So I've read through https://www.physicsforums.com/showthread.php?t=32409" thread on valence electrons but it doesn't really answer my question. I think I am in the right forum, I was debating between here and chemistry but it deals more with atoms. What I want to know is this:

How do I figure out (/calculate) the chance of an atom (ion) losing an electron? Is it based on how many electrons in the valance shell? As it's shell is closer to being filled, or closer to being empty how does it effect the ability to loose an electron?

What I am trying to figure out is how many amps are possible in a flowing fluid.

What I did was I found the total number of valance electrons contained in all of the negative ions in regular tap water. I found out the ions, then I looked up the model, to find out how many valence electrons. From there I divided the weight of the valance electrons by the weight of the entire atom. I then found out what % of weight was valence electrons. Knowing that, I think just multiplied out the % by the mg/l in tap water. I then had several smaller mg/l number which is about equal to the total weight of valence elections per mg/l. From that I divided by the weight of an electron, finding the number of valence electrons per liter. Multiplied by the charge of an electron to get charge / liter. Then I multiplied by my flow in liters / second and I got charge / second or amps. I know the formula for amps is I = q * n * v * A, but v * A = flow, so I can keep that.

Bicarbonate = 113 mg /l *.000134 => .0151 mg/l

Sulfate = 25 mg/l *.000137 => .00342 mg / l

Chloride = 21 mg / l *.000124 => .00260 mg/l

Total valence electrons are 0.0212 mg /l . That is .0212 * 10^-6 kg / l.

Mass of electron is 9.1093818 * 10^-31 . Divide .0212*10^-6 kg/l / 9.1093818 *10^-31 kg = 2.318 * 10^22 electrons/liter. Charge of an electrons is 1.60217646 * 10^-19. q * n = (1.60217646 * 10^-19) * (2.318 * 10^22) = 3,714.63 coulombs/liter

I = q * n * flow = 3,714.63 coulombs/liter* .667 liters/second = 2,477.6 amps.

I have no clue if this is right, or not. Can anyone give me some pointers?

How do I figure out (/calculate) the chance of an atom (ion) losing an electron? Is it based on how many electrons in the valance shell? As it's shell is closer to being filled, or closer to being empty how does it effect the ability to loose an electron?

What I am trying to figure out is how many amps are possible in a flowing fluid.

What I did was I found the total number of valance electrons contained in all of the negative ions in regular tap water. I found out the ions, then I looked up the model, to find out how many valence electrons. From there I divided the weight of the valance electrons by the weight of the entire atom. I then found out what % of weight was valence electrons. Knowing that, I think just multiplied out the % by the mg/l in tap water. I then had several smaller mg/l number which is about equal to the total weight of valence elections per mg/l. From that I divided by the weight of an electron, finding the number of valence electrons per liter. Multiplied by the charge of an electron to get charge / liter. Then I multiplied by my flow in liters / second and I got charge / second or amps. I know the formula for amps is I = q * n * v * A, but v * A = flow, so I can keep that.

Bicarbonate = 113 mg /l *.000134 => .0151 mg/l

Sulfate = 25 mg/l *.000137 => .00342 mg / l

Chloride = 21 mg / l *.000124 => .00260 mg/l

Total valence electrons are 0.0212 mg /l . That is .0212 * 10^-6 kg / l.

Mass of electron is 9.1093818 * 10^-31 . Divide .0212*10^-6 kg/l / 9.1093818 *10^-31 kg = 2.318 * 10^22 electrons/liter. Charge of an electrons is 1.60217646 * 10^-19. q * n = (1.60217646 * 10^-19) * (2.318 * 10^22) = 3,714.63 coulombs/liter

I = q * n * flow = 3,714.63 coulombs/liter* .667 liters/second = 2,477.6 amps.

I have no clue if this is right, or not. Can anyone give me some pointers?

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