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How is chemical energy stored in carbon
I'm fairly new to chemistry, and I have a hard time understanding how chemical energy is stored in carbon (I'm aware that the question can be generalised to 'how is chemical energy stored', but I'm trying to understand it in the example of carbon). In the example in my book, carbon and oxygen react to form carbon dioxide, releasing energy in the form of warmth in the process.
My book says that the energy released in this process mainly comes from carbon, which is originally the solar energy stored by plants. I have tried to look for explanations how that energy is stored (not how the process of photosynthesis works, but in what form or what is the meaning of 'stored energy'), but I still can't really wrap my head around it. The best answer so far was What is the nature of chemical energy?
If I understand the answer there correctly, the bond potential energy of a carbon dioxide molecule is lower than the combined bond potential energy of a carbon atom and an oxygen molecule. Given that, there are still two things I don't understand:

Why is the bond potential energy in the carbon atom and/or oxygen molecule higher than that of a dioxide molecule? How can that be explained? Is this directly related to the chemical traits of the elements carbon and oxygen and the compound carbon dioxide?
Why, or better said how is said bond potential energy converted to warmth in the process?

I'm aware that there might be concepts in physics here I don't know or understand to understand a possible answer, and I would be thankful for any pointers to what subjects I should be first reading to understand such an answer.
Edit: I would like to quote here something I came across on Wikipedia. I'm not providing it as an answer nor claiming it is correct, just trying to illustrate how confusing it is to find a satisfying answer for this matter. This comes from https://en.wikipedia.org/wiki/Chemical_bond#Overview_of_main_types_of_chemical_bonds:

In the simplest view of a covalent bond, one or more electrons (often a pair of electrons) are drawn into the space between the two atomic nuclei. Energy is released by bond formation. This is not as a reduction in potential energy (sic), because the attraction of the two electrons to the two protons is offset by the electron-electron and proton-proton repulsions. Instead, the release of energy (and hence the stability of the bond) arises from the reduction in kinetic energy due to the electrons being in a more spatially distributed (i.e. longer de Broglie wavelength) orbital compared with each electron being confined closer to its respective nucleus.

Italic and bold added by me. Note that Wikipedia itself has a '[clarification needed]' tagging in the middle of that text, right after 'two protons', which I removed.
So if I understand that correctly and assuming that info is correct: any formation of a covalent bond would result in energy being released. But that still doesn't explain why my book is claiming the energy is particularly stored in carbon (and not in oxygen), and why it is released as warmth. In addition, the explanation on Wikipedia would contradict what I referred to earlier, namely that the energy released comes from differences in bond potential energy.

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"My book says that the energy released in this process mainly comes from carbon, which is originally the solar energy stored by plants." - what is this book, and is that a direct quote?

One of the most important issues in understanding bond energy in chemistry is the question: energy relative to what?
The formation of bonds from (mostly hypothetical) atomic elements releases energy (or they wouldn't be bonded.) Solid carbon (graphite or diamond) has less energy than a cloud of carbon atoms so it could be said graphite has less energy relative to a cloud of atomic carbon atoms. We would describe this energy as "bond energy". But this is the energy released by the hypothetical process of forming lists of C-C bonds from isolated carbon atoms.
But there is another subtle complication if we want to understand reactions. sometimes, when chemists write a reaction scheme they don't specify all the detail. So C +O2->CO2 doesn't actually mean atomic carbon atoms plus oxygen molecules give carbon dioxide. Most chemists would assume that the carbon starts in the state it is normally found as in laboratories: graphite (so not isolated atoms but a molecular solid containing lots of C-C bonds). This terminology seems to have confused some people. But chemists don't like to waste space in formulae redundantly specifying things most of their colleagues will already know. The default state of an ingredient in a reaction is assumed unless specified otherwise and this usually means the form the compound or element takes in a bottle on the lab shelf.
So how can we say that carbon stores energy? Well we have to ask the question, relative to what? In the case of the question the answer is relative to carbon dioxide gas. The point here is that the double bonds in CO2 are stronger than the combined bonds in oxygen gas and graphite (so they are lower energy) This means that the reaction between carbon and oxygen releases energy (it takes less energy to break all the O-O and C-C bonds than is released when new C=O bonds are formed. So elemental carbon "stores" energy relative to carbon dioxide.
The claim that carbon "stores" solar energy absorbed by plants is a little simplistic (mainly because plants don't store carbon but usually store more complicated molecules like sugars or poly-sugars like lignin or cellulose). Dead plants are converted to more carbon like compounds (oil is mostly hydrocarbons, but coal is mostly carbon) under some geological conditions. What plants actually do at the start of this process is to convert carbon dioxide into sugars using energy input from sunlight. They do this to store energy and to grow by converting the sugars into the structures of their leaves, stems and trunks.
We can reverse that process by burning the plants or their degradation produce like oil and coal in air to reproduce the carbon dioxide and release energy. In this highly simplified sense, carbon is a store of "solar" energy.
But the key to understanding what is meant is to understand the chemical reaction involved in the creation or burning of the carbon. Then you can understand what energy is stored and how it is stored.

Just to make sure I understand you correctly: in my (rudimentary) understanding of a reaction, C + O2 -> CO2 means that one carbon atom and one oxygen molecule form together one carbon dioxide molecule. Maybe I don't understand the nature of a chemical reaction, but in that context, how can we speak of C-C bonds?

Also, how should I understand the idea that a cloud of carbon atoms (maybe you meant a C-C molecule or otherwise carbon-based bond but I took it at face value?) has more energy relative to solid carbon? I might be understanding things a bit too literally here, but I'd like to know what exactly your usage of bond refers to in this context. Is this purely the forming of a molecule (as in covalent or otherwise any form of bond), or is it also simply "a lot of carbon atoms together at a given place" (= a piece of graphite).

@user2311517 carbon, as normally found in nature as graphite, consists of a molecular solid with every carbon bonded to several other carbons. The reaction C + O2 is shorthand for solid carbon plus oxygen gas, so the reaction involves breaking lots of C-C bonds and O=O bonds resulting in CO2. You have to account for all the bonds broken and formed to know the energy difference.

@user2311517 The "cloud" of carbon atoms is a hypothetical concept which we use to understand the relative bond energy of solid carbon: we don't find clouds of carbon atoms in nature: so C-C bonds exist in the natural state of carbon.

In this case, it begs the question of why we are taught that pure O always comes in the form of a O2 molecule, while the form of C in the reaction can be pure, where in reality it is not? In other words, in this reaction, it is visible that we have an O2 bond which is broken, but it is not visible that we have a C bond with three other Cs which is also broken. That seems somewhat non-intuitive to me.

@user2311517 Your question is phrased such that I have interpreted your reaction to literally mean the reaction of 1 free carbon atom and 1 free oxygen molecule reacting with each other. As that was the context, I took your question as a purely theoretical one, as did the others who posed their answers.

@TanYongBoon yes, the question was posed as a theoretical one, trying to understand where the energy is coming from in a very small system composed of just this reaction. However, this answer went on to speak about carbon bonds (as in graphite), and I think my question as to why the implicit breaking of these bonds when you ignite a graphite stick is not explicit in the reaction formula, while the breaking of the oxygen molecule bond can be pretty much 'read' from the reaction. I think it is a valid (follow-up) question.

I realise that you are not interested in the mechanisms of photosynthesis or metabolism, however, as I am a biochemist, it will be easiest for me to explain this from a biochemical perspective. I will try to avoid talking about specific mechanisms. Regardless, I do believe that this may be able to help you understand this problem.
Bond potential energy is used to provide a means of "paying" for the cost of cellular processes that have no entropic gain. And as you have stated in your question, this all begins with photosynthesis.
Consider the carbon dioxide molecule. It consists of two $\ce{C=O}$ bonds - each of which contains a bond energy of $\pu{799 kJ/mol}$. This inorganic form of carbon is the most stable of all carbons in a biological system. To break this bond, an input of energy greater than, or equal to $\pu{799 kJ/mol}$ must be provided; in plants this energy is provided through photosynthesis.
So, to avoid going through how a plant builds energy-rich molecules, let us skip to the more interesting part: how is energy stored and utilized in carbon bonds created by photosynthesis?
As you don't seem to care much for mechanisms, I would like you to know that what I am about to say is a vast over-simplification. It will also be helpful to note that a $\ce{C-C}$ bond has an energy of $\pu{346 kJ/mol}$, a $\ce{C-S}$ bond has an energy of $\pu{272 kJ/mol}$, and a $\ce{S-H}$ bond has an energy of $\pu{363 kJ/mol}$.
During metabolism, sugars will eventually be converted to a molecule called pyruvate. This pyruvate molecule will react with a thiol containing molecule called coenzyme A (CoA) to form acetyl-coenzyme A as follows:

The enthalpy of formation can be calculated as follows:
$$\Delta H = \sum \text{Bond energy reactant} - \sum \text{Bond energy of products}$$
For this reaction to occur, the following must happen:

Break the $\ce{S-H}$ bond of the thiol of CoA
Break the $\ce{C-C}$ bond of pyruvate
Form a $\ce{C=O}$ bond of carbon dioxide
Form a $\ce{C-S}$ bond of acetyl-CoA

This means that the enthalpy of formation:
$$\Delta H =  ( 346 + 363) - ( 799 + 272) = \pu{-362kJ/mol}$$
That is, this reaction releases $\pu{362 kJ/mol}$ worth of energy.
The best way to think about bond energies, is to consider that the "bond energy" is actually the amount of energy that is released (as heat or light) upon a bonds formation, or the amount of energy that must be provided (by heat or light) in order to break a bond.
When you have molecules, such as sugars, they are eventually broken down in such a way to form carbon dioxide - which we have seen to contain very high energy bonds. It is in the process of forming these $\ce{C=O}$ bonds, that the "stored" energy can be released.
I should also point out that the $\ce{O2}$ that enters a cell never actually directly participates in any reactions with carbon. $\ce{O2}$ will however be converted to two molecules of water, by reacting with an acid called NADH.

