This page continues from last week's update at the link above.

The Weakest Link in a Bell

The youtube video below shows that a bell's sound becomes more bell-like when the bell is heated, and I think you can make out that it's due to the atoms vibrating while further apart per increasing heat. The heat is acting to unlock the merged atoms to a degree, allowing them to vibrate more when struck by an object. But nobody should be able to argue that tapping the bell slightly with a light object, as we see in the video, should be able to create sound while the kinetic atoms that supposedly vibrate at many hundred of miles per hour make no sound at all. That's because the atoms are not vibrating until the bell is tapped, duh:
https://www.youtube.com/watch?v=id-u9ro50TM

By more "bell-like," the tone has longer vibrations in combination with higher pitch, suggesting that higher pitch is more-forceful sound waves, which is attested to where high pitch is more painful to eardrums rather than low pitch. Sound waves form when atoms emit outward into the sound-wave "aether" (the air), and light waves form when electrons emit into the light-wave aether. Stronger, more-frequent light waves hurt the eyes.

Lead is not a good bell material, suggesting the lead atoms are strongly bonded / deeply merged such as to be unable to vibrate much. The sound or vibratory abilities of tapped materials could therefore systematically reveal how deeply and/or strongly atoms are merged.

AI: "Pure copper alone is generally not considered a good material for traditional, resonant bells because it is too soft and dampens sound, resulting in a dull 'thunk'". Then, "Pure tin is a very soft metal that would easily deform upon being struck. Tin has low tensile strength, making it unsuitable for the force required to create a resonant sound in a bell."

On the other hand, "Traditional high-quality bells are made from bell metal, which is an alloy of roughly 80% copper and 20% tin (a form of bronze), rather than pure copper or pure tin alone." What can we learn from this? I'm seeing that tin has the weaker atom of the two, allowing atoms of a copper-tin alloy (bronze) to merge with less bonding strength at 20-percent of its atoms, and consequently allowing for a higher-pitch when struck.

The good thing for this discussion is that tin and copper essentially have the same boiling points. The definition of boiling point is when the number of heat particles going into the air equals the number of heat particles entering the liquid. Someone needs to grapple with the reason that, suddenly turning up the heat source at boiling point cannot raise the temperature. I saw, I grappled, I explained.

The boiling point is when heat particles have finally spread certain (not all) liquid atoms apart, or the point where they have weakened some liquid bonds sufficiently, to allow heat particles to move through the liquid such that they exit the surface, mainly in boil bubbles, in numbers equal to the number entering at the heat source. Most of the water molecules in boiling water still resist heat passage at boiling point, but some do not.

When boil bubbles travel through the liquid with zero resistance, that's boiling point. That is the best and truest definition of boiling you have ever heard. The main transport of heat particles through the liquid is from within the boil bubbles. Heat-particle buses. Boil bubbles are formed from the outspreading of dense heat particles under their inter-repulsion.

At boiling point, no matter than we turn up the gas flame under a pot of water, rather than increasing the liquid temperature, it only increases the size and speeds the rise of boil bubbles, which can only mean that they are passing through the liquid with zero resistance to upward flow, at the boiling point. Only the atoms in contact with the bubbles are conquered of all resistance. The rest of the liquid yet offers resistance to upward heat-particle flow. The bubbles take the same paths because they first work to reduce restriction in various paths, and then naturally keep the same paths once they enlarge because they insert themselves into paths of least resistance.

When people go to slowly through a self-closing door, the door will be closing by the time the next person arrives, and that person finds some resistance because the door needs to be fully opened again to get through it. But when people file through the door so quickly that it has no time to close, work is no longer needed to keep it open. That's boiling point for boil bubbles, defined as passageways fully open, no restriction.

I'm open to another definition, where some small resistance yet exists, but where bubble size increases with rise distance such that the numbers of heat particles exiting the liquid equals the number entering. As soon as bubbles pop out into the air, they release their passengers into the air. As the bubbles increase in size with upward flow, it's clear that they are picking up more passengers along the way. The bubble has no wall to keep passengers from escaping. Heat particles do escape, yet if the bubble picks up more than escape, the bubble grows in size.

Passage makes more passage easier, allowing bubbles to have more passengers. That's it, the true mechanics of boiling point, bubbles versus the liquid's desire to close the bubbles. The bubbles win thanks to increasing heat weakening the atomic bonds. Bubbles can become larger in higher temperatures, it's clear.

My all-atoms-weigh-the-same law predicts, on the one hand, that tin atoms are slightly larger than copper atoms, for a unit volume of copper solid is 1.22 times heavier than for tin, and copper liquid is 1.14 times heavier than for tin. The heavier the solid or liquid per unit volume, the more atoms it has according to the law above, and consequently the first order of suspicion is that it has the smaller atom.

However, this gets nasty for making predictions for all types of atoms, because atomic-merger depth also plays a role in explaining the relative weights of materials. For example, tin and copper atoms could be the same size, yet if copper atoms are merged more deeply, it can explain why copper weighs more. The relative weights of solids and liquids needs to consider both density of numbers and merger depth, and that frankly stinks for anyone wanting to know which atoms are larger.

There are different ways to get a handle on relative atomic sizes, but if two or more things are going on simultaneously, out the window goes an easy means to determine atomic sizes. The good news is that one can know the relative size of atoms from the evaporation rate of a liquid if one knows the atomic-bond strength of the liquid while evaporation is taking place. And that's the bad news.

Evaporation, which occurs below the boiling point, is when the atomic-bond strength is overcome by the upward flow of heat particles that serve to knock liquid atoms into the air. The atomic bond is broken more easily when atoms are larger, for as they all weigh the same, the larger atom needs less rising heat particles to get knocked into the air. That is, larger atoms can be expected to evaporate at lower temperatures, yet sizing them is not so simple as proportionality with the temperature at which evaporation first occurs.

If someone can know that two liquids have equal atomic-bond strength when one suffers initial evaporation at 300 degrees while the other suffers initial evaporation at 200 degrees, then one can know that the material evaporating at 200 degrees has the larger atom. Evaporation is a battle against atomic-bond strength, and so it's possible that the material with initial evaporation at a higher temperature has the larger atom. Though larger, it needs a higher temperature to break the stronger atomic bond as compared to a material with smaller atoms and suffering evaporation at a lower temperature. That is the bad news.

However, I think it's a given that the material with the lower evaporation point has the larger atom when two materials initiate evaporation several hundred degrees apart. Safe to say, we can know which is the larger atom, but until one knows the precise atomic-bond strength of both liquids, the perfect relative sizing of their atoms escapes us...by using this method alone. The good news is, there are other methods contributing to the sizing of atoms. The bad news is, they too are complicated.

Some possible good news: the relative atomic-bond strengths of all materials may be proportional to their melting points. The lower the melting point, the lower the atomic-bond strength. If this proportionality is general true, one can probably size most atoms fairly well. It would be a new science. Need a new hobby? Go for it. It's better for you than watching youtube to waste your time away.

More bad news was, there's no tables I could find listing or detailing evaporation rates, the last time I checked (before having AI access) at google. The good news is that google AI answers my question with: "Based on studies of evaporation from liquid metal, specifically in vacuum conditions used for metal purification, liquid copper has a higher evaporation rate than liquid tin at comparable temperatures." The first order of suspicion is that copper has the larger atom, thus getting the stronger heat-particle lift when it and tin are at equal temperatures.

However, as copper weighs about 1.25 times more than tin, the prediction -- from that consideration alone -- is that copper has the smaller atom. But not necessarily, for copper atoms could be more-deeply merged, which can explain why copper is harder to bend or deform i.e. why its atoms bond stronger.

However, if we are tempted to say that copper and tin atoms are roughly the same size, but with copper merged more deeply, then we'd expect tin to evaporate more easily because evaporation is a battle against atomic-bond strength. If indeed copper evaporates more easily than tin, I think we can conclude that, in spite of copper atoms merged bore deeply, they are also larger such that extra heat-particle lift (due to size) more than counteracts their stronger bond.

The more strongly copper atoms are bonded as compared to tin atoms, the larger we can expect the copper atom as compared to the tin atom. The low melting point of tin versus the same for copper could reveal that copper has a large atomic-bond strength by comparison, compelling to size the copper atom as the larger one by a significant degree. We could get a better grasp as to precise sizing if we could get detailed evaporation data.

I asked google: "evaporation rate of copper versus tin", and part of the response was insightful: "Start Temperatures: Under electron beam evaporation, the starting temperatures for evaporation are lower for copper than for tin, indicating that copper begins to evaporate at a significantly lower temperature." We want the beginning of evaporation, when the heat-particle lift becomes strong enough to lift JUST ONE liquid atom into the air. At higher temperatures, heat-particle lift likely sends two or more attached liquid atoms into the air. So, if you want to know the start-temperature of evaporation, task AI with "electron beam evaporation" in your query.

I'm providing an example of a revolutionary, or at least brand-spanking new, set of methods of finding relative atomic sizes and merger depths / bond strengths. I really like this science when my mind is up to the challenge.

Both tin and copper have roughly the same boiling temperature, which presents such a grave problem that only one explanation seems suitable. The density of heat particles in both liquids is rather identical at their boiling points, and both liquids have upward electrons passing through at the same rate.

google AI: "Tin generally produces larger boiling bubbles than copper at their respective boiling points." It's hinting that tin has the weaker atomic-bond strength, not surprising, but tin also has less liquid pressure, and that too contributes to larger boil bubbles. When a combination of factors are involved, it means, BRAIN FOG. We now can't know how much of bubble-size difference between tin and copper is in regard to tin's atomic-bond strength alone. But, in the midst of the fog, we can get a ghostly glimpse of the situation, or a feel for it.

Asking for relative bubble sizes, AI had a response: "At their respective boiling points, tin's bubbles are approximately 2.3 to 2.5 times larger than copper's in diameter. This difference is primarily driven by tin's significantly lower surface tension..." The latter point (probably incorrect) reveals that these bubbles are the semi-spherical ones with a liquid skin, existing above the liquid surface while sitting on the surface, after flowing through the liquid. I watched water's boil bubbles grow smaller quickly on the surface until bursting. That's because heat particles in the bubbles go fast through their water skin.

Nobody can see boil bubbles in tin or copper, and so the best AI could do when asking for in-liquid bubble size is a math solution they call the "Fritz correlation." You should be wary of this math, not only because it finds tin's in-liquid bubbles only .7 the size of copper boil bubbles, but because the quackers don't understand atomic mechanics in the first place due to their view on kinetic atoms and no heat particles. As the bubble sizes they see on the surface don't jibe with their hallucination of liquid mechanics, they just fake in-liquid bubble sizes with math, and then claim that the bubbles grow or shrink when reaching the surface due to some off-the-wall, liquid-tension condition they perceive at the surface.

In my mind, it seems that, if the copper has the smaller bubbles at the visible surface, they are the best indication of relative, in-liquid bubble sizes. With my heat-particle-bus model, it makes sense that tin has the larger in-liquid bubbles for two of three possible reasons. I'll get to the third reason, atomic shape, below, which I think causes tin's bubbles to be smaller than they would otherwise be if atomic shape were not a factor in bubble passage.

HEAD ACHE: weaker atomic bonds for tin expects heat passage easier through tin liquid, and thus a lower boiling point for tin. Instead, tin has a slightly-higher boiling point than copper. I can show that copper atoms have bonds stronger than tin atoms where tin has the lower melting point by a whopping 850 degrees C (232 C versus 1,085). Yet tin, even though it's lighter than copper (less liquid pressure), has a higher boiling point than copper????? C'mon, make this work.

Melting point can be defined as when the bond strength of the atoms of solids has been weakened to the point that gravity can pull those atoms downward toward itself. Melting point is clearly the point when added heat is finally able to dislodge solid atoms in combination with gravity force. There we have some hope of clearing some fog, especially if the battle is only atomic-bond strength versus gravity. Alas, liquid pressure pushes atoms inward toward each other, acting against the heat that wants to weaken the bonds. We need a name for when both bond strength and liquid pressure work simultaneously to keep atoms from putting distance between each other. I know of no name for this total resistance to atomic spreading, which includes air pressure acting on the liquid surface, though we can ignore it for this discussion because it's always at the same pressure. Let's call it TOTAL bond strength.

There is great news if only it's true that all materials have the same total bond strength at their melting points. Gravity force alone cannot dislodge the atoms at the melting point, but contributes to weakening atomic bonds to the degree that atoms roll on each other while bonded. That rolling is the definition of a liquid.

No matter how many contributors there are total atomic-bond strength, melting point is when the latter equals the gravity force. To come to this conclusion, imagine the melting taking place on the SIDE of a solid. As soon as the first melting takes place, gravity is able to pull the liquid to itself. It can only mean that every liquid at its melting point has atoms resisting separation from each other by a force equal to the downward pull of gravity upon those same atoms. With a little extra temperature, gravity wins the tug of war.

Therefore, tin has less atomic-bond strength by far as compared to copper atoms, for far less heat is needed to loosen/separate tin atoms to the point of their having attraction equal in force to gravity force.

From the mechanics involved here, we can glean that atoms attract each other with less force the more that they are spread apart, for if their attraction grows larger, or if it remains the same, as heat spreads them apart, their total bond strength could not go weaker to equal the pull of gravity.

To put it another way, liquid pressure continually applies to resist atomic separation, and it applies with exactly the same force no matter how separated atoms become. Therefore, some other force(s) or "glue" resisting atomic separation must be weakening during the partial-spreading apart of solid atoms.

