See how Jehovah's Witnesses slander Jesus by denying their members the bread and wine of Jesus. I've never heard of this trick before:
https://www.youtube.com/watch?v=OEWxjSBb3J4

You may not have heard that the British prime minister announced that, "by the end of this parliament," you won't be able to work without the digital ID that the government will make free. Wow, Starmer, it's even free of charge, way to go....wicked fool. It sounds like he'd want you to starve if you think you'd rather be in the old days. He's the type who'd like to kill you if you don't take the jab.

Some temporary good news: "As of May 2026, Sir Keir Starmer's digital ID plan has shifted from a proposed mandatory scheme to a voluntary 'one-stop' app following significant public backlash, a large petition, and political pressure." Is the anti-Christ in Britain?

Although Starmer was forced to back down with the mandatory iron fist, everyone saw the true colors of those who forced him to come out with that plot. The ones who forced him are probably not all known, if any of them. The plot may have been to allow rife, illegal immigration, portray it as dangerous due to dangerous types, as Trump has done, and finally to justify the compulsory ID program.

Take Kineticists to Court

The kinetic goof at Action Lab typically denies his audience the realities when they stare him in the face, when he can't explain them well. Shortly after 3 minutes below, he shows that the weight of a falling feather transfers to a weight scale when it's a few inches above the scale, but as he can't explain it, he says nothing to attempt it. Not even a bona fide attempt. The only explanation is that air atoms REPEL each other such that their weight transfers through the repulsion forces, for those forces are in literal (not physical) contact with all atoms while the lowest atoms are in contact with the ground. There is NO OTHER explanation, but his science teacher taught him that air atoms don't repel, and so he says nothing:
https://www.youtube.com/watch?v=EFhMcL3Kf1c

One could sue kineticists on this basis because they are dishing out false information to students, to all of humanity, justified because they cannot explain how kinetic atoms, which they say don't repel each other, can transfer both the feather's weight, and all the collective weight of air atoms, to the ground. If gases do not exist as kinetic atoms, then some judge should find it within himself to order the educational departments of governments to cease that teaching. Don't you think?

It can be demonstrated that non-repelling air atoms cannot transfer weight as they merely move through space. Nor can they transfer any force apart from collisions. It can be demonstrated that atoms, as they collide, cannot transfer weight to the ground because, no matter that three, six, or ten atoms collide simultaneously, they are not in contact with their neighbors at the moment of collision, wherefore neither can their weights transfer to the ground due to collisions. The weight needs to travel ATOM-TO-ATOM, duh.

How long have kineticists understood that air weight needs to move ATOM-TO-ATOM, and yet they have all kept their mouths shut, as if keeping a well-guarded secret to promote a false gas theory, as if something big depended upon it? Great shame. What are they protecting? Why don't they teach, if only as a good option, the simple alternative, that gas atoms repel? Why is that reality so destructive or disruptive to them that they would carry a lie for a century and counting? What do they understand about the kinetic model that is so dear to them that they protect it at all costs?

EVERY gas atom is either radiating more positive force than negative, or vice-versa, and it starts to inform us that there is some common entity that makes all atoms either net-positive or net-negative.

If one places a sealed container filled with air on a weight scale, the gas' weight pressure is not its gas pressure. There is almost no gas weight in the container, yet the gas pressure is huge by comparison. The gas has both weight and pressure, distinct and different, and the weight pressure CONTRIBUTES to, but cannot define the whole, gas pressure.

If air is pumped into the container until the weight scale registers twice the weight, there will be twice the pressure too, not at all meaning that gas weight is equal to or defines gas pressure. The doubling of the extra pressure is due to atoms forced closer together such that their repulsion increases, and so we learn that twice the atomic density per unit volume of space produces twice the repulsion force. It is what it is, and it is folly to argue that doubling the weight force is what doubles the pressure, for even these imposters argue that gas pressure is due to atomic weight x velocity (the two combined), not weight alone.

I asked google: "is it true that a liter of air, after pumped into a vacuum at STP, has a little more weight than a liter of air at STP in the open atmosphere?" I expect the answer to be yes, because the air in the vacuum does not possess the significant pressure due to the collective weight of air all the way up to the atmospheric ceiling. To compensate, the air in the vacuum needs more air pumped in (than normal air has per unit volume) before the pressure equals the pressure in open air. AI responds correctly with "yes," but also with error:

Yes, that is true. A sealed container holding a liter of air that has been compressed into a vacuum (or otherwise increased in pressure) will weigh more than a container holding a liter of air at normal STP atmospheric pressure [this is probably new to most physics buffs].

...Air in the atmosphere is subject to buoyancy — the surrounding air pushes it up slightly, reducing its apparent weight. By compressing the air and closing it off, you are...reducing the buoyant force acting upon it, resulting in a higher measured weight on a scale.

There is not a word in that response about the air weight. Instead, it argues on the basis of a buoyancy principle (it's based on weight pressure) that is fallacious, impossible. There cannot be a buoyant force on the air from the air, are we going nuts? How can lone air atoms force other air atoms upward due to air-weight pressure, especially with the bang-bang kinetic theory in view where atoms are banging with equal force in all directions, never more upward than downward? Are we nuts? Not me, but they are absolutely dishing out false information.

It is very clear to the baboons that air weight contributes to air pressure, yet they did not program AI's response above with collective air weight as the reason that a vacuum needs more air atoms to achieve 14.7 psi than open air. Why not?

When they teach that the weight of the open air transforms (like magic) into the kinetic energy of the air, these absolute goons will not tell you that a liter of vacuum, when filled with air, has so little weight that it can't possibly account for the gas pressure. It is easy for them to understand that the weight of atmospheric air is NOT transforming to air pressure such that the entire weight is transferred to the ground via atomic collisions (which they envision).

It shouldn't be me, the buff, to expose these flaws; it should be they, the baboons, the trolls, the goons, the imposters, the wicked. They know full well that if atoms are not in kinetic bang-bang mode, the only alternative explanation for gas pressure is inter-repelling atoms. It's not like a big brain is necessary to reach this easy conclusion. All I needed to start investigating their atomic model was the Christian-based knowledge that wicked evolutionists had overtaken science departments. Expect wickedry from them.

I asked google to comment on heat in a vacuum equipped with a filament: "In a 'best-possible' vacuum (near-perfect vacuum, such as Ultra-High Vacuum), a filament can achieve temperatures exceeding 2,500°C to 3,000°C, even if the exterior walls of the container are near absolute zero or liquid nitrogen temperatures (-196°C)." This wording leaves open the possibility that the filament itself is achieving those temperatures, not the space, for kineticists have AI brainwashed into thinking that the space of a vacuum cannot hold any heat.

I asked: "how hot can the space in a vacuum get when equipped with a filament? I don't want the temperature of the filament." AI stays loyal to kineticists with this answer from someone at reddit: "In a vacuum, the space itself has no temperature because temperature measures the average kinetic energy of particles (matter), which are absent in a vacuum. However, a hot filament inside a vacuum will radiate heat, and the surrounding environment — the chamber walls and any contained air molecules — will absorb this energy until a thermal equilibrium is reached." They bank on radiated heat but ignore the bulk heat that moves through space, which obvious-to-them exists in the vacuum, which they do not want to admit to in front of their students.

I re-phrased: "how hot can the space in a vacuum get when equipped with a filament? I don't want the temperature of the filament, just the temperatures measured by experiments, not kinetic-heat theory." "Based on experimental measurements in high-vacuum environments, the 'temperature of the space' (meaning the temperature of a substrate or sensor placed near an active filament) can reach over 1000°C if placed in close proximity, due to intense radiative heat transfer." It at least admits to extreme-high heat moving through the space, and anyone with a working brain knows it's not due to RADIATED heat, which is nothing more than light, which they wrongly view as non-existent photons flying through the space because they decided a century ago not to include the aether that carries light waves through space, and which itself defines heat in space.

I re-phrased: "how hot can the space in a vacuum get when equipped with a filament that is surrounded by an opaque material such that radiated heat cannot get into the vacuum? I don't want the temperature of the filament." It gave this double-speak response below where it first remains loyal to the baboons, then admits that there is indeed some FEELY heat, but AI wants me to view it as non-real heat, just some ghost of heat that somehow originates from the radiated heat:

In a perfect or near-perfect vacuum, the space itself cannot have a temperature because temperature is defined by the average kinetic energy of particles, and there are no particles.

However, if you are referring to the thermal energy density (how "hot" it feels) within a vacuum chamber that has an internal, shielded filament, the temperature is dictated by the thermal equilibrium between the radiation emitted by the filament and the emissivity of the opaque shielding surrounding it [i.e. the radiated heat alone transfers through the opaque material and into the vacuum].

...The temperature of the space [vacuum] will rise until the rate of heat lost through the shield equals the rate of heat produced by the filament [i.e. heat passes through the opaque material, into the space normally, as it would into a space filled with air].

... In high-vacuum furnaces (which use opaque shields to prevent radiation from reaching the outer walls), the "hot zone" can easily operate between 800°C and over 1,000°C, as evidenced by brazing processes (830–860°C).

There, it admitted that non-radiated heat is in the vacuum. But I had to pull teeth. The best they can do is say that the heat in the vacuum started off as radiated heat, enters the light shield, then passed through it as non-radiated heat. But they miss the 800-pound elephant, the fact that the filament pours out free electrons, the true definition of both the aether and heat.

Note that it puts "hot zone" in quotation marks as if it wants us to believe that the vacuum is not really hot, but only feely hot, or filled with some form of ghost heat. It's not supposed to be there, but it is there because a vacuum is always filled with heat, filament or no filament. This is a second means by which one can sue kineticism in court.

There are endless examples of people measuring heat in a vacuum that they cannot adequately explain by the kinetic model predicting absolute-zero temperature. The buff at Action Lab found plenty of heat in the vacuum, but he wrongly explained its origins because his thinking has been twisted by the kinetic model.

A jar filled with air is primarily a wall-to-wall jar of evenly-spread, stationary, heat-particle electrons with some air-atom "pepper" sprinkled in. The air atoms are likewise evenly spread, and stationary unless moved by a force.

I watched a video where Action Lab took a container filled with air, at normal air pressure, down to a near-perfect vacuum, while showing the temperature drop in the vacuum from 17 C to 11 C, which is not much of a drop to speak of, yet he refused to assail the kinetic theory that demands absolute-zero temperature in a perfect vacuum. There was no filament in the vacuum to create radiative heat. He didn't have the fortitude to say, "I've been brainwashed. They said it was absolute zero in there." He didn't have the kindness to alert his audience, but instead remained faithful to his fellow baboons.

With zero pressure registered in a vacuum, a caloricist like myself could reason with solidity that the heat particles in the vacuum offer no pressure upon the pressure gauge, and therefore no pressure on the wall atoms of the container. But is it true that there is no pressure in the vacuum just because the pressure gauge reads zero? Not necessarily, because the heat particles could be making donkeys out of kineticists.

A vacuum container, and therefore the pressure-measuring device of a pressure gauge, is filled with innumerable holes that electrons can move through. That's known as heat transfer. When the heat-particle density on both sides of the pressure-gauge's diaphragm is equal, the diaphragm can no longer measure pressure because it's going to register as zero due to equal pressure from both directions. To measure pressure, it must be pushed in one direction only, but when heat particles stream through it to get on its backside, the push is equal upon it from both directions, duh.

The heat particles in boiling water poured into a mason jar can penetrate the jar within a couple of seconds sufficient to be felt. It would penetrate much faster if the jar is made of metal, and almost instantly if the metal were extremely thin. "A typical [metal] diaphragm in an air-pressure gauge is designed to be extremely thin to ensure sensitivity, with thicknesses often ranging from 0.02 mm to 0.4 mm (approximately 0.0008 to 0.016 inches)." Now you know why heat particles in a vacuum have made goofballs of the goofballs.

I have never had access to certain information as I now do with google's AI. I can better prove my atomic model this way. I thought to ask it a trick question: "can high heat in a vacuum crack or explode a thin-walled container sitting in normal air pressure?"

Yes, high heat applied to a sealed, thin-walled container in a vacuum can cause it to explode or rupture when surrounded by normal atmospheric pressure. While a thin-walled container might be expected to implode due to the high vacuum outside and low pressure inside, the application of high heat changes the dynamics, causing internal pressure to build rapidly and potentially exceed the container's structural strength.

You see? AI, now betraying its programmers by admitting a reality it should have kept hidden from us, was able to fetch the reality against the general programming that orders it not to.

To explain the high heat in a vacuum in the thousands of degrees, the stupids get stupider by using the only means they can, the falsification that tungsten atoms off of the filament are filling the vacuum as a gas...such that it explains heat attainment in the thousands of degrees. But, donkeys, there are not enough tungsten atoms in the entire filament to create a gas such that it could explode the container walls. Yet they will claim that a very-few tungsten atoms off of merely of the filament's surface can fill the vacuum to explosive pressures. I'm being very kind to call them jackasses. They should be in jail.

Thanks to AI's response above, I will contest that there is indeed pressure in a vacuum at standard temperatures, though I can't know how much, and neither can a pressure gauge know how much.

Increasing heat in any sealed gas doesn't bring atoms closer center-to-center. Therefore, the increased pressure due to higher temperatures must be due to something besides atoms moving closer. Doubling the temperature roughly doubles the gas pressure, and as gauge-registered pressure can be due only to atom-applied pressure, how can we explain the doubling unless credit goes to the heat particles in some way?