Although this is not the actual reaction that living organisms utilize to produce $\ce{CO2}$, I think you have the reaction of formation of carbon dioxide in mind, which can indeed be explored to understand bond energy.
Let's have a box with pure carbon in some allotropic form, say graphite, immersed in an pure oxygen atmosphere. As time goes on - and it may indeed take a lot of time - the insistent collisions of oxygen molecules with the solid lattice will produce a chemical reaction. Think of throwing a bunch of keys randomly at a surface full of compatible locks. The keys aren't necessarily moving at the same speed or direction, so some of them will certainly bounce back from the surface - if they even get there, or they might not have sufficient speed to push itself into the lock. Nonetheless, some keys may have the exact appropriate motion to insert itself into a lock. But what does this keys unlocks? To proceed the analogy, the key unlocks that lock from the other ones, releasing it from that surface and bringing a key-lock pair onto the keys' flight path.
Now, remember from physics that when you presence a collision in which things stick, you probably have an inneslatic collision, where momentum is conserved but kinetic energy isn't. Some of that energy becomes heat. You seem to want to know how. But the real question should be why not?. Molecules move incessantly at random directions, exchanging momentum and energy chaotically. If you could canalize all molecular velocities to follow a main direction, you would have macroscopic motion and the energy to produce that phenomenon is called work. That's what's going on when you push or pull an object. When you stop pushing it, we observe that friction slows it down until it stops. Microscopically, the object molecules are colliding with the surface molecules over which it is sliding and also with the air molecules, both of which are somehow static in this reference frame, so they share momentum and energy until the object is also static and the molecules once again have no distinct preferential direction to follow.
The former kinetic energy of the object now belongs to the microscopic motion of molecules, which is randomly distributed. This is heat. The fact the same energy once visible at macroscopic motion is transferred in totality to the molecular realm is the first law of thermodynamics. The fact that energy in a certain way prefers to become heat is the second law of thermodynamics. Pull a string and release and it will oscillate for while with a decreasing amplitude, as its mechanical energy turns into thermal energy. This also happens with the key-lock lair. You could think of the mechanism as something like: key hits the lock, pushes itself in, transfers some kinetic energy and momentum with a normal direction with respect to the surface, the energy now belongs to the nearby surface elements, which then share this energy and distribute it along the other directions. If you're confused by this, don't worry. The underlying thermodynamics are mechanism-independent. So no matter how the phenomenon happens, we can conclude that for a sufficiently large amount of particles interacting, any energy that can become thermal energy certainly will within enough time.
The key-lock pair only fails to be a good analogy to the electronic energy of molecules. You might think of keys translating and rotating, just like molecules do and both certainly store energy within those degrees of freedom. But chemicals also present electrons, which move and interact, so they contain energy. Different atoms or molecules have different amount of electrons, with different levels of energy, which is related to all the quantum and electromagnetic interactions. This is hard stuff. We are better off thinking about chemical reactions simply as the movement of electrons - with the nuclei maintaining structure and "following" the electrons - towards more stable entities, with a lower energy, like the key inside the lock, and the energy is released as heat, not because there is some special mechanism to produce heat, but because heat is simplify unorganized energy, and it's the most likely product with septillions of molecules reacting altogether.

Note that this is an edited answer. I have realised that there was initially a very severe error in my answer. Please re-read if you have not read this edited version.
In chemistry, we purely look at the energies in the process of bond breaking and bond formation when we look at energetics, as Zhe aptly pointed out in the comments. Thus, free atoms, such as a carbon atom, do not "store energy". As you have clarified in the comments, you wish to know, "in the context of the reaction $\ce{C + O2 -> CO2}$, where the released energy comes from, how it is stored in carbon, and how does it become warmth". In my response, I will tackle this 3-part question. Note that I will not consider nuclear chemistry and the interactions of subatomic particles, other than the particle chemists are most concerned about - the electron.
In chemistry, energy changes arise due to two key processes: the formation of a chemical bond and the breaking of a chemical bond, where the term "chemical bond" refers to any of the following: ionic bond, covalent bond, metallic bond and intermolecular force of attraction (although some may dispute this). Bond formation is an exothermic process while the breaking of a bond is endothermic. This is rather intuitive as energy is needed as input to overcome the force of attraction between atoms. The exothermic bond making process may not be so intuitive. However, you can simply think of it as an application of the conservation of energy: If bond breaking is established to absorb energy, the opposite would give off an equivalent amount of energy. Another intuitive explanation why this is so comes from a very important chemical principle: Chemical systems, in general, prefer to be in the more stable state (i.e. a lower-energy state). Thus, bond formation and exothermic reactions, in general, are favoured thermodynamically.
Part 1: In the combustion of elemental carbon, energy is given off by the exothermic formation of two strong carbon-oxygen double bonds (endothermic), while the oxygen-oxygen double bond is broken (exothermic). The overall enthalpy change of the reaction is exothermic as the sum of the bond energies of the bonds formed is greater than that of the bonds broken. Thus, energy is released from the reaction.
Part 2: No energy is stored in a carbon atom, in a chemical sense. Now, this is what has come upon me just this morning as I was reflecting on this question:
What stores energy in chemistry? It came upon me that the statement "bonds store energy" is very flawed. Thus, I am afraid I would have to say that Zhe is wrong in his comments. But neither is it correct to say "free atoms (or bonded atoms) store energy". As much as I would not like to use this analogy to answer this question for possible points of contention regarding the relevancy, I feel nevertheless that it is appropriate: Consider two simple bar magnets (i.e. magnetic dipoles) which are attracted to each other. What holds them together is a magnetic force. To pull them apart, we input energy into them as work in order to overcome the force. This system of magnetic dipoles is analagous to how atoms are bonded to each other, just that the force of attraction between the two atoms are much more complex and that the nature is electrostatic, not magnetic (of course, we now know that magnetism is very closely linked to electric charges in physics but that is not important for this discussion).
So going back to relevant examples, which would illustrate my next important point as to explain where energy changes comes from. Consider the chemical energy stored in food. We say that there is a chemical potential energy (at a very basic level of physics). But is that true? Does food and many fuels store energy? That may be correct from the point of view of the fuel undergoing reaction. Why do I say so? Remember: It is the process of bond formation and bond breaking that give off energy. Bonds themselves do not store energy. Only from reactions would we observe energy changes as energy is given off or taken in by the reactants to form products.
Let's go back to the combustion of elemental carbon. Using this perspective, neither carbon, nor carbon dioxide, nor oxygen store energy. The substances in the final and initial state do not store energy. In the process of bond formation and bond breaking, that energy is given off or taken in by the system.
Why do bonds not store energy? Well... bonds are merely electrostatic forces of attraction, in the most elementary sense. How can a force store energy?
Even if you don't believe the rest of my low-level explanation, please at least take away one thing: Bonds don't store energy.
Part 3: In chemistry, the energy is very often given off as heat in exothermic reactions. However, they can also be given off as light energy or sound energy.
Regarding your more difficult query about why energy is given off during bond formation, you can refer to my very brief explanation in Para 2. However, I fail to give you a detailed answer. Although I find no fault in Wikipedia's explanation based on the greater delocalisation of electrons upon bond formation, I do not believe that it is the main reason. Certainly, the strong attractive forces of the nucleus must be involved as well.

Matter is energy. Everything that exists subnuclear particle, nuclear particle, atom, molecule, aggregate, probably even black holes is at a local relative energy minimum surrounded by an activation energy barrier. This barrier can be as high as the stability of atoms or as low as the entropy of separation. With a possible mechanism, the correct stoichiometry, and sufficient energy of the correct entropy anything can be changed into anything. A simple example: Sodium metal can easily be turned into hydrogen gas and then water; simply toss it into water and stand back and watch the sodium disappear the gas form, and then burn. Chemists finally learned that sodium reacted with water to give H2 that reacted with O2 to give water. It took centuries to realize that atoms, elements, could react to form compounds but not easily react to form different elements and to again realize that they could. Any entity is energy at a certain level This depends on the arrangement of particles, molecular structure, its surroundings, its position in a gravitational or electronic field. These are explained in nuclear, atomic, molecular, intermolecular, thermal, and field energies and are compartmentalized more than they should be.
Elemental Carbon exists in 2 very stable molecular forms: graphite, the more stable and diamond. Both are very stable because of their strong bond structure and consequent high activation energies. Graphite is more reactive because of the pi bond structure. Graphite exists in its relative energy minimum modified slightly by its physical conditions of pressure, temperature, gravitational and electronic and magnetic fields etc. [The question makes a point of C being a carbon atom. Elemental carbon has molecular bonds that require energy to break. If broken independently the energy of combustion of the atoms will be much greater than that of the graphite or diamond.] The calculation or measurement of absolute energy levels is very difficult or next to impossible. To surmount this difficulty it was decided to define chemical potential and internal energies of each element in its most stable form at 0. under standard conditions of temperature, pressure and Earth surface.  Standard energies, enthalpies and free energies are zero for pure elements in their most stable states at standard conditions, usually STP sometimes 298K. This does not
apply to entropies these can be calculated from absolute zero using the Third Law of Thermodynamics. Each element is in its relative energy minimum that is defined to be zero, but each element is at a different energy level in the grand energy spectrum. This means that when they are in contact and there is sufficient activation energy and there is a possible mechanism they will react to reach a different relative energy minimum or energy level. This is not always a lower energy because it depends on both energy and entropy and the precise conditions. What happens depends on the change in the free energy functions: the Helmholtz free energy, A = U -TS and the Gibbs free energy, G = H - TS not on U or H alone.
Chemistry usually involves the outer electrons, the so-called valence electrons and the rest of the atom or molecule is just an interested bystander altho sometimes it can be very interested. The arrangement of the valence electrons determines the shape, size and the chemical and physical properties of the substance. While a miniscule part of the total energy of matter the electrostatic attraction of electrons for nuclei is almost the total of the energy involved in chemical reactions. The electrons in energy levels attracted to 2 or more nuclei will be of lower energy provided energy can be removed from the molecule either by collisions with other molecules, forming heat, or by radiation. This means that successful bond formation involves a loss of energy to the environment, a lowering of entropy of the system and an increase in entropy of the environment. This is conveniently explained in the differences in bond dissociation energies and the requirement for removal of energy and the consequential decrease in entropy for an exothermic reaction is implicit. [regardless of exo or endothermic reactions the entropy of the universe increases]
Bond energies depend on the environment of the bond, they are influenced by the overall molecule. My initial premise each substance is in its own relative energy minimum. It can definitely be said that CO2 is in a lower energy plateau than the graphite + O2 mix, but more difficult to determine whether C or O2 are lower in absolute energy.
Plants with the appropriate enzymes and correct light frequencies can convert CO2 and H20 into carbohydrates Cn[H2O]n and also glycerides, hydrocarbons, with nitrates, sulfates, and phosphates, aminoacids, nucleic acids and many other compounds. It is lazy and presumptuous to call any or all of these wonderful compounds "carbon". Likewise CO2 in the atmosphere, not carbon, is the compound driving global warming.


