text:
What happens to the argon in the Haber-Bosch process?
The Haber–Bosch process is used to synthesis ammonia from atmospheric nitrogen and hydrogen derived from a hydrocarbon such as methane. The process requires high temperatures and pressures and also requires repeated cycling of the gasses.
I would have thought that atmospheric argon and other noble gases would have slowly built up in the reactor. How are these inert gases removed?

Inert gaseous components such as methane and argon are indeed should be eliminated from the system in order not to lower the partial pressure of the reactants too much. Technically there is usually a standalone gas separation plant where the extraction of argon from the recycle gas is performed basically using modified Linde process [1 , pp. 428–430].

Suitable cryoprocesses [...] are available for production of noble gases from synthesis purge gas. Initially, extensive separation of nitrogen, argon, and methane is carried out by partial condensation. Purge argon is then recovered from the condensate in a two-stage condensation process. If helium is to be recovered it can be concentrated by liquefaction of the hydrogen in the hydrogen-rich gas phase, followed by purification. The heavier noble gases, krypton and xenon, pass into the methane fraction.
The purge gas at pressures of up to $\pu{70 bar}$ is ﬁrst introduced into the adsorber (a) where traces of water and ammonia are removed. It is then transferred to the heat exchangers (b1) and (b2) for cooling. The gas is then fed into separators (c1) and (c2) where separation of liquefied nitrogen, argon, and methane from gaseous hydrogen takes place, and dissolved hydrogen ﬂashes off. The liquid bottom product is fed into the fractionating column (d1) where methane (bottom) is separated from the nitrogen fraction (top).
A methane-free nitrogen – argon mixture (liquid) is withdrawn from the middle of this column and fed as reflux into the $\ce{Ar}$ purification column (d2), where nitrogen is separated from argon. The bottom product is argon of product purity and is transferred to a vacuum-insulated storage tank.
Both columns (d1) and (d2) are operated within an $\ce{N2}$ cycle in which cold is produced by expanding high- or medium-pressure nitrogen.
The $\ce{CH4}$ bottom stream from (d1), compressed by liquid pump (g), is evaporated against feed gas and normally led to battery limits as fuel gas. The higher boiling noble gases krypton and xenon are contained in this stream and, in principle, can be isolated.

Reference

Häussinger, P.; Glatthaar, R.; Rhode, W.; Kick, H.; Benkmann, C.; Weber, J.; Wunschel, H.-J.; Stenke, V.; Leicht, E.; Stenger, H. In Ullmann’s Encyclopedia of Industrial Chemistry; Wiley-VCH Verlag GmbH & Co. KGaA, Ed.: Weinheim, Germany, 2001. DOI: 10.1002/14356007.a17_485.

Andselisk's answer focuses on the way the ammonia plant can be used to "produce" argon byproduct for sale, but this is by no means essential.
The ammonia reaction cycle is continually purged to remove argon and other undesirable substances. So what do we do if there is no argon plant fitted?
The hydrogen used in an ammonia plant is made by steam reforming. Further H2 is made by the shift reaction:
CH4 + 2H2O --> CO + 2H2  (steam reforming, strongly endothermic, 1000C approx)
CO  + H2O  --> CO2 + H2  (shift reaction, exothermic, 300-400C)

(CO2 removed in older plants by ethanolamine scrubbers,
and in newer plants by pressure swing absorption)

The steam reforming reaction uses an enormous amount of heat, some of which can be supplied by combustion of process air to generate nitrogen:
(Air) + CH4 --> CO2 + 2 H2O + trace argon + x N2 (x = 8 approx)

However there is insufficient heat from the partial combustion of air to generate all the hydrogen needed to consume the nitrogen produced. Therefore the plant must include a fuel fired furnace using additional air to produce more heat to drive the steam reforming reaction. The exhaust gases of this furnace are sent to atmosphere in the normal way.
The purge steam from the ammonia synthesis loop consists of some argon, with the rest being combustibles or inerts (H2, CH4, N2). If an argon recovery plant is not fitted, the purge gas is simply fed to the furnace and used as a supplementary fuel. The argon is therefore returned to the atmosphere via the chimney.

The short answer is that some of the recycle gas is vented at a rate to keep the Ar below about 5% ( I am not a process person so the 5% is a guesstimate ).


text:
Polycyclic nitrogen network?

So today I suddenly stricken with this structure and started to wonder if this polymeric structure of nitrogen is possible or not? If possible then what's its stability?

With N, it will collapse instantly. With P, it might stand a chance.

It's not "graphene-with-nitrogen" exactly, but I see what you mean. It's a planar tiling (like graphene) but with mixed 12-ring/3-rings as opposed to a hexagonal tiling

you could also go with squares/octagons like https://en.wikipedia.org/wiki/File:1-uniform_n2.svg (see https://en.wikipedia.org/wiki/Euclidean_tilings_by_convex_regular_polygons) - although as other commenters point out, these are extremely chemically unlikely! (it's this one https://en.wikipedia.org/wiki/File:1-uniform_n4.svg)

Yikes!  Where was this dreamed up?

Looks like someone tried out his chemicals sketching software. ? Polmeric nitrogen exists (at very high pressures), but the structure is quite different. https://phys.org/news/2004-08-polymeric-nitrogen-synthesised.html

What do you mean by "possible" and "stable"?

That would make quite a BOOM as it collapsed...

@Ivan Ah? Why so? O:)

After reading your comments and a bit of googling I found this : This polymeric structure of nitrogen isn't possible but cg-n structure at very high pressure (100 GPa) is possible. This structure is highly unstable and upon release in open atmosphere it would be highly exothermic.

@Mithoron By possible I meant that can this structure be formed anywhere in the universe without decomposing to another structure and by stable I meant that is it stable at STP?

I searched CCDC for the infinite structures with the following building block:

where where each vertex contains any non-metal atom linked by any type of bond. The only known crystal structure that would meet these criteria is $\ce{[B12]2[CBC][C2]Mg_{1.42}}$ where $\ce{B12}$ icosahedra are linked via $\ce{B–C}$ and $\ce{B–B}$ bonds forming the moieties of the previously shown topology:

Figure 2. Projection of the $\ce{B25C4Mg_{1.42}}$ structure onto the $0ac$ plane (the block of icosahedra of boron carbide structure is marked by shading).

Unfortunately, no similar structures are found for nitrogen yet. Also, the structural motif breaks on boron clusters, but this is the closest as I was able to get.
Reference

Ponomarev, V. I.; Konovalikhin, S. V.; Kovalev, I. D.; Vershinnikov, V. I.; Borovinskaya, I. P. Mendeleev Communications 2014, 24 (1), 15–16. DOI: 10.1016/j.mencom.2013.12.004.

After a bit of research i found http://www.sciencedirect.com/science/article/pii/S0927025614008209, but ccdc structure of nitrogen isn't possible/found yet

@G.OVed Ah, the mighty USPEX – this program predicted numerous structures at extreme conditions. Actually, you can answer your own question with this, I think the article is relevant.

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text:
How can a methyl halide react by intramolecular SN2?

Out of the following, $\ce{CH3Br}$ doesn't react by which mechanism?

SN2
SN1
E2
Intramolecular SN2
E1

My textbook gives 2, 3 and 5 as the correct answers, which means it does react by 1 and 4.
But how can a methyl halide react by intramolecular SN2, without a nucleophile present in the molecule? There's just a leaving group ($\ce{-Br}$).

I agree. There is no way that a one-carbon and one-heteroatom compound could react by intramolecular substitution. All intramolecular nucleophilic substitutions will require some ring-like transition state. The smallest ring you can build is a three-membered ring — but bromomethane does not have three atoms that could build up that ring (hydrogen does not participate).

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text:
Why does benzaldehyde undergo Cannizzaro reaction?
Benzaldehyde has an aromatic ring in it which has a pi cloud of electrons surrounding it. Would they not repel the nucleophile when it attacks?

If your theory held true, then benzaldehyde would be relatively unreactive to nucleophiles however the opposite it true. This should give you a big clue about your argument


text:
Why is ZnO reduced in this reaction?
Consider the following Ellingham diagram.

At 673K, the Gibbs energy value of oxidation of Zn is more negative than the Gibbs energy value of oxidation of C (coke). So, Zn should be oxidised to ZnO. But in various sources like the textbooks of class 12 of CBSE (India), it is mentioned that ZnO is reduced to Zn and C (coke) is oxidised to CO at 673K in the reaction between ZnO and C. Looking at the Ellingham diagram, coke should not reduce ZnO to Zn at 673K.

So why is ZnO is reduced to Zn in this case?

NCERT contains incorrect information. The temperature is 1673 K. Verified from earlier versions and other sources

This method works because one of the reaction products, $\ce{CO}$, is gaseous (at the stated reaction temperature) and is continually removed. Following Le Chatelier's Principle, the equilibrium:
$$\ce{ZnO(s) + C(s) <=> Zn(l) + CO(g)}$$
is pushed to the right and the reaction proceeds.
There are numerous of these reactions where Ellingham data don't tell the full story, notably also the reduction of magnesia ($\ce{MgO}$) by coke, is in accordance with the same principle.
One of the most interesting cases is a lab preparation of caesium metal that takes place according to:
$$\ce{CsCl(s) + Li(l) <=> Cs(g) + LiCl(s)}$$
The reaction is carried out under vacuum and at about $970\ \mathrm{K}$, conditions under which $\ce{Cs}$ is much more volatile than $\ce{Li}$. The former thus distils off and is caught in a condenser, pushing the reaction equilibrium to the right.

For us to remove CO on formation, shouldn't ZnO get reduced first? But the oxidising gibbs energy change of ZnO is greater than coke, so how can the reaction move in the forward direction? Also why is it an equilibrium reaction?

ALL reactions are in essence equilibrium reactions, this one is no exception. By removing one of the reaction products continually, its concentration (chemical activity) is lowered, thus pushing the equilibrium to the right. Have you read the linked page on Le Chatelier's Principle?

Somebody mistyped something. We do require a higher temperature than that quoted in the question.
Zinc is a chalcophilic element, generally available as a sulfide ore which is first roasted to the oxide. The oxide may then be reduced to zinc metal by coke (carbon), but it requires a higher temperature than $400°\text{C}$. From Wikipedia:

For further processing two basic methods are used: pyrometallurgy or electrowinning. Pyrometallurgy reduces zinc oxide with carbon or carbon monoxide at 950 °C (1,740 °F) into the metal, which is distilled as zinc vapor to separate it from other metals, which are not volatile at those temperatures.[1] The zinc vapor is collected in a condenser.[2] The equations below describe this process:[2]
${\displaystyle {\ce {ZnO + C ->[950°C] Zn + CO}}}$
${\displaystyle {\ce {ZnO + CO ->[950°C] Zn + CO2}}}$

The Ellingham diagram then shows that this temperature is sufficient to give carbon monoxide formation up to one full atmosphere carbon monoxide pressure; the carbon/carbon monoxide line is clearly below the zinc/zinc oxide line. We do not oxidize coke to carbon dioxide at that partial pressure (the carbon/carbon dioxide line is slightly above the zinc/zinc oxlide line), but because the free energy difference is small we do get a significant partial pressure below one atmosphere which is made up by the carbon monoxide.
The carbon monoxide can be burned to remove the toxic gas and provide some of the heat energy needed to run the process.
Cited References

Bodsworth, Colin (1994). The Extraction and Refining of Metals. CRC Press. p. 148. ISBN 978-0-8493-4433-6.

Porter, Frank C. (1991). Zinc Handbook. CRC Press. ISBN 978-0-8247-8340-2.

