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What drives a molecule/atom towards stability?
I'm very familiar with the concept that chemical reactions occur because the reactants a state of stability via the reaction. It's a very satisfying and fulfilling statement. But my question has always been - "How do the molecules know what to do? Do they have a certain amount of intelligence? Are they living?" I'm really sorry that the question isn't properly framed because this topic is just a mess inside my mind.
It's like the reactants already know what to do in order to attain stability.
The question arises because we are taught chemistry in a way relatable with moral values and actions humans perform under a given circumstance. So who is to be frowned upon - the way we are taught, or the lack of understanding within the students?
Thanks for going through the mess :)

Does a ball need "intelligence" to fall down if dropped?

Physics and Chemistry are two different scenarios. We've seen a ball fall numerous times in our lifetime and it comes with no surprise that gravity is the culprit behind it. Newton must have also thought "Why only downwards, why sideways or upwards?" He might have also speculated the presence of "intelligence" in a falling object. Only after years of hard work did he come up with a possible explanation that we call the "Gravitational Theory". Much to our surprise, it is being criticised by modern physicists. My question was out of sheer curiosity too. I had no viable explanation for the above.

It boils down to thermodynamics. No, molecules do not possess intelligence. But they respect the laws of thermodynamics, which in a general look state:

Energy of a closed system is constant.
The entropy of the universe is always increasing.
You can't get to absolute zero temperature! (Ok, this one is less important for this analysis)

The second law is the key: in order to augment the entropy of the universe a certain transformation can do one of two things: increase its own entropy or the entropy of its surroundings. The latter can be achieved by releasing heat, which is something exothermic reactions do. The sum of these effects is described by the system's Gibbs free energy:
$$G = H - TS$$
For a chemical reaction occurring at a constant temperature and pressure, $G$ has to decrease and becomes minimized at equilibrium. We then say the system is (thermodynamically) "stable". So when you hear someone say "chemical reactions minimize their energy", it would be better spoken as "... minimize their free energy".

Now you might wonder why the second law is true. Why does entropy has to increase? Why does the free energy has to be minimized? It certainly doesn't want to.
The truth lies in the molecular interpretation of entropy: it measures the amount of micro-states at which system can be found. Put together a bunch of blue balls with another bunch of red balls in a box, shake it all and look at how they end up. They surely won't be found in separate sides of the box, they will mix! The scenario where they are separated corresponds to a very low-entropy state: you can think of the red ones to one side or vice-versa: not that many possibilities of arrangements. The mixed state though has a much higher entropy: think of how many arrangements are possible by exchanging the positions of red and blue balls!
Since molecules in a closed system are very very much like tiny balls being shaken inside a box - not by a mechanical source but by their own thermal energy -, their behavior with respect to entropy is the same. Again, it's not that they want to increase the entropy of universe, there just is a higher probability that they do rather than don't. Given that we are usually observing systems of sextillion molecules or more, statistics kicks in and tell us probabilities mean more or less certainty: we are pretty much certain that chemical reactions will keep on trying to maximize the entropy of the universe.

Ok, so what you're trying to explain is that in the massive ocean of atoms, the random microstates are more possible that the non random ones?

Yes. That's the second law of thermodynamics. Now to truly understand how the entropy of a chemical system can be increased, you have to check the shape, size and interaction energies of the molecules, while also keeping track of the overall homogeneity in the case of a mixture. They all contribute to the total amount of possible micro-states.

@Godim What about Formation of NaCl from NaOH and HCl? What about the entropy change here?

$\ce{NaCl}$ is not a chemical species in solution made of $\ce{NaOH}$ and $\ce{HCl}$. The ions stay dissolved as $\ce{Na+}$ and $\ce{Cl-}$. If you start from a solution of each reactant, the only true reaction taking place is water self-ionization $\ce{H2O <=> OH- + H3O+}$. With neutralization, the only proceeds to the left and presumably  induces some entropy loss, but the total entropy is still increased by the other ions remaining in solution.


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Why are acetylene and air commonly used for AAS?
I see that this fuel/oxidant combination is most commonly used, but it's never stated why.

Because it gives a very high temperature: 2200-2400 $^\circ$C.
This is due to its high heat of combustion, around 1300 kJ/ mol.
If still higher temperatures are required, as for Al, Si, Ti, etc., a combination of acetylene and $\ce{N2O}$ is used, which reaches 2600-2800 $^\circ$C.

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Shouldn't weak acids act as strong ones in alkaline solutions?
A weak acid in water dissociates reversibly as follows:
$$\ce{XH <=> X- + H+}$$
A strong base is one which dissociates completely.
Now, let's assume we're doing acid-base titrations, adding a weak acid to a strong base, trying to arrive at the titration curve.
My question is mainly why the $\mathrm{pH}$ at the equivalence point isn't $7$. Intuitively, it makes sense, one dissociates fully while the other doesn't. But if you dig into it a bit deeper you realise that that answer is incomplete.
Bear with me. You add the weak acid to the strong base until you reach the equivalence point. The strong base dissociates completely while the acid establishes an equilibrium as described above. However, thinking about it you realise that at no point should this be possible. An $\ce{OH-}$ ion from the base quickly reacts with any H+ ions produced meaning that $[\ce{H+}]$ is always near zero, meaning that more and more of the weak acid needs to dissociate to establish equilibrium. However, the $\ce{H+}$ concentration will always be 0 as any hydrogen ions are snapped up meaning that the acid will completely dissociate before it has a chance to establish equilibrium, at which point all of the $\ce{OH-}$ ions will also have been neutralised. The $\mathrm{pH}$ of this resulting solution should in other words at that point be $7$, because the acid, even though it's weak, acts as a strong one in an alkaline environment.
This however I know isn't correct. What's the issue with my logic above?

While titrating weak acid with strong base, the titration does occur like a normal acid, because the base added quickly takes the dissociated $\ce{H^+}$ ion of the weak acid and thus the equilibrium of the acid is distorted. To maintain constant $\ce{K_c}$, the weak acid also produces $\ce{H^+}$ instantaneously, and thus its behaviour almost becomes like a strong acid if base is added fast(which is normal in titration) to maintain equilibium.But the $\ce{pH}$ at end point is not $7$ not due to the acid but the salt formed as a product of the titration.Suppose the weak acid($\ce{HA}$) is being titrated with a strong base(say for example,$\ce{NaOH}$), and both are taken in same equivalents. Then the reaction does go until all the acid (though weak) is exhausted.$$\ce{HA + NaOH -> NaA + H_2O}$$At t= o, say both $\ce{HA }$ and $\ce{NaOH }$ has $c$ equivalents each. So, at the end of the reaction, both $\ce{HA }$ and $\ce{NaOH}$ will be totally exhausted and $\ce{NaA }$ will be $c$ equivalents.Now, in the medium only $\ce{NaA }$ is present which will control the $\ce{pH}.$ But $\ce{NaA}$ is actually "SALT OF A WEAK ACID AND STRONG BASE". So it hydrolyses as,$$\ce{NaA + H_2O <=> HA + Na^+ + OH^-}$$So  you can see as $\ce{HA }$ is a weak acid ,it's conjugate base (stronger) $\ce{A^- }$ takes the proton of water and leaves only $\ce{OH^- }$, which in turn is the reason for incresing the alkalinity of the medium.That's why generally, $\ce{pH }$ is above 7 at the end point. By doing mathematical calculation, it can be easily shown that if the weak acid's $\ce{pK_a}$ is given and the final concentration of the salt is say, $\ce{ c' M}$. Then the $\ce{pH }$ of the medium can be calculated as ,$$\ce{pH = 1/2 pK_w + 1/2 pK_a + 1/2logc'}$$. Assuming the titration is being carried at $25$ degrees, then$$\ce{pH= 7 + 1/2pK_a +1/2logc' >  7 (generally) }$$ if your acid is tending to strong i.e. $\ce{pK_a}$ is very less positive, and $\ce{c'}$ is close to $1 $, the $\ce{pH}$ can be close to $7$, but $\ce{pK_a}$'s  are normally of the order of $3$ or $4$. So, generally $\ce{pH}$ exceeds $7$.

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Using real wave functions for chemical bond explanation
My question related with another one here.
For explanation valence bond direction chemists are used real wave functions, for example $p_x, p_y, p_z, \ldots$. But this functions do not describe the preffred direction of angular momentum, they are not the eigenfunctions of $\hat L_z$.
The functions, that are describe the states with some preffered directions are not real, they are complex. For hydrogen, for example, they looks like:
If some other atom nearby first one, then there is some prefered direction, and good funtcions for this situation should be not real ones, but complex ones. Then why chemists used real functions for bond explanation?

Chemists use the linear combinations of hydrogen atom wavefunctions which are real because this is simply much more convenient to think about. Notice that for hydrogen, all the orbitals for a given $n$ are degenerate, so taking these linear combinations does not affect the energy, and thus thinking about the shapes of these orbitals is potentially more useful than thinking about the shape of something which is complex.
You are correct that taking the real-valued wavefunctions causes them to no longer be eigenfunctions of $L_z$. This is a bit frustrating, but in practice we never use these orbitals for anything, so it does not really matter. I used to think organic chemists took the shape of orbitals a bit too seriously (and they maybe do), but the notion of the shape of an orbital is actually recovered in a more rigorous manner.
That is, molecular orbital theory is just a qualitative form of Hartree-Fock (HF) theory. In HF, the Fock matrix is unitarily invariant, so we are free to transform our orbitals by any unitary rotation. These orbitals can be written as functions in $\mathbb{R}^3$ as desired. Thus, there exists schemes by which molecules orbitals can be "localized" and they often take the shape of orbitals which look a whole lot like the real-valued hydrogen atom orbitals you mention.
Thus, the use of these real-valued orbitals as a notion of shape is justified after the fact from HF theory and beyond.
In practice, this never really matters if one is being rigorous because any attempt at saying exactly which orbitals are involved in a bond is thwarted by unitarily transforming those orbitals. Thus, many chemists use the shape of an orbital as a heuristic because thus far, it has proven to be a very useful conceptual tool. I think pretty much all chemists know this is not completely correct, but neither is anything in physics.
Basically, the shape of orbitals provide a useful model.


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Counting number of stereo isomer
I am having trouble counting the stereo-isomers in this structure:

There are no chiral carbons.
The double bonds means there maybe cis-trans geometrical isomerism but the groups are different. So that approach failed.
Answer given is: 4

The only thing that is different when the groups are different is that cis/trans becomes known as Z/E. Other than that, it is very much still there.

Though all the four groups around double bond are different, the geometrical isomerism is still present. If the groups are not same then the highest priority group on one side of the carbon is determined through CIP sequence rules.
Now if two highest priority groups are on the same side of double bond (syn position), the isomer is called Z-isomer. If two highest priority groups are on the opposite side of double bond (anti-position), that isomer is called E-isomer.
Here there will be four combinations, i.e (2Z,6Z); (2Z,6E); (2E,6Z); (2E, 6E). Thus, it gives four possible geometrical isomers.