It's my task to explain how heat particles make atoms push atoms more forcefully. There are two ways that could work in combination to produce what I'll call "heat-push." One of the ways is heat-particle push, and another is increased atomic repulsion force with increased heat. Together, the two produce TOTAL heat-push, but if atoms do not gain more repulsion force with added heat, then I would need to define total heat-push as heat-particle push alone. Heat-push causes out-spreading of atoms, and so does atomic repulsion. As they are likely working together, it's tricky to discover how much of each exists.

It's a fact that weight of gases in the open atmosphere counters out-spreading, but I've been wondering whether it should apply also in a sealed container where the gas is inserted into a vacuum. For as long as the gas exists (versus liquid formation), it's obvious that the out-spreading force is greater than the gravity force acting on the atoms. Gravity makes liquids (though liquids can form in other ways), and it's gravity that made all atoms weigh the same by granting them all the same net-positive charge.

If all atoms retain net-positive charge to high temperatures, it's expected to contribute to gas pressure, but cannot contribute extra pressure unless atoms are brought closer, for only then can the positive force increase. Therefore, adding heat alone to a container can increase pressure due to heat-particle push but not due to net-positive forces. It's possible that heat, because it's made of electrons, has caused all atoms to surpassed their net-positive charge into net-negative zone, in which case the addition of heat to a gas does increase the repulsion forces of atoms i.e. that contributes to out-spreading = gas pressure.

A water molecule is not an H2O molecule, but rather an HO8 molecule, with eight oxygen atoms merged into a giant H atom. The all-atoms-weigh-the-same fact demands that this molecule is an HO8; it's not optional. They tell us that steam weighs nine times more than hydrogen gas, and so it's very fortuitous that the steam molecule has nine atoms. That is, not only is steam nine times heavier than H gas, when both are in a sealed container at the same temperature and pressure, but the HO8 molecule weighs nine times more than one lone H atom because all atoms weigh the same. Thus far in my adventures into chemistry, I say that the H gas is made of lone H atoms, not the diatomic molecules that physicists imagine.

What should be the distance between steam molecules (at the same temperature and pressure) as compared to the distance between H atoms? This is where the fortuitous comes in. The only way for the steam to weigh nine times more, where its molecule happens to weigh nine times more too, is for the centers of the steam molecules to be identical in distance to the centers of the H atoms. For every H atom at the same location in its container as a water molecule in an identical container, the steam molecule weighs nine times more such that the whole gas weighs nine times more as well.

When seeking the reason that both gases have particles at the same spacing, it's fortuitous that both gases have a single H atom. The steam molecule is larger than one lone H atom, and so the steam molecule gets more out-spreading force due to heat-push, and yet as the two particles have the same distance between them at STP, it can suggests that the higher weight of the steam molecule acts in the other direction, making the molecules come as much closer to each other as the out-spreading force makes them go further apart.

I'm thinking that, by exactly as much as the O atoms of the steam molecule increase its cross-sectional area to more than the cross-sectional area of the lone H atom, that's the specific area upon which heat-push acts to perfectly counter one atom of weight force. On the other hand, there is reason to believe that atomic weight plays almost no role in small-container gases at standard pressures. For this quest, the only thing I have to work with is the assumption that both atoms are spherical. Maybe I should just go fly a kite and have less of a waste of time.

As O gas weighs 16 times more than H gas, while both O and H atoms weigh the same, it's clear that H atoms receive 16 times more heat push, because there are 16 times fewer H atoms (in an equal volume at STP). The question is: must the H atom also have 16 times less cross-sectional area, in order to get 16 times the heat-push? Yes.

If the only thing at work is heat-particle push, the H atom would need to be 4 inches round, which has 50 square inches of surface area. It would need to be 4 inches round because that's four times more cross section than the one-inch O atom, and because heat-particle push pushes on the cross section. Heat-particle-push math needs to use cross-sectional area, not diameters. Logically, 16 times the area gets 16 times the push. This is why the H atom needs to be the largest of all.

As both the H atom and O atoms weigh the same, there's no need to know anything else but the extent of out-spreading force, because both gases get the same amount of in-spreading due to weight. If the out-spreading force on the O atom is 1, the H atom gets 16 units, but I prefer to number the H atom with 1, and the O atom with 1/16. For spheres, the cross-sectional area of a 4-inch sphere is 16 times greater than for a 1-inch sphere. That's why, if heat-particle push were all there is (no extra atomic repulsion simultaneously at work), I could tell you with confidence that the H atom is a 4-inch ball while a 1-inch ball is the O atom.

If the addition of heat to a gas increases the atomic repulsion, the H atom might be smaller than 4 inches because the inter-repulsion provides some of the 16-times more push. One can imagine various force levels from atoms at any size. The size wouldn't dictate the force level. At 4 inches, the heat-particle push alone exercises 16 times more push, but I'm thinking that this size can be the case only if the inter-repulsion is from net-positive charge in all gas atoms. If all gas atoms possess net-negative charge, then the H atom might not be as large as 4 inches.

Double the heat particles is expected to perform twice the heat-particle push, and where all gas atoms are net-positive in charge, increased heat adds zero atomic repulsion because heat increase cannot make the atoms more net-positive. Heat increase alone in a gas does not make net-positive atoms more repulsive due to closer distances, for added heat alone doesn't make the atoms go closer. But if heat makes the atoms more net-negative, then the addition of heat alone DOES make atoms repel harder with heat addition alone. This choice is my problem.

Just so you understand what I'm doing: H atoms are, as an indisputable fact, 16 times less sparse than the O atoms, center-to-center, which is why the H atoms must receive 16 time more push in order that both gases achieve the same gas pressure at STP. Every H atom forced onto a container is in the midst of the identical out-spreading force COMING AGAINST IT FROM THE SPACE as every O atom on the container wall, but every H atom receives 16 times what O atoms receive. That's the fact.

I'm stuck with a choice. Either I go fly a kite and be happy in the sun, watching the fluffy clouds go by, or I grin-and-bear a trial-and-error exercise starting with one of the two theories above, checking to see if I can make headway into revealing the true look of a few molecules, for starters. I'm going to start with the net-positive theory that requires the H atom to be 4 inches in diameter, which receives ONLY heat-particle push and zero atomic repulsion with added heat.

I can entertain the net-negative theory where it's atomic repulsion is proportional to temperature rise, as is heat-particle push. In this picture, the two in combination create a total heat-push indistinguishable from fully heat-particle force alone.

For example, at standard temperature, the combined heat-push from atomic repulsion and heat-particle push grants the H atom 16 times more push than upon the O atom, and at twice the temperature, both forces double to maintain proportionally with added heat. So, we might imagine a 3-inch H atom granting less than 16 times the heat-particle push (due to reduced size), yet the net-negative push force makes up for it such that it brings the score up to 16 times.

Chlorine's Mystery Atoms

It was like pulling teeth, but google AI did give me a table of real gas weights instead of the predictions of the ideal gas law. When I saw the 2.994 weight for chlorine gas, I asked google: "why do you give 3.17 g/l for chlorine while others give 2.994?" The response made me see red, because the ideal gas law is from the predictions of the kinetic model, which should have no default part in educating the world. Unless one requests the weights by ideal-law math, AI should always report the true weights only, but even when I ask for the experimental realities, it more often than not gives the ideal-law "realities" but with a most-minor change to make it appear that ideal-law math is appropriate:

The difference between 3.17 g/L and 2.99 --3.04 g/L for chlorine gas density arises from using...varying molar masses of Chlorine, and whether the calculation treats chlorine as an ideal gas or a real gas.

The 3.17 g/L Value (Common Educational/Ideal Standard). This value is commonly cited in textbook problems because it uses standard values that make calculations simple [this is baboon excrement dressed as acceptable science as per the prediction of kineticism]...

The approx 2.99--3.04 g/L Value (Physical Reality). More specialized chemical handbooks or engineering sources may give lower values (e.g., 3.04 g/L according to Air Liquide Air Liquide Australia) because chlorine does not behave as an ideal gas at 0 C and 1 atm. It is close to its boiling point -34C, meaning it is denser and more compressed than an ideal gas...

Who gives a tart about such excuses, whether it behaves like an "ideal gas" at O C? The world should have the experimental realities all the time, no monkey wrenches tossed in by the baboons. How can it be possible that after two centuries or more of toying with gases, they still can't decide how much a liter of chlorine gas weighs? What kind of apes are responsible for these sorts of "uncertainties"? I suggest apes who need to fudge numbers based on unjustified excuses, apes who need to create smoke-screen confusion to hide the numerous problems in their kinetic / Avogadrian model of the atom.

"Based on experimental measurements at standard temperature and pressure, hydrogen chloride (HCl) gas...[weighs] 1.64 g/L," which is 1.64 / .09 = 18.2 times more than H gas at STP. It means that HCl has 18.2 times as many atoms per liter. I'm reading: "To produce 2 volumes of hydrogen chloride (HCl), you typically need 1 mole [volume unit] of hydrogen and 1 mole of chlorine." The imposters need to tell us that it's exactly one-per-one liter because they think each liter has exactly the same number of atoms, because they are ultra-stupid, and there is no hope for them becoming wise to this stupidity anytime soon. In reality, the chlorine has roughly 34 times more particles than H gas.

Using the 3.0 g/l for chlorine gas instead of the idea-law calculation of 3.17, it turns out weighing 33.33 times more than H gas (.09 g/l), which probably exposes why they would want it higher than 3.0, for 33.33 is horrible for their Avogadrian atomic model...because they need whole numbers all the time. By using 3.17, the gas works out to be 35.2 times heavier, which is at least closer to a whole number (but no cigar).

Although AI insists that the HCl molecule weighs 18.2 times more than an H2 molecule even though HCl gas weighs about 33.33 times as much as H gas, AI fetched me something else that it probably wasn't supposed to, when I asked it: "how much more does HCl molecule weigh than the H2 molecule?" The response, "An HCl molecule weighs approximately 34.44 more than an H2 molecule," suggesting 1.0 H + 33.33 Cl, for they assign the H atom a weight of 1.0. Therefore, ignore the imposters when they use 3.17 and the corresponding 35.5, because they care not for the facts, but care only to promote the "ideal" expectations until they are factualized falsely.

As they haven't come to terms on exactly how much chlorine gas weighs at STP, where we saw above that they have it somewhere between 2.99 and 3.04 with the latter number stressed, I suggest that, due to every H atom merging with 17 to form H-chloride, it weighs 34 x .09 = 3.06 g/l, because H gas weighs .09 g/l. That is, instead of using 33.33 in the math, we must use a whole number, of course, and so 34 is the obvious choice. Therefore, I'll use 34 going forward.

Clearly, with 34 times as many atoms, there would be, in my model, 1 H atom per 34 when making a hydrogen-chloride gas, where roughly one liter of each is mixed to produce the chloride. However, this makes the chloride molecule weigh 34 times more such that one liter of STP chloride gas would weigh 34 times more instead of the 18.2 that it does weigh. I need to find a way to overcome this problem.

As they say that the result of the one-and-one liter mix produces TWO liters of the chloride at STP, they conclude that both the H and chlorine gas have diatomic atoms. That is, the H and Cl atom is a one-body twin that they envision separating into lone atoms to turn one liter of each into two liters of HCl molecules each having one lone H merged into one lone Cl. They can do that because they envision the same number of atoms per liter.

I can't do that because I see only one H atom per 34 Cl atoms. I need to find a way to split that number in half to 17. One way for me to do this is to argue that two liters of hydrogen mix with one of chlorine to produce two liters of hydrogen chloride. And so this HCl17 molecule works out to a total weight of 18 times more than 1 H atom, explaining why the chloride gas weighs 18.2 times more weight than H gas.

I'm learning something here because hydrogen and chlorine actually combust with flames but without need of oxygen, they say. I was under the impression that combustion always needs oxygen. I asked, "how many materials combust with flames without oxygen?" The response gives mainly oxides, or the need of oxidizers, and even says: "Fluorine and chlorine are powerful oxidizers that can cause many elements and compounds to burn, often forming flames." However, to my surprise, "oxidizer" is defined as "a substance...yielding oxygen [or] accepting electrons [italics mine]." The surprise is that "oxidation" is defined as taking electrons even from non-oxygen materials, suggesting that while the term was coined to define typical combustion with oxygen, it later came to define any chemical reaction yielding enough heat to be classified as "fast combustion."

The mind-cracker comes when they define oxidation as "stealing" electrons from a substance. No, STUPIDS, combustion is obviously to be defined as the RELEASE (not capture) of atom-captured electrons into the free state. The heat is free electrons, and the flame is filled with them at such a high rate that light is the result.

I'm reading that H and Cl burn "vicious." And "chlorine acts as a strong oxidizing agent by stealing (or pulling) electrons from other substances, such as hydrogen, to form HCl [gas]." This sentence kills the kinetic model. Let me show you how.

