What is the conversion ratio for electrostatic conversion of water to hydrogen?
405671
I assume you mean electrolysis?  I don't know the exact ratio of energy out to energy in, but I do know that it's not very efficient.  There will be a lot of lost energy.  What kind of H2 on demand system?  Are you trying to run an automobile?  If so, you'd need more than about 10 kilowatts of electricity.  Might as well just build an electric car if you have that much electricity to waste.
536437
In order to overcome the strength of the  oxygen-hydrogen bond of water, a force must be used to electrochemically decompose a sample of water.  This is called a decomposition voltage and it is roughly about 1.2 volts at 25 degrees C.  In the lab a little more voltage is necessary to have the reaction occur at an acceptable rate.\nThis requires a cathode and an anode to produce hydrogen and oxygen gas, and of course a voltage.  \nElectrolysis of water:\n4H+ (aq) +4e- ------ 2H2(g) reduction at cathode\n2H2(l) ------- O2(g) + 4H+(aq) + 4e- oxidation at anode\n\n2H2O(l) ------- 2H2(g) + O2(g) net reaction\n\nEnergy: DeltaGrxn = +474.4 KJ\n\nGood night and Good Luck.
750811
Your question makes no sense.\nIf I understand correctly you are making artificial sea water by making a solution of KCl, HNO3,Hg(+2), then you are adding Pb(+2) and Cd(+2) to known concentrations... measuring potential (what method?) and you want to determine the concentration of...what? You know the concentrations of all species (unless a reaction is taking place)
36210
water is Di pole material which means that it has some polarity due to partial charge distribution between oxygen and hydrogen . \n\nsince it is a di pole , it is divided into ions as \n\nH2O = OH-   + H+\n\nor it accepts an H+ and becomes  H20 + H+   = H3O+  \n\nso it seems like ions are the main thing to carry electrical current . so water can curry current providing a power source \n\n\nPURE WATER CANNOT HOWEVER CARRY ELECTRICAL CURRENT
374219
For the cell reaction: FeO(s) + H2(g) = Fe(s) + H2O(l) \n\nDG0(298 K) = DGf0 Fe(s) + DGf0 H2O(l) – DGf0 FeO(s) – DGf0 H2(g)\n\nFrom any table of Thermodynamic Properties:\nDGf0 Fe(s) = 0 KJ/mol\nDGf0 H2O(l) = -237.192 KJ/mol\nDGf0 FeO(s) = -251.4 KJ/mol\nDGf0 H2(g) = 0 KJ/mol\n\nThen: \nDG0(298 K) = 14.204 KJ/mol\n\nFurthermore: E0 = -DG0/(nF) = -14204/(2 x 96500) = -0.074 V vs NHE (normal hydrogen electrode), where n: number of electrons in each half reaction (FeO/Fe and 2H+/H2, n=2) and F is the Faraday constant (96500 C/mol).
827235
100% sure\n1. D\n2. B\n3. D (electrons become part of the WATER molecule which eventually reacts with hydrogen to form water)
406711
The Hydrogen atom is a donor/acceptor for the electron, it's kinda like a battery, or a hydroelectric energy plant. Imagine the electron is water it wants to get somewhere, and it turns a wheel (ATP Snythase) to create ATP to get there. That is a "big picture" the details you'll have to look up.
891313
The Cp of Hydrogen gas is 28.82 J/mol/k or 14.30 J/g/k\nTherefore, the Cv = Cp - R\n\nCv = 28.82- 8.3\n     => 20.52 J/mol/k\n\nTo covert this to B.T.U\nhttp://www.gordonengland.co.uk/conversion/specific_heat.htm\n\nRate mine as the best answer if you found this helpful.
801826
E=hf  where h = 6.6210^-34 Js\n\ne = 6.6210^-34*4.31*10^14 = 2.71 10^-19 Joule=1.7 eV
367911
For example, hydrogen burns in oxygen to form water vapor. \n\n2H2(g) + O2(g) -> H2O(g) \nHow much hydrogen must be burned to form 1.00 L of water vapor at 1.00 atm pressure and 3000C? \nTo work this problem, we need to figure out how many moles of water vapor we need to form. Use the ideal gas equation to compute this: \n\nPV = nRT \n1.00 atm * 1.00 L = n*0.0821 (L*atm/mol*K)* 473 K \nn = 0.0258 moles \nThus, we need to form 0.0258 moles of water. Now it's just a basic stoichiometry problem: for each mole of water formed, you need 2 moles of hydrogen gas, so 0.0258 moles water * 2 moles H2/1 mole water = 0.0516 moles of hydrogen \n\n=====================================================\n\nthis will help you a lot\n\nhttp://www.chem.hawaii.edu/asato161/Ch10NotesP2.pdf#search='Gas%20Stoichiometry%20PV%3DnRT'\n\nhttp://members.aol.com/profchm/gasstoic.html
248810
It's a matter of thermodynamics.  The end product of burning hydrogen with oxygen in an engine, or recombining it in a fuel cell, is water.  So, if you're going from water to hydrogen and oxygen, and then back to water when you burn it, the heat given off by the chemical reaction you're using to power the engine is the same as the amount of energy it took to split the hydrogen from the water in the first place.  Add to this the limitations of efficiency on engines, or any other device, and it ends up being impossible to have an engine that runs on water.  If you have some other power source, you can use this to split the hydrogen, and then burn it in your engine, but then the energy really isn't coming from the water being used as a fuel, you're just using the hydrogen to carry the energy from your power source to your engine.  \n\nNow, if you had a pure fluorine atmosphere, you could burn water in an engine, but I'm not sure what you could make an engine out of such that it wouldn't be destroyed by the fluorine!
198671
Electrolysis. By applying electricity, you can break H2O into H2 and O2 gas.\n\n2H2O --> 2H2 + O2
904431
Just the take the [H+] ion concentration divided by the initial [HF]:\nfrom your info i gather it should be (10^-1.59)/1.0 \n\nConvert it into a percentage.
311730
I think you should keep it simple yet smart. First get a piece of pure elemental Sodium and then let it drop on warm water. Afterwards explain how the chemical reaction released Sodium Oxide and Hydrogen Gas.
125657
No, sadly. When you break down the chemical compund water, you would get Hydrogen in a gaseous state. You would have to compress it or cool it to bring it down to the temperature needed for it to condense into a liquid state.\n\nOh yeah, and I'm not sure, but I think the process is called electrolysis.
919041
We're basically looking for a ratio of ionization energy E(n), where n =4 and n = 1 (ground state). Electron energy levels of a hydrogen atom is given by:\n\nE(n) = 13.6/n^2 eV\n\nTherefore,\n\nE(4)/E(1) =  1/16\n\nIt takes only 1/16 of 13.6 eV , or 0.85 eV, to ionize the ground state electron from a hydrogen atom on this very hot planet.
351009
Water with a high salt concentration, and plasma (electrically charged gas)