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Numbering bonds in cyclic compounds
Let's say I have a double bond between carbons 6 and 1 on a 6 carbon cycle (the numbering is imposed because of the positions of the suffix functional groups), is the bond numbered 1 or 6?
6 seems logical because it is not ambiguous (clearly if it starts on 6 it ends on 1), but usually you must minimize numbers (however numbering the bond 1 means there's ambiguity between it being between 1 and 2 and between 1 and 6).
I've tried finding the official rule but haven't managed to.

There is no ambiguity. If you number the bond 1, then it is the bond between 1 and 2, period. If you have a bond numbered 6, then so be it (though chances are you'd want to choose a different numbering in this case).

however numbering the bond 1 means there's ambiguity between it being between 1 and 2 and between 1 and 6

It might appear to you all of a sudden that there's an ambiguity, but actually there's none. Look at these two molecules carefully:

Note that the numbering for the molecule on the left will increase in the anticlockwise direction, while the numbering for the molecule on the right will increase in the clockwise direction. The pi bond in each case is named as a -1-en (and it should, because both the molecules are identical)
You seem to be assuming that the numbering for the molecule on the left also applies on the molecule on the right. But, that is not the case, as shown above.

The disubstituted cyclohexene ring complicates the scenario, but only slightly. Consider the case when functional groups of equal priority are present:

Now, recognize that - even before assigning a name - that both the molecules are different. Thus, their names will also be different. The molecule on the left begins numbering from the top $-\ce{COOH}$ and increases anticlockwise. The molecule on the right begins numbering from the lower $-\ce{COOH}$ and increases clockwise. The first pi bond is assigned a name "-1-en" while the other is assigned "-3-en".
If a lower priority functional group is present instead, like so:

the numbering will increase along the direction of the pi bond, beginning from the top $-\ce{COOH}$. Hence, their names will be 3-hydroxycyclohex-1-ene-1-carboxylic acid and 5-hydroxycyclohex-1-ene-1-carboxylic acid respectively.

A possible way to achieve a "-6-en" is to take equal priority suffix groups on positions 1, 2 and 4, and then place a pi bond between positions 1 and 6, like so:

It is named "cyclohex-6-ene-1,2,4-tricarboxylic acid" giving the suffix functional groups a lowest locant set of "1,2,4", consequently naming the alkene as a "-6-en". Actually, though, this is not 100% correct. A slightly technical detail is missing, which can be found in Loong's answer.

sir , in the last compound why shouldn't we start numbering from the lower carboxylic acid group. That would also give the unsaturation a lesser number and retain the locant set "1,2,4"

@vignesh when we are supposed to give the unsaturated bond a number, we cannot just "touch (its one end) and run". We have to completely move over the unsaturation at that locant in order to assign it that particular locant.   Hope you understood.

I understand what you are saying. But I can't seem to tell how we pass over the unsaturation when giving it a number of 6

I've tried finding the official rule but haven't managed to.

Names with the highest locant for the double bond such as ‘cyclohex-6-ene-1,2,4-tricarboxylic acid’ are formed analogous to the names of usual unsaturated cyclic compounds. This might be considered common sense. Also, most known chemical structure drawing programs understand and generate this name for this example.
Somewhat surprisingly, however, such names are not in accordance with IUPAC recommendations. And this does not only concern the last (2013) edition but also the now obsolete previous editions. The corresponding rules in past and present editions are as follows.
IUPAC recommendations 1979:

When, in cyclic compounds or their substitution products, the locants of a double bond differ by unity, only the lower locant is cited in the name; when they differ by more than unity, one locant is placed in parentheses after the other (…).

IUPAC recommendations 1993:

Only the lower locant for a multiple bond is cited except when the numerical difference between the two locants is greater than one, in which case the higher-numbered locant is cited in parentheses (…).

IUPAC recommendations 2013:

Only the lower locant for a multiple bond is cited, except when the numerical difference between the two locants is greater than one, in which case the higher locant is enclosed in parentheses.

Explicit examples given for the case when locants differ by more than one are bicyclic compounds; nevertheless, the rules are unambiguous also for simple cyclic compounds.
Therefore, the correct name for ‘cyclohex-6-ene-1,2,4-tricarboxylic acid’ according to the current IUPAC recommendations actually is cyclohex-1(6)-ene-1,2,4-tricarboxylic acid. Since the double bond is between the atoms ‘1’ and ‘6’, which differ by more than one, the lower locant ‘1’ is cited and the higher locant ‘6’ is enclosed in parentheses.

Remarkably, I found only one program (MarvinSketch) that automatically generates this name.

I understand the purpose of the IUPAC is to assign a unique unambiguous name to a compound. I do not understand why this rule (2013) that you cited - "Only the lower locant for a multiple bond is cited, except when the numerical difference between the two locants is greater than one, in which case the higher locant is enclosed in parentheses."  - was brought into existence. Are there any corner cases of organic molecules where the ambiguity in the location of the pi bond was resolved through exactly this rule?

I don't think that ‘cyclohex-1(6)-ene-1,2,4-tricarboxylic’ acid name is correct, no naming software is perfect. See my answer (I do somehow refer to yours, but not explicitly).

When using general IUPAC rules for locant number minimization, one might be tempted to use the compound locant for some rather simple cases like multi-substituted cyclohexenes, like

‘cyclohex-1(6)-ene-1,2,4-triol’ instead of
cyclohex-6-ene-1,2,4-triol
This (x)y type of locant is called compound locant in IUPAC Nomenclature of Organic Chemistry (Preferred names (2013)) terminology:

P-14.3 LOCANTS
P-14.3.1 Types of locants
Traditional types of locants are arabic numbers, for example, 1, 2, 3;
  primed locants, for example, 1′, 1′′′, 2′′;
  locants including a lower case Roman letter, for example, 3a, 3b;
  italicized Roman letters, for example, O, N, P;
  Greek letters, for example, ‘α-, β-, γ-’
  and compound locants, for example, ‘1(10)’ and ‘5(17)’.
(…)

Why ‘cyclohex-1(6)-ene-1,2,4-triol’ is most likely wrong might be concluded from the following rule

P-31 MODIFICATION OF THE DEGREE OF HYDROGENATION OF PARENTHYDRIDES
(…)
P-31.1 THE ENDINGS ‘ENE’ OR ‘YNE
(…)
P-31.1.4 Bi- and polycyclic von Baeyer parent hydrides
(…)
P-31.1.4.2 If there is a choice of names and numbering, the following criteria are considered in order until a decision is reached:
(1) a minimum number of compound locants. A compound locant is used for a double bond if the locants of the atoms at each end of the bond do not differ by a value of one. When a compound locant is required, the higher locant is cited in parentheses. A benzene ring is shown and described as a cyclohexatriene corresponding to the Kekulé structure. Other aromatic rings are treated similarly, When required;
Note: The seniority of single locants over compound locants was established in 1989 for numbering steroids [see S3-2.5(2), ref.
  16] and extended to von Baeyer systems in the 1999 publication on von Baeyer nomenclature (see VB-8.3.1, ref. 7).
Examples:

  bicyclo[4.2.0]-oct-6-ene
  [not bicyclo[4.2.0]-oct-1(8)-ene]
(…)

which however applies for polycyclic rings.
The mentioned ref 16 (IUPAC Nomenclature of steroids) section:

3S-2 Fundamental carbocycles, unsaturation and alkyl substitution at C-17
(…)
3S-2.5 Unsaturation
(…)
2) If there is a choice of locants, single ones are preferred to compound locants, e.g. estra-5,7,9-triene is preferred to estra-5(10),6,8-triene.

So, IUPAC organic nomenclature rules tend to avoid these compound locants where it's possible. But, as Ivan Neretin has stated in his comment, for monocyclic compounds, there even is no ambiguity at all.
I'd add: … if there was a rule stating, that the compound locants for ‘ene’ position can be applied only for branches.

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Will additional ascorbic acid preserve hemp milk for a longer duration?
Ascorbic acid is known as a natural preservative and many food manufacturers add it to their food to prolong its shelf life.
Hemp milk is a drink that is created by blending 1 part hemp seeds with 2/3 parts water at a high speed. Usually, hemp milk will stay fresh if refrigerated for about 4-5 days.
Hemp seeds naturally contain ascorbic acid. However, how much additional ascorbic acid is required to extend the shelf-life of refrigerated hemp milk by a given amount of days?
For example, to extend the shelf-life for an extra 5 days.


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Why are tertiary alcohols more reactive towards Lucas reagent but show low rates of esterification?
Generally the order of reactivity of alcohols follows the order tertiary > secondary > primary. But why is it reverse in case of esterification.

Probably because esterification does not involve forming a cation from the alcohol.

Do you know the mechanism of esterification? The first step - which is also the rate determining step - involves the nucleophilic attack of the lone pair on the alcohol's oxygen atom on the electrophilic carbon atom of the carboxylic acid. Hence, the more nucleophilic the oxygen atom, more will the equilibrium shift to the right, favoring esterification.
Now, can you tell which degree alcohol is the most nucleophilic? And can you tell now which degree alcohol will favor esterification the most?

For Lucas reagent, the reaction involves the formation of a carbocation (at the position where the hydroxyl group originally was) by elimination of the hydroxyl group as an oxonium ion. Thus, the more stable the carbocation, the more the Lucas reagent favor the white precipitate.


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Why can ethanoic acid not be used in the production of paracetamol?
Ethanoyl chloride or ethanoic anhydride can be used but not ethanoic acid. Why is this?

In the preparation of paracetamol, the last step is the nucleophilic attack of the $\ce{-NH2}$ group of p-aminophenol on ethanoyl chloride or ethanoic anhydride. The last step is a selective attack by $\ce{-NH2}$ group and not by $\ce{-OH}$ group , which has to be taken care of.
Now you should agree that the ease of nucleophilic acyl substitution follows the trend as:
$$\ce{CH3COCl > CH3COOCOCH3 > CH3COOR > CH3COOH > CH3CONH2}$$
(simply compare the electrophilicity of carbonyl carbon and strength of the bond between carbonyl carbon and $\ce{-OH}$ or $\ce{-OH}$ substituted groups or stability of the leaving groups.)
So, the nucleophilic acyl substitution happens more easily in ethanoyl chloride or ethanoic anhydride than in ethanoic acid. As the transition state has much lower energy in the former two cases, hence the kinetically favoured attack (by $\ce{-NH2}$; as this is better nucleophile than $\ce{-OH}$)  is more feasible. But as the difficulty in substitution increases, the transition state energy increases and the competition between $\ce{-NH2 }$ and $\ce{-OH}$ increases and the yield of paracetamol will decreases.
That is why easily substitutable reagents will enhance the yield and hence should be used.


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Acidity comparison of charged group 16 hydrides
March's Organic Chemistry (7th ed - page 337 under "Periodic Table Correlations" in chapter "Acids and Bases") states that:

$\ce{H3O+ > H3S+ > H3Se+}$ is the acidity order of these charged group 16 hydrides

I am trying to make out why this is true. I have seen the related questions but they are about the neutral hydrides and not charged ones.
My working was that the order of acidity should be the reverse order of the basicity of their conjugate bases. The conjugate bases are $\ce{H2O,H2S, H2Se}$. We know that basicity decreases down the group, as the lone pair becomes more dispersed over the large surface area  (corroborated for group 15). Hence, according to me, order of basicity order is $\ce{H2O>H2S> H2Se}$. Hence, order of acidity should be $\ce{H3O+<H3S+<H3Se+}$. But, in March, the reverse order is given.
What is the reason for this order given by March? And what is wrong in my logic?