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text:
Birch Reduction of Benzene
According to my Organic Chemistry Textbook:

Benzene can be reduced to 1,4-cyclohexadiene by treating it with an alkali metal (sodium, lithium, or potassium) in a mixture of liquid ammonia and an alcohol. This reaction is called the Birch reduction.

Why the conjugated 1,3-cyclohexadiene is not formed in this reaction (as conjugated molecules are more stable)?

https://chemistry.stackexchange.com/questions/35621/why-does-the-birch-reaction-not-yield-fully-saturated-products

https://chemistry.stackexchange.com/questions/48545/mechanism-in-birch-reduction https://chemistry.stackexchange.com/questions/61545/birch-reduction-of-anthracene

According to the article on Birch Reduction on the Organic Chemistry portal:
The question of why the 1,3-diene is not formed, even though it would be more stable through conjugation, can be rationalized with a simple mnemonic. When viewed in valence bond terms, electron-electron repulsions in the radical anion will preferentially have the nonbonding electrons separated as much as possible, in a 1,4-relationship.
http://www.organic-chemistry.org/namedreactions/birch-reduction.shtm


text:
Electrolytic silver deposition

A current is applied to two electrolytic cells in a series. In the first, silver is deposited; in the second, a zinc electrode is consumed. How much $\ce{Ag}$ is plated out if $\pu{7.5 g}$ of $\ce{Zn}$ dissolves?

What I ended up doing was
$$\frac{\pu{7.5 g}~\ce{Zn}}{\pu{65.382 g mol}~\ce{Zn}} \times \frac{\pu{2 mol}~\ce{Ag}}{\pu{1 mol}~\ce{Zn}} \times \pu{107 g mol-1}~\ce{Ag}$$
but I'm not even sure I'm doing this right.

I'd suggest to start with writing down the processes occurring during the electrolysis:
\begin{align}
&\text{Cathode (–):} &\ce{2 Ag+ + 2 e- &-> Ag^0} \\
&\text{Anode (+):} &\ce{Zn^0 &-> Zn^2+ + 2e} \\
\hline
&\text{Net} &\ce{2 Ag+ +Zn^0 &-> 2 Ag^0 + Zn^2+}
\end{align}
Applying the unified equation for Faraday's laws of electrolysis:
$$m = \frac{QM}{zF}$$
for each cell, and taking into account that in series circuits current is the same for every element ($I = \text{const}$):
$$\frac{Q}{F} = \frac{z_i m_i}{M_i} = \text{const}$$
$$\frac{z(\ce{Ag}) \cdot m(\ce{Ag})}{M(\ce{Ag})} = \frac{z(\ce{Zn}) \cdot m(\ce{Zn})}{M(\ce{Zn})}$$
The valency numbers of ions are 1 and 2 for silver and zinc, correspondingly. The last thing to do is to express and calculate $m(\ce{Ag})$.

 $$m(\ce{Ag}) = \frac{z(\ce{Zn}) \cdot M(\ce{Ag})}{z(\ce{Ag}) \cdot M(\ce{Zn})} \cdot m(\ce{Zn}) = \frac{2 \cdot \pu{107.87 g mol-1}}{1 \cdot \pu{65.38 g mol-1}} \cdot \pu{7.5 g} = \pu{24.7 g}$$


text:
Trends of electron affinity
The trend in electron affinity is to increase negatively across a group.
Does this mean noble gases' electronegativity is more negative than halogens?
Noble gases should have a positive electron affinity as adding an electron would make it unstable and thus will not happen.
But for halogens the attraction for that one electron to fill up the $\mathrm{p}$ subshell is very strong.
But given the theory that electron affinity increases negatively across the period, does this theory make an exception for noble gases?
Next, what about $\ce{Be}$ and $\ce{Mg}$? Their electron affinities are not negative.

Not exceptions, but see, theories say that going $L→R$ electron affinity increases up to halogens.
For the noble gases, $\Delta_\mathrm{eg}H$ is positive (energy required). While electron affinity is $0$. Which implies no tendency to lose electrons.
For $\ce{Be}$, the small size and nuclear charge are responsible, which is related to $\ce{Mg}$ due to a diagonal relationship.

Be and Mg are not diagonally related, and in any event diagonal relationships do not work for electron affinities.  Mg lacks an exothermic electron affinity because like Be, it has a closed valence shell and an electron would have to be added to a much higher-energy orbital.


text:
Rate of reaction for hundred reactants
I'd like to programm a toy chemical reaction program. Rough simplifications that deviate from industry and academia are totally acceptable. Not very educationaly accurate, but fun.
Are there methods to calculate rate of reaction for dozens and hundreds of reactants in a closed system?
Or I can blatantly use original r = k(T)[A][B] for every pair of reactants while selectively ignoring presence of other molecules?
Or should I use something else to evaluate what will happen with hundred of reactants after two seconds?
Sample gas composition below:

As a programmer, be aware that you're running straight into a combinatorial explosion of epic magnitude. Dozens of reactants means hundreds of possible pairs, many of which will not react, but many will. (How are you going to find that out, to begin with?) Some reactions may yield different products in different conditions. Then the products might want to react with something else. Then some of them might form another phase and bring an entirely new level of complexity (which is the case in your example, unless T is pretty high, and if it is, that would cause other issues). Then...

Start with a much simple system, I say! Then see how you get on.

You may be interested in the program Cantera, which does precisely what you seek to accomplish, provided you have estimates for the reaction barriers, rate constants, and thermochemistry. Looking up any examples on microkinetic modeling will lead you to plenty of information. Another example, but in the context of heterogeneous catalysis, is CatMAP. The corresponding paper is a fairly nice reference too.

Allright, guys. All I wanted happens to be stochastic simulation.
Here is the list of sample algorithms:

Gillespie SSA - original.
Tau-leaping - evaluate time step and then out of that reactions and substances amount.
R-leaping - vise versa. Good for big systems with only a few relatively highly reactive species.
S-leaping - hydrid of R-leaping and Tau-leaping.

Tau-leaping can be explicit and implicit. The second one is better for stiff systems. Adaptive Tau-leaping switches between one and the other.
Many algorithms for Tau-leaping time step evaluation exist:

The first original algorithm to evaluate time step in Tau-leaping is flawed in performance -
Cao, Y.; Gillespie, D. T.; Petzold, L. R. (2005). "Avoiding negative populations in explicit Poisson tau-leaping"

Upgraded one is better all around -
Cao, Y.; Gillespie, D. T.; Petzold, L. R. (2006). "Efficient step size selection for the tau-leaping simulation method"

Don't bother with Binomial distribution. No gain over Cao-Gillespie-Petzold-2006 upgrade but greater algorithm complexity.

There is also Hybrid Chernoff Tau-leaping paper

And Anderson, David F - 2005 and 2018 papers

Or should I use something else to evaluate what will happen with hundred of reactants after two seconds?

The great mistery is why none of you have mentioned a single algorithm above.

The questions has at least two aspects to address.
The first thing is, there is no guarantee that the reaction is first-order with respect to each of the reactants in first place. In other words, without further investigation you can not just take $r=k[\ce{A}][\ce{B}]$ because $r=k[\ce{A}][\ce{B}]^2$ or $r=[\ce{A}]^2[\ce{B}]$ or something more exotic may happen, depending on the nature of the reaction.
OK. So let's say for simplicity we take $r=k[\ce{A}][\ce{B}]$ when the system has only $\ce{A}$ and $\ce{B}$, as we don't need to be quite educationally correct. Then yes, after mixing many reactants, the equation still holds for each pair of the reactants, provided they are elementary reactions. It will be a completely different story if, for example, neither of two of the reactions are elementary and they share the same reaction intermediates.
I also have a reminder that when stepping from the expression of the rate to the expression of concentration with respect to time, you have to be careful and do the integration properly instead of just naively extrapolating the concentration expression when only two reactants exists. Let's say that we don't consider chemical equilibrium for the moment (i.e. consider all the reverse reactions to be prohibitively difficult to happen), then as far as I can see, the $r-t$ curve will not be smooth throughout the time and will have some turning point every time a reactant is fully consumed.
Side note: I suggest that you do your formatting properly in your program for the superscripts and subscripts.

In general, assuming that reactions are at least first order in all their reactants, the concentrations will be smooth functions of time. At least if you zoom in close enough. One can have pretty sharp-looking corners e.g. if a catalytic decomposition reaction is normally rate-limited by the saturation of the catalyst, but at some point before the reactant concentration would hit zero, the catalyst will stop being saturated and the reaction rate will become proportional to the reactant concentration.

... That said, I certainly agree with you that one has to do proper numerical integration, rather than just linearly extrapolating the concentrations.  In fact, for these kinds of problems, usually one should use an implicit solution method that avoids numerical instability caused by extrapolation within a timestep.

For truly elementary reactions, the law of mass action can generally be assumed to hold fairly accurately, as predicted by the collisional theory of reaction kinetics. (This is something of a tautology, since a deviation from the law of mass action generally indicates that the reaction isn't really elementary.)  In particular, this implies that:

elementary unimolecular reactions occur at a rate proportional to the concentration of the reactant (or more generally to its activity, if we're treating e.g. the effects of a solvent on the reaction rate implicitly);
elementary bimolecular reactions with distinct reactants occur at a rate proportional to the product of the concentrations (or activities) of the reactants;
elementary reactions between two molecules of the same species occur at a rate proportional to the square of the concentration (or activity) of the reactant; and
elementary reactions involving more than two reactants don't occur, except at negligible rates, since they would have to involve a perfectly simultaneous collision of three molecules.  (Any such reactions that are observed to occur must actually proceed via one or more ephemeral transition states produced in bimolecular collisions, and are thus not actually elementary.)

Thus, in principle, any* such system of simultaneous elementary reactions can be modeled using a system of ordinary differential equations of the form:
$$
\frac{d}{dt} [A_x]
= \left( \sum_{y=1}^n k_{xy}[A_y] \right)
+ \left( \sum_{y=1}^n \sum_{z=1}^y k_{xyz}[A_y][A_z] \right),
$$
where $[A_1]$ to $[A_n]$ are the concentrations of the $n$ distinct chemical species present in the system (some of which might only exist as temporary transition states), $k_{xy}$ is the sum of the rate constants for reactions of the form $A_y \to A_x + (\text{other products})$, and $k_{xyz} = k_{xzy}$ is the sum of the rate constants for reactions of the form $A_y + A_z \to A_x + (\text{other products})$.**
So yes, if you know the rate constants for all the elementary mono- and bimolecular reactions your reactants can undergo, then you can indeed plug them into a differential equation like the one above and solve it.
However, as a fair warning from someone who's worked on numerically solving these kinds of differential equation systems, they are notorious for often exhibiting high stiffness and consequent numerical instability.  This is particularly often the case if some of the reaction constants are much larger and/or some of the reactant concentrations are much higher than others.  Efficiently solving such stiff ODE systems is a hard problem, and there's a whole bunch of advanced (and often jargon-filled) mathematical literature on it.
In the end, the practical ways of solving such systems usually come down to timescale separation or other stiffness-reducing approximations, where we essentially assume that certain "fast" reactions occur so rapidly that the concentrations of their products and reactants are always at a local quasi-equilibrium, where the overall net reaction rate of the fast reactions and their reverse reactions is zero.  This allows us to then only focus on the remaining "slow" reactions, which are presumably the ones we're actually interested in.
Alas, as a side effect, these approximations also often tend to destroy the mathematical simplicity of the basic elementary reaction model by reintroducing non-elementary reaction laws and/or nontrivial dependencies between the reactant concentrations.  Also, unfortunately, at least I know of no general method for automatically constructing such approximations, just as there's no fully general method for efficiently solving the original stiff ODE systems in the first place.