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Polarity of xenon fluoride
The structure of xenon fluoride is a capped octahedron. The lone pair is stereochemically active, i.e. it will rotate about.

The dipole moments of the axial fluorides will get cancelled, and so will those of the four equatorial fluorides will also cancel each other. Thus, overall, the molecule should be nonpolar.
But what effect will the stereochemically active lone pair have on the polarity? Will the polarity keep on changing?

Molecular Structure of XeF6 and IF7. J. Chem. Phys. 1968, 48, 312

https://chemistry.stackexchange.com/a/50189/9961

https://chemistry.stackexchange.com/questions/59238/help-regarding-stereochemically-inactive-s-orbital

strongly related: https://chemistry.stackexchange.com/questions/64550/structure-of-xenon-hexafluoride

The picture used in the question is exactly same as that of the picture used in the question I linked.

Also see this: https://chemistry.stackexchange.com/questions/34073/what-are-the-structures-of-potassium-superoxide-and-xenon-hexafluoride/34075#34075

First of all, the point group for $\ce{XeF6}$ is disputed. The latest I saw was $C_\mathrm{3v}$, not $O_\mathrm h$ but the $O_\mathrm h$ is so energetically close to its $C_\mathrm{3v}$ that it is likely in a perpetual state of transition between the two. As for polarity, due to this it is hard to think it would have a permanent dipole. One would assume it would be easy to induce one, though.
$\ce{XeF6}$ is not the typical molecule the different models were made for, so YMMV.

First of all, the point group for XeF6 is C3v, not Oh as you have shown || AFAIK there is no clear consensus on it.

I have no idea about point groups . Is there some other way to explain its polarity ?

@permeakra I accept that there is no clear consensus. It is somehow relevant though to mention that it is either highly transitory or doesn't allow for polarity. Should I rephrase?

Yes please rephrase.

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What is the simplest explanation of the ASTM G71 procedures that can be done in a laboratory?
In my corrosion testing, I am to use the galvanic corrosion testing by ASTM G71. But due to limited knowledge about this field, I can't understand fully the procedures needed to conduct the test in a laboratory scale. What is the simplest explanation of the ASTM G71 procedures that can be done in a laboratory?

ASTM maintains a variety of publications. To the practitioner, usually the most useful are the test Methods. I have also found the guides (such as G71) and glossaries to be useful background. G71 is not a test method, and it lacks any "procedures", the only answer to your question is: "There aren't any procedures in ASTM G71." This may seem confusing since the Guide contains a diagram which seems to imply that a laboratory test involves two equally sized sample rectangles partially submersed in an electrolyte and connected via an external wire.
Since you are unfamiliar with ASTM publications, allow me to use an analogy which may be more intuitive. Take for example a "Guide for baking bread" which recommends that only a circulating air oven, with an air flow of at least 4 cu. meters per minute and a volume of less than 0.1 cu. meter capable of controlling internal temperature in the range of 50° to 250°C to within ± 2° and that the baker should wash his/her hands prior to starting while keeping the tools, ingredients and preparation surfaces at 20°C. You'll note that my guide (I made it up, so don't take it too literally) completely avoids telling you:

what the ingredients are,
How to mix and process those ingredients, and
how to bake bread.

That is, it is not a "method"; it is a series of best practices (or preferred, consistent practices). When is something like G71 appropriate? It is appropriate when the test needs to be customized to fit the problem. The guide recommends the metal be "as large as possible" (meaning up to full commercial size) while being practical in either lab or field test conditions.
There is also the need for the exposed surfaces to be in the same area ratio, and to avoid edge effects, especially those which will not or do not appear on the "real" assembly. The guide mentions 4 techniques to evaluate corrosion: visual (and photographic), mass loss, current, and potential (voltage). Which of those to use (if any) is up to the individual. It is vague because it can't be specific. It can't be specific because it is intended to apply to a great variety of different galvanic corrosion scenarios.

For future reference, please format your answers more legible, i.e. by use of spaces, sections, enumerated lists, etc.  This is not (only) about aesthetics, yet structures and eases comprehension of your answer by a reader.

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Why should the temperature be maintained at 0–5 °C in a diazotisation?
I know that in a diazotisation reaction, the temperature should be carefully maintained within the range of 0–5 °C by use of an ice bath. Why is this so? How does the temperature affect the stability of the aryl diazonium salt?

Diazonium salts are not stable. If the mixture is warmed or just let stand over time, decomposition will occur. If the decomposition is not controlled, it will be explosive.

Warming aryl diazonium salts in the presence of water is a well-known method for synthesizing phenols (with moderate yields).

The first step of this reaction is probably the reason why aryl diazonium salts are unstable at higher temperatures.

"The first step of this reaction is probably the reason why aryl diazonium salts are unstable at higher temperatures." I fail to see how the first step in the picture above is explaining why the diazonium salt is unstable. Please could you explain? Thanks!

The first step shows the decay path of diazonium salts. The driving force is the stability of the formed nitrogen. The above procedure is sometimes called phenolverkochung (german for to cook something till decay) which suggests that it is favored by high temperatures.

I understand, but: your answer correctly identifies that the diazonium salt decomposes at higher temperature, but does not identify why it was stable at low temperatures as well. If formation of dinitrogen leads to stability, why couldn't it have formed at $\pu{0-5^\circ C}$ as well? (Sorry if the point is already explicit, but I am not able to see it :( )

If you heat an organic substance, it eventually will decompose. Some substances sooner, some later. As it happens, diazonium salts are somewhat stable at around 0°C. I'm quite sure that they decompose even at 0°C, but the reaction rate is low enough to let handle them as intermediate products e.g. in dye synthesis.

The temperature at which many reactions are carried out is often arbitrary, with certain values being favoured due to the ease at which cooling baths can be prepared. As an example  -78°C is highly prevalent, and a question about this 'magic temperature' has been asked here previously.
In the case of reactions forming diazonium salts, there has been a common rule of thumb that they must be kept somewhere between 0 - 5°C, which is even quoted on Wikipedia:

In aqueous solution diazonium salts are unstable at temperatures above 5 °C  (Ref: Wikipedia)

At temperatures above 5 °C, diazonium salts (in aqueous solution) have been known to decompose, often explosively, to liberate nitrogen gas, temperatures too low however can cause crystallisation, which causes issues later on (diazonium salts should never be dried upon isolation - they explode).
Although people often keep to this rule of thumb, theres no real body of evidence for its generality (i.e. different diazonium species will have vastly different decomposition temperatures).
Often the only time people carry out in-depth studies of a particular reaction is when the reaction needs to be scaled up, such as in the the pharmaceutical industry. Process chemists go to extraordinary lengths to study their reactions to avoid any possible safety issues (or even unintended side reactions).  In Organic Process Research & Development 2004, 8, 1059−1064, a team of process chemists at Lundbeck carried out extensive thermal studies on a diasotisation reaction they were doing as part of a Sandmeyer reaction, and they found no issues well above 5 °C (though as a precaution they still kept the reaction below 15 °C).

+1 "run between x and y" usually means "it failed outside, but we were not keen on finding out why exactly". Or even "that's what we tried and it worked". It's a thankless job, pinning down the limitation of a reaction.

0 °C and 5 °C are both "magic" in the same sense of -78 °C; 0 °C is obviously the temperature of an ice water bath, while 5 °C is about the temperature used in most refrigerators, where diazonium salts are often stored. So basically, the range means "diazonium salts can be stored in fridges for a while without major decomposition, and reactions need to be done with mild cooling".

Obviously temperature affects decomposition rates, but maintaining generation and stability of the nitrous acid is key to actually achieving aryl diazonium salt production. Different counterions impart lesser or greater stability of the aryldiazonium salt, to the point where certain counterions, such as tetrafluoroborate or acetate are stable at room temperature, so would be of lesser concern in the reaction.

Aryldiazonium salts are unstable at high temperature and decompose to produce hydrogen chloride, nitrogen gas and chlorobenzene.
From nature journal:

The decomposition of aromatic diazo compounds is non-ionic in
  mechanism. When suspended in an organic liquid like benzene,
  diazonium chloride appears to melt at about 50° C and then immediately a violent decomposition sets in. There is great heat
  evolution and except on the small scale, the reaction tends to
  become explosively violent. Often hydrogen chloride is evolved,
  and whenever its formation has been observed, chlorobenzene has been
  found amongst the reaction products.

Decomposition of aryldiazonium salts has also been studied in other solvents like water(to produce phenol) and alcohol(to produce aldehyde and alcohol).
Hence, the salts are never dried and always kept at low temperature. It has also been studied that among all diazonium salts, benzenediazonium fluoroborate is by far safest to use. It is not known to explode unless heated at high temperature. It is stable and can be stored for more than a month at room temperature or for a few years at -20 °C under $\ce{N2}$ in the dark. Like other diazonium salt, it decomposes but slowly and should not be prepared well in advanced and should be stored in refrigerator.
Why benzenediazonium fluoroborate is stable is discussed in another question.


text:
Briggs-Rauscher numerical model
I want to use the Briggs-Rauscher experiment to demonstrate oscillating reactions, and then explain it using a numerical model.  How much is known about this reaction?  Is it possible to write down a set of differential equations for the concentrations of each chemical, and solve it to find oscillatory behaviour?
If so, what is a fairly simple but realistic set of differential equations for Briggs-Rauscher, and what are good parameters to find oscillations?
(By fairly simple I mean that it could be handled in an Excel spreadsheet or similar by talented school students.)

https://en.wikipedia.org/wiki/Brusselator

Is this supposed to model Briggs-Rauscher?  If so, what corresponds to A, B, D, E, X and Y in the wiki article?

See this for detail. It contains all you need to solve your problem.

@IvanNeretin Brusselator is too simple.


text:
Would copper sulfate corrode copper metal over time?
I'm performing electrodeposition experiments with copper electrodes (anode and cathode) and a copper sulfate electrolyte solution. For my project, the cells assembled would need to be stored for a period of months. I was wondering if the copper sulfate can corrode the copper metal and how long it would take? The copper sulfate concentrations range from 5 mM to 100 mM.