The order is correct.

@Mithoron What's the reason behind that?


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Why is ortho-hydroxybenzoic acid more acidic than its para-isomer?

Why is ortho-hydroxybenzoic acid $(\mathrm{p}K_\mathrm{a} = 2.98)$ more acidic than its para-isomer $(\mathrm{p}K_\mathrm{a} = 4.58)$?

March's Advanced Organic Chemistry (7th ed) gives the reason to be intramolecular hydrogen bonding between the $\ce{OH}$ and $\ce{COO-}$ groups of the conjugate base.
However, note that intermolecular hydrogen bonding (between the $\ce{OH}$ and $\ce{COO-}$ of two different molecules) is possible in the para-isomer. Interestingly, this question also states that intermolecular hydrogen bonding is relatively stronger than intramolecular one. So, shouldn't the para-isomer be more acidic?

I don't think that in the first link it is stated that the intermolecular $\ce{H}$-bonding is stronger than the intramolecular one. On the contrary, the accepted answer starts as:
"I don't think there is much of a difference between the strengths of intramolecular or intermolecular hydrogen bonds."  With which I agree.
In that post they discuss the effect of intermolecular and intramolecular $\ce{H}$-bonding on the melting and boiling points of the compounds.
The answer by @AllYouCanEat86 that follows in the same post I think is more relevant to your question.
Therefore in our case there are intermolecular $\ce{H}$-bonds in p-hydroxybenzoic acid but they don't last for long due to the thermal motion and higher degrees of freedom of the $\ce{COO-}$ and $\ce{OH}$ groups. In contrast, in the ortho-isomer the $\ce{COO-}$ and $\ce{OH}$ groups are locked in close proximity due to the aromatic ring and therefore are forced to maintain the $\ce{H}$-bond for much longer. This could be thought as an equivalent to the chelation effect in metal complexes. Lastly, one could also argue that in the case of o-hydroxybenzoinc acid, there is both the option of intermolecular and intramolecular $\ce{H}$-bonding compared to the para where there is only intermolecular $\ce{H}$-bonding.

Also, usually the pKa of a substance is determined in dilute solution, isn't it?  So just due to dilution, there won't be many / any inter-molecular H bonds, but there can still be plenty of intra-molecular ones.

Just curious, does steric inhibition of resonance also play a role in increasing the acidic strength of Ortho?

The intramolecular hydrogen bond in ortho-hydroxybenzoic acid between the $\ce{COO-}$ and the -$\ce{OH}$ groups forms a six membered ring.
Since this structure is very stable, the deprotonation of the acid is favoured and the compound is more acidic.
In the case of the para-isomer, there is hydrogen bonding between different molecules which is much less stable. Therefore it's acidity is lower than that in the ortho-isomer.
In the link you have provided, they explain the lower melting and boiling points of the ortho-isomer. That is again because of the hydrogen bonding, but it does not give an indication is the relative strength of the intra and intermolecular types.
To melt a compound, the intermolecular forces that pack them together must be eliminated. In the ortho-isomer, an intramolecular hydrogen bond is formed.
But this bond does not maintain the molecules together. When the molecules are separated, this hydrogen bond goes with each molecule in which it occurs.
So it does not matter if this bond is strong or weak. It has not to be broken.
On the other hand, in the para-isomer, the intermolecular hydrogen bonds maintain the different molecules together, so more energy is required to separate them.
The hydrogen bonds must be broken, and as a consequence, the melting point is higher.
This is the case in ortho- and para-hydroxybenzoic acid as well. The former has a melting point of $\pu{158 ^\circ C}$ and the latter has $\pu{215 ^\circ C}$.

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In Finkelstein reaction, how does precipitate formation favor forward reaction?
In my textbook it's stated that:

$\ce{NaCl}$ and $\ce{NaBr}$ thus formed is precipitated in dry acetone. It facilitates the forward reaction according to Le Chatelier's principle.

Is this related to common ion effect in any way? I'm not familiar with Le Chatlier's principle. Please explain this in simple words as I'm in high school.
How does formation of $\ce{NaCl}$ and $\ce{NaBr}$ (both insoluble) prevent backward reaction from taking place? Does it mean $\ce{NaI}$ will be produced again?

If NaCl and NaBr are insoluble in the reaction medium then they are not available for the backward reaction because they are not in solution which is a prerequisite for reaction

@Waylander Explained it without Le Chatelier. Nice!

https://chemistry.stackexchange.com/questions/6947/why-are-nacl-and-nabr-not-soluble-whereas-nai-is-soluble-in-acetone-in-the-fin

$\ce{NaCl} $ and $\ce{NaBr }$formed as products are polar compounds which are vey less soluble in non- polar solvents like dry acetone, but $\ce{NaI } $due to high polarizability of $\ce {I^-} $ is soluble in acetone. So the products precipitate.  If a reaction is in equilibrium in constant temperature  and it is distorted somehow, it goes in the direction to nullify the effect of distortion.  This is the Le-Chattelier's principle.  If some of the products are removed to keep the $\ce{K_c}$ constant , some more reactants will react and convert into products. Thus constant removal of products by precipitation helps to dissociate reactants into products and thus favour the forward reaction.


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Book recommendation for studying chemistry through scientific method
I really like physics and I think the reason is that I enjoy studying the process of discovering rather than reading known facts. I've tried to study chemistry but I feel introductory texts on chemistry doesn't focus on the process of discovery and how one could reach certain conclusions after following such and such thought processes. Instead they state facts.
For instance, while studying chemistry the author would tell you that a water molecule is $\mathrm{H_2O}$ (two atoms of hydrogen and one of oxygen) but doesn't explain how early scientists reached that conclusion.
Can someone recommend any introductory textbook/book that'll teach chemistry that way?

You're looking for resources on the history of science.

@Zhe Not really. Motivation and line of reasoning to reach a certain conclusion need not be how it was discovered historically.

Line of reasoning is the most recent discovery.


text:
Pressure dependency in Haber Bosch ammonia synthesis
Haber-Bosch ammonia synthesis reaction:
 $$\ce{3H2 +N2 -> 2NH3}$$
According to the ideal gas law: $pV=nRT$ where constant volume implies $\displaystyle \frac nV=\frac p{RT}$
Then $\displaystyle K_\mathrm c=\frac{[\ce{NH3}]^{2}}{[\ce{H2}]^3\cdot [\ce{N2}]}$
When I substitute $\displaystyle \frac nV=c=\frac {p_{\text{gas}}}{RT}$ I get:
$$ K_\mathrm c=\frac{\left(\frac{p(NH_3)}{RT}\right)^2}{\left(\frac{p(H_2)}{RT}\right)^3\left(\frac{p(N_2)}{RT}\right)}=\frac{R^2T^2}{p^2}$$
But according to Le Chatelier's principle, increased pressure should result in an increased $K_\mathrm c$, but my $\displaystyle K_\mathrm c\approx \frac 1{p^2}$.
Where is the error? Is is impossible to combine the partial pressures like that because they scale differently with total pressure? Can one derive from the gas law the pressure dependency of this reaction?

Edit after solved:

First mistake was that the formula had $K_c$ (constant) instead of Q,
$$ Q=\frac{\left(\frac{p(NH_3)}{RT}\right)^2}{\left(\frac{p(H_2)}{RT}\right)^3\left(\frac{p(N_2)}{RT}\right)}=\frac{R^2T^2}{p^2}$$
where Q is the current value/ratio of concentrations, so the reaction tries to get Q back to $K_c$ so for greater pressure p, Q is smaller so the reaction goes in the direction of NH3...

How do you know that $p_{\ce{NH_3}}$, $p_{\ce{H_2} } $ and $p_{\ce{N_2}}$ are all same and all equals $p$. You can't just substitute same $p$ for all of them because according to Le-chatterlier's principle, the increase in total pressure of the system, distorts the equilibrium, but your $p$ is not at all total pressure. So, you can't just say anything from your result. By the formula, $$K_p = K_x (P)^{\Delta n}$$ you can say something. Here $P$ is the total pressure of the system. (This is because the formula is derived by using Dalton's law of partial pressure which says, $p_i = x_i\dot P$ where $p_i$ is the partial prssure of the $i^{th}$ gas, $x_i$ is its mole fraction and $P$ is the total pressure of  the system.)
Here $\Delta n = -2$, so ,$$K_p = K_x/P^2$$ If you now increase $p$ at constant temperature, $K_p$ remains constant as it depends only on temperature. So, $K_x$ also increases to make $K_p$ remain constant, but $$K_x = x^2_{\ce{NH3}}/x^3_{\ce{H2}}\cdot x_{\ce{N2}}=n^2_{\ce{NH3}}\dot N^2/n^3_{\ce{H2}}\cdot n_{\ce{N2}}$$ where, $N = n_\ce{NH3} + n_\ce{N2} + n_\ce{H2}$
$K_x$ increasing means that the moles of $\ce{NH3}$ increase while the moles of $\ce{H2}$ and $\ce{N2}$ decrease, which is the favour of forward reaction and as $$K_p = K_c (RT)^{\Delta n}$$ here $\Delta n = -2$ and the temperature doesn't change so $$K_c= K_p (RT)^2 -> K_c \propto K_p$$
As $K_p$ increases, $K_c$ also increases.

According to Le Chatelier's principle, a pressure increase should result in the equilibrium shifting to the right, but it should not - at least to first approximation - affect the equilibrium constant.
$$K = \frac{\ce{[NH3]}^2}{\ce{[H2]}^3 \ce{[N2]}}$$
Using partial pressures, this can be rewritten as
$$K = \left(\frac{RT}{P}\right)^2 \frac{y_{\ce{NH3}}^2}{y_{\ce{H2}}^3 y_{\ce{N2}}}$$
Then for $K$ to be constant, an increase in the denominator, given by the pressure for example, has to be followed by an increase in the numerator, which means the reaction working in the direction of formation of ammonia.

Oh I see, so if I have two containers where one has double the pressure of the other, then at the start of my reaction c(H2) and c(N2) are increased (aswell as c(NH3) but the total fraction of the momentarily reaction "fraction" is changed by 1/4 and so the system tries to get back to equilibium / to Kc by synthesising NH3. Thank you @Vinícius.