*) At least assuming that the reactions occur in a homogeneous well-mixed environment.  Things like reactions occurring on the surface of a substrate can be included by treating the substrate-bound and free forms of a molecule as distinct species and, if necessary, also treating free surface bonding sites as a separate species with its own concentration.  Similar tricks can also be used to model e.g. reactions occurring in two immiscible solvents, provided that mixing within each solvent phase can be assumed to be rapid.  However, if the reactant concentrations may vary spatially to a significant degree, then one would have to use a reaction–diffusion model instead.
**) If the products of a reaction include $m$ molecules of species $A_x$, that reaction's rate constant should be included $m$ times in $k_{xy}$ or $k_{xyz}$.  Also note that, to avoid double counting $A_y + A_z \to A_x + \dots$ and $A_z + A_y \to A_x + \dots$ as distinct reactions, the last sum in the differential equation is only taken over $z \le y$.


text:
Why is phenol more acidic than aniline?
I find that there are two contradicting factors at play here:
1) On removing proton, negative charge on oxygen is more stable than it is on nitrogen (because of greater electronegativity of oxygen). Therefore, phenol is more acidic than aniline.
2) +M effect of nitrogen is more than oxygen, so negative charge (after deprotonation) in the case of aniline will be more delocalized making the anion more stable. Therefore, aniline should be more acidic than phenol.
So is that a factual question or am I missing something?

What does point 2 have to do with this? I don’t see the link.

@orthocresol After deprotonation, the negative charge will be more delocalized in the case of aniline because of greater +M effect of nitrogen. Since the anion is more stable, aniline should be more acidic.

Ok, I see now. Electronegativity is usually the main factor. The pKa difference between alcohols and amines is ~20 which is quite large; a difference in the extent of resonance delocalisation isn’t likely to affect pKas by that much. (Compare for example methane (pKa 50) vs toluene (41). In your example both molecules are resonance stabilised so the difference due to this should be much smaller.)

Is the difference between their electronegativities too large to explain the pka difference of 15. Nitrogen is 3 and oxygen is 3.5 on Pauling scale of electronegativity.

The electronegativity scale is essentially arbitrarily constructed, there’s little reason to expect that there should be a quantitative correlation. Maybe there is, I don’t know, but you can’t say that an electronegativity difference of 0.5 “looks small to me” so the pKa difference “shouldn’t be so large”, or anything along those lines.

Same reason that water is more acidic then ammonia.

The hydrogen of phenol is more acidic than that of aniline due to greater electronegativity of oxygen

Taken from my answer here:

Comparing methane and acetylene shows a difference of $\approx 20$ $\mathrm pK_\mathrm a$ units
Comparing $\mathrm p K_\mathrm a$ of phenol($10.0$) and methanol($15.5$), we see that the presence of one resonating ring only produces a 5.5 $\mathrm p K_\mathrm a$ difference. Resonance in triphenyl system is effectively only one phenyl group at a time due to the steric effects due to the bulkiness of the phenyl groups.

Here methane and acetylene form anions where the negative charge on an $\mathrm{sp^3}$ and $\mathrm{sp}$ carbon respectively. Here, there is an electro negativity difference between the two carbons which leads to a very high difference in $\mathrm p K_\mathrm{a}$.

The difference between their electronegativities too large to explain the $\mathrm p K_\mathrm a$ difference of $15$. Nitrogen is $3$ and oxygen is $3.5$ on Pauling scale of electronegativity.

This small amount makes a big difference in $\mathrm p K_\mathrm a$ values. I say this because, if methane and acetylene have a difference of 20 units where the only difference is their hybridization and therefore electronegativity, oxygen and nitrogen will have a big difference in $\mathrm p K_\mathrm{a}$ values.
This can be seen in the actual values as referenced in Evan's $\mathrm pK_\mathrm a$ table, where aniline has a $\mathrm pK_\mathrm a$ value of $30.6$ and phenol has a $\mathrm p K_\mathrm a$ value of $18.0$ both referenced in DMSO.


text:
Role of Lewis acid in synthesis of adamantane from bicyclic species
What is the role of $\ce{AlCl3}$ in the following synthesis of adamantane? I'm guessing it functions as a Lewis acid, but I am unsure how the mechanism of the second step works, as there is no region of high electron density in the bicyclic species.
If a mechanism for the second step exists, I would greatly appreciate if someone could propose one.

Image source: Wikipedia

Tada, found it: https://chemistry.stackexchange.com/questions/57826/isomerisation-of-alkanes-under-lewis-acidic-conditions Not the same question, but related. The reference in my first comment (Chem. Soc. Rev. 1974, 3, 479) hopefully contains more than enough info for an answer, if anybody is keen on writing an answer.

The carbocation rearrangement mechanism is here: Schleyer, et al., J. Am Chem. Soc., 1973, 95, 5769. Too much to draw and I don't want to lift it from the paper.


text:
A chamber to control concentration of a vapor: Is there a name for this equipment?
Mechanical engineer here so please excuse me if my question seems dumb to you!
I want to have a chamber $(\approx 40 \times 30 \times \pu{30 cm})$ to control concentration $(\pu{ppm})$ of a flammable material (e.g. alcohols or ketones) at a specific temperature $(T\approx \pu{24 \pm 10 ^\circ C})$ and pressure $(P\approx \pu{1 \pm 0.5 bar})$

What should this chamber be called? I need the the name to be able to search for it and also mention in my publication.
Candidate names so far:

Incubator --> mostly for biological entities
Reactor
Environmental chamber --> too expensive
Atmospheric controlled chamber
Autoclave --> for sterilization(high temperature/pressure)
Flash drum
Vapor inhalation chambers --> for animal testing

P.S. I have posted the question also on Reddit here and here. There are some good points mentioned there.

Maybe environmental chamber? Welcome to Chemistry.SE by the way:)

@andselisk thanks for not burning me to the ground :) other SO communities are really hostile to newcomers. Environmental chamber seems very close to what I need. but it seems very expensive. Are there any smaller versions ($\approx 40 \times 30 \times 30 cm$) with not too many requirements?

Good question; I don't really know, I'm sorry. I used a big one only once a decade ago, and yep, those were and are remaining expensive. Probably you can find a good deal on LabX, it's like an eBay for lab equipment.

they are indeed expensive! Thanks. I will keep searching.

I would probably also add the dimensions or the volume of the chamber to the question. Also, the information whether the liquid and the vapors are corrosive might come in handy.

@andselisk are Alcohol or Acetone corrosive?

Not really, they can dissolve some plastics though. If you intend to use them, I would just add both as the examples to the question.

@andselisk I also mentioned that they are flammable.

I think the term is autoclave. This is normally for high pressure processes. If you are running at about atmospheric P and just controlling the partial P of substances I would suggest controlled atmosphere chamber.

There is a book The Design of Controlled-atmosphere Chambers for the Study of Oxygen Toxicity, so I suggest "contolled-atmosphere chamber".

Thanks @davephd , I'm not able to find the source on the internet. is it a theses or something? do you have the PDF? It seems there are no off the shelve solution for me and I have to make anyway!

Searching controlled atmospheric chamber on google mostly refers to glove boxes made by Plas-Labs.  that's not what I'm looking for.

from this comment: typically atmospheric tends to be reserved for situations where atmospheric gases are used instead of organic vapors


text:
How to name Hg[Co(SCN)4]?
I have basically written the entire name except for the oxidation state of $\ce{Co}$.

Mercury tetrathiocyanatocobaltate(  )

to know what is the oxidation state of cobalt, I must know the oxidation state of the mercury cation. But mercury can have $+1$ and $+2$ oxidation states. How do I know which one is it in this compound?

You don't (that is, unless you know the crystal structure and stuff). In fact it is +2.

So, I cannot write the name? It's a question from my homework.

Then write (II) and be done with it. I don't see how you were supposed to figure that out.

It's Mercury(II) tetrathiocyanatocobaltate(II).

It couldn't be Hg (I) Co (III) compound. There would need to be Hg2 group and cobalt is too strong oxidant for it to be present.

These are certain exceptions which need to be remembered I guess.

TL;DR: There is no inner- or outer coordination spheres in this complex. Both cobalt(II) and mercury(II) have nearly ideal tetrahedral coordination environment with $4$ $\ce{N}$ and $4$ $\ce{S}$ atoms, respectively, hence the proper name would be cobalt(II) mercury(II) tetrathiocyanate. Mercury(I) readily undergoes disproportionation, so assuming mercury(II) is the safest option.

Actually, denoting an inner coordination sphere is not correct for this compound. First crystal structure [1] has been assigned to cobalt(II) tetrathiocyanatomercurate(II) $\ce{Co[Hg(SCN)4]}$:

It appears reasonably certain that the $\ce{S}$ atom is attached to the $\ce{Hg}$ in tetrahedral co-ordination, with $\ce{S-Hg-S}$ angles of 120° and 104°.

But two decades later the structure has been refined again, by the group including the author of the original publication. This time the geometry has been established more precisely (ICSD#36062), and it turned out that there is an infinite network of $\ce{Co^2+}$ and $\ce{Hg^2+}$ cations cross-linked via thiocyanate ligands, and the new suggested name was cobalt mercury thiocyanate $\ce{Co(SCN)4Hg}$ [2]:

The combination of two tetrahedrally coordinated atoms, $\ce{Hg}$ and
  $\ce{Co}$, has produced a most unusual arrangement in which the $\ce{Hg}$ and $\ce{Co}$ atoms are held apart by four spirals, each containing $4$ $\ce{SCN}$ bridges, which are interlinked so that any one $\ce{SCN}$ bridge takes part
  in eight spirals.
[...] Each such spiral is a spring holding the $\ce{Hg}$ and $\ce{Co}$ atoms apart and straining the bonds in the process. It is almost certainly this strain which flattens the tetrahedral coordination round $\ce{Hg}$ and $\ce{Co}$ in the $c$ direction. If the arrangement could be reproduced mechanically it would probably provide the ideal spring mattress!

References

Jeffery, J. W. Nature 1947, 159 (4044), 610. DOI: 10.1038/159610a0.
Jeffery, J. W.; Rose, K. M. Acta Cryst B 1968, 24 (5), 653–662. DOI: 10.1107/S0567740868002980.

