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Explain Meyers Electrons electronVolts Covalent bond theory

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massive:

charge on a capacitor

https://en.wikipedia.org/wiki/Coulomb
https://en.wikipedia.org/wiki/Coulomb%27s_law

if electrons are removed from a fixed amount of water, then there is a majority of protons left , + charge.
the Oxygen wants electrons so it has to suck them from some where

tektrical:

--- Quote from: Steve on December 01, 2019, 12:36:53 pm ---Stan tried to explain us his theory about using Voltage for breaking down the water molecule.
He wanted to overcome the attraction force of the bonding electrons.
Sofar nobody really explained this to me in a normal way.
I did some research and i want to share that with you, here on our forum.

The photoelectronic spectrum of H2O reveals four different energy levels that correspond to the ionization energies of the two bonding and two nonbonding pairs of elections at 12.6eV, 14.7eV, 18.5eV, and 32.2eV.

So what does that mean now?
It means that you need somewhere between 12,6 and 32,2 volts per bonding electron.

Now it becomes massive and tricky.

If you have two electrodes and you put 12,6v, you never ever get 12,6V on 1 molecule of water.
I explain: You dont have 1 electron on that electrode. You have a lot of them. And you have to divide the applied voltage per electron going to the other electrode.
If you have two electrons moving to the other electrode, you have 6,3V per electron.

So, back to Stans theory.
He needed to to use small electrodes, which he did. He had to use high voltage (lots of electrons)
Otherwise he would not be able to do any covalent bond breaking at all.

So, now it is you turn.
If we take 1 square inch of flat electrode.
How many electrons do you have on it when you apply 20kv?

--- End quote ---

Thanks for posting those photo electron voltages, Steve.  I wasn't aware that Stan was into that aspect.  One value he left out is the 6.1 ev for breaking out a single Hydrogen, leaving hydroxyl.  But keep in mind that these are kinetic energy values, representing precise electron velocities to produce impact dissociation.  A cloud of electrons traveling at one of the indicated velocities will produce the desired effect when injected into some water mist.  Vibrating free electrons in a water bath, at the exact velocity to cause the molecules to split, would be a good trick, and might be a more complex approach than is needed.  This is a quantum effect - a tiny bit too much voltage, and it won't work.

Steve:

--- Quote from: tektrical on December 10, 2019, 15:18:01 pm ---
--- Quote from: Steve on December 01, 2019, 12:36:53 pm ---Stan tried to explain us his theory about using Voltage for breaking down the water molecule.
He wanted to overcome the attraction force of the bonding electrons.
Sofar nobody really explained this to me in a normal way.
I did some research and i want to share that with you, here on our forum.

The photoelectronic spectrum of H2O reveals four different energy levels that correspond to the ionization energies of the two bonding and two nonbonding pairs of elections at 12.6eV, 14.7eV, 18.5eV, and 32.2eV.

So what does that mean now?
It means that you need somewhere between 12,6 and 32,2 volts per bonding electron.

Now it becomes massive and tricky.

If you have two electrodes and you put 12,6v, you never ever get 12,6V on 1 molecule of water.
I explain: You dont have 1 electron on that electrode. You have a lot of them. And you have to divide the applied voltage per electron going to the other electrode.
If you have two electrons moving to the other electrode, you have 6,3V per electron.

So, back to Stans theory.
He needed to to use small electrodes, which he did. He had to use high voltage (lots of electrons)
Otherwise he would not be able to do any covalent bond breaking at all.

So, now it is you turn.
If we take 1 square inch of flat electrode.
How many electrons do you have on it when you apply 20kv?

--- End quote ---

Thanks for posting those photo electron voltages, Steve.  I wasn't aware that Stan was into that aspect.  One value he left out is the 6.1 ev for breaking out a single Hydrogen, leaving hydroxyl.  But keep in mind that these are kinetic energy values, representing precise electron velocities to produce impact dissociation.  A cloud of electrons traveling at one of the indicated velocities will produce the desired effect when injected into some water mist.  Vibrating free electrons in a water bath, at the exact velocity to cause the molecules to split, would be a good trick, and might be a more complex approach than is needed.  This is a quantum effect - a tiny bit too much voltage, and it won't work.

--- End quote ---

Thanks Tek,

Good point.
An electronvolt is the amount of kinetic energy gained or lost by a single electron accelerating from rest through an electric potential difference of one volt in vacuum.

So, if i have a potential difference of 1 volt across our waterfuelcell, electrons only flow from the cathode to the protons.
And also electrons from the water molecule flow to the anode.
The atoms feel the potential difference across a wfc.
Meaning that the higher the voltage potential across the electrodes, the more kinetec force electrons get?

Steve:
I just try to understand Stans process.
He clearly is pulsing DC in stead of strait DC.
I know from my own experiments that the type of gas changes to the better for my test engines when i used unregulated DC.
Stans always claimed that his process is about de-stabilizing the gasses he produced.
So, what happens with the proton, swimming from the anode to the cathode? His bond with the oxygen atom was terminated when the oxygen let go some electrons at the anode.
The proton still feels the kathode and all electrons on it, so it moves towards the cathode. Right?
But, now we terminate the DC pulse. We remove the electrons from the kathode.
What will the swimming proton do? Go up and releases it self from the water bath?

Steve:
There must be a difference in weight between a hydrogen atom and a proton because the 1s electron shields the nucleus of the proton and neutralizes same charge of it.
However, when i google, most of the dummy´s give the same mass for a proton as for an atom.
And the molecule is of course times 2.

Ill guess and i read it also somewhere else that protons have more mass and more energy.

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