I had found a quote saying that one liter each of H and Cl gas makes ONE liter of the chloride at STP, but when you ask google how many liters the burning of one-and-one liter H and Cl produce, it will say two. But look at the wording: "One liter of hydrogen gas (H2 [twin]) and one liter of chlorine gas (Cl2 [twin]) react to produce 2 liters of hydrogen chloride (HCl) gas at STP, ACCORDING to the balanced chemical equation: (H2(g[as]) + Cl2(g) > 2HCl(g)." The latter's '2' translates to 2 liters. Yes, ACCORDING to that set-up, it predicts 2 liters at STP, but is this the experimental reality, or just decades of factualizing the false expectations?

Let's put the one-and-one liter mix at STP into a two-liter container. We now envision both atoms in the same container before something ignites them to merge them. The H and Cl atoms are both twice as far apart as they were in their one-liter containers. Yes, because they both have twice the volume of space to spread out in. Though twice as far apart, the pressure remains at standard pressure because the H atoms produce half standard pressure while the Cl atoms likewise contribute half standard pressure.

Although I oppose their invention of diatomic H and diatomic O atoms because they were invented to invent a fallacious H2O molecule, I am not opposed to the existence of diatomic atoms if the evidence requires them in some cases. In the two liters above, kineticists envision both diatomic H and diatomic Cl atoms, and they think that both are separated into lone atoms, at the moment of ignition, that then become two liters of HCl molecules at standard pressure instead of two liters of H2C2 molecules at half-standard pressure (same as one liter at standard pressure).

What happens to the standard pressure in the two liters of mixed H and Cl atoms when the atoms form the chloride molecules? In their atomic model, where they can't satisfactorily explain (though they try) where the heat comes from that performs the combustion, they are required to claim that the pressure increases wildly due to the increased heat. Instead, they claim that H and Cl merge to create 1 atmosphere of pressure, which is a lie. And they say that two-and-one liters of H and O gas merge and combust to form two liters of steam at 1 atmosphere of pressure, because this is the expectation of their kinetic mechanics, only they leave out the pressure formed from combustion.

Sure, if they leave the container sit for a couple of hours to cool, the pressure and temperature will come back down to STP, but only after ALL the formed heat has escaped through the container walls. In that case, the molecules must have less energy than prior to combustion because they just lost a whole whack of heat. How can the morons justify an end result with exactly the same motion energy remaining both before and after combustion when they define heat as the motion speed of molecules?

Clearly, they need to devise a way for the PURE CREATION of motion energy in the combustion process, and I've done battle against how they do it, finding it to use crock. In reality, merger releases the electrons in the merged regions, and freed electrons are heat particles.

They are required by the laws of physics to envision H atoms travelling very fast, on average about 34 (or their 35.5) times faster than the Cl atoms. But they obviously found a problem with that expectation because, when I ask the following question, it tells me that the speed differential is only 5.93 times: "as the Cl2 molecule weighs about 34 times more than the H2 molecule, doesn't the H2 molecule need to travel about 34 times faster than the Cl molecule to produce the same pressure?" No, not 5.93, but 34 times, because a law of physics demands it where force = mass (weight) x velocity.

However, the morons have devised other mathematics because that simple law doesn't work for them. They have another math formula for figuring out kinetic energy as if it and force are not the same thing. Of course they're the same thing, for the only thing kinetic atoms do it speed about, having the same weight at all times, wherefore the only math called for is weight x speed. That is, where something is twice as heavy, it needs to go twice as slow, and vice versa, in order for both to possess the same force of collision.

AI is programmed to teach the public that "According to Graham's law of effusion, the speed of molecules is inversely proportional to the square root of their mass." What? Are you kidding? There's nothing wrong with weight x velocity, and so this must be a gimmick, a means to drastically reduce the 34 times expected to about 6 times (they use the square root of 35.5, = 5.95). In the meantime, the square root of H's 1 unit of weigh = 1. This is utterly boneheaded, you should be able to spot why it's trickery simply because it violates weight x velocity.

Ask yourself what moron believes that the H atoms will continue travelling 34 times faster for as long as one keeps the gases mixed and not ignited. But even if we entertain with their 6-times differential, how does one decide how fast the molecules will travel once they are formed from a attraction-attack collisions. The morons insist that both atoms carry the same energy exactly, even though they crash about, and when they collide head-on, instead of both stopping in their tracks, as the laws of physics demands, these magical atoms bounce off at the speed at which they collide. The laws of physics demands that collisions from any angle, from opposing directions by parties having the same energy level, slow both parties involved, because the energy of one counters and slows the other.

They argue that the H and Cl atoms, prior to ignition, carry exactly the same energy such that they form exactly the same gas pressure. They have the problem of finding how fast the resulting molecules will travel in that scenario. How can one go about figuring this out, with all atoms moving in random directions such that the average direction is zero direction? This task looks like a nightmare.

Instead of seeking molecule's speed with the expectations of the laws of physics in mind, they use the expectations of Avogadro. They say that, since there are the same number of HCl molecules, in the two liters above, as there once were either H atoms or Cl atoms, the molecule's speed needs to conform to the newly-formed weight of 36.5 units. They assign the Cl molecule a weight of 35.5 times that of H atom, then add one unit for the H atom that joins the Cl atom, and so the new weight is 36.5. They then apply their square-root-of-mass math for finding how fast it goes in mph, only they use the weight-of-gas differential (not weight-of-molecule), which in this case is 18.2 times (half of 36.5) more than H gas.

So, even though the H atom is travelling much faster than the Cl atom, in their moronic minds, the newly-formed molecule takes on a speed that is essentially the speed of the Cl molecule. Does that sound correct to you? They would say that the mph speed of the Cl atom is the square root of 35.5 while the speed of the molecule is the square root of 36.5, and so they have a speed of 5.93 units versus 6.05. How can that make any sense at all since both the H and Cl possessed the same energy level? If anything, shouldn't the molecule's speed be midway between the two original speeds (we are entertaining that the particles never slow down upon collisions)? Or, why should the molecule's speed end up almost the same of Cl's speed and not almost the same as H's speed? To be sure, we are dealing with wicked people who invent math when necessary to protect their big-bang cosmos.

I asked, "what's the basis for the square root of mass calculation of atomic speeds?" Answer: "The square root arises because kinetic energy (KE) depends on the square of the velocity, KE = 1/2 x mass x velocity-squared." Look at what they did. The tricksters changed the correct mass x velocity math to 1/2 x mass x velocity-squared. For the H atom, mass x velocity is done in kilograms x meters per second, and of course their math is faulty if only because they don't know the true weight of the H atom, or any atom. They claim to know, but they are imposters because they are Avogadrites. There is no speed in gas atoms. They sit motionless in the container because they repel each other.

Using their KE math above, the speed of the H2 atom works out: 1/2 x 2 weight x 1-squared speed = 1 speed. To show how this math is bogus, we'll use their speed for the Cl2 atom of 3,789 mph / 639 mph = 5.93 times the difference: 1/2 x 35.5 weight x (1 / 5.93)squared = .5 speed. That is, half the speed if H atoms. TRASH MATH.

The math for their HCl molecule needs 3,789 / 919 = 4.12 times slower than the H atom: 1/2 x 36.5 weight x (1 / 4.12)squared = 1.08. That is, this math makes the resulting HCl molecule 1.08 times the speed of the H2 atom. TRASH MATH.

The reason that kineticists latched onto to Avogadro's theory (or they may even have invented it) is highlighted by this H-Cl mix of gases, but also by the H-O mix and others. They first view both the H and Cl atoms with exactly the same spacing, because Avogadro says that both gases have the same number of diatomic atoms per identical volume. Then, when the H and Cl atoms merge, the resulting molecules maintain the same spacing because there is no other choice, for both had the same spacing before merger. This is how they roll, saying that all molecular gases have exactly the same number of particles as all other atomic or molecular gases at STP. And they all, every one of them, possess the same energy level at STP, which is such a cosmic coincidence you can bet your house it's not correct.

The average speed claimed for the diatomic H2 and Cl2 atoms at STP is 3,789 and 639 mph, a difference of 6 (5.93) times. Thus, they derived those speeds by their square-root-of-mass math. Throughout the whole world, they have taught this trash.

They say: "The root mean [average] square (RMS) speed of a hydrogen chloride (HCl) molecule at Standard Temperature and Pressure is approximately 411 m/s," or 919 mph. Look at how ridiculous this is, for even though the formation of H2 and Cl2 (weight 35.5) makes the HCl molecule (weight 36.5), they increase the speed from 639 units to 919, a difference of 280 units just because one wee-wee H atom comes on-board the Cl atom. Thus, while the product's weight increase is 36.6 / 35.5 = 1.03 times (3-percent increase), they claim that the speed increase is 919 / 639 = 1.43 (43-percent increase), which greatly violates the correct mass x velocity math which demands that 1.03 increase in weight requires 1.03 reduction is speed.

In other words, while they claim from one side of their mouths that the merged HCl molecule has exactly as much energy as the diatomic H or Cl atom prior to merger, their math confounds that claim by granting the HCl molecule more than 43 percent energy. Without doing a deep dive into how exactly they create this mess, I assume it's because they have need to make the math conform to their chosen weights and speeds for all atoms...yet the reality is that all atoms weigh the same, and all have zero speed.

In whatever ways they try to squirm out of this problem, it's their lunatic problem, not yours. The easy prediction is that their math MUST go lunatic at some points as they try to juggle Avogadro with the experimental realities and the expectations of kineticism. Conformity of the realities to their unrealities has become their demon work because they should know better by now. We can excuse the investigators in the 1800s and 1900s. The would-be mystery buster lacking the proper knowledge was at least not the imbecile who replaced proper math with Avogadro-serving math.

I'm now feeling helpless before their claim that one-and-one liter of H and Cl make two liters of H-chloride, for who's going to believe me if I claim that two liters of H need to mix with one of chlorine to make two liters of the chloride? I can tell you this, that two liters of hydrogen is known and accepted to mix with one liter of oxygen to make two liters of steam. Why should it be different with the H-Cl mix? It shouldn't. I think they lie.

Once we assail Avogadro, to put his theory into the grave where it belongs, the only other theory that even begins to work is mine. I've been showing my readers how well my theory works, so well that I see it as the reality, because there's no other way to approach the experimental realities aside from all-atoms-weigh-the-same. Nothing else is going to work. We need to replace all-gases-at-STP-have-the-same-number-of-atoms with all-gases=at-STP-have-atoms-with-identical-weight. Molecules at STP go by that weight scheme too, i.e. if HCl gas weighs 18.2 times more than H gas, then the chloride has 18 times as many atoms, 18 per 1 H atom, and slightly less than a liter (18 / 18.2 = .99) of Cl gas is needed for each perfect liter of H has.

If one-and-one liter of H and Cl makes only one liter of H-chloride, then the hydrogen chloride molecule would be an HCl34 and thus weigh 34 times more than an H gas. But where two liters of H gas are needed instead of one, then each H atom merges with half of 34, and thus the H-chloride molecule weighs 1 + 17 = 18 atoms, predicting the reality that H-chloride gas weighs 18.2 times more than H gas.

If there is absolute proof that one-and-one liter becomes two liters of HCl17 molecules, then I would need to admit that both H gas and Cl gas have diatomic molecules. I would need to admit that igniting the two gases splits each H2 into lone atoms, and each Cl2 into lone atoms. In this picture, there are 17 Cl molecules, not 34, weighing 34 atoms. Each grouping of lone 17 atoms merges with each lone H atom.

I can tell you this, that hydrogen and oxygen gas can exist in the same clear-bottle jar for a century, and not merge, even in sunlight, until someone gives it a spark. In the case of hydrogen and chlorine gases, they are more volatile because even sunlight (or a spark) causes them to ignite. Just think about that. What could be taking place that merely the weak energy of sunlight could change the electrical charge of either the H or Cl atoms? We know that sunlight doesn't make H atoms suddenly attract O atoms, and so it seems undeniable that sunlight affects chlorine atoms.

In fact, the Cl and O atoms merged as chlorine dioxide are broken apart by mere sunlight. It suggests a weak bond due to very-slight merger. And, by the way, they say that two liters of O atoms are needed with one liter of chlorine atoms to make two liters of chlorine dioxide. This is the 2:1:2 ratio that could be the fact also for the making of hydrogen chloride.

The point is, as it's hard to believe that sun alone could cause a H-to-Cl attraction, if they were lone atoms, one solution is that chlorine atoms could be diatomic and barely clinging to one another, which doesn't sound correct. "It is the sunlight itself — specifically the Ultraviolet (UV) light component — that initiates the reaction between Hydrogen and Chlorine, not merely the heat from the sun." Why does AI capitalize "hydrogen" and "chlorine," because it bows down to them?

AI says that chlorine gas breaks down into lone-atom Cl in sunlight, whether it's mixed with hydrogen or not. It claims that one liter of Cl2 becomes 2 liters of Cl gas. Having my doubts, and thinking that this is just another example of how AI is programmed to respond as per Avogadrian expectations, I tricked it by asking, "what is Cl gas, not Cl2 gas, used for in industry?" The response mentions Cl only once like so, at the start of the response: "Elemental chlorine (Cl2) or, in general context, Cl gas) is a highly reactive, greenish-yellow oxidant primarily used for large-scale chemical manufacturing..." Ya-but, it's got Cl2 inside it brackets too, and I didn't ask for that.