There are two little clues you can use.
First, mercury(I) isn’t actually present as $\ce{Hg+}$; the cationic species is actually $\ce{Hg2^2+}$ with a $\ce{Hg-Hg}$ bond. Thus, if this were a mercury(I) compound, the sum formula would be $\ce{Hg2[Co(SCN)4]2}$.
This obviously only helps you if the correct formula was supplied as part of the question but the second clue is independent:
Look up a table of standard electrode potentials. There are two redox reactions that can occur, they are reproduced below:
$$\begin{align}
\ce{2 Hg^2+ + 2 e- &<=> Hg2^2+}&&&E^0 = \pu{+0.91V}\\[1.0em]
\ce{Co^3+ + e- &<=> Co^2+}&&&E^0=\pu{+1.82V}\\
\end{align}$$
The half-cell with a higher standard electrode potential will oxidise a half-cell with a lower standard electrode potential which means that, in solution, cobalt(III) and mercury(I) should react as follows:
$$\ce{2Co^3+ + Hg2^2+ -> 2Co^2+ + 2 Hg^2+}$$
As
$$E^0_\text{cell} = E^0_\text{red} - E^0_\text{ox}$$
we get a standard cell potential of
$$E^0_\text{cell} = \pu{+1.82V} - (\pu{+0.91V}) = \pu{+0.91V}$$
which is clearly positive, indicating a spontaneous reaction.
Of course, this is using aquacomplexes of both cations rather than whatever structure they would have in the complex you are supposed to name. Different chemical environments will modify a reaction’s electrode potential. However note that the uncorrected potential difference is so large – $\pu{+0.91V}$ – that it will be very hard to stabilise these otherwise incompatible oxidation states in solution.
On the other hand, a compound made up of mercury(II) and cobalt(II) is redox-stable.


text:
Water cooling by evaporation

For many years of drinking water has been cooled in hot climates by evaporating it from the surfaces of canvas bags or porous clay pots. How many grams of water can be cooled from $\pu{35 ^\circ C}$ to $\pu{20 ^\circ C}$ by the evaporation of $\pu{60 g}$ of water?
Heat of vaporisation of water in this temperature range is $\pu{2.4 kJ g-1}$. Specific heat of water is $\pu{4.18 J g-1 K-1}$.

I'm pretty confuse by what the question is asking for. This is my attempt for the evaporation of $\pu{60 g}$ of water:
$$q = \pu{60 g} \cdot \pu{2.4 kJ g-1} = \pu{144 J}$$
I then subbed this energy unit ($\pu{144 J}$) to the heat change formula to find the mass of water. Why am I wrong?

What answer did you finally get and why do you think it's wrong?

@IvanNeretin I think it's wrong because 144J is the energy needed overcome the forces to form the bond at 100 degrees C (evaporation), but I have no idea on how to find how many grams can be formed from 35 to 20 degrees C

You said something about some heat change formula, so in fact you did have an idea. What made you abandon it?

@IvanNeretin I abandon it because after looking through the question again and again, I cannot substitute 144KJ into the heat temperature change formula as it make no sense. With that, this means I need to find the amount of heat energy (q) to sub it in to find the mass which is what I am unclear about .

144 J is the amount of heat energy; what's wrong with that?

@IvanNeretin but 144KJ is the amount of heat needed for the water to undergo a phase change from liquid to gas . But in order to find the mass of water , I need to substitute the heat energy value that is needed to decrease the temperature from 35 to 20 degrees C , which again , I feel that 144 KJ cannot be that energy value .

Two questions that might bring you closer to solve this yourself: First of all, check the units. Is it J or kJ? Second, how much energy does water need to be heated from 20 to 35 °C?

@user175089 That's an awfully shaky foundation. You feel it can't be right, do you? All physical laws and equations lead you to plug in the values and get the number, but some kind of premonition prevents you from doing that? Too bad.

It helps to write down the formulas, before you plug in the numbers.
Heat you need to extract to cool the water:
$$Q=m_1 c_p\Delta T$$
Heat needed to evaporate water:
$$Q=m_2L_v$$
So:
$$m_1 c_p\Delta T=m_2L_v$$
So:
$$m_1=\frac{m_2L_v}{c_p \Delta T}$$
Now plug in the numbers.

I'm pretty confuse by what the question is asking for.

Because you are confused, I'll try to explain without any formulas. Please follow this step for step. Do not continue with the next step if you did not understand the previous one.

When water evaporates, it needs energy for that.
The task assumes that water evaporating from a clay pot will cool the water in clay pot. Part of the energy of the water in the pot (part of its 35 °C) will be used for evaporation of the 60 g.
The task simplifies things enormously. We have no indication that we would take care of the pot itself, of the surrounding air or anything.
How much energy does the evaporation of 60 g of water need? You already calculated that. But you need to double check your units! Is it J or kJ? It is 144 kJ.
Now, this energy is taken from the water inside the pot. The water has some energy (because it is 35 °C). Can we calculate that? We could, it we knew how much water it was, but we don't. But you could write down a formula that takes the mass of the water and results in the energy of that water. What would this be?
We can do the same thing with water of 20 °C.
We could then create the formula for "How much more energy does 35 °C water have in comparison to 20 °C water.
Okay so far? But we are lucky the energy difference between water of 35 °C and 20 °C is the same as between water of 15 °C and 0 °C. There is a constant that sais "How much energy is needed to heat one gram of water by 1 °C". You are lucky, you already have it mentioned in the original question.
So, if we need 4.18 J to heat one gram of water by 1 °C, how much energy do we need to heat one gram of water by 15 °C? Do you have that number? It is 15 times 4.18 J = 62.7 J.
The energy needed to heat water by 15 °C is the same as we need to remove if the want to cool water by 15 °C.
Okay, we know how much energy is needed to evaporate the 60 g. That is 144 kJ.
We know how much energy we need to remove from the inner water to cool one gram of it by 15 °C. That is 62.7 J. If there were 2 grams in the pot, it would need double that, 125.4 J. If there were 3 grams in the pot, it would need 188.1 kJ to cool it down by 15 °C.
How many grams of water could we cool by 15 °C with the removal of 144 kJ? That is how often does 62.7 J fit into 144 kJ? Be careful with the units! How often does 62.7 J fit into 144,000 J?

If you understood every step of this "manual", you should try to answer the same question with formulas. This will be much faster. But, of course, you need to understand the underlying concept first. This is why I wrote this step by step instructions.


text:
Heating an ideal under constant pressure

To what Celsius temperature must a gas sample initially at $\pu{−20 {^\circ}C}$ be heated if its volume is to double while its pressure remains the same?

What I've tried: first of all change to Kelvin, then I figured this out $V_1/T_1=2V_2/(2T_2)$
Since its volume is doubled so the temperature should be doubled. Am I right? The answer is supposed to be 233.

You seem to already have your answer, to double the volume you need to double the temperature (in Kelvin). Then just convert back to celsius.

You're almost right.  You applied the gas laws properly, but ... You must include units in your answer.  You need 233 °C not just 233.  You may also want to include intermediate steps where you convert the temperature to absolute temperature and tgen double it, the teacher may look for that to grade your understanding of the problem.


text:
A mole of atoms is the number of atoms whose total weight equals the atomic weight?
My textbook says, a mole of atoms is the number of atoms whose total weight = the atomic weight. There's something in the definition that's confusing me.
Now what is confusing me is:
Atomic weight = weight of 1 atom
Therefore, number of atoms whose total weight is atomic weight = 1.
Going by definition then, 1 mole atoms=1 atom. Which is incorrect.
Because 1 mole atoms = $\pu{6.022*10^23 atoms}$.
So from the definition apparently, weight of $\pu{6.022*10^23 atoms}$ = atomic weight(weight of 1 atom)
Which is contradictory.
Could someone clarify where am I wrong and explain the same?
PS: There are no units in the provided definition in the book.

1 mole atoms = 1 atom? No! Read https://en.wikipedia.org/wiki/Mole_(unit)

Also, Wikipedia says: Atomic weight: the ratio of the average mass of atoms of an element (in a given sample) to one unified atomic mass unit. The unified atomic mass unit, symbol u, is defined being  1⁄12 of the mass of a carbon-12 atom.

This is the problem when people don't use units...

Your textbook is inaccurate.
First of all, it should be total mass, not total weight.
The following statement is correct: A mole of atoms with an atomic weight of 12 has a total mass of 12 unified atomic mass units.
But you cannot say: A mole of atoms is the number of atoms whose total mass (not weight!) is the atomic weight. (wrong!) although it might look very similar.
The atomic weight is an alternate name for relative atomic mass. It is "The ratio of the average mass of the atom to the unified atomic mass unit." (source) It has no unit.
A mole is a unit of amount. "The mole is the amount of substance of a system which contains as many elementary entities as there are atoms in 0.012 kilogram of carbon-12. When the mole is used, the elementary entities must be specified and may be atoms, molecules, ions, electrons, other particles, or specified groups of such particles." (source)
So your textbook sentence should end: A mole of atoms is the number of atoms whose total mass is 0.012 kg if they were all carbon-12 atoms.
The unified atomic mass unit (see definition of atomic weight) is "one twelfth of the mass of a carbon-12 atom in its ground state" (source).
So, a mole of atoms is the number of atoms whose total mass is the atomic weight (relative atomic mass) multiplied by one twelfth of the mass of an carbon-12 atom.
Or, a mole of atoms is the number of atoms whose total mass is the atomic weight (relative atomic mass) multiplied by the unified atomic mass unit.
I find these statements much harder to understand than the original definition ("a mole is as many elementary entities as there are atoms in 0.012 kilogram of carbon-12").
What might have some application: One mole of atoms with the atomic weight $x$ have a total mass of $x$ amu.

If you are looking for a simple answer:
1 mole of atoms ($6.022*10^{23}$ of atoms) = the Relative Atomic Mass of the atom in grams.
For example, looking at $C-12$, it has a relative atomic mass of 12.000 as shown on the periodic table. One mole of $C-12$ atoms has a mass of 12.000g, hence equivalent to the relative atomic mass of $C-12$.

It has a relative atomic mass of 12.000 ~DOT~


text:
Synthesize camphene
How does one easily synthesize camphene in the modern day? The old 1902 patent says to combine ammonia and pinene hydrochloride in an autoclave at 210-220C for 20 hours. Would alpha-pinene work with that process?

alpha-pinene can be isomerized using acid treated clay as the catalyst to produce camphene. (see references in the comments below)
Isolated Camphene can be naturally sourced from valerian (Valeriana officinalis) extracted via steam, expression, or cryogenic methods, and fractionally distilled for purity http://www.tandfonline.com/doi/abs/10.1080/10412905.2008.9700079

Could you provide a reference or additional details regarding your answer?

Alpha-pinene isomerization on acid-treated clays
Nurgün Be˛sün a,∗, Fehime Özkan b, Gönül Gündüz a
a Ege University, Faculty of Engineering, Chemical Engineering Department, 35100 Bornova, ˙
Izmir, Turkey b Izmir Institute of Technology, Faculty of Engineering, Chemical Engineering Department, 35230 Çankaya, ˙
Izmir, Turkey
http://www.sciencedirect.com/science/article/pii/S0926860X01008523

Isomerization of Alpha-pinene Over Acid Treated Natural Zeolite
Ercan Ünveren , Gönül Gündüz  & Fehime Cakicioğlu–Özkan
http://www.tandfonline.com/doi/abs/10.1080/00986440590477773


text:
Determining the reaction time of water synthesis by use of differential equations
How do I determine the reaction speed?
I know you need a value, k, and order, which is only gained through experimentation, but how do the differential equations play in? How do I plot the result?
Assuming it's the first order ...
How do I proceed? I now have this function but I am not sure what to do with it. I do not have the means to observe the reaction and figure out its k-value or order. Is this reaction a known one, though? Is the order and k-value something you can find on the Internet?:
$$\ce{2H2 + O2 -> 2H2O}$$

You either guess at the elementary steps and derive the rate law from that, or you perform the measurements and determine an experimental rate law. There is no other way. You can find pretty much anything on the internet.

Unfortunately, I have been unable to find the order/value on the internet

Because there's no "order/value"? Mechanism of this reaction is quite complicated, assuming it could be first order was overoptimistic as there's no order, only in specific situation it may seem there is.

Assuming you somehow can get the air resistance and other stuff out of the way, you would, in effect, be measuring the thrust (force provided by the reaction) vs time. This would depend on the exiting reaction mixture. In this case, it is hot steam. (Also, if you react gases in an otherwise uncontrolled manner, you'll get an explosion, no sustained thrust as such). You can calculate how much of the reaction must have taken place to deliver that thrust.
So, I'd say it is possible, but only if the errors in measurement are low enough...

