Take Kineticists to Court

The 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, 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 air atoms cannot transfer weight, nor can they transfer any force, as they merely move through space, 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 want to teach the reality, 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?

As kinetic atoms do nothing but either move alone through space or collide, it cannot be true that the air is made of kinetic atoms that either attract each other or have a neutral charge. The only solution is that gas pressure forms from the inter-repulsion of gas atoms. It informs them that EVERY gas atom is either radiating more positive force than negative, or vice-versa, and it challenges them to understand how that reality can be true. 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 (is 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 its pressure such that the entire weight is transferred to the ground via the 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)." It leaves open the possibility that the filament itself achieves those temperatures, 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." This time, AI betrayed its programmers: "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 high heat, but only mentions the space heat near the filament, and moreover credits only radiated heat, 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 through space, and which 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).

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.

AI slipped up to admit the above, because, when I re-phrased the question, it once again became a 2100-percent stooge of the baboons. I asked: "according to actual experiments, not kinetic-heat theory, 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 now bites down and clamors out the kineticist line fully:

Based on experimental data regarding vacuum environments, a "vacuum" itself has no temperature because temperature is defined as the kinetic energy of atoms, which are absent in a vacuum...

...Experiments in vacuum chambers with thermal heaters show that heat transfer occurs only by radiation, and the residual "space" remains at an extremely low temperature, often close to absolute zero [bunk, easily proven untrue], regardless of the heat passing through it [it admits to heat passing through, but it's just a ghost; it really is very cold in there, cold enough to freeze your finger instantly].

Effect of Shielding: An opaque material surrounding a filament...is designed to block radiation (photons) from being re-absorbed by the surroundings. While the filament might be at thousands of degrees, and the opaque shield might become quite warm due to absorbing radiation, the evacuated space between them does not develop a temperature. The vacuum stays cold, behaving as an almost perfect thermal insulator because there are no molecules to carry thermal kinetic energy. If a sensor [thermostat] is placed in that vacuum space, it will only gain heat by intercepting radiated heat (photons), but the vacuum medium itself holds no heat.

The space has heat, but it has no heat, from the mind of a lunatic plagued by his own stupidity. AI admits to heat in the vacuum even when there is a shield blocking radiation, but it lies to us this time, saying that the heat is from radiation anyway. It lies because it's programmed to lie. It failed, this time, to tell us of the high-vacuum furnace receiving heat from normal heat transfer through the light shield.

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.

Heat in a vacuum is from the electrons streaming out of the filament. If you are unwilling to admit it, there is likely an infection in your thinking injected into your brain by the kineticists. You need healing. Once you know that a filament emits free electrons, it's a no-brainer to realize that heat in a vacuum, the aether, is defined as the physical presence of free electrons.

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.

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? No, 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 again 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 possibility is increased atomic repulsion force with increased heat, though I may reject the latter. Together, the two could 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.

When it comes to various molecules, they don't all weigh the same, as all lone atoms do. I think it's fair to hypothesize that atomic weight, even in a sealed container, goes counter to out-spreading of the atoms/molecules because gravity pulls downward while out-spreading goes in all directions, including up. Both "in-spreading" gravity upon the atoms, and out-spreading due to heat-push, cause gas pressure.

It's a fact that weight of gases in the open atmosphere counters out-spreading, but I think 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 effect 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 charge, in which case the addition of heat to a gas does increase the repulsion forces of atoms i.e. that contributes to gas pressure.

It would be essential to discover how much heat-push equals (perfectly counters) the force of one atom of weight, and I think the water molecule can help to discover this roughly. That's the information I'm after. One needs first to know that a water molecule is not an H2O molecule, but rather an HO8 molecule, with eight oxygen atoms merged into a giant H atom.

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. 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 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. Is that a fortuitous picture or what? In this case, the out-spreading force equals the weight force. What lucky thing can be had by this?

My hope is this: 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. Is that a lucky strike or what? Not so much yet, first because I don't know what the cross section of a steam molecule is, and can't know it until I know the relative sizes of its two atoms. Plus, 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, but only if total heat-push is due to the effects of heat particles alone.

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, four times the area gets four times the push.

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. 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 versus a 1-inch ball for O atoms.

If the addition of heat to a gas increases the atomic repulsion, the H atom could become smaller than 4 inches because the inter-repulsion provides some of the 16-times more push. At 4 inches, the heat-particle push alone exercises 16 times more push, but this size can be the case only if the inter-repulsion is from net-positive charge in all gas atoms. But if all gas atoms possess net-negative charge, then the H atom might not be as large as 4 inches. I'll explain.

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 wall faces the same out-spreading force COMING AGAINST it 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.

However, I can also start with the net-negative theory where it's a mix of atomic repulsion, but where the latter 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, and therefore indistinguishable to the net-positive situation too, except that the H atom must be less than 4 inches in diameter (for the task at hand, the relative size of the H atom doesn't matter).

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, the total push doubles on both atoms because the two forces rise 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.

I have the additional problem of not knowing how deeply merged the O atoms are into the H atom (to form the water molecule), and so I'll need to ballpark it to 25-percent merger to see how well, or not well, it works.

When 8 O atoms are one-quarter merged into the H atom, the full out-spreading force upon this steam molecule is 16 units upon its H atom, plus .75 (percent) x 8 = 6 units for all the merged O atoms combined, for a total of 16 + 6 = 22. That is, out of 8 merged O atoms, the equivalent of 6 are left protruding out of the H atom. This resolution is the gift to us from the fortuitous water molecule. We can do the easy math as per 9 / 6 = 1.5 times where 9 atoms of the water molecule goes against its 6 O atoms of out-spreading force.

In other words, the out-spreading force upon one O atom, when steam molecules are immersed in heat material, is equal to 1.5 atoms of in-spreading weight. This is an important point for what's to come, so remember it. Every atomic area equal to the area of one O atom gets 1.5 times the total heat push one atom, any atom, gets in in-spreading force.

However, this 1.5 number is dependent on how close to reality O atoms are merged 25 percent into H atoms, for the 1.5 was obtained by a 25-percent merger. If they are barely merged at all, the math would be more like 9 / 8 = less than 1.5. This is trial-and-error by assuming 25 percent, by which I mean that 75 percent of the O atom's cross section remains unmerged.

Another way to do the math, which one can do using any percentage of merger, is the following:

When 8 O atoms are one-quarter merged into the H atom, the full out-spreading force upon this steam molecule is 16 units upon its H atom, plus .75 (percent) x 8 = 6 units for all the merged O atoms combined, for a total of 22. In an H gas, each H atom receives 16 units of force, and that gives the steam molecule 22 / 16 = 1.375 times the out-spreading force. As the in-spreading force due to the 9-atom weight of the steam molecule exactly counters that extra .375 of force, each atom of weight accounts for .375 / 9 = .041 of it. As the extra .375 is due to 6 full O atoms, each one accounts for .375 / 6 = .0625 of it, which is 1.5 times more than .041.

We can even learn that the out-spreading on one H atom works out to 16 x 1.5 = 24 times more than the weight of one atom. Or, where nitrogen gas weighs 13.9 times more than hydrogen gas, the out-spreading force on one nitrogen atom is 16 / 13.9 x 1.5 = 1.72 times more than upon one O atom. This math method reveals that the N atom is 16 / 13/9 = 1.15 times larger in cross section than one O atom. The fortuitous water molecule grows more than plants, it can also grow our knowledge a little on what molecules and atoms look like.

Chlorine Dioxide

I'll have the rest of this update out soon, after making some adjustments.

At STP, chlorine gas weighs 3.17 g/l, 2.2 times more than oxygen gas, wherefore there are 2.2 times as many chlorine atoms per unit volume. Asking google, "what is the diameter of a ball having 2.2 times less cross section than a one-inch ball?" It gives .674 inch i.e. .674 the diameter of an O atom at 1-inch round.

I'm reading: "one volume of chlorine gas reacts with two volumes of oxygen gas to produce two volumes of chlorine dioxide (ClO2) gas." In my model, where there are 2.2 times more Cl atoms per liter, it appears that less than one equal volume of Cl gas mixes with two of O gas such that there will one O atom merged per one Cl atom. The science I'm trying to explain can reveal whether the molecule has one-and-one, or two-and-two, atoms. We need to know the average cross section of the chlorine-oxide molecule in order to find out.

After trial-and-error, I learned that a Cl2O2 or a Cl4O4 molecule cannot work as per the cross-sectional math, but when trying one-and-one as per a ClO molecule, the math was perfect. However, I have a hard time understanding why an O atom should receive just one Cl atom.

The cross section of the molecule changes depending on whether we: 1) look straight on seeing both atoms at once, 2) versus looking toward one end where the smaller molecule is obscured by larger O atom. Whatever can be seen from either direction, and the top view too, is what heat-push sees and acts upon. In the case of this molecule, the top view is identical to the straight-on view.

A good assumption to begin with is that the one O atom is barely merged into the chlorine atom because sunlight alone dislodges them. I'll therefore treat them as not merged at all. The O atom has a cross section of .785 square inch, wherefore the Cl atom has .785 / 2.2 = .36 square inches. We add the latter to .785 for a total cross section of 1.16 square inch for the perspective 1) above.

When doing perspective 2), we can see only one O atom for a cross section of .785. Finally, we need the average of the two perspectives, the average of 1.16 and .785, that is, which is .97 square inches for the full but average area upon which heat may push from all directions.

I now need to find how many O atoms are in .97 square inches: .97 / .785 = 1.23. This means that the molecule receives 1.23 times as much heat-push as an O atom in an O gas, but there is also the weight factor of the molecule to reckon with which counteracts some or all of the out-spreading.

