text:
What is the "curse of dimensionality" in molecular dynamics?
The 'curse of dimensionality' or the 'bottleneck problem' in molecular dynamics is explained in page 5 and 6 of Ab Initio Molecular Dynamics: Basic Theory and Advanced Methods by Dominic Marx and Jurg Hutter .
They proved that Car-Parrinello MD outperforms classical MD with a computational advantage growing as $\sim 10^N$ with system size, which I am unable to understand, especially the last paragraph of page 5.
To be more specific, I am copying the relevant contents:

In the case of using classical mechanics to describe the dynamics - which is the focus of the present book - the limiting step for large systems is the first one, why should this be so? There are $3N-6$ internal degrees of freedom that span the global potential energy surface of an unconstrained $N$-body system.  Using, for simplicity, 10 discretization points per coordinate implies that of the order of $10^{3N-6}$ electronic structure calculations are needed in order to map such a global potential energy surface. Thus, the computational workload for the first step in the approach outlined above grows roughly like ∼$10^N$ with increasing system size ∼$N$. This is what might be called the curse of dimensionality or dimensionality bottleneck of calculations that rely on global
  potential energy surfaces.

What is needed in ab initio molecular dynamics instead? I am confused about the parts placed in bold:

Suppose that a useful trajectory consists of about $10^M$ molecular dynamics steps, i.e. $10^M$ electronic structure calculations are needed to generate one trajectory. Furthermore, it is assumed that $10^n$ independent trajectories are necessary in order to average over different initial conditions so that $10^{M+n}$ ab initio molecular dynamics steps are required in total. Finally, it is assumed that each single-point electronic structure calculation needed to devise the global potential energy surface and one ab initio molecular dynamics time step require roughly the same amount of cpu time. Based on this truly simplistic order of magnitude estimate, the advantage of ab initio molecular dynamics vs.  calculations relying on the computation of a global potential energy surface amounts to about $10^{3N-6-M-n}$. The crucial point is that for a given statistical accuracy (that is for $M$ and $n$ fixed and independent of $N$) and for a given electronic structure method, the computational advantage of ``on-the-fly” approaches grows like ∼$10^N$ with system size. Thus, Car–Parrinello methods always outperform the traditional three-step approaches if the system is sufficiently large and complex. Conversely, computing global potential energy surfaces beforehand and running many classical trajectories afterwards without much additional cost always pays off for a given system size $N$ like ∼$10^{M+n}$ if the system is small enough so that a global potential energy surface can be computed and parameterized.

How do the bolded points prove that fact that "computational advantage of 'on-the-fly' approaches grows like ∼$10^N$ with system size" over classical MD?

It's certainly more of a "back of the envelope" calculation than a rigorous proof. The main point is, that for a certain system, the cost of obtaining a set of trajectories with a given statistical accuracy is fixed with ab initio MD, as only the dynamically relevant parts of the PES are calculated.
With what is here called "classical MD" (but also applies to "quantum dynamics" and "wave packet" methods), the whole PES is computed in advance, and this cost is system dependent (scales with $3N - 6$ for nonlinear systems); the individual trajectories over this PES are then calculated comparably quickly.
The ratio of the two scalings is $\frac{10^{3N-6}}{10^{M+n}}$, increasing with system size (making ab initio MD favourable) and decreasing with required statistical accuracy (making classical MD favourable).


text:
Comprehensive resource on diazo coupling?
I am starting to do research on diazo coupling reactions and am looking for a book or literature review that is a comprehensive exploration of everything diazo coupling.
I found one book which is pretty amazing: https://ia802302.us.archive.org/8/items/aromaticdiazocom031270mbp/aromaticdiazocom031270mbp.pdf
It has a whole chapter devoted just to diazo coupling with various classes of compounds like phenols, amines, phenolic ethers, hydrocarbons, etc (which is exactly what I am looking for).  BUT, it's from the 1930's! It is quite challenging to read because it mostly describes reactions rather than showing them pictorially. Also, some of the theory is a bit dated!
My question: Does anyone know of a more modern resource?

If you prefer books as secondary reference, the The chemistry of diazonium and diazo groups in the Patai series (back 1978) may be one source.  More recently, the two books about diazochemistry by Zollinger, 1994 here and here.
Maybe not as comprehensive as the two sources above, but offering an overview are Merino's critical review about diazocoupling reactions in Chem. Soc. Rev. (doi 10.1039/C0CS00183J) in 2011, and Oger's overview about their application in flow chemistry in Organic Chemistry Frontiers (doi 10.1039/C5QO00037H) in 2015.

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How do you recognize a carbohydrate molecule?
I am studying carbohydrates in organic chemistry and I am confused a bit on what they are and how you recognize whether a molecule is a carbohydrate or not. For example, will a carbohydrate always have an $\ce{OH}$ group with either an aldehyde or a ketone group? Can it also contain other atoms like nitrogen or bromine? Can it have ether, ester?

