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Aleilius
05-22-2009, 11:16 PM
I've been thinking about the bonding that happens with m-state gold.


Common oxidation states of gold include +1 (gold(I) or aurous compounds) and +3 (gold(III) or auric compounds)

These are the bonding schemes I've come up with: NaAu (Au is +1; electronegativity = 3.47v), Na3Au (Au is +3; electronegativity = 5.33v), and I think Na2Au-Au (is it Au[Na2Au] or Na2Au2?; electronegativity = 4.4v or 6.94v [not sure which one is correct]).

The Essene mentions the white lion is Na3Au. He also explained the green lion, but the interview I have of this explanation isn't too clear. He said the green lion was Na2Au-Au. One gold atom is bound to another gold atom, this gold atom is in the +1 state. The gold atom bound to the sodium is in the +3 state. Is this correct or am I missing something? The interview I have doesn't mention the red lion, but I assume the red lion is NaAu, right? If this is the case then the red lion has the least electronegativity out of all the gold m-states, and would be the easiest to convert back to metal.

What happens when we precipitate these? Is this correct: Na[NaAu], Na[Na3Au], Na[Na2Au-Au]? Am I totally wrong here? I don't think the alkali will be able to break the bond because of the high electronegativity of m-state, so I think we're seeing something else happen. The Essene reported that electrolysis, radiation, or a laser beam would be the only practical way to break the bonds. NaAu is the least electronegative out of all the gold m-states. Is this why it's so reactive? Does our body break this apart (I'm not sure if the electronegativity is still too high to be broken apart by chemistry)? Maybe if that's not the case, then I suppose it's so reactive because it's the smallest out of all the gold m-states.

Vlad
05-22-2009, 11:35 PM
Here is some info I got on this nearly 6 years ago on ORMUS_SWG:

Vlad,

The process does not yield Na3Au. Let's do the calculations.

The Essene uses 1 troy ounce of gold and 1.5 grams of sodium chloride.

1 troy ounce = 31.1 grams = 0.1579 moles of Au

1.5 grams = 0.0257 moles of NaCl

As you can see, the ratio of Au to Na is not 1:3. It is 6:1 (0.1579/0.0257 =
6.14).

Let's look at it another way - 31.1 grams Au and 27.7 grams NaCl give an Au
to Na ratio of 1:3.

This also means that Barry's box theory is incorrect.

Hudson has the best theory. The chemistry in his patent is consistent with
experiment and theory.

The chemical structures proposed by the Essene are inconsistent with known
and theoretical chemical pathways. The chemical structures proposed by the
Essene are inconsistent with the quantities of NaCl and Au used in the
process. Na3Au can not form under the conditions employed. Sodium auride is
hydrolyzed in water to form hydrogen auride; this is supported by theory and
Hudson's experiments.

There is a tendency to believe the Essene because he willingly shares his
process with most everyone.

Hudson, on the other hand, is secretive.

Hudson, the Essene, August, Don, Joe and others have made the Green Lion.
The Green Lion is monatomic gold chloride. This conclusion can be drawn from
the chemical process used to make it. The gold in the Green Lion has an even
number of electrons.

According to the Essene's process, the White Lion is made by adding NaOH to
the Green Lion. The White Lion is monatomic gold hydroxide. This conclusion
can be drawn from the chemical process used to make it. August posted xrf
data on Barry's website identifying it as such. August is a physicist, but
did his work at the chemistry dept of Cal Tech. The scientists at Cal Tech
are very interested in m-state. August is collaborating with them. The gold
in the White Lion has an even number of electrons.

The Red Lion is made by adding a little more NaCl to the Green Lion. The Red
Lion is monatomic gold chloride. This conclusion can be drawn from the
chemical process used to make it. The gold in the Red Lion has an even
number of electrons. Why is it red? What is the difference between the Green
and the Red? The most plausible explanation is that the nucleons in the Red
have moved into higher nuclear shells. The Red Lion and the Alkahest (of
Paracelsus) will yield the true Aurum Potable.

Gold ORME may be monatomic gold. Since it is 5/9 here and 4/9 somewhere
else, we can not know how many electrons are here and how many electrons are
somewhere else. Gold ORME may be monatomic mercury 198. Theoretically, this
could occur by the fusion of the H and the Au during the annealing of the
HAu.

Gold S-ORME would be the above only with the nucleons in higher nuclear
shells.

The Hudson theory was not developed by Hudson. It was developed by many top
level scientists in government, industry and academia.