It is relatively easy to work out what happens in a gas phase reaction when the total pressure is changed by considering the degree of dissociation $\alpha$. $K_p$ may be defined in terms of free energies of the species in their standard states, i.e. 1 bar pressure thus it is independent of pressure.
The changes due to dissociation are shown under the equation for the Haber process
$$3H_2 + \;\;N_2 \leftrightharpoons \;\;2NH_3\\
3(1-\alpha) \;\; 1-\alpha \qquad 2\alpha$$
As $\displaystyle K_p= \frac{p_A^2}{p_H^3p_N}$, where subscript A is for ammonia. To work out the partial pressures $p_A$ etc. in terms of the total pressure $p$ and $\alpha$ recall that each partial pressure is the extent of dissociation over the total (in this case $3-3\alpha+1-\alpha+2\alpha=4-2\alpha=f$) which  are $\displaystyle p_A= \frac{2\alpha}{f}p, \; p_N=\frac{1-\alpha}{f}p,\; p_H=\frac{3-\alpha}{f}p$  which produces the equilibrium constant in terms of $\alpha$ as
$$K_p=\frac{4\alpha^2p^2 (4-2\alpha)^4}{(4-2\alpha)^2\cdot 3^3(1-\alpha)^3p^3(1-\alpha)p}= \frac{4\alpha^2(4-2\alpha)^2}{3^3(1-\alpha)^4}\frac{1}{p^2}$$
To find the pressure dependence, rearrange the equation as $\displaystyle p=\sqrt{ \frac{4\alpha^2(4-2\alpha)^2}{3^3(1-\alpha)^4K_p} }$ and workout what the pressure is at small and large $\alpha$. In this case when $\alpha =0,\; p=0$ so as pressure increases $\alpha$ increases as $\alpha$ cannot be negative. If you plot a graph this is clear also. (Incidentally it does not matter what value $K_p$ has as it only multiples the curve by a constant value). Thus in this case as pressure is increased the extent of dissociation increases, and more product is formed. le-Chatelier's principle is in accord with this conclusion.

Thank you for your answer @porphyrin , but why is the formula for the degree of dissociation alpha for $3H^2=3-a$ and not $3-3a$ or $3*(1-a)$? Degree of dissociation is an unknown topic for me, so I'm sorry if it is obvious... I found https://chemistry.stackexchange.com/questions/531/how-do-i-calculate-the-degree-of-dissociation-in-equilibrium and if I understand it correctely alpha is the fraction of (reacted substance/substance to start with (constant)). So it should be between 0 and 1...

Thank you for noticing my error. I have corrected the equations. The conclusions are unchanged.


text:
Why is telluric acid dibasic?
$\ce{Te(OH)6}$ has an octahedral geometry with $6$ $\ce{-OH}$ groups.

Is it a protonic acid (like $\ce{HCl}$) or a Lewis acid [like $\ce{B(OH)3}$]?
What makes $\ce{Te(OH)6}$ dibasic?

Related: https://chemistry.stackexchange.com/questions/43960/what-is-the-molecular-structure-of-polymetatelluric-acid

For further deprotonations, would it be that they are rarely observed since oxidation to a higher oxidation state for Te is easier?

Two may be as many protons that a Bronsted-Lowry base as strong as hydroxide ion can remove.  Building up more than two negative charges than that on the ion is a tough job.  Possibly a stronger base, such as amide ion in ammonia, could do that.
That the basicity of the medium could impact how many protons can dissociate is nicely demonstrated by citric acid.  Ordinarily it is triprotic in aqueous solution, but spectroscopic evidence indicates that at high alkali concentration (pH>14.4) a fourth, alcoholic proton can be removed.  There would be no doubt with a base such as amide in ammonia.

Possibly.  The point I was trying to make is there is no law on how many protons can come off.  It likely depends on conditions.

Oscar's answer covers the realistic part perfectly, that there is no exact law on how far deprotonation can occur; the exact answer would depend on the conditions. That said, however, if you're given that telluric acid is dibasic (probably the question assumes mild conditions), and are then asked to justify why, the following is a theoretical explanation.
Telluric acid has an octahedral geometry. The first two negative charges (on deprotonation) would be formed on the two oppositely directed alcohol groups (to minimize repulsion) and be far from each other.
The third negative charge instead would be much closer to either of the two existing negative charges. This would cause repulsion and destabilize the anion. Hence, telluric acid won't deprotonate a third hydroxyl group under mild conditions.


text:
Reaction of sulphur and nitric acid
I would like to ask a question about Sulfur in the reaction with Nitric Acid, $\ce{HNO3}$ to produce Sulfuric Acid, $\ce{H2SO4}$
I read that as Nitrogen and Sulfur are reduced and oxidised respectively, the reaction equation can be constructed as follows:
$\ce{S + 6HNO3 -> H2SO4 + 6NO2 + 2H2O}$
Now, when I read an article on wikipedia regarding sulfur, the introductory paragraph states the following:

Under normal conditions, sulfur atoms form cyclic octatomic molecules with a chemical formula $\ce{S8}$.

So why is sulfur represented in the equation as $\ce{S}$ and not $\ce{S8}$?

Because that doesn't make much of a difference. Sure, you may multiply everything by 8, if you'd like.

Just call $\ce{S}$ an empirical formula.  Empirical formulas work perfectly well in chemical reaction equations.

See this: https://chemistry.stackexchange.com/questions/8438/reaction-between-zinc-and-sulfur

The most stable allotropic form of sulfur exists in the formula of S8. In other words, sulfur occurs naturally in S8 formula. Multiply all the stoicheomtric coefficient by 8 (other than S) to obtain the balanced chemical equation

Balance the equation by taking S8.
Yes, it is true that naturally, sulfur does exist in S8 (crown sulfur) form.


text:
Bond angles in a oxonium cation
What are the bond angles in protonated alcohol? The bonds in the carbon chain are presumably unaffected by the $\ce{-O^{+}H2}$ group and so in their usual tetrahedral arrangement; what are the angles $\ce{ROH}$ and $\ce{HOH}$?

In $\ce{NH3}$ there are 4 substituents around the nitrogen atom (3 hydrogens plus 1 lone pair); what are the approximate angles? Analogous to methane, also with 4 substituents, we'd estimate ~109°.  In $\ce{H2O}$ there are again 4 substituents (2 substituents and 2 lone pairs); what are the approximate bond angle here (see this earlier answer for more detail)? ...continued

continued... In $\ce{R-O^{+}H2}$ there are again 4 substituents (1 alkyl group, 2 hydrogens and 1 lone pair) around the oxygen atom.  Based on the previous examples, what would you estimate the bond angles to be in this case?

Thank you for the edits and answer. I understand the principles behind VSEPR, but not the mathematics. How would I calculate an exact angle? Presumably it, like ammonium, is a little less than the standard tetrahedral 109.47.

You can search Chem SE for "Coulson's Theorem" which relates angle and hybridization.  If you know one, then you can calculate the other.


text:
Amount dry acid or sodium bisulphate needed for varying hardness values
My question is regarding the use of sodium bisulphate to lower the hardness of water. I am attempting to come up with a label to let consumers know how much product (dry acid or sodium bisulphate) is needed for varying hardness values.
Using the equation $$\ce{CaCO3 + 2NaHSO4 -> Na2SO4 + CaSO4 + CO2 + H2O}$$I have found the following ratios:
$\pu{120 mg/L}$ $\ce{CaCO3}$ requires $\pu{0.288g}$ $\ce{NaHSO4}$,
$\pu{180 mg/L}$ $\ce{ CaCO3 }$ requires $\pu{0.432g}$ $\ce{NaHSO4}$,
etc.
My question: How much calcium will react with bisulfate? And how can I use this information to determine the change in hardness? Will starting pH of the water affect the calculation?

Why is this on hold? This is not a homework problem, this is a valid softening question. @Mithoron

You are in equivalents already with 120 mg/L as CaCO3 - that's your hardness and the amount you need to remove. The EW of SB is MW/2 or 60 g/eq. The equivalent weight of CaCO3 is 50. Thus, for 120 mg/L as CaCO3 of hardness, you will need 120*(60/50) of SB or 144 mg/L. OR a better way to look at it, you need 1.2 times the SB (mg/L) that you have in hardness (mg/L CaCO3).
So, if you add 288 mg/L, you should have zero hardness left because it's ~2x the stoichiometric requirement. The pH will drop and hinder solubility, which may be where the 2x comes from. I'll ans. better when unlocked.


text:
How can we separate carbon monoxide from wood gas?
I am thinking about using open source wiki designs from allpowerlabs to construct a wood gas generator.  The motivation came from experiencing a 10 day power outage after super storm Sandy.  The utility company could not provide power due to damaged transformers, power lines, and other grid infrastructure.  Most of the gasoline service stations did not have backup power to run the pumps and meters and those that did had huge queues to wait through.
Meanwhile there was tons of biomass available in my 2 acre yard.  I used some of the wood in our wood stove for heating and some cooking but did not have a way of getting the energy stored in wood and other biomass to electrical energy.
The solution is a wood gas generator that converts the stored solar energy in wood to a gas that can be combusted in a hacked generator to provide electrical power.  Allpowerlabs.com has a turn key solution all ready to go but it is beyond my budget, but they offer the open source build it yourself GEK (Gasifier Electricity Kit), mentioned earlier, that I would like to build.
Instead of using the wood gas to run a hacked Briggs and Stratton  generator, I would prefer to route it to a BIY (Build It Yourself) fuel cell.  Wood gas consists of hydrogen, carbon monoxide, carbon dioxide, water vapor, and small amounts of other.  For fuel considerations hydrogen(H2) and carbon monoxide(CO) are usable.
While molten carbonate and solid oxide fuel cells can use CO, the operating temperatures are too high to be practical and we want to sequester the carbon instead of releasing to atmosphere.  I would like to separate the carbon oxides from the wood gas and use the H2 directly into a fuel cell or store it for later use.
I know that we can remove CO2 from the wood gas or exhaust with low cost or Brew it yourself sodium hydroxide (NaOH), via caustic scrubbing.
My question is: How can we cost effectively separate the CO from the wood gas?  I am not sure how NaOH reacts with hydrogen gas and CO.

Have you considered using a Stirling engine to convert some of the heat produced from your stove into electrical energy.  That might be a lot more feasible.

This master's thesis by Dahiru Rufai Ahmed at the university of Oulu provides a comprehensive background into the removal of CO2 from wood gas.
On pages 36-37 it mentions the water-gas shift (WGS) reaction as a method of adjusting the hydrogen/carbon monoxide mixture or removing CO entirely to get pure H2.  "The WGS reaction is an exothermic reaction
(∆Hr = -41.1 kJ/mol) that converts excess carbon monoxide into hydrogen."
CO + H2O -->  CO2 + H2 + heat
"At low temperature the reaction can be catalyzed by catalysts based on copper, aluminum and zinc oxides. (Basu, P. (2010). Biomass gasification and pyrolysis: Practical Design and Theory. 1st ed., Elsevier Inc. 365pp. ISBN: 9780123749888.)
However, in recent times, interest has been shifted to a catalyst-membrane system that incorporates the WGS reaction and H2 separation via a hydrogen selective membrane (Mendes, D., Chibante, V., Zheng, J-M., Tosti, S., Borgognoni, F., Memdes, A. and Madeira, L. M. (2010). Enhancing the production of hydrogen via water-gas shift reaction using Pd-based membrane reactors. Inter J Hydrogen Energy. 35 (22), 12596–12608.)."
This article describes a membrane based catalyst for the WGS reaction.  The catalyst in the membrane uses Palladium alloys with Silver and Copper due to their higher selectivity and permeability to Hydrogen.
Since Palladium is an expensive noble metal and since catalytic converters for automobiles already use Palladium for CO oxidation, I am considering using an automobile catalytic converter (CAT) for this purpose.  The CAT uses oxidation rather than WGS to convert CO, so I am concerned that it would also oxidize the hydrogen.  However, I think the CAT would perform WGS if the input gas mixture is lacking oxygen which would be the case for wood gas.  Any useful comments and feedback appreciated.