I showed above, from the water molecule, how each O atom receives 1.5 times as much force as one atom of weight. Therefore, this 2-atom molecule has 1.23 x 1.5 = 1.85 units of out-spreading force, versus 2 atoms of in-spreading weight force. I can tell you this, that if the 1.85 out-spreading force matched the 2-atom weight, the chlorine-dioxide molecules would have the same spacing as O atoms in an O gas, and consequently the chlorine-dioxide gas would weigh 2 times as much as oxygen gas. But as the weight force is a little greater than the out-spreading force, the chlorine-dioxide gas is predicted to weigh a little more than 2 times, perfect because, at STP, chlorine dioxide is said to weigh 2.1 times more than oxygen gas.

When doing this math for a Cl2O2 and a Cl4O4 molecule, the chlorine-dioxide gas came out way too heavy.

It seems that chlorine-dioxide molecules have a little less numbers per unit volume than O atoms in O gas. We learn here that, when an atom having .67 times the diameter of an O atom is added to an O atom, the push-apart (out-spreading) force predicts to be almost as much, but not quite, the weight force of one atom. It suggests that the amount of chlorine atom left unmerged is not enough to equal the in-spreading force.

The percentage of O-atom cross section needed to produce one atom of weight is 1 / 1.5 = .67 atom. We therefore multiply the latter by the .785-square-inch cross-sectional area of the O atom to get .53 square inches. Therefore, the cross section of the chlorine atom needs to be .53 to match the in-spreading force of one atom. But the chlorine atom has a total of only .36 square inches.

[Update -- I should probably re-do the math in this section because the 3.17 g/l used is probably incorrect, for I've just found this:

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...

I should soon re-do the math with 3.0 to see what happens. End update]

The Look of the Ammonia Molecule

Ammonia becomes a gas at -33 C, not far below the gasing temperature of chlorine dioxide at 11 C. They call the ammonia molecule, NH3, but it's not that. Instead, that symbol refers only to the volumes of gases involved. However, I asked google: "is it correct that EXACTLY three volumes of hydrogen gas mix with exactly one equal volume of nitrogen gas to produce exactly two volumes of ammonia gas, or how exact are those volumes in practice/reality?"

It is theoretically correct that exactly three volumes of hydrogen gas react with exactly one volume of nitrogen gas to produce two volumes of ammonia gas. This is based on Gay-Lussac’s Law of Combining Volumes and Avogadro’s Law [trash, sham], which state that at the same temperature and pressure, equal volumes of gases contain an equal number of molecules.

...The 1:3:2 (Nitrogen:Hydrogen:Ammonia) ratio holds true only if all reactants are converted to products, which does not happen in industrial production.

The response was not able to tell me what the exact volumes are, but my point is that the mix is likely such that the molecule works out to one H atom into which five N atoms are merged. The reason I take that view is that nitrogen gas weighs almost 14 times more than hydrogen gas, revealing that there are 14 times more atoms in an STP liter of gas than in an STP liter of hydrogen gas. As 14 / 3 is less than 5, I assume that slightly more than one volume of nitrogen gas is used with three volumes of hydrogen gas such that there are five N atoms parked into one H atom. It makes sense where a steam molecule is 8 O atoms into one H atom, for the N atom is larger than the O atom.

Steam at weighs 9.0 times as much as hydrogen gas at the same temperature and pressure. Ammonia gas (.765 g/l) weighs 8.5 times more than hydrogen gas (at STP), meaning that there are 8.5 times as many ammonia atoms (not molecules). My ammonia molecule weighs 6 atoms, which is 6 times more than a hydrogen atom. In order for ammonia gas to weigh more than 6 times as much, the ammonia molecules must be closer to each other, center-to-center, than the H atoms in an H gas. I therefore reason that the out-spreading force of the ammonia molecule is less than the in-spreading force of the molecule's weight, for this is my way to explain how it weighs 8.5 times as much.

The other conclusion is that the steam molecule is larger in stuff than the ammonia molecule, for I determined that the steam molecule has exactly as much out-spreading as in-spreading. Steam weighs 9 times more than H gas, and the steam molecule has 9 atoms. Therefore, I conclude further that there is more space between each of the five N atoms when merged into an H atom than when 8 O atoms are merged into an H atom. There is less cross-sectional area, that is, in the ammonia molecule. The ammonia molecule might have a larger diameter thanks to N atoms being larger than O atoms, but in any case it doesn't quite have as much stuff.

If ammonia gas weighed 6 times more than hydrogen while the ammonia molecule weighs 6 times more than the H atom, then the distances between H atoms and ammonia molecules would be identical. But where the gas weighs more times heavier than the molecule weighs more times than an H atom, it tells me that the ammonia molecules are the closest together of the two gases.

Everything checks out very well, or I wouldn't be saying so, with my view of these gases, tending to confirm that my atomic model, based on all atoms weigh the same, checks out.

You need to be warned against google's AI, because when I ask it for the enthalpy for making a liter of ammonia, it gives about 2,500 kJs, but if I ask it not to include the heat needed to make ammonia: "The enthalpy released (exothermic energy) when producing 1 liter of liquid ammonia from its constituent elements, excluding the energy required to drive the process (heat/work), is approximately -1,800 to -1,900 kJ/L." However, as I didn't ask for a liter at STP, that high-heat must be for the high pressures at which ammonia is made. If I ask for the enthalpy at STP: "The enthalpy change for making one liter of ammonia gas at Standard Temperature and Pressure (STP) is approximately -2.05 kJ/L (released), excluding the heat needed to operate the reactor."

Then, "The enthalpy of combustion for hydrogen, when producing water vapor (steam) at standard temperature and pressure (STP), releases approximately 10.8 to 11.5 kJ per liter of hydrogen gas consumed." This, about five times more than heat release from ammonia's formation, tells me that the eight O atoms are merged significantly deeper into the one H atom and than the five nitrogen atoms, because total heat production in chemical reactions is a measure of the number of captured electrons that go free due to mergers. Still, even though the five N atoms are less merged, their unmerged exteriors still don't quite match the total cross section, available to heat-push, of the eight O atoms.

I hold a theory that nitrogen is a multi-atom molecule that is not of nitrogen atoms. That is, what science thinks is the N atom could be one or more merged but unknown, and undiscoverable, atoms. In that case, all the math in this ammonia section is moot, unless for every extra atom, in its make-up, it receives as much out-spreading as it does in-spreading. Otherwise, the math needs to be redone, which is impossible if we can't know what type atom makes the N molecule.

The Look of a Sodium-Chloride Molecule

I'm going to start here with what I found recently:

I'd like to add that, at 883 C and one atmosphere of pressure, the reported density of sodium gas is approximately .242 g/l, or 3.1 times lighter than chlorine gas, and 1.39 times lighter than oxygen (.337 g/l), when both are at the same 883-C temperature (boiling point of sodium).

For me, this means that the sodium atom has 1.39 times more cross-sectional area than the O atom, and 3.1 times more than the chlorine atom. The sodium atom therefore has a cross section of .785 x 1.39 = 1.09 square inch where the one-inch-round O atom has .785 square inch. It makes the sodium atom 1.18 inch in diameter.

The chlorine atom's cross section becomes 1.09 / 3.1 = .35 square inch, with a diameter of .67 inch. The chlorine atom's cross section is 2.2 times less than for the O atom. I'll use 1.1 instead of 1.09.

As sodium CHLORITE (a solid) doesn't liquefy or gasify with added heat, but decomposes partly to sodium CHLORIDE instead, there's no use seeking the cross section of the chlorite molecule. The chloride, which they view with a NaCl molecule, where "Na" is the sodium atom, does have a boiling point of 1413 C, and so I'll seek its cross-sectional area.

The true meaning of "NaCl" is that one liter of sodium gas mixes with one liter of chlorine gas, when both are at the same temperature and pressure, to produce one liter of sodium chloride. The question is, how many more sodium atoms are in its liter versus chlorine atoms in its liter? It's not one per one, as the fools of Avogadro claim, and therefore the sodium-chloride molecule is not an NaCl molecule.

We saw above that the sodium atom has 3.1 times more cross section than the chlorine atom, which, in my model, means that there are 3.1 times as many chlorine atoms in its liter versus sodium atoms in its liter. Therefore, my expectation is that, in reality (not ideal-gas expectations), 1.0 liter of sodium gas mixes with a little less than 1.0 liter of chlorine gas, leaving .1 of the 3.1 parts as left-over gas.

Thus, the sodium-chloride molecule must either be a 4-atom NaCl3 molecule, or an eight-atom Na2Cl6 molecule. It doesn't make much sense to me for a lone sodium atom, at 1.18 inches in diameter, having only three chlorine atoms, each at .67-inch diameter, merged upon it. There is much-more room upon the sodium atom than for three saddled atoms.

If two sodium atoms are merged, there is a circular nook between them all around where they are merged, same as the nook you see in this 8 shape. I suggest that there are six chlorine atoms merged in that nook. That's a much-better picture. The following indentation has important math to prove that six atoms can fit into the nook. You can bypass the indentation anytime to go to the final math result:

Back-to-back, with all six in contact with each other in circular form (making a ring), the circular ring would be .67 x 6 = 4.0 inches long. Is there room for all six, if the sodium atoms are merged 20 percent? I asked google: "what is the diameter of the ring where two balls make contact when merged into each other by 20 percent?" The response, which comes with a drawing to verify what it's talking about, is: "When two identical spheres merge into each other by 20% of their diameter, the diameter of the contact ring is 1.2 times the radius (.6) times the diameter of the individual balls."

That response is not for the circumference around the ring which I'm after, but only the ring's diameter, given as .6 the diameter of the ball sizes. The ring above is exactly the nook, by the way. As my "ball" is the sodium atom at 1.18" diameter, the ring's diameter is 1.18 x .6 = .7 inch, and so its circumference works out to .7 x pi = 2.2 inches.

However, we are not yet done, because the circumference I'm looking for is further out from the ring. I'm looking for the circumference running through the center of all six chlorine atoms, while the ring/nook above is altogether outside of the chlorine atoms. I can hardly believe that google AI answered the following question just as I want it answered, and once again includes a drawing to verify it correctly understands me. The question is: "if a ball .67 inch in diameter is in the nook of two merged balls 1.18 inch diameter, and in contact with the two balls that are themselves merged into each other by 20 percent, how far will the nook be to edge of the small ball?" With explanations, it says: "The distance from the nook to the nearest edge of the small ball is approximately 0.107 inches."

