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Is 1,2-dihydronaphthalene "partially aromatic"?
I've learned that if any one ring of a structure like the below is aromatic, the whole compound is aromatic.

But one of my friends suggested that such compounds cannot be aromatic and gave the following resonance structures as the reason:

I feel that the non-aromatic/anti-aromatic resonance structures wouldn't make significant contribution. Is my reasoning correct?

Antiaromatic species are unstable. No molecule would attempt to take up an antiaromatic contributor.

I edited your post to make the picture more readable. I used Web Chemdoodle for this purpose. See my guide on how to format chemical compounds and mechanisms. I used Microsoft Paint to write "Aromatic" and "2e" and stuff.

I think I address this in the answer to the very similar (possibly duplicate) question: Is 1,2-dihydronaphthalene aromatic?

Aromaticity is a quite difficult concept but in this case we can answer the question by simply looking at similar compounds and experimental data.
What you got here is a styrol derivative, the pi-system (which we need to look at if we are talking about aromaticity) is the same. So if this compound is not aromatic then styrol shouldn't be either.
But styrol is in fact aromatic (or better: the benzene ring in styrol is). It reacts like an aromatic compound should, it got the right shift in NMR, the bond lengths are in accordance with aromaticity and so is the UV/VIS spectrum.
So a conjugated double bond clearly doesn't make a benzene (significantly) less aromatic.

According to Huckel Rule the compound is aromatic only if the conjugation is spreaded all over the structure and the compound is flat or nearly so, so that the sideways overlapping of pi orbitals can take place for conjugation and there should be conjugation of (4n+2)pi electrons on the whole compound and as can be seen from your provided structure the second(right hand sided) ring does have conjugation on the whole structure.

So, is the compound aromatic or not?

I don't think so!

The way you describe the rule implies that substituted benzenes would not be aromatic as electrons are not delocalised over the substituents. That's just wrong. This compound contains an aromatic ring.

This answer is plain wrong.

As the question asked the whole compound is not aromatic but can be said because of the presence of the benzene ring. I actually know this question which is from the IIT-JEE examination paper!

@VaibhavDixit if a chemist says that a compound is aromatic it means that there's at least one aromatic structure in it.

OK sir! Thanks for correction. I appreciate that!

@VaibhavDixit You realise this answer doesn't comply with the official key.

For a compound to be aromatic or anti aromatic there should be conjugation throughout the peripheral , this compound lacks conjugation throughout the peripheral so it falls under non-aromatic category.

Huckel baba fails many time for prediction of aromaticity of a compound, and the chart for checking if the compound is aromatic or not goes like this: 1- we check if compound is cyclic and presence of conjugation , 2- planarity of molecule,3- 4n+2 pi electrons .There are many examples in which some of double which can conjugate with the ring doesn't conjugate and resides localized on two carbons just to maintain aromaticity  like Pyrene ,in this exmple also the double bond should be localised on those two cabon atoms and not participating in resonance and then only  it only it can be aromatic

@KushalTripathi I don't know which "chart" you are speaking of but you should only count the pi electrons in the conjugated ring system, not every pi-system in the molecule and also not pi-systems conjugated to the cyclic one. If we do that here we got a 6-membered ring with presence of conjugation and 6 pi-electrons, which fulfills the hückel rule. And for the second part: I don't nderstand what you mean at all, that's wild talks. And also Pyrene is completly aromatic.

@DSVA yes , Pyrene is completely aromatic . I was relating this molecule to the Pyrene because in pyrene also the total no of pi electrons are 4n and not 4n+2 . But the pi bond at the center of pyrene is not considered to be conjugating with other pi bonds of molecule, so we do not count that pi electron and that's why we say it has 4n+2 pi electrons, simply in this example double bond in the non benzenoid ring should not be disturbing the conjugation of the benzene ring and electrons of benzene are only counted. It is an aromatic compound ,my answer is wrong and I was explaining this only

@KushalTripathi the hückel rule doesn't work for fused polycyclic compounds, in fact it does only work correcly for monocyclic systems. The middle bond in Pyrene is in conjugation with the others...

@DVSA It may be wrong whether the middle pi bond conjugate with rest of the double bonds, but to the best of my knowledge that bond is not considered as pi bond and this is why we say that no. of non pi electrons in pyrene is 2. You can confirm this from the fact that reaction of pyrene with Br2/CCl4 gives the addition reaction at the middle pi bond.


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Extimating the repartition of water between hygroscopic compounds
Sulfuric acid is sometimes used as a drying agent.
According to literature, one possible application is, for instance, in the drying of bromine.
Some questions came to my mind: which concentration of sulfuric acid would be suitable for such a task? Under which concentration would the acid stop absorbing water from bromine, and there would then be an "equilibrium" between the water bound to bromine and that bound to sulfuric acid?
To put the problem in a general and less specific way: is there a way to estimate the capability of a compound to "catch" water from another compound, and in case of liquid compounds or of hemihydrated molecules, to estimate the maximum amount of water present before an equal repartition of water between the two substances occurs?
I was thinking about the enthalpies of hydration, but I honestly didn't come up with an idea for using them in such calculations.

It's characterised by equilibrium water vapor pressure over agent.

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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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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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Terms related to the structure of atom
In my chemistry textbook, the chapter "Structure of Atom" deals with the old Bohr's model and the new Quantum mechanical model. But when I read the contents of the new model of atom, there were terms like energy levels, shells being repeated. I failed to understand the context. Moreover, Bohr's model has been discarded so why do we still talk about energy levels/ orbits in an atom?
Basically, I want a satisfactory explanation of these terms: energy levels, orbits, sub orbits, shells, sub-shells, orbitals and principal quantum number in accordance with the Quantum mechanical model of the atom.
I tried to understand this from two other books but there too the distinction was vague.
I also read the contents of:
Is there a difference between energy levels and electron shells?
https://www.khanacademy.org/science/chemistry/electronic-structure-of-atoms#orbitals-and-electrons
PS: Please provide an answer keeping in mind that the questioner is a Grade 11 student.

There are no orbits, suborbits, and subshells in QM.

What is the problem with the common QM definition of shell, subshell etc?

So, does the book not provide definitions of those terms? Or are you confused about those terms?

https://chemistry.stackexchange.com/questions/32860/quantum-mechanical-model-of-atom-and-the-quantum-numbers https://chemistry.stackexchange.com/questions/18466/difference-between-shells-subshells-and-orbitals

@Greg I can't tell if that comment is directed at my comment or the OP.