The problem is that the water gas shift reaction requires steam and does not have a strong driving force, you would need to provide a source of steam. You also need to scrub the CO2 out and then you will need some methanation catalyst to convert any residual CO to methane which is less able to poison the Pt containing fuel cells. I think it would be better for you to choose a more simple electric power production system than a Pt based fuel cell system plus high purity H2 production from wood gas.


text:
Deciding more electrophilic centre between ester and thioester for SN type reaction

My question: (though it won't alter the product)

Which carbon is attacked first, the one adjacent to sulphur or the one adjacent to oxygen?

The solution says that the carbon adjacent to sulphur is attacked first. I don't understand why.
According to me, sulphur is less electron withdrawing than oxygen (electronegativity decreases down the group), so the carbon adjacent to oxygen must be more electrophilic.
Could someone please explain as to why attack should first occur at the carbon adjacent to the sulphur atom? (or is the solution incorrect?)

P.S. Here, due to syn stereochemistry (NGP is not favourable) choice of electrophilic centre undergoing attack first doesn't influence the product at all. However, in the case where the sulphur and oxygen atoms are anti with respect to the ring, and NGP is a possibility, and then the products would greatly differ (depending on whether attack first happens on carbon adjacent to oxygen or sulphur)

Source: MS Chouhan; Advanced Problems In Organic Chemistry; 11th ed; Q35 in Chapter Alcohols, Ether, and Epoxides

There are several questions to ask here.
The first is why does it matter?
The more technical answer involves first asking what the reaction mechanism is. Cleavage of both the thioester and the ester involve addition of hydroxide to the carbonyl and collapse of the tetrahedral intermediate. What step is the rate determining step? Your answer here impacts what factors could be relevant to the this question.

There is also another effect of delocalisation of the electron pair on $\ce{O}$ and $\ce{S}$ atom adjacent to the cabonyl carbon, which is dominant over the electron-withdrawing effect ($\ce{-I }$ effect).
In both the parts i.e. $\ce{OCOCH_3}$ and $\ce{SCOCH_3}$, there are two lone pairs on each of $\ce{O}$ and $\ce{S}$, which can delocalise with the carbonyl carbon. Thus, the bond between $\ce{O}$ and $\ce{C}$, and $\ce{S}$ and $\ce{C}$, will get a partial double bond character. Due to this delocalisation, the electrophilicity of the carbonyl carbon is decreased.
When the delocalisation happens in case of $\ce{O}$, the overlap between  $\ce{O}$ and $\ce{C}$ is $\ce{2p$\pi$ - 2p$\pi$}$ overlap, which is a strong overlap and thus the double bond character between $\ce{O}$ and $\ce{C}$ is more. Thus, electrophilicity is reduced to a great extent. But in case of $\ce{S}$, there is a relatively weaker $\ce{3p$\pi$ -2p$\pi$}$ overlap which introduces very feeble double bond character between them so, the electrophilicity is reduced to very less extent and thus the first electrophilic attack will happen at the $\ce{C}$ adjacent to $\ce{S}$. This is the reason to your question.

While I agree with your answer, how would you account for the fact that, since the $\ce{S}$ atom is less electronegative than the $\ce{O}$ atom, it should should donate its lone pair to a greater extent than oxygen (contrary to your answer)?

Since the ovelap with $\ce {O}$ is very good, its resonance hybrid's  stabilisation energy will be very high and thus a very large fraction of  the molecules of  compound will undergo that stabilisation because by that the overall $\Delta $H will be negative even if we consider the low positive  $\Delta$H of the lone pair of oxygen undergoing delocalisation.


text:
Electrolysis of Potassium Iodate/Hydrogen Peroxide in Briggs-Rauscher Oscillation
I would like to investigate the effect of adding a voltage in the Briggs-Rauscher oscillation reaction. I will examine the oscillations visually.
The ingredients I will use include potassium iodate, sulfuric acid, acetone, manganese sulfate, starch indicator and hydrogen peroxide.
There are a number of complex reactions which have not to this day been completely analysed. Hence I will not attempt to explain all of them.
To the point. What electrodes should I use for electrolysing potassium iodate? Just like the saltwater circuit, would electrolysis will proceed as iodine carries charges towards the aluminium anode, and as potassium obtains them in the copper cathode? Could I electrolyse hydrogen peroxide alternatively?
I would like to examine reduction potentials, comparing them to what extent the electrolysis of potassium iodide prevents the behavior of potassium iodide in the other equilibria, as possibly seen visually.
Is it also possible for the manganese metal to conduct with copper just as potassium might?

The experiment has been attempted. Electrolysis through the use of copper electrodes in fact "kills" the reaction. This means that the whole system is immediately turned dark-blue, and eventually made more clearer, and eventually light blue indicating the presence of copper ions.
It is suspected that not only is the sulphate ion electrolysed, but perhaps the copper iodizes the potassium and other compounds.


text:
Finding density of the gas mixture

A gas in a vessel is under pressure of $\pu{1800kPa}$. The design temperature of the tank is $\pu{423K}$. The gas consists of (by volume) $20\ \%$ of $\ce{CH4}$ and $80\ \%$ $\ce{N2}$. Estimate the density (in $\pu{kg/m^3}$) of the gas.

I can't just use the ideal gas law to find out the density of the mixture as there are two components, right?
$$\rho=\frac{pM_\mathrm r}{RT}$$
From the ideal gas law, I can also predict that vol% equals mol%. But with this, am I supposed to find the average value of $M_\mathrm r$ and then substitute it into the equation above?

Hi, welcome to Chem.SE! Is the last line actually "...of the gas mixture" in your question?

@GaurangTandon nope.

Yes.  What is the molar average molecular weight?  (Hint:  when we say 20% by volume, we mean that the mole fraction is 0.2)

You originally said that:

The gas consists of 20vol%,CH4 and 80% N2...

This is a fault in the question. A gas cannot consist of two other gases and still remain a gas. Instead, it is now a gas mixture. (hence I had added the term "mixture" into your question, which I've now removed to retain the original meaning)
Once this is out of the way, you can think of density as: $$\begin{align}
\text{density}&=\frac{\text{total mass}}{\text{total volume}}\\\\\
&=\frac{n_1M_1+n_2M_2}{(n_1+n_2)RT/P}\\\\
&=\frac{P(n_1M_1+n_2M_2)}{(n_1+n_2)RT}\\\\
&=\frac{PM_{\text{avg}}}{RT}
\end{align}$$
and recall that volume percent is equivalent to mole percent, as you correctly did already. Substitute the correct values into this equation and you're good to go.
PS: I do not understand what you're denoting by $M_\mathrm r$. But, the formula you were using is identical to mine if you swap $M_\mathrm r$ with $M_\text{avg}$.

I think there is a mistake in 2nd equation, where you took $P$. You should have taken individual pressure of the gases. Hence the equation should be $\frac{n_1M_1+n_2M_3}{RT/P}$.

What are n1 and n2 in your equation? Is it would be a molar fraction which is equivalent to volume fraction and hence it should not be a mass fraction, right?


text:
Is it absolutely necessary to use centrifuge to separate ethylene glycol from precipitate?
I've been trying to synthesize intermetallics of $\ce{XY}$ type using transition metal salts and main group element salts. For my system, EG, DEG, and TEG are good solvents. However, because of them being highly viscous, I've experienced normal drying doesn't do the trick and that the precipitate remains puffy even after drying (which shouldn't be the case) before sintering but by using centrifugation, I've been successful at removing the puffy texture of the precipitate.
Recently, I came upon this paper where they haven't used centrifugation for the main sample. Could anyone please explain how does this work?

What they do in the paper is what i was going to suggest even before reading it: Addition of ethanol in the ethylene glycol suspension lowers the viscosity of the solution making it more filterable. Since their nanoparticles are insoluble in ethanol they can add as much is necessary. After filtering the solid out they wash with plenty of ethanol to remove the leftover ethylene glycol (which since it has a high boiling point cant be dried easily). Hence they end up with their product contaminated with ethanol left over from the washings. However, ethanol is very easy to get rid off by applying vacuum to the powder due to being reasonably volatile. So the ethanol contaminated solid can be dried in a schlenk line or even a rotary evaporator overnight to remove any ethanol left. Note that you can do this not only with ethanol but any solvet that is volatile (b.p up to 110 deg C, you can do more but is a pain), miscible with the solvent you are trying to get rid of, and wont react with the compound that is being isolated. Therefore the answer to your question is: no, in general it is not necessary. However, if your precipitate is too fine for the filter you have available, and you cant get it to crash out as larger particles or if it is a very small amount and you cant scale it up you might actually need to use one.


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Combustion at 298K
Question: (source IITJEE 1992)

An aliphatic thiol undergoes combustion at 298K to give carbon dioxide, water and sulfur dioxide. What are the physical states of the products?(solid/liquid/gas)

I am not able to understand what combustion at 298K means, do they mean the temperature of all reactants and products is 298K ?  In that case, I think the answer would be gas, liquid, gas respectively.
But the answer given is all are gases. Do they mean the room temperature is 298K and the temperature at actual site of the reaction is higher? Is it because it is not possible for combustion to take place at room temperature?

Your answer is correct. The answer given in my copy of 38 years’ IIT-JEE chemistry has the answer as
$$\ce{CO2 (g), H2O (l), and SO2 (g)}$$
The answer given in your book might be wrong. I don’t think iit released any official answer keys in the subjective days of jee so there may be discrepancies in the answers in different sources.


text:
Why Bond Angle is different?
H2O has two lone pair electron and H2S also have two lone pair electron but Bond Angle in H2O is 104.5° but H2S has 92.12°bond Angle .
Both H2O and H2S has same hybridization ,which is sp3 and i know that the four hybrid orbitals of sp3 hybridization are directed towards the four corner of a regular tetrahedron making Angle of 109.5° .
My teacher taught me that for every lone pair you have to subtract 2.5° from 109.5° so if that's the case then both of them should have same Angle as both of them have 2 lone pair electron ?
If that's not the case then what is the exact way to determine Bond Angle ? And the way my teacher taught me is it ok ?

Again Bent's Rule does the job.

In H20 it is sp3 hybridisation, but in H2S it is purely p3 orbital , there is no hybridization in H2S, it cannot afford the cost of hybridization you may recall that all the p orbitals I.e. Px,Py, Pz are 90 degree to each other as along x y and z axis respectively and if u compare this with the Bond angle of H2S which is 92.12 degree( almost 90 degree) which clearly shows that it is formed by p orbitals only, hybridization is a costly process , when sulfur can easily bond with p orbitals itself , why will it go for expensive process of hybridization unnecessarily isn't it?

Hybridisation is a purely mathematical model; energy cannot be gained nor lost due to it. This is wrong.


text:
Factors that influence the kinetics of an irreversible exothermic reaction
I'm trying to understand what are the main factors that influence the reaction rate of an irreversible exothermic reaction.
I think these could be the main factors:

chemical nature of the reagents and their concentration;
nature of the solution (homogeneous or heterogeneous)
presence of catalysts
temperature.

(Let me know if I missed some factors or added one or two that had nothing to do with what I'm looking for.)
But I struggle understanding how the temperature influences an exothermic reaction. I can't use Le Chatelier's principle since the reaction is not reversible, so how do I know what is the effect of (for example) an increased temperature on such reaction?
I'm prone to think that an increased temperature will bring more products, but then again I'm not so sure since is an irreversible reaction.

