n1) 4Li + O2 ---> 2Li2O\n\n2) 4HF + SiO2 ---> SiF4 + 2H2O\n\n3) It is simpler if we eliminate the K and the sodium and write the equation ionic because it is a redox reaction and we need to balance.\n\nThe overall redox equation will be:\n\nCr2O7(=) + SO3(=) + H(+) ---> Cr(+3) + SO4(=).\n\nNow you balance by the half reaction method.  The two reactions are:\n\n6e(-) + Cr2O7(=) + 14H(+) ---> 2Cr(+3) + 7H2O\n3SO3(=)  + 3H2O --->  3SO4(=) + 6H(+) + 6e(-)\n\nEliminating the electrons:\n\nCr2O7(=) + 3SO3(=) + 8H(+) ---> 2Cr(+3) + 3SO4(=) + 4H2O\n\nThis one was tough!!!  Thumbs up!\n\nIf you want you could put the K and the Na back.\n\nBest answer?
951534
Ca(OH)2 + H2SO4 = 2H2O + CaSO4\n\n1 + 1 + 2 + 1 = 5
929685
When you have weak electrolytes, (equillibria in general) you should make the following table.\n\n1. write the reaction\n\nHNO2 <=> H+ + NO2-\n\n2.Put the initial amounts (expressed as concentration) of each molecule. In this case we have only the concentration of HNO2, so\n\n.. .. .. .. .. .. HNO2 <=> H+ + NO2-\nInitial .. .. .. .. 0.5\n\n3.Assume that an unknown amount "x" of the reactants react.\nThen write also how much of each product you expect to be produced according to the stoichiometry. In this case 1 mole of HNO2 gives one mole of H+  and NO2-, so x mole/L HNO2 will  give x mole/L of each and our table becomes\n\n.. .. .. .. .. .. HNO2 <=> H+ + NO2-\nInitial .. .. .. .. 0.5\nDissociate .. ..x\nProduce .. .. .. .. .. .. .. ..x .. .. .x\n\n4. Ka is an expression for the concentrations at equilibrium. So let's see what are those. For the reactants you will have initial-what reacted. We have only HNO2 with 0.5-x. For the products it will be the initial+what was produced. We didn't have any intial so it is only what was produced (x for both of the products). Thus the table in its final form is \n\n.. .. .. .. .. .. HNO2 <=> H+ + NO2-\nInitial .. .. .. .. 0.5\nDissociate .. ..x\nProduce .. .. .. .. .. .. .. ..x .. .. .x\nAt Equil. .. .. 0.5-x .. .. .. x .. .. .x\n\nKa= [H+][NO2-]/[HNO2] = x^2/(0.5-x)\nNow you either solve the quadratic equation and find x, or we assume that 0.5 >>x and practically 0.5-x=0.5.\nThen Ka=x^2/0.5=4*10^-14 =>\nx=squareroot(2*10^-14)= 1.4*10^-7. It is much smaller than 0.5 so our assumption is valid (otherwise we would have to solve the quadratic and get an exact solution).\n\nFor a simple level of chemistry we would now say\npH=-log[H+]= -logx =-log(1.4*10^-7) = 6.85\n\nHowever for more advanced and accurate expectations we see that the H+ coming from HNO2 is found to be in the same order of magnitude as that coming from the self dissociation of water (10^-7 for distilled water). Therefore in this case we cannot ignore the self dissociation of water (as we usually do) for calculating the pH. \nThere are 2 approaches for this case, here is one of them.\nRemember that we didn't put an initial amount for H+ since in most cases we ignore the self dissociation of water. Now we will consider that we had pure water, so the initial concentration of H+ is 10^-7 and we want to see how much the HNO2 will dissociate under these conditions so our table becomes\n\n.. .. .. .. .. .. HNO2 <=> H+ + NO2-\nInitial .. .. .. .. 0.5 .. .. .. 10^-7\nDissociate .. ..x\nProduce .. .. .. .. .. .. .. ..x .. .. .x\nAt Equil. .. .. 0.5-x .. 10^-7+x. .x\n\nand Ka=(10^-7+x)x/(0.5-x)\nNow we have to solve the quadratic so we can't do any simplification\n\n0.5Ka-Kax=10^-7x +x^2 =>\nx^2+ (10^-7+Ka)x-0.5Ka=0\n\nThe only acceptable solution is the positive one\nx=0.5(- (10^-7+Ka)+ squareroot (D))\nwhere D=(10^-7+Ka)^2-4*(-0.5Ka)=\n=(10^-7+4*10^-14)^2-4*(-0.5*4*10^-14)\n=9*10^-14\n\nand x=0.5*(-((10^-7))+4*10^-14) +3*10^-7) =10^-7\n\nso now pH=-log(10^-7+x) =-log(2*10^-7) = 6.70
734862
HgO ------> Hg + O2 at about 400 C\n\nMg + CuCl2 ---------->  MgCl2 + Cu(solid)\n\nMg, being higly electropositive than Cu, replaces Cu.