If you think about this longer, you can realize that, at some point above the melting point, gravity force is able to fully dislodge liquid atoms, yet we can't witness it because the bottom of liquids are always upon a material surface that forbids the atoms from flying off toward gravity.

However, as we can see in the case of a water droplet hanging off of a vertical surface, gravity probably causes (indirectly) its evaporation before gravity can pull its molecules fully apart and toward itself. We imagine gravity coming close to fully pulling a water molecules apart from the drop, but just before it happens, the rising heat particles sweep the molecule into the air as part of evaporation. Gravity has sufficient power to pull the evaporated water molecule downward as dew, if the air is cool enough i.e. if the upward heat-particle flow is weak enough. In that picture, I see as possible that gravity could steal molecules, from a water droplet, downward to itself. The problem is, the colder the droplet, the less that heat weakens molecular bonds such that gravity could take advantage as a robber.

We naturally want to know why tin needs the same heat density to reach boiling point if tin atoms are easier to dislodge from each other. If I have all the facts correct to this point, something else, besides atomic-bond strength, resists the upward passage of heat particles and boil bubbles in the case of tin. I can't appeal to the smaller tin atom because, in spite of its possible smaller size, it has less liquid pressure to combat upward heat rise. Smaller atomic size acts to increase liquid pressure, yet tin liquid has less liquid pressure than copper because tin atoms are less-deeply merged.

How in tarnation can smaller, more-weakly bonded tin atoms resist upward heat more than more-deeply merged copper atoms do? Grave problem. But wait. One of the reasons can logically be atomic shape. Tin atoms must have some sort of protonic shape that creates more difficulty for rising heat particles than spherical protons. For example, L-shaped protons that hook a little into each other.

Why is it that copper and tin both make thuds rather than ringing sounds, yet when 80-percent copper is mixed as an alloy with 20 percent tin, suddenly, the bell sound is nice? I argued that bell sound needs the atoms to be less-strongly bonded, more capable of vibrating against air atoms with greater force, which must be the case where copper atoms merge with tin atoms. But why?

I hate to toss a monkey wrench into the machinery, but it now looks like a must-do. Judging by the low melting point of tin as compared to copper, one might get the impression that tin has far-smaller atoms than copper, which are therefore far-less merged to explain the lower melting point, yet the feel I got above was that tin could have atoms approaching the size of copper atoms. A way to satisfy all the requirements is for tin to have a giant but weak proton. The large size of the proton gives the tin atom more bulk size, but in possessing weak positive force, it's electron atmosphere is shallow, allowing for only weak bonding. The monkey wrench, at your service. It doesn't ruin the machinery, but can help to solve a problem, where needed.

We now ask: would heat have a greater resistance to passage through a tin liquid having large protons and shallow atmospheres, or through a tin liquid having small protons and deeper atmospheres where both atoms weigh the same, and where both atoms have the same merger force of attraction (i.e. same bond strength)? As protons are brick walls while captured-electron atmospheres have much space, the first option is expected to offer more resistance to heat rise, and so that picture, ganged up with an atomic shape capable of hooking atoms to some degree, can explain the unexpectedly high boiling point of tin.

I dislike it when there two options as possible explanations, for it drastically complicates the hunt for the true look of the microscopic situation. There's no way I can see how anyone can know whether an atom has a large-but-weak proton versus a smaller-but-stronger proton. The reason I'm presenting these two options is that both can provide EQUAL inter-attraction because the large-but weak proton can allow a larger merger region while the smaller atom has higher attraction force but with less merger region due to the smaller size of the atom.

Therefore, even if one can get a good feel for the specific bond strength of atoms, it doesn't tell which of the two options are in play, but if the one better satisfies the need, for example the need to explain tin's high boiling point, then use it. In this picture, tin's atomic size can be roughly that of copper's, and I suspect that large-but-weak protons allows more malleability of the metal than smaller-but-stronger atoms. The larger the atoms, the fewer "hinges" to bend.

In an 80-20 mix, there would be eight copper atoms per two tin atoms if both tin and copper solids had the same density of atoms. As we saw above that copper atoms are more numerous per unit volume, we might imagine ten copper atoms per two tin, or five per one, in the 80-20 mix.

An alloy is formed when the two liquids form a solution, defined where two or more types of atoms spread out as far apart as they can be such that they are evenly spread (same distances apart per tin atoms). We can easily imagine five copper atoms with a tin atom at their center, a pattern repeated throughout the liquid, and of course this pattern remains when the liquid solidifies.

We might imagine that when we tap the copper-tin bell, five copper atoms jump out rather freely toward the air because the one tin atom, to which the five are attached, has a shallow electron atmosphere. The bell sound is, therefore, predicted to be from copper atoms setting up the sound wave as they bounce up from weak tin atoms.

I asked google a question, and it told me what I expected: "Here is what happens to bronze with 40% tin: Extreme Brittleness: At this high percentage, the material becomes brittle, often compared to glass or cast iron. It is prone to fracturing rather than bending when under stress." Too many weak links exist from too many weak tin atoms.

We are hearing the sound of vibrating copper atoms, and the sound, per one strike of the bell, doesn't go on forever because attraction forces bring vibrating atoms to a motionless state, a thing the kineticist denies purely because he's an ignoramus in his treatment of physics. He clearly sees that atomic vibrations come to an end, but rather than admitting that all atoms in a solid tend to non-motion (i.e. no such thing as kinetic atoms), he argues that every strike to a bell scatters the motion energy thin throughout the bell material such that sound is no longer audible.

According to the kinetic model, a continual striking of any material should increase its temperature because the kineticist thinks that the atomic vibrations of solids defines heat. He'll then argue that striking a bell does increase its heat, but I'll argue it's from frictional heat, not due to faster-vibrating atoms. As the vibrations he imagines are not audible, each hard striking of a material should greatly increase temperature simply because its audible.

Ask thineself: if atomic crashes at thousands of mph make no sound, how can a straw tapped at 5 mph against your table make noise?

He's got the atoms of solids vibrating at hundreds of mph, yet if we strike a bell with a metal ball at just five mph, the sound rebounds within the bell, striking all the atoms inside of the bell multiple times before sound dies off fully. Yet his atoms at hundreds of mph make zero sound??? C'mon, only an ignoramus argues in this way.

His vibrations are purely mechanical, atom-to-atom force, as is the striking of any material with any object. In both cases, nothing is going on aside from atoms striking atoms, yet his atoms make no noise because they are not in fact vibrating. The more he makes excuses as to why they don't make noise, the more you should view him as a fraud. It breaks the laws of physics for atoms to be under attraction forces to one another while never ceasing their vibrations. Only an ignoramus tries to wiggle out from such an obvious fact.

He knows that striking an object in a vacuum causes vibrations for only so long. As the motion energy can't get past the vacuum for lack of atoms there, he's required to explain how the vibrations cease, but rather than argue that attraction forces between the atoms of the object bring the vibrations to a halt, he'll give some excuse or argumentation as to why total motion can never be killed. It obviously can be killed, duh.

Room temperature is about 300 K degrees. He claims that the vibrations in a sheet of metal are sufficient to keep it at 300 K. But we can pound every square inch of that sheet 3,000 times per minute, yet though the vibrations are now very audible and therefore of much-greater energy than the inaudible vibrations he imagines, the sheet will not increase in heat by much, and moreover it's obvious that there's frictional heat involved at the points where the sheet is struck that have nothing to do with his imagined scattering of heat-motion to every atom in the sheet. There's no doubt that all the pounding energy spreads to every atom in the sheet, but it's not heat energy, and whatever vibrations are started quickly cease.

Kineticists pick and choose their science "facts" according to their need for fixes of erroneous theories. In this case, we can predict that, should I ask google, "are air atoms in contact with objects," the response will be: "No Physical Contact: Atoms are surrounded by electron clouds rather than solid shells. When one object (like your hand) approaches another (like a table), the electron clouds of the atoms in your hand repel the electron clouds of the atoms in the table." I've never heard this silliness before, but AI was able to find it for us. Who in tarnation would create such a theory, and why, that gas atoms do not touch objects due to inter-repulsion at close range?

It's the expected explanation as to why the imagined bombardment of air atoms against everything in your room make zero sound. These same imposters argue that gas atoms attract each other at a small distance to form liquids, and here they now need to argue that atoms repel each other when drawing close. PICK-AND-CHOOSE IMPOSTERS.

We can now know partly why they frame their atomic model such that liquid and solid atoms are not in physical contact, certainly not merged into one another. They are geared to inventing reason for lack of noise in kinetic atoms, even though they full-well know that kinetic atoms do make noise. They are willing to go as far out on a limb as laughably claiming that solid atoms make no contact, for noise requires atom-to-atom contact.

MADNESS: they claim that gas atoms make physical contact some ten-billion times per second, at something like 1,500 mph or better. Yet we hear no noise.

In order for condensation to take place on a can of frozen orange juice out of your freezer, water molecules need to make contact upon the atoms of the can in spite of their repelling each other. They make no noise while coming into contact with the juice can, for obvious reason that the contact speed is slow.

Yes, air atoms do repel each other, which is exactly why they make contact with EVERYTHING. They force each other onto the surfaces of liquids and solids, and it's called air pressure. I had to grapple with this particular question as to whether gas atoms contact objects physically or only via their inter-repulsion forces. Condensation taught me that physical contact is made and maintained by attraction forces, otherwise the first water molecules to condense near a cold surface would float away in the air before it could build up on a surface.

But even if no physical contact is made by air atoms upon solid surfaces, the fact that vibrating solid atoms cause sound in spite of it only serves to show that fast kinetic atoms are expected to make sound whether they make physical contact or not against anything they strike.

Even if I take the position that gas atoms do not physically touch objects, kineticists do not have the same luxury because their atoms are screeeching toward materials at many hundreds of mph. Sorry, baboons, you don't have that option. Ask the wind about that.

It's nonsense for them to claim that atoms repel each other from a close distance only to then say they attract each other into a bond at yet a closer distance. That's called brain infection by the moron bug. The big bang exploded the moron particle into the universe, which has an affinity for the brains of the Godless. Not all Christians are immune, for some not only subscribe to the big bang, but are blind to the should-be-obvious wacko physics taught everywhere.

Kineticists need gas atoms to attract from a distance in order to form condensation upon a surface. If they attract from a distance, it can only be by the positive charge at the atomic nucleus attracting another atom's electrons. Therefore, these monkeys cannot be permitted to turn and say that electron-to-electron repulsion is in effect when gas atoms draw near to the atoms of solids. Therefore, in not having the luxury to deny physical contact, their speedy air atoms are expected to crash deeply into the atoms of all solids. Forced "mergers" are the expected result. CHAOS at the atomic level.

They teach that tapping an object lightly with your finger nail makes many times more noise than every air atom in the room crashing into each other zillions of times per second. And they make this claim knowing that sound is a wave through air atoms. How can a reasonable person wiggle out and excuse themselves from incrimination by this laughable nonsense? If you really want to know the answer, just ask google, "why don't air atoms make noise as they crash into each other?" The response is funnier than I thought it might be:

Air atoms (molecules) do not make audible noise when they crash into each other because sound requires a synchronized, collective movement of billions of molecules, whereas individual molecular collisions are random, extremely small, and happen at too high a frequency for human ears to detect.

There's two excuses thrown in for good measure. If you don't bite at the first excuse, you can choose the inaudible noise excuse. If we cool air, the too-high-to-hear sound should become audible at some low temperature. Yes, for if the sound is too high in pitch for the ear to register as audible sound, and as it can only be too high due to the fast velocity of the atoms, then, at some velocity closer to zero velocity, the sound should become audible. Therefore, this excuse isn't valid.

In what other way should the pitch be too high to hear if not for the fast velocity of the atoms? They cannot argue that the smallness of the atom is responsible because, alas, the schemers know that moving atoms do make sound. Yes, even though they know that wind makes audible sound, they argue that atoms in collision can make no audible sound. That's called, craft. The crafty kineticist, the same sort of people who clump together in liberal politics, who are showing the world, as we speak, that they're willing to cling to their own fabrications for the sake of profit. Humanity is so capable of deception when it comes to protecting the empire.

As they know, wind, moving a lot slower than 2,000 mph, makes noise when it collides with leaves, for example. Those are lone atoms each contributing to the sound level. Yet, the quote above claims that no sound is expected from lone-atom collisions because "molecular collisions are RANDOM, extremely small." Smallness isn't an excuse because it needs to be combined with great volume of lone atoms. No matter how slight the noise from two atoms colliding, all the atoms in your room multiply that sound by zillions of times. The RANDOM direction of atomic collisions seems immaterial to me as to whether or not the collisions should produce noise.

The quote claims that sound MUST be from "a synchronized, COLLECTIVE movement," which refers to objects as opposed to lone atoms, a weak excuse. A guitar string is a collective movement with far fewer metal atoms striking air atoms than the kinetic theory has in all your room COLLECTIVELY striking each other. I don't see what synchronization has to do with creating sound. Anything striking air atoms, including air atoms to air atoms, if they strike without synchronization, will make a noise that doesn't sound synchronized.

AI makes it easy to make deceivers out of kineticists. Just ask: "does air striking air make noise?" "Yes, air striking air makes noise," it says. You can hear the roar in the sky when winds are in the range of 50-60 mph. AI: "During severe weather, the extreme turbulence and high-speed, moving air can produce a sound often described as a 'freight train' or a continuous, deep roar." That's got to be from faster air striking slower air, or from wind striking wind.

That is, as a fast gust of wind scrapes along the "walls" of slower air currents, noise happens. At the root, it's air atoms striking air atoms...at far slower speeds than kineticists imagine them to be striking each other in windless air. Therefore, kineticists are bound by this fact to cease teaching kineticism.