$\ce{CuSO4}$ will not corrode pure copper appreciably over a few months (though there is some dissolution and redeposition through random processes, such as density gradients and thermal motion, it might just reduce a mirror finish to slightly roughened).
However, if the electrodes or wires are brass, bronze or other alloy, or if there are other metals in contact, such as a tin-lead solder joint, there is a chance of some corrosion. In fact, copper sulfate is used to test passivation of stainless steel, and visible corrosion of iron can take place in minutes.


text:
Why is the E2 protein important for protein degradation?
In the ubiquitination process, the E2 protein transfers ubiquitin from E1 to E3. Why can't E1 work with E3 directly to tag some protein with ubiquitin? Going further, why can't one of the E(1-3) proteins just do it all by itself?
E1: ubiquitin-activating enzyme
E2: ubiquitin-conjugating enzyme
E3: ubiquitin-protein ligase

Pretty hard question, "why", as we've not been watching for the millions of years it took to build that system... I think the better way to look at it is to name the advantages this system has and hypothesize on it's selection. BTW in biology a good answer to why is always because.
I would guess though that given the function of ligases, it has been selected against due most probably to lack of selectivity. E1s are very small and already need to do both activation of Ub consumming atp and doing a somewhat major conformation shift to accept ligation of the activated Ub. The amount of Protein Protein surface usually required to maintain the exquisite selectivity that is required for such a system can be assumed to be too great for the size of the E1s.
Then if you ask why it did not evolve in some kind of huge multidomain protein; I would probably argue to the point that you need to produce a very large amount of it all the time, and that it must be somewhat difficult to target that protein for proteosomal degradation / make sure it's folded properly. Having a smaller footprint might help in keeping the domains ordered and reducing entropy hence making sure the protein stays functionnal longer.
Then why not the E2 doing it directly, and why do you need the E3 in between... Well again, I would assume this is also a question of selectivity. If you need to regulate thousands of cytosolic proteins and control their regulators at the same time, you need some modularity, you need the ablity to create multi protein complexes that bring different layers of regulation and cannot assume that one protein can do that. The term E3 is also some horrible bin where people put anything. E3 are very often large and sometime huge multiprotein complexes, think of the APC complex, or in the most well documented cases, the Cullin complexes.
The way E3s evolved is pretty cool actually, to recruit E2-Ub a protein just needs to get a domain through some mutation (RING, U-Box, HECT...) then it can target whatever it used to bind to a degradation complex. That is some pretty neat and quick trick to create diversity rapidly if a system actually does something that brings an evolutionary advantage.
I see you left out the E4 which may play a role in Ub chain propagation but as they pretty much function as E3 anyways, there is debate as to why we would need to use that terminology.
This is by no mean some definite answer since I could be wrong on most of these things. I hope I helped you to further your reflection on the subject.

Welcome to Chem SE. A great first answer I must say. Keep contributing to this site!

Agreed this is good!

I do agree with @Dan as to “why” the ubiquitination proteins are arranged the way they are and are able to complete the process together.
It is easier to answer the question after examining the role each protein works.
The main reason is that these proteins vary in structure and selectivity and therefore have different functions in the process on of ubiquitination, thus not one or two proteins are essential but all of them.
Firstly, these proteins are very much different in terms of structure, selectivity, abundance and variation.
To understand this difference lets consider the x-ray structures of these proteins:

Let’s discuss a bit on the functionality of these proteins on named subheadings:
Specificity
The E1 protein (ubiquitin-activating enzyme (E1), a homodimer of ~1050-residue subunits), initiates the ubiquitination process through conjugation by forming a thioester bond. The ubiquitin is then transferred to a specific Cys sulfhydryl group on one of numerous ubiquitin-conjugating enzymes E2.
Most organisms, including yeast and humans, have only one type of E1. Thanks to the role of E1 which play a part in the committed step of the process. Even if proteins were marked for degradation, they would not be processed unless the enzymes system is activated. Basing on structure, I don’t think E2 or E3 would be a good activating enzymes.
The abundance of catalytic cysteine sulfhydryl groups on E2 makes it a good transporting medium for ubiquitin., which make them specialised transporters of ubiquitin.
To expand more on specificity, each of the many E3s present in eukaryotic cells mediates the ubiquitination of a specific set of proteins and thereby marks them for degradation. Different E2 and E3 enzymes exhibit different specificities for target proteins and thus regulate different cellular processes. Some E2 and E3 enzymes are highly localized in certain cellular compartments, reflecting a specialised function.

Selectivity and Cooperation mechanism
In protein degradation, ubiquitin is not just transferred to any protein, in fact proteins only marked for degradation are the ones that get the ubiquitin.
One signal which controls whether a protein becomes ubiquitinated is the half-life of a cytosolic protein mainly determined to a large extent by its amino-terminal residue (this dependency is referred to as the N-terminal rule). Not surprisingly E3 plays a central role in recognising and selecting proteins for degradation. E3 selects proteins by the nature of the N-terminal amino acid.
E2-S ubiquitin transfers ubiquitin to free amino groups on proteins selected by E3, the ubiquitin-protein ligase.
Although the E3 component provides most of the substrate specificity for ubiquitination, the multiple combinations of the E2-E3 complex allow for more finely tuned substrate discrimination.

Faster processing

More than one ubiquitin may be attached to a protein substrate, and tandemly linked chains of ubiquitin also occur via isopeptide bonds.

The attachment of a single molecule of ubiquitin is only a weak signal for degradation. However, the ubiquitination reaction is processive: chains of ubiquitin can be generated by the linkage of the e-amino group of lysine residue 48 of one ubiquitin molecule to the terminal carboxylate of another. Chains of four or more ubiquitin molecules are particularly effective in signalling degradation. These polyubiquitin (polyUb) chains, which can reach lengths of 50 or more ubiquitin molecules, are generated by the E3s (although how they switch from transferring a ubiquitin to the target protein to processively synthesizing a polyubiquitin chain is unknown).
The use of chains of ubiquitin molecules may have at least two advantages. First, the ubiquitin molecules interact with one another to form a binding surface distinct from that created by a single ubiquitin molecule. Second, individual ubiquitin molecules can be cleaved off without loss of the degradation signal.

Summarising the discussion, as we saw since humans have one type of E1 enzymes, if that enzyme alone would have been responsible for the all the processes, then selectivity of proteins would have been very low (recognition and marking for protein degradation or other cellular compartments lacking the enzyme would not have ubiquitination at all). This is one of the reasons why E2 is necessary - thete are several forms of E2 with different specifities (some are about to react with RING domain E3 while others with HECT), which would not be the case if E2 weren't present, E1 alone would not accomplish this either.
E2 is able to recognize the proteins which must “receive ubiquitin”, E1 alone would not achieve this neither the two of them because, they must be first marked by E3.
Now the supposedly unanswered question is since E3 is more selective and abundant wouldn’t it carry the process alone? . Well this is a tough question but I guess the answer lies in the discussion above: Although the E3 component provides most of the substrate specificity for ubiquitination, the multiple combinations of the E2-E3 complex allow for more finely tuned substrate discrimination. These are some points which might address the importance of these enzymes, I believe there are deeper explanations to this but this what I managed to dig in my research.
References

Biochemistry Jeremy Berg
Biochemistry Voet and Voet
Biochemistry Grisham

Thanks for the fanstastically detailed answers. The x-ray structures contribute a ton to the discussion.


text:
How can I calculate the standard reaction enthalpy if I lack one formation enthalpy?
I have this reaction, for which I need the standard enthalpy of reaction:
$$\ce{CuO (s) + H2S (g) \rightarrow CuS (s) + H2O (g)}$$
This is the data from the NIST web book (standard formation enthalpies, $\mathrm{kJ}\cdot\mathrm{mol}^{-1}$):

$\ce{CuO (s)}$: -156.60
$\ce{H2S (g)}$: -20.5
$\ce{H2O (g)}$ -241.83

There are no data tabulated for $\ce{CuS (s)}$, so how can I compute the standard reaction enthalpy lacking this data?

To put it bluntly, you can't.

The standard enthalpy of formation of $\ce{CuS}\,\mathrm{(s)}$ is -48.4 $\mathrm{kJ}\cdot\mathrm{mol}^{-1}$. See this reference  for your requested data and more.


text:
How do I convert an alkane to an alkene?
I was doing some conversions from my book and I was stuck on one question, where I need to convert propane to prop-1-ene. Help me with the concept behind it.

As to your query. There are probably hundreds of possibilities at varying conditions to convert alkanes to alkenes. This question is therefore very broad and you should probably try to narrow it down to add context and add your own research on possible reactions. Otherwise its score will drop, or it gets closed (or both) and you won't get an answer. The easiest (but also very unspecific) I can think of is alkane -> alkyl halide -> alkene.

There are numerous methods to convert propane to prop-1-ene. Some of them are:

Using boron nitride($\ce{BN}$)

The exothermic oxidative dehydrogenation of propane reaction to
  generate propene has the potential to be a game-changing technology in
  the chemical industry.  Here, we report that hexagonal boron nitride
  (h-BN) and boron nitride nanotubes (BNNTs) exhibit unique and hitherto
  unanticipated catalytic properties resulting in great selectivity to
  olefins. As an example, at 14% propane conversion, we obtain
  selectivity of 79% propene and 12% ethene, another desired alkene.
  Based on catalytic experiments, spectroscopic insights and ab initio
  modeling, we put forward a mechanistic hypothesis in which
  oxygen-terminated armchair BN edges are proposed to be the catalytic
  active sites.

Using cobalt and magnesium molybdates

Catalytic performances of various metal molybdates were tested in the oxidative dehydrogenation of propane to propene with molecular
  oxygen under an atmospheric pressure. Most of the molybdates tested
  promoted the selective oxidative conversion of propane to propene and
  among them cobalt and magnesium molybdates were found highest in the
  activity and selectivity. It was also found that their catalytic
  activities were highly sensitive to the catalyst composition, and it
  turned out that $\ce{Co_{0.95}MoO_x}$ and $\ce{Mg_{0.95}MoO_x}$
  catalysts which have slightly excess molybdenum showed the highest
  activity in the oxidative dehydrogenation of propane. Under the
  optimized reaction conditions, higher reaction temperatures and lower
  partial pressures of oxygen, these catalysts gave 60% selectivity to
  propene at 20% conversion of propane. Since the molybdates having the
  surface enriched with molybdenum oxide tended to show high activity
  for the propane oxidation, surface molybdenum oxide clusters supported
  on metal molybdate matrix seem to be the active sites for the
  selective oxidative dehydrogenation of propane.