I re-phrased: "what is Cl gas used for in industry? Don't mention Cl2 gas because it's confusing me. Just Cl gas please." "Chlorine gas (chemical symbol Cl) is a highly reactive, yellow-green elemental gas widely used in industry as a powerful oxidant,..." Ya-but, AI appears to be trying to trick me because lone-atom Cl is NOT chlorine gas, because they view that gas with CL2 atoms. The response goes on long, and not once does it say that Cl gas is used. Suspicious, I asked: "is CL2 or Cl the oxidant?" "Cl2 (chlorine gas/dichlorine) is the oxidant." I said: "are you saying that Cl is not an oxidant, only Cl2?" "Yes, elemental chlorine (Cl2) is the powerful oxidising agent frequently used in chemistry, not the chloride ion (Cl-) or neutral atomic chlorine (Cl) in isolation."

? You see, it did try to trick me. It wasn't able to list one thing that lone-atom Cl is used for in industry, because, I think, it doesn't exist. It's in their Avogadrian imaginations. It had said, with "Cl" in its brackets), "Chlorine gas (chemical symbol Cl) is a highly reactive, yellow-green elemental gas widely used in industry as a powerful oxidant, DISINFECTANT,..." I therefore asked, "is it Cl2 gas, or Cl gas, that's a disinfectant?" As part of the response, it told me, don't worry, Cl gas exists, even though I didn't ask whether it exists:

CL2 gas (chlorine gas) is a well-known, highly effective disinfectant...While elemental chlorine (Cl) exists, it is unstable as a single atom and rapidly forms the diatomic molecule (Cl2), which is the form used for disinfection.

Ya-but, "unstable" can mean it exists only in the unstable minds of the baboons, and disappears into Cl2 before anyone can shake an Avogadrian wand at it to make it appear.

In other words, I'm not convinced that sunlight and chlorine gas makes lone chlorine atoms. Therefore, it could be that the chlorine atom is really a molecule of 1 H atom with 34 unidentified atoms merged into it, which is an alternative way to have chlorine gas (and the molecule) weighing 35 times more than hydrogen. I'll call it the HM34, where the "M" is for the mystery atom.

When one adds one liter of lone H atoms to one liter of this HM34, the loners attract half of the mystery atoms, same as how hydrogen peroxide, which they call H2O2, is somewhat unstable such that it loses half (8 in number) its O atoms to become water i.e. what they call H2O. Hydrogen peroxide is truly an HO16 molecule, becoming HO8, same idea as HM34 (chlorine) becoming HM17 (hydrogen chloride). The latter, when dissolved in water, is called hydrochloric acid, which is roughly as unstable as hydrogen peroxide. In this picture, a diatomic H atom is not needed.

Unless they can split their Cl2 molecule into lone Cl atoms, they can't disprove my theory that chlorine is HCl34. If they say that, yes they can split the Cl2, but then admit that the lone Cl atoms instantly revert to Cl2, then I can say, no, the Cl2 molecule does not exist, it's instead an unstable HM34 that's heavily loaded with mystery atoms. It has an affinity to load only 17, resisting an 18th, but there is a way(s) to force an additional 17 upon it.

"Chlorine (Cl2) is produced from hydrochloric acid (HCl) mainly through oxidation processes that remove hydrogen, either via catalytic oxidation (Deacon process) or electrolysis." The latter process simply runs electricity through the acid, and out pops chlorine gas (HM34) from the acid's surface. As the acid is made of HCl17 and water, we learn here that the mystery atom must be in the water. But it can't be oxygen because that would make the HM17 nearly the HO16.

The apparent way to solve the mechanics of this chemical reaction partly lies in this: "Hydrogen gas is produced at the cathode (negative electrode) during the electrolysis of hydrochloric acid." Another clue is in this: "The primary formation of chlorine during the electrolysis of hydrochloric acid does not free oxygen atoms from water (H2O). Instead, chloride atoms from the acid form chlorine gas." If it's not sending out O atoms, neither is it breaking the H-O bonds of water molecules...which it normally does in electrolysis, but not when the water has CHl acid in it. Therefore, the freed H atoms (the freed gas above) is from the breakdown of the acid itself. What does this mean?

It means that the electricity pops out all 17 mystery atoms out of SOME (not all) the acid molecules, and as each of the lone H atoms wave bye-bye, the 17 atoms are free to re-marry. They obvious re-marry other acid molecules such as to turn them into HM34 (chlorine). But how? One way: the electrical flow adds electrons to the 17 mystery atoms on the outer perimeter of the HM17 molecule, and being made negatively charged, they attract the freed mystery atoms because they become positive in charge as soon as they get unmerged from their central H atom.

To possibly explain why 17 are attracted while 17 are already saddled on the molecule, chances are that the saddled 17 attract one each. In fact, as hydrogen peroxide gets 8 more O atoms upon a molecule already having 8 O atoms saddled, it could be that the latter attract one each. Once attracted in merger, it's a good bet that each newcomer rolls down on the already-saddled atoms until it contacts the H atom and enters it too, though not as deeply. In the case of the peroxide, if they do enter the H atom, don't enter very deeply, because they are eradicated from the peroxide molecule even at room temperature. However, in the case of the chlorine, my theory is that the 17 newcomers don't come out even with added heat to the highest-possible levels. They can come out when chlorine burns in the presence of H gas.

There is also this: "Hydrogen Chloride fumes come out along with evaporated water vapor, and they are essentially 'attached' in a technical sense, forming an acidic aerosol mist." I'm reading that a 10-20-percent solution of hydrogen chloride in water sees its hydrogen-chloride molecules streaming out of the water in roughly the same number of months as a 3-percent solution of hydrogen peroxide loses its extra O atoms to revert to pure water. That's what I meant by "unstable." However, the HCl molecules are not unstable when in the air.

However, there is a difference because the one loses only its extra compliment of O atoms from the water surface while the chloride is lifted out of the water fully intact, not loosing merely its mystery atoms.

I asked: "what is the gas above liquid chlorine in a sealed, transparent container, Cl2 or Cl>" As expected: "The gas above liquid chlorine in a sealed, transparent container is Cl2 (molecular chlorine), not Cl (atomic chlorine)." There you have evidence that the imposters are imposters when claiming that sunlight turns diatomic chlorine atoms into lone Cl atoms. In fact, I was hoodwinked into believing that sunlight breaks chlorine atoms apart, but now learn that, no not really:

Chlorine molecules are not inherently unstable in a general sense; rather, they are highly reactive. The phenomenon where they break apart in sunlight but quickly reform illustrates a dynamic equilibrium governed by the high energy of UV light and the extreme reactivity of individual chlorine atoms... This recombination is almost instantaneous...

I was hoodwinked. I now find that sunlight only makes the chlorine's atoms bob up and down, jiggle around, almost take flight, but they manage to stay on. During this jiggling, the molecules are able to join other atoms, for the response adds: "While they can react with other substances (like water or organic matter), if there is no other substance to react with they will collide with another chlorine atom and reform a stable Cl2 molecule." That's trash, there are no collisions, and the molecule is never a lone Cl atom as is implied. It's unstable only in as much as excitation by sunlight allows it to capture some other atoms -- if they are next door -- between its jiggling atoms.

What happens as the HM34 (chlorine) molecule has its 34 atoms bobbing up and down as sunlight excites them? I suggest that the outer 17, those saddled on top of the underlying 17, are the highly-bonkers (unstable) ones, not so much the lower ones that are stable in merger with the H atom. Sunlight alone ignites the chlorine in hydrogen gas, and so I'm suggesting that the outer 17 are pulled away to the lone H atoms to start the combustion process. The 17 unmerge a little from their O counterparts, which absorbs a little heat (i.e. makes the environment colder), but merge deeply with the lone H atoms to produce much heat (which continues the combustive chain reaction).

You may think that, if hydrogen chloride is truly a molecule of 17 mystery atoms merged into a hydrogen atom, then, sooner or later, with high heat, the mystery atoms should pop out to reveal themselves. But not necessarily. There could be quite a few types of elemental "atoms" that are in reality molecules which do not break apart due to heat reinforcing the mergers rather than undoing them. This can be the case when the atoms are so deeply merged that heat particles can't get into a nook between them to spread them apart. Instead, with the nook not present or barely present, heat presses the atoms into deeper merger. Possibly, the 17 mystery atoms could fully submerged.

If correct that chlorine has 34ish mystery atoms, a clue of their identity is that they are small enough to pack in that high number upon an H atom. Small atoms into a giant H atom could explain unusually-deep merging.

I asked google what happens to the HCL molecules as they enter the air. It told me that they do not become chlorine molecules. It told me that they are very stable and go from water molecule to water molecule, thanks partly to wind breaking them apart whenever they bump into and attach to water molecules. It then said that they get washed out into the soil by rain.

I then came across a contradiction, where AI said: " In water, HCl is a strong acid that dissociates completely into H+ and Cl- ions at about 25C." What? This is trash, same as when they claim that table salt (sodium chloride) is broken apart while dissolved. AI says further: "When hydrogen chloride (HCl) gas is dissolved in water, it forms hydrochloric acid. Water acts as a powerful solvent that rips the HCl molecule apart,..." TRASH. If that were true, the H-chloride could not remain stable in the water for months. It not only sits there not splitting apart, but goes up into the air not split apart.

I asked: "how can you be so contradictory as to say that HCl disassociates wildly to H+ and Cl- atoms in water while also saying that HCl can remain months in water, and eventually exit as a stable HCl molecule?" It didn't let up, but continued to claim the same: "The apparent contradiction between HCl dissociating completely in water and its ability to exist as a molecule [in water] comes down to the difference between aqueous ions and gaseous hydrogen chloride." Illogical. They are defining hydrochloric acid as part chlorine gas in solution, and so they then need to invent a way for the two atomic parts to come together again before they leave the acid as evaporated HCl fumes.

I pressed: "are you suggesting that chlorine as Cl- exits HCl acid?" "Yes. When hydrogen chloride (HCl) is dissolved in water to form hydrochloric acid, it dissociates (splits) completely into hydronium ions (H3O+) and chloride ions (Cl-)." Never mind that - sign with the Cl, it only means a slightly-negative charge on chlorine, but chlorine it is. AI is going bonkers on this one, as if the baboons need this theory to solve yet another problem with their model.

I was finally able to fish out AI's admission to guilt with: "are you suggesting that chlorine gas as Cl- exits HCl acid without electrolysis?" "Yes, chlorine gas can be produced from hydrochloric acid without electrolysis, but it requires the addition of a strong oxidizing agent to oxidize the chloride ions (Cl-) into chlorine gas (Cl2)." Ahh, suddenly, it admits that an oxidizing agent is needed, whereas it said nothing of the sort previously, but instead told me that the water itself is what tears the H and Cl atoms apart. Oxidizing agents include manganese dioxide or a copper catalyst.

I went in for the kill: "as you insist that adding HCl to water tears the H atoms from the Cl atoms, how much of the HCl remains in the water?" "When hydrogen chloride gas is added to water, it acts as a strong acid and dissociates into ions (hydronium H3+) and chloride Cl-) almost completely. Under standard conditions in a dilute or moderately concentrated solution, the amount of intact HCl molecules remaining in the water is considered negligible, effectively zero." This is crazy because the acid would no longer exist.

It even admits: "Hydrochloric acid (HCl) does not have a strict expiration date and can last for years (often 2–15+ years) if stored in a properly sealed,..." What? No tearing apart of the molecule as long as the lid's on? And: "HCl is a gas dissolved in water. If left open, the hydrogen chloride gas will evaporate into the air]. The concentration of the acid will decrease over time." There you go, it goes into the air, and does not turn to H3O and chlorine ("chloride").

"Why did you say that chlorine can be made from HCl acid only with a catalyst, and then say that the water itself can make H3O and chlorine?" "It appears there was a misunderstanding caused by using the same term ("chlorine") for two different chemical scenarios: elemental chlorine gas (Cl2) and chloride ions (Cl-)." Do you see the double speak that most people do not understand? Cl- is nothing but pure chlorine, but this AI imposterism would have you think that there's a difference. AI is programmed by imposters who juggle facts to suit their repair needs. They really don't have even a good handle on the facts, but are desperate to convince you that they are super-duper, almost never wrong.

In this case, they imagine Cl- atoms in the acid that don't come out, or they do come out, depending on the need. They know full well that chlorine evaporates wildly from water, and even the HCl is able to evaporate out though it's heavier than chlorine. They might say that it evaporates from the acid under discussion, but when they do say that, they need to explain how the Cl- atoms remain in the water, because they must remain there in order for the acid to exist in the first place, for they define the acid as part Cl-. In reality, there is no negatively-charged chlorine in the acid. The truth is: they are boneheads waving the flag of high intelligence.

They define a lot of typical acids as having the H+ atom, just crock. They haven't got a clue what happens on the electron-proton level in an acid. "An acid is a substance that releases hydrogen ions (H+) when dissolved in water, characterized by a sour taste, a pH below 7,..." There you have the crocky reason for their insisting that HCl acid must have the HCl molecule disassociated in water. It means that they have the wrong definition of an acid. Just because acids can break down metals doesn't mean they need positive H atoms in solution. Another definition is in order.