Zinc Versus Tin

The copper-tin alloy makes bronze while a copper-zinc alloy makes brass. Zinc and tin have near-identical weights per unit volume, which on that basis makes them appear the same diameter, unless one is more-deeply merged than the other. The one more-deeply merged should be the larger atom, because, if it were not, its material would weigh the lesser of the two, but if it's more-deeply merged, its material will weigh the most. How can we discover which of the two materials have deeper-merged atoms?

Zinc has a melting temperature of 419.5 C versus 232 C for tin, suggesting that zinc has atoms more strongly or tightly bonded, which in itself could predict deeper mergers. "More strongly" and more tightly" are not necessarily the same. By more strongly, I'm referring to the inter-attraction, but more tightly can refer to the clinging together due to protonic shape.

If start off hypothesizing that zinc and tin have same-sized atoms both merged by near-identical depths, while tin likely has the most "hooks" due to its high boiling point, then these hooks do not raise tin's melting point above that of zinc's, a good thing where I'd like to find that atomic shapes have nothing or little to do with setting melting points. In any case, it seems certain to me that zinc atoms are more-strongly bonded than tin atoms, judging from their melting points. The chief suspect is by far zinc's greater merger attraction, for zinc and tin both have the same liquid pressure due to weighing the same per unit volume. Therefore, I can judge already that zinc has the larger atom, but, in being more-deeply merged, zinc material weighs as much as tin material. If tin has atoms approaching the size of copper atoms, then we tentatively mark the zinc atom down as at least as large as the copper atom.

Zinc boils at 907 C while tin at 2,602 C, which means that heat is able to pass through zinc much easier than through tin, which looks like a contradiction to zinc atoms having the stronger bond. No matter whether we view zinc as having a stronger bond by hook or by attraction, yet heat gets through it much more easily than through tin. Problems are the stepping stones to solutions, if we can find the next stone to step on.

Why does zinc have such a lower boiling point than copper (2,561 C). Off the cuff and following the rules I've been presenting, it could be that zinc has a spherical shape (no hooking possibility) while copper has snarled or tangled atoms, though not as tangled as tin atoms. We can adjust the zinc atom at any time, where warranted, to near-spherical with low hooking ability.

google AI: "Pure zinc is generally brittle at room temperature, but it becomes ductile and malleable when heated to temperatures between 100C and 150C, allowing it to be rolled into sheets or drawn into wire. Below this range, it is hard and breaks easily, while above 200C, it becomes brittle again." It sounds like zinc has a stick-together problem, perhaps due to it's sphere-like shape.

I've read that zinc coatings on iron sacrifice zinc atoms to the bare iron such that it protects the latter from rusting as badly. I could imagine zinc atoms slipping away to the iron more easily if spherical.

Thus far, looking at copper, tin, and zinc, there hasn't been an explanatory problem that cannot be overcome from the available options: 1) atomic size; 2) protonic size; 3) depth of merger / atomic-bond strength; 4) material weight-density; 5) atomic shape. Did I miss anything?

Can we fathom the reason that bronze has a boiling point of 2,230 C, below the 2,562 of copper and the 2,602 of tin? We need a logical reason as to how mixing tin with copper, both having higher boiling points than bronze, provides a lower boiling point for bronze? How do we make sense of it? What might make heat rise a little more efficient when copper and tin are mixed?

Classic bronze is about 12-percent tin and 88 percent copper. Why should this small addition of tin to copper bring the boiling point down of the bronze result? Well, as bronze is a little lighter in weight (than copper) due to the addition some tin, the liquid pressure of bronze is less so that it restricts heat rise less. But there's more going on because heat plows faster between the atomic spaces where copper and tin atoms meet due to tin atoms having weak bonding ability. Those two reasons in combination well explain the lower boiling point.

I'm reading: "A standard bronze, such as 88% copper and 12% tin, usually melts around 950°C." I'm reading that the melting temperature goes down when more tin is added. The logic here is that tin, having a weak bond, becomes the weakest link in the metal, allowing the copper atoms to fall to gravity power with less heat as compared to copper with less tin mixed in. Copper's melting point is 1085 C, which fits nicely into this explanation.

With more tin mixed in, less temperature needed to make the total atomic bond strength equal to the force of gravity. I imagine gravity dragging toward itself trains of some seven copper atoms connected to one tin atom. Gravity can't cause the copper atoms to roll on each other at temperatures under 1085, but it can cause the copper atoms to roll where they connect to tin atoms.

Therefore, if melting can be defined as gravity dragging trains or clumps of atoms, as opposed to pulling along single atoms, then it can explain why hooking may not raise the melting point, where gravity attracts clumps of atoms away from their weakest hooked link which itself does not resist being pulled off much more than the inter-attraction forces do.

This may appear to be far-reaching speculation for someone who's never stood on a proton to see what really takes place, but it's a vital point: gravity probably attracts clumps of atoms more efficiently / easier than single atoms because the clump weighs more. When the clump becomes disconnected from the solid at the weakest atom(s), it falls to gravity with larger force than a single atom does. I don't think that materials forming clumps or trains necessarily have lower or higher melting points. That point is determined by the weakest link that gravity overcomes.

"The zinc content [of brass] dictates the melting point. Higher copper content (like in Red Brass) typically increases the melting temperature." That is, the more zinc, the lower the melting temperature, and thus the weaker the weakest link (the zinc atoms) in the metal. It's telling us that zinc is the weakest link in brass, just as tin is in bronze.

The 950-degree point for the melting of bronze is slightly more than 900 for brass. We would like to know why heat gets through a copper-zinc soup easier than through a copper-tin soup. If tin has the weaker attraction bond, shouldn't heat get through the copper-tin soup easier? Brass weighs ever-so-slight lighter than tin, but I suggest that the main culprit for the lower boiling point of brass is: "Typical brass is an alloy primarily composed of copper and zinc, generally in a ratio of 60–70% copper and 30–40% zinc." Ahh, the brass has much more zinc than bronze has tin. We are still winning, still able to explain the properties as we meet them head-on.

It's a very big deal if this new science I offer can work for all materials in as much as there are various ways to test it. If it proves that liquid and solid atoms are merged, which is by far the better option, it proves, as expected, that the monkeys who created the orbiting electron are out in left-jungle with the claim that liquid and solid atoms are NOT merge. With no atomic mergers, explaining these hard-to-explain property specifics becomes a nightmare to them.

For example, they claim atomic weights of 118.7 units versus 65.4 units for tin and zinc. Right out of the gate, they wrongly have the tin atom weighing 1.8 times more than the zinc atom. They're never going to explain the specifics with this extra error marked into their books. BLOTCH.

Then, zinc and tin solids have roughly the same density as solids. How can they explain it if they don't even allow for atomic mergers? How can they have the one atom weighing 1.8 times more yet the solids weigh the same? They need anti-mergers, which is to say solid atoms further apart than others. Solid atoms not touching, are you kidding me? No, stupids, go the opposite direction. Go the way of mergers to rightly explain the specific densities of solids.

It's a nightmare that you are not notified of. They keep you in the dark as to such critical problems. They don't have any stepping stones to solutions. All they have are booby-trap doors to hell. Have the wisdom not to hang on to their hands as they fall. They give tin a cluster of 50 protons at its core, and 30 for zinc. They're science lunatics. Atoms that are in contact, they say are not in contact, and protons that cannot be in contact, they say are in contact. At least they give me some good laughs, otherwise this topic could become dry.

Not only are they afraid to admit that atoms are merged for fear that we would expect crashing orbiting electrons, but they have the atoms vibrating wildly simultaneously, which, if you just imagine the situation as clearly as you can, makes that atomic model so incredibly unlikely as to be utterly laughable. Whenever anyone shows the orbit-model atom in a computer-animated video, two or more atoms bonded are never shown vibrating wildly faster than you can imagine. Who can believe in such an atom? The entire establishment, that's who.

I'm offering an alternative atomic model with stationary, hovering electrons around lone (non-clustered) protons of various shapes, sizes and protonic force. Merger is impossible without hovering electrons, and the hover feature is LOGICAL, EXPECTED, COMPELLING. The hovering electrons can circle protons, but only if moved to do so. Otherwise they tend to stationary while hovering.

One can cause magnets to hover over a magnet, there's no mystery about it. When the downward gravity pull on magnets matches their upward inter-repulsion, they hover. Protons offer downward "gravity" force to hovering electrons, and some of these captured electrons (not all) inter-repel each other upward from the proton.

The entire protonic surface is littered with non-hovering electrons to an unknown number of layers, with more layers per stronger proton. LOGICAL, EXPECTED, COMPELLING. But electrons further out from the proton must hover, or God would not have been able to bond gas atoms. The hovering electrons of one atom fit like fingers into the hovering electrons of another atom. No orbital madness to complicate mergers. Mine is not a laughable atomic model.

This short video gives you an idea of how electrons hover over the protonic surface. In this set-up, the central magnet attracts the large magnet that you can view as one of many electrons coming in to become captured even while partially repelled away from the ring. The small six magnets can be viewed as electrons on the protonic surface:
https://www.youtube.com/watch?v=LyvfDzRLsiU

Mercury Mystery

Mercury has a melting temperature significantly colder than the melting point of ice, and a boiling point of 356.7 C, far less than the boiling point of tin. It doesn't take much heat to allow gravity to roll the atoms of solid mercury upon each other, and as even the boiling point is on the very low side for metals, it seems that mercury may have the weakest proton of all metal atoms.

However, although I'm tempted to view mercury as one of the smallest atoms, the following suggests significantly larger: "Mercury evaporates at room temperature, but it begins to evaporate profusely and rapidly at temperatures of 40 C and above." Any atom that gets evaporative lift at merely 40 C cannot be a highly small atom. On the other hand, its extra-low melting point suggests such a weak atomic bond that it helps to explain the low evaporation temperature.

The fact that mercury beads up when allowed to spill out of a bottle can indicate either than its atoms have a strong attraction force for each other, and/or that most materials repel it so strongly that its atoms huddle up and stick together. I tend to think that the first option is correct, and that the second option has no bearing.

This is very mysterious. We have a material the atoms of which are pulled down by gravity at -39 C, and yet gravity cannot pull mercury atoms individually down to a table, to spread them out upon the table in a thin layer, but instead this material beads up (sits on the table as small spheres). God throws us so many riddles for our entertainment, and they would not entertain Him unless He stumps us for a long time. He's watching me now, checking out how dumb I am as I try to explain mercury's mechanics.

Just because the table atoms do not attract mercury atoms, as do gold, silver and copper atoms, does not necessarily mean that the table atoms repel mercury atoms. As you can understand, the only necessity for beading is inter-attraction force of the beading material. It's not the only material that sets itself up in this formation. There are other atoms that are liquids at normal temperatures: "Gallium: A metal that is solid at room temperature but melts at approximately 30°C (86°F), allowing it to form silvery, mercury-like beads when slightly warmed or handled."

Beading is not the same as when water or other substances form droplets, for they are attracted by a surface whereas mercury clearly is not. Mercury does NOT form a droplet, but rather a crisp sphere even on its underside.

None of the metals with low melting temperatures approach a weight as heavy as mercury. Mercury is such a heavy metal that we expect it to have a high boiling point, yet, in spite of being more than 13 times heavier than water, it comes to a boil rather easily. I would assign it low hooking ability, and therefore a spherical atomic shape. It must have weak atomic bonds to boot, but I'm waking up to God's trick here, possibly. I'm thinking that mercury has a wild/rare large size, as far as metal atoms go, but with a super-weak proton.

Yes, an atom on the large side for metals would evaporate at a low temperature as compared to temperatures of other metals, and yet if it's also a weak proton, it explains its low melting temperature. Once I grasp that it's a largesse atom, I can explain beading, not from the high inter-attraction of its atoms (which I do not see as high), but from heat-particle push. Yes, a new discovery right here: heat-particle push.

Gravity repels heat particles upward, and they push atoms upward, for example whole water molecules. Heat particles repel each other in all direction, and they push water molecules onto a cold surface when they themselves inter-repel each other in the cold material. Therefore, why can't heat particles push mercury atoms in all directions such as to make it form a sphere?

Yes, and the weaker it's own atoms attract each other, the easier that the heat particles can be the boss to form mercury's sphere shape. Zowie, this works, and is even to be expected, though only for largesse atoms not attracted by whatever they are in contact with.

I suggest that, when mercury is poured from a bottle, the heat particles in the air invade it, slice it up into small sections, and immediately form those sections into spheres. Due to the lack of another viable explanation, I think this is the reality. Again, gravity is expected to pull mercury atoms down to a table top, where mercury is able to cause mercury to melt at -39 C, and so one very-good explanation for gravity's inability to lie down on a table is where heat particle demand that mercury stands on its feet as a sphere.

If the atoms of a liquid are too small, heat-particle push on its atoms may not be capable of forming a sphere or droplet. Gravity then gets the upper hand to force the liquid to wet a surface, to make the liquid lie flat on the surface.

Yes, heat-particle push can even explain the formation of droplets, especially if modern physics does not have full/wide success in explaining the mechanics of droplets. Yes, their explanation may work at times, for some substances, but if it doesn't always work under scrutiny, perhaps the explanation is erroneous to begin with. I'm seeing that droplets have the combination of heat-particle push and ordinary, non-surprising (not special) attraction to surfaces. For example, the less water is attracted to the surface it's on, the more it takes the sphere/bead shape.