I assume by kinetics you mean the rate of the reaction (i.e. how fast it happens), in which case the factors you mention are fine.
Regarding the temperature: the rate constant and therefore the reaction rate at a given concentration always increases with temperature as you can see in the Arrhenius equation. The fact the the reaction is exothermic means that it releases heat which, in the case where no adequate temperature control is applied (such as an ice bath), will increase the temperature of the solution and therefore the rate of the reaction releasing even more heat and so on, resulting in a runaway exotherm. This is the reason why exothermic reactions must be well temperature controlled especially when performed in large scale.
If your reaction was reversible, then heating it would not only accelerate the reaction (both forwards and backwards at equilibrium) but also shift the equilibrium towards the reactants.


text:
Why do tungsten and chromium have different electronic configurations?
Why is the outer structure of tungsten $\mathrm{5d^46s^2}$ instead of $\mathrm{5d^56s^1}$?
I try to explain this in comparison  with outer structure  of chromium $\mathrm{4d^55s^1}$  taking into consideration  5d orbitals have lesser energy than 6s orbitals in tungsten by saying that the size of the energy gap between the 3d and 4s orbitals lesser than size of the energy gap between the 5d and 6s orbitals, and the repulsion between electrons in the structure$\mathrm{5d^46s^2}$  more than  the repulsion  in the structure $\mathrm{5d^56s^1}$ which can push the electron from 5d to 6s to produce the lowest energy structure.
I need qualitative answer?

also https://chemistry.stackexchange.com/questions/41304/why-tungsten-does-not-have-stable-configuration

Cr is 3d5 4s1, not 4d5 5s1.

@orthocresol thank you i  edit and  correct  the structure of Cr.

Firstly note that the energetic differences between $\mathrm d^m\mathrm s^n$ and $\mathrm d^{m-1}\mathrm s^{n+1}$ configurations are generally not large. Sometimes these are not amenable to simple qualitative rationalisation.
However in this case it is possible to argue along two lines.

Relativistic effects lead to the stabilisation of the 6s subshell and destabilisation of the 5d subshell, which directly favours population of 6s over 5d.
Electron-electron repulsions in the 3d/4s orbitals are generally larger than repulsions in the 5d/6s orbitals, because the latter are more diffuse. Hence chromium chooses to adopt a configuration in which exchange energy is maximised (i.e. electron-electron repulsions are minimised).

The energetic differences between 3d and 4s  lesser  or larger than  energetic differences between 5d and 6s   ?

Electron-electron repulsions in the$\mathrm{5d^46s^2}$ structure are larger  or  lesser than repulsions in the $\mathrm{5d^56s^1}$ structure?


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Acidity order of 4-halophenols
How do I theoretically predict the acidity order of the following para-halophenols?
(experimental $\mathrm{p}K\mathrm{a}$ values beside each)

parafluorophenol (9.92)
parachlorophenol (9.41)
parabromophenol (9.17)
paraiodophenol (9.30)

Experimental acidic strength order: parabromophenol > paraiodophenol  > parachlorophenol  > parafluorophenol
At first, I thought of simply using the -I effect to determine the stability of these phenol derivatives. Since -I effect is shown best by fluoro group, therefore charge delocalisation would be best in the fluoro derivative's anion. But then at the same time, there's +M effect of fluoro group too. This answer emphasises on the dominance of inductive effect of chloro group over its mesomeric effect. Is this applicable to fluorine too?
I also thought of intermolecular hydrogen bonding that would be strongest in parafluorophenol decreasing its acidity.
Quora answers here associate the higher acidity of 4-chlorophenol to the presence of vacant d- orbitals in chlorine atom, but "d-orbitals'" usage by a third period element is debatable.

https://chemistry.stackexchange.com/questions/76611/relative-acidity-of-halo-substituted-benzoic-acids-at-para-position

I think that, this acidity order can be justified by considering a combined effect of +R effect, -I effect and intermolecular hydrogen bonding (where it is dominant).
For 4-fluorophenol, the +R effect of $\ce F$ is significant but weaker than -I effect. The electron pair on the $\ce{2p}$ orbitals of $\ce F$ can easily delocalise with the benzene ring because of the stronger $\ce{2p\pi -2p\pi}$ overlap, which makes $\ce {+R} $ effect significant . That's why it increases the electron density on the carbon near the $\ce{-OH}$ group to such an extent which decreases its ability to donate protons, greatly decreasing
its acidity.
Considering intermolecular hydrogen bonding also, the acidity of the compound decreases further due to strongest intermolecular hydrogen bonding between the halogens.
For 4-chlorophenol, the +R effect is negligible due to much weaker $\ce{3p\pi -2p\pi}$ overlap and thus the $\ce{-I}$ effect becomes more prominent. But $\ce {Cl}$ is electronegative enough to form strong intermolecular hydrogen bonding($\ce{HO-C6H4-Cl...H-O-C6H4-Cl...)}$. That's why, it becomes difficult to remove hydrogen atoms from $\ce{-OH}$ group, decreasing its acidity.
For 4-bromophenol and 4-iodophenol, there is no strong hydrogen bonding due to much less electronegative $\ce{Br}$ and $\ce{I}$ atoms, and also +R effect is ineffective due to very weak $\ce{4p\pi -2p\pi}$ and $\ce{5p\pi -2p\pi}$ overlap. Only -I effect is acing for these two compounds in favour of acidity. As -I effect of $\ce{Br}$ is larger than $\ce{I}$, the acidity of p-bromophenol is greater than the p-iodophenol.

What about HOMO/LUMO iodine as we go down the group the HOMO of the halogen increases in energy, thus decreasing the energy gap. Of course there is something wrong with my reasoning as the pKa values will not match the order through HOMO/LUMO,, but what's wrong ?

@Avatar Shiny,Can you please write your logic properly in the comment box. I am not getting it properly what you are thinking about HOMO-LUMO .

@Soumik what I mean is that, as we go down the the group the halogens become softer their HOMO's are less strongly attracted down the group this leads to an increase in the HOMO's energy . For a good molecular orbital to form the HOMO/LUMO gap must be small. LUMO's are high in energy as they are unoccupied. So a high energy HOMO helps in reducing the gap . So by HOMO/LUMO we are getting an opposite trend to the observed acidity order of $\pu{+R}$ effect . So how is it that while seeing the nucleophilicity order we see iodene's HOMO energy and ignore the size of orbitals and over here

We are supposed to see the size of orbitals

@AvatarShiny, See, Nucleophilicity and the $\ce +R$ effect are different cases. because in Nucleophilicity , the bond has not been formed and it is going to be formed.So, the electron is going to be placed in a $\sigma$ -bonding molecular orbital. But in case of $+R$ effect, there is already a $\sigma$-bonding molecular orbital and there is a new $\pi$ bonding molecular orbital is going to get formed.But due to huge energy difference in 5p and 2p, the energy of the $\pi$ bonding MO is not very decreased i.e. stability is much less.I hope, that might distinguish Nucleophilicity and $+R$ effect.

@Soumik I didn't get that but I will ask a question on this by tomorrow . Maybe you will be able to explain there better. Thanks for your time!

I effect is stronger than R effect for all halogens. Therefore all halogens are deactivating.


text:
Are there any compounds with these properties?
I'm looking for a chemical compound that satisfies the following criteria:

Solid at room temperature
Decomposes into gases, primarily oxygen when supplied with an electrical current, preferrably exothermically.
Kinetically stable

Any ideas?

Will potassium ozonide do?

@mcchucklezz Sorry. I withdrew the comment. Thought this was a duplicate.


text:
Why are most atoms solids or gases but not liquids at room temperature?
Of all the elements of the periodic table, "most" elements with valences of four or less are solids, with most of these being either "metals" or "semiconductors."
On the right hand side of the periodic table, most of the elements are gases at room temperature.
There are only a "handful" of elements that are liquids (at room temperature). The one I can think of is mercury, and maybe there are one or two others that I have missed. Is there a reason why there are so few?
The reason I'm asking is because I expect the qualities of elements to fall on a spectrum. 1)At low temperatures, I expect a lot of solids. 2) At high temperatures, I expect a lot of liquids or gases. At room temperature, most items are solids (case 1). A fair number are gases (case 2). But few are the "intermediate" (liquid) case. Why is that?
The gist of the other question is, what qualifications cause an item to be a solid, liquid or gas at room temperature. My question is, why are these "qualifications" distributed as they are, in a seemingly "bipolar" manner? It would be like asking, why at a certain university, most GPA's are around 3.0, the second most around 4.0, but there are very few 3.5s (the intermediate case).

related https://chemistry.stackexchange.com/questions/30976/what-elements-and-or-substances-without-water-are-liquid-at-room-temperature https://chemistry.stackexchange.com/questions/65416/is-there-any-substance-that-is-liquid-at-room-temperature-and-pressure-that-does

@Mithoron How does that answer the question "Why are most atoms solids or gases but not liquids at room temperature?" Can you quote a specific part from the accepted answer you linked to that answers this question.

My question was edited to show why it is not a duplicate.

Like you said, the qualities of elements fall on a spectrum. And the spectrum is simple: a huge field for solid from minus infinity to certain temperature, then a narrow field for liquid (sometimes missing altogether), then another broad field for gas, from certain temperature to plus infinity. Where would you end up?

sigh And maybe you'll ask why is our "room temperature" typically defined as 25 °C? Or maybe why are metals high melting? Or why are there so many of metals? How do you imagine your desired answer really?


text:
Substitution of a cyclic diether by a halogen

I’m in an organic synthesis class and I have to draw a mechanism for the reaction in the picture. At first I thought it was just forming a free radical but I haven’t been able to find any way to do that without UV light, which isn’t shown anywhere in the given reaction. How do I go about solving this?

Try protonating oxygen, break C-O bond, and form an enol ether. Go from there for bromination.