Possible duplicate of Difference between shells, subshells and orbitals

@JonCuster both


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How to deal with solving reactions with not enough background
The book I'm studying with often confronts me with exercises that involve solving reactions such as $\ce{CA(OH)_2 + CO_2 -> ?}$.
Now, most of the time, the questions are just repetitions of material that has been taught, that reaction in particular has been shown many times in the chapter before the questions, $\ce{... -> CaCO_3 + H_2O}$, but not explained.
When I look up such reactions on the Internet to try to understand, I mostly come across people who ask others to balance the equations for them. But that is not really interesting for me, as I find balancing a trivial mathematical task.
I am, however, failing to understand why this reaction (or many other reactions) happen the way they do. My book has also mentioned that you can combine two materials, like $\ce{CO + H2}$, in different ratios, under different temperatures, pressures, and with different catalysts, to get different results (methane, methanol, and many others).
So my question is -- when asked to solve a reaction, and given only the reactants (or possibly also other factors such as mentioned above), is there a mathematical principle (for example, relying on differences in electronegativity) to solve the reaction besides being completely familiarised with molecular orbital theory and quantum mechanics? Because so far, the only alternative I can see is memorising the reactions.
I realise this question is rather broad, and possibly there is no satisfying answer to it or any answer besides to go study quantum mechanics, but to provide some context, I'm a student at high school studying chemistry not as my main subject (so on a rather elementary level), and I'm looking for a perspective how to understand the material I learn and not memorise it.

There are some general principles, so after a while you would begin to make your own generalizations and recognize patterns, but initially there is quite a lot to memorize. E.g. organic chemistry strongly relies on polarity, which is indeed related to electronegativity, but steric effects and much more complex quantum mechanics gets involved at some point, so much so that people often just have to run an experiment and see what happens. In inorganic chemistry, how would you guess that iron (2+) ions react with sulphide to give an insoluble iron sulphide precipitate? It's just experimental...

And in your example, why does calcium hydroxide react with carbon dioxide to give calcium carbonate? We can explain it a posteriori, but you could not safely generalize to similar systems. E.g. I am pretty sure that silicon dioxide and magnesium hydroxide would not react in the same way. So no, it's not like mathematics where an all-encompassing rule can be stated and applied universally: chemistry is still an experimental science. At least, that's the way I see it.

possible duplicate of https://chemistry.stackexchange.com/questions/50684/how-can-i-predict-if-a-reaction-will-occur-between-any-two-or-more-substances

Possible duplicate of How can I predict if a reaction will occur between any two (or more) substances?


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How does P4O10 make sense?
I'm trying to understand the gains phosphorus has from forming $\ce{P4O10}$. The molecular structure is such that every oxygen atom forms two bonds (either two with one phosphorus atom, or one with two), while every phosphorus atom forms five bonds.
I'm clear about the gains the oxygen has from this -- it fills its outer shell to the octet rule, mimicking neon effectively. But what about phosphorus? We're talking about a covalent bond, so forming five bonds would leave it with five extra electrons, which is two more than it needs to reach an optimal configuration.
I remember reading somewhere (possibly on this site) that ionic bonds can be regarded as an extreme form of covalent bonds, where the difference in electronegativity is so great, that the metal gives its extra electrons to the non-metal. Can we assume that the difference in this case is also large enough, such that phosphorus effectively 'gives' away five electrons, therefore reaching neon too? Or am I oversimplifying it?

The octet rule isn't universal, atoms don't have to satisfy it all the time.

sigh These are single dative bonds between P and O, it's just dumb archaic way of drawing compounds.

Related: https://physics.stackexchange.com/questions/308494/how-does-covalent-bonding-actually-work

"ionic bonds can be regarded as an extreme form of covalent bonds." They "can be regarded" as magic, too. What's your point? In what specific context is such regard useful? (I'm not questioning that there may be some specific context that you could regard them that way, rather I AM saying that it is not generally useful.)
 The octet rule is used to predict valence.  Obviously H doesn't obey it. It's quite useful for the chemistry of C,N, O, and F.  But consider ClF5. The Rule is not going to get you there. Nor is it useful, imho, for P4O10 --- but there are ways to bend the meme to "explain" the deviancy. see for example the discussion of hypervalent compounds in https://en.wikipedia.org/wiki/Octet_rule#Hypervalent_molecules. Once you get away from the 2nd & 3rd Period and the 13th - 17th Groups, it shouldn't be relied upon. So, its applicability is limited, but fortunately it's applicability is for a lot of elements which are interesting for us in our environment here on Earth. Since d-orbital (and f-) contributions to hybridizations is generally small, the octet rule is generally useful. What sort of confidence should you have in using it to predict valence? Well, there's two ways to answer this: 1. as applicable to a student learning basic chemistry or 2. as applicable to experts who have in addition to the rule a vast literature on the subject. It is useful for the first and nearly useless for the 2nd.


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What determines "polarizability" of an element?
In this question, the answer given discusses that fluorine has a good orbital overlap with hydrogen and also that fluorine is not very polarizable. However, is it always true that strong bonding is inversely related to polarizability?

Fluorine is not very polarizable because it is small. Its electrons, therefore, are all close together. A polarized atom has shoved all its electrons to one side. Since they are close together in Fluorine, the negative-negative repulsion is too large for that. In another halide such as Iodine, whose valence electrons are much farther from each other, much better polarization can be achieved.
More polarizability does NOT always mean better or weak bonding. Generally, polarizable atoms like to bond to each other, and non-polarizable atoms like to bond to other non-polarizable atoms. For more info, see:
http://www.adichemistry.com/inorganic/cochem/hsab/hard-soft-acid-base-theory.html

The polarisability is a measure of how an atom's electrons can be displaced from equilibrium by a nearby electric field.
The polarisability is approximately inversely proportional to the ionisation energy (taken across periods in the periodic table); the easier it is to ionise the easier it is to alter the average electron's positions because they are less energetically bound to the nucleus.
Forming a chemical bond is an all together different thing and depends on the balance of the electron's kinetic and potential energies and these are determined by electron-electron repulsion, electron-proton attraction and proton-proton repulsion energies. These energies are calculated using a quantum mechanical approach that involves Coulomb and related integrals.

It is not true that the strength of a bond is inversely proportional to the polarisability of the atoms involved in the bond.
Bond strength involves various factors, including the electronegativity difference between the atoms, the orbitals involved in bonding and electrostatic interactions. The list is non-exhaustive and I don't think all the factors can be easily summarised in a list even.
In the post you have linked in your question, "low polarisability" of the fluorine atom was discussed as a reason for the strong hydrogen-fluorine bond. Now, let us make sense of that...
When an atom is not very polarisable, we are saying that the electron cloud of the atom cannot be easily distorted to give an uneven electron distribution, such that there are partial charges induced on different sides of the atom. Such atoms usually have high electron density. A good example often mentioned is the trend in polarisability in the halogens: Fluorine is the least polarisable while iodine is the most polarisable. This is due to the different sizes of the atom. Iodine having a larger and more diffuse electron cloud, allows for ease of electron movement within the electron cloud.
The effect of this is significant in bonding with hydrogen. When a halogen atom is bonded to hydrogen, there is evidently a partial positive charge on the hydrogen atom and a partial negative charge on the halogen atom. The polarisation of the iodine atom in hydrogen iodide is particularly significant due to its polarisable nature. The very uneven electron distribution of the iodine atom causes the hydrogen-iodine bond to be slightly unstable. On the contrary, since there is not so much of polarisation within the fluorine atom and thus, the hydrogen-fluorine bond is much more stable.
However, that is not the most important explanation for the bond strength to be as such. In fact, the "polarisability reason" as far less important, compared to the "orbital reason", which is that the fluorine atom has more defined orbitals used in bonding with hydrogen, compared to iodine. The interaction of more defined and electron-dense orbitals used in bonding gives rise to a greater accumulation of electron density in the bonding region. The attractive forces of the nuclei acting on the bonding electrons is thus stronger. Thus, there is a stronger bond.


