How many Amperes is needed to produce an x amount of hydrogen  in electrolyse?.?
112430
193,000 Amperes per second to produces 1 mole of H2 or 2.016 grams of x.
885670
Get the chemical equation first:\nHCl + NaOH --> NaCl + H2O\n\nMoles of HCl used = (50/1000)dm3 X (1.0mol/dm3) = 0.05 mol\nMoles of NaOH used = (50/1000)dm3 x (1.0mol/dm3) = 0.05 mol\n\nFrom chemical equation, 1 mole of HCl needs 1 mole of NaOH for complete neutralization.\nFrom what you are given, you have the exact amount for complete neutralization (meaning, there is no excess reagents).\nFrom chemical equation again, 1 mole of HCl or NaOH produces 1 mole of water.\nThus you had exactly 0.05 mole of water produced here.\nHope this helps. :)
1045656
An electron passing through a potential difference of 150 kV picks up 150keV of energy.  So:\n150keV=h*c/lambda\nlambda =h*c/150keV.\nh=6.626*10^-34 J*sec\nc=3*10^8m/sec\n1eV=1.6*10^-19 J\n\nSo, lambda=((6.626*10^-34)(3*10^8))/((150000)(1.6*10^-19))\n\nlambda=8.28*10^-12 m, or 8.28 nm
490327
solution A has a hydrogen ion concentration of 10^-5\nsolution B has a hydrogen ion concentration of 10^-3\nAll you have to do is 10^-ph\n\nIs this homework?
198671
Electrolysis. By applying electricity, you can break H2O into H2 and O2 gas.\n\n2H2O --> 2H2 + O2
984026
You need to find the heats of formation of each species (will come from a standard table) and use this equation to get the heat of reaction:\n\n(delta)H = sum of heat of formation of products - sum of heat of formation of reactants\n\nEach heat of formation will be multiplied by its stoichiometric ratio, and the physical state will be important (especially that of water, as the heat of formation of steam is higher than that of liquid water).  You will not find a heat of formation for oxygen gas, because by definition there is no heat of formation for an element.
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)
479909
Electrolysis is the process that would produce hydrogen gas from water by sends electric currents through water, breaking its bonds and thus separating the two hydrogen atoms from the oxygen.
515359
1/1000th of an ampere
947695
H2S + 2NaOH --> Na2S + 2H2O\nMoles of H2S given = (1.19/34) = 0.035\nMoles of NaOH given = (4.91/40) = 0.12275\nFrom the chemical equation, 1 mole of H2S needs 2 moles of NaOH for complete reaction. Hence for 0.035 moles of H2S, you need only (0.035 x 2 = 0.07) moles of NaOH.\nThus the limiting reagent is H2S.\nFrom chemical equation, 1 mole of H2S produces 1 mole of Na2S. Thus moles of Na2S produced = moles of H2S used = 0.035\nMass of 0.035 mole of Na2S = mole x Mr of Na2S = 0.035 x 78 = 2.73g\n\nHope this helps :)
411418
Ox reaction:  Fe ---> Fe(2+) + 2e\nRed reaction:  Ni(2+) + 2e ---> Ni\n\nGiven that Fe/Fe(2+) is +0.44V and Ni(2+)/Ni is -0.26V, the net voltage is +0.18V.
948313
http://www.chemguide.co.uk/physical/catalysis/hydrolyse.html   \n                                                                                                    this will help ya
328575
reduction:  2e- + Cu2+ --> Cu      Eo = +0.34 V  cathode (+)\n\noxidation:   Zn   --> Zn2+ + 2e-   Eo = +0.76 V    anode (-)\n________________________________________\n\nZn  + Cu2+ --> Cu  +  Zn2+         Eo = +1.10V therefore cell spontaneous as written.
514134
i think this is the equation\nE=VxI\nE is energy. V is volts . I is current which is measured in Amps\nE= 120 x 12.5\nE=1500Watts
561926
XeF  + e-  ============> Xe  + F-      E°/V = 3.4 with reference to the standard hydrogen electrode
981166
I assume that you mean that the resistance of the solution kkeps rising with time. If you are using a Dc based system, then depending on the conductivity (1 over the resistance) electrolysis products may be building up at the electrodes and increasing the resistance. A yourself a good conductivity meter and this should eliminate the problem.\n\nThe resistance of a solution will dpend on the area of your electrodes, the distance bewteen them ,the concentration of dissolved salts in the water and the solution temperature. You will have a few things you will need to keep standard.
339643
You need a reference table and some instruction. Go to this site and scroll down to electrochemistry. Good luck!\nhttp://spiderpong.sitesled.com/chemistry_12_handouts.htm