One can then argue that mercury usually takes the bead shape precisely due to its weak attraction forces. It works. It's logical. Heat-particle push is da boss of mercury.

Mercury puddles can be formed when much mercury is poured from a bottle. For me, this means that heat particles are unable to invade the puddle to turn it into hundreds or thousands of beads. Logic tells me that, if the mercury is placed into a hotter environment, heat particles can then turn a puddle into beads. The puddle does not have affinity for a table surface. The edges of the puddle are rounded, not wanting to merge with the table atoms, as if wanting to get away from the table. This is not a proper puddle in a dip or basin, but rather the "puddle" exists on a perfectly-flat surface, and elevated from it about 1/6th of an inch. Heat particles are pushing in on the puddle's sides, countering the pull of gravity that wants to flatten the puddle as much as possible.

google AI: "Yes, a mercury puddle tends to break into small, rounded beads (droplets)...a behavior that becomes more pronounced or continues to be observed as it is heated."

The mercury atom cannot be a giant because mercury's evaporation doesn't become mentionable until well above water's freezing temperature, yet water molecules evaporate from ice. However, they leave ice easily because ice expands in size with lower temperatures, and therefore something's going on to unmerge water molecules with decreasing temperature (opposite of how other materials behave). The water molecule is a giant, but it weighs nine times more than a single atom, meaning it weighs nine times more than a mercury atom. It's telling me that the mercury atom looks to be at least nine times smaller than the cross section of a water molecule.

As oxygen has a low evaporation point of -219 C, the mercury atom must be significantly smaller than an oxygen atom. Just want to put the "large" size of the mercury atom into context.

Although mercury evaporates significantly, slightly above room temperature, it does not rise, as do water molecules at the same temperature. Mercury gas atoms fall to the floor at room temperature, but then water molecules can fall to the ground at that temperature too. It's suggesting that water molecules get about nine times more lift than mercury atoms, or perhaps less.

Therefore, as atomic lift by heat particles concerns the cross-section of an atom/molecule, I suggest that mercury's cross section is about nine times smaller than the water molecule's. The width of a water molecule is the width of a hydrogen atom plus one oxygen atoms. A water molecule is not an H2O. I'll show the true water molecule below.

The all-directional, heat-particle-push process can now explain why ice grows smaller with increasing heat, even though heat in the atomic spaces wants to expand the size of ice. Simultaneously, the heat on the outer sides of the ice pushes it inward. Why this happens to water below 2 C, but not to other materials at any temperatures, I don't know for sure.

The weight/density of a liquid is crucial to a boiling point. The heat particles not only need to slither their way past liquid atoms that, to some degree, refuse to open a track for them due to their bond strength, but they refuse to open a track even more as the liquid is heavier. The more atoms per unit volume of liquid, the more the upper atoms weigh down on the lower atoms, squeezing them together more than is the case for liquids having fewer atoms. Liquid weight is known also as liquid pressure.

Strangely, mercury, much heavier than iron, and even heavier than lead, has a very-low boiling point for a metal. How can such a heavy metal have a low boiling point? It's weak atomic-bond strength, in combination with its inability to hook atoms, must greatly or even fully compensate for its heavy liquid pressure. I expect a heavy liquid, with weakly-attracting atoms, to have a mid-range boiling point, and so mercury must have an extra-weak bond strength and virtually no atomic tangling.

Mercury is 1.5 times heavier than copper, creating a liquid pressure 1.5 times stronger against heat rise, yet as copper's boiling point is much higher, it underscores how weakly mercury atoms are bonded. As uranium, gold and tungsten are about 1.4 times heavier than mercury, and especially because none are expected to have deep merger due to the small sizes of their atoms, I think that all three have about 1.4 times as many atoms per any unit volume.

While tin atoms are significantly larger than gold atoms, tin has a boiling point approaching that of gold (2800 C). Tin weighs more than 2.5 times less than gold, meaning that gold has more than 2.5 times as many atoms per same volume. Gold therefore has much greater liquid pressure to frustrate upward heat rise, suggesting that gold atoms are either more weakly bonded and/or less tangled than tin atoms, otherwise gold would have a much higher boiling point than tin. If the extreme weight of gold is due to deep merger, then I would nix the weak-bond theory and go all-out with entangled atoms.

As gold's melting point of 1064 C shows that its atoms are bonded much stronger than tin atoms (232 C), the only conclusion I can make is that gold's boiling point is too low in comparison to tin's because gold's atoms are not nearly as entangled. Both gold's higher density and deeper merger call for boiling point much higher than that of tin, but tin's high entanglement by comparison is what gets its high boiling point.

How do we explain that both gold and tin can be easily scratched? Apparently, the high density of atoms does not produce surface hardness. I'd expect weak atomic bonds to produce soft surfaces, and while gold has a higher melting point such that we could expect it to be less soft than tin, it proves correct: "Yes, tin is softer than gold. On the Mohs scale of mineral hardness, pure tin has a hardness of approximately 1.5, while pure gold is harder at about 2.5."

Or: "Cesium (Cs) is generally considered the softest solid metal, featuring a Mohs hardness of approximately 0.2, which allows it to be cut with a butter knife. It is an alkali metal that is highly reactive, melts just above room temperature (28.5 C)..." Potassium, sodium, lithium, lead and tin have hardnesses of .4, .5, .6 1.5 and 1.5 respectively, and they have well-jibing melting temperatures of 63.5, 98, 180.5, 327.5, and 232. These pieces of data are thus-far confirming that relative melting points do in fact reveal relative atomic-bond strength, but also show surface softness directly related to atomic-bond strength. If any metal does not jibe, one could expect some other factor involved as the monkey wrench.

For example, tin has a slightly-lower melting temperature than lead, where both have roughly the same softness, wherefore the monkey wrench looks like lead's much-higher weight (higher liquid pressure), which contributes to total atomic bond-strength, the thing that sets the melting point, and thus requiring a higher melting point for lead.

A bigger monkey wrench is where gold is slightly softer than both silver and pure aluminum (can be made hard with alloy), yet silver has a melting point of 962 C, and aluminum 660, both lower than gold's 1064. This could be a case where entanglement raises the melting point of gold above what it would be if the entanglement was not involved. On the other hand, I would lay that theory aside because gold's higher melting point can be explained purely from gold's higher liquid pressure.

The gold solid was once a liquid, you see, and the gold atoms solidified while under the greater weight of its more-dense atoms such that the atoms were merged more deeply by the greater weight. That can well explain why gold has a higher melting point than its specific softness expects.

Silver has a weight-density roughly four times greater than the 2.7 g/cc of aluminum, and the latter's density is some seven times less than gold's. With these numbers, it makes some good sense that aluminum should have the lowest melting point of the three.

As silver's melting point approaches that of gold's while silver's weight is far below gold's, I reckon that silver's atomic-attraction force is greater than that of gold's, for the latter's total atomic-bond strength comes significantly from gold's extra weight. I'm suggesting that, if gold weighed as low as silver, gold would have the lower melting point. As weight contributes to the depth of atomic merger, it can explain why gold is expected to have more merged volume than silver even though silver is expected to have more atomic attraction.

According to their weight comparison, there's almost twice as many gold atoms per unit volume of material, suggesting that gold atoms are far more merged than silver's, and thus revealing that material weight (not the same as atomic weight) could play a major role in extent of merger depth, not at all meaning that material weight is proportional to extent of deeper merger, for atomic merger has a limit where even high mechanical pressure cannot shrink liquid volume by much. But where atomic attraction is very weak (very shallow merger), a little added material weight could as much as double the merger depth (just guessing at the extent to make a point).

Silver's density of 10.5 g/cc is below mercury's 13.6, yet as mercury has a much-lower melting point, it compels me to see mercury atoms with extremely-low atomic attraction. If not for mercury's high weight, it's melting point would be lower still.

Another monkey wrench is where gallium metal has the same hardness as lead and tin, yet melts at merely 29 C. I expect gallium's melting point to be higher, in the range of lead's or tin's, and so to explain gallium's lower melting point, I appeal to it's weight of 5.9 g/cc, lower than both lead (11.3) and tin (7.3).

Therefore, it seems certain that the setting of melting point is determined by the total bond strength that definitely includes both inter-attraction force and material weight, but might yet include some atomic entanglement (the more entanglement, the higher the melting point).

I hate the tangle, because, if not for tangle, I'd be able to find easier, and with more certainty, the near-exact relative sizes of atoms.

Revelations From Heat of Liquid Formation

Another way to discern (not without headaches) how deeply atoms are merged is by the "heat of condensation" that all substances release when forming liquids from gases. THE LAW: the greater the merged regions and the denser their electrons, the greater the heat output.

When it comes to liquid formation, merged sections cannot hold half their electrons; exactly 50 percent go missing as heat particles. With absolute certainty, liquid formation exactly doubles the electrons in the merged sections, and the situation quickly reverts to the density existing prior to merger. It's that simple to explain; it's that certain. I've got the candy, the imposters have brain pollutants.

There's a question as to which material puts out the most heat, the one with smaller or larger atoms, if both atoms have equal volume of merged sections? It depends on which of the two options have denser electrons in their merged sections, and larger atoms have a thicker population of less-dense electrons at their upper atmospheres, which is where merger begins. In some cases where merger is shallow, the upper atmosphere is all that's partaking in merger. I'm just pointing this out to show that neither merger depth nor volume of merger are expected to be perfectly proportional to heat output.

Here's the heat of condensation for three metals, not to be confused with "heat of liquefaction":

Gold 334 units (kJ/mol)
Tin 296 units
Zinc 115 units

Gold puts out more heat when liquefying from a gas, meaning that gold atoms, as a lot anyway, merge more deeply or voluminous than tin and zinc atoms. We gathered that from gold's higher melting point, which revealed a stronger bond i.e. assuming deeper merger. The order of total bond strength from weakest to strongest is tin, zinc, gold, yet zinc puts out less heat of condensation than tin, which is not the expectation unless there's yet another thing going on.

With a stronger bond than tin, and where both metals weigh roughly the same, zinc should have more merger volume. Might the zinc atom be so much larger than the tin atom as to have a less-dense outer atmosphere? Can that explain zinc's slightly lower heat output? The problem is, less heat output means fewer electrons released from the atmospheres. How could zinc have the stronger bond if zinc protons attract fewer electrons than tin's protons do? Easy answer: zinc has the stronger proton, which can explain why zinc has the larger atom.

As temperature in a gas goes up, the electron atmosphere of each atom is expected to grow in density, and feasibly also in size, because the free electrons surrounding it invade it from all directions, press in on it, and add themselves to it. The density of the electron atmosphere definitely and logically goes up with higher temperatures. Zinc's melting point is higher than tin's, which can suggest that zinc has the denser electrons in its electron atmosphere, wherefore you might think that zinc should put out more heat than tin.

But watch out, because heat of condensation occurs at the boiling point, having nothing to do with the melting point. The latter only tells us which of the two atoms has the strongest bond strength at the melting temperature, though I would assume that zinc yet has the stronger bond even at boiling point.

By the time of the boiling points, both atoms grow larger with electrons, and as tin's boiling point is higher than zinc's, that alone can explain why tin unexpectedly puts out more heat of condensation even though it has the weaker bond. Where tin has the smaller atom, the density of its atmosphere increases more per less depth toward the proton.

Yes, the larger the atom, the greater the depth of sparser, outer electrons, because every outer edge of every type of atom has the same protonic force level, and that force increases faster in the downward direction for smaller atmospheres such that they go more quickly from sparse to dense electrons. I'm sure that most people would rather be eating popcorn than engaging this complication, but you're different if you're still reading; you can eat popcorn and chew atomic physics at the same time. There's lots to celebrate when over a century of quack atomic physics is finally corrected, but I'm in my late 60s. When I die, this all dies in my dead head. Who's going to take my torch and burn the imposters as they deserve?

Tin atoms merge less deeply but either have a higher-density of captured electrons at the boiling point than zinc at its boiling point, or have larger electron atmospheres at tin's boiling point than zinc's at its boiling point. Therefore, variations in boiling-point temperature matter when comparing the heats of condensation of various materials.

Tungsten has the highest melting point at 3,422 C, and the highest boiling point, 5,555 C (it's used in light-bulb filaments). Tungsten has a heat of condensation of 384 kJ/mol, not much higher than gold at 334. Both metals weigh the same, meaning that both have the same number of atoms per unit volume, wherefore the heat-of-condensation logic suggests tungsten atoms to be more-deeply merged, and thus larger in size than gold atoms.

If both gold and tungsten were merged with identical depths while tungsten has the larger atom, a tungsten solid would come out weighing less than a gold solid, meaning that tungsten needs to be more deeply merged to create more weight per unit volume. A way to figure out a reality is to tentatively create some equality between atoms, and then logic things out from there, as I did in this paragraph. If such-and-such is equal, then so-and-so would be the result, wherefore this one material must have more such-and-such than the other.

The high heat of condensation alone of tungsten does not argue for huge metal atoms as compared to other atoms. Small atoms and large can both become deeply merged as to put out the same heat when merging. But where tungsten and gold both weigh the same, that's where we discover that tungsten does not have smaller atoms than gold. If tungsten atoms were smaller than gold atoms and more-deeply merged to boot, tungsten material would weigh more than gold material.

Gold boils at 2,856 C, far below the 5,555 of tungsten, and so we expect a larger-sized tungsten atom if only for that high temperature, for temperature grows atomic sizes. Tungsten looks like it has both deeper merger at melting point and larger atomic size at boiling point. We can safely assume that tungsten's deeper merger remains the fact all the way to its boiling point. The deeper merger is at least partly the reason for the higher boiling point, for deeper, stronger mergers require more heat particles to open up passageways for boil bubbles. However, as tungsten's boiling point is so high, we can start to suspect that atomic tangling (hooking) plays a role in keeping boil bubbles smaller such that more heat is needed to get things to the boiling point. We can start to suspect that tungsten has more entanglement than gold.