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Atoms or molecules with spin 1 in the ground state?
Is there any atom or molecule that has spin 1 in its ground state?
Do Hund's rules keep this from happening for an atom?
The reason I'm curious is that it would be nice to have a spin-1 example for use in pedagogical discussions of the Stern-Gerlach experiment.
[EDIT] Clarification: when I say "spin," I mean the total angular momentum, not just the sum of the spin-1/2's. (The total angular momentum is what you are seeing in the Stern-Gerlach experiment.) I deleted the part of the question about ions, because, as pointed out by Orthocresol, they won't be usable in a normal Stern-Gerlach spectrometer.

Atoms
Atomic carbon with its $\mathrm{1s^2 2s^2 2p^2}$ configuration has a triplet ground state ($S = 1$), precisely because of Hund's first rule.
However, in the context of the Stern–Gerlach experiment, you might run into a problem with orbital angular momentum, as carbon's ground state also has nonzero orbital angular momentum ($^3\mathrm{P}$ ground state, $L = 1$). The behaviour in a magnetic field will be rather more complex and you probably need to take into account spin-orbit coupling.
At the moment I can't think of any atoms with a ground state term symbol of $\mathrm{^3S}$. I actually suspect that it's impossible, but I'm not really up to proving it right now.

Molecules
Triplet dioxygen (as Zhe mentioned) has no orbital angular momentum ($^3\Sigma_\mathrm{g}^-$ ground state, $\Lambda = 0$), but I'm not sure if the inhomogeneity of the electron density would have any impact. (as in, I'm genuinely not sure.)

Ions
The $\ce{^2H}$ nucleus (which us chemists usually refer to as $\ce{D+}$) is lightest stable nucleus that has a spin of 1.
The only issue is that you then have a moving charge in a magnetic field. If you could ignore it, or take it out of the equation, somehow...

The behaviour in a magnetic field will be rather more complex and you probably need to take into account spin-orbit coupling. I don't think the behavior of a neutral particle can be any more complicated than what it normally is in a Stern-Gerlach spectrometer. The Hamiltonian is $z\mu_z=zgJ_z$, and spin versus orbital angular momentum would just have an effect on $g$. I'm not sure if the inhomogeneity of the electron density would have any impact. For similar reasons, I don't think this matters.

In an S state (L=0) you only have room for 2 electrons, one with spin up, one down.

@BenCrowell It is slightly complicated (imo, at least!). The coupling of the angular momenta $S$ and $L$ gives rise to the overall angular momentum $J$, which takes values $0, 1, 2$ (Clebsch-Gordan series). Of the three resulting terms, Hund's third rule dictates that $\mathrm{^3P_0}$ is the ground state (which unfortunately isn't magnetic). $\mathrm{^3P_1}$ (the one we're probably interested in) and $\mathrm{^3P_2}$ are then $16$ and $43~\mathrm{cm^{-1}}$ above the ground state.

@orthocresol: You're talking about carbon, right? All I'm saying is that a Stern-Gerlach experiment only measures the component of the total magnetic moment along the field axis. There is no other, independent parameter that is available to measure. The angular momentum isn't independent of the magnetic moment. Since the first two excited states are so close in energy to the ground state, they would presumably be equally populated in an atomic beam. Therefore you would probably see a superposition of results corresponding to the three energy states, each with its own quantized magnetic moment.

@BenCrowell Yes, I agree entirely. By "complicated", what I meant was that if you performed a S–G experiment on a beam of carbon atoms, the result wouldn't be a clean separation into three different beams. If you consider carbon to be a good enough example, though, then I guess we're done here? I was hoping to find an example that doesn't have a mixture of multiple states, but I don't think it's forthcoming.

Triplet oxygen has two unpaired electrons with the same spin, and a total spin value of 1.
In fact, by Hund's rule, the triplet states are preferred over the singlet states which have two electrons with opposite spins.
Further reading:

https://en.wikipedia.org/wiki/Triplet_oxygen
https://en.wikipedia.org/wiki/Singlet_oxygen

Rubidium 87 is one candidate, if you take into account hyperfine splitting, the ground state (which is part of hyperfine manifold of 5$S_{1/2}$) has total angular momentum $\vec{F}=\vec{J}+\vec{I}$ momentum of $F=1$.

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


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Does a positive electron affinity actually correspond to the release of energy?
I read about the periodic properties — ionisation energy and electron affinity — in my book. It was quite intuitive that energy is required to remove an electron from the atom but I wonder why energy is released when an electron is added to an atom? Is it actually released or is it because of our convention that energy of electron at infinite distance from nucleus is 0, that we consider it to be released?

It may help to take a look at the two processes you are comparing in terms of reactions. Ionisation follows the following reaction:
$$\ce{A ->[][$\Delta E_\mathrm{i}$] A+ + e-}\tag{1}$$
This process involves tearing charges apart, separating a positive and a negative one. Even if the original atom was somewhat unstable, it will be unfavourable to remove that electron simply due to electrostatic interaction. You write in the question that you were able to deduce this already.
The electron affinity on the other hand describes the process in this following reaction:
$$\ce{A + e- ->[][EA] A-}\tag{2}$$
Here, a charged particle is approaching a neutral particle and we have the same number of charged particles (and the same charges) on both sides of the equation. This means that we don’t have a strong, obvious physical effect that will make all electron affinities sign-equal for very simple reasons. Instead, we need to consider the details more.
The process often involves a release of energy because most atomic configurations have space for extra electrons at little or no extra cost. Except for the noble gases and the second group, atoms don’t have completely populated sub-shells so there should always be a position available in a subshell at an acceptable exothermic energy level. While adding an electron will increase electron-electron repulsion and increase all energy levels by a certain amount, this increase is typically less than the first ionisation energy which corresponds to the energy level of the highest occupied atomic orbital. Thus in electron affinity, we in principle gain the entire ionisation enthalpy minus the destabilising contribution of the new electron. (Caution! This is a very simplified picture and not suitable for generalisation or even strictly for explanation. It gets the basic point across though.)
Therefore, most electron affinities are exothermic with the notable exceptions of group 2, group 18 and nitrogen.

Energy has to be released when electrons are added to atoms in most cases.
Otherwise, why would they ever form covalent bonds?
The whole point of a covalent bond is that electrons are stabilized by the attraction to multiple nuclei, which helps to hold the system together. Therefore, the system must be more stable when you add an electron to an atom.
Note that in the case of something like a neon atom, where adding an electron is not favorable is not the case, we don't form covalent bonds.
From the atomic perspective, even for a neutral atom, the nuclear charge is not completely shielded by all of the electrons. Some electrons are going to be on the other side of the nucleus from the new electron you're adding.

This is because potential energy is more in this case when electron is further from nucleus as the forces between them are attractive. As the distance is decreased between nucleus and electron, potential energy decreases. Since energy is always conserved, the net decrease in potential energy is released as any form of energy. However the energy is not released in all the cases. It depends on the net effect of the interactions of the electron with the nucleus and the electrons that are already present in the atom.