Using gallium oxide

Dehydrogenation of propane to propene in the presence or absence of
  $\ce{CO2}$ over four polymorphs of gallium oxide was investigated.
  $\ce{β-Ga2O3}$ exhibits the highest activity among the polymorphs, and
  it is even more active than chromium oxide catalyst in the presence of
  $\ce{CO2}$. H2-TPR and XPS studies show that gallium oxide is hardly
  reduced below 600 °C. The dehydrogenation reaction is suggested to
  proceed through a heterolytic dissociation reaction pathway, and it is
  enhanced by $\ce{CO2}$ because of the existence of the reverse water
  gas shift reaction and the Boudouard reaction. The high catalytic
  activity of $\ce{β-Ga2O3}$ is probably associated with an abundance of
  surface medium-strong acid sites related to the coordinatively
  unsaturated $\ce{Ga^3+}$ cations and the conjugated effect of proton
  and oxide. Furthermore, increasing the reaction temperature
  facilitates the activation of $\ce{CO2}$ over $\ce{β-Ga2O3}$. The
  promoting effect of $\ce{CO2}$ on $\ce{β-Ga2O3}$ catalyst is more
  evident above 550 °C.

Using platinum and platinum-gold alloys

The rates of dehydrogenation of propane to propene over platinum and very dilute platinumin-gold alloys have been measured. In the
  composition range of 0.5–14.0 atom % platinum, the rates per unit
  surface area of the alloy powders vary linearly with the bulk platinum
  concentration in the alloys. From this it is concluded that only one
  platinum atom is involved in the rate-determining step.
For both platinum and the alloys, activation energies of some 29 ± 2
  kcal/mole were measured. The reaction rate order in hydrogen, however,
  is different. It is proposed that propane dehydrogenation over
  platinum and over the alloys occurs via the same reaction mechanism,
  namely, dissociative chemisorption of propane on a single platinum
  atom to which two adsorption sites are associated, one of which
  carries a hydrogen atom. The subsequent conversion of the propyl
  radical into π-bonded propene via β-hydrogen elimination appears to be
  rate determining. The last step, desorption of π-bonded propene, has a
  comparatively low activation energy. The difference in negative
  reaction order, with respect to hydrogen, between platinum and its
  diluted alloys reflects a lower steady-state θH on Pt atoms surrounded
  by Au atoms vis-à-vis Pt atoms on a Pt surface.

Attribution to used sources is important. Additionally, to prevent invalidation of links and as a backup mechanism, we prefer having human-readable citations and (if possible) stable links. If you would like to learn more, see meta.chemistry.se.


text:
Different chemoselectivity of transition metals in catalytic hydrogenation
I am quoting my Clayden et al 2nd edition:

Some catalysts are particularly selective towards certain classes of compound – for example, Pt, Rh and Ru will selectively hydrogenate aromatic rings in the presence of benzylic C-O bonds, while with Pd catalysts the benzylic C-O bonds are reduced faster.

My question is, why $\ce{Pt, Ru, Rh, Pd}$ all being transition metals still show different chemoselectivity in hydrogenation reactions?
1st edit: (After reading ringo's comment I think I should add something more) Well, I first thought that the number of $\ce{d}$ electrons is the cause. But, both $\ce{Pd}$ and $\ce{Pt}$ have same number of $\ce{d}$ electrons. Yet, they show different chemoselectivity. I think the difference is because $\ce{Pt}$ has an $\ce{f}$ orbital filled up.
And, I guess as $\ce{Ru, Rh, Pd}$ belong to same period, they show similar chemoselectivity.

Because they all have different numbers of valence electrons.

Well, Palladium and Platinum have same number of $d$ electrons, as they both belong to group 10, but they show different chemoselectivity.

@Ringo And talking of valence electrons, all transition metals have valence shell configuration $ns^2$.

It’s because each element is subtly different. All four will interact with both functional groups in some way but some interactions will be stronger due to the intrinsic electronic properties. And the stronger, i.e. better interactions will drive the reaction. Sometimes even, seemingly random things happen, like my colleague who managed to completely reduce an N-protected 3-acyltetramic acid to the fully saturated compound leaving only the amide carbonyl functionality. That was funny.

@Jan Your comment seems to tell me that there isn't much research about why certain catalysts act in a certain way.

The actual problem is that I’m just the organic chemist who uses the catalysts in published ways and that I have little clues as to what has been done or not. It could well be that someone has already checked out that question and it is published somewhere.

@Jan Then I will have to search...

One important piece that's missing here is what ligands each metal is coordinated to (or if it's the plain metal).

Plain metal @Zhe

Still nobody answering...

@Mockingbird, it's an interesting question, but searching the literature doesn't yield much. The problem with a lot of heterogeneous catalysis (bulk metals absorbed on surfaces) is that often its not entirely clear what the active catalytic species is (or more generally whats going on)

@Not Evans Has anyone tried shed some light in this scenario?

@Mockingbird not that I can find

Unfortunately, this is extremely complicated and the topic of active research in both computational and inorganic chemistry and chemical engineering. There is no simple answer to your question. We can discuss various forces that go to determining chemoselectivity though.
1) Crystal structure: You said that you are primarily focused on bare metal catalysis, so I will restrict myself to those systems. Not all surfaces of even the same material behave the same. These are commonly described by Miller indices (1), which describe the surface by how a plane cuts through the material's unit cell. The reactivity and selectivity are heavily determined by the exact surface structure (think dangling bond sites, binding sites, multicenter reactivity, all of which are influenced by the surface atomic structure both local and in its immediate environs). Different materials will have different crystal structures based on both electronic effects (like how many valence electrons there are) and size effects (how big atoms are, although there is relatively less variation on this in the 2nd and 3rd rows).
2) Chemistry of binding sites: Some materials are more likely to bind molecules than others. The Pt-H bond is extremely strong, much more than a Pd-H bond (these trends down a period are often explained in terms of the diffuseness of the 5d orbitals vs 4d which lead to better overlap with the main group orbitals of hydrogen and other light elements). This leads to greater coverage of a Pt surface with hydride equivalents than a Pd surface (generally, caveats apply). On the other hand, Pd tends to bind oxygen atoms more strongly than Pt does (one can think about hard-soft acid base theory here to explain that, but the situation is probably more subtle and complicated than that). One can easily imagine a benzylic C-O oxygen bind to the Pd surface which holds the site for hydrogenation close the the surface until a hydride equivalent can diffuse over to react (hydrides on surfaces are quite mobile and diffusion is often quite rapid, of course, hydride diffusion rates are also strongly determined by surface structure and material properties (because almost nothing can be simple)). On the other hand, $\pi$-system can interact strongly with metal surface (especially of soft (again in the HSAB sense) metal). So, for Pt, the aromatic system is brought close to the surface, allowing hydrides to migrate over to it.
So we can hand-wavily discuss why Pd is more likely to hydrogenate benzylic C-O bonds and Pt is more likely to hydrogenate aromatic systems as we did above, but the nitty-gritty detail is quite complicated. This is a massive area of research. Inorganic chemistry is its own field not because everything is so similar, but because metals have hugely varying behavior. The fact that these are "all being transition metals" should NOT imply that their reactivity is similar in the slightest. Being in the same period as a transition metal tells you as much about being in the same period in the main group. C and O and N have totally different chemistry, as do Rh and Ru and Pd.
P.S. f orbitals do not figure into this.


text:
Bicarbonate decomposition and pKa
I want to understand why the p$K_{\mathrm a}$ of $\ce{HCO3-}$ is approximately 10.2 and not less.
My hypothesis is this :
Bicarbonate could react in two ways in an aqueous solution:
$[1]\qquad\ce{HCO3- + H2O -> CO3^2- + H3O+}$
$[2]\qquad\ce{HCO3- + H2O -> H2CO3 + OH-}$
Since $\ce{H2CO3}$ is more stable than $\ce{CO3^2-}$ (because it has a lower charge and all the octets are completed), the p$K_{\mathrm a}$ must be high (more alkaline than acid).
In [1], the equilibrium will be shifted to the left and in [2] the equilibrium will be shifted to the right.
Is my line of thought correct?

No, it isn't. You can't meaningfully compare neutral molecules to ions in terms of stability. Also, bringing up completed octets doesn't help much, because all molecules and ions mentioned in your post share this feature.

Your second equation is unbalanced. You have added an extra hydroxide ion to the right-hand side.

Carbonic acid is not necessarily more stable than the carbonate ion.
Why carbonic acid ( p$\ce{Ka = 6.3}$ )  is not stable?
The reason for this is same as the one given for the instability of aluminium carbonate, iron (III) carbonate etc. That is, the carbonate ion has a highly polarisable electron cloud, which can be distorted by densely positively-charged species, such as the hydrogen ion, in this case. Thus, carbonic acid is not a very stable species and compared to the bicarbonate ion, which only has one densely-charged hydrogen ion distorting its electron cloud, it is definitely more unstable. Thus, the position of the 2nd equilibrium mentioned in the question should be to the left, favouring the more stable species.
Why is the bicarbonate ion ( p$\ce{Ka = 10.3}$ ) more stable than the carbonate ion  ?
This is due to there being a smaller negative charge on the bicarbonate ion and there being still 2 equivalent resonance structures to spread out the negative charge. Compared to the carbonate ion, which has twice the charge density but yet only 1 more resonance structure, the bicarbonate ion is more stable. Therefore, the position of equilibrium in the first reaction mentioned in the question should be to the left, not to the right.
Therefore, in solution, the dominant species among the three, should be the bicarbonate ion. This should be the correct line of thought.

"That is, the carbonate ion has a highly polarisable electron cloud, which can be distorted by densely positively-charged species, such as the hydrogen ion, in this case." I understood that , but i didnt get the realation between what you said and why H2CO3 is more unstable than HCO3-. For me you explained why CO3-2 is more unstable than HCO3-.Could you try explain to me again ? I aprreciate your anwser though. @Tan Yong Boon

@user158657 Your first query can be answered very logically. Since in bicarbonate, there is only one H+ distorting the electron cloud while in carbonic acid, there are two H+ producing a much greater amount of distortion, carbonic acid is definitely much less stable due to the electron cloud of its carbonate ion being extremely polarised.