That seems very correct to me, meaning that the center of the six atoms are .107 + .33 = .44 inch from the nook (.33 is half of .67). The point in asking that question is to find how far above the nook the center of chlorine atoms would be if they are merged 25 percent into the chlorine atoms. Twenty-five percent of .67 inch is .17 inch, meaning that the center of the atoms move .17 inch toward the nook when merged by 25 percent. Therefore, after merger, the nook is .44 - .17 = .27 inch from the center of the atoms.

Finally, we add .27 to both sides of the .7 diameter of the nook to find that the circumference of a ring through the center of the six atoms is: ((.27 x 2) + .7) x pi = 3.9 inches.

We needed 4.0 inches (.67 x 6) to fit all six atoms into that nook, and it looks like we can get it with slightly less than the 25-percent merger I used because it increases the 3.9 result. Assume 20-percent merger.

The sodium chloride molecule I've just described has different heat-push force from two directions. From one angle, where we see both sodium atoms with the ring of chlorine atoms at its center, the cross-sectional area includes the full two sodium atoms minus 10 percent each due to their 20-percent merger: 1.1 x 2 x .8 (80 percent) = 1.76 square inches.

The full cross section of the molecule is as much larger than 1.76 as two chlorine atoms merged 20 percent into the sodium atoms, and so we do: (.35 x 2) x .8 = .56 square inch. We can see about three chlorine atoms from this line of sight, but roughly one of them is in front of the sodium atoms such that we don't count it for heat-push calculations. In fact, a little less than .56 is closer to the reality. The total cross section here is a little less than 1.76 + .56 = 2.3 square inches.

Then, for the other view, perpendicular to the one above, we can see all six chlorine atoms, but only one sodium atom because the second is behind the first. The heat-push will therefore go against one sodium atom only (1.1), and about .6 of each chlorine atom because all six are partially obscured within the nook. The math: 1.1 + (.35 x 6 x .6) = 2.2. We add the latter to 2.3 above for a grand tally of 4.5 square inches.

We then do 4.5 / .785 to find that 5.7 oxygen atoms are equivalent in cross section to the 4.5 square inches, meaning that the sodium-chloride molecule is, on the one hand, predicted my this math to have 5.7 times more out-spreading force than upon one O atom in an O gas. On the other hand, the molecule has eight atoms and thus weighs 8 times more than an O atom.

However, the 5.7 yet needs to be multiplied by 1.5 as per my tentative finding that one O atom of out-spreading force is equal to one atom of in-spreading weight: 5.7 x 1.5 = 8.6 units of out-spreading. This 8.6 versus 8.0, with out-spreading dominant, means that the sodium chloride molecules will have more distance between each other than O atoms in an O gas, where both gases are at the same temperature and pressure. It means that the sodium chloride gas is predicted to weigh less, and in fact AI reports that, at the 1465-C boiling point of sodium chloride, it weighs .177 g/l versus .224 for O gas at the same temperature (the boiling point is sometimes reported as low as 1413 C).

As the difference between 8.0 and 8.6 is 1.075 times the force, it may always be true that 1.075 units of extra outspreading versus atom weight gives a gas lighter by 79 percent, for .177 g is .79 of .224. For example, for a four-atom molecule having 4.3 units of out-spreading, the extra .3 is 1.075 time more than 4.0. The 79-percent amount means that the molecules are .79 times more sparse than O atoms. When it gets down to 50 percent, the molecules are twice as sparse (half as dense by volume).

I doubt anyone will take the time to understand what I'm about to say, but for my records, 50 percent is 1.58 (.79 / .5) times less than .79. Then, I want to know .6 x 1.58 = .95 because 8.6 is .6 larger than 8.0. I therefore GUESS that when a gas with an eight-atom molecule comes out with an outspreading force of 9.55 (8.6 + .95) units, it's going to weigh half as much as oxygen gas (at the same temperature and pressure).

This is a new science for me; I'm not familiar with it. Thus far, my atomic model seems to be working dandy to explain the look of molecules and the expected weights of various gases.

AI: "Gaseous sodium chloride (NaCl) thermally decomposes into sodium metal and chlorine gas at temperatures roughly around 8,000 K (approximately 7,700°C)." On the one hand, it suggests that the two sodium atoms of this molecule are DEEPLY merged, because 8000 K is hot-hot, with lots of heat-push seeking to squeeze the two apart.

However, there's more to it when we consider the ring of six chlorine atoms circling the nook where the two sodium atoms are merged. We can view them as glue that helps to bond the two, for all six are merged into both sodium atoms. Any heat getting between the two must simultaneously squeeze the six chlorine atoms deeper into each other. See that? The chlorine atoms will resist, to a degree, the invasion of heat between the sodium atoms.

If there are too-few atoms in the nook of some other two-atom molecules, where the ones in the nook are not in contact, then heat invasion doesn't merge them, wherefore they don't resist heat invasion as much. But if six chlorine atoms are tightly packed around the nook of a sodium pair, such that they are in contact or even merged to begin with, then heat pressing into the nook finds them to act as an "arch" resisting inward-coming force. The heat now needs to detach one sodium atom from the six chlorine atoms, and ditto on the opposite side for the other sodium atom.

Once the sodium atoms are dislodged from the six, heat easily invades between the chlorine atoms because they no longer have the help, to resist heat-squeeze, of their merger into the sodium atoms.

I suggest that the sodium-chloride molecule, which is table salt, was glued together hard by God because He didn't want it to separate in the body. The goofballs are STUPID when claiming that water alone pulls the chlorine atoms apart from the sodium atoms. They make that claim knowing that 8000 K heat is needed to pull them apart. STUPIDS. The fact is, all salt that enters the body comes out, which is not expected if body water pulls the molecules apart into separate atoms. Besides, do you want free chlorine in your body? They are STUPIDS.

As their atomic model doesn't always work, they defined chlorine gas as a Cl2 diatomic molecule, which doesn't likely exist. I suspect that all of their diatomic molecules, named for elements, are false. They therefore argue that the chlorine in table salt becomes a safe "chloride," which they view as a lone chlorine atom. I think they are simply stupid-fied by their erroneous atomic model.

As opposed to sodium chloride, they also named sodium chlorite, which they call, NaClO2. It adds an extra liter of material to the two of sodium and chlorine gas. If one simply heats the chlorite, it breaks down to sodium chloride and what they call sodium chlorate, which they call, NaClO3. Not only are they wrong in pegging how the molecules look, but they magically make extra oxygen appear, because, when asking, "how many liters each of sodium chlorate and sodium chloride are made when two [theoretical gas] liters of sodium chlorite decompose due to heat alone?", the AI response decided to use three liters instead of two, for "3NaClO2" means three volumes.

The response is: "the stoichiometry based on the disproportionation reaction 3NaClO2 > 2NaClO3 + NaCl implies that 2 moles of NaClO2 produce 4/3 mole (1.33 liters) of NaClO and 2/3 mole (.67 liters) NaCl." In other words, every two liters of sodium chlorate robs the diatomic O atom from the remaining liter -- which is the same as saying that every two molecules robs one molecule, not expected because where heat causes one molecule to lose its oxygen, all identical molecules should too such that free oxygen and NaCl (salt) should be the result.

Then, once robbed, each diatomic O atom is split into two lone O atoms, one going to one of the two surviving NaClO2 molecules, and the other O going to the second surviving NaClO2 molecule to turn them both into NaClO3 molecules while leaving behind the robbed one that becomes the NaCl molecule. This occurs with heat alone at about 200 C, but at 300 C, and with heat alone, the chlorate decomposes to free oxygen and salt.

Unfortunately, neither the chlorate nor the chlorite turn to gas. If they did, I could venture a good or even easy guess as to what those molecules look like. I don't know whether to trust AI when saying that, according to their molar science, a theoretical liter of chlorate gas weighs 4 g/l while a theoretical chloride gas weighs 1 g/l, both at STP.

I asked: "is it correct to say that one liter of sodium gas, one liter of chlorine gas, and one liter of oxygen react to form two liters of sodium chlorite?" The response: "Stoichiometry & Formula: The formula NaClO2 indicates 1 sodium (Na) atom, 1 chlorine (Cl) atom, and 2 oxygen (O) atoms. The volumes of gaseous reactants (if they were gases) would need to be in a ratio of 1 : 0.5 : 1 for Na : Cl2 : O2 to fit this stoichiometry, not 1 : 1 : 1." If their ratio is correct, I can do some calculating.

It works off of their ideal-gas calculation that, at the same temperature, sodium gas is 3.1 times lighter than chlorine gas, and 1.39 times lighter than oxygen (these numbers were presented shortly above with more details). If only half as much chlorine gas, as sodium and oxygen gas, goes into the making of sodium chlorite, then there are 3.1 / 2 = 1.55 times fewer sodium atoms than chlorine atoms in that mix. The ratio thus far is 1.0 sodium atom per 1.55 chlorine atoms. As there are equal volumes of sodium versus oxygen gas in that mix, then the full atomic ratio is 1.0 : 1.55 : 1.39.

However, if the gas-volume ratio is 1.0 liter of both sodium and chlorine, instead of a half-liter of chlorine, the atomic ratio becomes 1 : 3.1 atoms instead of 1 : 1.55. That's a big deal, and it could be that the goofs claim a half-liter of chlorine gas just because they wrongly view the chlorine atom as a diatomic molecule, Cl2. In other words, their half-liter of Cl2 can really be a liter of my Cl (i.e. with lone chlorine atoms).