The thing is, they view the H atom as the smallest of all, but as they give it just one electron, because we are dealing with wackos, they define the H+ atom as a naked proton without its electron. WACKOS. It is impossible to have a bare proton. AI admits: "There are many materials that break down or dissolve metals that are not classified as acids." You don't need an H+ to drive atoms out of metals, and you certainly don't need a baboon on your back either. Hydrogen makes acids probably because it's the largest of all atoms, having the largest "bucket" (electron atmosphere) into which it can attract other atoms during merger. That makes sense.

"Chlorine reacts vigorously with most metals at elevated temperatures, often causing them to burn or melt." This is a good example of both the large "bucket" and deep-merger ability of chlorine. The chlorine atoms pull the metal atoms numerously from other metal atoms, which absorbs heat at the unmergers, yet more heat is created than absorbed when the metal atoms sink deep into the chlorine molecule. The numbers pulled are so numerous that "burn" takes place with "smoke." The latter are molecules or lumps of molecules that go up into the air as merged chlorine and the metal particles, but only due to the lift granted by the high heat temporarily in their midst. At cooler temperatures, they fall to the ground/floor. Chlorine particles alone (almost four times heavier than water molecules, 34 versus 9 units of weight) fall at room temperature.

Checking the Identical Spacing

"Pure hydrogen chloride gas has a boiling point of -85.05°C" This molecule has 18 atoms total, and 9 for water, which, by itself, predicts that water should have the lower boiling point (which it doesn't), because boiling point is when rising heat particles, inside boil bubbles, overcome all resistance to upward flow such that there are as many particles leaving the surface as are entering at the heat source (when the liquid cannot rise in temperature anymore). Part of the overcoming of this flow resistance is against the weight of the atoms, and the hydrogen chloride molecules weigh twice as much.

Therefore, as the HCl17 liquid has the lower boiling point, it tells me that the atomic bond between its molecules is weaker than for HO8 (water) molecules, for another part of overcoming upward heat flow is how strongly the atoms are bonded, for heat particles flow between atoms. To achieve boiling point, they need to open up paths between atoms until there is zero resistance to bubble flow. Atomic weight defines water pressure, and atomic-bond strength is expected to strengthen with deeper mergers. We can therefore assume that water molecules get the deeper merger of the two liquids.

The weaker the atomic bond, the easier it is for heat (i.e. the less heat needed) to lift the liquid molecule out as evaporation. Thus, it seems that the H-chloride molecule has a weaker bond that the water molecule.

By the way, as a liter of water gas has about 1,600 times less weight than a liter of water when both are at 100 C, it can be calculated that water molecules are about midway between 8 and 16 diameters apart center-to-center as a gas at that temperature. AI tells me, according to the kinetic model: "At 100C and standard atmospheric pressure, steam molecules are approximately 12 to 13 diameters apart."

If we start with back-to-back water particles at one diameter apart center-to-center, we have liquid water. To get the density of that picture 1,600 times lower (gets the volume 1,600 times larger), we need to double the volume about 10.5 times. For each time that the volume is doubled three times, the atoms go twice the diameters apart such that they will be 2 x 2 x 2 = 8 diameters apart after doubling the volume 9 times, at which time the volume has increased 512 times. After doubling it 10 times, the atoms are 10 diameters apart (according to AI), and the volume has gone up to 1,024, but we want it at 1,600, and AI says that they will then be 11.7 diameters apart at that time.

The difference in density between an H liquid at its boiling point and H gas at 100 C is about 1,075 times, meaning that H gas atoms would be significantly less than 11.7 diameters apart at that temperature. At first, I thought that, to do a proper comparison, we need the weight of a theoretical H liquid at 100 C, but then I thought, no, we can use the boiling-point weight versus 100-C weight because the atoms do expand from one condition to the other no matter which method we use. I hope I'm correct about that. I think I am.

If H liquid weighs 70 grams (boiling point), the math: 70 / .065 = 1,077 volume increase.

I use these numbers as per my question, "with back-to-back balls 1 diameter apart center-to-center, but then spread into a volume 1,600 times more voluminous, how many diameters apart will they then be?" The 1-diameter apart is with atoms kissing at the edges, zero merger. When I re-phrase with 25-percent merger of the balls, the wording is: "with back-to-back balls merged into each other and therefore .75 diameter apart center-to-center, but then spread into a volume 1600 times more voluminous, how many diameters apart will they then be?" The water molecules are now said to be 8.8 diameters apart instead of the 11.7 above.

When asking 10-percent merger for H atoms, but at 1,077 times more voluminous, they are said to become 9.2 diameters apart. One can now say that, if H atoms have less diameter than water molecules, then they need more diameters-apart in order that they and water molecules end up the same mileage-distance apart. However, I don't think the 1,077 is a correct figure, but I don't know what is.

Although I'm cherry picking the merger depths, I'm doing so to show that the possibility exists for the two particles to have the same mileage distance of spacing. Even if I use less merger for the water molecules such that I get them the same diameters apart as H atoms, one can yet justify things, for when two water molecules are in contact, there's room for the O atoms of one to squeeze in between the O atoms of the other such that both of their central H atoms will make contact. In this picture, the center-to-center distance is identical with the center-to-center distance of lone H atoms, in which case we need to treat the molecules as having the same diameter as H atoms.

I claim that water molecules and H atoms have the same spacing as hydrogen chloride gas. There are 1,190 grams/liter of pure H-chloride liquid at its boiling point, and the gas weighs 1.19 g/l at 100C. The difference in liquid-versus-gas weight is 1190 / 1.19 = 1,024 times, and so I asked for a 25-percent merger, and a volume spread of 1024 times, to find them 7.6 diameters apart, or 9.0 at 15-percent merger. This is looking good for confirming that H-chloride molecules are the same mileage-distance apart as H atoms and water molecules, where H-chloride molecules have a larger diameter, and they very well can. If its 17 mystery atoms don't allow the central H atoms of the molecules to make contact when the whole molecule merges as liquid, it has a larger diameter than the H atom, and as such it needs to be less than 9.2 diameters apart found above for the H atom to get equal spacing.

[The Chlorine section that was here was moved to two sections below.]

Oxygen liquid weighs 1,141 g/l at its boiling point of -183 C. So, 1,141 / 1.046 (gas weight at 100 C) = 1,090. At 20-percent merged, AI calculated 8.2 diameters apart. Where 1-inch balls are 8.2 diameters apart, they are 8.2 inches apart, and where 4-inch balls are 9.2 diameters apart, they are 36.8 inches apart. To scale, that's 1 unit apart per 4.5 units apart, or 4.5 times further apart for H atoms. That's more than the reality. The 9.2 figure is the diameters-apart for H atoms.

The huge problem is, particles 16 times more numerous in the same volume become only 2.51 times further apart, not 4.5 times. The only solution I can see is that the H atom needs to be viewed as less than 4 inches round, which can be justified where total heat-push is due to a combination of heat-particle push and net-negative atomic repulsion (that increases in negativity with increasing temperature). With that type of atomic repulsion in the picture, the H atom doesn't need to have 16 times more cross-sectional area than the O atom.

I asked: "if balls are 3 inches apart in a liter, how far apart will they be when 16 times more numerous?" The 1.19-inch response is correct because it's 2.51 times larger than 3. It's a fact that when a volume is 8 times more packed with balls, they are 2.0 times closer, and 2.51 times closer when 16 times more packed. So, in order to get 16 times more O atoms in a liter than H atoms in a liter, the H atoms need to be 20.6 inches apart where the O atoms are 8.2 inches apart, because 8.2 x 2.51 = 20.6. Now, instead of 3 inches round, it has been found that the H atom is about 2.24 inches in diameter because 2.24 x 9.2 dia = 20.6 inches apart.

How large are the O atoms by comparison? One inch. What was said above, "Where 1-inch balls are 8.2 diameters apart, they are 8.2 inches apart, and where 4-inch balls are 9.2 diameters apart, they are 36.8 inches apart," can be changed to: "Where 1-inch balls are 8.2 diameters apart, they are 8.2 inches apart, and where 2.24-inch balls are 9.2 diameters apart, they are 20.6 inches apart." That 2.24 is not written in stone, because I don't know the true merger depths that were assigned above. Someone would need to spend a lot of passionate time seeking the perfect reality. This math is as per the H atom found 9.2 diameters apart when 10-percent merged; the 2.24 becomes larger if the merger is made deeper.

In order for the H atom to become less than 2.24 inches, the diameters-apart for O atoms needs to be less than the 8.2 used for it in the math. That in turn requires deeper merger of liquid O atoms than the 20 percent I've used. Nobody can tell us how deeply they are merged. The stupids who accept orbiting electrons like cookies and tea don't allow them to merge at all as liquid particles.

If we change the 8.2 to 7.5, the H atom comes out to 2.0 inches in diameter. I asked for the result when 25-percent merged instead of 20: "with back-to-back balls merged and therefore .75 diameter apart c-to-c, but then spread into a volume 1090 times more voluminous, how many diameters apart will they then be?" The response is 7.72. The H atom's diameter now becomes: 7.7 x 2.51 / 9.2 = 2.1 inches.

That's how sure or unsure I can be as to how much larger the H atom is as compared to the O atom, by this math method. The better method is if someone can find the percentage of atomic repulsion versus heat-particle push, per degree of temperature increase. If we can know that atomic repulsion in the H atom is 60 percent of the heat-push force, we can then know that heat-particle push accounts for 40-percent, in which case the H atom's cross-sectional area is: 16 x .4 = 6.4 times larger than for the O atom. As the cross section is 12.56 square inches when the H atom has 16 times more cross section than the 1-inch O atom, it makes the diameter of the H atom work out to: 12.56 / 6.4 = 1.96 inches.

By the way, AI may be reporting, not the true weight of H liquid at its boiling point, but the ideal-gas math that aligns with their view of the number of H atoms in an STP gas combined with their claims that liquid H atoms are not merged at all. AI quotes someone as saying: "Liquid hydrogen atoms, typically found as diatomic molecules (H2) in a condensed state, are separated by a distance that is roughly 2 to 3 times the diameter of a single hydrogen atom." To me, this is ludicrous, but who of you can cool things to 20 degrees above absolute zero to check the true weight of H liquid?

The reasoning in the quote above may be due to hydrogen liquid's assigned weight of 14.2 times lighter than water liquid at 0 C, and so not knowing how to rectify this super-lightness, they conclude that liquid H atoms don't make contact. The irony is that the only way to begin to explain that 14.2 difference is by mergers.

As the H atom (and H gas) weighs 9 times less than water (and water gas), an over-simplification is that liquid H should weigh about 9 times less than liquid water, or about 110 grams instead of the 70 I used in the math above to get the 9.2 diameters. The weight difference depends on how deeply merged the two are, and so as there is good evidence that water molecules are merged deeper than any other liquid, the weight of 110 above will move toward the 70 by an unknown amount.

Therefore, until I can get a true H-liquid weight, the 9.2 diameters-apart I obtained for H atoms above has got to be wrong. But, I'll continue to use it for lack of anything else at this early time. Make adjustments as you think necessary. The diameters-apart will become larger than the 9.2 as the H liquid becomes heavier per liter. Keep in mind too that water is a molecule while hydrogen has a lone atom so that the math above requires multiple adjustments. I don't yet know how to make those adjustments, but they may not be needed if the true size of the H atom can be obtained by another method, by the method I'm in the process of showing you.

AI reports: "At 100 °C and standard atmospheric pressure, hydrogen atoms are approximately 34 diameters apart center-to-center." I assume they need that high number because they make the H atom the wee-weeist of all atoms. They have everything backward in the atomic size department; they put the biggest for the smallest and the smallest for the biggest. Trying to rectify things in such a confusing mess makes me the stupid just for trying in vain. But, the more confusing the jungle, the more satisfying when I see daylight.

I don't know by what logic the H and O atoms should have the same diameters apart (if indeed they do) unless the combined heat-particle push and atomic repulsion arranges it according to their sizes. It's feasible. In other words, regardless of the atom's specific diameter or proportional cross-section, every part of one atom's cross section tends to increase in heat-push by the same rate, per degree of temperature, as all other atoms, predicting some type of uniformity, but only because all atoms weigh the same. If they all increased in the same heat-push per given area, but were at variance in weight, they would not all keep some type of uniformity.

I'm not saying that O atoms and H atoms have the same heat-push force regardless of size, but that each piece of them can. Each piece upon the giant H atom that has the area of the O atom gets as much total heat-push (atomic repulsion included) as the full O atom. Where the O atom is smaller, it gets less overall heat-push, yet the uniformity might cause all atoms to space out by the same number of diameters. Where H atoms spread in to the number diameters by midway between 8.2 and 9.2, perhaps O atoms spread out by 8.7 diameters too (when the two gases are at the same temperature.)

I asked: "if a pack of 1-inch balls receive electromagnetic repulsion force equal to the same force upon a pack of 2-inch balls, will both sets of balls spread out by the same number of their diameters?" I hope AI is not merely being kind to me when responding: "In a standard physical scenario where the balls are charged to the same electric potential, the answer is Yes — both sets of balls will spread out by the same number of their diameters." This "little" thing could go a long way to explaining Gay-Lussac's realization that gases tend to react in full volumes.

I now need to redo the math where both O and H atoms have the same diameters apart: 8.7 x 2.51 / 8.7 = 2.51 inches in diameter. That was easy. But is it correct? A keener mind than mine is needed here.