How do I determine the reaction speed?

The rate of reaction can be determined experimentally or theoretically, the latter case requiring that a reaction mechanism be known. Proposing a reaction mechanism requires knowledge of the reaction at multiple regimes and some guesswork. Evidence for a certain reaction mechanism is usually indirect.
Theoretically speaking, once the reaction mechanism and rate constants are known, we can write down a coupled system of kinetic rate equations and solve them simultaneously to find the rate of reaction.

I know you need a value, k, and order, which is only gained through experimentation, but how do the differential equations play in? How do I plot the result?

The differential equations govern the reacting system, and their solution completely specifies the system, enabling you to find the rate of reaction. You would then simply plot, say, the rate of reaction vs. time.

How do I proceed? I now have this function but I am not sure what to do with it. I do not have the means to observe the reaction and figure out its k-value or order.

I don't believe you can proceed, because your initial assumption was wrong. The combustion of hydrogen is far more complicated than a first-order reaction, which you have assumed; you should not, therefore, expect that your first-order analysis should describe your system well.

Is this reaction a known one, though? Is the order and k-value something you can find on the Internet?

It has certainly been studied. I will refer you to Reaction mechanism of combustion of H2 and the links therein.

Thank you. What about the combustion of solid sulphur to form sulphurdioxide? I found out that giving a rate constant to the combustion of hydrogen is incredibly complex. Is it the same for sulphur?

@javanewbie, essentially all combustion reactions will have very complicated mechanisms.


text:
Why is formic acid more acidic than benzoic acid?
Why is formic acid(pKa=3.751)  more acidic than benzoic acid(pKa=4.204)?
Here's what I understand:
1) Phenyl ring is electron withdrawing therefore it should decrease electron density on oxygen thus stabilizing the anion obtained after deprotonation. Considering this argument, benzoic acid should be more acidic than formic acid.
2)
It looks like resonance of phenyl ring with carboxylate should destabilize the anion and therefore, formic acid should be a better acid. But the "aromatic ring is so thermodynamically stable that the phenyl ring donating electrons to the carboxylate is not a major resonance structure".
Considering the two above factors, benzoic acid should be stronger than formic acid. Then why it's not?

Related: Comparing acidities of substituted and aromatic carboxylic acids


text:
Is it a "Bary center" or a "Bari center"?
(Studying Coordination Chemistry, "Crystal Field Theory" at school)
Apparently, different sources spell this differently.
The Wikipedia article on Crystal Field Stabilization Energy, calls it the "Barycenter" (as does my school textbook)

However, the actual Wikipedia article on the "Barycenter" refers to, (as far as I can tell) the center of mass of two mutually orbiting bodies. Whereas the "Bary/Baricenter" I'm talking about is the zero-energy reference level in octahedral complexes.
Another book I have at hand (Concise Inorganic Chemistry Lee, J.D), prominently uses the term "Bari centre".

The center/centre distinction aside, what I've been able to glean from my books (and the internet), is that this term is spelled in four different ways (Wah! Such gross inconsistency in literature!):

Barycenter
Bary center
Baricenter
Bari center

But which one's correct (or at least, a more preferred term in Chemistry-literature)?

"barycentre" is the original spelling of the word.  https://books.google.com/books?id=fIRAAAAAIAAJ&pg=PA88&dq=%22Barycentre%22&hl=en&sa=X&ved=0ahUKEwi9gIjfh_vXAhXkSN8KHaFpAcUQ6AEIKjAA#v=onepage&q=%22Barycentre%22&f=false so none of the above

My Greek roots say barycenter, though I'll allow DavePhD's variant.

@DavePhD +1 for insisting on th BE version. ;-) And Bari is a town in Italy. Looks to me like Lee tried to be original.

But here it is written "bari"center :(.

@Nilay Yup, I saw that too. Judging from the other comments, apparently "barycentre" works (but I'll wait till someone posts an answer). O:)

https://en.oxforddictionaries.com/definition/barycentre , https://en.oxforddictionaries.com/definition/barycentric , https://www.merriam-webster.com/dictionary/barycenter

@Karl Yep, it seems now that "barycenter/re" is the more widely used term; the question now, would be whether it applies to the Crystal Field Theory as well as two-body systems (the links suggest the latter). Thanks for the link! O:)

As far as I can tell Barycentre is the English spelling, Baricentre the French (and maybe Italian?). So maybe the version with an "i" is from francophones?

"bari" or "bary" are not English words, so the versions with two words are  just plain wrong.

It is safe to say that Barycenter is the correct spelling for the sought word. The other versions are artifact of the original word as some authors preferred to name (especially in the internet).
Barycenter (or Barycentre in British English) is a commonly used term in astronomy and in chemistry (in Crystal Field Theory) as well. In both fields, it is used to explain center of mass balance and most of the literature has used the aforementioned spelling. For example, NASA explains the question "What Is a Barycenter?" as:

We say that planets orbit stars, but that’s not the whole truth. Planets and stars actually orbit around their common center of mass. This common center of mass is called the barycenter. Barycenters also help astronomers search for planets beyond our solar system!

In Crystal Field Theory (CFT), electrons in the $\mathrm{d}$-orbitals are repelled to different degrees by the ligands. The degeneracy of the $\mathrm{d}$-orbitals is lifted. The Barycenter is the average energy of the $\mathrm{d}$-orbitals (Coordination Chemistry II: Theories of Electronic Structure and Crystal Field Theory):


text:
Lewis Structure of SO2
What is the structure of SO2? I have seen two different ways the Lewis Structure is written:

The formal charges of the SO2 with the single bond and a double bond is larger than the SO2 with two double bonds. So I would assume that the one with two double bonds is the correct structure. But chemistry books I have looked at (Zumdahl Edition 5 and 7) says that it is the opposite.
Which is the correct Lewis Structure?

There is such thing as a "correct Lewis structure."

SO2 Lewis Structure - How to Draw the Lewis Structure for SO2 (Sulfur Dioxide)

The Lewis structure most closely resembling reality consists of two resonance structures: the first one posted in the question and its mirror image. The reason is that the octet rule is observed this way (no hybridization of d-orbitals for main group chemistry necessary or possible) as well as the symmetry of the molecule: the two bonds are identical.

I had the notion that sulfur had expanded valence electrons. So why wouldn't sulfur be able to use them and make the formal charges get to 0 for all three atoms? Also, how does the molecule have symmetry? The bonds aren't identical, one is a single bond and the other is a double bond.

The identical bonds are a physically observable reality. Expanded valence electrons are an outdated concept refuted by modern quantum chemistry.

How does sulfur make a molecule such as Sulfur Hexafluoride in modern quantum chemistry, since expanded valence electrons are outdated?

@Hawkeye check out molecular orbital theory.  You can easily hold sulfur hexafluoride together without using "outer $d$ orbitals".


text:
Determine rate law from experimental data
For the reaction A -> B + C we have the following experimental data.
\begin{array}{cccc}
\text{Experiment} & \ce{P_A} &\ce{P_B} &\ce{P_C} & \text{Reaction rate}\ce{}\\\hline
1 & 1 & 1 & 1 & 3.5\\
2 & 5 & 1 & 1 & 1.1\\
3 & 10 & 1 & 1& 0.7\\
4 & 2 & 5 & 1& 1.9\\
5 & 2 & 10 & 1& 1.6\\
6 & 3 & 0 & 5& 1.5\\
7 & 3 & 0 & 10& 1.2\\
8 & 0 & 5 & 8& 0\\
9 & 3 & 3 & 3& 1.4\\
\hline
\end{array}
I have been searching around the internet, but it seems like many of the examples that I found include fewer experiments and/or different chemical reactions.
I am thus wondering what method one can use to find the appropiate rate law, and if one uses all of the experimental data? Looking at all the data at once it is harder to see the pattern.
Supposedly this should be possible to do without using regression with Excel or other computer software.
Any suggestions are highly appreciated!

There are no good answers here for an arbitrarily complicated system. Usually, some sort of curve fitting for each variable with possibly brutal application of Okkam's razor. Here, though, it's not even clear your data make sense. Why is the reaction rate inversely proportional to the reactant? Something isn't right.

That is a good observation. Maybe the whole point is that the data doesn't make sense or that some of the data should be disregarded (maybe due to human errors ).


text:
Acid-base titration: Calculate pKa with only three values given

There's an unkown acid, diluted with an unknown amount of water and titrated with $\ce{NaOH}$.

After adding $\pu{10.00 mL}$ $\ce{NaOH}$, a $\mathrm{pH}$ value of $4.65$ is measured.
After adding another $\pu{12.22 mL}$ ($\pu{22.22 mL}$ $\ce{NaOH}$ in total), the equivalence point is reached.

Calculate the $\mathrm{p}K_\mathrm{a}$ of the acid!

I'm currently facing this uncommon task regarding acid-base titration. There's quite few information in this text, only three values are given:
\begin{align}
V_1(\ce{NaOH}) &= \pu{10.00 mL}: &\quad &\mathrm{pH = 4.65} \\
V_2(\ce{NaOH}) &= \pu{22.22 mL}: &\quad &\text{equivalence point reached}
\end{align}
What I've got so far:
Before adding any $\ce{NaOH}$, the reaction should look like this:
$$\ce{HA + H2O <=> A- + H3O+}$$
I expect the neutralisation reaction to be:
$$\ce{A- + H3O+ + Na+ + OH- <=> NaA + 2 H2O}$$
In general $n = c V$. At the equivalence point $n(\ce{NaOH}) = n(\ce{HA})$, so:
$$C(\ce{NaOH}) \cdot V_2(\ce{NaOH}) = n(\ce{HA})$$
$$C(\ce{NaOH}) = \frac{n(\ce{HA})}{\pu{22.22 mL}}$$
Henderson-Hasselbalch:
\begin{align}
\mathrm{pH} &= \mathrm{p}K_\mathrm{a} + \log{\frac{C(\ce{A-})}{C(\ce{HA})}} \\
\mathrm{p}K_\mathrm{a} &= \mathrm{pH} - \log{\frac{C(\ce{A-})}{C(\ce{HA})}} \\
\mathrm{p}K_\mathrm{a} &= 4.65 - \log{\frac{C(\ce{A-})}{C(\ce{HA})}}
\end{align}
I'm unsure how to express this equation as a function of $V_1(\ce{NaOH})$. I may be on the wrong track as well. Any help is appreciated!

Who marked this off-topic as homework? There is clearly an attempt to solve the poblem...

@Zhe It's not a mandatory homework but a voluntary exercise. My fellow students also had problems with it, so I decided to ask for help here.

That's not what I meant. Someone voted to close as homework, but there's no reason to do that because you showed a lot of work towards solving the problem.