Some drastic thing keeps tungsten atoms from boiling at lower temperatures. It can't be its mere bond strength, especially as metal atoms as a near-whole have smallish atoms. We can't have materials so heavy as tungsten and gold unless they have the smallest of all atoms too. Therefore, when I say "larger" for tungsten, don't imagine large.

I suggest that the tungsten proton could be a C- or U-shape, like horseshoes interlocked. I think tungsten well argues that atomic shapes do play to lowering or increasing boiling points. If the entanglement's ability to resist the spreading apart of atoms is consistent throughout the material, I think that entanglement will also raise the melting point. The only way to avoid the increase in melting point due to entanglement is where the latter has strong and weak points, and where the weak points are roughly equal to the inter-attraction force of the atoms.

The release of heat of condensation is equal to the level of heat absorbed by the heat of vaporization. Heat of condensation is when gas atoms merge to form liquid. If they become unmerged by evaporation, the atoms take back all the electrons lost when merging, meaning they now absorb the heat from the gas. You cannot find a better explanation in all the earth for heat of condensation and vaporization. The kineticist hasn't got a clue as to how the mechanics of both processes work.

A liquid atom jumps off the liquid surface and into the gas. How does it steal heat from the gas, mechanically speaking, by the kinetic theory? I've just asked google, "how does an evaporated liquid atom take heat from the gas?" But it responds with a trick:

The escaping molecule [from the liquid surface] takes that extra energy with it. Because this energy was absorbed from the immediate surroundings, the surrounding air molecules lose kinetic energy, resulting in a drop in temperature [in the air].

Let me explain why this is double-speak hocus-pocus. It first says that, in order for a liquid atom to escape the liquid surface, it needs to become fast and hot enough, but rather than saying that this fast and hot atom adds fast and hot heat to the gas above the liquid, it tries to laughably make you think that the gas lost heat by making the pre-evaporated jumper atom fast and hot in the first place. No good, that's cheating deception because the heat of vaporization can't occur in the gas until AFTER the jumper leaves the liquid.

Now you know why these imposters insist that the gas has faster atoms than the liquid, which is a farce. The reason is to cool the gas with a jumper that's moving slower than gas atoms. But it's a farce because there's no logic that the liquid atoms should have less energy than the gas atoms, for the latter spread out into the air and cool, duh.

The liquid receives heat from the heat source, not from the gas, duh. But if the gas is hotter than the liquid, then the gas helps to heat the liquid. Only a moron, or a deceiver, makes this cheating argument...just because they have no way to explain the heat of vaporization.

Besides, it doesn't matter where the surface atom gets its heat sufficient to jump out into the gas. The question needing an answer is: why does the gas lose heat as a direct effect of this jumper? The quote above does not answer it. It tries to make us think that the gas provided the jumper with heat to make it jump, and thus the gas is said to lose heat BEFORE it jumps. Haha, no good, try again. In reality, the air above the liquid does not lose heat to the liquid surface unless someone puts a lid on the pot, and we can't allow that when trying to explain heat of evaporation.

The liquid throws heat into the air, not vice-versa, because the liquid is the heat source, duh.

It's the process of evaporation that causes the heat absorption, and so we cannot claim that the heat ABSORPTION is the heat LOSS from gas to liquid BEFORE evaporation takes place. If I didn't explain this, you would have believed the imposters with what's said in the quote above. This is an example of how they deceive students all over the world with law-breaking nonsense.

We want to know why the jumper steals heat from the gas, after it's jumped out. As kineticism only has this jumper to work with which is the hottest part of the liquid surface, according to his own claim, it's incapable of explaining how it steals heat from the air/gas.

In reality, the jumper doesn't exist because water molecules don't jump out. They fly off under free-electron propulsion under their butts. In reality, this flyer or rocket is adding heat to the gas unless it can steal more heat from it than it's adding. The flyer flies off because heat particles underneath it push it out, and consequently the heat particles enter the air-gas mixture too. But the number of heat particles in the merged sections of the liquid atoms must be greater than the number of heat particles entering the air-gas mixture, or there would not be a heat-of-evaporation amount.

I asked, "is there an exothermic [heat-creating] example of heat of vaporization?" AI does not know of one case, suggesting that heat released from a liquid's surface is NEVER more than the heat absorbed by the simultaneous evaporation process. This makes sense even with liquids at super-high temperatures, which release super amounts of heat from the liquid surface. However, increased temperature also increases the density and size of electron atmospheres such that they will absorb more heat to be satisfied (to become fully loaded) once a flyer flees the liquid surface.

Even tungsten with the highest boiling point has a high heat-of-evaporation absorption of 824 kJ/mol. That's a huge net-excess of heat absorbed versus heat granted to the gas from the liquid.

The flyer loses heat because it's deficient in captured electrons. It sucks electrons from the gas in order to re-load with them to the satisfaction of the proton's attraction. The flyer was merged with other liquid atoms prior to flying off, but as it was sharing electrons with the atoms it was merged with, it comes out of the liquid deficient in electron numbers. It can't fly off and take all the electrons it needs because the other atoms are pulling on them too. It needs to rip away from the other atoms, and therefore both it and the merged region it leaves behind become deficient in electrons, and so both re-load. It means that the liquid absorbs some heat too, not just the gas.

The flyer needs to return to its full / maximum load prior to flying off. The more electrons the flyer takes with it, the fewer it needs to re-load with, the less cold the gas becomes. The more atoms there are attracting merged-section electrons, the fewer electrons, I figure, the lone flyer is apt to take with it. But, for all I know, evaporation involves batches of rocketed atoms, not lone flyers only.

In reverse, when a gas atom merges with a liquid body, the merged regions involving multiple atoms suddenly have double-density electrons, but as none of the atoms can hold that many electrons, the double-density electrons push each other out, not only from the merged regions, but from the atoms completely, into the gas, adding heat particles to the gas. That's heat of condensation...always equal to the amount of heat of condensation for the same material.

For mercury, the heat of condensation and evaporation is just 50 kJ/mol, attesting to the shallow merger of its small atoms, though we need to keep in mind that mercury has a "cold" boiling point to boot. What would be the heat release / absorption if it's boiling point were as high as that of tungsten? In other words, don't size atoms according to their heats of condensation / evaporation, because the heat involved goes up as the boiling point goes up.

Nor can we assume that the heat creation / absorption is a mirror reflection of the merger depth. Atoms at higher temperatures will have more heat to give off or absorb even if merged by an equal distance with atoms having a lower boiling points. See the problem? The good news is that materials with similar boiling points can reveal which of them are more deeply merged. We even saw that, although gold and tungsten do not nearly have the same boiling point, we were able to establish tungsten as the one more-deeply merged.

The good news is that, for materials with similar boiling points, we can better guess at the difference of merger depths, not merely guess at which are merged more deeply. That in turn nets a better idea of relative atomic sizes. Tin boils not much below the boiling point of mercury, and sulfur's boiling point (444.6 C) is not much above mercury's (357 C). Tin has a heat of condensation of about 48 kJ/mol, and sulfur and mercury both about 50. It tends to verify that heat release from mergers is somewhat proportional to temperature.

As the density of sulfur is only 2.1 grams/cc, it's more than seven times lighter than mercury, for which reason I would claim that sulfur has the larger atom by far, for as both materials appear to have roughly the same merger depths (as per their identical heat of condensation), I gather that there are eight times as many mercury atoms per unit volume, and this necessitates much-smaller mercury atoms. The melting point of sulfur is only around 115 C, indicating that sulfur atoms are slightly more merged than mercury atoms. Note only .9 kJ/mol for the H atom, the largest atom. The reason for such little heat release for such a large atom is it's very low boiling point of -253 C. Clearly, H atoms barely merge when forming a liquid, yet when they bond with oxygen atoms, the two together release 286 kJ/mol, about 300 times more than when hydrogen forms liquid. Just try to imagine how much deeper the merger is when 300 times more heat leaks out.

Oxygen's heat of condensation is 6.82, significantly more than H atoms even though oxygen atoms are the smaller of the two by far. Yet oxygen's boiling point is much higher than that of hydrogen, explaining at least some of the higher heat release. I'm becoming fast convinced that much of the heat of condensation for all materials is due to free electrons squeezing inward on electron atmospheres and thus raising the numbers of electrons within them.

Behold, while neon gas has a boiling point of just seven degrees above hydrogen's, and while argon gas has a boiling point just 10 degrees above nitrogen's, neon's heat of condensation at 1.7 kJ/mol is slightly higher than the .9 of hydrogen, and argon at 6.4, is slightly higher than the 5.6 for nitrogen. It starts to appear that the heat of condensation is roughly, though not exactly, proportional to the temperature, at least for materials having low boiling points.

It can only make sense where heat is a major player in forming electron atmospheres. It begs me to ask what percentage of captured electrons upon atoms at room temperature had been foisted upon atoms by heat rather than by protonic attraction.

If you can imagine, protons would be force-loaded with electrons even if protons had zero attractive force. A proton is a "wall" onto which heat particles press from all directions, and thus heat particles naturally push each other more thickly upon the proton wall in higher temperatures. Protons need to have positive force in order to make liquids.

The proportionality of increasing temperature with increasing heat of liquid formation is not expected to be exact because merger plays a role, and of course merger depth is dictated by the proton's positive force. One could say that, where the proportionality is off, it reveals the relative depths of merger. The amount of heat release is by two things alone: captured-electron density, and merger depth. I can't think of a third thing.

How much would an atom grow if its small proton were to be taken from 100 C, and placed at 1,000 C, versus an atom with a larger proton? Which atom will develop the greater atmospheric depth due to the addition of heat particles in their midst at 1,000 C? At first glance, I think the protons need to be treated as flat walls, and I think that both will gain the same extra thickness in electron depth, yet the atom with smaller proton will always have fewer electrons and smaller atomic size regardless of the particular temperature increase for both (or all) atoms.

Therefore, if relative atomic size can be found in any ways, it's likely those same relative sizes will persist at any one temperature. That's good news if correct, otherwise I'm throwing in the towel.

The boiling point of argon is three degrees above oxygen's, but argon has a higher heat-of-condensation output of 6.4 versus 3.4. In such a situation while the two are yet gases, we deem that both of their electron atmospheres get invaded by heat particles equally (not much force at that temperature) such as to adopt the same added density, and so the only ways to explain that argon has twice the heat output is by a deeper merger, and some extra electron density it may have had to begin with from protonic attraction. Starting with two materials with nearly the same boiling point, the comparison of heat outputs at liquefaction gives us a feel for the variation in depths of merger.

Oxygen liquid has a density of 1.14 g/cc, and argon 1.4. It means that there's more atoms in a unit volume of argon, which can make sense if argon is more-deeply merged, though it's not the only possible explanation. Possibly, on this basis alone, the atoms could be merged identically deep, in which case argon atoms are the smaller of the two. But as the heat of condensation suggests more-deeply merged argon atoms, the sizes of the two atoms work out closer to each other than when merger depths are equal. Thus, by comparing two materials with nearly the same boiling temperature, we can find their relative sizes not-bad.

Oxygen gas weighs 1.43 g/l versus 1.78 for argon (both at STP), meaning that argon has 1.24 times the atoms in its gas. Liquid argon weighs 1.22 times more than liquid oxygen.

Technically, we can't use "merger depth," as I've been doing. "Merger volume" might be better, but ultimately, by merger depth, I'm implying the number of electrons in the merged sections, more numbers per deeper merger.

Every type of atom has an atmosphere with steadily increasing density of electrons from top to bottom of their atmosphere. I claim with certainty that the increasing density is more gradual for larger atoms and vice-versa for smaller atoms. That is, the latter will go from sparse to thick with less depth.

The closer the electrons to the protonic surface, the stronger the proton can bring the electrons closer to it, and therefore closer to each other. The outer electrons are always just barely hanging on, but in deeper regions, they are fixed in place more strongly, with less ability to move around if jolted or nudged.

This picture tells you why merger occurs at all, but it also tells that atoms can be held with low bond strength not only due to shallow merger, but to the looseness of electrons at the outer layers. The deeper the merger, for example when hydrogen and oxygen merge deeply to make water, the much higher the temperature needed to get the atoms to split apart. It takes thousands of degrees to get the H and O atoms to separate from water, but very low temperatures to get hydrogen or oxygen liquid to unmerge as gas. You get a very good feel here for depth of mergers in each case.

Just when we thought things couldn't get more complicated, enter gravity force. The temperature at the liquefaction of hydrogen is so slight that one wonders whether hydrogen liquid forms, not from atom-to-atom attraction, but solely from gravity force merging the atoms as far as possible against the repulsion forces working between atoms.

Very apparently, the slight heat at the boiling point of hydrogen is able to give lift-off to H atoms, but when the temperature drops below the boiling point by just one degree or less, gravity is able to bring the atoms downward, and as such it's going to bring them together at the bottom of a container. No H-to-H attraction is necessarily needed to make hydrogen liquid. Just a thought.

The best way to explain how temperature unmerges atoms is to imagine free heat particles invading this '8'. The heat particles will work into the middle region of the 8, and separate (unmerge) the top half from the bottom half until, when fully separated, they fly away from each other. They steadily re-load with electrons as they steadily unmerge, and when fully separated they repel each other again, as all gas atoms do. You can see my logic that attraction between gas atoms can only exist if they first merge a little. It allows one proton's positive charge to grab hold of the other's electron atmosphere.