The predicament of molecules is that they don't all weigh the same, and so the same uniformity, if it exists with atoms, shouldn't apply to them. Avogadro and Lussac would have done much better had they confessed what should have been obvious from Galileo: all atoms weigh the same.

The logic is that a 2-inch ball under x amount of repulsion force receives twice the out-spreading force as compared to a 1-inch ball under x amount of repulsion force, wherefore a pack of 2-inch balls will be pushed away twice as far such that they move apart the same number of diameters and a pack of 1-inch balls. If the 2-inch balls move 10 inches apart, that 5 diameters apart, and where the 1-inch balls move 5 inches apart, that's 5 diameters apart for them too. If this is how atoms behave at the same temperature, due to receiving the same heat-push force, then H atoms will move apart by the same number of diameters as O atoms do. See that?

Therefore, where it's known from the relative weights of H and O gas that O atoms are 2.51 times further apart, it seems full-proof that the H atom is 2.51 inches in diameter i.e. 2.51 times greater than the 1-inch O atom. The H atoms get 2.51 times more heat-push such that they move apart 2.51 times as far as O atoms do.

Believe it or not, if we can depend on this uniform out-spreading principle, we can also know the exact merger depths of all atoms, if the true merger depth of just one atom can be found! All other atoms will fall in line. The math I did above would be correct only when ALL atoms have the same diameter-apart! I'll soon re-do that math to see if it can work. If not, this uniform theory can't be quite correct. It could be close, but not perfect due to some monkey wrench in the machinery.

If you want AI to report the merger depths instead of the diameters-apart, change this question, "with back-to-back balls merged and therefore .859 diameter apart c-to-c, but then spread into a volume 1077 times more voluminous, how many diameters apart will they then be," to: "how much less than 1 diameter apart, center-to-center, must a pack of balls be if they grow to 8.8 diameters apart when expanded to 1077 times their initial volume?" It responded with: "1.0 diameter - .8585 diameter = .1415 merger depth. The .8585 response is my .859 in the first question, and the first question answers, 8.8, same as I have in the second question. All's good.

The thing is, the 2.24-inch diameter found above for the H atom has a cross section of 3.94 square inches, exactly 5.0 times more than the cross section of a 1-inch O atom. That's interesting, but why should a 5 figure be magical? I may need to get used to viewing the water molecule now as 8, 1-inch O atoms merged evenly around roughly an H atom in the ballpark of 2.25-2.5 inches.

There are 15.9 square inches of surface area on a 2.25-inch H atom, and the 8 O atoms merged only partially need a lot less than 8 square inches of it. There's yet room for 8 more squishers-in for forming hydrogen peroxide. I hope you like my Greek. With O atoms merged 25 percent into this 2.55-inch H atom, the diameter will be 3.75 inches, though two molecules in contact as liquid molecules don't necessarily extend 7.5 inches (twice 3.75) across, because the O atoms of one molecule can fit between the O atoms of the other such that the H atoms could make contact, or nearly so, meaning that two water molecules could be as small as 2.25 x 2 = 4.5 inches across.

I asked: "if 8 1-inch balls sat over the entire surface of a 2.25-inch ball, and each were merged 25-percent of their diameter into the large ball, how far apart would their edges be if evenly spread?" The response is .6 inches of distance, but there is significantly more distance between their edges at their farthest points. There's room for 16.

Where nitrogen liquid weighs 808 grams per liter at boiling point, and as a gas at 100 C weighs .908 grams, volume has increased by 808 / .908 = about 900 times from one condition to the other. Google's AI tells that N atoms will become 7.45 diameters apart if we start them 20-percent merged as liquid. That sounds about right where they are a little larger than O atoms, for in being larger, they get more heat-push such as to take less distance-apart in spacing, i.e. which requires the N atom to have less diameters-apart than O atoms. The latter worked out to be 8.2 diameters apart starting with 25-percent merger.

The O atoms at 1-inch diameter reach 8.2 inches after 8.2 diameters, and where I figure that N atoms are a little less than 1-inch in diameter, I'll use 1.08 x 7.45 = 8.05 inches. Therefore, the 7.45 for the N atom looks about bang-on.

My Model's Explanation for Avogadrian Suspicions

At the same temperature and pressure, they tell us that steam weighs nine times more than hydrogen gas. Plus, the HO8 molecule weighs nine times more than one lone H atom, because all atoms weigh the same. The only way for the steam to weigh nine times more, where its molecule happens to weigh nine times more too, is for the centers of the steam molecules to be identical in distance to the centers of the H atoms.

Where 8 O atoms bring 8 times more weight to each H atom in forming a water molecule, and 1/32 the out-spreading force of an H atom, it follows that the H atom has 1/8 the weight and therefore must have 32 times the out-spreading force in order that water molecules take up the same spacing as H atoms in an H gas. There is no extra out-spreading or in-spreading force loaned from the 8 O atoms to the H atom, and so the water molecules end up with the same spacing that the H atoms had prior to merger.

Atomic weight is a downward force that causes in-spreading of atoms in all directions. However, I'd like to say up front that, while I was recently entertaining atomic weight as a formidable in-spreading force, what I'm about to show suggests that the weights can be ignored. Last week, I made a colossal oversight when deciding that the O atoms on a water molecule are countered significantly by their weight of 8 atoms. I now recant: out-spreading alone is the concern for understanding molecular spacing.

Where I said that the 8 O atoms grant 1/32 the out-spreading (as an H atom), I can't just pull the latter number out of my hat. I need a reason for it, which I'll show below. In this picture, where and if the weight of 8 atoms counter their 1/32 out-spreading force, 1 atom of weight (of any atom) is countered by (or counters) 8 times less than 1/32, or 1/256ths of the out-spreading force. Now you see why weight is almost of zero concern.

As we will see, the cross-sectional area creating the total out-spreading is equivalent to .5 O atom (if AI gave me the right number). As 16 O atoms equals the outspreading force of 1 H atom, 1 O atom performs 1/16 of it, and so the .5 O atom is 1/32nd. This is actually incredible. It comes as one big math surprise because I did not choose the 1/32 in order to get that surprise. I'll show you how it was obtained.

I need to show why 8 O atoms lend the H atom a mere 1/32 the out-spreading as compared to the out-spreading of an H atom, in spite of 8 O atoms having 1/2 the cross-sectional area of 1 H atom, i.e. which amounts to 1/2 the out-spreading force. When 8 O atoms are merged into an H atom, they contribute 32 times less force than is available from the full cross-sectional area of all 8 O atoms. It makes good sense because any O atom that is in front of, or behind, an H atom, adds zero cross section to it. Only the parts of O atoms that protrude (stick out) from the H atom, as seen from any one direction, add cross section to the H atom.

The 1/32 figure is what justifies (requires) the H atom to exert 32 times the out-spreading force of the O atom. That is, once we find the 1/32, we have also obtained the 32, because water molecules have the same spacing as H atoms in its gas. If they did not have the same spacing, we could not pair the 32 with the 1/32. They need to balance out (counter each other exactly) when the same spacing is maintained.

[I did this section before finding above that the H atom is likely only 2.6 inches round.] After several re-wordings, I finally got AI to understand me, and it actually gave an answer. The final question was:

Do you understand the difference between balls distributed over the 50-sq in surface of a sphere versus around its perimeter? I don't want the latter, but you keep giving it to me. I don't want balls sitting on a circle's circumference.

In a two-dimensional drawing of 8 one-inch balls equally dispersed over the entire 50 sq in SURFACE (not around the perimeter) of a 4-inch ball, and each merged 20 percent into the larger ball, how much cross section will the total 8 balls possess that stick out past the 4-inch circle of the larger ball, IF VIEWED FROM ONE DIRECTION where most of the balls do not stick out past the large circle? Why do your drawings not show evenly-dispersed balls?

Although the drawing that comes with the response shows the eight balls not evenly dispersed, the response does say "equally dispersed":

To answer your question directly: if eight 1-inch balls are evenly dispersed over the entire surface of a 4-inch ball (merged 20% by depth), and viewed from a single direction where the dispersion minimizes protrusion, the total protruding cross-sectional area is approximately 0.38 square inches.

That seems a little too small, but I'll need to go with it because I can't do the calculation myself in one lifetime. It had given me 1.56 square inches without the 20-percent merger, and so .38 is possibly correct. The total cross section of the 8 balls is .785 x 8 = 6.28 square inches. When we divide that by AI's .38, the outcome is 16.5 times less than all 8 balls, or about .5 ball of extra cross section. As it takes the combined cross sections of 16 O atoms at one-inch round to equal the cross-sectional area of an H atom 4 inches round, the .5 atom's cross section is 32 times smaller, and therefore lends 32 times less out-spreading force than is granted by the H atom. That's how I got the 1/32 in the math above.

It's tending to confirm that the math-method is correct, and moreover might confirm that the H atom is spherical, otherwise the .38 square inch in the response becomes something else.

The day after writing that, I found the reason further above for shrinking the H atom to 2.6 inches. I'm torn because the .38 above seems perfectly expected, making a compelling case for a 4-inch H atom. None of that AI tabulation, however, considered atomic repulsion, but only the heat-particle push on the cross section. Besides, as AI's number can't be fully trusted, I think I'll fully ignore that exercise.

AI gave me more problems with this task, and can we blame it? It's not an easy task. I'm surprised it tackles it at all. I said:

You once gave as little as 1.15, then .44, and now 1.85, for the task below. Are you sure you understand me. Can't you supply a proper drawing showing the spread of the 8 balls properly? Understand the difference between balls distributed over the surface of a sphere versus around its perimeter. I don't want the latter, but you keep giving it to me. I don't want balls sitting on a circle's circumference. Where 8, one-inch balls are equally dispersed over the entire 21.2 sq in SURFACE of a 2.6-inch ball, and each merged 20 percent into the larger ball, how much cross-sectional area, of the 8 balls, will stick out past the 2.6-inch circle, IF VIEWED FROM EVERY DIRECTION but then averaged out as if from one direction.

Response: "The average cross-sectional area, viewed from all directions, of 8, one-inch diameter balls on a 2.6-inch diameter surface, where 20% of the small ball's radius is merged inside, is approximately 1.14 sq in" Whether that's correct or not, I'll go through the math ropes to share what can be accomplished, though once again I'm not including the atomic-repulsion heat-push. We do 1.14 / .785 = 1.45 O atoms of total cross-sectional, stick-out area. Aren't you glad I'm not using science Latin?

I don't know what to do with the 1.45 stick-outs where the H atom has a 2.6-inch diameter, no longer 16 times less than the cross section of an H atom. The 2.6-inch H atom yet exerts 16 times more out-spreading force than the O atom, but only .42 percent of the 16, or 6.7 times, was found above to be from its new cross section. The remaining 9.3 times is from atomic repulsion. There is a secret to be discovered in these numbers; I just don't have a fresh-enough brain at this time to know what it is, and I can't trust the AI's response.

As the cross section of a 2.6-inch sphere is 5.3 square inches, the water molecule has a cross section of 5.3 + 1.14 = 6.4 square inches, about the same as the 6.28 for a full 8 O atoms combined, suggesting that the water molecule gets as much heat-particle push as half an H atom gets in both heat-particle and atomic-repulsion force. The water molecule therefore gets atomic repulsion force calculated as: (1.14 / 5.3) + 1 = 1.2 times what a full H atom receives. This type of math is for hairy scientists only, not potato-chip lovers. One can do that math easier as 6.5 / 5.3 = 1.2. The water molecule gets 1.2 times more heat-particle push, and MORE than 1.2 times more atomic repulsion. Does that sound fair, Mr. Hairy?

Sixteen O atoms yet produce 16 times the heat-push (total out-spreading force) as one H atom, regardless of the latter's size. The atom represented by the 5.3 figure alone produces 16 times more out-spreading force than the O atom, and the 1.45 O atoms represented by the 1.14 figure add another 1.45 force in heat-particle push for a sub-total of 17.45 heat-push versus 16 for the lone H atom. Yet the 17.45 does not include the added atomic-repulsion force from the 8 O atoms.

For the life of me, I can't figure out why 6.4 / 5.3 x 16 does not equal 17.45, but gets 19.3 instead. I've spent an hour on it, but no luck finding the reason. The extra above 16 out-spreading force, in whatever extra out-spreading force is obtained by the water molecule, ought to be equally countered by the in-spreading force of the 8 O atoms...such that the water molecule and the H atom take up equal spacing at STP. That's how I'm seeing things.

I've suggested that the ammonia molecule, what they name, NH3, is really a central H atom with 5 N atoms merged into it. I'm reading that ammonia gas at 100 C weighs .55 g/l, while the liquid at its boiling point (-33) weighs 680 grams per liter. At 100 C, it would expand to 680 / .55 = 1,236 times its volume. AI says that these atoms, beginning at .9 diameter apart (10-percent merged) as liquid, will expand to 9.7 diameters apart as a gas.

While ammonia expands 1,236 times, water (steam) expands 1,600 times, both at 100 C. Why? It gives appearances that the water molecule gets significantly more heat-push, for there's no other conceivable way to explain atomic spreading (where kineticism has been discredited). The 1,236 versus 1,600 is about expected where the heat-push on ammonia's 5 N atoms is expected to be less than the heat-push on the water molecule's 8 O atoms.