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Silicon organic chemistry
I've heard that in some alien planet there might be life based on Silicon chemistry, but have we ever discovered molecules that are similiar to carbon organic compounds? Something like chains or rings...
P.S. I'm not asking if silicon-based life is possible, i'm just asking if there actually are silicon organic compounds(in the sense that they are similiar to their carbon counterparts)

Yes there are chains and rings of silicon, though they are quite totally different from their carbon counterparts. All in all, the question is way too broad to be answered.

@Ivan The OP's asking if molecular silicon rings/chains have been synthesized (Yep). Though the first line in the post seems to be asking, "Is silicon-based life possible?", the OP doesn't explicitly request that. So...not all that broad ;)

Some silicon based molecules are similar to the carbon analogs. However, the full richness of organic chemistry that we observe as necessary for life on Earth is not expressed in silicon-based chemistry, so silicon-based life is extremely unlikely.

This should be well suited in worldbuilding.SE.


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Valence Bond theory, VSEPR theory and predicting the shapes of the molecules
I read about the above mentioned theories then used them to predict the molecular geometry. However, I found that the shape/geometry of a molecule obtained from both the theories was same. For example, water has bent/angular shape according to VSEPR theory as well as Valence Bond Theory. So I wonder if there's any difference between the two theories in predicting molecular geometry.
Do these two theories ever yield different results?

I'm not the downvoter. But, you were probably downvoted because this is not a very good question. Why would you expect the two theories to differ? What research have you done on your own to try to solve your own question?

Yes they do yield different results. What do you predict the shape of $\ce{PH3}$ to be?

@bon as they do yield different results, on what occasions we employ each separately?

VSEPR theory is a purely observational model, based on empirical knowledge. Many of the "rules" were added to fit the theory to what has been observed. It has quite a lot limitations and may only provide an initial guess for the structure of a molecule.
According to VSEPR, the structure of water is bent. While we would expect $\angle^\mathrm{theo}_\mathrm{bond}(\ce{HOH})=109.5^\circ$, it is indeed smaller $\angle^\mathrm{expt}_\mathrm{bond}(\ce{HOH})=104.5^\circ$. This discrepancy is explained with the reasoning that lone pairs require more space. This is wrong, and a caveat added to make the theory fit.
It would also predict the lone pairs of water to be equivalent. A statement which has experimentally been disproved. Quoting myself:

One popular example is water. In many texts the central oxygen is described as having (approximately) four sp3 orbitals, which essentially makes the lone pairs equivalent. That view, however, is not in agreement with the photoelectron spectrum, which clearly shows that the lone pairs are not equivalent. At this point I'd like to refer you to Michael Laing's article: "No rabbit ears on water. The structure of the water molecule: What should we tell the students?" (J. Chem. Educ. 1987, 64 (2), 124.).

The theory sometimes predicts correct structures, but that has no theoretical foundation. It is basically educated guessing.
When it comes to more complicated molecules, for example hyper-coordinated compounds like $\ce{IF7}$, most explanations (the involvement of d orbitals) break completely down and have been disproved many times. However, just by minimising the inter-ligand distances, it yield approximately the correct shape. This is a nice coincidence.
Valence Bond theory is on the other hand a highly sophisticated and complex approach, deeply rooted in theory. If applied at a modern, computational level, it will always yield the correct molecular geometry for the correct reasons (within its other approximations).
While we can approximately use a piece of paper and a pencil to do these calculations for simple molecules, it really is an approach at formulating a wave function and approximately solving the electronic Schrödinger equation. That requires a lot of maths and integral solving. It is not so much a theory that can be done on-the-go.
As it is such an extensive and accurate theory, it is complementary to molecular orbital theory. That means that they will yield the same results at their respective accurate (infinite) expansions.
For the example of water, VB theory would predict a bent structure. It would also predict non-equivalent lone-pairs, which agrees nicely with the photoelectron spectrum.
In conclusion; A sad fact is that both theories, VSEPR and VB, are usually (more than just often) taught completely wrong at below university levels (sometimes even there) and will introduce many artefacts into young scientists brains. Your statement, that you "used" valence bond theory is a reminder of that.
These theories could not be more different from each other. In most cases they predict structure on completely different grounds. It is basically a coincidence if they predict the same thing.
Take home message: VSEPR theory is a rule of thumb, VB theory is an accurate description of physics (within its approximations).

@Karsten I'm not sure I can agree with this edit. I really personally do not like to include these kind of images if I have not verified them myself. As they currently stand, without the necessary details of how they came about, I find them distracting. The left image is definitely incomplete, the right image is just trickery. In any case: images like these must be cited accordingly; a link hidden behind the word source is the worst possible way of attribution.

Not a problem, I will revert it. I liked the web site that had the interactive Jmol scenes for different molecules, from this collection: https://chem4.cns.uaf.edu/orbitals/index.htm


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How to rationalise the increasing bond length order in the carbonate ion, carbon monoxide, and carbon dioxide?
I am unable to rationalise the order of increasing bond length in $\ce{CO3^2-}$, $\ce{CO}$ and $\ce{CO2}$.
Having gone through the factors affecting bond length in two different books, my approach to the problem was:

$\ce{CO}$ has bond order 3 while carbonate and $\ce{CO2}$ have bond order 4 so
$\ce{CO}$ should have longest bond length.
$\ce{CO2}$ is sp hybridized while $\ce{CO3^2-}$ is sp2 hybridized.
Since carbonate has lesser s character and therefore it should have a greater
bond length than $\ce{CO2}$

Thus,
  $$\ce{CO2  <  CO3^2- < CO}$$
However, I am unsure about my method. What would be the best approach to determine the relative bond length order?

Actually, in regard to bond length, you should be looking at the bond order of a single bond within the molecule. The bond order is 2, not 4, for $\ce{CO2}$, and the bond order for the carbonate ion is somewhere between 1 and 2 due to resonance. Thus, the carbonate ion has the longest bond length, followed by carbon dioxide, and finally carbon monoxide.


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Color difference among oxovanadium species
Why dioxovanadium $\ce{[VO2]+}$ is pale yellow, whereas orthovanadate $\ce{[VO4]^3-}$ is colorless?
Both of them should be colorless if we ignore charge transfer for a moment, because in both cases vanadium is at +5 oxidation state.
Now considering the charge transfer is the reason for the pale yellow color of dioxovanadium species, why orthovanadate does not show charge transfer band?

The colour will depend on several factors. You've identified the oxidation state (+5 in both cases) as one.
Other factors are:

Ligands
Complex shape

With $\ce{[VO2]}^{+}$, the actual complex ion is $\ce{[VO2(H2O)4]}^{+}$ (Taken from ChemGuide). So our ligand is water, and the shape is tetrahedral (or square-planar).
The $\ce{[VO4]^{3-}}$ ion usually only exists at high pH in solution but polymerises at lower pHs. These polymeric ions can have colour (e.g. $\ce{[V10O28]^{6-}}$ is orange).(Link for colour / polymerisation)
However, I could find no details as to the ligands (I would assume that they would be hydroxide ions at high pH) nor shape.
If we are talking about the solid, then the same applies... does the solid have any water of crystallisation. (Think copper sulfate... blue when "wet" and white when "dry").


