what is diffusion coeffeceint in cyclic voltammetry?
535845
A solute cannot undergo reversible oxidation/reduction if it diffuses away from the electrode. The diffusion coefficient simply measures how quickly this diffusion happens. \n\nWhen deciding how reversible a particular redox reaction is by CV, you have to consider both the intrinsic reactive properties of the species in solution as well as how much material is lost to analysis as it diffuses away from your electrode.
1125729
The index-of-reaction equation (i.r.e.): dF/dtZ = kZ(l - βF)m is compared with a generalized Prout-Tompkins equation (g.P.T.e.): dF/dt = kn(l - βF)y With the boundary condition F = 0, t = 0 the equations are equivalent for z = 2 or m = 0 or β ≤ 1 and the i.r.e, is closely approximated by the g.P.T.e. for other values of the parameters by use of the following approximation ∫(1- F)-m dF ≈ AF(1-BF)5 The parameter s is a function of m. Both A and B are close to unity for 1/3 < m < 3. The i.r.e. is recommended as being more useful than the g.P.T.e. because the i.r.e. has a better theoretical basis and because it is more readily integrated and fitted to experimental data.
988889
understanding the reaction meachanism \neffect of catalysts etc....\ndetermining the amounts needed
1133718
A. Yes, the use of ATP against the gradient is active\nB. This is diffusion or osmosis\nC. Nope\nD. This is passive transport
562274
Diffusion of a Dye in Hot and Cold Water\n\nPurpose:  The purpose of this experiment is to demonstrate the principle of diffusion.  Diffusion is the net movement of particles (atoms, ions, or molecules) from an area of greater concentration to areas of lesser concentration.\n\nMaterials and Methods:  For this experiment, the student will require a projector, two petri dishes, two 150 ml beakers, hot plate, potassium permanganate (purple dye), and a spatula.  Obtain about 50 ml of cold water in one beaker and about 50 ml or boiling water in the other.  Place the petri dishes on the projector and add cold water to one dish and boiling water to the other.  Allow the water in the dishes to become still.  Using the spatula, add a crystal of potassium permanganate to the center of each dish\n\nExperiment:  Observe the experiment for several minutes and note what happens to the purple dye in each petri dish.\n\nConclusion:  Do you observe diffusion in each?\n\nHow does the rate of diffusion differ? \n\nWhy is there a greater rate of diffusion in one
994673
I am not sure of the final answer you are looking for.  If you are looking for the number of molecules transferred after a certain time t, the answer is simply n=-DAt(dc/dz).  This of course assumes that dc/dz remains relatively constant.  If not you will need a dc/dz as a function of t.
845374
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
592402
I agree with the first poster on C and E, but I'm not sure if A is also.  A solute dissolving in a solvent seems like a chemical process to me, but I don't remember for sure.
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
446643
The movement of an electron from one place to another.
830323
yes size matters in determining rate of diffusion.bigger the size  more the time taken for diffusion.
270139
If you mean conductometric TITRATION, it won't work.  Assuming they're both at the same concentration, you'll get a flat line.  Conductance is a rough measure of the ion concetration.  The start (H+ + Cl-) has the same number as equivalence (Na+ + Cl-).  Adding more (Na+ + OH-) past equivalence will not change the concentration either.
467240
Given that a certain type of molecule is small enough to permeate through the cell membrane, the most important factor influencing the rate of diffusion is the concentration gradient.  This is simply the difference in concentration levels of the substance between the outside and the inside of the cell.  The larger the difference in concentrations, the greater the rate of diffusion.
784896
fractional distillation
825689
CuCO3 + H2SO4 --> CuSO4 + CO2 + H2O\n\nCu(OH)2 + H2SO4 --> CuSO4 + 2H2O\n\nFor the electrolysis:\n\nCuSO4 = Cu+2 (aq) + (SO4)-2 (aq)\n\nCu+2 (aq) + 2 e-1 --> Cu (metal)
1027970
Simply that during the induction period, the actual concentrations may not be 100% homogeneous and the rates will increase of decrease from theoretical.  This is much more important with macromolecules, but can be seen in other cases.  See\n\nwww.uem.mz/faculdade/ciencias/ ciencias_revista/textos/elalves.pdf\n\ndx.doi.org/10.1063/1.1954769\n\nwww.platinummetalsreview.com/ dynamic/article/view/50-2-54-63\n\nwww.citadel.edu/chemistry/scas2005.pdf
679108
voltametrty is measurement of voltage (which control type of reaction)\nwhile coulommitery is measuerement of electricity quantity (which control the time of  reaction)