You've done quite well there. You correctly calculated the number of moles of acid in the original solution as:
$$n(\ce{HA}) = C(\ce{NaOH}) \times V_2$$
After the first addition of base, the number of moles of conjugated base is:
$$n(\ce{A}) = C(\ce{NaOH}) \times V_1$$
And by stoichiometry, the remaining number of moles of acid:
\begin{align}
n'(\ce{HA}) &= n(\ce{HA}) - C(\ce{NaOH}) \times V_1 \\
&= C(\ce{NaOH}) \times V_2 -C(\ce{NaOH}) \times V_1 \\
&= C(\ce{NaOH})(V_2 - V_1)
\end{align}
Because the volumes the remaning acid and conjugated base are contained are the same:
$$\mathrm{p}K_\mathrm{a} = 4.65 - \log{\frac{C(\ce{A-})}{C(\ce{HA})}} = 4.65 - \log{\frac{n(\ce{A})}{n'(\ce{HA})}}$$
The $C(\ce{NaOH})$ drop out and you have your $\mathrm{p}K_\mathrm{a}$:
$$\mathrm{p}K_\mathrm{a} = 4.65 - \log{\frac{V_1}{V_2 - V_1}}$$

I brushed up the formatting a little. \ce{...} from mhchem is used for chemical equations and chemistry-related notations. Also, generally I wouldn't recommend putting chemicals in super- or subscripts; this would make indices in chemical formulas hard to read and to parse (compare $X_\ce{H2O}$ vs $X(\ce{H2O})$). Aside from that, looks great:)

Actually, I missed the connection between $n\ce{(HA)}$, $n(\ce{A-})$ and $n\ce{(NaOH)}$ expressed in your first two equations. From my neutralisation reaction, the mole ratio is pretty obvious, but I didn't know that it's allowed to use it to equate $n(\ce{A-})$ and $n\ce{(NaOH)}$ this way. Thank you @Gert!

@andselisk: personally I prefer the subscript notation but each to their own. Thanks!

@Candor: yes, that's where you went wrong. Easy when you know it, I guess!

First, we note that this is not a strong acid, since when it is only partially neutralized, the $\mathrm{pH}$ has already risen to almost $5$.
At that point, we know that $\frac{10}{22.2}$ of the total acid has been neutralized. Therefore, the ratio
$$\frac{\ce{[A-]}}{\ce{[HA]}} = \frac{10}{22.2} \cdot \left(1 - \frac{10}{22.2}\right)^{-1} = \frac{1}{1.22}$$
Substitution into Henderson-Hasselbalch gives that the $\mathrm{p}K_\mathrm{a} = 4.73$.
Note that I assumed that the solution wasn't super dilute. If that were the case, I couldn't plug directly into the Henderson-Hasselbalch equation because the acid-base equilibrium would need to shift, but the change in the concentration of hydronium would be dominated by auto-ionization of water.
And in general, the total volume of solution does matter. You could consider a solution of hydrochloric acid that were dilute enough that even the first assumption of a weak acid were wrong. It's pretty easy to construct the solution such that the partially neutralized solution had a $\mathrm{pH}$ of $4.65$. I'll leave that as a follow-up exercise for you to explore.

My calculation looks like this: $\ce{{pKa} = 4.65 - log(\frac{1}{1.22}) = 4.73}$. Eventually you used addition instead of substraction? Anyways, thanks for your explanation!

@Candor Yes I did that in my head and didn't add a minus sign for the log.


text:
Could detergents dissolve cell membranes?
Animal cells have a lipid membrane and detergents are great at emulsifying lipids. Yes, there are proteins, cholesterol and other numerous substances in the lipid bilayer, but the lipids make up the majority of it. If we dipped a bacterium or an isolated animal cell in detergent, would their membrane be dissolved?

Yes they can dissolve membranes and are even used to isolate active forms of membrane proteins by solubilization. The product are micelles of a lipid-detergent mix and a protein-lipid-detergent complex. Maybe you find this short abstract interesting.

A critical step in any in vitro analysis of membrane proteins is the
  solubilization of the membrane to extract the protein of interest in
  an active form to obtain an aqueous solution containing the membrane
  protein complexed with detergents and lipids in a form suitable for
  purification and further analysis. This process is particularly
  delicate as the aim is to maximally disrupt the lipid components of
  the membrane while putting the protein components in an un-natural
  detergent environment without perturbing them. Looked at this way, it
  is remarkable that it ever works. Although the process is difficult
  and hard to master, an increasing number of membrane proteins have
  been successfully solubilized in active forms, allowing some general
  principles to be established that we illustrate in the method
  developed in this chapter.

Duquesne K., Sturgis J.N. (2010) Membrane Protein Solubilization. In: Mus-Veteau I. (eds) Heterologous Expression of Membrane Proteins. Methods in Molecular Biology™ (Methods and Protocols), vol 601. Humana Press

So if our skin wasn't protected by the keratin layers, it would dissolve with detergents as well? What about GI epithelia?


text:
Stripping zinc electroplating with phosphoric acid - hydrogen danger?
I'll start by saying that I'm not a chemist, more of an artist. I'm stripping zinc electroplating from some regular hex bolts, nuts, and washers. I'm using a product called Ospho. It's basically phosphoric acid with a surfactant and a couple other additions, from what I understand.
I've read that zinc and phosphoric acid react to produce zinc phosphate and hydrogen gas.
My unknown is whether this hydrogen gas is emitted in great enough quantities to be dangerous / explosive when this reaction is carried out indoors. I don't know how much zinc is being reacted, but electroplating leaves an extremely thin coating and I'm talking about small quantities of fasteners.

Yes, mineral acids like $\ce{H3PO4}$ react with electropositive metals like zinc to produce dihydrogen gas ($\ce{H2}$).
The quantities generated of course depend on how much metal is being dissolved that way. About $65\ \mathrm{g}$ of zinc would produce about $44.8\ \mathrm{L}$ of hydrogen (in Standard conditions).
This may seem like a lot but your nuts and bolts are covered with only a very thin layer of zinc.
Hobby chemists dissolve metals in mineral acids all the time, with minimal danger. Just avoid open flames (use an electrical hot plate) and make sure your work place is well ventilated and you should be fine. Consult any safety advice supplied with the Ortho.

Yep, there's no danger of any problem, provided there are no flames in the immediate vicinity of the reaction.
$\ce{H2}$ is lighter than air, so it disperses itself and doesn't pool like most flammable gases.
If you were dissolving zinc powder, then you'd have to know that most zinc powder contains zinc sulfide contaminants, leading to toxic $\ce{H2S}$ emissions.
But that's not an issue for you.


text:
XeF4 - why are there only 2 dihedral mirror planes?
I started studying molecular symmetry and I became somehow confused about dihedral mirror planes. I found a definition, which says,  that it is a vertical mirror plane, which bisects two $C_2$ axes.
But when I look at $\ce{XeF4}$ molecule, I see, that it has 4 $C_2$ axes and 4 vertical mirror planes. All of those planes are bisecting 2 $C_2$ axes and both all of them also go through some atom.
Still, there are only 2 dihedral mirror planes $\sigma_\mathrm{d}$ and the remaining two planes are just common vertical ones $\sigma_\mathrm{v}$.
So, could you please tell me what's the real difference between vertical and dihedral mirror planes?
OK, in a link given in comments by Tyberius is said, that dihedral planes are such planes, which bisects as many bonds as possible, while "normal" vertical planes bisects as many atoms as possible. Then I don't get, why ethene (see the picture below) doesn't have one dihedral plane, as one of them clearly bisects its central bond and we can't symmetrically bisect more bonds in this molecule.

$\ce{XeF4}$

Ethene

http://www.chembio.uoguelph.ca/educmat/chm2060_preuss/L3-2013.pdf

@Tyberius Ok, now I see "bisects as many bonds as possible"... Could you, please, tell me then, why there is no $\rho_d$ in Ethene? I think, that one of the vertical planes should be dihedral, as it bisects the central double bond and there is no larger number of bonds to be bisected symmetrically.

It's in the definition you quoted: 'bisects two C2 axes'.

@orthocresol I thought about that, but while I didn't find it in the provided slideshow, I thought it's not correct...

It probably depends on the text, I'm not sure if there's one single authoritative source. Cotton's Chemical Applications of Group Theory (3rd ed., p 53) says that the use of $\sigma_v'$ vs. $\sigma_d$ is "somewhat arbitrary", but other textbooks say that $\sigma_d$ planes need to bisect C2 axes. Incidentally, if you look up the character table for $D_\mathrm{2h}$ you'll probably find that all three mirror planes are conventionally labelled with just $\sigma$.

@orthocresol Ok, thank you very much. Could you, please, write a short answer, so I could accept/upvote it?

@Tyberius Could you, please, write a short answer, so it could be accepted/upvoted?

I took the liberty to rearrange the structures on the pictures horizontally so that users with smaller or wider screens could compare them both without scrolling back and forth; plus I changed the notations for mirror planes from $\rho$ to more common $\sigma$ (see How are point group character tables typeset correctly?). If you are unhappy with this, please feel free to rollback to the original version:)

Strongly related: https://chemistry.stackexchange.com/questions/23954/why-does-symmetry-have-to-be-maintained-in-molecular-orbitals

So it seems that the use of $\sigma_\mathrm{v}$ vs $\sigma_\mathrm{d}$ is basically a matter of convention when both terms can be applied. Really, even $\sigma_\mathrm{h}$ vs $\sigma_\mathrm{v}$ is somewhat arbitrary if there isn't a single clear principal axis.
By a certain convention $\sigma_\mathrm{d}$ are vertical mirror planes that bisect $2C_2$ axes. When there are multiple mirror planes that fit this criteria, those that cut through the most atoms are referred to as $\sigma_\mathrm{v}$, while the others are called $\sigma_\mathrm{d}$.
This again is somewhat arbitrary and can be ambiguous. In particular, for $D_\mathrm{2h}$ molecules like ethene any of the mirror planes could fit the description for $\sigma_\mathrm{h}$ depending on which $C_2$ you say is the principal. For this point group, you can unambiguously label the axes and planes using the Cartesian axes

This is a situation of coordinate systems. In the $D_\mathrm{4h}$ point group, as you've noticed, there are 4 $C_{2}$ rotation axes in the plane. The easiest way to set the coordinate system is to have one pair be coincident with the $x$ and $y$ axes and the other pair coincident with the functions $x=y$ and $x=-y$.
Herein lies the issue: which pair is which? I think most people would agree that a reasonable choice is that the primary set should be coincident with the Cartesian axes. Indeed, an example character table for this group makes this choice as well (these are the $C_{2}'$ elements). This then defines which pair of mirror planes are $\sigma_\mathrm{d}$ and which pair are $\sigma_\mathrm{v}$.
This covers $\ce{XeF4}$ only, not ethene, because ethene belongs to the $D_\mathrm{2h}$ point group.
Here's the character table for that group. There's no dihedral mirror plane because the mirror planes perpendicular to the plane of molecule are coincident with the rotation axes.


text:
How do you remove residual Lawesson's reagent after thionation?
I am currently synthesizing a peptidomimetic bioisostere that replaces an oxygen with a sulfur. I use lawesson's reagent to achieve this, and althought the reaction works and despite running it through a column ( 80 mL silica, 10:1 DCM:MeOH) There is still a lot of residual lawesson's reagent left over. Does anyone have experience with this reagent and removing it? Any suggestions are appreciated

I would suggest recolumning with a gradient of MeOH in DCM starting at zero% MeOH.

Interesting, my current gradient is 30:1, 20:1, 10:1..so youre suggesting  starting at 0 % MeOH ? What would a typical gradient that follows 0% look like? I'm new to chemistry

My recollection is that Lawseson's residues are fairly non-polar and elute early so I would go with several column volumes of neat DCM before starting at 1% MeOH for a couple of column volumes then 2% etc. This means a much shallower gradient and more fractions than you've previously uxed.


text:
electron counting for transition metal-oxo ligand compound
Using donor-pair method, how many electrons would an oxo ($O^{2-}$) ligand contribute?  I cannot decide if it is 2 or 4.