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Protonation of compounds with two or three neighboring nitrogens
I am working with organic compounds. Many of them are similar to this ones:

The R group (which I paste for completeness) is:

I am a bit confused about the protonation state of those amine-compounds at pH=7.4, in aqueous solution.
N-substitued amines are more basic than the other ones. For example methylamine is more basic than ammonia.
But the protonation state of the ring is more complex than that, and I can't understand how to protonate it.
I suppose that in a kind of analogy with the guanidine (product) we could protonate again the double bonded nitrogen (imine).
Will amine be protonated? Will that compound has $+2$ charge?
Guanidine
It will possibly be protonated in the double bounded nitrogen, because it has more negative density charge and resonation is not broken. Any other point of view will be welcome.

These are guanidines and it's imino group which gets protonated - it's not "secondary amine".

This is quite similar to the question that I am trying to reason as well.

" with two or three neighboring nitrogens" There's no two nitrogens right next to each other in these structures...


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Determining the structure of N2O using hybridization
I have learnt how to determine the structure of molecules where only the central atom is hybridised like $\ce{ClF3}$, $\ce{C2H2}$, $\ce{PCl5}$, but in $\ce{N2O}$ it seems as though both nitrogen and oxygen have hybrid orbitals. How do I find the hybridization of oxygen and nitrogen in $\ce{N2O}$ and finally determine its structure?
Research Effort: I watched some lectures on hybridization, but all of them included cases in which only the central atom had hybrid orbitals. I also read the topic in my book but this issue wasn't addressed there too.

I don't see how the hybridsation model could yield a structure prediction. In my understanding hybridisation can be used a post Lewis/VSEPR justification of a structure prediction. In that line you 1. draw a proper Lewis representation (possibly even some mesomeric forms) 2. apply VSEPR and 3. then explain it using hybridisation.

Usually you would approach this problem from the other direction. First, find the most likely structure with the correct number of bonds, then deduce the hybridisation. This is also how it works in quantum chemistry: hybridisation is deduced from geometry, not the other way around.
If you try to write the structure of $\ce{N2O}$ you should come up with two equally like structures; these are mesomeric structures as given in the spoiler tag below.

 $$\ce{\overset{-}{N}=\overset{+}{N}=O <-> N#\overset{+}{N}-\overset{-}{O}}$$

From this, we can deduce the most likely hybridisation which will result in a linear molecule. The central atom in a linear molecule typically features two $\mathrm{sp}$ hybrid orbitals and two unhybridised $\mathrm p$ orbitals.
So what about the outer atoms? Well, we cannot say for sure because we do not have enough geometric information. We can tell by the type of bonds that we again need two unhybridised $\mathrm p$ orbitals on the end atoms (remember that the orientations of the double bonds are equivalent!) but we don’t know whether it is a better description for the other two orbitals to be seen as $\mathrm s$ and $\mathrm p$ or two $\mathrm{sp}$ hybrids. In these cases, I would tend to go with ‘unhybridised’ until an experimental or calculative result proves me wrong.

After we have done this, we can indeed look back and realise that we could have determined the structure a priori like this from the general rules; we would just have had to ignore the outer atoms and discuss hybridisation only for the central atom. Indeed, ignoring the hybridisation of terminal atoms gets you very far even in the simplified theory that puts hybridisation first (and is practically wrong).

I am/ was a bit hesitant to up-vote this, but given the two wrong answers, this is much better. Okay, back to nitpicking: The only correct way to use hybridisation is to use it post-geometry determination (so it's not only practically incorrect). In principle there is no reason to assume any other external field than local linear coordination at terminal atoms. Constrained to this sp hybridisation is the max you can get.

To do this, you will need to follow the steps mentioed by R_Berger in the comments. I will use $\ce{CO2}$ as an example.

Draw a Lewis diagram.

Look at bonding & lone pairs on each atom.

Carbon has 2 double bonds which means 2 $\times \ \sigma$ bonds and 2 $\times \ \pi$ bonds.
Therefore, it has sp hybridization (2 $\times$ sp orbitals and 2 $\times$ p orbitals)
Each oxygen has 1 double bond and 2 lone-pairs which means 1 $\times \ \pi$ bond and 3 $\times \ \sigma$ bonds (actually 1 bond and 2 orbitals with lone pairs). Therefore they are sp2 hybridization. (3 $\times$ sp2 orbitals and 1 $\times$ p orbital)

Determine the shape

This can be done directly from the Lewis diagram using VSEPR theory. However, we only need to consider the central atom as that's where the bond angles will be! (You need two bonds to measure an angle...).
So for carbon dioxide, we have got sp hybridization at the central atom (or two areas of electron density with no lone pairs), so the shape will be linear.
Aside: You probably were already fine with the determining hybridization part. You just need to use the same rules on outer atoms!

Oxygen does not have sp² hybridisation. Terminal atoms have at most sp hybrid orbitals. Also note that atoms are never hybridised, only orbitals are. It is perfectly valid, to have a variety of hybrid orbitals at one atom. Just because there is one sp² orbital, it does not mean that there need to be two more. (Have a look here for more on carbon dioxide.)

That link is not one I've found before so I've just spent some time reading it. A few notes (please correct me if needed)... if I understand this, the method I described above (and is used in a majority of places if I search) is outdated? Where do the lone pairs reside? From your link, I see the central argument is that the electron density map is a cylinder which could not happen with my sp2, but does fit an sp hybridisation. Do you have any links for those maps? (Aside: my knowledge is from teaching IB level Chemistry, so much of the advanced theories has gone past me!)

After reading around the topics more, I believe that my answer is correct AND Martin's answer is also correct! I have answered the question using Valence Bond theory (where the hybridisation sp² describes what happens on the oxygen atom itself) whlie Martin used Molecular Orbtial theory (where the sp hydridisation is between the carbon's s and oxygen's p orbitals) (See here)

I am sorry, but you have not answered the question with Valence Bond theory, which is essentially an equivalent theory to molecular orbitals. The sp² orbitals for such atoms used to be part of a VB description, but (at the latest) with the formulation of Bent's rule that basically became obsolete. Mathematically this would still work, physically it simply does not make any sense. For the simple case of water it can be proven, that the lone pairs are not equivalent; this can be extended to more complex molecules. Side remark: MOT does not use or need hybridisation.

Thank you for the above. I'm now reading about Bent's rule... there's a lot to take in / relearn here. I've mainly been involved at secondary/ high school level where the older hybridisations are still considered as the correct answer. Once I've done more reading I'll look at updating my answer.