Mike Halat


Mike,

There is something else to consider concerning The Essene's process.
You said:

The process does not yield Na3Au. Let's do the calculations.
The Essene uses 1 troy ounce of gold and 1.5 grams of sodium chloride.
1 troy ounce = 31.1 grams = 0.1579 moles of Au
1.5 grams = 0.0257 moles of NaCl
As you can see, the ratio of Au to Na is not 1:3. It is 6:1 (0.1579/0.0257
= 6.14).
He uses an ounce of gold dissolved, but that does not mean it is all ment to
be taken to m-state.
It is easily assumed though. After all I don't readily understand why he
would use more gold than needed.
But perhaps the reason might be because not all of the dissolved gold would
be able to get to the monoatomic state.
In any case, considering the amount of sodium present, only about 1/18th of
total amount of gold could get used up to form Na3Au. Depending on how much
sodium is around the green and red, some more gold when these get formed,
but still very likely not all of the gold at all. At best 1/6th.

There is something else to consider. The volume increase.
Avi said that when doing a sodium burn on an ounce of gold, it yields two
gallons of white precipitate, while still returning 80% of the gold as
metal. Thus 1/5th of an ounce of gold yields about two gallons of hydrated
Na3Au precipitate.
If the full dissolved ounce would get used up in The Essene's process, the
flask required would probably need to be huge, even considering the white
precipitate is hydrated while in solution it wouldn't require the same
volume, but still...

Vlad

Vlad,

Gold is not detected by ICP-MS in the green gold chloride solution. It is
assumed that all material has been converted to m-state.

Na3Au does not form. Gold will not give a minus 3 valence ion. To do so
would mean a partially filled 6p orbital. The Essene is not infallible, and
this is one that he got wrong. Why do you try to rationalize the Essene's
folly?

Mike

Vlad
05-22-2009, 11:50 PM
More:

I do not understand the exact role that sodium plays in all this. Hudson says that the molar ratio of sodium to gold should be 20:1, Jim says 3:1, and the Essene uses 1:6. All three give the grass green solution of gold chloride. The only explanation that I can come up with has to do with a change in the electron configuration. The outer orbitals of metallic gold are 5d^9 6s^2; that is why the preferred valence of gold is +3. The aqua regia digestion leaves a positive gold ion with a 5d^8 outer orbital. The purpose of the repeated boildowns is to break up the microclusters and to bring the gold back to a 5d^10 outer orbital. I'm guessing that the sodium somehow facilitates this.

This leads to other questions though. Hudson says that NaAuCl2 is formed and can be oxidized to NaAu by HNO3. Both Hudson and Jim use enough sodium for this to occur. With the Essene's way, there is not enough sodium to form NaAuCl2. So what would happen if the Essene took his material through the rest of the Hudson process?

When the green gold chloride solution is made properly, a white flocculant forms. The Essene has seen it; Knowland has seen it; I have seen it. It is hydrophobic and rises up the sides of the beaker about an inch or so above the fluid level at 50 degrees C. When washed down with water, it just rises up again. At 100 degrees C the stuff clumps together and floats on the surface. When the temp is allowed to drop back to 50 degrees C, it rises up again. Unfortunately, there is not enough material for assay.

Aleilius
05-23-2009, 12:00 AM
Hi Vlad, thanks for posting that. NaAu in water forms NaOH and HAu. I'll get deuterium auride if I use heavy water instead of normal water. Let me get this straight, m-state refers to gold (or any other metal that can form m-states) bound to any element in group 1 of the periodic table. These would be acceptable m-states in theory: HAu, NaAu, KAu, RbAu, CsAu, and FrAu. Am I right?

EDIT: I just realized I know less than I thought I knew on this subject.

Aleilius
05-23-2009, 12:27 AM
Gold ORME may be monatomic mercury 198. Theoretically, this
could occur by the fusion of the H and the Au during the annealing of the
HAu.
I thought m-state mercury drops to gold when heated?

Vlad
05-23-2009, 12:32 AM
That's what Hudson says. It's his theory.
Alchemical theory speaks about the seed(s) of metals and mercury and sulfur, totally different. ORMEs may be a proto metallic matter, like the seed mixed with some mercury. I don't know if monoatomic gold is really monoatomic gold or monoatomic mercury. Mike Halat could be right too.

Aleilius
05-23-2009, 12:39 AM
This new information really opens up a whole slew of new ideas for me!


I don't know if monoatomic gold is really monoatomic gold or monoatomic mercury. Mike Halat could be right too.
I think Mike Halat indicated that HAu might undergo nuclear fusion during the annealing process, and that it might be an isotopic m-state form of mercury. I don't think a fusion occurs without energy input from the annealing process. I've been working on a theory myself about low energy nuclear transmutations. In order to see low energy nuclear transmutation one needs to add energy to the system. The nuclear reaction only takes place as long as energy is being inputted. There's something about m-state matter that allows nuclear transmutations to occur with relative ease.