DrDoggy

but my equation just told me that i only need 1.5v * s to do a split, for each single molecule

also this is important to me:

"with 1.5ev = V * s is there a minimum number s(econds) it can be, ie what if s drops below the period time of emission, ie (1/frequency)??"

would a single xray emission @ 2480eV shot at the salt then react to produce 2480/1.5 = 1653x salt molecules broken up?

that makes sense Thanks,

also I have been reading this:

http://hyperphysics.phy-astr.gsu.edu/hbase/molecule/nacl.html

and i think i get this mostly now, but there is still one issue, if 5.1(Na) - 3.6(Cl) = 1.5eV , then why do i need to add so much heat???, plugging it in to this chart

http://www.colby.edu/chemistry/PChem/Hartree.html

1 ev = 1.60217657 × 10^-19 joules

or

1.5ev = 2.4 e-19 joules

2.4 e-19 joules = V * I * s

2.4 e-19 joules = V * 1.6e-19amps * s

2.4/1.6 = V * s

1.5 joules/amp = V * s

but NOW the part i dont get is why do i need to melt the salt, if all i need is 1.5v, which is my minimum for 1 second to split 1 molecule, as per calculation, also why cant I use 0.8Volts for 2 seconds to get the same effect,

also i wonder in the case examples of lasers and ionization shells and similar, in o2 plasma i see whiter shades of pink as i get closer to the electrode(anode), can i presume these are different energy level emissions? is there a way to determine what emissions i will get when adding energy to my elements/compunds?

for example i was told that i need to input around 2480ev to generate xrays, so thats a minimum of 2480volt for 1second.

but does that mean that if i apply 24.8 volts for 100seconds i will still get an xray out of it?

Also what happens if i need 2480 but only put in 2400V for 1second, will that energy get dumped as heat and/or light?

also back to the salt, if i apply 5 volt instead of 1.5volt what will happen, will the energy go towards splitting 3x fast , and 0.5 dumped to heat/light, also how do we know when it will be dumped to light or heat? or any other harmonic emission,,

Also with 1.5ev = V * s is there a minimum number s(econds) can be, ie what if s drops below the period time of emission, ie (1/frequency)??

I will confirm that about anything, including electric current which i discovered when i stuck my finger in a wall wart,

So let me get this straight, every time someone wants to break up a salt crystal they put it in a oven before they electrocute it?

I need to know : specifically, who would waist time doing this>? Please explain how much energy they waste doing this?

also why dont they just dissolve the salt into water for electrolysis?

also please , tell me how much eV's do THEY need to break down that 1 mol of salt and how do I translate that to electric power?

also how/where do i find the electrical resistance of this salt stuff.?

also please explain why it smells like chlorine up in here?

you will want to throw them out if you leave the clothes wet too long and they go mouldy

If you wanted to try this (I wouldn't recommend it) you would need to melt the NaCl (by heating it to about 800C) and then use electrolysis to separate the sodium and chlorine. Both of these are dangerous chemicals so this is not advisable unless you know what you are doing. (And it doesn't seem as if you do.)

Also , I wonder what would happen if i invented a machine that could electrolysis salt without heating, would this be a hit item in the industry? i wonder who would buy it?

just wondering , cos i am able to,... also during my tests a purple residue showed up on my salt, what could be purple?

ok ya got me, so i may not be completely fully trained, but i do know:

to vent the chlorine

the rectivity to skin of sodium

not to lick my electrodes

and to do other safe things aswell,

I am fully equiped to run this expirment if i decide, my interest lies in how it works behind the scene, and i am too lazy to physically do it, espically if math can do it for me,i have been eager to learn this since I was a pup!

so if that is true:

2H2O = 2H2 + O2 - 5.7eV

then to generate that 5.7eV or 500Kj/mol I need to run electrolsys 1.6kw for 5 minutes to get 1 mol?

OK I get it!~ so it is all about the bond Enthalpy , so really to break up water i add 5.7eV and to put it back together again i loose 5.7eV, correct?

but how many eV does it take for my bbq piezo lighter to do its trick to start the process, 5.7eV?

Also back to the salt:

Cl:Cl = -242 kj/mol bonding energy to split, but now what? i could not find a bond energy chart for sodium, to see Enthalpy, or does it need since Na is just Na and not Na2? also in the link i noticed that they did not add in the energy requeired to split Cl2.

Of course im fully trained in safty, and this salt idea is just for theory, but wait so dissociation energy is less than ionization?

also i am aware of second comment , but i am more interested in the energies!

Also true i know only basics on the chemistry of things but would like to understanted better about the energy levels involved, very nice link! but could you explain about where these energy values come from, and how to calculate?

in the h20 example:

A)where does 5.7eV come from?

B)and does this work? -----> 2H2O = 2H2 + O2 - 5.7eV (cos im putting sum in)

If you could explain how to do the math for this whole process from a-z-a for the water example and where you get those numbers , and then i would like to try for myself in the salt example

a-z-a = h2+o2 , ignition energy, energy out, water, energy requeired for back to h2+o2

THanks!

A similar thing happened with me ....did u use copper electrodes,if yes then probably its a cu+1 compound which is red brown

or green even

HI ALL!

Im trying to understand ionization energy, which as i understand is the amount of energy it takes to bump these electrons out of orbit, first i have problems measuring a gram of O2, so lets start simple with NaCl:
Na = 500 KJ/mol = 5eV
Cl = 1250 KJ/mol = 13eV
so if i want to break down this compound I would go by the Na value rite? since it is the first in the compound?
1 mol of Na requires 500KJ so thats what i need to dump in to my pot, but how?

according to this: http://www.rapidtables.com/calc/electric/Joule_to_Watt_Calculator.htm
If i am to dump 500KJ to get my mol of na i will need to insert electric power at 1.6KWatt for 5 minutes? or 800watts for 10 minutes?
in order to set that value though i need to know 2/3 of the following values: voltage, current, resistance, but what is the best way to do this? could I use a volt divider to bring down the KV value for my voltmeter...?
but how should i get my other value(s) , what would be easiest method? is it safe to put kv(low amps) on my ammeter? and how do I measure high resistance like that?

Also what is the point of stripping a second electron?
and with this energy level as an example what temperature can i expect in my core(between anode and cathode), so far the pot gets warm to the touch, but at the same time i show evidence of temperatures of over 800degrees on the inside?

am I right about these asumptions?

THANKS!