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oozing on the muggy shore of the gulf coast
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--- Begin Message ---Greeting to all astute silver makers and users, A friend of mine asked,"How much silver is in that spoon of CS." Told him I didn`t know ,but would calculate it for him. This is what I did and you can too. This is going to be technical but try it any way, your probably smarter than you thought. We will start buy determining what a *chemical equivalent* is. The chemical equivalent is the atomic weight of an element divided by its valence number. That will be (w)/ (z). atomic weight =w valence no. =z Look at this Table: atomic weight in grams valence number Cu copper 63.57 +2 Ag silver 107.88 +1 Au gold 196.97 +1 This will give us : Cu copper 63.57/2 =31.785 ( one chemical equivalent) Ag silver 107.88/1 =107.88 ( one chemical equivalent) Au gold 196.97/1 =196.97 ( one chemical equivalent) The quantity of electricity to liberate *one chemical equivalent*of a substance can be found this way. We will keep two phyical constants in mind as we do this. 1. A coulomb of electricity will deposit 6.25x10+18 atoms in one second of time. 2. 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.) 6.06x10+23 atoms in 196.97 grams of gold. (one chem. equiv.) 6.06x10+23 atoms in 63.57 grams of copper. (two chem. equiv.) To liberate the 107.88 grams of silver will require: 6.06x10+23 divided by 6.25x10+18 = 96,500 coulombs of elec. atoms in silver atoms in amp. Remember that 6.25x10+18 is a *coulomb* or *one ampere* of current. To reduce 107.88 grams of silver will require 96,500 amps of current over a period of time. 96,500 coulombs of electricty will liberate one chemical equivalent of any substance, or in our case, 107.88 grams of silver ,or 6.06x10+23 silver atoms. Now we will make a table of currents vs * chem. equiv.* : A current of TEN amps for ONE hour will be 36,000 coulombs. and we need 96,500 coulombs to reduce *one chem. equiv.*. So we will have: 10 amps for one hr. = 0.373 chem equiv (36,000/96,500) 5 amps for one hr. = 0.186 chem equiv 1 amp for one hr. = 0.037 chem equiv 0.5 amp for one hr.(500ma) = 0.0186 chem equiv 0.1 amp for one hr.(100ma) = 0.00373 chem equiv 0.01 amp for one hr.(10ma) = 0.000373 chem equiv 0.009 amp for one hr.(9ma) = 0.0003357 chem equiv 0.008 amp for one hr.(8ma) = 0.0002984 chem equiv 0.007 amp for one hr.(7ma) = 0.0002611 chem equiv 0.006 amp for one hr.(6ma) = 0.0002238 chem equiv 0 005 amp for one hr.(5ma) = 0.0001865 chem equiv 0.004 amp for one hr.(4ma) = 0.0001492 chem equiv 0.003 amp for one hr.(3ma) = 0.ooo1119 chem equiv 0.002 amp for one hr.(2ma) = 0.0000746 chem equiv 0.001 amp for one hr.(1ma) = 0.0000373 chem equiv Remember that the chem. equiv. of a substance equals the ratio of atomic weight in grams divided by valence number, and the chem. equiv. for silver is 107.88 . If we ran 7ma of current for one hr. we would liberate 107.88 x 0.0002611 =0.02816 grams of silver in our water. Keep this number in mind,(jot it down). Now we will look at the Faraday Laws of Electrolysis. The one which concerns us says **THE MASS OF A SUBSTANCE LIBERATED IN AN ELECTROLITIC CELL IS PROPORTIONAL TO THE QUANTY OF ELECTRICITY PASSING THROUGH THE CELL.** The amount of material liberated for each coulomb is called the *Electrochemical Equiv.* of a substance. This is also *Chem. Equiv.* divided by 96,500 or = Electrochem. Equiv. Electrochem equiv is called *k* The k for silver is 107.88 /96,500 or =0.001118 k for gold is 196.97 / 96,500 =0.002041 k for copper is 31.8 / 96,500 =0.0003295 Faradays equation for electrolysis is m =kIt That is mass in grams = electrochem equiv times current times time m = k x I x t or m = k x coulombs , I x t = coulombs Now we have m= mass in grams k= electrochem equiv (atom weight divided by valence times 96,500)I= current in amps t= time in seconds Now for 7ma of current for one hr. we have: 0.001118 x 25.2 coulombs =0.028159 gm of silver round 0.028159 to 0.02816 gm of silver ( now look at the number you jotted down before. Viola ! We have done it! We ran a currunt of 7ma for one hr. and got 0.02817 gm of silver in the water we are using. Now if it was 8oz of water the silver is in the 8oz of water. Thats 0.02817 gm in 8oz of water. there are 48 teasoons in 8oz of water. So we divid 0.02817 by 48 and get 0.0005868 gm of silver in one teaspoon of CS. That also 0.5868 milligram (mgm) of silver. If we had a current of 3ma for one hr. it would be 0.01207 gm in 8oz of water or 0.0002514 gm in one teaspoon or 0.2515 mgm of CS. A little over 1/4 mgm of silver. But my current starts low and slowly gets higher as it goes. The way I do it is to read the current every 5min and addup the readings and take the average. If my current started at .5ma and stopped at 7ma after an hr. the ave would be about 3.5ma. Will stop here and give it a rest. Tomorrow will tell you how to find how many atoms of silver are in your CS. Bless you all, especially if you read this long post. Bob Lee Hi Ken, Repost of long ago subject which may be of interest to you -- oozing on the muggy shore of the gulf coast [email protected] -- The silver-list is a moderated forum for discussion of colloidal silver. 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