It's a big deal because the ratio 1 : 3.1 means one sodium atom per three chlorine atoms, what I proposed above where the sodium-chloride molecule has two, merged sodium atoms ringed by six chlorine atoms. To make sodium chlorite (NaClO2) from sodium chloride (NaCl), there is only an addition of oxygen, and where they would claim the addition of one volume of O2, I'm claiming one volume of O, i.e. normal oxygen gas as I see it. They really do make a confusing state of things.

The atomic ratio found above for sodium chlorite, 1 : 3.1 : 1.39, could suggest that a little less than one liter of oxygen is in the mix such that the 1.39 comes down to 1.33 atom. In this picture, the sodium-chlorite molecule looks like a 16-atom molecule, 3 : 9 : 4, when all numbers above are tripled, and when one liter of sodium mixes with slightly less than a liter of both chlorine and oxygen.

I asked: "how many liters of sodium chlorite does 1 liter each of sodium and O gas make with a half-liter of chlorine gas?" Just keep in mind that while AI agrees with that 2.5- liter mix, I'm viewing them as roughly three liters total. As chlorite doesn't become a gas, the response gives the total volume of chlorite product as a solid: "Based on stoichiometric calculations at Standard Temperature and Pressure (STP), 1 liter of sodium metal, 1 liter of oxygen gas, and 0.5 liters of chlorine gas will produce approximately 0.0016 liters (or 1.6 milliliters) of solid sodium chlorite." I'm told that a milliliter (= 1 cc) of sodium chlorite weighs 2.47 grams, wherefore 1.6 milliliters weighs 4.0 grams.

Now we know how they get their 4 g/l of a theoretical sodium-chlorite gas, for they calculated the 1.6 milliliters of the solid as all the stuff needed to make a theoretical gas filling one liter, though they can't prove it by making it a gas (because it doesn't become a gas). Their problem is that their math is based on Avogadro's theory in which all gases at the same temperature and pressure have the same number of atoms. If that theory is an ingredient in the calculation to 4 grams, it's a wrong calculation. Plus, they calculate how fast they think the molecules are colliding, which they think tells them how far apart the gas molecules are, and when they have just the right speed to become as far apart as Avogadro wants them, that's how the stupids operate, too stupid to realize that Avogadro cannot be correct, and too big-bang wicked to abandon kineticism.

We see their hidden claim here that the 2.5 liters (what I view as 3) of the sodium-chlorine-oxygen mix makes only one liter of sodium chlorite. Therefore, my molecule with ratio, 3 : 9 : 4, stands, because if the mix produces two liters instead of one, that atomic ratio would be forced to become 1.5 : 4.5 : 2, which we can't have because we need full atoms in all cases.

Besides, if we imagine 3 sodium atoms merged in a spherical shape as opposed to a linear shape, three nooks are formed, each a semi-circle. In the case of 2 sodium atoms merged with one nook, it was fully circular, fitting 6 chlorine atoms, but if we have three semi-circular nooks, they can be expected to fit 3 chlorine atoms each for a total of 9 in this molecule. I assume that the 4 O atoms are 2 each on top and on bottom of the triple-merged sodium atoms. The top and bottom are where the three nooks meet. Possible, they calculated twice as much oxygen as actually exists.

To show you how ignorant the evolutionists made their stooges, they devised Avogadro's theory such that the speeds of atoms/molecules are inversely proportional to their weights. If an atom is three times heavier than another, it must be three times slower at STP because both then apply the same pressure when colliding with objects. And so we need to ask what they might not want us to ask: how did atoms take on different speeds after the big bang spit out the protons that made all the atoms and molecules?

The Big Bang Makes Avogadro's Theory Impossible

Not knowing whether they view the big bang as providing all the same speed for all protons, I asked google: "how did atoms take on their different speeds after the big bang provided all protons with the same speed?" The response makes Avogadro's theory look ridiculous:

While early protons and nuclei were part of a hot, dense plasma, they were not all moving at the exact same velocity — they had a "speed distribution" (a Maxwell-Boltzmann distribution) determined by the extremely high temperature, meaning many were already moving at different speeds.

Of course, they speak it as a fact as if they were there to see it all happen. But these lunatics will tell us with a straight face that all types of atoms, each with a different number of protons, took on speeds, by sheer coincidence, exactly as needed to provide the same gas pressure at any one given temperature. If you can't see how laughable that is, you ain't no physicist. You become merely a loyal, trusting stooge. The response goes on, as if the clowns were at the event, watching and recording everything:

Thermalization and Interaction: In the first few minutes, protons, neutrons, and electrons constantly collided. These collisions resulted in a range of speeds (a "thermal distribution") rather than a single, uniform speed. The hotter the plasma, the faster the average speed.

You're being duped because it would have been impossible for the particles exploded from the big bang to strike each other, for, with distance from the explosion, they all move twice as far apart with twice the distance from the explosion. The goofs will therefore invent some mechanism to make the particles change their straight-line paths in order that they can collide and bond, to make star formation "possible" by those collisions. I therefore asked: "can a non-magnetic object without propulsion be made to curve through pure space by only the thing that causes its motion? No gravity and spacetime trash please." Response: "According to classical mechanics and Newtonian physics, no. A non-magnetic, non-propelled object in pure space cannot curve its path if it is only acted upon by the initial force that caused its motion." You should therefore not imagine the curve of a soccer ball, for example, when struck.

I asked: "if a cannon ball is sent out with a spin in pure space, no air, will it move in a curved path?" "No, a cannonball sent out with spin in pure space (a vacuum with no air) will not move in a curved path. It will travel in a perfectly straight line while spinning."

Therefore, the clowns have no way, short of a lunatic invention, to make big-bang particles come together in collisions. They had Einstein as their favorite lunatic: "According to General Relativity, an object can indeed curve through space without an external, sideway force acting on it, provided it is following the curvature of spacetime itself."

As you can glean from above, the wackos came to teach that, due to RANDOM collisions with each other, where the big-bang-birthed particles had different masses and therefore different kinetic energies, they all took on RANDOM speeds. Then, billions of years later, the hydrogen atom weighing 16 times less than an oxygen atom happened, by pure chance, to be moving 16 times faster at 0 degrees, or when the comparison is made at any other temperature. And a nitrogen atom, weighing 14 times more than a hydrogen atom, happened, by pure chance, to be moving 14 times slower...etc., etc., for all the atoms and molecules.

Then, when hydrogen atoms come together with oxygen atoms to form steam, the latter molecule suddenly changes speed at the moment of formation, not going 16 times faster than the O atom, not going the same speed as the O atom, but 9 times faster just because the steam molecule weighs 9 times more than hydrogen. This follows Avogadro, not reality.

It's just a fluke in this case that 9 is almost midway between 1 and 16, but even so, nobody who has integrity thinks that objects with different speeds, and bonding at a collision, end up with a speed exactly midway between their original speeds, in order that no kinetic energy is lost. Everyone with integrity accepts that collisions slow the total velocities of the colliding objects. When steam is formed, H and O atoms COLLIDE due to electromagnetic attraction.

I asked: "is it true that, in the real world, that two colliding objects in pure space will lose some total velocity due to the collision?" "Yes, that is true for most real-world scenarios." They admit the reality, but do not explain why that reality happens, for if they admit it, their kinetic theory of atoms cannot be substantiated. The response goes on: "It is essential to distinguish between momentum and velocity. Total Momentum is Conserved: In a closed system, the total momentum before and after the collision is the same."

In other words, while the objects slow down upon colliding, ALL the initial motion energy still exists as heat, sound or internal vibrational energy. They deliberately omit the real cause of the slow-down: energy transfer. In fact, they have programmed AI to fetch a lie, for I've just asked: "if a pool cue or second ball strikes a first ball dead-on with ten pounds of momentum, in a direction opposite that the first ball itself moves with ten pounds of momentum, will the first ball's momentum remain 10 pounds of momentum, or significantly less?" Instead of telling the obvious truth, than ten pounds of momentum will cancel the ten pounds of momentum, the response (from someone at quora) caters to the kinetic theory of atoms:

In a head-on elastic collision between two objects of equal mass (like two pool balls), they will exchange their velocities and momenta. Since both started with 10 units of momentum moving toward each other, the first ball will rebound with 10 units of momentum moving away from the point of impact.

This is the kind of wacko science the kineticists have foisted on the world. If two objects each having the same momentum strike in a head-on collision, and if they both bounce away with the same momentum/velocity by which they struck, then there can be no scenario that stops the ball. I rephrased in order to trick AI: "how much momentum force, by a pool cue or second ball, is needed to stop a second ball moving with 10 pounds of momentum, if there is a head-on collision from opposite directions?" Response, haha:

To stop a second ball moving with 10 units of momentum in a head-on collision from the opposite direction, the pool cue or the first ball must provide an impulse of 10 units.

...To bring a ball from 10 units of momentum to a complete stop (0 units), you must apply a change of exactly 10 units in the opposite direction.

AI just heaped condemnation upon its bosses. It's so logical, so correct, yet there are wackos denying it for fear that the kinetic theory will be undermined. They use tricks to deny it. The reality is exactly as I say, that, if atoms are racing at several hundreds of mph from big-bang momentum to this day, they would slow with at least half of all their collisions to a stop within a second, due to the expectation of striking billions of times per second.

AI admitted above that the ball's slow-down to a stop is due to the TRANSFER of 10 pounds of momentum. It is to be assumed that the first ball moves with 10 units because 10 units were transferred to it while it was not moving. Therefore, 10 units coming against it from a second ball will bring it back to a stop. So logical, so correct. The only explanation is that there was energy TRANSFER, meaning that energy transfer slows things down. Yet the kineticist wants you to believe that energy transfer cannot slow things down because the energy needs to survive as some sort of motion, whether heat or sound or vibrational energy.