502184
5H2O2(aq)+2KMnO4(aq)+\n3H2SO4(aq)-->\n2MnSO4(aq)+K2SO4(aq)+\n8H2O(l)+5O2(g)
47432
3AgC2H3O2 + Na3PO4 -----> Ag3PO4 + 3NaC2H3O2
612107
a. 2k+ = .20moles , SO42-=.10moles\n\nb.Fe3+=.10moles,3NO3-= .30moles\n\nc.2Al3+=.20moles,3SO42-=.30moles\n\nd. Ni2+=.10moles,SO42-=.10moles
803450
MgCl2(aq) + 2NaOH(aq)  --> Mg(OH)2(s) + 2NaCl(aq)\n\nNet ionic equation:\n\nMg2+(aq) + 2OH-(aq) --> Mg(OH)2(s)
832296
A) Ag2O (s) >>> 2 Ag(s) + 1/2 O2 (g)\nB) Nickel (II) Carbonate >>>>>Nickel (II) Oxide + [Carbon Dioxide], NiCO3 -> NiO + CO2\nC) Ammonium Nitrate >>>>> Dinitrogen monoxide + Water,  \nC does not balance as is... but this does:\nNH3NO3 -> N2O +  1.5 H2O + 0.5 O2\n\n48) A) 2 HCl (aq) + Ca (OH)2 (aq) >>>> 2 H2O + CaCl2 (aq)\nB) 3 Ag2 SO4 (aq) + 2 AlCl3 (aq) >>>> 6 AgCl (s) + Al2(SO4)3 (aq) (silver chloride is a precipitate)\nC) H2C2O4 (aq) + 2 KOH (aq)>>>> 2 H2O + K2C2O4 (aq)\nD) CdBr2 (aq) + Na2S (Aq)>>> CdS(s) + 2 NaBr (aq) (Cadium sulfide is a precipitate)\n50 Write a Balanced equation for the complete combustion of each compound:\nA) Butene (C4H8): C4H8 + 6 O2 -> 4 CO2 + 4 H2O\nB) Octane (C8H18): C8H18 + 12.5 O2 -> 8 CO2 + 9 H2O\nC) glycerol (C3H8O3): C3H8O3 + 3.5 O2 -> 3 CO2 + 4 H2O\nD) Acetone (C3H6O): C3H6O + 4 O2 -> 3 CO2 + 3 H2O\n52: Complete these:\nA) Al (s) + H2SO4 (aq)>>>> AlSO4(aq) + H2(g)\nB) 2 HCl (aq) + Ba (OH)2 (aq)>>>> BaCl2 (aq) + 2 H2O\nC) Au (s) + 3 HCl (aq) >>>>> AuCl3(aq) + 1.5 H2 (g)\n54) EACH IS BALANCED BUT "INCORRECT", so correct these:\nA) Cl-2 + NaI(aq) >>> NaCl2(aq) + I-\nB) NH3(aq) >>> N- + H+3\nC) Na (s) + O2(g) >>NaO2(s)\nD) 2Mg (s) + H2SO4(aq) >>Mg2SO4(aq) + H2(g)\nE) MgCl2(aq) + Ca(OH)2(s) >>> Mg(OH)2(s) + CaCl2(aq)\nF) H2 (g) + Cl2 (g) >> H2Cl2 (g)
906651
2KOH=>K(2)O+H(2)O\n\nFeO+C=>CO+Fe\n\nCa+2AlCl=>CaCl(2)+2Al\n\nP(2)O(5)+3H(2)O=>2H(3)PO(4)\n\n2KI+Pb(NO3)2=>2KNO(3)+PbI(2)