What is the source of this question?


text:
Two different methods to express the equilibrium constant give different results

$\pu{2.0M}$ of $\ce{I2}$ and $\pu{2.0M}$ of $\ce{Br2}$ are initially present in a reaction mixture. Calculate the equilibrium concentration of $[\ce{I2}]$ , $[\ce{Br2}]$ and $[\ce{IBr}]$ at $\pu{425K}$. $K_c=100$
Equation: $\ce{I2 + Br2 -> 2IBr }$

1st method

Equation:$\ce{I2 \ \ \ + \ \ \ Br2 ->\ \ \ \ 2IBr}$
Initial: $\ \ \pu{2.0M}\ \ \ \ \ \ \pu{2.0M} \ \ \ \ \ \ \ \ \ \ 0$
Change:   $\ \ x \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x \ \ \ \ \ \ \ \ \ \ \ \ \ 2x$
Equilib:     $\ \ 2-x \ \ \ \ \ \ \ 2-x \ \ \ \ \ \ \ \ \ \ 2x$
$K_\mathrm c =\frac{(2x)^2}{(2-x)^2}=100$

2nd method what if I separate $2\ce{IBr}$

Equation: $\ce{I2 \ \ \ + \ \ \ Br2 ->\ \ \ IBr \ \ + \ \ \ IBr }$
Initial:     $\pu{2.0M}\ \ \ \ \ \ \ \pu{2.0M}\ \ \ \ \ \ \ \ \ \ \ \ 0 \ \ \ \ \ \ \ \ \ \ \ \ 0$
Change:    $\ \ \ x \ \ \ \ \ \ \ \ \ \ \ \ \ x \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x \ \ \ \ \ \ \ \ \ \ \ \ \ x$
Equilib:       $2-x \ \ \ \ \ \ 2-x \ \ \ \ \ \ \ \ \ \ \ \ \ \ x \ \ \ \ \ \ \ \ \ \ \ \ x $
$K_\mathrm c =\frac{(x)(x)}{(2-x)^2}=100$
So which one is actually right and which one is wrong? Why?
The idea to use the second method came to my mind because of the the definition:

The equilibrium constant is a ratio of the concentration of the products to the concentration of the reactants

If $\ce{aA +bB \ —> cC + dD}$ (1)
$K_\mathrm c = \frac{[\ce{C}]^c.[\ce{D}]^d}{[\ce{A}]^a.[\ce{B}]^b}$
If I separate : $a\ce{A} = \ce{A}+\ce{A}+...+\ce{A}$ ($\ce{A} a$ times )
Same thing with $b\ce{B}, c\ce{C}$ and $d\ce{D}$, the equation (1) becomes:
$(\ce{A}+\cdots+\ce{A})+(\ce{B}+\cdots+\ce{B}) \ce{—>}(\ce{C}+\cdots+\ce{C})+(\ce{D}+\cdots+\ce{D})$
$K_\mathrm c = \frac{([\ce{C}]\cdots[\ce{C}])([\ce{D}]\cdots[\ce{D}])}{([\ce{A}]\cdots[\ce{A}])([\ce{B}]\cdots[\ce{B}])}$=$ \frac{[\ce{C}]^c.[\ce{D}]^d}{[\ce{A}]^a.[\ce{B}]^b}$ (Proved)
So, by using the second method, how and why is it not right ?

Both methods are right and both are the same, except that you are not using the 2nd one right. Just where did those little $x^2$'s come from?

Oh it’s actually (x)(x)  ! Am confused

No, it's actually (2x)(2x), because the concentration of IBr is 2x.

What I think is 2x is the concentration of 2IBr ( 2 molecules of IBr ) not just IBr because in the “ change “ section we already balance the molecule.

There is no such thing as concentration of 2IBr.

So which one is actually right and which one is wrong? Why?

The second method is wrong.  In your second method, you've written: $\ce{I2 \+ Br2 ->IBr  + IBr }$ and then assigned the concentration $x$ and $x$ to each of the two (identical!) $\ce{IBr}$'s. Till now, it's fine.
The $K_\mathrm c$ has been used by you as:
$$K_\mathrm c=\frac{[\ce{IBr}][\ce{IBr}]}{[\ce{I2}][\ce{Br2}]}$$
which is okay but a better way to write it would have been
$$K_\mathrm c=\frac{[\ce{IBr}]^2}{[\ce{I2}][\ce{Br2}]}$$
as you've taken the single concentration term for $\ce{IBr}$ (=$2x$) from the correct formula, split it up into two parts ($x$ and $x$), and then used those two parts separately in $K_\mathrm c$, as if there were not one but two distinct $\ce{IBr}$'s being formed! It is just confusing.
Anyways, the first formula will still work. The main problem here is that $[\ce{IBr}]=\color{blue}{\text{total }}\text{concentration of}~\ce{IBr}$. When you put any reactant/product inside the square brackets, you're supposed to write their total concentration, even if they have been split up in parts. So, when in the derivation of $K_\mathrm c$, you write:
$$K_\mathrm c = \frac{([\ce{C}]\cdots[\ce{C}])([\ce{D}]\cdots[\ce{D}])}{([\ce{A}]\cdots[\ce{A}])([\ce{B}]\cdots[\ce{B}])}= \frac{[\ce{C}]^c.[\ce{D}]^d}{[\ce{A}]^a.[\ce{B}]^b}$$
for the reaction: $$\ce{aA + bB —> cC + dD}$$
each of those individual $[\ce{C}]$'s will refer to the total concentration of $\ce{C}$. That means, as $\ce{C}$ was separately written $c$ times by you in the reaction, the final $[\ce{C}]$ that you use will be the sum of all those $c$ $[\ce{C}]$s.
So, in the end of your second method, while applying the correct formula for $K_\mathrm c$, you must add those $x$ and $x$ together, and then write:
$$K_\mathrm c=\frac{(x+x)\times(x+x)}{(2-x)\times(2-x)}$$
into the formula you were using.


text:
Atomic and molecular orbital
I have read in my book that ,energy of atomic orbital is greater than that of molecular orbital . I know the fact that lower energy corresponds to higher stability .
But I can't figure out the reason of why does atomic orbital possess higher energy ?

See Linear Combination Of atomic Orbitals


text:
Why is acetyl cyanide not obtained from acetyl chloride?
Why doesn't acetyl chloride or any other aliphatic acyl chloride react with cyanide ions and undergo nucleophilic addition elimination to form aliphatic acyl cyanide, even though this reaction occurs for aromatic acyl chloride?
This was the related question in my workbook:

Which of the following compounds will react with ethanolic $\ce{KCN}$?
  A) ethyl chloride B) acetyl chloride C) chlorobenzene D) benzaldehyde

My answer is A, B, D undergoing SN2, nucleophilic addition-elimination, and benzoin condensation respectively.
Answer is given as only A,D.
Searching the internet led me to this article, but I am not able to understand why.

Are you sure about Chlorobenzene reacting with cyanide under these conditions?

Acetyl chloride in ethanolic KCN will give ethyl acetate

@Waylander Oh I wanted to type B. Thanks for pointing out.

@Waylander But isn't CN- a stronger nucleophile than ethanol?

It is - but if acetyl cyanide is formed at all under the reaction conditions that just makes it more reactive than acetyl chloride (better leaving group) so it forms ethyl acetate even more readily.

@Waylander Thanks. That helps me understand a lot. I think you can post as answer

@Waylander Isn't Cl- better leaving group than CN-

Depends on what they are are attached to. When it comes to acyl cyanides/chlorides the cyanides are more reactive.

Acetyl chloride on reacting with $\ce{KCN}$ in ethanol will give ethyl acetate. If acetyl cyanide forms at all, it will immediately react with the ethanol solvent as it is a more reactive species than acetyl chloride.
Acetyl cyanide is known and may be formed by catalytic reaction of hydrogen cyanide and ketene at elevated temperature. Details.
Aromatic acyl cyanides are less reactive than alkyl acyl cyanides much as aromatic acyl chlorides are less reactive. They may be formed by reaction of the neat aromatic acyl chloride with cuprous cyanide (details) and are stable to a wash with cold aq. sodium bicarbonate.

See the last step in this mechanism where an acyl cyanide reacts with water. https://chemistry.stackexchange.com/questions/86061/what-is-the-mechanism-of-the-reaction-between-hexane-2-5-dione-and-nitrous-acid/86090#86090

Waylander, could you please elaborate why aromatic acyl chloride undergoes nucleophilic addition elimination with KCN, while aliphatic acyl chloride does not? (this is mentioned by OP in first paragraph of his post)

I have edited my answer to include comment on aromatic acyl cyanides

Actually, my question was whether or not aromatic acyl cyanides can react with KCN. I don't see how you've answered that :/

The paper cited by the OP http://www.journal.csj.jp/doi/abs/10.1246/bcsj.60.1085 details reaction of aromatic acyl chlorides  with sodium and potassium cyanides impregnated on resins and contains other references to their formation. It also notes that aromatic acyl cyanides in liquid solution are prone to dimerise.


text:
How sure can we be about the structure of purine?
While answering the question Basicities of nitrogen atoms in purine, I noticed, that there might be a rapid interchange of the hydrogen bonded to the nitrogen atoms.
I have found a crystal structure in D. G. Watson, R. M. Sweet and R. E. Marsh, Acta Cryst. 1965, 19, 573-580. (DOI: 10.1107/S0365110X65003900, CCDCID: 1239746, 1239747), that resolves it as 1. However, at least in solution I would assume that there should be a distribution of structures according to the scheme below. In all cases the aromaticity is retained.

I am not an expert on crystal structures, and I don't know how to read the .cif file correctly. I remember from my course on crystallography, that the positions of hydrogen are notoriously hard to confirm.
Can we therefore really be sure about this structure? Is there evidence in the crystallography data for a large disordering?
To provide a bit more context, I am asking this, because it is likely to change part of the answer I have given to the linked question. If there exists an equilibrium between the above protomers, then that would certainly make it even harder to find the most basic nitrogen in purine.

I'm not sure about crystallography, but $\ce{^{1}H}\text{-}\ce{^{15}N}$ HSQC experiments should be able to tell  which nitrogen atom is associated to a proton (for example, here) and the extent of the contribution of each tautomer.

For a small moleculesuch as this, crystal structure determination will give the positions of hydrogen with high certainty. However, equilibrium constant in the crystal might be different than in aqueous solution. For example, protonation of one nitrogen might be favored over another in the crystal because there is a hydrogen bond acceptor positioned in a favorable spot.

Is the hydrogen atom jumping from one nitrogen atom to another? If so, how? Would it still be called "purine" if the hydrogen atom disappeared completely?

@MatthewChristopherBartsh that's an acid base reaction, protons tend to be quite mobile. It they are "lost" it's just the conjugate base then.

The article text (Ref.1) itself does go into details. The first thing to note: while aromaticity on the larger scale is retained in your structures 3 and 4, this is only true for the 10-π-electron system as a whole. In many of these types of 10 π systems made up of a six-membered and a five-membered ring, the two rings are not equal with the double bond in the five-membered ring often being much more susceptible to hydration. This prior thought should bias us against structures 3 and 4.
A pretty obvious first thought would be to take a look at the arrangement of individual molecules in the crystal structure. The paper does this in figure 3 which I have reproduced below. Wherever we expect the single proton, it should surely form a hydrogen bond to another purine molecule, meaning that these two nitrogens should be sufficiently close and in a favourable angle arrangement.

As the figure shows, the two nitrogens of the five-membered ring are in an arrangement that would suggest a hydrogen bond between them. The cut-off bits outside the unit cell are fragments of the next purine molecule not completed for brevity. This arrangement rules out 3 and 4 on the basis of no available hydrogen bonds and limits the discussion to 1 and 2.
As for the question whether the hydrogen prefers structure 1 or 2 this is admittedly more complex. As the authors note:

Since there is a possibility of tautomerism of the hydrogen between N(7) and N(9), particular attention was given to this problem. Two least-squares cycles, not quoted in Table 2, were computed in which a half-hydrogen was assigned to each of N(7) and N(9). Refinement of H(10), H(ll), H(12) and the two half-hydrogens resulted in the half-hydrogen attached to N(9) shifting to a position about 1 Å from N(7) of the neighbouring hydrogen-bonded molecule.