NO STUPIDS. Motion energy counters motion energy, period. If equal motion energy collides head-on, they cancel each other. If they quite strike head-on, total velocity slows. I re-phrased the question to check whether AI has the integrity left to admit the full reality as per the finally condition of both balls: "how much momentum force, by a second ball, is needed to stop a second ball moving with 10 pounds of momentum, if there is a head-on collision from opposite directions? And what will be the momentum of the second ball after the collision?" The response even did me a favor by ignoring the bounce (elastic) factor:

To stop a moving ball with 10 units of momentum in a head-on collision, the second ball must provide a momentum of 10 units in the opposite direction, assuming the collision is perfectly inelastic (they stay together and stop).

...After Collision Momentum: 0 units for both balls, as they are now stationary.

The bounce factor is evidence that a secondary thing takes place at the collision of solid balls (i.e. not like a tennis or ball). The motion energy moves the atoms and electrons out of their natural resting places such that they rebound back to their natural resting places, causing some bounce for the balls. The bounce is not sufficient to send the balls away from each other with the speed they had prior to collision. ALL of the energy transfer goes into cancelling all motion, but in the meantime the atoms do their rebound thing to create NEW motion energy. The collision triggers the formation of NEW motion energy. The atoms have natural resting places because they are caught in equilibrium, between their inter-repulsion and inter-attraction forces. It's those very forces that bring them to a complete stop; they are not eternally vibrating as the big-bang lunatic envisions them. Shame on Creationists for being duped by them.

I was lucky to get AI to agree with my claims by changing "velocity" to "momentum" in my questions. When I use "velocity," the response can toy with me by assuming the balls might not be identical. It is programmed to differentiate between velocity and momentum, though I see no difference between them for the purpose of this question when the two balls (or pool cue) are said (in the question) to carry the same force (i.e. momentum is equal for both). As part of the question above, AI tells the truth yet again:

If the second ball is stationary (0) momentum) and has the same mass as the first, a head-on collision will transfer all 10 units of momentum to the second ball. Initial: Ball 1 = 10 units; Ball 2 = 0 units. After Collision: Ball 1 = 0 units (stops); Ball 2 = 10 units (continues the motion)

Excellent, so logical. This is a proper example of proving that energy cannot be destroyed. And when two balls stop each other, no energy has been destroyed, but rather it all gets used up to form the new motion energy, zero motion. What can't the goofs understand about this? Why do they want you to be hung-up by the fable that the result of zero motion, or the result of slower motion, per collision, is "destruction of energy." It's not destruction; it's energy in action that causes inaction. It's just a simple fact that needs to be accepted at the expense of the kinetic model.

If I've convinced you that it's simply impossible for Avogadro's fantasy to be real, because it's impossible for 100 different types of atoms to have the exact-same momentum at STP, then the only alternative is that all atoms weigh the same. It's the only other way to explain the experimental, gas realities. On top of the cosmic-sized coincidence that Avogadro requires, it's impossible for atoms to remain at the same total velocity through billions of collisions per second. Plus, the big bang is a cosmic-sized fraud. Therefore, all atoms weigh the same. Accept it, get used to it; it's a remarkable reality that God has devised.

Carbon Oxides

We must not trust them when they convert a solid to a theoretical gas at STP. It requires their estimating how many atoms are in a certain volume of the solid enough to make one liter of gas at STP. The problem is worse than their not knowing how many atoms are in a gas though they think they do, but they moreover do not know that a solid weighing twice as much as another solid has twice the atoms. They don't know how many atoms are in any volume of the solid, wherefore they cannot be correct when determining that a carbon gas would weigh six times more than an H gas, if the estimation is done by mathematically converting part of the solid into a carbon gas, especially as this gas doesn't exist.

Carbon wood or charcoal does not turn to pure-carbon gas with added heat, even in a vacuum (doesn't allow them to burn). I asked AI what happens to wood in a vacuum in the range of 800 C: "The wood releases flammable volatile compounds (wood gas, including CO, CO2, H2, CH4) and liquids (wood tar, methanol, phenols)." The rest of it, the solid, goes black. When heating carbon monoxide and carbon dioxide with the help of catalysts, the C and O can be separated, yet the C does not gasify, but instead turns to solid-carbon powder. It therefore seems that carbon atoms are so strongly bonded that heat in it's midst cannot cause it to go up as lone gas atoms.

High heat and catalysts (such as iron, cobalt, or nickel) typically convert CO and CO2 into solid carbon, such as powder, graphite flakes, or carbon nanofibers/nanotubes. It does not turn them into individual, loose carbon atoms that accumulate on the floor. Instead, the process — often referred to as catalytic decomposition or coking — causes carbon atoms to bond together and deposit onto the surface of the catalyst.

They can't take that coke and turn it into a gas by heating it. The rising force of heat, which evaporates most substances, will not lift carbon atoms out of carbon atoms. There's two possible explanations: the atoms are so incredibly small that insufficient heat lift applies to them to blast them off into space, and/or they have such a strong bond (by deep merger and/or shape-based entanglement) that heat lift can't overcome it. It very much appears that carbon atoms yanked out of CO or CO2 (by the metals above) have an affinity for each other, and a bond strength too strong to make a liquid, yet in powder form they do not constitute a true solid if they can't first become a liquid.

Roughly the maximum thickness of coke that forms on the metals above, by surrounding them with carbon dioxide, is .15 millimeter. It suggests that the metals electromagnetically attract the carbon atoms to themselves, upon robbing them from the dioxide molecules. Why should the metals have that capability while high heat alone can't break the C-O bond of the dioxide? I suggest that, when molecules are made of smaller atoms filling the nooks formed by larger atoms, heat can both press or squeeze atoms deeper into the nooks while seeking to squeeze between the two types of atoms to pop them apart. Both processes take place simultaneously with rising heat, and while I assume that heat can pop/free some molecules out of the nooks, it seems that it continually presses atoms deeper onto the nooks, when it comes to CO and CO2 molecules, wherefore there's no getting between the C and O atoms to pop them apart.

I'm therefore of the mind that carbon oxides have deep nooks such that the atoms in the nooks tend to go deeper into them with added heat. Every atom in a nook is merged into, and therefore attracted by, both atoms creating the nook. It's double-bond strength.

The left side of this 'B' can represent two atoms merged so deeply that the nook disappears. The right side of the B can act as two atoms merged not as deeply such as to create the nook you see between them.

They say: "Based on the ideal gas law at standard temperature and pressure (STP, 0°C and 1 atm), where one mole of an ideal gas occupies 22.4 liters, the weight of a theoretical pure-carbon gas depends on its molecular form. Monatomic [= lone atom] Carbon (C): 0.536 g/L." It's important that it's six times heavier than H gas at .09 g/l. How do you think they arrived at .536 when they don't have a carbon gas to work with?

The carbon-monoxide molecule, which they view incorrectly as a CO molecule, represents one volume of a theoretical carbon gas mixed with an equal volume of O gas to make one volume of carbon monoxide. The latter weighs 1.25 g/l, or 13.9 times more than H gas, while carbon dioxide gas weighs 21.9 times more than H gas. (I can know that there are 13.9 and 21.9 times more atoms for those gases, than H atoms in an H gas.)

If true what they say that carbon dioxide has exactly twice as much oxygen as carbon monoxide, then the 21.9 weight, being EXACTLY 8.0 units more than 13.9, allows us to reduce the 13.9 by 8 more units such that the only thing left is 5.9 units of carbon atoms. And so this is the true way in which they estimate that a theoretical carbon gas weighs 6 times more than H gas. That's how they arrived to .536 grams.

In the past, due to their assigning the carbon atom a weight of 12 times more than the hydrogen atom, I made the mistake of thinking that a theoretical carbon gas weighed 12 times more than hydrogen. But as they claim that the diatomic hydrogen atom (which makes hydrogen gas in their view) has a weight of two H atoms, then it's only six times heavier than hydrogen gas.

My task is to see if some configuration of the carbon monoxide and/or carbon dioxide molecule predicts certain out-spreading versus in-spreading forces explaining why these gases are roughly 14 and 22 times heavier than hydrogen. I'll start with the monoxide, which has 14 times as many atoms as an H gas simply because it weighs 14 times as much. As it seems that the difference between the monoxide and dioxide gas is that the latter has 8 more O atoms, it seems correct that for every 8 O atoms in the monoxide, there are 6 C atoms, and so the molecule could be 4 O atoms per 3 C atoms.

If the 3, merged C atoms form a sphere at the center of the molecule, they create three, semi-circular nooks. But how can we explain just 4 O atoms attached to this three-atom core with three nooks? Makes no sense. If we assume 4 merged O atoms forming the core, having four nooks, how can we explain 3 C atoms attached to a four-atom core with four nooks? Makes no sense.

It makes more sense that, rather than spherical, there are 3 C atoms merged in a LINEAR shape, for this forms only TWO full-circle nooks that can sensibly accommodate either 4 O atoms in the case of the monoxide, or 8 O atoms in the case of the dioxide. In order to prove this molecule shape, I've got to prove that each nook creates a large-enough circle to accommodate four O atoms.

As the way to find the relative size of the O atom is 1 / 16 = 1/16th the cross-sectional size than the H atom, the way to find the cross-sectional size of the C atom is: 16 / 6 = 2.67 times larger than the O atom...IF the C atom is indeed 6 times smaller than the H atom. In my books, a theoretical carbon gas weighing 6 times more than a hydrogen gas will have atoms 6 times smaller.

AI has this correct: "The diameter of a ball with 2.67 times the cross-sectional area of a 1-inch ball is approximately 1.63 inches." In this picture, the O atoms are 1.0 inch round. Can four, 1-inch balls fit around the circular nook of two merged balls each 1.63 inches round? Looks feasible, and a fifth may not physically fit to explain why the carbon dioxide molecule has a maximum of 8 O atoms (4 per nook).

I've got to hand it to AI for some information that I would not be able to get on my own without much effort. I asked: "if two balls 1.63 inches round were merged 20-percent into each other, what would be the circumference of a circle .4 inch away from nook where the balls meet, but not .4 inch inward into the balls but rather outward into the space "above" the nook?" The response, which includes a drawing at the bottom to verify that it understood my question, is: "The circumference of a circle .4 inch away from the nook (the crevice where the two balls meet), moving outward into the space above it, is approximately 5.59 inches."

