url: http://escribe.com/health/thesilverlist/m60532.html
CS>Calcs and dimensions
From: Tony Moody
Date: Sun, 22 Jun 2003 07:29:13
> Hi Mike,
> Interesting posts.
> I have tried to go through the dimensions in your calculation
> below but there is not enough information. Perhaps you could post
> the dimensions of Faradays electrolysis equation? I was wondering
> about the volume in milliliters. The standard I would use would be
> litres of course but if you are being consistent then millilitres
> would be OK.
> Regards,
> Tony
Hi Tony,
Thanks for the comments. I use the Faraday equation in a number of
different calculations, and I'm not sure which one you are referring
to.
Bob Lee posted the original analysis in "How much silver did I
drink" on April 11, 1998. Here's the url:
http://escribe.com/health/thesilverlist/m1659.html
The basic equation is
gm = k * I * T
where
gm = number of grams transferred
k = 107.88 / 96500 = Coulombs required per gram of silver
I = current in Amperes
T = time in seconds
The constant k is defined as
Atomic weight of silver = 107.88 grams
Coulombs required to liberate 107.88 grams of silver = 96,500
Coulombs required to liberate 1 gram = 107.88 / 96500
Here is a further explanation from Bob Lee's post:
Avogardo`s constant states the number of atoms in a substance is
always 6.06x10+23 atoms in a * gram-atom * of the substance. This
means that there will be: 6.06x10+23 atoms in 107.88 grams of
silver. (one chem. equiv.)
To liberate the 107.88 grams of silver will require: 6.06 X 10+23
divided by 6.25x10+18 = 96,500 coulombs of electricity.
Yes, you are right about litres to calculate ppm. Some measuring
cups are graduated in millilitres. This requires converting to
litres somewhere in the calculation.
Is this what you were looking for?
Best Regards,
Mike Monett
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