Some sugars do have nitrogen in them. See https://en.wikipedia.org/wiki/Amino_sugar

Ok then. How can you recognize a carbohydrate molecule, without running into so many exception?

You've asked a great and challenging question, especially for people just learning this material.  The other answers are useful and provide good background.  They also illustrate how the great structural diversity of carbohydrates quickly muddies the waters.  I'll offer an answer that is less technical and gets at the "recognition" part.
The trouble with simple carbohydrates (5- or 6-carbon monosaccharides specifically, but we should avoid that term since we haven't defined it here) is that they can be drawn in an an open chain form (as a Fischer projection) or in a cyclic form (as a Haworth projection, or in some cases as a "chair").  So one commonly encounters two or even three "pictures" of the same thing, which hardly helps (examples below).
In my courses I nearly always use the cyclic forms since for 5- and 6-carbon sugars they are the dominant and biologically relevant forms.  If one sticks to this, a pretty simple rule of thumb can be stated:  carbohydrates are compounds with a 5- or 6-membered ring containing oxygen, and decorated extensively with hydroxyl (or alcohol) groups.  This "scales" to polysaccharides, and accommodates deoxy and amino sugars, so it works a great deal of the time (and is enough for non-experts in many cases).  When and if someone needs to understand more, it's not too hard to transition to the Fischer projections by focusing on opening (breaking open) the ring.

The name "carbohydrate", which literally means“carbon hydrate,” arises from their chemical composition, which is roughly $\ce{(C.H2O)_n}$, where $n\ge 3$. The basic units of carbohydrates are monosaccharides.
Monosaccharides or simple sugars are aldehyde or ketone derivatives of straight-chain polyhydroxy alcohols containing at least three carbon atoms. Such substances, for example, D-glucose and D-ribulose, cannot be hydrolyzed to
form simpler saccharides.
Examples of monosaccharides:

Classification
Monosaccharides are classified according to the chemical
nature of their carbonyl group and the number of their C atoms.
If the carbonyl group is an aldehyde, as in glucose, the sugar is an aldose. If the carbonyl group is a ketone, as in ribulose, the sugar is a ketose. The smallest monosaccharides, those with three carbon atoms, are trioses.
Those with four, five, six, seven, etc., C atoms are, respectively, tetroses,
pentoses, hexoses, heptoses, etc. These terms may be combined so that, for example, glucose is an aldohexose, whereas ribulose is a ketopentose.
However, there are similar molecules with different atoms such as nitrogen which are also considered carbohydrates. A class of polysaccharides known as glycosaminoglycans (unbranched polysaccharides of alternating uronic acid and hexosamine residues).Practical examples are Hyaluronic acid(linked disaccharide
units that consist of D-glucuronic acid and N-acetyl-Dglucosamine) and Heparin (sulfated disaccharide units):

Nevertheless, when carbohydrates bond with other molecules to form other newer bonds, they are given different names and are not necessarily carbohydrates but are given new names - a good example are glycoproteins (proteins covalently associated with carbohydrates.)
Reference
Biochemistry Voet and Voet, Grisham

Thanks for the reply. The reason why I ask this question was because the books definition of carbohydrate was too basic. For instance, do carbohydrate have to follow the strict ratio of 1:2:1 ratio for C:H:O atom. IF not then can 3hydroxybutanal be considered a carbohydrate? It has a aldehyde and oh group? Also, in carbohydrate, does every carbon atom in the carbohydrate molecule have to be attached to 1 OH and 1 H atom? Can you have exception where one carbon atom is attached to both 2 OH group and other carbon atom attached to 2 H atom as in CH2? These are just my curiosities.

Very good question! Only the simple sugars, or monosaccharides, fit this formula exactly. The other types of carbohydrates, oligosaccharides and polysaccharides, are based on the monosaccharide units and have slightly different general formulas, as for the other part, its not necessarily each carbon is bonded to OH, H , a good example to explain this is deoxyribose , it has a carbon bonded  to H,H

ahh... Got it thanks... If i could ask just one more question, since the previous comment wasn't allowing me, I was reading other resources and was reminded that carbohydrates are polyhydroxy aldehydes and ketone, which thus means that 3 hydroxybutanal cannot be carbohydrate since it does not have multiple OH groups. So my other question is that can simple sugars have poly-adehydes or ketone groups, or can each molecule just have one of the other? Or can a simple carbohydrate contains both aldehyde and ketone groups together in one molecule? Sorry if this seems a lot. But thanks anyways.

I'm sorry, the last part i'm not really sure so I can't answer now, would it be fine if I just confirm and get back to you later ?

Sure no problem.