I asked using .4 because that would be roughly the distance to the nook (the deepest part of the nook) from the center of a one-inch ball when it's slightly merged into the carbon balls. If the O atoms are more-deeply merged, the 5.59-inch circumference becomes shorter, and likewise it becomes shorter again if the C atoms are merged less than 20 percent. The 5.59 inches is a circle passing through the centers of the O atoms. It becomes 4.96 inches when asking the same question with .3.

Therefore, five O atoms can fit into each nook, but with a slight change in merger depths, a fifth one cannot fit. But even if five could fit, the inter-repulsion of 4 saddled atoms could keep a fifth from squeezing in and taking a seat.

Besides, even if we merge the carbon atoms deeper than 20 percent such that there is a nook 6 inches round or more, the space between neighboring O atoms will be very small. I asked: "if a ring through the center of one-inch balls was 6 inches long, what would be the distance from edge-to-edge of neighboring balls?" The response is .123 inches. There would need to be a lot of attraction between the carbon core and a fifth free-floating O atom to drag it in through a crack .123 inch wide while pushing the four in the nook simultaneously closer, forcing them to merge.

I would even argue that due to the inability of heat to separate the C atoms when squeezing into the O-filled nooks, the C atoms could be as much as 40 percent merged. AI says that such a 40-percent merger, with the center of O atoms .3 inch from the nook, the ring would be 5.98 inches long. Therefore, I do think I am correctly describing the look of the carbon dioxide molecule.

With 4 O atoms being so close together in the nook, added heat climbing to the range of 1,000 C serves to push them deeper into the nook, reducing the length of the circle, therefore, until the O atoms first make contact, and afterward merge into each other. However, that's not going to be the case for the carbon-monoxide molecule, if I'm correct in describing it as only two O atoms per nook.

The only way I can explain the difficulty in tearing the O atoms from the carbon-monoxide molecule is either by very-deep O-atom mergers into the nooks, or by some feature, perhaps atomic shape(s), that creates a rare, deeper merger with increasing heat rather than less merger (which is more-often the case by far).

I'm reading: "The Carbon-Oxygen (C-O) triple bond in carbon monoxide (CO) is one of the strongest chemical bonds known, meaning it requires extremely high temperatures to break apart (dissociate) into individual carbon and oxygen atoms. The thermal dissociation of CO into C and O atoms becomes significant only at temperatures above 4700 C." If correct that the C-O bond can be broken with heat (the liars can't be trusted with anything), I suggest that the O atoms are first pressed into the nooks more deeply, meaning they are forced to merge with the C atoms more deeply such as to defy the in-squeezing heat particles. But, finally, the in-squeezing separates the C atoms, at which time, yikes, the O atoms don't have the nooks any longer to take refuge in. They all scatter.

It begs why the monoxide seats only two atoms per nook when four are able to get in. Perhaps the two repel other O atoms away as they near, unless the temperature is hot enough (when the monoxide combusts) to allow two more to get in. The combustion of carbon monoxide means only that heat turns it into carbon dioxide. The ignition temperature of the monoxide is about 610 C. The clue here is that heat alone forces two more O atoms into each nook, for the very definition of burning monoxide is the merger of more O atoms into it. It's those extra mergers that release the heat of the combustion.

With 2 O atoms at opposite sides of a C-atom nook, such that a circle through their centers is 6 inches long (all around the nook), there is two full inches between the outer edges of the 2 i.e. lots of room to fit more O atoms. There's two inches of open space both on the left side of the molecule, and two inches on the right side of the molecule. It's very likely that one O atom can get into each space if only the C atoms could suck them in.

I would therefore claim that the repulsion between the 2 seated O atoms toward roamer O atoms is greater than the attraction between the roamers and the C atoms. Apparently, at 610 C, that changes: 2 roamers are sucked in such that, now, 4 atoms saddled in the nook, now with only .5 inches between their outer edges, have enough extra repulsion not to allow a 5th to be sucked in.

Then, at 1700 C, the heat disassociates (separates) the carbon dioxide molecule, turning it back to the monoxide. The only explanation I can see is that added heat above 610 C starts to unmerge the two roamers per nook that created the dioxide, until, at 1700, they pop out of the carbon atoms and become free roamers again. It suggests that the two roamers, though sucked into the nook at 610 C, are not permitted by their O neighbors to merge as deeply into the nook as they themselves are. While added heat above 610 pushes the original two deeper into the nook due to their deep merger to begin with, the same heat starts to pop the two roamers out because they are in a shallow merger.

The nooks I've not talked about until now, between the O and C atoms, is large where the roamers are merged, but small where the originals are merged. The larger the nook, the greater the numbers, and greater the out-spreading force, of the invading heat particles.

So far, this new view of the shape of the two carbon oxides has me excited. I've not had this shape in mind before. It's gets me more excited because it was always been a little unexpected that the monoxide should be spherical with three C, and four O, atoms combined. It was fuzzy to explain how/why the dioxide molecule could seat four more O atoms on top of the first four of the monoxide molecule when the first four were attracted to the C atoms.

I can tell you another thing, that the O atoms in the nooks serve to keep the line of 3 C atoms straight, especially when there are 4 O atoms per nook. If something bends the molecule into a C-shape, for example, it will squeeze four O atoms into a deeper merger on the inner side of the C-shape, such that they will want to return to their previous, default merger depths, which acts to straighten the C-shape back to an I-shape. And, the four O atoms on the outer side of the C will be forced to unmerge a little, wherefore they will seek to merge deeper to their default position, and once again this acts to straighten out the C-shape. Merger depths have a default position because that's where the attraction between merged atoms equals their repulsion of one another.

There's a good question as to why a string of 3 C atoms should not be 4 or 5 or more. I suggest that coke formed on metals when carbon dioxide is disassociated do in fact create linear molecules of more than 3 atoms, perhaps of many numbers, which can explain why high heat won't evaporate them: they are too heavy when in strings of more than 3. I assume that a molecule of four C atoms with three O-filled nooks is just too heavy to float in the air via heat-particle lift alone. Such molecules may exist, but they fall to / stay on/in the ground.

I'm not out of the woods because I've yet to test the out-spreading versus in-spreading forces to see if the carbon-oxide molecules predict the known weights of the monoxide and dioxide gases. As the monoxide burns with much heat, it attests to a deep O-into-C mergers. I'll assume 35 percent, and 20 percent for the C-into-C mergers.

I had shown above that the cross-sectional area of the C atom is 2.67 times more than for the O atom, meaning the that C atom has .785 x 2.67 = 2.1 square inches. When three such atoms are merged linearly, we use the full 2.1 for the first one, then remove 30 percent for each of the next two, for a total of 2.1 + 1.47 + 1.47 = 5.0 square inches.

AI made a drawing to scale of the dioxide molecule (scroll down), at my request; just remove two O atoms in each nook to see what the monoxide molecule generally looks like. The drawing has the O atoms too deep into the nooks, however. It won't give me a second drawing for the monoxide, too stingy, I suppose.

To this picture of the monoxide molecule, where we are viewing the molecule such that all three C atoms can be seen, we need to add the cross sections of roughly 65 percent of four O atoms 1-inch round, each having a cross section of .785 square inch prior to a deep 35-percent merger into the C atoms. However, from a top view, while still seeing all three C atoms, only two O atoms are visible that add zero cross section of the C atoms, wherefore, as the side view sees four, we use their average, two (midway between 0 and 4), for the math: (2 x .785) x .65 = 1.0 square inches. We add the latter to 5.0 to get 6.0 square inches total for this one perspective of the molecule.

However when viewing this molecule from its other perspective, from one end, where only one C atom can be seen, we begin with the 2.1 square-inch cross section of one C. In this view, heat-push can see only 2 of the 4 O atoms because 2 other are behind the first 2. Not only are they 35-percent merged into the C atoms, but some of the 65-percent remainder cannot be seen due to being in a nook. After making a drawing to scale, I saw that only about 55 percent of the O atoms can be seen. The math: 2 x .785 x .55 = .86 square inch of additional area on top of the 2.1 for this perspective, a total of 3.0.

The average heat-push is midway between 3.0 and 6.0, which is upon 4.5 square inches upon this monoxide molecule. This is equivalent to the heat-push on: 4.5 / .785 = 5.7 O atoms. We need to multiply the latter by 1.5, which gets 8.6 of out-spreading force as compared to the 7.0 in-spreading weight force of the 7 atoms that make the monoxide molecule. As the out-spreading supersedes the in-spreading, it expects that the monoxide gas is lighter than oxygen gas, and in fact it's 1.25 g/l as compared to 1.43 g/l for oxygen.

We can now do the same math for the dioxide molecule, but with 4 extra O atoms. However, from the first perspective, these extra atoms add zero cross-sectional area because there are two in front of, and two behind, the C atoms. Therefore, the dioxide's cross section is identical to the monoxide's as found above with (2 x .785) x .65 = 1.0 + 5.0 = 6.0.

For the other perspective, viewing from either end (but not both), we change the 2 x .785 x .55 + 2.1 above to: 4 x .785 x .55 + 2.1 = 3.8 square inches. We find the average of 3.8 and 6.0, which is 4.9 square inches. We then divide the latter by .785 to find 6.24, and finally we multiply it by 1.5 to find an out-spreading force of 9.4 versus an in-spreading force of 11.0 due to 11 atoms in the dioxide molecule. This expects the dioxide to weigh more than oxygen, and in fact it weighs more, 1.96 g/l versus 1.43.

I'm happy with this outcome, that while the monoxide needs to weigh less, the math got less, and while the dioxide needs to weigh more, the math got more.