Ok
So a very large ligand like ETDA are hexadentate
But
I thought of different metals
Different metals can accept different amount of pairs in 1 oxo ligands
For example MnO4-
Yes, it can hold total of 7pairs, and accept 4 electrons per oxo.
And then for W
Tungsten is very large, and can therefore accept a lot of electrons from one O2-, in fact, it can handle 3 pairs, or 6 electrons


text:
Do all the angles change in cyclopropane and cyclobutane or it is only the C-C-C angles?
Are the angles other than the C-C-C angles in cyclopropane and cyclobutane still 109.5° as the acyclic compound, or what?

The other bond angles can't all be $109.5°$ if the angles in the ring are different from that.  Bond angles around an atom are not independent variables because the amount of space to be spanned is fixed.


text:
Dumb Question about Oxidation
This is probably a really dumb question, but I can't find a clearly spelled out for me yes or no answer by searching through google, so hopefully I'll get a yes or no answer her. Can metals, for example lead, be oxidized by being submerged in boiling water?

No, it's not a dumb question (I'd suggest to edit the "dumb" part out). Yes, metals can be oxidized in boiling water (and also at lower temperatures, e.g. alkali metals).

(1/2) Long story for my concern, told by hypochondriac) I wanted to shrink a section of my hat that I bought from Bangladesh, because it had been was protruding outward slightly. I boiled water, collected it with a dropper and then fed that just boiled water to the section for 20 minutes. Some of it spilled onto the metal button thing that all the seems meet at in the middle, on the inside of the hat during the process. Left it on my bed to dry. Later saw on a small part of the bed where the hat was, a faint orangish color surrounded by the circle dampness that was left.

(2/2) Paranoid self went to the thought that the metal button was lead for maybe a less regulated country like Bangladesh could let stuff like that slide even though the hat was by a western brand (Abercrombie and Fitch), and that it had deteriorated while under the boiling water and bled into the soakage of my hat, later soaking into my bed, leaving the faint oragish mark because lead oxide can be red apparently. There may be no connection between the mark and the metal button at all for all I know. And I may be making a mountain out of a molehill. But in your opinion is my concern plausible.

Nope definitely not lead, but yes for a few of the metals.
There are different reason why things oxidize and reduces
To have a thing oxidize, you need another thing that reduces
To oxidize, the thing (like Pb you say) has to lose electrons and become more positive. Reduction is vice versa.
Water has to reduce in order to oxidize Pb.
However, water is a very persistent compound, it doesn't change chemically after heated to high temperatures (boiling). Neither does it want electrons. And so without doubt water is very useless here.
Pb on the other hand, is not very keen on giving out electrons (at least compared to water), both of them (even when heated) does not interfere with each other and no chemical change is observed.
But.
Some metals have a grudge on electrons and would lose it very rapidly. For example Group 1 metals like sodium
Compare to sodium, water's resolve to reject electrons is very weak, and such, when Na is in contact with cold water, Na rapidly provides electrons to water.
This process is very quick and it produces hydrogen gas, and we can see it catching on fire.
Caution: do not touch the liquid after having water react with sodium.
So, yes, some metals do oxidize when contact with water, but also, no, some don't, at least for lead.


text:
What does an equal sign mean in chemical equations?
I have been doing a bit of literature review, and have come across a few equations which use an equal sign (=) instead of an arrow, for example:
$$\ce{Fe^2+ + 1/2O2 = 2Fe^3+ + H2O}$$
What does the equal sign mean? Is the same as a single arrow $\ce{->}$ or the double arrow $\ce{<=>}$?

the gist is that it means a net or overall reaction.

The IUPAC Green Book (Quantities,
Units and Symbols in Physical Chemistry, Third Edition, 2007) reads:

2.10.1 Other symbols and conventions in chemistry
  [...]
(iv) Equations for chemical reactions
(a) On a microscopic level the reaction equation represents an
  elementary reaction [...] A single arrow is used to connect reactants
  and products in an elementary reaction. An equal sign is used for the
  "net" reaction, the result of a set of elementary reactions. [...]
$\ce{H + Br2 -> HBr + Br} \quad$ one elementary step in HBr formation
  $\ce{H2 + Br2 = 2 HBr} \quad$ the sum of several such elementary steps
(b) On a macroscopic level, different
  symbols are used connecting the reactants and products in the reaction
  equation, with the following meanings:
$\ce{H2 + Br2 = 2 HBr}\quad$ stoichiometric equation
  $\ce{H2 + Br2 -> 2 HBr}\quad$ net forward reaction
  $\ce{H2 + Br2 <--> 2 HBr}\quad$ reaction, both directions
  $\ce{H2 + Br2 <=> 2 HBr}\quad$ equilibrium
The two-sided arrow $\ce{<->}$ should not be used
  for reactions to avoid confusion with resonance structures [...]
2.12.1 Other symbols, terms, and conventions used in chemical kinetics
  [...]
(ii) Composite mechanisms
  A reaction that involves more than one elementary reaction is said to occur by a composite mechanism. The terms complex mechanism, indirect mechanism, and stepwise mechanism are also commonly used. Special types of mechanisms include chain-reaction mechanisms, catalytic reaction mechanisms, etc.

Examples
    A simple mechanism is composed of forward and reverse reactions
    $\quad\quad\quad\quad\quad\quad\ce{A -> B + C}$
    $\quad\quad\quad\quad\ce{B + C -> A}$
    It is in this particular case conventional to write these in one line
    $\quad\quad\quad\quad\quad\quad\ce{A <--> B + C}$

However, it is useful in kinetics to distinguish this from a net reaction, which is written either with two one-sided arrows or an "equal" sign
  $\quad\quad\ce{A <=> B + C}$
  $\quad\quad\ce{A =  B + C}$
When one combines a composite mechanism to obtain a net reaction, one should not use the simple arrow in the resulting equation.
  $\quad$Example
  $$\begin{align}
\ce{A &\to B + C}\quad\text{ unimolecular elementary reaction} \\
\ce{B + C &\to D +E}\quad\text{ bimolecular elementary reaction} \\
\hline
\ce{A &= D + E}\quad\text{net reaction (no elementary reaction,no molecularity)}
\end{align}$$

Déjà Vu

I'd say, the core fundamentals warrant a repetition if that makes them easier to find for future questioners.

I just like when it's dry, that's all:)

Rest assured. We covered arrows and we covered equals. I think there cannot be any more questions that could be answered by this quote. :-)

Right, this sounds fair, +1:)


text:
What is a 'triplet biradical transition state'?
My text book just referred to this term. I have absolutely no idea what it is. Articles on the internet are too difficult for me to comprehend.

In some alkenes, steric crowding in the planar state may be so severe that the latter is no longer an energy minimum but becomes an energy barrier with the ground state becoming twisted. In this context the compounds in figure are of interest. The drop in rotational barrier from stilbene ($\pu{42.8 kcal mol-1}$, $\pu{179 kJ mol-1}$)to compound $\ce{P}$ ($\pu{21.1 kcal mol-1}$, $\pu{88.3 kJ mol-1}$) to compound $\ce{Q}$ (“small barrier”, cf.) may be explained by increasing stabilization of the triplet biradical transition state. However, the “negative” barrier in $\ce{R}$, that is, the fact that the stable ground state of $\ce{R}$ is non-planar and that the planar conformation represents the transition state for rotation requires a different explanation; the difference between $\ce{Q}$ and $\ce{R}$ is presumably due to the steric interaction of the four chlorine substituents in the planar conformation of $\ce{R}$. This situation resembles that in o,o'- tetrasubstituted biphenyls to alkenes with low or “negative” barriers.

This is the question given: for which the above is the solution

Source: "Advanced Problems in Organic Chemistry by MS. Chouhan 9th Edition", Chapter GOC, page 53

For example, when the overlap of p orbitals are not sufficient to form a double bond, therefore the two electrons behave like radical electrons weakly coupled to each other instead of an electron pair in bonds.

A triplet biradical state simply implies that the bonding pair will be cleaved homolytically .  For example the triplet biradical transition state for

Will be


text:
How to calculate the pH of a buffer after adding HCl?

In a certain $\pu{1 L}$ buffer with $\pu{0.25 M}$ $\ce{NH3}$ and $\pu{0.4 M}$ $\ce{NH4Cl}$, $\pu{0.1 mol}$ of $\ce{HCl}$ has been added to the solution. What is the $\mathrm{pH}$?
$K_\mathrm{a}(\ce{NH4+}) = \pu{5.56e-10}$, and $K_\mathrm{b}(\ce{NH3}) = \pu{1.8e-5}$.

The answer says that at last there will be $\pu{0.5 mol}$ of $\ce{NH4+}$ formed from the equation:
$$\ce{NH3 + H+ -> NH4+}$$
So $\pu{0.1 mol}$ of additional $\ce{NH4+}$ is formed and the total amount of $\ce{NH4+}$ is $\pu{0.5 mol}$.
However, I do not agree with this statement and I believe that there is still a limited power of $\ce{NH3}$ accepting $\ce{H+}$ (from $\ce{HCl}$).
My theory is that $\ce{NH3}$ accepts $x$ amount of protons and will form $x$ amount of $\ce{NH4+}$, with the original $K_\mathrm{b}$. Meaning
$$K_\mathrm{eq} = \frac{[\ce{NH4+}]}{[\ce{NH3}][\ce{H+}]}$$
which I thought was $1/K_\mathrm{b}$. Is my theory correct?

Reasonably concentrated buffers like this one can be analyzed directly with Henderson-Hasselbalch assuming that we're in the buffered range (which we are).

I'm not sure I follow your argumentation, but I guess you don't really need $K_\mathrm{b}$ at all since addition of strong acid $\ce{HCl}$ influences the dissociation of the one buffer component — the weak acid $\ce{NH4+}:$
$$\ce{NH4+ <<=>[$K_\mathrm{a}$] NH3 + H+} \qquad K_\mathrm{a} = \frac{[\ce{NH3}][\ce{H+}]}{[\ce{NH4+}]}\tag{1}$$
Henderson–Hasselbalch equation applied to this buffer system before the addition of acid allows to find initial $\mathrm{pH}$ (not required by the problem, I do this solely for demonstration):
\begin{align}
\mathrm{pH} &= \mathrm{p}K_\mathrm{a} + \log{\frac{[\ce{NH3}]}{[\ce{NH4+}]}} \\
&= -\log{(\pu{5.56E-10})} + \log{\frac{\pu{0.25 M}}{\pu{0.40 M}}} \\
&= 9.05\tag{2}
\end{align}
Once the strong acid $(\ce{HCl},$ assuming complete dissociation) is added, the equilibrium shifts accordingly:
\begin{align}
\mathrm{pH_1} &= \mathrm{p}K_\mathrm{a} + \log{\frac{[\ce{NH3}] - [\ce{HCl}]}{[\ce{NH4+}] + [\ce{HCl}]}} \\
&= -\log{(\pu{5.56E-10})} + \log{\frac{\pu{0.25 M} - \pu{0.10 M}}{\pu{0.40 M} + \pu{0.10 M}}} \\
&= 8.73\tag{3}
\end{align}
You would've needed $\mathrm{p}K_\mathrm{b}$ though when a strong base (e.g. $\ce{NaOH})$ were added.


text:
Rate the following ions in increasing ionization energy
Sr2+, Rb1+, S2-, P3-
I believe it is P3- < S2- < Rb1+ < Sr2+ because it should be easier to take an electron away from the negatively charged ions, but I'm not sure about the Rb and Sr because they both have their electrons taken out of their outer shell.


text:
Why is 1-bromotriptycene inert to nucleophilic substitution?
According to my textbook, 1-bromotriptycene is inert to any nucleophilic substitutions.