Based on hybridisation theory, the hybridisation states of the atoms are $\mathrm{sp}$, $\mathrm{sp}$ and $\mathrm{sp^3}$, for the atoms N, N and O, which are connected to each other in this order:

Conventionally, as most textbooks suggest, hybridisation states are derived from electronic geometry of the atoms (i.e. how the electron density regions are arranged around the atoms). For 2 electron density regions, it is sp-hybridised. For 3 electron density regions, it is $\mathrm{sp^2}$-hybridised. For 4 electron density regions, it is $\mathrm{sp^3}$-hybridised etc.
From the Lewis structure I have drawn, we can derive the results as such.


text:
About the protonation of guanidines
I am working with the protonation of guanidines.
The structure of these compounds looks like this:

I know that possibly the protonated-nitrogen will be the double bonded one (that structure conserve resonance), but i am looking for some experimental or theoretical proof, and it is not being easy to find a nice one. Ideas or links about this problem (maybe a paper) will be helpful.

All other lone pairs are delocalized into the $\pi$-system. Try drawing resonance structures.

In solid state, yes: http://scripts.iucr.org/cgi-bin/paper?a04854

Also in solution, there's just one signal by proton NMR: http://www.rsc.org/suppdata/cc/c3/c3cc46518g/c3cc46518g.pdf


text:
Free software to draw Newman projections
I need to draw Newman projection of a molecule. Particularly 1,2-dicloroethane rotamers, but have no idea if there is any program to do that kind of things.
Do you know about some program available for Linux?
I refer to this kind of sketch:

Do you use Latex? If so, the chemmacros package can do this.

Sure! You can make a standalone document, then convert the pdf to png using ImageMagick (convert file.pdf file.png). There is also a way to compile latex directly to png (run latex, then dvipng), but I have never used it.

My guess is, not enough people need this feature for the developers to implement it. It is however present in proprietary softwares like ChemDraw (which is, of course, not available on Linux).

Some versions of ChemDraw work on Linux via Wine.

@santimirandarp Wine is under GNU LGPL, ChemDraw is obviously proprietary.

@santimirandarp How do I put this - if you have ChemDraw already, you are good, Wine is free:)

You can use GChemPaint, which is available on the repositories of many popular distributions. Among its many components for drawing structures, it has a specific tool for Newman's projections.
You can add H, O and many other elements by simply right-clicking on the molecule, and it even allows you to search your molecule in databases like PubChem and ChemSpider.
A useful video tutorial can be found in spokentutorial.org

3.11 To Add a Newman Projection
To add a Newman projection, use the following tool:

This tool allows to add a Newman projection. The changes become permanent only when the mouse button is released.
To add a Newman projection, click on the background. Contrarily to other tools existing atoms are never used by this tool. It creates new atoms of the current element.
When the Newman projection tool is active, the toolbox looks like:

Options for the main bond

Bond length
The default length of the new projected bond. This has no influence ion the drawing, but might be important when exporting to a 3d model.

Bond order
The bond order.

Options for each atom

Bonds number
Number of other bonds for the atom.

First bond angle
The direction of the first bond. Angles are measured counter clockwise from horizontal right.

Angle between bonds
The angle between two consecutive bonds.

Warning
This version of GChemPaint does not correctly support Newman projections with eclipsed bonds.

Source: GChemPaint manual, pages 41-43.

You can visit ChemistrySteps.com and use the free drawing tool embedded in the text editor. It doesn't have the ring but you can try using the text and add "O" as the ring. It may not be the best drawing tool but it is free and you can definitely draw many different structures.


text:
Hybridisation and bond strength
I'm a high school student with minimal mathematical knowledge about the theory of chemical bond. (I just have a bit of theoretical background). I came across a statement which stated that "The more is the s character in a hybrid orbital, the stronger will be the bond. "
I also came across a statement the "The extent of overlap increases with increase in directionality of bond." (Example : p-p > s-p > s-s reflects the order of the extent of overlap of orbitals keeping the internuclear distance same). What I think is that the more will be the overlap , the stronger will be the bond formed. How can I account for these contradictory statements? I'm confused. Thanks in advance.

Hello and welcome to Chemistry.SE. Feel free to take the short [tour] or visit the [help] if you have any questions about how this site works. Best of luck with your question!

The statements are not contradictory when read both carefully. One is about overlapping and other is about strength. Also, you might be getting confused that more overlap leads to more strong bond but the actual statement is for a particular type orbital(s, p, d, etc.) greater extent of overlapping leads to more strong bond.

Thanks shubham for your consideration. So should I conclude that the second statement is only valid for pure atomic orbitals in reference to their bond strength?


text:
pKa of imidazoles
I was looking for the $\mathrm{p}K_\mathrm{a}$ of imidazole and found this $\mathrm{p}K_\mathrm{a}$ table [1]:
\begin{array}{ll}
\text{Compound} &\mathrm{p}K_\mathrm{a} \\
\hline
\text{Imidazolium} &7.38 \\
\text{2-Aminoimidazolium} &8.46 \\
\text{1-Methyl-2-aminoimidazolium} &8.65 \\
\text{4,5-Dimethyl-2-aminoimidazolium} &9.21
\end{array}
Wouldn't you predict totally inverse order of $\mathrm{p}K_\mathrm{a}$ values?
For example, $\ce{-NH2}$ is a donor by resonance, it will make the cation more stable than imidazole, as far as I know. So, how to understand those values of $\mathrm{p}K_\mathrm{a}$?
I upload the first two compound structures:
IMIDAZOLE

AMINOIMIDAZOLE

Storey, B. T.; Sullivan, W. W.; Moyer, C. L. The Journal of Organic Chemistry 1964, 29 (10), 3118–3120. DOI 10.1021/jo01033a537.

Edit
Only for completeness I should add that pka values are of the protonates species, I mean imidazolium, 2-aminoimidazolium, and so on..

Amino groups being "donors" is not nearly specific enough. Is it via resonance or inductive effects?

Right, I will edit. It is by resonance @Zhe

My point was that it is inductively withdrawing. Give the nature of the cation, resonance donating is probably not super helpful.

Another good question is what resonance of the amino lone pair in the guanidine moiety does to the aromaticity of the ring. Interesting competing effects for sure.

Everything's all right. You've got confused but what you pointed out in your edit. Cation becomes weaker acid - "more stable" just like you say. Higher pKbH+ means weaker acid - stronger neutral base.

I don’t understand where the problem lies. A higher $\mathrm pK_\mathrm{a}$ means less likely to get deprotonated. The more electron-donating groups, the more a positive charge is stabilised and the less likely it is for a proton to leave — exactly what the data tells you. Am I missing something?

@Jan right. No, I was missing the final part..thanks a lot.

@Mithoron yes you are right,thanks for the comment.

Ah. Do you want to write up a self-answer to your question? =)

@Jan I can try.., please pay attention to correct me if am wrong..(and my English is quite bad also)

That’s what we’re here for =)

As pointed out by @Jan and @Mithoron in comments, part of the answer was on the OP. I will complete it.
If we compare acidity of imidazolium and 2-aminoimidazolium (the cations), we find that 2-aminoimidazolium is a weaker acid because of the bigger $\mathrm{p}K_\mathrm{a}$ value.
The reason is that the amino group ($\ce{-NH2}$) is donor by resonance, stabilizing the cation more than imidazolium is, because it doesn't have any extra donor group.
The same reasoning can be done for the remainder compounds of the table. The more inductive or resonance donors, the bigger stability of the cation and weaker acid (bigger $\mathrm{p}K_\mathrm{a}$).
One last thing
As @Zhe observed, the pKa values of all those compounds are really different of guanidine($\mathrm{p}K_\mathrm{a}$=13.6). In the paper below, authors make the same observation. I haven't found an easy answer for that. Maybe it can be done with computational calculations.