All of this math is based on a sizing of C atoms as per the prediction that they are 1/6th the size of H atoms as per the predicted weight of a theoretical carbon gas of 6 times more than H gas. It's what gave the C atom the cross section of 2.1 inches. I use inches because the reader doesn't like microscopic dimensions. You can easily understand these sizes to scale, all relative to a one-inch O atom. It's a lot easier for me to do the math this way.

As the monoxide gas weighs 13.9 times H gas, and as that number is twice as much as the number of atoms in the monoxide molecule, I can glean that the molecules are twice as dense, per unit volume of space, as H atoms in an H gas. That is, twice as many molecules each seven times heavier makes a gas 14 times heavier.

If these were lone atoms, I would say that the molecules have half the cross section of H atoms, but as they are molecules, it doesn't apply because molecules weigh more than one atom. That is, the cross sectional out-spreading force is not all there is, but there is specific weight as well, and so that's why the math above was done with both under consideration. If they were lone atoms with twice the density per unit volume, I would say that these molecules get half as much heat-push (out-spreading) as H atoms, in order to find themselves at twice the density, because H atoms are likewise lone atoms i.e. likewise weighing one atom. But we cannot compare the cross sections of atoms versus molecules.

To put it another way, the carbon monoxide molecule, if it weighed one atom, would have a density-per-volume (or numbers-per-volume, same thing) almost the same as, or even less than, H atoms in an H gas, but due to weighing seven times more, the molecule's final density becomes twice as dense as H atoms.

The average cross section of the monoxide worked out to 4.5 square inches, and 4.9 for the dioxide. The latter is not much larger than 4.5 due to the placement of the extra 4 O atoms in such a way that don't add much to the cross section. The O atoms can be counted as extra cross section only if they protrude (stick out) from the C atoms, not if they are in front or behind the C atoms. As the monoxide must have less than half the cross section of an H atom, this 4.5 number seems to be verifying that the H atom is four inches round when the O atom is one inch round. A four-inch sphere has a cross section of 12.56 square inches, and so the 4.5 is indeed less than half of that, as it should be. The dioxide's 4.9 cross section is yet less than half of 12.56.

A 3-inch ball has 7.1 square inches of area, and so the theory that the H atom might be a sphere in the ballpark of 3 inches is ruled out here where 4.5 is larger than half of 7.1. It tends to suggest that O atoms, when merged into atoms, retain a strong inter-repulsion (to each other) because it's the only explanation I can see as to why the combustion of hydrogen allows only eight, 1-inch spheres to merge into a gigantic 4-inch sphere. The latter has 50 square inches upon its perimeter, yet the maximum area taken by eight O atoms, when merged 50 percent deep, is 8 x .785 = 6 square inches.

I asked: "if eight one-inch spheres are merged 25 percent into a 4-inch sphere, and spread evenly upon its perimeter, how far apart will the edges of the 1-inch sphere be?" The response is 1.6 inches. This is not a terrible situation at all. There is ample room between the eight to fit many more, but that 1.6-inch distance is relatively close such that the repulsion forces can conceivably keep other O atoms from squeezing in.

I asked: "what is the maximum number of one-inch balls, each merged 50 percent into a 4-inch ball, before the one-inch balls make contact? " "The maximum number of 1-inch balls that can be 50% merged into a 4-inch ball without touching is 32."

Hydrogen peroxide has twice the oxygen as does water, and the same amount of hydrogen, wherefore the peroxide is one H atom with 16 merged O atoms. To make this product more effectively, PRESSURE is used, which I assume squeezes the extra O atoms between the eight that are already there. With 16 merged, there is less than an inch between the edges of O atoms.

As the dioxide gas weighs 21.9 times H gas, and as that number is twice as much as the 11 atoms in the dioxide molecule, it tells me that the dioxide molecules are likewise, same as the monoxide molecules, twice as dense per volume as H atoms. I detect a problem in the math because, with four extra atoms but with little extra cross section, it should end up more dense than the monoxide. My math gave it only a little more cross section, not enough to counter 4 atoms of weight. I need 4 / 1.5 = 2.7 square inches more of cross section for the dioxide versus the monoxide.

Part of the solution can be where I suggested, with a reason, that the additional 4 O atoms are not as deeply merged as the first 4 O atoms (of the monoxide). The math above did not figure that in, and so that will increase the cross section.

Another part of the possible solution is that the additional 4 O atoms in the two nooks pulls the three C atoms closer i.e. forces them to merge more deeply, which then reduces the depth of the nooks, causing the first four O atoms to protrude further out from the C atoms than the 55 percent I used in the math.

I had said: "For the other perspective, viewing from either end (but not both), we change the 2 x .785 x .55 + 2.1 [for the monoxide] to: 4 x .785 x .55 + 2.1 = 3.8 square inches." But where the nooks are decreased in depth to the point where 70 percent of the monoxide's O atoms can be seen, and where even more of the dioxide's additions can be seen, I can average it out to 80 percent, in which case the math changes to: 4 x .785 x .8 + 2.1 = 4.6. On the other hand, the monoxide was found to be the above: 2 x .785 x .55 + 2.1 = 3.0. We have an extra 4.6 - 3.0 = 1.6 square inches when the monoxide becomes the dioxide, though we need to yet reduce that number due to the C atoms merging more closely, meaning it's not going to get to the target of 2.7.

If instead of their being merged 30 percent (as I have them in the math), we may assume 40 percent due to the addition of 4 more O atoms, the math changes from the 2.1 + 1.47 + 1.47 = 5.0 used above to: 2.1 + 1.26 + 1.26 = 4.6 square inches, for a loss of .4, bringing the 1.8 above down to 1.4, only half of the 2.7 target. But I think this is a pretty good theory, that the dioxide molecule grows in size sufficient to compensate for the addition of 4 more atoms of weight, so as to explain why the dioxide molecules don't position closer to each other than the monoxide molecules do.

The fact that the 1.4 is about half of the 2.7 target has got me wondering whether the cross section, when the molecules are viewed from the end, should include ALL the O atoms, in BOTH nooks. I've included the O atoms only in one nook because the other atoms in the second nook are behind them, not visible to the eye. However, if the O atoms in both nooks are "visible" to heat-push, then the 1.4 advantage that the dioxide has over the monoxide becomes 2.8! That works.

When we multiply 2.8 by 1.5 -- because out-spreading / heat-push on .785 square inch had worked out to 1.5 times stronger than the in-spreading of one atom of weight -- we get 4.2, perfect because the in-spreading is 4.0 due to the extra 4 atoms (in the dioxide molecule).

This means that whenever there is a linear-shaped molecule with atoms in nooks, ALL nooks should be considered as part of the cross section.

You may have realized that heat-push works from all directions such that there is net-zero heat-push in any direction. Yes, this is true for atoms in space, but not for atoms that are pushed against the walls of a sealed container, for example. They are the ones which receive the heat-push under discussion, and they are the ones technically causing gas pressure. The thing is, as atoms/molecules are allowed into a vacuum within the container, we want to know what the relative numbers-per-volume is when the gas pressure reaches STP. I'm teaching that the larger the atom, the fewer of them proportionally there will be in the container when the pressure reaches "standard pressure" at STP.

But when it comes to molecules that don't all weigh the same, it's not their size alone that matters in the resulting numbers-per-volume, because gravity pulls them harder as the number of atoms increases per molecule, and downward pull brings atoms closer together i.e. more numerous per volume. Thus, as it was shown above why there are the same number of monoxide and dioxide molecules at STP, the dioxide molecule must receive as much extra out-spreading force, due to larger size, as the extra downward force from gaining four extra atoms.

By the way, google AI may be as generous to you as it was to me when I asked the following question to find the depth of a nook. I had to re-phrase the question three times to get it to understand me: "when one circle 1.63 inches in diameter is superimposed on an identical circle such that the latter has only 70 percent of its cross section showing, how far is it from a line across the tops of the circles to the nook formed by the circles' intersection?" The response included a drawing at the bottom to verify it understood me. The distance given is .154 inches, meaning the nook is that deep.

However, a nook becomes much deeper with less merger. The same question posed with 90-percent of the cross section remaining after merger, the nook is said to be .33 inches deep, more than twice as deep, and also much wider so as to fit more of an atom into it. I asked for the situation with a circle 1.63 diameter because that's of the carbon atom in this discussion.

The Vatican Refused to Repent

Mike Winger's "good" for opposing catholic heresies because he's too kind. I'm not as kind to catholics. I wish them well, but for as long as they don't mind the horrible vatican history such that they yet call the vatican the true Church of Jesus, or the only Church of Jesus, I say they are worthy of dire warnings, even contempt when they become vatican evangelists. Mike is apt to making catholics feel too comfortable. In his video below, he shows why it's not enough for catholics to excuse Mariology by saying they merely honor her, and that's all that's going on. They go much further, especially into Biblical mysticism, making ridiculous claims, hoping to capture willing heretics or Bible fumblers:
https://www.youtube.com/watch?v=5p7DIQnRYO8

The heresies in Mariology alone is sufficient reason for all Bible-honoring catholics to abandon catholic churches. This is the biggest cult ever, perfectly defined because the pope thinks he's much weightier in holiness than most leaders of other cults. There's nothing to be afraid of in abandoning this cult. There's no reason to believe that it's the true Church of Jesus. Free yourselves. And if you really respect the Bible, then make light of catholic heresies, because, if you don't, how can Jesus become proud of you? How can you say with one side of your mouth that you are a Bible-based believer, but then allow vaticanites to freely, without challenges from the top, promote age-old heresies? You're being a Bible hypocrite to simply shun the Bible-jilting traditions of men that popes either created or blessed. Jilt the vatican, be free.

It makes sense that Revelation's Babylon the Great, which sat on the seven hill of Rome, touches upon the vatican. I suspect that vaticanites had to do with changing "Passover" to "Easter," and so it's conspicuous that: "Ishtar (or Eshtar) is primarily the ancient Mesopotamian goddess of love, war, fertility, and sexuality, serving as the Akkadian counterpart to the Sumerian Inanna. Known as the 'Queen of Heaven'..." You can clearly see that the vatican wanted to replace her with Mary. Babylon was in Mesopotamia. The vatican is on a hill just outside of the seven hills of Rome.