@user158657 Regarding your second query, we would have to assess stability based on both charge density (i.e. charge of the ion) and charge distribution (i.e. number of resonance structures). Although the carbonate ion has 3 equivalent structures, compared to the the bicarbonate ion, which only has 2 equivalent ones, the charge density of the ion is twice that of the bicarbonate ion. Thus, even though it can delocalise the charge over one more oxygen atom, it does not compensate for the extra unit of negative charge. Thus, the carbonate ion is more unstable compared to bicarbonate.

It makes no sense to ask why a compound has a specific value of pka. Values are meant to compare things together especially in terms of pka.
When you claim that one product is more stable, you need to mention in which condition otherwise we cannot make any prediction.
Furthermore, when talking about carbonates, you need to consider not only pKa of solution but pressure and proportion of carbon dioxide in gas phase. With no pressure applied, the amount of carbonate in solution if very low. Sodium bicarbonate is not as soluble as one might think it is.
The reason for it comes to what I assume is your question, which I will rephrase as such: I expect that second acidic proton to behave like a carboxylic acid because of it's structure but it appears to be much more basic than anticipated.
The answer to that question is to reverse it, why does the carboxylic acid in the carbonate at a pka of 3.6 (which is ballpark carboxylic acid). The answer to that question is obviously resonance structure of the carboxylic acid that stabilises the negative charge. Problem is that this system of resonance cannot stabilise two charges.
Then what remains of the other OH in the carbonic acid, what is it? Well it's an OH linked to an SP2 carbon. Another OH on an sp2 carbon is phenol, and funny enough phenol pka is 9.95.
Evan pKa table is the perfect tool to answer most structural pKa questions, work by comparing species together.

In fact, the $\mathrm{pK_a}$ values themselves are telling you that the most stable species changes with pH, and these values are boundary values.


text:
Why does whipped cream use nitrous oxide instead of nitrogen gas?
It seems that nitrous oxide $(\ce{N2O})$ is frequently used to create whipped cream. But why can't just regular nitrogen gas $(\ce{N2})$ be used instead?

Probably because of the much higher boiling point, i.e. you need to apply less pressure to store it. But that's just a wild guess that sprung to mind.

Whipping cream  is a surface tension issue so there might also be that oxides work better, @Martin.

If you make the whipped cream yourself, air and a whisk works fine. It gets more complicated if you want to create a spray bottle.

There are two ways to efficiently make an aerosol product:

Use a gas that liquifies under the pressure inside the can. For example, butane lighters. Nitrogen is one of the "fixed gases", meaning it's a gas under most conditions (but take a look at the temperatures and pressures needed for liquid nitrogen—it's not going to ever be found in consumer products).

or

Use a gas that is highly soluble in the liquid (carrier) and that will "substantially" vaporize when the higher pressure inside the can is reduced to atmospheric.

The US government restricts the pressures that can be used in aerosol cans (and requires 100% quality control testing—when's the last time you heard of an aerosol can exploding? [although it does happen]). If you cut up a can, you'll notice that it's pretty flimsy. The higher the pressure (and a gas that has been dissolved in a liquid doesn't exert much pressure), the more expensive it will be to build the container (aerosol can).
Thus, the fixed gases are almost never used, except in some medical products. Why? Because they just aren't soluble enough to help move the liquid (or, in other cases, solid) out of the can and also to disperse it into a very fine mist. The customer wants basically one thing when using an aerosol: uniform, consistent spray from first to last drop. Using a very soluble gas helps, and using one with a boiling point near room temperature also helps. But the laws of thermodynamics say that the temperature of the can will drop as you spray out its contents. This may dramatically interfere with the liquid-to-gas phase change, while solubility is less sensitive to temperature. The trick is to make a product (and I've made some) that sprays out consistently and also doesn't leave so much left behind in the can that the customer feels ripped off.

According to Nitrous oxide, carbon dioxide, and their mixtures as propellants, Proc. Chem. Specialties Mtrs. Assoc., June 1950, page 45, William Strobach, $\ce{CO2}$ and $\ce{N2O}$ are both suitable because they each have some solubility in both the aqueous and oil phase of the cream emulsion.  $85\%~\ce{N2O}/15\%~\ce{CO2}$ has the most preferable taste, with too much $\ce{CO2}$ being too acidic. Other references say that the $\ce{N2O}$ provides a sweet taste, which is partially offset by the sour taste of the $\ce{CO2}.$
Freon- C318 Propellent for Aerosol Food Products, F. W. Blodgett and R. C. Webster, in a 1958 issue of the same journal, reports that adding perfluorocyclobutane to $\ce{N2O}$ resulted in a firmer, longer lasting cream.

Ahh, so just plain nitrogen gas (N2) would be "too inert" for that purpose?

I don't think it would be soluble enough.  It seems to be solubility that is desirable.

Well, what I meant was, is the nitrous more soluble possibly because it may somewhat react more with the cream?

@ManRow I don't think "react" is the right word, maybe "interact".  There is no chemical reaction.

Well, if not react (even if ever so slightly!), I wonder what physical reasons may explain nitrous's "greater interaction" with the constituents of cream as opposed to plain nitrogen gas (granted, both are probably very inert compounds anyway!). Also, Henry's Law data indicates much greater solubility for nitrous oxide than nitrogen as well (though, I'm not sure if that would be relevant here!)

Ahh, key factor I seem to have missed -- nitrous oxide is probably a lot more "polar" than N2! And that then probably explains the differences in their respective (even somewhat inert) chemistries as well!

Solubility of CO2 in water is enhanced by CO2 being converted to H2CO3 (carbonic acid) that drives the equilibrium towards shoving more CO2 into the water. Probably something similar for N2O?

It would be cool to have the real citations. Google books results are dependent on the country where you access them. I can't access any of the links, and a human-readable citation would help finding other means.

@Martin-マーチン I edited to be specific about the titles and authors of the articles.  The is the journal http://www.worldcat.org/title/proceedings-of-the-annual-meeting-chemical-specialties-manufacturers-association/oclc/3589813 Relatively obscure.

The critical temperature of nitrous oxide is 36 Celsius.  That means under pressure and shaken well it is distributed thruout the cream as a dissolved or nanodroplet liquid. As the pressure is released microbubbles form thruout the cream forcing a foam emulsion out. Similar propellants are iso and neopentane that happen to be flammable, not a good idea about birthday candles or food; I think isopentane is used in shaving cream. Aerosol cans are again a chemical art form to maintain stability, homogeneity and a pleasing product. An alternative to N2O might be CO2 but that that forms dry ice crystals upon releasing the pressure. N2O has a lower FP and probably tastes better.

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


text:
How are phosphoglycerides different from phospholipids?
(Currently studying high-school Biochemistry)
I'm unable to distinguish between phospholipids and phosphoglycerides from their "definitions" (Courtesy: Wikipedia)
On phospholipids:

Which suggests that phospholipids necessarily contain a glycerol "bridge" that connects the fatty acid tail to the phosphate head. To me, this looks like,

On phosphoglycerides:

"Glycerol-based phospholipids"?! But aren't phospholipids already 'glycerol-based'?
If both phospholipids are glycerol-based, and if phosphoproteins are "glycerol-based phospholipids"...how are they any different?
What am I missing here?

https://en.wikipedia.org/wiki/Sphingosine

Phospholipids of a lipid bilayer can be divided in to two groups:

Those derived from glycerol -  phosphoglycerides.
Those derived from sphingosine -  sphingomyelin.

Phosphoglycerides are indeed a type of phospholipid.


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Functionalizing CTAC stabilized nanoparticles
I am working on functionalizing silver-gold bimetallic nanoparticles. My nanoparticles are stabilized by CTAC(cetyltrimethylammonium chloride).The most articles which I found, are about the functionalizing of citrate stabilized nanoparticles.
Is functionalizing CTAC stabilized nanoparticles more difficult than citrate stabilized nanoparticles?


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Deviation in bond angles (from standard $109°28'$ ) in hydrides of group 15
Here are the approximate bond angles for first three members of group15.
$\ce{NH3}$: $107°$
$\ce{PH3}$: $94°$
$\ce{AsH3}$: $92°$
As we move down the group the bond angles approach towards $90°$. I don't understand why there is such a major deviation from the standard angle of $109°28'$.
In the case of $\ce{NH3}$, the deviation is justified because of the presence of lone pair(lp-bp repulsions) , but for others, this much deviation is quite unexpected.
What causes this large deviation in angle?

In the same direction ($\ce{N -> P -> As}$), not only the atoms include more nucleons and electrons (increase of mass), but the atomic radii increase, too ($65, 100, \pu{115 pm}$) which contributes to the reason. (consulted source: https://en.wikipedia.org/wiki/Atomic_radii_of_the_elements_(data_page))

Also incorrect is the assumption that VSEPR works in general.


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Buffer (buffer capacity)
I know that there already exist multiple questions concerning buffers and buffer capacities, but I would nevertheless like to clarify two things with your help:

Adding to a buffer $n$ moles of an acid of negligible volume or a solution of the same acid (also containing $n$ moles of the acid) both produce the same change in $pH$. Correct?
However there is a difference between the two, as the latter diminishes the buffer's capacity. How can I observe, from a mathematical standpoint, the decrease in the buffer capacity for the second approach; more precisely:
Given,say an acidic buffer solution of volume $V_i$ with $HA$ being the weak acid and $A^-$, its conjugate base (with concentrations $c_a, c_b$ respectively), how does the buffer capacity change upon the addition of a solution of volume $V_s$, containing $n$ moles of a strong base?

Any help would be appreciated.

Both scenarios decrease buffer capacity if you're consuming the conjugate base. The dilution effect might be important but it depends the concentration of your buffer (if it's dilute, then higher order effects will matter).


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Why is the notation for reactions not explicit?
I hope that the title of the question is not completely wrong or misleading, however I couldn't think up of a better one.
I have a follow up question to this one. If we look at the following reaction, $$\ce{C + O2 -> CO2}$$
we are taught that gases like oxygen (looking at the reactants), when not part of a compound, always appear in molecular form.
But if to follow one of the answers in the referred question, the reaction has a little more to it. I quote:

carbon, as normally found in nature as graphite, consists of a molecular solid with every carbon bonded to several other carbons. The reaction C + O2 is shorthand for solid carbon plus oxygen gas, so the reaction involves breaking lots of C-C bonds and O=O bonds resulting in CO2. You have to account for all the bonds broken and formed to know the energy difference.