Storey, B. T.; Sullivan, W. W.; Moyer, C. L. The Journal of Organic Chemistry 1964, 29 (10), 3118–3120. DOI 10.1021/jo01033a537.

With regard to the comparison to guanidine, see "The pKa Values of Some 2-Aminomidazolium Ions"J. Org. Chem. 1964, 29, 10, 3118–3120 https://doi.org/10.1021/jo01033a537


text:
Remove only salt from salt water
I have a bottle of salt water I collected on a local beach. I want to see what is in it, apart from water and salt. If I boil the whole bottle and wait for it to evaporate, mostly salt will remain. I want to see everything except salt that is still in the pot after I boil the water. Is this possible with common household items and commonly available (and safe) chemicals?
Disclaimer: I know nothing about chemistry.

How do you know there is anything other than water and salt in the first place?

The water is often pretty dirty, supposedly because criminal organizations dump waste at night. I'm curious to see if there is anything visible to the naked eye after evaporating a bunch of jugs filled with water.

Is there anything visible now, before any evaporating?

I'm getting ready to go to the beach right now! I'll report later.

Sea water contains lots of salts not just sodium chloride. Are you interested in the others? or do you just want to remove the sodium chloride? And you are equating whether household items can replicate what analytical chemical labs with lots of dedicated equipment: that could be a stretch.

Evaporating won't give you much information, especially if there are traces of organic pollutants, therefore water samples are analysed as is.  Elemental analysis, photometry and chromato-mass spectrometry would be ideal tools to pinpoint the exact composition, but you need an analytical lab to do it for you (some will accept water samples and do analysis for free, in return you allow them to use the data).

Before boiling the water, just let the water sample sit over an extended period of time in a clear container and see what settles to the bottom.

You could remove the bulk of the sodium chloride (and most other salts) by evaporation and crystallization. Most of the impurities you are looking for are going to stay in solution or suspension when you filter off the salts.
You could also use dialysis to remove the remaining salts: this requires a bag or length of thin-walled tubing made of a common polymer such as cellulose acetate. Tie off one end (the bottom), put the concentrate from the above step into the tubing, and tie off the top end. Then place it in a bucket of distilled water. The salts and other small molecules will diffuse out into the water, whereas the higher molecular weight materials like toxins will stay in the bag. You may need to change the water in the bucket a few times to get rid of all the salts.
You will still need some chemistry to figure out what substances remain in the bag. With seawater, it could be a very complex mix. However, looking at it with a fluorescent light might show something interesting. Good luck, and hope you get inspired to study chemistry!

Exactly what I wanted! :) I am inspired to study chemistry already, but right now I have a bit too much on my plate, especially with music software programming (I'm making my own DAW with most of my free time). Chemistry is one of the first things I want to learn when I have more time.

You could use ion exchange resin to remove the salt, I would use DOWEX-50 in the protonated form. I would wash it several times with water to remove the yellow water soluble impurity which is normally on brand new DOWEX-50- I would mix it with the water at the same time. I would then add to the water IRA-67 (weak base resin) in the free base form. I would have washed it first. This combination will remove the sodium chloride from water.
However if I wanted to see what was in the water, then this desalination treatment with resin would be counterproductive.
If I wanted to find out what organics were in the water, then I would extract with diethyl ether or ethyl acetate. The sodium chloride in the water will reduce the amount of the organic solvent which dissolves in the water. It will also sometimes increase the partioning of the organic solute into the organic phase. I would be then wanting to inject the organic extraxct into a gas chromatography machine after dilution with hexane and drying with sodium sulfate.
If I was to want to isolate metals from the water, then I would be looking to do solvent extraction. Depending on the metal I am interested in I would either use 30 % DEHPA in kerosene or 30 % aliquat 336 in toluene.
The DEHPA (Di- 2-ethylhexyl hydrogen phosphate) will extract many divalent metals from the water, depending on the pH it can extract calcium, strontium, barium and lanthanides. I would strip the metals using nitric acid.
The aliquat 336 is able to extract from sodium chloride solution things like copper, zinc, cadmium, iron and cobalt. I would want to strip (back extract) the metals from the aliquat phase into dilute nitric acid, then I would want to use ICPOES or AAS to measure them.
The general idea is that if you take a clean glass bottle and put in it 200 ml of the water sample, 20 ml of the organic phase. You then seal it and shake it. Then take a 15 ml sample of the organic phase to a smaller vial and add 5 ml of the strip solution. This has the potential to increase the concentration of the metals.
The problem is that if I was to use the DOWEX-50 resin on the orginal water sample then it would remove many of the metals which we want to measure using the solvent extraction pretreatment method.
In general if you want to measure things in salt water, I would try to devise methods of recovering the anaylte without needing to remove the common salt.


text:
Estimating standard Gibbs free energy of a molecule
In order to estimate the standard Gibbs free energy of a molecule at room temperature and normal pressure, we use Maxwell Boltzmann average over  rotational conformers.
Question
Should we consider cis/trans isomers besides rotational conformers? Is there any criterion for it?

Why, it's simple: those species which are there can contribute. Those which aren't, can't. Suppose you have a cis isomer; does it convert to trans at room temperature? No.

@IvanNeretin After a long time the two isomers trans/cis should equilibrate. So why we should no consider them?

@adosar Well, yeah, they will, but not before hell freezes over, so for all practical purposes we may safely assume they won't.

You will need to calculate the partition functions for each type of motion and then calculate the thermodynamic quantities from these.
As to internal rotation, if the energy for this is far greater than $k_BT$, such as with a C=C bond, then treat it as a torsion and use this frequency. If the barrier to rotation is small compared to $k_BT$ then treat it as a free rotor but reduce the partition function by n if there are n barriers. In the intermediate case you will have to find the barrier (say by spectroscopy) and fit the data to a model of the potential (usually of a cosine form)and then numerically solving the Schrodinger eqn. to find energy levels.
If your temperature is so high that cis-trans reaction occurs then you should probably have to treat this as a mixture.


text:
Etymology of alanine
Oxford dictionary online gives etymology of alanine as:

Coined in German as Alanin, from aldehyde + -an (for ease of pronunciation) + -ine.

But I see no resemblance to the aldehyde structure in alanine. Is this etymology incorrect, or am I mistaken about the resemblance?

In the original German paper [1] Adolf Strecker used Aldehyd-Ammoniak or aldehyde-ammonia as a precursor, that's where the name derives from:

Vor einigen Jahren habe ich gezeigt, daſs Aldehyd-Ammoniak und Blausäure beim Erwärmen mit verdünnter Chlorwasserstoffsäure sich zu einer schwachen Basis, Alanin genannt, vereinigen [...]:
  $$\ce{\underset{\text{Aldehyd-Ammoniak}}{C4H4O2 * NH3} + HCl + \underset{Blausäure}{C2NH} +2 HO = \underset{Alanin}{C6H7NO4} + NH4Cl}$$

As David Richerby mentioned in the comments, Strecker's brutto-formula ($\ce{C6H7NO4}$) deviates from the moden one ($\ce{C3H7NO2}$), also the reaction scheme is a bit different.