Hydrogen and oxygen liquids have atoms barely merged, and nitrogen has slightly deeper merger because its heat of condensation is 5.6 kJ/mol (more than for O atoms). Nitrogen and oxygen atoms are nearly the same size. At standard temperature and pressure, there are 14 nitrogen atoms in its gas as compared to 16 in the oxygen gas of the same volume. The nitrogen atom is slightly larger. I know these details because nitrogen gas weighs 14/16ths as much as oxygen gas, and as all atoms weigh the same, that fraction reveals the density of the two gases: 14 atoms versus 16.

Hydrogen gas weighs 1/16th of O gas, meaning that, at STP, there is 1 H atom in its gas per 16 for oxygen. The big bangers are very wrong when they claim that all gases at STP have the same number of atoms, for there is no logic to this old theory, and, besides, it violates the all-atoms-weigh-the-same law. That error birthed another error where they designed the water molecule as an H2O, but the reality is that it's an O8H molecule. But this is another story for some other time.

As krypton gas has a boiling point 30 C higher than oxygen's, we expect the heat of condensation of krypton to be higher than oxygen's, and it is, 9.0 versus 3.4. If you care to figure out how much of the 9 units of heat is from additional heat particles at 30 C versus the merger depths of krypton atoms, be my guest, I'm going on vacation.

By this point I'm getting the impression that oxygen may be the igniter of combustion due simply because it has an unusually-shallow merger ability. The only thing I can fathom that stops merger once it sets in is proton-to-proton repulsion becoming equal to the attraction of one proton to the other atom's electrons. It works just dandy in theory. If there's something else going on, I don't know about it.

As xenon gas has a boiling point of 45 degrees warmer than krypton, it tends to reveal why xenon's heat of condensation is 12.6 kJ/mol versus 9.0 for krypton. Over and over again, the higher the boiling point, the higher the heat of condensation by roughly a proportionate degree.

If the temperature figures given to us are an accurate reflection of temperature increase, and if 6 degrees above 0 K has half as many heat particles as 12 degrees above 0 K, then I suppose that the amount of heat of condensation attributable to the temperature can be calculated for any specific temperature. One can then know how much of the heat of condensation, for any material at it's boiling point, belongs to the temperature factor, and how much (the remainder) belongs to merger. It can give a fairly accurate picture of how relatively deep all atoms are merged.

For example, if 6 of the 9 units of heat from the merger of krypton is from the temperature factor, then 3 of those units, or 33 percent, is due to the merger factor. One can find the percentage of merger for every atom in this way.

Helium has been assigned the lowest heat of condensation (less than .1), and the lowest boiling point at the cusp of absolute zero temperature. Modern science has obviously known of this trend we're seeing, but I've not known about it until now (March of 2026). It'll take me some time, if ever I get to it, to understand this trend better. I have no idea how the kinetic model of heat explains it. I've shown that I do have an explanation: electron atmospheres grow larger and more dense with increasing temperature, and consequently it predicts more heat of condensation released per higher temperature. It matches the trend we're seeing.

Another problem that kineticism can't explain on a mechanical level is that some chemical reactions absorb heat while others throw some out. As long as only two types of atoms are merging in chemical reactions, they must throw out exactly half all of the electrons in their merged regions, but when three or more atoms are in a molecule undergoing a chemical reaction with another molecule, the resulting product(s) may create or absorb heat depending on whether there's more merger volume formed versus merger volume unmerged.

For example, mixing vinegar with baking soda starts off with two products at room temperature, and so how possibly is kineticism going to lawfully explain that they produce coldness when reacting, just because they form new products when reacting? They'll have an explanation, but it is lawful or just another bad joke?

Carbon-dioxide gas is released in the reaction, and gas formation from non-gases always absorbs heat. There you have the explanation as to why coldness results, because the newly-formed carbon dioxide attracts electrons from the air. Duh. But that's not going to be the establishment's explanation.

Vinegar combines carbon, hydrogen and oxygen atoms, and baking soda does the same but with some sodium atoms thrown in. When mixing the two substances, atomic electrical forces are set up. In all "chemical" reactions, which should be viewed as electrical reactions, any atom in one material can steal one or more atoms from the other material. Whatever happens, the full reaction can absorb or release heat depending on the final merger depths as compared to the original merger depths. Release of atoms as a gas ALWAYS absorbs heat, not in evaporation only, but also in chemical reactions.

Kineticism sees that combustion produces much energy that was not present in the original atoms that cause combustion, but they refuse to define that newly-formed energy as captured electrons coming out of atomic mergers.

For example, when hydrogen gas burns, there's nothing more complicated taking place than hydrogen atoms bonding with oxygen atoms. They attract one another electrically as soon as a spark is introduced in their midst. Attraction force doesn't add motion-energy to particles because whatever motion it causes is canceled / counteracted by the head-on collision of the attracted particles. How, then, does kineticism explain the fire that comes from burning hydrogen and oxygen atoms? In realty, both are burning, not just the H atoms, because electrons are released from both types of atoms.

Let's ask google: "where does the fire come from when H atoms merge with O atoms in hydrogen combustion?" The response "The 'fire' (heat and light) in hydrogen combustion comes from the release of chemical potential energy when hydrogen and oxygen atoms break their original bonds and form new, much more stable bonds in water molecules. Essentially, the atoms 'fall' into a lower energy state, and the excess energy is released as heat and light." IT EXPLAINS NOTHING.

They define the resulting water molecule as a lower-energy condition, but they do not tell where the excess energy comes from just because the water molecule loses energy. I do tell where it comes from, the released electrons, but they have no such option, especially with their lunatic H atom having only one electron.

One needs to know that the baboons needed to invent a diatomic H atom as well as a diatomic O atom, because their atomic-weight scheme in combination with their atomic model didn't otherwise work with the known weights of gases. So they created diatomic atoms defined as two atoms in one. They needed twice the weight for H and O atoms, so, poof, they turned them into twins attached at the hip.

Therefore, before oxygen and hydrogen unite as a water molecule, the baboons think that both the diatomic H and O are ripped apart by a spark such that two separated H atoms, and two separated O atoms, are the result. Then, they wrongly envision two H atoms uniting in MERGER with one of the O atoms, which ends the chemical reaction as "H2O". The latter is what they call the "low energy state" above, because it's the part that loses the high-energy flame.

Later in the same response as above, google AI says: "When the H and O atoms merge to form water, they create new, stronger bonds. The energy released in forming these strong H-O bonds is much greater than the energy required to break the original bonds." You probably have no clue what that means. It means that the original diatomic atoms came apart rather easily, not taking much energy because they are not as strongly bonded as the molecules that make up the water molecule. The latter is said to be a literal merger of three atoms combined. If the baboons had been able to avoid a merger, they would have on behalf of their electron-orbit model, but, alas, they had to confess that mergers do happen.

When you read above, "create new, stronger bonds," it means that the water molecule is more-strongly bonded than the original diatomic atoms. But then, without explanation as to how energy is released, or in what form, it just adds that the energy released from the water molecule is much greater than "the energy required to break the original bonds" of the diatomic atoms. Ya? So what? Who cares how much energy is needed to break the bonds of the diatomic atoms? The question is, why does the formation of the water molecule make fire? It's irrelevant whether the original atoms that make the water molecule come from a bat cave or a wishing well.

The baboons are correct in saying that the water molecule releases energy, but it's plain stupid to define those flames from the idea that it takes less energy to split diatomic atoms apart. SO WHAT? It's irrelevant. It doesn't matter whether one needs to boot the atoms in the arse up a hill, or let them roll down a hill, whatever velocity they have prior to becoming a water molecule isn't going to explain the flames that, in the case of hydrogen, are hotter (make more heat) than any other combustion process. You can't say that a tiny spark needed to separate the diatomic atoms is going to create flames and flames and more flames. The spark is irrelevant, even if diatomic atoms exist.

BONEHEADS! The flames are from electrons streaming out of the merged sections of the newly-forming water molecules. They come out while the molecule is forming, while its parts are merging.

Oxygen and hydrogen co-exist together without a reaction until there is a spark, flame, or high temperature in both of their midst. The spark is enlightening because we imagine a negative force blowing away some electrons from the outer edges of both atoms, and somehow it causes the atoms to attract. I suggest that, due to the presence of oxygen in all combustion processes with other materials, the spark makes the O atom highly positive in charge, possibly making it capable of attracting the outer atmospheres of other atoms from a small distance.

In any case, whatever happens to the O atoms in the midst of the spark, the simultaneous explosion (that is the spark) physically blows O and H atoms together such that they merge upon contact. The rich release of electrons from each water molecule formed probably sets up a chain reaction of water-molecule formation because heat can ignite this gas mixture.

At about 550 C, H and O atoms will ignite apart from a spark, from the temperature itself. I do not know the mechanics here, can't even offer a clue. My understanding is that increased heat compresses all around all atoms such as to make them more dense in captured electrons, and to make the atoms repel each other with stronger force, but I can't see why O and H atoms should attract each other in that picture. I suspect that something unique takes place with the O atoms, because mixing two or more flammable gases apart from oxygen never gets them ignited with high heat. The O atom is what causes the sudden attraction to the atoms of combustible materials that then sets up a chain reaction of combustion.

In all three cases of the formation of carbon monoxide, carbon dioxide, or ammonia, there is a release of heat (called "heat of formation") because in all three cases gas atoms merge to make the material, with zero atoms unmerging. You will never find a chemical reaction absorbing heat (colder environment as the result) if the only thing going on is atomic mergers. It's not coincidental.

However, the cheats go to bat trying to dissuade the public of that fact. While they admit that most atom-merging reactions release heat, they don't want it to be a law, and so when I ask for an example of merged atoms absorbing more heat than releasing it, the response is laughable: nitric oxide. It forms from nitrogen and oxygen atoms merging, but when I ask for the heat of formation, it gives the answer as +90.3, where the plus sign indicates absorbed heat rather than released. Cow patties.

They cheat, because they include the heat needed to get the reaction to occur. That is, they need the reaction to occur around 1,600-2,000 C, and so they include the heat energy needed to get the temperature that high, then subtract it from the true heat of formation, and their result is that more heat is "absorbed" such that they assign the heat of formation with a plus sign, dirty rats. They are defining USED heat as absorbed heat, dirty skunks.

I therefore asked: "what is the heat of formation for nitric oxide if we ignore the high temperature needed to cause that reaction. Just the merging of the atoms, please." The response is now with a minus symbol: "The enthalpy change for the direct formation of nitric oxide from ground-state nitrogen and oxygen atoms (i.e., the merging of the atoms) is approximately -632 kJ/mol, representing the energy released upon forming the bond." Look at that massive number as compared to gas atoms merging merely as liquids at low temperatures. It's more than twice as large as the 286 kJ/mol for water formation, and thus the N and O atoms are far more deeply merged than the atoms of a water molecule.

The mergers of the atoms in nitric-oxide are so deep that 1,300 C or better is needed to unmerge them. The water molecule needs 3,000 C, not necessarily meaning that the water molecule has deeper mergers, because the water molecule is in reality one H atom into which 8 oxygen atoms are merged. You can imagine that it's going to be easier for heat, i.e. at a lower temperature, to push apart one N atom merged into one O atom as compared to persuading 8 O atoms sunk into one giant H atom. That's because increasing heat all around the molecule is going to push the O atoms into the H atom a little deeper such that much more heat will be needed to get both atoms so highly negative in charge that the O atoms repel completely out of the H atom. That's my guess as to what takes place.

The high temperature needed makes me think that roughly half of every O atom is sunk into the H atom. That way, it's hard for heat particles to get under the O atom to push it out because the underside of the O atom is inside the H atom.

Gas Weights Versus Atomic Weights

This section should be up by tomorrow.

When asking, "how much gas at STP would 1.36 grams of frozen oxygen make if weighing that much at -225C," the response is likely according to the unreliable ideal gas law...which kineticists conveniently make for themselves. The response is that "1.36 grams of frozen oxygen will expand to occupy .95 liters of gas at STP."

I'm warning you, the look of molecules claimed by modern science are usually never the correct ones. An H2O molecule is not correct, I'm warning you. However, I think they luckily got the nitric oxide molecule by viewing it as an NO molecule. They claim that one volume of nitrogen gas at STP mixes with one equal volume of oxygen gas to produce nitric oxide. Idiots that they have been, they believe that both volumes have exactly the same number of atoms. It is that belief which makes them claim that a nitric-oxide molecule is one N atom merged with one O atom.

They luck out on this one because nitrogen gas weighs 14/16ths as much as oxygen gas, when both are at STP. The reality is that, for every 14 N atoms in its gas, there are 16 O atoms in its gas, almost the same number. That's why they luck out when calling a nitric-oxide molecule an NO. My guess is that one volume of N gas mixes with slightly less than the same volume of O gas in order to result in NO molecules.

You can understand why they would be loath to admit that, when molecules form, it's not from one volume of gas mixing with one perfectly-identical volume. If it's close to 1.0 volume, they will just round it off to 1.0 to keep their brainwashed students from asking questions. If they admit that it's .9 volume of oxygen gas that mixes with 1.0 volume of nitrogen gas, they have a problem on their hands when claiming that nitric oxide is an NO molecule...because it should instead be each N atom in the volume merging with .9 of an O atom. Wooops, doesn't work.