Why is this fact not accounted for in the reaction $\ce{C + O2 -> CO2}$, in other words, why do we 'allow' carbon to be a free atom (although it's not, in the form of graphite) and 'demand' that the oxygen is in a molecular form? I do assume that chemical reactions are a form of abstraction, that is, we say that regardless of where that carbon atom comes from, it will react with the oxygen molecule into a carbon dioxide, but I still don't understand why we then apply those molecule constraints to oxygen.
To be more explicit about it: the energy spent to break the C-C bonds in the graphite crystal is not easily readable from the reaction, while the fact that the O=O bonds are split, and therefore energy is input into the system, is.

You are right in that it's kinda illogical. Then again, what notation would you propose for covalent solids like graphite? $\rm C_\infty$?

Chemistry is complicated. You want to capture all of the nuances of a reaction with a few scribbles?

@Zhe I'm not really out to capture all of the nuances of a reaction with a few scribbles. But I do try to point out inconsistencies where I see them, and I think the 'handling' of the C and the O elements in this equation is inconsistent. Also, although when getting into the topic of energy, discussion usually moves away from chemistry into physics, I think (and my book seems to point it out whenever possible) that the flow of energy involved in a reaction is a very important part of it, which led me to wonder why this is not notated in a better way. I think this is a justified consideration.

@Zhe also, excuse me if I'm being somewhat direct here, there's not much I can learn from "subject x is complicated". That's a given. Every subject is complicated as you delve more into it. As someone with background in computer science, when I don't understand a problem, I try to break it down to simpler problems for which I can then fashion solutions and combine into a solution of the larger problem, which would be the resulting algorithm. I don't know how much that approach is applicable to chemistry, but I am trying to understand things by breaking them down to "why is X the way it is".

OK, but generations of chemists have settled on this notation as a compromise between what is feasible/simple and what is correct. I understand that you might not agree with that, but you shouldn't definitely go with it for now. Physics does that, too, since in general physical systems in real life are a lot more complicated than the simple model systems we study.

Actually, there's a lot to learn from "x is complicated." Science is about reducing complexity, but that doesn't mean the complexity will disappear. The interesting phenomena occur because of complexity, not in spite of it. Therefore, you should learn to be comfortable with and operate around complexity. We try to reduce it for understanding basics, but it is wrong to try to eliminate it.

@Zhe I meant that in the sense that it's not really pedagogical to tell someone that something is complicated, since more or less everything is complex. Tell them how, break it down into the individual parts that make it complex so that they go on and explore them individually, to understand that complexity. A human mind cannot grasp something complex in a single thought process, which leads to us reducing it to problems we can work with.

Comments are not for prolonged discussions, you can continue in chat

https://chemistry.stackexchange.com/questions/9627/c-o2-is-equal-to-c-o-how-is-that-possible

Your notation of equation can be as specific as you see it fit, therefore this is opinion based problem.

@Mithoron I beg to differ. Of course notation is in itself arbitrary; in maths you don't have to use x or whatever symbol to express something -- you could speak about pink elephants. The point is that within a system of symbols, there should be consistency, and that's what this question aims at.

The reaction, described in the form given in the original question, i.e.
$$\ce{C + O2 -> CO2}$$
simply conveys that each formula unit of carbon reacts with one formula unit of oxygen to yield one formula unit of carbon dioxyde. As such, for example, useful to balance how many grams of carbon may be burnt with $\pu{22.7 L}$ of oxygen.  As such, yes indeed, it is a limited representation.  In an analogy, it is like book keeping; (within reason) is does not matter how you paid at a garage (cash, debit or credit) to get your car back -- but to satisfy both sides, it has to be balanced.
This view contrasts with many organic reaction equations as reaction mechanisms which are intentionally drawn with further detail like this:

(source)
Here, the how is important, and consequently outlined with more attention.  As if you would need to explain an apprentice how the teeth of a gear have to be adjusted in face, pitch, flank, etc. to transmit the force from the engine to the wheels.


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polymer/resin/material that melts around 40 C
Does anybody know of a polymer/resin/material that melts around 40 C and has a high heat of melting, thanks

Welcome to ChemSE.  For future reference, aim for a more specific question, which in your case could be "what are the conditions you want to use the material, what are chemicals (?) the material would be exposed to". Or, if something like $\pu{50 ^\circ{}C}$ already is outside your tolerance / window of application.
Because of the sparse information the question (currently) provides, the answer barely may be more specific than this suggestion:  paraffin waxes.  Among the chemical suppliers, Sigma-Aldrich's online catalogue, for example, lists several of them:

(source).


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How to determine interaction of photon with a hydrocarbon Bixin
Bixin is an apocarotenoid found in annatto with 25 carbons with  carboxylic acids at each end of the chain. It is an insoluble orange red pigment.
I am interested to study photon interaction with this molecule. Please, give me some head start on what to read. It is a red pigment, therefore it absorbs photons below 670 nm or so. How can I determine if this molecule allows luminescence in any degree? Electronic transitions?  or just any refraction or scattering.

Wkikepedia has an article on this molecule, it is a derivative of a carotenoid. You should be able to find lots of data on these molecules , for example the pi-pi absorption transitions in these will be similar to that  in Bixin. Usually these compounds are not luminescent to any great extent as non-radiative processes dominate. Often the cause is due to the lowest lying transition being symmetry forbidden .

You should add this as an answer . It was very helpful.

In a first step I would widen the available search criteria.  While accessing the English entry of wikipedia, collection of the chemical name, CAS number, smiles code (copy/paste), pubchem and chemspider ID are presented.  These identifiers may become helpful in the subsequent searches in other databases.  Nor would I hesitate (as long as my web browser allows proper display .and. there are at least some words barely understood in reading) to access the other versions (Spanish, Portuguese, ...), too.
In the public libraries of US' NIH, just the keyword "bixin" yields 150 (pmc) or 116 (pubmed) hits, that after some narrowing yield entries like this abstract in Spectrochimica Acta or this about triplet states of bixin, published in J. Agric. Food Chem.
While the British chemspider does not yield spectra directly, the literature references listed by this database include even surprising finds like doi 10.1039/C4CS00309H (2015ChemSocRev3244) about Vegetable-based dye-sensitized solar cells.
Do not omit collections like researchgate, as here, or figshare. If your institute has subscriber access, chemical databases like Reaxys (Elsevier) or Scifinder Scholar (ACS) may indicate primary literature including spectroscopic data, too.  No longer chemistry/physics centred, a search in Thomson Reuters' Web of Science, and more engineering-oriented Scopus (by Elsevier) may round the picture.


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Oxidation - Reduction Ionic Half Reaction
Trying to figure out how to balance the following molecular equation:
$$\ce{KMnO4 + Na2C2O4 + H2SO4 -> MnSO4 + CO2 + Na2SO4 + H2O}$$
I have compiled all of their oxidation states
$$\ce{KMnO4 + Na2C2O4 + H2SO4 -> MnSO4 + CO2 + Na2SO4 + H2O}$$
$$\ce{(+1)(+7)(-2 * 4) (+1 * 2)(+3 * 2)(-2 * 4) -> (+2)(-6)(-2 * 4) + (+4)(-2 * 2) + (+1 * 2)(+6)(-2 * 4) + (+1 * 2)(-2)}$$
Now I know that theoretically Mn changes from +7 to +2, so it gains 5 electrons. I also know Carbon in this situation goes from a state of +3 to +4, so it loses one electron.
However, after this I am drawing blanks on how to setup my ionic half reactions with these polyatomic ions, and I am suspicious that I may have set this up incorrectly. Any help at all would be appreciated, I can not find much to explain this type of equation with this particular molecular formula.
The idea of Redox reactions make sense when talking about individual atoms or more clearly depicted changes, but I am having trouble figuring out what is even changing into what in this example.

Welcome to ChemSE.  Be invited to familiarize yourself with formatting this site allows (especially when dealing with math, phys, chem), briefly outlined here: https://chemistry.meta.stackexchange.com/questions/86/how-can-i-format-math-chemistry-expressions-here/88 Secondly, both reaction equations are not yet balanced (two atoms of carbon on the left, but only one on the right; potassium vanishes entirely).  Hence let me suggest you start formulating the balanced (atom- and charge-wise) half reactions, e.g. $\ce{MnO4^-}$ is reduced to $\ce{Mn^{2+}}$. Then put the equations together.

@Buttonwood I noticed the equation seemed a bit off with the Potassium vanishing and the lack of atomic balance, but I thought I couldn't derive the half reactions without the equation first being balanced.

Credits to my awesome teacher for this method.
(I had to add a potassium ion to the right hand side, there was no other way)

Please typeset this.

@ToddMinehardt Can't, it's a pain

@ToddMinehardt So you downvote my answer for this, huh?

Nope - that's not me. You can confirm that I'm telling the truth by looking at my profile, under Activity and All Actions. Here's a link. Before writing this comment, the last thing I did 23 hours ago was make comment above about typesetting.


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Which of the following Lewis Structures is NOT correct?
Which of the following Lewis Structures is NOT correct?

Homework policy. This site does not exist to do your homework. However, if there is any specific concept that you are struggling with, we would be glad to help you.

I'm voting to close this question as off-topic because no effort has been shown towards solving this question from an assignment.

The first one, Fluorine does not have a full octet, is electron deficient and would have a charge. Also, the Nitrogen has too many electrons in its valence. And by the way, this is not a site to post homework questions.


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How to calculate mass produced when a given gas is burned?

The equation below represents combustion of methane ($\ce{CH4}$, $\pu{16.04 g/mol}$). Balance the equation, and calculate the mass of water ($\pu{18.02 g/mol}$) formed when $\pu{40.0g}$ of methane is burned.
  $$\ce{CH4(g) + O2(g) -> CO2 + H2O(g)}$$

I have started the problem by balancing:
$$\ce{1CH4(g) + 2O2(g) -> 1CO2 + 2H2O}$$
I then continued by calculating the amount of substance of methane:
$$\ce{C ($\pu{12g/mol}$) + H ($\pu{4 \times 1g/mol}$) = $\pu{16 g/mol}$ \implies ($\pu{40 g}$ methane) / ($\pu{16 g/mol}$) = $\pu{2.5 mol}$}$$
From here I wold like to know how to convert or get to amount of substance of $\ce{H2O}$ to get the mass of $\ce{H2O}$ produced in this reaction.

Well, how many moles of water have been produced?

Welcome to SE! feel free to take a [tour] of this site.