When I calculate the cross section of either the ammonia or water molecule, I don't include any atomic portions in front of or behind the H atom, as seen from any direction. I include only the portions sticking out past the circle of the H atom, which is the proper way to tabulate total cross-sectional area for the whole molecule. I don't include the portions in front of and behind the H atom because the cross section of the H atom had already been included in the tabulation.

However, I cannot exclude the atomic repulsion of the N and O atoms in front of and behind the H atom. That force is FULLY active from every atom, and is expected to radiate equally in all directions from a spherical molecule.

Where N and O atoms are roughly the same size, they EACH will effect roughly the same heat-push, but where the ammonia molecule has only 5 atoms, it's going to get little more than 5 / 8 = .625 as much. There's then a question as to why 1236 / 1600 does not equal .625, but rather .77. In order to get .625, the 1,600 needs to become about 1,980. The lower 1,600 figure shows that the spread of the water molecule is not as great as the difference in 5-versus-8 atoms requires, and so the solution here is that the water molecule gets 8-versus-5 in-spreading force to counter some of the ammonia's out-spreading advantage. The in-spreading doesn't counter it all, otherwise the figure for the O atoms would itself become 1,236 too.

The number distance between 1,236 and 1,980 is 744, and the number distance between 1,980 and 1,600 is 380. See anything suspicious? The latter number is almost half of 744. One might conclude that, in this case, anyway, each O atom of in-spreading force is equal to half the out-spreading force of an O atom, which is a lot more in-spreading than I thought it would be by comparison.

In other words, above and beyond the net-out-spreading force of the ammonia molecule, the water molecule has 8 x 1 = 8 units of out-spreading force and 8 x .5 on in-spreading force. If it only had out-spreading force, it would increase its gas volume to 1,980 times (illustratively speaking) the volume of a liter of water liquid. Instead, it spreads out only 1,600 times as much (to 1,600 liters, that is).

These volumes of out-spreading under discussion have nothing to do with the merger depths I assign. These volumes of out-spreading can be deemed facts in so much as I'm being fed the correct liquid and gas weights. This math method is full-proof, only the weights may be a little off.

If you think that these numbers, in combination with my interpretation of these ammonia and water molecules, looks compelling, then you have reason to trash their H2O and NH3 molecule, and go with the HN5 and HO8 instead.

We then might figure that the 5 N atoms upon an H atom gives the ammonia molecule an out-spreading advantage of 5 atoms of out-spreading minus 5 atoms of in-spreading, but where the in-spreading may again be only half as strong. The out-spreading I assigned the H atom above was 1,077, and comparing it to the 1,236 of the ammonia molecule, the difference is about 160. We could hypothesize that the 5 N atoms give the ammonia molecule an out-spreading advantage of 320 force units over lone H atoms in an H gas, but that the 5 atoms of weight reduce it to half, 160. But keep in mind that the 1,077 figure is unreliable because it was obtained from the H-liquid weight at just 20 degrees above absolute zero, where it's viewed as a semi-gas.

The above suggests that ammonia atoms are more sparse than H atoms in an H gas, and that water molecules are more sparse than ammonia molecules, but this is problematic for the necessity of H atoms having the same spacing as water molecules. With use of the 1,077 figure, H atoms turned out 9.2 diameters apart, and ammonia atoms turned out 9.7 diameters apart, which again suggests that ammonia molecules are more sparse than H atoms. Water molecules worked out to 8.8 diameters apart, but that was with 25-percent merger; they would become 9.3 diameters apart with 20-percent merger (into each other as liquid). The 9.7 becomes 9.1 with 15-percent merger of ammonia atoms. It's very do-able to get these three particles spaced to whatever the reality is. Hydrogen atoms are permitted to have the greatest diameters-apart because they might have the smallest diameter of the three.

The point was that the 5-versus-8 atoms predicts that the ammonia out-spreading should be 5/8ths of the 744 obtained above for the 8 O atoms, which is 465, yet we saw only 320. It's close, but I don't know why it's not more accurate.

The lower out-spreading for the ammonia molecules jibes with the fact that ammonia gas weighs 8.5 times more than H gas, while steam weighs 9 times more than H gas, when all are at the same temperature of 100 C. With the gas weighing a little less, the ammonia molecules are not necessarily predicted to be a little further apart than water molecules, for we need to include the lighter weight of the ammonia molecule such that they can be the factor in making ammonia gas a little lighter in spite of the water molecules having greater out-spreading force.

The out-spreading war is won by water molecules by a score of 1,600 to 1,236, meaning water molecules are more sparse per liter, yet the lighter weight of the ammonia molecule wins the gas-weight war if the winner is the lighter gas.

Quoting two or three sources, AI says: "The formation of ammonia (NH3) from nitrogen and hydrogen is an exothermic process, releasing heat. The net energy gain, defined as the heat released minus the energy needed to activate the reaction, is typically around 46–46.1 kJ mole of NH3 formed." That's 46 / 22.4 liters/mole = 2.05 kJ/liter.

Then, "The combustion of hydrogen and oxygen to form water vapor (steam) releases approximately 10.8 of energy per liter of steam formed (measured at 1 atmosphere pressure and 100 C)." As this process releases about 5.25 times the heat that ammonia formation releases, the O atoms are merged deeper into an H atom than the N atoms into an H atom. On this aspect, the N atoms get some out-spreading advantage that would not exist if both they and O atoms were merged equally (by total atomic volume) into their respective H atoms.

As there are 8 / 5 = 1.6 times as many O atoms for molecule, heat from H-O combustion would be 1.6 times more if both the N and O atoms were equally merged by volume. As 5.25 times the heat is produced, O atoms are sunk significantly deeper. I'm showing the real scientists, if they are reading, how to do real science to make baboons of the baboons.

Discover Relative Numbers of Molecules Via Merger Depths

Chlorine gas at 100 C weighs 2.32 g/l, and 1,560 g/l for the liquid at its boiling point of -33 C. The volume increase is 1560 / 2.32 = 672. In being a heavy gas, chlorine has a lot less volume spread than water molecules (1,600 versus 672), which seems very problematic for my guess that the chlorine molecule is 34 atoms merged into a central H atom.

Chlorine gas weighs 3.9 times more than steam. Chlorine liquid weighs 1.56 more than water, meaning there are 1.56 times as many atoms per liquid liter. Then, the numbers above, 1,600 / 672 = 2.38 means that the chlorine gas particles spread out 2.38 times less, not to be confused with 2.38 times more numerous molecules per liter. There are 1.56 times as many chlorine atoms per liter of liquid, and so when that material spreads out 2.38 times as much, there will be 3.7 times as many atoms in the gas.

However, as there are instead 3.9 times as many atoms, there must be a problem with the 1,600 and/or the 672 figure(s). They are not my figures, but the figures dictated by the AI-stated differences of the weights of the liquids versus the gases. I prefer to use 2.5 rather than 2.38 because 3.9 / 1.56 = 2.5.

As water molecules have 9 atoms each, there are 9 x 3.9 = 35 atoms per liter of chlorine gas. If water and chlorine molecules in their respective liquids were both unmerged, only making contact, the chlorine molecules would be known to weigh 1.56 times more. If the water molecules are merged deeper, as they should be, the chlorine molecule goes up in weight.

Plus, the more water molecules there are in its liquid as compared to the chlorine molecules in its liquid, the more the water molecules will spread out to create gas volume in comparison to the out-spreading of chlorine molecules. Thus, the greater numbers of water molecules can partially explain the greater out-spreading.

I asked: "if a pack balls [atoms] are merged by 25 percent their diameters, how much volume will they lose?" Response: "The balls will lose 57.81% of their total volume" That's a surprising GAIN of material.

Yes, gain. When a molecule in a liter volume shrinks by 58 percent via merger, it allows additional material to become a part of the liter to fill the void. Each molecule ends up weighing 1.58 times what it does unmerged. This is a new realization for me. If it seems surprising to find that 58-percent of the material goes into "hiding" with only 25 percent mergers, it must be due to each molecule merged in several spots (from all directions) by several neighboring molecules. Each one gets merged by more than 25 percent. The latter number can be viewed as 12.5 precent per atom per two-atom merger.

To verify I'm not making a mistake, I asked, "if a pack of balls in any volume weigh 1 unit, and are merged by 25 percent their diameters, how much will they weigh when merged?" The response is .42. That is, they lose 58 percent their weight and end up at 42 percent of the original. They lose 58 percent their volume and need an additional 58 percent their volume and weight to bring the volume back to the original.

However, for liquids, we don't want just the reduction in molecule volume, but also the spaces between them. After being told that back-to-back balls merged 25-percent have their spaces reduced by 87.5 percent if the balls begin and end as tightly packed as possible, I asked: "When balls back-to-back balls in 3 dimensions, and packed as tightly as possible, are merged by 25 percent of their diameters, when you say they lose 57.8 percent of their volume, and the spaces between them lose 87.5 percent of their volume after merger, because balls are yet again packed as tightly as possible, what is the TOTAL volume loss balls and spaces together?" The answer is "The total volume loss of the balls and the spaces together is 65.51%."

In a response, AI showed how it makes the calculation for 25-percent merger. It merely multiplies .75 three times (cubes it) to get .422 volume left over, wherefore the volume loss is 1 - 42.2 = .578 = 57.8 percent. For your information. AI then said that the spaces lose more volume by percentage during merger than the ball-merger amount.

I re-phrased: "when 42.2 percent of volume remains of back-to-back balls merged by 25 percent, that's the volume lost by the balls only. How much total volume is lost amongst the balls, including the spaces between them, where the balls are packed as tightly as possible?" This time: "The total volume lost amongst the balls, including the interstitial spaces between them, is 68.75% of the original total volume. This accounts for both the 57.8% reduction of the spheres' own volume and the collapse of the empty space between them as they merge." Why is that 68.75 not the same as the 65.5 above?

I did my own calculations, knowing that the closest-packed balls have 74-versus-26 percent ball-versus-space volume. Where 87.5-percent of the spaces are eradicated, the math is .26 x .875 = .23. As 74 / 26 = .35, meaning there is .35 space versus .65 molecule, we need to do: .23 x .35 = .079, meaning 7.9 percent in addition to 57.8 percent, which is 65.8 percent, roughly the 65.5 AI gave.

However, AI has wasted a good day giving me conflicting responses galore, like it's a lunatic. When I asked the same question above with 30-percent, it gave me the same 65.7 percent answer as only the balls alone lose. I asked, "When balls back-to-back balls in 3 dimensions, and packed as tightly as possible, are merged by 30 percent of their diameters, when you say the balls alone lose 65.7 percent of their volume, what is the TOTAL volume loss balls and spaces included when the balls are again as tightly packed as possible after merger?" This time:

Interestingly, while the individual balls lose 65.7 of their volume due to the overlapping "caps" being removed (as you noted), the overall container size required for the stack also shrinks by exactly 65.7% because the dimensions of the lattice scale directly with the distance between the sphere centers.

The way I translate that is: the spaces between balls likewise shrink by 65.7 percent. I therefore asked the lunatic: "When balls are back-to-back in 3 dimensions, and packed as tightly as possible, and after they are merged by any percentage, do the spaces between the balls shrink by the same volume as the balls lose in volume, if, after merger, the balls are again as tightly packed as possible?" The lunatic responded:

No, the spaces do not shrink by the same volume. In fact, if the balls shrink in size but remain in the tightest possible packing, the total volume of the empty spaces actually decreases by less than the volume lost by the balls

AI can be a lunatic because it has different sources who are not God, who often conflict and/or make errors in their claims. The only good news is that, whether I use 57.8 percent versus 65.5, it doesn't make a big difference in the number of chlorine molecules versus water molecules. I'm going to use 65.5 percent, but if you prefer the 57.8, know that AI gives 65.7 percent with mergers of 30 percent, meaning that you can view/treat my 25-percent merger below as 30 percent.

When the water content increases by 65 percent (by 1.655) its weight by molecular merger, where molecules are partially superimposed on others, each place/space the size of a molecule then weighs 9 atoms x 1.655 = 14.9 atoms. Therefore, we need to view every water molecule as weighing 14.9 instead of 9.0.

If the weight of the water is represented by 14.85 units/atoms of weight, the chlorine liquid, with its atoms at an as-yet unknown merger depth, weighs 14.9 x 1.56 = 23.2 units/atoms.

To discover something, we first realize that, if the chlorine molecule has 23.2 atoms, there would be the same number of molecules per either liter. As I'm tentatively viewing the unmerged chlorine molecule as 1 H atom surrounded by 34 mystery atoms, the math is 23.2 / 35 rather than vice-versa, to discover that, if the chlorine had zero merger of its molecules, there would be .66 as many molecules as compared to the 25-percent-merged water molecules. Each chlorine molecule must weigh more than 35 atoms when merged, wherefore the math above tells that there will be less than .66 times the molecules. This assumes that the chlorine particle is a molecule. If it's a lone atom, I don't think these numbers apply.

When the two liquids become gases, the 14.9 atoms per water molecule spread out by 2.5 times (above) more volume than the chlorine molecules. The spread of the water (into 1,600 liters) is represented by 1 unit of weight, and if the chlorine spread out by the same volume, it would weigh 1.56 unit by comparison. But as the chlorine spreads 2.38 times less, it instead weighs 2.4 x 1.56 = 3.9 times more than the steam. As 14.9 atoms can represent the full volume of steam, where there would be 23.2 chlorine atoms if it too spread out as much as the steam does, we find that the chlorine gas, in spreading only .4 as far (2.5 times less), has 23.2 x 2.5 = 58 atoms versus 14.9 for the steam. It checks out where 58 / 14.9 = 3.9.