So basically, during refinement Watson attempted to put the hydrogen on both nitrogens equally. However, the refinement would show that the partial hydrogen on $\ce{N}$9 was moved sequentially along the $\ce{N\bond{...}H\bond{...}N}$ distance towards $\ce{N}$7. Taking the final $\ce{N-H\bond{...}N}$ distance of $\pu{285pm}$ and the $\ce{N-H}$ bond at $\pu{85pm}$, this means that there are approximately $\pu{115pm}$ between the two possible hydrogen positions on $\ce{N}$7 or $\ce{N}$9—and most of that was covered by the moving hydrogen. However, Watson also notes that the electron density corresponding to that $\ce{N-H}$ hydrogen was ‘much more smeared’ that those of the other hydrogens.
Sweet and Marsh arrive at the same conclusion using a different method that I did not fully understand.
All things considered, this gives us a pretty strong case that purine is best represented by structure 1 and that the principal tautomerisation structure would be 2 as that is where the hydrogen bond is directed.

Note: The authors state in the introduction to their paper:

This paper presents the results of two entirely independent investigations of the structure of purine (I).
The duplication of effort was discovered only when both three-dimensional studies were almost complete (Watson, 1964). Since different experimental techniques had been used, it was considered appropriate to publish both sets of results in the same paper and to make the necessary comparison between them. It has been a comfort to both parties that, except for systematic differences in the temperature factors, the two sets of results differ by less than their standard deviations.

This is why I focused so much on Watson and ‘ignored’ the other authors as the principal arguments are from Watson’s part of the paper.

Reference:

D. G. Watson, R. M. Sweet, R. E. Marsh, “The crystal and molecular
structure of purine,” Acta Crystallographica 1965, 19, 573-580
(https://doi.org/10.1107/S0365110X65003900).


text:
Is the first ionization energy in oxygen slightly more than nitrogen?
Why is the first ionization energy in oxygen slightly more than nitrogen?

In nitrogen: $\ce{[He] 2s^2 2p^3}$
In oxygen: $\ce{[He] 2s^2 2p^4}$

This tells me that it should be easier to remove an electron from oxygen than it is for nitrogen as the electron in oxygen is slightly further away from the nucleus meaning lesser nuclear charge.
But why is it harder to remove an electron from oxygen, i.e. why is the first ionization energy of oxygen higher?

Actually, it is the other way around.

https://chemistry.stackexchange.com/questions/40044/why-does-f-have-a-higher-ionization-energy-than-o

Warning to readers: This entire question is wrong, the first IE of oxygen is smaller than the first IE of nitrogen. Two of the answers are wrong, too.

Does this answer your question? Why doesn't ionization energy decrease from O to F or F to Ne?

You see from the electronic configurations:

nitrogen: $\ce{[He] 2s^2 2p^3}$
oxygen: $\ce{[He] 2s^2 2p^4}$

In reality, the first ionisation energy of nitrogen is greater than the first ionisation energy of oxygen because nitrogen, in a stable half filled orbital state, is comparatively more stable than oxygen. Oxygen, on the other hand, would tend to lose an electron easily to achieve it's more stable half filled orbital state.
Also, as a rule, half filled and fully filled orbital states are more stable as compared to other configurations because they attribute to maximum exchange energies.

Are you sure "Oxygen, ..., would tend to lose an electron easily"? I am asking because $\ce{O+}$ is really uncommon...

p4 has the same exchange energy as p3. It is due to repulsion.

(-1) Despite being accepted, this answer is wrong. Exchange energy is only a small consideration (if it is one at all), the correct reason is repulsion between paired electrons.

Oxygen has a lower first ionization energy as the electron that is removed is coming from a paired orbital.

Electrons within the same orbital experience maximum repulsion as the distribution of their wavefunctions is the same, so the probability density distribution is the same and the electrons can be thought of as occupying the same space. This maximizes their repulsion and increases the potential energy of the electrons in that orbital, making the electrons easier to remove. This is despite the increased effective nuclear charge experienced by the electron in oxygen and the decreased radius of the orbital.
See: "Physical Chemistry", Atkins, P.W. Section 13.4, p.p.370 (4th edition) - sorry, I have an old one!

This is the correct answer!


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Why must the ground state of hydrogen molecule be gerade?
As we know, the electronic ground state of hydrogen molecule is $\Sigma_g^+$. My question is, how do we know it from the theoretical perspective?
The nuclear potential in the hamiltonian of hydrogen molecule is of the inversion symmetry, so the electronic wave function of the system must be gerade or ungerade. However, how can we derive that the ground state is of gerade symmetry?

Check the answer to this question : https://chemistry.stackexchange.com/questions/42877/how-to-determine-gerade-ungerade-symmetry-of-a-mo-orbital


text:
Matching the NMR to a compound
I was given a molecule that I had to identify the NMR of. The molecule is 2-methylpent-1-ene (shown below):

The correct NMR spectra for this molecule is:

My question is, why is this the right spectra for this molecule. Particularly, how and why did the single peak come up? As far as I know, the only two single peaks are the two that are downfield (aka closest to the $\ce{sp2}$ hybridized carbon).

The singlet peak is obviously due to methyl group which is closer to the double bond. See, it is attached to carbon with no hydrogens on it, hence no splitting.

This spectrum is too perfect (simulated?). See what it looks like at 400 and 90 MHz. [http://sdbs.db.aist.go.jp/sdbs/cgi-bin/direct_frame_top.cgi]

If you look at the structure closely, you'll find that there are no Hydrogens on the C with the double bond attached. Adjacent to this C is the methyl group. Due to the n+1 rule, the singlet should be due to that Methyl group connected to the "Hydrogenless" Carbon. Hope this simplified answer helps

I think you are correct to wonder why that peak is singlet. The truth is that it depends on the conditions that the spectrum was acquired such as spectrometer frequency, solvent, etc.
Due to the double bond, there is a small $^4$J$_H$$_H$ coupling between the CH3 and olefinic CH2. If you look carefully, you can see in your spectrum  that although it says singlet it is actually a triplet due to the tiny shoulders that are barely visible about a bit more than half way on the peak! However, for a weak spectrometer (200-300 MHz) and depending on the solvent and conditions the measurement takes place, it can appear as a slightly broad singlet. Similarly, the olefinic peaks that are stated as doublets are not real doublets but multiplets due to multiple small coulings with the aliphatic CH3 and CH2. They can appear as slightly broad doublets though if resolution is not great or even a broad singlet. You would be able to see all these finer details clearly with a high filed NMR instrument (e.g. 700 MHz) and zooming in the aquired spectrum.

Could an un-coupled 1H NMR spectra look like the one above? Since as you said coupling could affect the peaks given

If it was 1H decoupled then you wouldn't see multiplets so no. I think the above spectrum might be computed rather than actually measured judging from how perfect it looks but it is hard to be 100% sure.


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Periodic boundary conditions in DFT for cylindrical system
As I understand it, a DFT calculation uses periodic boundary conditions, whenever you want to simulate a bulk material. This is convenient because all information can then be extracted from the fundamental unit cell.
Currently I am faced with the problem of calculating the band structure of a carbon nanotube, which has a cylindrical geometry, i.e. only bulk in only one direction, say z, and circular in the x and y directions. As my unit cell I am using a ring of C-atoms surrounded by vacuum and I then employ periodic in the z-direction. Obviously, for large diameters, this approach involves a rather large unit cell, which makes the calculation slow.
My question is: Is there an alternative way to do this, which exploits the cylindrical symmetry of the system? Specifically what I have in mind is to also apply periodic boundary conditions in the other directions, but where the Schrödinger equation is expressed in cylindrical coordinates with the middle of the nanotube as the origin.
Another, less important, question is also: Do you ever NOT use periodic boundary conditions in DFT? For example, when you do a calculation for an atom, it always seems like you have to add vacuum around it, which is essentially, because you are applying pbc, and do not want it to interact with other unit cells. Is there another way to solve this problem, where you assume the density to vanish far away?

Your question seems to be related to some particular software which you do not specify. That being said, rotational symmetry in an infinite cylindrical system is the same as rotational symmetry in some finite molecule, and should be treated the same.

I am using VASP. I do not understand why it is software related question. My question is simply if there is a way to simplify the unit cell of a nanotube by taking the cylindrical symmetry into consideration.

IIRC, VASP will always use periodic boundary conditions (PBC) in all directions. CRYSTAL is different in this regard. (So it clearly is program-dependent.)  PBC are necessitated by VASP's basis set (plane waves). For atom-centered basis sets (such as GTOs, STOs) the situation is very different.

Let's dissect this:

As I understand it, a DFT calculation uses periodic boundary conditions, whenever you want to simulate a bulk material. This is convenient because all information can then be extracted from the fundamental unit cell.

One important point is that you employ periodic boundary conditions (PBC) only when you are treating a periodic system, i.e. something with translational symmetry. And in those cases you use plane waves as a "basis set".
In the ordinary molecular case you use atom centred basis functions, which fulfil the basic requirements of a wave function (in short): orthonormality, continuous, continuous differentiable, vanishing at infinity.
Therefore a quantum chemical calculation electronic energy is always in vacuum at "0 K".

Currently I am faced with the problem of calculating the band structure of a carbon nanotube, which has a cylindrical geometry, i.e. only bulk in only one direction, say z, and circular in the x and y directions. As my unit cell I am using a ring of C-atoms surrounded by vacuum and I then employ periodic in the z-direction. Obviously, for large diameters, this approach involves a rather large unit cell, which makes the calculation slow.

This approach actually only makes sense if you are treating a very, very long nanotube. I'd believe a molecular approach using appropriate symmetry, Dnh would be far superior.

My question is: Is there an alternative way to do this, which exploits the cylindrical symmetry of the system? Specifically what I have in mind is to also apply periodic boundary conditions in the other directions, but where the Schrödinger equation is expressed in cylindrical coordinates with the middle of the nanotube as the origin.

A nanotube is discrete in two directions, i.e. the wave function has to vanish in these directions, egro apart from applying vacuum as PBC, there is no other choice in this framework.
An alternative approach is to treat your nanotube as a discrete molecule and cap it off at the end with hydrogen, then increase the length of the tube until you have sufficient convergence, obviously using Gaussian or Slater type (atom centred) orbitals.

Another, less important, question is also: Do you ever NOT use periodic boundary conditions in DFT? For example, when you do a calculation for an atom, it always seems like you have to add vacuum around it, which is essentially, because you are applying pbc, and do not want it to interact with other unit cells. Is there another way to solve this problem, where you assume the density to vanish far away?

For discrete molecules: always. For periodic systems: only sometimes not (cluster approach).
Prior to the golden 90s, almost every calculation was a molecular calculation. In the early stages of quantum theory nobody ever dared to dream that we would eventually able to treat a huge stone quantum chemically (a priori) with a computer.
A nanotube is typically right in the middle, between a bulk material and a molecule, and you should either rely on the experience of others, maybe you have an advisor, through the literature, or built it yourself.
There are quite a few groups (I suppose) who work with nanotubes, and some of them have surely tried computations. Do a thorough literature review and apply common practices.
I fear that you are getting your knowledge about DFT calculations from working with a calculation suite designed for bulk materials. You should probably brush up on some of the fundamental theory. I recommend Frank Jensen's book for a rough overview, and Attila Szabo & Neil Ostlund's book for some better maths  (See resources for learning chemistry for more details).
A fairly comprehensive "review" of DFT can be found here: DFT Functional Selection Criteria and in the links within.