You have already balanced the equation, so we can get the stoichiometric  coefficients. From the equation we can say that 1 mole $\ce{CH4}$ gives 2 moles of $\ce{H2O}$ because their coefficients are 1 and 2, correspondigly. So 2.5 moles of $\ce{CH4}$ would give 5 moles of $\ce{H2O}$, that is $\pu{90 g}$ of $\ce{H2O}$.


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Corrosive? Aluminium Oxide?
I moved into a house that has a light fitting that has been made out of an old food 'tin' can.
The light hadn't worked since moving in. On trying to remove it the glass of the bulb pulled straight out into my hand. There is a white powder on the bulb and also the wire towards the bulb.
It also looks in a structure like mould or mildew, as the 'white' is also up the wire and on the beam that the light lead is hanging from.
What is the powder? Is it dangerous, as in corrosive to other materials as it fell onto the floor and myself? Or is it possible that it is more a mould?

IMO this might be tungsten(VI) oxide, $\ce{WO3}$, which is fairly non-toxic, unless you consume it spoonwise.
From the look of this DIY setup I would doubt the quality of the light bulb itself was superb, and most likely one wouldn't use a fancy lamp anyway. So, the glass enclosure could be breached and some point, and the inner inert atmosphere (Argon/Nitrogen) was substituted with air. When hot tungsten filament oxidizes, it produces tungsten oxides which evaporates/scatters from the filament as white smoke:
$$\ce{2W + 3O2 -> 2WO3}$$
Actually, bulky $\ce{WO3}$ is yellow, but being dispersed in a form of tiny crystals, tugsten(IV) oxide has pale-yellow or white color depending on the fraction.
Anyway, I would get rid of this can enclosure and buy a decent light source, mainly from an aesthetic point of view.
P. S. Here is a funky experiment showing what could happen in your case: https://www.thenakedscientists.com/get-naked/experiments/light-bulbs-liquid-nitrogen

If it is tungsten(VI) oxide, WO3, and fairly non-toxic. Are there any corrosive properties or further damage that could ensue from dispersed crystals? Much shook loose freely whilst getting rid of the light set-up.

Well, $\ce{WO3}$ is considered to be mild irritant, for more info you can check MSDS (e.g. here: http://www.sciencelab.com/msds.php?msdsId=9925359). I think if you don't try to eat or inhale on purpose stuff from that can you are gonna be fine. Damaged fluorescent lamp would be a totally different story due to potential mercury contamination, but conventional filament lamps are pretty much safe from this point of view.

I must add though that I don't have medical degree, therefore I can only speak for what I know is applicable to an average healthy person. If you have specific allergies, or any other pathologies, or unsure how this could influence your well-being, please contact your doctor directly. When it comes to health issues, internet might not be the best place to stumble upon.

@andyselisk Correct. I was not so much questioning harm to health, but the possibility of damage to other materials crystals have come in contact with (fabric, nylon etc).

I wouldn't expect more damage from tungsten oxide than from mud or dust.


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Curious as to what CorA1 is and what it does?
First time Chemistry SE poster,
Was wondering what actually CorA1 is and what it actually does?
I ask because I found out I was allergic to a incredibly long list of items and this popped up to be a high allergen as I work through the list.
Google defines it as a: Magnesium transport protein CorA and it contributes to Ion transport?

Welcome to Chem SE. Feel free to take a [tour] of this site, so you'll have a better idea of who we are and how we run.

I'm voting to close this question as off-topic (even though it has an accepted answer) because it's about the biological function of an enzyme, and is thus better asked at Biology.SE.

You see, CorA and CorA1 are two different things. CorA is a protein, while CorA1 is the gene that codes for that protein.
It is the genes that when translated give rise to the protein.
This gene was isolated from the Rhizobium meliloti, a bacterium commonly known for forming leguminous nodules in certain plants.
As for it's function, I'm not really a biologist, but I know that magnesium is a component of chlorophyll (It's coordinated at the center of the porphyrin complexes).

So you could say magnesium is rather important for plants. The $\ce{Mg^2+}$ ion needs to be transported from the soil to the leaves, and this protein facilitates this transport.

@RushRed Most welcome!


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How exactly do we superimpose molecules while checking optical activity?
I am struggling with these type of questions where in some cases we rotate a molecule's mirror image by 180 degrees to check superimposability.
Like when checking any tetrahedral chiral compound, we draw its mirror image and rotate it by 180 degrees and say that it's not superimposable.
Suppose the mirror image of the left-handed version is the right-handed version, hence both are not superimposable. If we rotate the right-handed version by 180 degrees and then try to check, will it be superimposable now? Similarly, trans 1,2 dimethyl cyclopentane is chiral but if we rotate its mirror image by 180, is it superimposable?

There is probably some misinterpretation or error.. if you talk about deciding whether a compound with chiral center(s) is optically active, you take the two stereo isomers, an just try to superimpose them (to check if they in fact are the same), NOT mirroring one of them beforehand. They already are mirror images. These might OR might not be the same. Hands are not. E.g. eyes are (in the first simplified approach, when ignoring anatomical details).

You're being confused by the rotation. You take the mirror image. Any rotation or sequence of rotations that superimpose the mirror image over the original means that the molecule is achiral. It's not just a 180-degree rotation.

Nice Explanation, that was my confusion too. Many thanks

But I have one question: Does it also apply on enantiomers ? Please take a look here: https://youtu.be/PoeNd6YhMuA?t=93

@VKatz Does what apply? Does this statement apply to enantiomers? Sure, it does. Enantiomers are not super impossible on their mirror image, no matter how you rotate the mirror image.

That is what my confusion is. In the link: https://youtu.be/PoeNd6YhMuA?t=103 If I rotate that the compound which is, on right has side will be superimposable. And the video that is saying is totally opposite. Can you please clarify .

@VKatz That is not a rotation. Remember that a projection on a flat surface is a representation of a 3 dimensional structure. You can "flip" a drawing of a person's face, but does that look exactly like the back of their head?

Okay, So What I understood is from the link: https://www.youtube.com/watch?v=PoeNd6YhMuA&feature=youtu.be&t=103 1: Both structures including Left and Right are 3D representation for a molecule. 2: We should try superimposable on 3D structure, not 2D. Is that correct? Is there anything which I should aware of? Can you share some resource if possible?

There are several options. You can build the structures and compare. Or you can identify the stereochemistry at each center and compare. The latter requires that you understand how to identify the stereocenter at a center.


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Why are metalloids found in a zig-zag line on the periodic table
I am confused about the zig-zag pattern of metalloids in the periodic table. Why are metalloids arranged in a zig-zag line? Can anyone answer my question?

They're just sometimes put between metals and non-metals which are in the corner.

In addition to Pritt Balagopal's answer, in which he has clarified the trends in metallic character across the Periodic Table, I would also like to explain why the metalloids occur in a zig-zag fashion.
Firstly, I would like to again clarify that it is possibly a coincidence that the metalloids start from boron, not carbon or any other non-metal in that period. My guess could be the start of electron occupation of the p subshells. Note that boron has one electron in its p subshell. Thus, its chemical properties start to change to become more non-metallic in nature.
Diagonal relationships is a common occurrence in the Periodic Table. However, it is often used to describe the relationship between two elements which are diagonal to each other, not a whole line of them. Thus, I would like to clarify again that this is just my conjecture, not what is the true reason. But I believe this should be part of the correct explanation as it is logically sound.

This site offers a more in-depth explanation as to why diagonal relationships occur:
https://chem.libretexts.org/LibreTexts/University_of_Denver/Chem_2132%3A_Chemistry_of_the_Elements/Unit_3%3A_Descriptive_Chemistry/09%3A_Building_a_Network_of_Ideas_to_Make_Sense_of_the_Periodic_Table/9.3%3A_The_Diagonal_Effect
In short, the diagonal relationships are based on two effects: the increasing effective nuclear charge across a period and the decreasing effective nuclear charge down a group. Effective nuclear charge increases across a period because the number of shielding electrons in the inner shells is constant while there are additional protons in the nucleus of the atoms of the elements across the period. Effective nuclear charge decreases down the group as there is an addition shell of shielding electrons from one element to the next in the group. (These are similar to the trends in metallic character pointed out in Pritt Balagopal's answer.)
These two competing effects cancel each other out and thus, chemists believe that the chemical properties of elements in a diagonal should be similar. The image shows some examples.

Well, the answer is quite simple:
Elements on the periodic table tend to follow this trend:

Metallic character increases as you move down a group: This is because every step you move down the group, an extra shell of electrons is added, along with a bunch of protons that increase nuclear attraction. The extra shell factor plays the dominant role, making the outermost electron easier to remove, thus attributing to increased metallic character.
Metallic character decreases as you move across the period along the right: As you move along the period, one extra electron is added to the outermost shell, and a proton is added to the nucleus. Since the extra electron doesn't considerably change the location of the shell, the extra proton brings the shell considerably closer to the nucleus, making the outer electrons more tightly bound, decreasing the metallic nature.

Keep this in mind.

Start with boron, you see that all thats below it are metals. Carbon is a non-metal, Silicon and Germanium are metalloids, and the rest below are metals. I think now you can see the relationship here. Going right from a metalloid gives you a non metal. To find another metalloid, you gotta look down. What you get in the end is a diagonal cluster of metalloids.

This answer would be improved with some explanation of why metallic character behaves as it does - that is, why does metallic character increase as you move down a group, and why does it decrease as you move across a period to the right.

You are not explaining the trend. You merely describing the trend.

It's done @R.M.

What I meant by my previous comment is that what you are explaining is the trend in metallic character across the Periodic Table and not the diagonal trend in the metalloids. Therefore, the explanation you have provided explains why do metalloids exist and why they are found at that position in the table but not why they follow a diagonal trend.

@TanYongBoon read my last paragraph.

Your last paragraph describes the trend. You do not seem to be explaining, using chemical reasons, why the metalloids occur in a zig-zag fashion. I would think an answer to this question should also feature "diagonal relations in the Periodic Table".

Perhaps, the link wasn't made clear enough. But now that you've pointed it out, it does make sense. I didn't ask for chemical reactions, but chemical reasons. I guess the reasons you provided are sufficient, after clarification.

But if you think about it, what is explained is why it occurs in  a diagonal, not zig-zag fashion. Note that zig-zag means to include the element which is below one element and to the left of the other. I seemed to have not realised this until just recently. So I have missed this out in my answer as well.