So, when you see their NH3 molecule for ammonia, what they really mean is that one volume of nitrogen gas mixes with three volumes of hydrogen gas, but we really don't know how exact those volumes are, and they will probably never tell us because they badly need us to believe that these volumes are perfectly proportional to the NH3 molecule they invent.

In real science, the look of the ammonia molecule must be assessed from 14 N atoms in its volume merge with every 3 H atoms in three volumes of H gas, if indeed perfectly-identical volumes are fully used in the reaction. My guess is that three equal volumes of H gas mix with slightly more than one equal volume of N gas such that every 15 N atoms merge with every 3 H atoms, such that an ammonia molecule can be expected to be 5 N atoms merged into each H atom.

My thinking is that a water molecule is an 08H because only eight O atoms can fit merged into one H atom, with no room for a ninth, and so the ammonia molecule can be an N5H molecule because six N atoms do not fit merged into one H atom. It's known that steam at STP weighs nine times more than hydrogen gas, and indeed the 08H molecule has nine atoms such that it weighs nine times more than a hydrogen atom.

Nitrogen atoms are slightly larger than oxygen atoms, explaining why there's only 14 in an STP volume per every 16 of oxygen. My law: the more a gas weighs at STP, the smaller its atoms but the higher the number, because all atoms weigh the same.

As helium gas weighs exactly twice as much as hydrogen gas, I often wonder whether it's a molecule of two H atoms merged. My argument is that, if helium were truly an atom, it would merge with other gas atoms, but if it's already merged, that's why nothing merges with it.

If helium is two H atoms merged, a problem arises on their spacing such that I can't get the gas to weigh twice as much as H gas unless helium molecules have identical spacing as H atoms...which is not predicted because two H atoms merged should have greater spacing than H atoms. But if there is greater spacing with the gas having the double-H atoms, it won't weigh twice as much as H gas. It seems I've got to abandon the theory that helium "atoms" are double-H atoms.

I therefore have reason to believe that helium is best viewed as three H atoms merged with non-problematic spacing at further distances than H-atom spacing. I'm thinking triple-H atoms with spacing of about .25 times greater than H-atom spacing, for the helium gas would then weigh twice as much as H gas. It works.

As all gas atoms in a sealed container take up equa-distance, one can discover that O atoms are about 2.5 times further apart than H atoms, when both are at STP, for objects are 2 times further apart when eight times less dense, and 2.52 times when 16 times less dense. This is how it's done, by weighing gases at STP, and knowing that all atoms weigh the same. It's easy and reliable.

You can ask google, "how many times closer together are balls if 14 times more dense," and it will tell you, 2.41 times. It tells you that N atoms are that much further apart at STP than H atoms. Don't ask "how many times closer," but "how many times closer together," or "how many times further apart if less dense." google AI is not always reliable with such questions, depending on how they're framed.

However, while I can trust the weights of atmospheric gases as presented by modern science, and while they are proportional to their so-called "atomic weights," I can't trust their atomic-weight figures for metals because the gases of metals are at high temperatures well above standard temperature (0 C). Atmospheric gases exist at standard temperature, but one cannot have metal gases at that low temperature, and so there is room for the imposters to fudge their atomic-weight numbers for metals if they need to protect some erroneous tenet(s), for this is what the imposters do regularly, cheat to protect their own erroneous models. They use their "ideal gas" for their gas math, which is a fantasy gas as kineticism needs it.

One way to know that they have erroneous atomic weight numbers for metals is by asking google how they are arrived to. Part of the response is: "For metals, the atomic weight is often confirmed by measuring specific heat capacity, as this value is inversely related to atomic weight." As all atoms weigh the same, that statement is false.

As oxygen can be frozen aside from compressing its gas, one can find the relative weights between oxygen and iron atoms by bringing iron to some temperature less than the freezing point of oxygen. The weight of each solid, per unit volume, tells you the relative number of atoms in each case. However many times the iron weighs is a direct indicator of how many more atoms it has. One can then bring any metal to the same low temperature to find the relative number of atoms for all metal solids.

Then, once one has the number of metal atoms relative to solid oxygen atoms, one needs to find how much metal solid is needed to make one volume of gas at iron's boiling point, for comparison with the same volume of oxygen gas at the same boiling-point temperature. But one must not use ideal-gas math to find how much oxygen weighs at iron's boiling point. Rather, one needs to find the correct figure from an actual experiment.

Metals do not become much smaller in volume from room temperature to, say, -225 C. Besides, we can now ask AI how much any metal decreases in size from 0 C to -225 C. For example: "how much smaller does a block of iron become between 0 C and minus 225 C?" The response is .81 percent smaller, almost nothing.

When asking, "how much gas at STP would 1.36 grams of frozen oxygen make if weighing that much at -225C," the response is likely according to the unreliable ideal gas law...which kineticists conveniently make for themselves. The response is that "1.36 grams of frozen oxygen will expand to occupy .95 liters of gas at STP." That's correct because O gas at STP weigh's 1.43 grams per full liter.

I then asked for the theoretical weight of iron gas at STP: "Theoretical Weight of Iron Gas: .00246 grams per cc," the same as 2.46 grams per liter, which is only 1.72 times more than the weight of oxygen you see above. Right away, I know that this 2.46 weight for the theoretical iron gas is incorrect, for even their own atomic weight for iron is set at 3.5 times the weight of an oxygen atom, yet the math got us 1.72, which happens to be exactly half of 3.5. In reality, all atoms weigh the same.

Put it this way, that if the true weight of an iron gas at its boiling point is extrapolated to weigh just 1.72 times more than the known weight of O gas at STP, they are obligated, by their own atomic model, to set the atomic weight of the iron atom at 1.72 times more than their O atom. They knew that's too light for the iron atom, and so the way they got out of the problem is to make the O atom diatomic, which doubles its weight such that the iron atom thus comes out 3.5 times as heavy as "one" O atom. That is, two diatomic O atoms, viewed as "one" atom, is what they perceive to weigh 1.72 times less than an iron atom.

As you can see, the numbers here match perfectly with how they view the situation, meaning that they have the "ideal" gas law rigged to ideally reflect their erroneous atomic model. MATH MONKEYS. It means that they chose exactly 3.5 because it's the roundest number they could get away with.

They always want round-number multiples when it comes to relative atomic weights, because all their weight figures are based on the concept that all atoms have multiples of the same proton (always weighs the same per proton). They once reported only round numbers in the school textbooks, but the Internet has forced them to cough up some real numbers that are not round multiples.

We ask how much frozen oxygen weighs, per cubic centimeter, at -225 C. The response is 1.36 grams per cc. We then ask how much one cubic centimeter of iron weighs at -225 C. The response is 7.90 grams, which checks out with 7.85, the known weight at standard temperature. To find how many more atoms are in a gram of iron at -225 C: 7.9 / 1.36 = 5.8 times. The reality is that there are 5.8 times more iron atoms in a cc of iron as opposed to a cc of oxygen. That's true science.

When asking google, "what is the experimental weight of iron gas at its boiling point," it won't give it to me, even though the figure must be all over the Internet. Instead, it's programmed to give me an estimate based on the ideal gas law. This is called brainwashing by the establishment, and hiding realities. When asking the same question as well as asking it not to give me the ideal-gas math, it included the following in the response:

While the ideal gas law predicts 217 g/L, experimental deviations (compressibility) at these extreme temperatures are documented in high-temperature pulse heating experiments.

Yes, but AI did not share those documented figures with me. Why not? Inside secret only?

You can ask for the weight of any metal gas at its boiling point, and when AI converts that weight for a theoretical gas weight at STP, it'll be worked out to jibe best as possibly with their chosen atomic weights. It's all systematic trash, jibing only with itself but not with reality.

However, there is something suspicious happening when I ask for the "weight of iron gas per liter at iron's boiling point." It's as if the establishment doesn't want us to know, or as if there's disagreement in chemistry circles, because the response is: "The density of iron gas (vapor) at its boiling point of approximately 2861°C to 3134 K is very low compared to its solid form, estimated to be in the range of 0.5 to 1.5 grams per liter, depending on pressure calculation." As I didn't stipulate a specific pressure in the query, I assume that one or the other, .5 or 1.5, is the gas weight at 1 atmosphere of pressure, and therefore at STP.

Then, when I change the query slightly to, "weight of iron gas per liter per one atmosphere at iron's boiling point", the response is not nearly .5 to 1.5, but: "The weight (density) of iron gas (vapor) at its boiling point of approximately 2861°C (3134 K) and a pressure of one atmosphere (1 atm) is approximately 0.217 grams per liter." It's the same number as above as per their ideal-gas math. The physicists that google permits AI to ape are messing with AI's head by not agreeing with each other. AI might just have accidentally snitched, but I can't be sure. The reality might be .5 at STP.

The .217 figure can be shown to jibe with ideal-gas math where I ask: "where iron gas weighs .217 grams per liter at one atmosphere at iron's boiling point, how much would it weigh per cc at STP if possible?" The response is .00249 grams per cc (= 2.49 per liter), essentially the same as the .00246 we got above when AI was sticking to the ideal-gas math.

The weight of .217 per liter (at iron's boiling point) is less than half the .5 grams per liter, and seven times lighter than the 1.5 grams. I welcome iron gas weighing more at its boiling point than the ideal-gas math predicts. I think we're on the right track.

AI had said: "1.36 grams of frozen oxygen will expand to occupy .95 liters of gas at STP." As 1.36 grams fills one cubic centimeter while a liter has 1,000 CC, the .95 liter above tells us that one cc of frozen oxygen becomes 950 cc's of gas at STP. One cc has spread out by 950 times the volume, wherefore I asked: "if balls begin in contact with each other, how many diameters apart will they be when filling 950 times the volume?" The response: "When balls expand or are spaced out to fill 950 times their original collective volume, they will be approximately 9.83 diameters apart (measured center-to-center)."

AI shows, in the response, that it understood the initial balls to be one diameter apart. I have no complaint with the finding of 9.83 diameters apart, but, as oxygen atoms in a solid are merged to some degree, and therefore less than one diameter apart center-to-center, the gas will see them less than 9.83 diameters apart.

I've just shown you a method for finding the maximum distance between gas atoms in atomic diameters, in case you need it. Take a volume of the solid, and find how much gas volume it makes at STP. But it only works if the gas volume is the reality, not the ideal-gas "reality." The 9.83 as the maximum distance of O atoms is correct because the math started off with the correct weight and volume of O gas at STP. on

Checking, I asked point-blank: "how far apart are O atoms at STP, measured in atomic diameters?" Although the response includes someone's quote claiming 25 diameters apart, the top of the response has a quote merely from Quora, but looking correct: "At Standard Temperature and Pressure (STP), the average distance between oxygen molecules is approximately 8 to 10 molecular diameters apart."

I now want to compare the weights of oxygen gas with iron gas when both are at the boiling point of iron. I asked: "how much gas, at the boiling point of iron, would 1.36 grams of frozen oxygen make if weighing that much at -225C?" The response now is: "Starting with 1.36 grams of frozen oxygen, heating it to the boiling point of iron (approximately 2862 C) would produce approximately 10.94 liters of gas at standard atmospheric pressure" (standard pressure is 1 atmosphere). That number is incorrect because it's the ideal number, but it's all I have to work with.

As iron weighs 7.85 grams per cc, I ask: "how much gas, at the boiling point of iron, would 7.85 grams of iron make: "At the boiling point of iron (2861 C), 7.85 grams of iron would produce approximately 36.71 liters of iron gas at 1 atmosphere of pressure." I've got no evidence that one cc of iron solid makes 36.7 liters of gas, nor do I have evidence that one cc of oxygen solid makes 10.9 liters of gas, when both are at iron's boiling point. Just the same, this exercise can reveal that the modern atomic model is incorrect, because it was reliably found above that one cc of solid oxygen has 5.8 times fewer atoms than one cc of solid iron, wherefore the goofer should expect that 38.7 liters of gas must have 5.8 times as many atoms as the 10.7 liters of gas.

Ya-but, if we squeeze the 36.7 liters down to 10.7 liters, I know that the iron gas will then have 5.8 times as many atoms. Wherefore, as 36.7 / 10.7 = 3.4, it finds that there are: 5.8 / 3.4 = 1.7 times as many iron atoms per liter of each gas at the boiling point of iron. According to the ideal gas law, both liters will continue to have the same 1.7 difference in numbers no matter what the temperature, and thus the goofers must now claim that iron gas has 1.7 times as many atoms when both liters are at STP. See any problem there? It violates their claim that all gases, at the same temperature and pressure, including at STP, have the same number of atoms.

I've stubbed them with an alternative atomic model. I'm stepping on the baboon's toes. Earlier, we found that, according to their ideal-gas math, iron gas at STP weighs 1.72 times as much as oxygen gas at STP, but here I've shown you, based only PARTLY on their ideal-gas math, that iron gas has 1.7 times the numbers of atoms. It proves that they confound atomic numbers with atomic weight, but it also proves my claim that all atoms weigh the same, for my claim is that a gas weighing 1.7 times as much has 1.7 times the number of atoms. They instead claim that both gases have the same number of atoms but where the iron atom weighs 1.72 times as much (as a diatomic O atom). They confound atomic numbers with atomic weight.

To refresh your memory, it was said above: " I then asked for the theoretical weight of iron gas at STP: "Theoretical Weight of Iron Gas: .00246 grams per cc," the same as 2.46 grams per liter, which is only 1.72 times more than the weight of oxygen you see above." They arrange the weight of iron gas at its boiling point such that, at STP, it comes to weigh exactly as much as they want it to, in order to make iron atoms weigh 1.72 times as much as their imagined diatomic O atoms.