Due to steric hinderance we can say SN2 reactions cannot occur, but I think SN1 reactions should be favoured as formed carbocation will be resonance stabilised.
Can someone explain why it doesn’t undergo SN1 reactions?

Hard to get resonance stabilization.  You can't align the vacant orbital on the carbocation with the $\pi$ electrons on the ring.

Think of the consequences of an SN1 or SN2 reaction.

Sn1 reaction proceed through carbocation intermediate which has planar structure. The structure of 1-bromo triptycene  is cage like structure so the bridgehead carbon cannot assume planarity , hence,the formation of a carbocation at the bridgehead position does not take place. Consequently 1-bromotriptycene is inert to sn1reaction.

It would be better if you make your answer more interesting by describing elaborately or adding some images. You could also mark some important sentences as bold, so that it becomes catchy.

Bredt's rule is a consequence of the fact that having a double bond on a bridgehead or carbocation  is not stable for small rings (fewer than eight atoms) due to a combination of ring strain, and angle strain (nonplanar alkene).Since SN1 reaction involves carbocation intermediate, and such intermediate is not possible here, hence SN1 IS RULED OUT.SN2, reaction involves a backside attack from the leaving group.Due to stearic hinderence to the backside attack, such a reaction is also not possible.Totally nucleophilic substitution is not possible.


text:
Resonance in Organic Compound
Is the pi bond of a $\ce{sp}$ hybridised carbon in conjugation with another π-bond capable of resonance?
Examples:

$\ce{={C}=C-C=}$
$\ce{#{C}-C=}$

Is resonance possible?

Where would that charge come from? In both examples the carbon atom with the plus has a full octet it equally shares with its neighbour atoms, so four electrons. That makes it neutral.

If the charge is not present, is resonance possible??

Can you explain to me what exactly you think is "resonance"? Also please draw a complete molecule, and try it out. With the fragments in your examples, it's hard to tell.

It's delocalisation of pi electrons

Sorry that's an incomplete answer. Resonance means you can draw at least two resonance structures.

Yes, principally, any pi orbital can overlap with similarly oriented neighbouring pi orbitals. However the two orbitals at an sp hybridised carbon atom are orthogonal to each other, so only one of them can overlap with the same orbital at a neighbouring atom.
And you have to differentiate conjugation(=orbital overlap) from resonance(=total delocalisation=more than one structure formula for the same compound).
In the compound a) below, the double bonds are just conjugated. Resonance is typical for aromats, but you can imagine resonance structures here as well:
From a) to b) below, all pi electron pairs flip to the right, leaving a carbocation on the left, an a carbanion on the right. That is resonance, although the second version b) is not very favourable, energetically, as it requires a charge separation. You can draw more intermediate resonance structures where only one or two or three electron pairs flip. Under normal conditions, the energy for that charge separation is not available, so there will be no resonance..
a) $\ce{R-C=C-C=C-C=C-C=C-R}$
b) $\ce{R-\overset{+}{C}-C=C-C=C-C=C-\overset{-}{C}-R}$
In structure a) itself, there is conjugation, i.e. partial pi electron delocalisation. Resonance only comes in when you take into account further formula: $[a \leftrightarrow .. \leftrightarrow b]$
Now with sp hybridised carbon atoms (e.g. the two in the middle of the structures below), the aditional pi bond orbital is orthogonal to the other pi system, and does not take part in it:
$\ce{R-C=C=C=C-R}\leftrightarrow\ce{\overset{+}{R}=C-C\equiv C-C=\overset{-}{R}}$


text:
Quantity of water in a redox problem

$\pu{76 g}$ of $\ce{FeSO4}$ are oxidized by $\pu{0.04 M}$ solution of $\ce{KMnO4}$ in presence of $80\%$ $\ce{H2SO4}$. What quantity of water must be added to the final solution of $\ce{Fe2(SO4)3}$ for the concentration to be $\pu{0.25 M}$?

I wrote the reaction:
$$\ce{10 FeSO4 + 8 H2SO4 + 2 KMnO4 -> 5 Fe2(SO4)3 + K2SO4 + 2 MnSO4 + 8 H2O}$$
But I have no idea how to find the amount of water. I found the quantity of $\ce{Fe2(SO4)3}$ to be $\pu{100 g}$, but I have no idea to find all the water.

Could you please double-check the amounts? I might be wrong, but something seems a bit off here. Probably the concentration of KMnO4 is 0.4 M, not 0.04 M?

This is strange; in my opinion it's a typo, because this is a way too diluted solution to start with.

To achieve the necessary concentration, based on the water presence in the system the following condition must be fulfilled:
$$C(\ce{Fe2(SO4)3}) = \frac{n(\ce{Fe2(SO4)3})}{V_1 + V_2 + V_x} \tag{1}$$
where $C$ – concentration, $n$ – amount, $V_1$ – volume of water remaining from permanganate solution, $V_2$ – volume of water synthesized during the reaction, $V_x$ – additional volume of water to be found.
From the balanced redox reaction, which it seems you have established correctly:
$$\ce{10 FeSO4 + 8 H2SO4 + 2 KMnO4 -> 5 Fe2(SO4)3 + K2SO4 + 2 MnSO4 + 8 H2O}$$
one can express both $V_1$ and $V_2$, as well as the amount of iron(III) sulfate using molar ratios between reactants and products and density of water $\rho(\ce{H2O})$:
$$V_1 = \frac{n(\ce{KMnO4})}{C(\ce{KMnO4})} = \frac{2 \cdot n(\ce{\ce{FeSO4}})}{10 \cdot C(\ce{KMnO4})} \tag{2}$$
$$V_2 = \frac{m(\ce{H2O})}{\rho (\ce{H2O})} = \frac{8 \cdot n(\ce{FeSO4}) \cdot M(\ce{H2O})}{10 \cdot \rho (\ce{H2O})} \tag{3}$$
$$n(\ce{Fe2(SO4)3}) = \frac{5 \cdot n(\ce{FeSO4})}{10} \tag{4}$$
From (1):
$$V_x = \frac{n(\ce{Fe2(SO4)3})}{C(\ce{Fe2(SO4)3})} - V_1 - V_2 \tag{5}$$
Combining (2–4) into (5), and taking into account that
$$n(\ce{FeSO4}) = \frac{m(\ce{FeSO4})}{M(\ce{FeSO4})} \tag{6}$$
one can obtain the following expression for the volume of water to be added:
$$V_x = \frac{m(\ce{FeSO4})}{M(\ce{FeSO4})} \left( \frac{1}{2 \cdot C(\ce{Fe2(SO4)3})} - \frac{1}{5 \cdot C(\ce{KMnO4})} - \frac{4 \cdot M(\ce{H2O})}{5 \cdot \rho (\ce{H2O})} \right) \tag{7}$$
The problem is, plugging in the current values returns negative volume, which means the solution is already too diluted. That much water could've been introduced only by humongous amount of oxidant required due to low concentration, so I guess that the concentration of $\ce{KMnO4}$ is higher than $\pu{0.04 M}$ and there is a typo. Let's assume $C(\ce{KMnO4}) = \pu{0.4 M}$, then:
$$V_x = \frac{\pu{76 g}}{\pu{152 g mol-1}} \left( \frac{1}{2 \cdot \pu{0.25 M}} - \frac{1}{5 \cdot \pu{0.4 M}} - \frac{4 \cdot \pu{18 g mol-1}}{5 \cdot \pu{1e3 g L-1}} \right) = \pu{0.743 L}~\text{or}~\pu{743 mL}$$


text:
Why is butter harder if it once got warm?
I realized that butter, when kept cool after buying, can be easily cut and spread when taken out of the fridge. If it gets warm once, however, it will be less soft the next time I take it out of the fridge. If it reaches room temperature once, it cannot even be cut when cooled down again to fridge temperature.
What is causing this phenomenon?
(I refer to real butter from cow milk, not what might be sold as butter in some countries)

Even if this is not homework, you should be able to draw out a few ideas about what happens.

Butter is often sold "whipped", i.e. aerated, to make it softer (and to increase he volume, so it appears to be more for the money. Heat it, and your bubbles burst.

@DrMoishePippik I googled whipped butter. It's definitely not what I have here. Its a solid block. And there is no volume change.

Seems to be complicated. In this German FAQ they say that fatty acids have a large impact. With a special temperature treatment, they can make the "hard" fatty acids end up between the "soft" fatty acids, the so called "ball bearing effect". Also, mechanical treatment before packaging has an effect ("break some structures"). So, inverting this knowlege, my guess is that temperatures above fridge temperature allow the fatty acids to rearrange to harder structures.

Butter is a water in oil emulsion that contains butterfat that crystallizes at room temperature. The continuous phase of butter is a network of fat crystals. This network surrounds water and air droplets. One of the unique properties of fats are the interactions of fat crystals, and these interactions are directly related to product hardness. More fat crystal interactions lead to a harder fat. During storage, fluctuations in temperature from refrigerated to room temperature back to refrigerated temperatures result in postcrystallization. This postcrystallization strengthens the fat crystal network since more crystals are formed, leading to a stronger fat crystal network.


text:
Trend in the infrared stretching of molecular hydrogen to group 8 transitions metals?
While going through old notes, in relation to the Dewar-Chatt-Duncanson model, it states that the stretching frequency for $\ce{N2}$ to the group 8 transition metals ($\nu(\ce{NN})$) proceeds as:
\begin{array}{cc}
\hline
\text{Compound} & \nu,~\pu{cm-1} \\
\hline
\ce{[(SiP3)Fe(N2)]-}  & 1920 \\
\ce{[(SiP3)Ru(N2)]-}  & 1960 \\
\ce{[(SiP3)Os(N2)]-}  & 1931 \\
\hline
\end{array}
However, the notes fail to elaborate further.
I would have perhaps assumed that the trend was due to increasing size of the $\ce{3d -> 4d -> 5d}$ orbitals and thus increased pi-donation causing a weaker bond and thus lower frequency, however, this does not fit the trend observed. Thus leads me to my question: how does the Dewar-Chatt-Duncanson model relate to the observed trend?


text:
What is the order of acidic strength among phenol, catechol, resorcinol and hydroquinone?
This question is from a book that gives the answer as
$$
\text{hydroquinone} < \text{resorcinol} < \text{phenol} < \text{catechol}
$$
but does not provide any explanation for the answer. If I go by the stability of the conjugate base due to resonance, phenol seems to have the most number of resonance intermediates and seems to be most stable. Considerations about number of resonance intermediates of conjugate base gives an order that is nowhere close to that provided as an answers.
Please provide an explanation as to the provided answer or, if the answer is wrong, please supply the correct order with explanation.

A quick search on says that resorcinol is more acidic that catechol, followed by phenol.
Hydroquinone - strong +M effect and weak -I effect
Phenol - the normal resonance you'd expect in all phenols
Resorcinol - -I effect
Catechol - +M effect, stronger -I effect and hydrogen bonding
Taking these factors into account, it is easy to see why hydroquinone is least acidic. Hydrogen bonding and the vicinity of the $\ce{-OH}$ to the anion (for the stronger -I effect) make cactechol more aciic than the phenol despite its +M contribution. Resorcinol has only -I effect and no counteracting electron-donating effects, which makes it the most acidic.

Catechol is less acidic than resorcinol and quinol because in case of Catechol present hydrogen bonding and do not give easily H+ ions.. In case of resorcinol - I work but in case of Quinol work +M and - I but +M is dominant hence the acidity order of.     resorcinol) Quinol) Catechol.