The following is a good reason to suspect that liquid chlorine molecules are not deeply merged: "At normal atmospheric pressure and room temperature, liquid chlorine turns into gas at a very high rate." As chlorine is easily dislodged by sunlight, they could be merged as little as 5 percent or less. AI says: "If a pack of balls is merged by 5% of their diameters, the total volume of the pack will decrease by 18%." It means that, if chlorine molecules are merged by 5 percent, they would lose 18 percent their volume and weight when brought from kissing to merged, and consequently we assign them 1.18 times more weight when restoring them to a liter of volume.

Their merged weight is then 35 x 1.18 = 41.3 atoms, to be compared with the 14.9 atoms of weight for the merged water molecules, a difference of 2.77 times the weight in favor of the chlorine molecules. However, that math and 2.77 figure assumes both liquids having the same number of molecules, because it's one molecule compared to another. A zillion molecules in each liquid liter would yet have the 14.9 versus 41.3. It doesn't yet tell us the true difference in number of molecules.

As the chlorine liquid weighs only 1.56 times as much instead of 2.77, the apparent solution is that there are 1.56 / 2.77 = .56 times as many chlorine molecules in its liter. My math is checking out because I had said, "Each [kissing] chlorine molecule must weigh more than 35 atoms when merged, wherefore the math tells that there will be less than .66 times the molecules [when merged]." You just saw .56 the molecules when merged by 5 percent. The largest this number can be is .66, when the molecules only kiss. The deeper we merge the chlorine molecules beyond 5 percent, the lower the number goes from .56.

How do I know that the weight reduction to 1.56 times needs to be due to fewer chlorine molecules rather than fewer water molecules? Because, fewer water molecules makes the 2.77 larger.

We can attack this another way where the two unmerged molecules have a weight difference of 3.9 times in favor of the chlorine molecule. The latter then becomes 3.9 / 1.655 = 2.35 times heavier than a merged 25-percent merged water molecule. When the chlorine molecule is itself merged by 5 percent, it's 3.9 x 1.18 = 4.6 times heavier than an unmerged water molecule, and 4.6 / 1.655 = 2.77 times heavier than a 25-percent merged water molecule. We saw the 2.77 above. That's just the difference in molecular weight when both are merged; if there were the same number of molecules per liquid, the chlorine liquid would weigh 2.77 times more. But as the liquid weighs only 1.56 times more, there must be a reduction in the number of chlorine molecules to .56 times.

Doesn't it make sense that, if there are fewer chlorine atoms per liquid comparison, they should be larger than water molecules? On the one hand, yes, but only if they both have the same merger depth. Where there is 1.655 / 1.18 = 1.4 times more merger in favor of water molecules, the number of water molecules grows in comparison such that there needs to be as many more chlorine atoms, to maintain the same weight difference, and thus the latter need reduction in size accordingly. (I'm using 1.655, more accurate than 1.65).

Per each merged water molecule weighing 14.9 atoms, each merged chlorine molecule weighs 14.9 x 1.56 = 23.24 atoms, but with the 1.56 in that math, the 23.24 assumes the same number of molecules in both liquids. When we divide 23.24 by .56 to see how many atoms of weight the merged chlorine molecule really has, we get 41.5, almost the 41.3 above for chlorine merged 5 percent. Thus, it seems that this math is capable of predicting the merger depth of chlorine molecules according to the prior merger depth proposed for water molecules. That is, when using 25-percent merger for water, the chlorine molecule works out to very-slightly more than 5-percent merged (as much as 41.5 is larger than 41.3).

Therefore, we could know the relative number of molecules if only we could know how deeply both are merged. Thus far, I've seen nothing in the numbers to indicate the relative merger depths. But there is something else we could look at.

AI says that, if particles are merged by 5 percent their diameters, and then expand into a gas 672 times larger than the liquid volume, the (chlorine) particles will end up 8.3 diameters apart. On the other hand, when particles are merged 25 percent, and expanded by 1,600 times the volume, they (water molecules) end up 8.77 diameters apart (or 8.2 at 30-percent merger). We found .58 times fewer chlorine molecules per either liquid liter, and so I think we can now find the relative sizes of the molecules.

The question to ask for 25 percent merger is: "with back-to-back balls merged and therefore .75 diameter apart c-to-c, but then spread into a volume 1674 times more voluminous, how many diameters apart will they then be?"

As .56 times the chlorine molecules are increased by 2.5 the number, due to spreading out to 2.5 times less volume, they become .56 x 2.5 = 1.4 times more numerous than the water molecules in its expanded gas. Therefore, there are 140 molecules per 100 in their respective gases. I asked google: "In a 3-dimensional situation, how much larger or smaller are each of 140 balls 8.3 diameters apart compared to 100 balls 8.77 diameters apart, when the 140 fill a space 2.5 times smaller?"

Trust AI or not: "In this 3-dimensional scenario, the diameter of each ball in the second group (140 balls) is 0.696 times the diameter of the balls in the first group (100 balls)." I'll use .7 times, but then I asked the following question that gives a different answer: "how smaller must balls be when they are 8.3 diameters apart c-to-c while taking 2.5 times less volume than 1-inch balls 8.77 diameters apart?" The response is now .67 inches, or 6.7 times smaller instead of the .7 above. I asked AI why it has this difference, but I couldn't understand it.

A sphere with .7 the diameter has a cross section .49 less than a 1-inch ball. Therefore, I would say that this tentative chlorine molecule has roughly half the out-spreading force of a water molecule, minus however much more in-spreading force it has than a water molecule. Where it weighs 35 versus the 9 atoms, the chlorine molecule gets plenty more in-spreading.

The next section, to my great surprise, allows the chlorine atom to have a central H atom. Check it out.

A Nickel For Your Thoughts?

[This section was written before the math in the last section killed my suggested definition of chlorine and hydrogen-chloride molecules, both having the same mystery atoms merged into a central H atom. The theory was killed because I had chlorine's mystery atoms about .75 inches in diameter on the brain, making the chlorine molecule at least as large as an H atom where the latter was viewed as 4 inches in diameters. I can't remember how the .75 diameter was found, but I'm sure it used the 4-inch H atom as basis.]

The math had suggested that the mystery atom in H-chloride and chlorine, in having a diameter of .75 that of the O atom (1-inch round), has a cross section of .44 square inches. On that basis, the mystery atom needs to have 28.5 times less cross section than an H atom, for the latter (at 4 inches diameter) has 12.56 square inches, which, when divided by 28.5, gets the .44 above. I am therefore looking for an atomic gas weighing 28.5 times more than H gas. Google AI tells me that a nickel gas is 29.1 times heavier than H gas, cobalt gas 29.2 times, and iron gas 27.67 times (heavier than H gas), when all are at the same temperature.

It behooved me to ask whether there are natural chlorine-nickel, chlorine-iron, or chlorine-cobalt mixtures, and, yes there are, plenty from volcanic materials. "Volcanic Environments: Many naturally occurring metal chlorides, including those with iron, are found in fumaroles where hot gases interact with metals." "Yes, nickel chloride is considered a volcanic material...It is released into the atmosphere as a soluble aerosol or particulate during volcanic eruptions."

I'm reading, "Assuming a theoretical gaseous state for anhydrous nickel(II) chloride (NiCl2) at STP: the weight would be approximately 5.78 g/l" That's 64.2 times heavier than H gas (.09 g/l). They claim that: "1 mole [or liter] of nickel reacts with 1 mole [or liter] of diatomic chlorine [regular chlorine gas] to produce 1 mole of nickel(II) chloride."

Therefore, as nickel gas weighs 29.1 times more than H gas, it has 29 times as many atoms, and as chlorine gas weighs 34 or 35 times more than H gas (35 if it's 1 H and 34 mystery atoms), the total weight of nickel chloride is expected as 29.1 + 35 = 64.1, almost the 64.2 above. I'm therefore proposing that the chlorine "atom" is a central H atom studded with 34 nickel atoms that will not come out even in high heat. It can grab 29 more nickel atoms to become nickel chloride, but it can perhaps pick up more than 29 to become nickel chloride, whereas they report only 29 because they like to keep a public show of exactly one-and-one mixed liters. I'm reading: "Yes, nickel atoms have a strong affinity for each other."

Chlorine is greenish. "Many nickel(II) compounds are green...The most common green nickel compounds include nickel(II) chloride hexahydrate, nickel(II) sulfate, and nickel(II) hydroxide...Nickel(II) nitrate: An emerald-green,...Nickel(II) oxide: Specifically, the "green" form of this compound is a green crystalline solid..." From another AI offering: "Nickel corrosion products are often characterized by a bright lemon-green color." Looks like I may have found the mystery atom in chlorine and hydrogen chloride.

More: "Volcanic nickel chloride (NiCl2) is a highly soluble compound that readily dissolves in water. It is known to form green solutions of the metal aquo complex...Anhydrous nickel chloride (yellow) rapidly absorbs moisture from the air to form the green hexahydrate...It is formed by dissolving nickel in hydrochloric acid." The latter is hydrogen chloride in water. Nickle is thus attracted by hydrogen chloride, not surprising if the latter is made with nickel atoms.

So, while chlorine is named after "chloros," the Greek for "green," by what coincidence do we see all that green in nickel materials even when they do not include chlorine? It could appear that nickel atoms give chlorine its green color.

Where the nickel atom is .75 inches wide, it has a cross section of .44 square inches. The 2.6-inch H atom has a surface area of 21.2 square inches, which can fit 21.2 / .44 = 48 nickel atoms. There's no problem here, it can work. Saddled nickel atoms could then attract other nickel atoms on top of each other, but making the top layer so weakly held that sunlight can free enough of them to ignite H gas in their presence.

That was the end of this section, but I'm back to add comments fresh off of finishing the last section on chlorine molecules, which were found as .7 inches in diameter where water molecules are 1-inch in diameter by comparison. But now, with water molecules having a diameter equal to a 2.25-inch H atom surrounded by O atoms, it works out to 2.25 + (1 + 1) = 4.25 inches (doesn't include merger of O atoms into the H atom). That makes the chlorine atom's diameter 4.25 x .7 = 3.0 inches. There was ability to change the .7 to a little larger by changing merger depths of the molecules involved with the calculations.

Now that the H atom has been shrunk to 2.25 inches, nickel atoms are no longer in the range of .75 inches in diameter. They need to be re-sized. As nickel gas weighs 29 times more than H gas, the nickel atom has 29 times less cross section than the 4.0-square-inch cross section of the 2.25-inch H atom. That gives .138 square inches for the nickel atom, making it only .42 inches in diameter.

The diameter of a chlorine molecule, if stacked with a layer of 34 nickel atoms atop a central H atom, is 2.25 + (.42 + 2) = 3.1 inches, almost the 3.0 above! The reason that I was lamenting when writing the chlorine section is that I was envisioning chlorine's mystery atoms as .75 times as large as O atoms, but now that they can be only .42-ish, it works. Working out the .7-ish size of the chlorine molecules was apparently not in vain.

I asked: "If a ball 2.25 inches in diameter accommodates 34 .42-inch balls sitting on, and evenly spread upon, its surface, what is the nearest distance between the 34 balls?" AI responds with: "The nearest distance between the 34 balls is approximately 0.437 inches." There's lots of room such that one layer of nickel atoms is not a problem.

I repeat. Hydrogen gas burns with chlorine to form hydrogen chloride, which I expect to be a central H atom loaded with 17 mystery atoms i.e. possibly nickel atoms. The chloride gas weighs half as much as chlorine gas, and twice as much as water gas. Running electricity through hydrogen-chloride dissolved in water (it's an acid) frees H atoms from the hydrogen chloride, and released them into the air above the acid. No O atoms go out, meaning that water molecules are not being disassociated by the electricity.

The freed H atoms must therefore be leaving hydrogen-chloride molecules, wherefore each one leaving leaves 17 free, mystery atoms behind that are called "chlorine atoms," which I think could be nickel atoms if there's no way to remove them from the H atoms that has them captured. The free 17 load onto some hydrogen-chloride molecules, making molecules of an H atoms but with 34 atoms instead of 17.

I showed why they lie when telling us that chlorine gas can be disassociated with high heat to lone chlorine atoms from the diatomic chlorine molecules that they envision.

Moreover, AI cannot tell me how to make pure chlorine atoms aside from the electrolysis I mentioned above (which I'm interpreting as chlorine MOLECULES), or by using metals to attract chlorine atoms out of a substance to themselves, in which case metals could be attracting nickel atoms. What we need to see is free chlorine atoms that can be proven not to be nickel atoms, but apparently free and lone chlorine atoms are evasive.

Here's the product of evolutionists on earth:
https://www.youtube.com/watch?v=IN-1vRVcVVo

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NEXT UPDATE

Here's all four Gospels wrapped into one story.

For Some Prophetic Proof for Jesus as the Predicted Son of God.
Also, you might like this related video:
https://www.youtube.com/watch?v=W3EjmxJYHvM
https://www.youtube.com/watch?v=efl7EpwmYUs

Pre-Tribulation Preparation for a Post-Tribulation Rapture

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