When I ask google, "derivation of "easter" but no comments from google AI please," not only does AI stick its face into the question at the top of the page, as always, but it refuses to give me a long list of articles on the subject which google knows exists. This is the kind of treatment we are getting now, with google AI sticking it's face into every facet of education, including Biblical, in order that it might have the priority and meanwhile hide the many ideas of the many others. AI wants us to believe that "Easter" has origins in a German goddess, Eostre/Ostern.

AI hides that Inanna was called either the "great mother of the heavenly dragon," or "mother of the great heavenly dragon," for when I ask google for, "Inanna "great mother of the dragon"," AI hides that title from us. Even when I enter those two quotes at google, it refuses to bring them up in any article, even though I have them in my own files from some other writer. For example, in my 4th update in May, 2010: "The dragon cult starts at Inanna's circle because she was styled the 'great mother of the heavenly dragon,' or 'mother of the great heavenly dragon' (I prefer the first reading). The cult obviously honored the serpent of Eden."

I'm not suggesting that a pope will be the anti-Christ. I've never believed that. Revelation 17 speaks to the dragon, the Roman empire, and connects its end-time hurrah to some Babylonian entity that made its way to Rome. It's possibly the Hebrews of Akkadia's Habur river, tributary of the EUPHRATes, whom I think named APHRODite. She's generally thought to have morphed into the Roman Venus.

Aphrodite loved the god of war, and the anti-Christ is portrayed as a god of war in Daniel 11. He rejects all gods but honors his own god of war. Both Daniel 7 and Revelation identify him as a Roman ruler, yet Daniel points us to expect him in the Middle East. Isaiah calls him "the Assyrian," and the "king of Babylon." Therefore, we are to look for a Roman entity in the Middle East.

The Americans and the Brits have cleared a path for a Western leader in the Middle East. Trump is not closing that path. But we need to see a Roman leader that despises Israel, who utters threats against it, siding with Iran, and likely going to the aid of Gazans. Kushner and Trump could be the factor that makes him decide to strike out, from his small base, with a small army, but with great determination and zeal, provoked to action by God as His tool to punish Israel, to purify the nation in preparation for the Appearance of the King of the Universe.

This is not God's endless hatred for Israel, as the vatican would like to see it, because the vatican thinks it's at the right hand of Jesus upon his eternal throne. What a grand delusion if any catholic takes this seriously. I agree with catholics that there is no such thing as "replacement theology," for the Church of Jesus did not replace Israel, but was rather the continuation of Israel. The vatican may not approve that the very basis of the Church is Israel. Jesus merely took rulership of Israel away from the Sanhedrin, and gave it to fishermen and tax collectors obedient to (in favor of) His mission. Jesus is not going to rule eternally from Rome.

Paul even said that true descendants of Abraham include the gentile Christians. He also made it crystal clear that the nation of Israel grafted in gentile Christians. It is true that national Israel is under a curse foretold from ancient times, but for the sake of God's promise to Abraham, He will choose some of his bloodline, including the 12 apostles, to become preferred ("firstfruits" Revelation 14) over other Christians, a special class, or at least a special representation, perhaps even superior.

Jesus is Care incarnate. Nobody written in the Book of Life will become a peasant in the Kingdom, for Jesus is not the elite ruler, but Care is His middle name. Love your neighbors means to care for them when they need care. Biblical love is CARE with action. Everyone understands care. If we want to become good Christians, ask God how we can learn to be appropriately caring. Even pagans are caring; even pagans love each other, said Jesus. Even pagans will stop on the side of the road to help someone in need. Even pagans will pick up a hitch-hiker, but Christians have the treasure of Jesus to share. We have more to give. Too bad so few understand how much it means to them.

On Sunday's Baron Coleman show, he makes a pretty-good connection between Andrew Kolvet of Turning Point to the Zionist-billionaire Adelsons desirous of maneuvering Trump. Turning Point is a Trumpian, political bastion, and so we should likely call it fact that it and the Adelsons are involved in the Gaza-rebuild.

Baron's show above concentrates on Mark Burnett's links to SHARK TANK and The Apprentice, and thus links to Trump's TV fame and fortunes which catapulted him to the White House. I claim that Trump jumped into the swimming pool with shark, in my Sleeping Beauty dream, where he was half SWALLOWed by the nasty shark. I tend to identify the Shark surname with the SARACA's of RAGUSa, and therefore with the RAGGS' sharing the fleur-de-lys of BURNs, the latter first known in Cumberland with the SARACens, and with the Burnetts, and also with the Browns/Bruns while the Raggs', apart from their fleur-de-lys, share the Bruno Coat. English Bruns are first known in Middlesex with The Horns in the Burnett and Burn hunting horns.

The Burnetts even share the fesse in this Arms of Saraca (shows a fish). Plus, the swimming pool was kidney shaped while the Kidneys/Gedneys (Lincolnshire with the Fish's and Swallows) use "fish" while Fish's share the Scottish Brown/Brun Coat. The heraldry seems to fit the fact that Mark Burnett was the creator of Shark Tank and the Apprentice. I've never been on this heraldic turf before as per Shark Tank and The Apprentice.

There's also the Brunells, first known in Shropshire with Swallow-branch Sallows, and sharing the Coat of Adells/Adelmanns. The latter sounds like a Jewish surname, and as houseofnames has no Adelson surname coming up, it seems we can use the Adells, especially as they and Brunells share the lion of English Jeans while Jean-Luc Brunel was a partner with Jeffrey Epstein. Baron's video explains why the Adelsons might want to punish/sink Trump at this time, though it also gives hint that the Adelsons (and Rupert Murdoch, no surprise) are backing Trump's Iran war.

The shark represented Trump's demise, and so it seems he's on his way down at this very time in which he won't quit supporting Israeli gangsterism. Play with fire, get burnt.

he shark showed its nasty teeth, what I now think is a pointer to the Tooths because they share the Coat of Goose-loving Lauders in turn in the colors and format of the giant goose of Gaza-like German Gas'. Then, the proto-Washington Gaze's almost incorporate the Toot Coat. Toots love the Cressents/Craits who in turn share the Coat of NERETs (Brittany with Sarasins) while the Saraca's of Ragusa were near the NERETva river. The Craiths/Creights almost have the Washington Coat.

Creightons share the Coat of Adells/Adelmanns and Bruce's (Yorkshire with Toots and Belows), and while I trace Bruce's to the lion in the Arms of Brescia, it's a city beside Val TROMPia. The Tout variation of Toots is in the motto of English Belows while German Belows (Pomerania with Trumps/Tromps) essentially share the giant eagle of Dutch Tromps, but see also the Maxwells (ROXburghshire with Goose's) as per Ghislaine Maxwell and her father.

Irish Tute's/Tuits are first known in Norfolk with their Thwait branch who in turn share the fretty of English Belows. The latter use a "chalice" while Chalice's/Challes' get us to the Calles' (Wiltshire with English Neals) showing nothing but trumpets in ROACH colors and format. Irish Neals/O'Nails are first known in Tyrone with Sharks (share green trefoil with fish-using Rocks). German Neils/Nails have a saltire in the colors of the saltire-by-fish of Kidneys, and so the fish in the Arms of Saraca can be the fish of German Fishers, but also the one of Trump-connectable Carps (Mecklenburg with Trumps, in Trump colors and format).

Note that "Adell" is like the ADL, an Israeli organization thought to me a type of Mafia.

Beware Baron Coleman. He's not a good influence on Christians, but is more of a modernist/worldly vaticanite than full-true-blue to Jesus. He strikes me as the type who might turn and feel sorrow for the burning of Sodom.

Trump agreed to let Charlie Kirk feign his death, because it was time for him to get out of Trump's Turning Point, because Charlie was opposed to the Iran war and Gaza re-build. Charlie decided, rather than to go forward as Trump wished, he'd quit. But as quitting would look very bad for Turning Point without a reasonable explanation, the "Israeli lobby," with Trump's knowledge and support, convinced him (or vice-versa) to feign his murder, and to leave Erika too, perhaps with some compensation in return. Baron Coleman is such a conniving schmuck that he won't even present this possibility to his audience.

Everything screams crisis actors with Kirk's shooting event, but I'm now starting to believe that Trump has his fingers deep in some Intelligence-linked crisis-actor team(s) that he's been using to what he thinks is his advantage. The harder this dodge-the-voters game plays on his psyche, the more we should expect a tired-old man within the next year or so, especially if the year-end elections make him lame. Participating in faked events (intended to deceive the public) compromises a president, meaning he's under the thumb of Intelligence and others who could spill the beans. It's like making a blood-brother / Masonic pact, like being a slave to them, what a horrible end this imposter will have.

NEWS

Maybe too-little too-late with some light on the parties guilty of COVID-injection injuries, especially as Trump is protecting this crime because he proudly distributed the injections. The numbers are much higher than "the tens of thousands" that Ron Johnson mentions
https://www.bitchute.com/video/EkZkNroL7o2o

In the Jaxen Report above, we find an amendment to a farm bill allowing farmers to sell locally without the federal government saying "no" or sticking a hand or voice into the transactions. This is a good anti-globalist move.

I figure that Trump's only original concern in Iran is to get it to stop funding Hamas so that Kushner and company can advance the Gaza re-build sooner than later. However, Trump may have developed secondary goals that makes him just like any other treasury-abusing president who rules for fat cats in oil and related industries. The U.S. military is part of the oil racket, of course, but so is Israel these days, since the Iraq / Syria war empowered Kurdish oil.

As I've expected the anti-Christ to invade Egypt in the first half of the 70th Week, this lingering and budding Gaza situation looks like the magnet that brings him down. Trump's willingness to disappoint much of his voting base over this on-off war with Iran indicates his avid determination to fulfill the Gaza re-build.

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NEXT UPDATE Next Monday

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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