To help arrange this trick, they put out ideal-gas law "facts" until the whole world starts to use it to discover other "facts" by doing the math with their faulty numbers. In this way, someone could "find" that iron gas weighs 217 grams per liter at its boiling point, which is the number that AI fetched for me. But I also found AI fetching .5 and 1.5 grams for the same gas. If I use .5, it's 2.3 times heavier than .217, wherefore I'm now suspecting that there are 1.7 x 2.3 = 3.9 as many iron atoms in an iron gas as opposed to oxygen gas (both at the boiling point of iron). That sits much better with me, especially as I doubt very much that oxygen is a diatomic atom.

They can easily take a cc of solid oxygen, and turn it into a gas of oxygen's boiling point. They can then weigh a liter of the gas at that temperature and one atmosphere, and they can weigh a liter at any temperature above the boiling point. The higher the temperature, the more the gas grows larger such that its atoms are further apart, such that a liter of the gas weighs less at STP. But instead of AI fetching me the weight at STP from experimental results, it feeds us the prediction from the ideal-gas law, meaning that the 10.94 liters as the response above needs adjustment.

Where one cubic centimeter of solid oxygen is said to become 10.94 liters of gas at the boiling temperature of iron, it's grown in size by 10,940 times (because 10.94 liters is 10,940 times lager than a cubic centimeter). AI calculates that the original, back-to-back atoms of the solid are then 21.2 diameters apart, center-to-center.

Where iron gas from one cc of iron solid is said to fill 36.7 liters of gas at iron's boiling point, we can task AI to find how far apart back-to-back balls (don't mention atoms, or it might get suspicious) spread out when their volume increases 36,700 times. It says, 33.2 diameters apart.

However, iron atoms in a solid don't start back-to-back. If we assume them to be merged by a depth of 10-percent of their diameters, the 33.2 comes down to about 30 diameters apart. I think that's fair.

We thus have two gases at the boiling point of iron, oxygen gas with atoms 21.2 diameters apart, and iron gas with atoms 30 diameters apart. At first sight, it appears that the iron gas is more sparse in atoms than the oxygen gas, which can't be correct, but before we come to that conclusion, we need to realize that iron atoms are smaller, having less diameter than O atoms. I need to find their relative distances apart using the same measuring stick. I can't use oxygen diameters as the stick for both gases, nor can I use iron diameters as the stick for both gases, because I don't know the actual distance of either diameter.

If I ask the following question, I can find their relative distances apart using the 3.9 figure above for the greater number of iron atoms versus oxygen atoms. This question finds how far apart iron atoms are using the measuring stick of a partial oxygen diameter: "if balls are 21.2 diameters apart center-to-center in a sealed container, how many diameters apart would they be if 3.9 times more numerous?" Response: "If the balls are 3.9 times more numerous in the same sealed container, they would be approximately 13.5 diameters apart center-to-center." What this suggests is that the iron atoms are 13.5 / 21.2 = .637 oxygen diameters apart.

We are climbing the tree to the green top while the baboons are out on a dead limb hearing snapping sounds. We can either say that iron atoms are .637 times as far apart as O atoms in the O gas, or 21.2 / 13.5 = 1.57 times closer together. Checking the math, AI says: "If objects in a container become .637 times as far apart as they were originally, there will be approximately 3.9 times as many objects per container." The math checks out with the 3.9 used above.

We can test the .637 number by multiplying it by the 33.2 iron-atom diameters obtained above, to find 21.15, which is the number of diameters that oxygen atoms worked out to. In other words, we could have obtained .637 by dividing 31.2 by 21.2. We might want to change the 33.2 to 30, and the 21.2 to 20, to account for some merging of the atoms, and so we do 30 / 20 to get .667 instead of .637. If you find that a deeper merger is involved, the 30 / 20 number goes down accordingly

If we use their number of 1.7 times the number of atoms instead of 3.9 times, iron atoms come out only .825 times as far apart (or 1.19 times closer together), which for me seems ridiculous because we are comparing a gas with metal. Iron has such a high boiling point as compared to oxygen that it must have much-smaller atoms, and therefore a significantly greater numbers of atoms in its gas.

The reason that the iron solid has 5.8 times as many atoms while the iron gas has in the ballpark of 4 times as many is that the iron atom is smaller, receiving less inter-repulsion force from the heat particles in its midst. Iron gas provides less gas pressure per equal numbers of atoms such that there needs to be more of them to reach equal pressure.

Whether the .5 and related 3.9 figures are correct or not, I'm giving anyone a means/method for finding relative numbers of gas atoms, in case anyone has access to the real weights of gases at their boiling points.

To give an indication of how much smaller mercury atoms are than iron atoms, we need only ask for the weight of mercury gas at STP: "The weight of mercury gas (vapor) at Standard Temperature and Pressure is approximately 0.00895 grams per cubic centimeter, which is equivalent to 8.95 grams per liter." That's 100 times the weight of H gas, and 6.26 times heavier than O gas, both likewise at STP. It means that there are 6.26 as many mercury atoms as compared to O atoms such that mercury atoms are significantly smaller than iron atoms where the latter number only 3.9 times as many as O atoms.

To show how the ideal-gas math goes wrong, I first asked how much mercury gas weighs at mercury's boiling point. It says 3.97 grams per liter. I then asked: "if one atmosphere of mercury gas weighs 3.97 grams at mercury's boiling point, how much will it weigh at STP?" The response is 9.15 grams, which is at odds with AI's claim above that mercury gas at STP weighs 8.95 grams. The latter is probably the figure from experimental observation. There's not much of a gap between the two numbers, but neither is there much of a gap between standard temperature and mercury's boiling point.

Again, the reason that mercury atoms need to be smaller is that all gases at STP provide the same pressure, and so if there are more atoms in the gas, they need to be smaller, otherwise they would exert more than standard pressure. That's because gas pressure is from repulsion force, for the kinetic explanation of gas pressure is impossible. The smaller the gas atom, the less force it receives from added heat, for added heat is the inter-repelling electrons that repel between atoms to foist them further apart and against surfaces.

The larger the atom, the more that free electrons exist between atoms, the more the electrons push atoms apart, wherefore the law: there is greater gas pressure per the larger the atom. It's child's play. If only the establishment had chosen the correct way to interpret gas weights atomically, they would have engaged child's play. Instead, they have created a twisted monster ruled over by a nasty baboon preoccupied with monkey science.

I'm not speaking wee problem here. I'm saying that, if free electrons define heat, the imposters don't know what molecules look like. All or their intricacies in explaining or illustrating molecular configurations, blah-blah-blah, is worse than trash. It's harmful. Nobody can count how many wasted hours there have been as industrial departments racked their brains trying to understand why the physics "facts" couldn't help them with their problems. Who's going to pay them for lost time?

Unfortunately, there's probably two repulsion forces acting in unison, the other being the atom-to-atom repulsion from the net-positive force that radiates beyond atomic perimeters from protons. I've reasoned that this force must exist in order that gravity can attract all atoms.

This net-positive force may be so small, however, that it may not apply enough to make much of a difference for finding the relative sizes of atoms. As it can never become stronger, it's, with certainty, continually of less importance with rising temperatures. The repulsion from rising temperatures continually builds.

I suggest that, in ignoring the proton-to-proton repulsion, we can reason like so: if mercury has 100 times the number of H atoms at STP, its cross-section must be close to 100 times smaller than the cross-section of the H atom. For heat particles push the cross-sections of atoms. I ask: "how much diameter does a four-inch ball have with 100 times less cross section?" The response is .4 inches. Where the H atom is imagined at 4 inches round, mercury atoms are looking less than a half-inch round. Asking the same question for O atoms: "A four-inch ball with 16 times less cross-sectional area has a diameter of 1 inch." Mercury atoms are thus expected to be about 2.5 times smaller than O atoms. This math assumes spherical atoms, not necessarily the reality.

The hydrogen atom has the fewest atoms per gas at STP, because it's the largest atom, meaning that O atoms are amongst the largest.

If the iron-gas atom (at theoretical STP) is 3.9 times more numerous than O gas, the O atom has 3.9 times the cross-section, meaning that the iron atom works out to about .5 inch wide where the O atom is one inch wide. That doesn't make the mercury atom much smaller than the iron atom, but then I don't know how more-numerous iron atoms are than oxygen atoms because google refuses to tell me reliably the true weight of iron gas at its boiling point.

The entire molecular structure in all of chemistry is based on a mere whim, a guess that suits the big-bang imposters. This is criminal. Sound the sirens. Let the lights flash. Citizens arise with whips to the posteriors of the baboons. How dare they.

How does it suit the big bangers? By making the H atom the smallest, the most rudimentary, suitable for star formation in the EARLY stages of big-bang outspreading of the cosmos. See that? This is not science, it's insidious, big-bang imposterism. One proton per one electron is quackery. The entire planet is under the thumb of folly. There has not been, nor is there now, even one respectable man in their ranks who takes it upon himself to expose this trash.

They calculated that iron gas would weigh 55.85 times more than H gas at STP, wherefore they assigned the iron atom a weight of 55.85 more than the H atom. When we take their 2.49 g/l weight for their calculated iron-gas weight at STP, it's 27.7 times as much as the .0899 weight of H gas per liter, meaning that they had a big problem because it forces them to claim that the H atom is 27.7 times heavier than the iron atom.

You can tell they cheated when choosing 2.49, or thereabouts, via their ideal-gas math, because it gets the 27.7 figure at half of 55.85 they assigned as the weight of their iron atom. Now you know why they poof the H atom into a diatomic atom so that it can appear to weigh 55.85 times less than the iron atom. There's no way for us to verify that the weight of iron gas is 55.85 times the weight of H gas. They can therefore fudge the numbers to make them work rather excellently.

It's all a sham to keep the big bang happier. Mythology is more correct to reality than atomic physics.

Imagine how many tables there are online showing weights of metal gases, yet when asking google for "table of metal gas weights from experimntal results," I get everything but what I asked for. You can't tell me that google doesn't understand what I'm after. Why do you think we're not allowed to know the true weights of metal gases? Why are the experts in this field blanked out by google? Don't you think there are many companies that would want to know the true weights of metal gases at various temperatures? Why is google so anti-educational?

Every atomic weight assigned to every metal atom is supposed to be perfectly proportional to the difference in weights between hydrogen gas and the metal gas. All we want to do is verify that this is true, but there's no way for me to do it in spite of having access to the Internet, at least not with google canada. The anti-Christian canadian government probably has wink-wink sway from the google corp. to block various topics, and to protect big-bang science too.

NEWS

This video reveals that there's been some serious and secret investigations against Maricopa's electrion fraud (start video about one0firth in), unless it's just for show only:
https://www.youtube.com/watch?v=cjbtNqMzooc

This is a good video in opposition to Israel, for the first 3/4s:
https://www.youtube.com/watch?v=8H-FQVX42oQ

Erika Kirk claims she first saw Charlie in Jerusalem, and that was in May of 2018 while Charlie was in Jerusalem with Candace Owens. I suggest Candace was sleeping with Charlie at that time. As Charlie got together with Erika later that year, doesn't it seem that Candace's vendetta against Erika these days looks like she's getting vengeance because she took Charlie away from him? google says that Candace met her current husband later in December of that same 2018.

It could be that Charlie was in Israel to set up some agreement between Turning Point and the Israeli government to push pro-Israeli propaganda amongst the American youth, but that when the Israelis got wind of Candace's not so pro-Israeli stance, they tried to put Charlie together with Trumpian Erika, and succeeded. Candace would thereafter work to turn Charlie away from Israel, it seems, because Candace blamed Israel for taking Charlie from him. Just a viable theory that looks feasible.

Baron Coleman treated that Jerusalem meeting on his Monday show this week. I think Baron knows that Candace was sleeping with Charlie, but, of course, he's not going to admit it, at least not now.

Midweek, Wednesday and Thursday, I think, Baron showed that the man who was asking the question to Charlie, when the gunshot went off, has a family tied to Hollywood, and moreover Baron showed more evidence of a staged operation, yet he can't bring himself to taking a faked murder seriously.

On his Sunday show, he shows evidence that Erika Kirk and/or her mother was involved in child trafficking in Jaco, Costa Rica, with a company, KidsFirst. He also expends passion giving an Iran-related reason for believing that Charlie was murdered, and while this has logic, it's not correct. Charlie is alive because he agreed to leave Turning Point when push came to shove with the war hawks:
https://www.youtube.com/watch?v=6hBq3dT-nrA

News this week is sicker than ever, I just can't stomach podcasters much anymore, and I still resist listening to leftist media. I need a new job. The same podcasters opposed to Gaza's re-build are also opposed to war in Iran. I welcome the overthrow of the religious Iranians, but I do not relish the West having its fingers in another Middle-East country.

The world's biggest threat to the world is Western billionairism. These few people gobble up all the wealth, and their having access to Iranian oil and other treasures there will only make them gobble up still more, until they have the sure powers to come down hard on the lowly people with wicked plots that they've held to secretly for decades. I think that Western billionairism, championed by Trump, is Babylon the Great. Iran is correct, America is the Great Satan. Not the people in general, but the Godless billionaire class and billionaire wannabees.

Here's a video producer who uses come pizzazz to expose evolutionists as long-standing frauds:
https://www.youtube.com/watch?v=7m7050jfAT4

This is a short video on the ancestral line of Jesus' mother:
https://www.youtube.com/watch?v=LGwmmFnFtys

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NEXT UPDATE

Here's all four Gospels wrapped into one story.

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

Pre-Tribulation Preparation for a Post-Tribulation Rapture

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