Strecker, A. Annalen der Chemie und Pharmacie 1854, 91 (3), 349–351. DOI 10.1002/jlac.18540910309.

@andselisk Easy-to-understand answers to easy-to-understand questions always get lots of votes. Think of it as compensation for the times when you've spent an hour writing a detailed answer to a difficult question and it scores +2 because nobody can be bothered to read it.

Having said that, I'm a bit confused. Isn't alanine $\ce{C3H7NO2}$?

@DavidRicherby I'm not 100% sure, but most likely it's brutto-formula was established incorrectly at that time. Many other sources around that period of time (mid 19th century) refer to alanine as to $\ce{C6H7NO4}$, or, to meet the era's spirit, $\ce{C^6H^7NO^4}$.

Ah, I see. Might be worth adding a note about that to the answer.

User364914 may be confused by the "odd" formulas in Strecker's 1854 paper. At that time there was no universal agreement on the masses of atoms. Was carbon 6 or 12? Was oxygen 8 or 16? Strecker used 6 for carbon and 8 for oxygen.
The aldehyde is acetaldehyde, $\ce{NH3}$ is OK, blausaure is hydrogen cyanide and $\ce{OH}$ is water. The corrected structures should have half the number of carbons and oxygens. The modern equation should read:
$$\ce{C2H4O.NH3 + HCl + HCN + H2O -> C3H7NO2 + NH4Cl}$$
You may want to read about the Karlsruhe Congress of 1860.


text:
Why does the addition of PCl5 increase the rate of dissociation of PCl5?

$\ce{PCl5 <=> PCl3 +Cl2} $

It's stated in my book that addition of $\ce{PCl5}$  to the equilibrium mixture increases the rate of forward reaction but no reason is mentioned for it.
However, before this the author derived $K_c = \dfrac{x^2}{(1-x)V}$ (using law of mass action at equilibrium ) where V is the volume of the container and $x$ is the number of $\ce{PCl5}$ molecules dissociated from $a$ number of moles initially.
I couldn't relate the "rate of forward reaction" statement to this equation while earlier in case of the reaction:
$\ce{H2 + I2 <=> HI}$
I was able to relate the equation I obtained through law of mass action with the effect of concentration on rate of forward or backward reaction.
Could someone please explain why addition of $\ce{PCl5}$ to the equilibrium mixture increases the rate of forward reaction?

Imagine a overpopulated kingdom suddenly gaining a large area of land (Start of Reaction). Immediately, groups of people start migrating to the area of land (Forward Reaction). However, as people migrate to this new land, some decide that, for whatever reason, they don't like this area (Reverse Reaction). As more and more people move there, more start coming back until there is an equal rate of people moving to the territory and people moving back (Dynamic Equilibrium).

To your question: To model this system using my example of immigration, we suddenly add an enourmous influx of refugees into the old kingdom (Increase in $\ce{PCl5}$).

So the land in the outer territory seems more valuable now (less crowding, more open space, better air, etc.). This causes a temporary increase in the forward direction until the reverse reaction increases to compensate. This is called Le Chatelier's principle in chemistry.
References: Le Chatelier's Principle, My Handy Old 2011 Introductory Chemistry Textbook.


text:
Calculating temperature change due to Joule-Thomson effect
I have a 10 L compressed argon cylinder with filling pressure of 10 MPa. According to Perry's Chemical Engineer's Handbook, argon's J-T coefficient (unit: K/MPa) is 3.7 at 0.1 MPa, 2.6 at 10 MPa. (Temperature is 300 K.)
When I open up the cylinder valve, 10 MPa argon in the cylinder will expand into 0.1 MPa atmosphere. So the $\Delta P$ is 9.9 MPa.

How much will the temperature change?
  The initial temperature is 300 K.

$9.9 \times 3.7 = 36.63$

(Which means the temperature of argon outside the cylinder will be 263.37 K)

$9.9 \times 2.6 = 25.74$
Something in between (like when the J-T coefficient in effect changes continuously as the gas expands).

It is going to be something in between.  The definition of the JT coefficient is $$\mu=\left(\frac{\partial T}{\partial P}\right)_H$$If the temperature-dependence can be neglected, then this gives $$\Delta T=-\int_{0.1}^{10}{\mu(P)dP}$$

Unless I'm mistaken, won't they need to evaluate this numerically (thus presumably needing more points then they have) since there isn't a known functional form for the JT coefficient in terms of pressure for a real gas?

For a large change like this, it can be done using generalized corresponding states correlations for the 2nd virial coefficient to determine the residual enthalpy at the initial and final states.  See Smith and Van Ness, Chapter 7, Throttling Processes.

The initial state of the argon gas is completely defined by the given temperature and pressure:

Initial temperature $T_1=300\ \mathrm K$
Initial pressure $p_1=10\ \mathrm{MPa}$

The corresponding values of other properties of the gas can be looked up in so-called steam tables. For example, in REFPROP – NIST Standard Reference Database 23, Version 9.0 we find for the given initial temperature and pressure:

Initial density $\rho_1=167.60\ \mathrm{kg\ m^{-3}}$
Initial specific internal energy $u_1=78.313\ \mathrm{kJ\ kg^{-1}}$
Initial specific enthalpy $h_1=137.98\ \mathrm{kJ\ kg^{-1}}$
Initial specific entropy $s_1=2.8716\ \mathrm{kJ\ kg^{-1}\ K^{-1}}$
Initial Joule–Thomson coefficient $\mu_{\mathrm{JT},1}=2.6066\ \mathrm{K\ MPa^{-1}}$

The final state of the gas is not completely defined by the given values since only the final pressure is given

Final pressure $p_2=0.1\ \mathrm{MPa}$

which is not enough to look up other values in steam tables.
However, we know that Joule–Thomson expansion is an isenthalpic process; i.e. the specific enthalpy $h$ remains constant. Thus

Final specific enthalpy $h_2=h_1=137.98\ \mathrm{kJ\ kg^{-1}}$

Now we have a second data point that can be used to look up other parameter values, for example

Final density $\rho_2=1.8105\ \mathrm{kg\ m^{-3}}$
Final specific internal energy $u_2=82.747\ \mathrm{kJ\ kg^{-1}}$
Final specific entropy $s_2=3.8155\ \mathrm{kJ\ kg^{-1}\ K^{-1}}$
Final Joule–Thomson coefficient $\mu_{\mathrm{JT},2}=4.5226\ \mathrm{K\ MPa^{-1}}$

and also

Final temperature $T_2=265.65\ \mathrm K$

Since the pressure is not changing gradually, in the sense that it is either $p_1$ or $p_2$, not in between (it is not a reversible process), the temperature drop should be $\Delta T=\mu_\mathrm{JT}(p_2)\cdot p_2-\mu_\mathrm{JT}(p_1)\cdot p_1$.

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