A carbohydrate is a biological molecule consisting of carbon (C), hydrogen (H) and oxygen (O) atoms, usually with a hydrogen–oxygen atom ratio of 2:1 (as in water); in other words, with the empirical formula Cm(H2O)n (where m could be different from n).This formula holds true for monosaccharides. Some exceptions exist; for example, deoxyribose, a sugar component of DNA,has the empirical formula C5H10O4.Carbohydrates are technically hydrates of carbon; structurally it is more accurate to view them as polyhydroxy aldehydes and ketones.
Carbohydrates often display chemical groups such as: ''N''-acetyl (e.g. chitin), sulfate (e.g. glycosaminoglycans), carboxylic acid (e.g. sialic acid) and deoxy modifications (e.g. fucose and sialic acid).This is from Wikipedia

They can also be defined as optically active polyhydroxy aldehydes and ketones.
Now about the general formula $\ce{C_m(H2O)_n}$ also have exceptions like $\ce{CH3-COOH}$($\ce{C2(H2O)2})$ but it is not a carbohydrate. Similarly, rhamnose($\ce{C6H12O5})$ and deoxyribose($\ce{C5H10O4})$ as mentioned in the above article does not fit into this general formula, but then too they are classified as carbohydrates.
Apart from that carbohydrates are also ether.

Source:Wikipedia
This is the open chain structure of glucose. Some of the properties of glucose was not able to be explained by this structure.

These are the two hemiacetal cyclic forms  $\alpha$ and $\beta$ of glucose also called as Haworth structures which remain in equilibrium with the open chain structure of glucose. It is an ether as you can see in the structure. Other examples of ether include cellulose, starch, sucrose, etc. Sucrose is a disaccharide which gives a glucose and fructose by hydrolysis. Starch and cellulose are polysaccharides. Starch is a branched polymer of $\alpha$ glucose units and cellulose is a straight chain polymer of $\beta$ glucose units.

Ok got it. So together with multiple OH groups, can simple sugars also have multiple aldehyde and ketone groups? Or can both aldehyde and ketone groups be found in simple sugars?

I don't think so, I have never heard of such molecule. You can ask this as a different question. @TLo

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Does temperature change pressure in isochoric system?
Consider the following equilibrium reaction:
$$\ce{2 NOCl (g) <=> 2 NO (g) + Cl2 (g)} \qquad  \Delta H > 0$$
My textbook has a statement that:

If you decrease the temperature when keeping volume constant, then reaction shifts to the reactants.

I know that since the enthalpy change $\Delta H$ is positive, we can think of the reaction as:
$$\ce{2 NOCl (g) + \text{heat} <=> 2 NO (g) + Cl2 (g)}$$
And according to Le Chatelier's principle we can say that if we decrease the temperature, then reaction favor the side have heat in it. This comes me so reasonable but there is one more thing we have to consider in my opinion: pressure.
While keeping volume constant, decreasing temperature will decrease also pressure. So reaction will favor the side that have more molecules in it. In this case, it is the opposite side of the previous one.
Is my reasoning true? If it is, how can I conclude the side determining job?

You need to look up the Van't Hoff Isochore; $d\ln(K_p)/dT = \Delta H^0/(RT^2)$. In an endothermic reaction ( $\Delta H^0 \gt 0$) , $K_p$ increases with increasing temperature, so more dissociation at higher temperatures and  less at lower temperatures as you initially state.

@porphyrin As far as i understand, even pressure decreases (means that $Q_p$ decreases), $K_p$ always decreases much more than $Q_p$, so, it ends up with always favoring reactants.(for $\Delta{H}>0$ of course.)

$K_p$ is constant with change of pressure  but does change with temperature.

The change in $K_p$ i mentiond is due to temperature change. So is it what is going on?

the answers here might help. https://chemistry.stackexchange.com/questions/73644/how-does-the-partial-pressure-of-gas-c-change-with-the-change-in-volume-of-the-c/73975#73975

Since it's given that ΔH > 0 (endothermic reaction) ,the reaction will proceed backward (towards the reactant). However as you know ,
Partial Pressure Of Gas= Total Pressure x Mole Fraction;
hence , even if the pressure is affected , the mole fraction increases/decreases in a way so that partial pressure remains constant.
If you calculate Kp ( Pressure Equilibrium constant) you will know that it depends on the partial pressure of the gases involved . Since partial pressure remains same no effect takes place. Hence , pressure  will have no effect on the equilibrium.
Also note that equilibrium constant is only affected by the change in temperature.
Hope this helps.
EDIT:-
In the light of the comments I would like to clarify ,
Partial pressure of a particular gas is the fraction of the gas multiplied by the total pressure which means it exerts a fraction of the total pressure. When I increase the total net pressure exerted by all gases , the fraction of moles of each gas decreases (according to the mathematical equation) thereby contributing lesser to the total pressure .
You can think of it as a ratio:-
If 1 mole X Gas exerts 10 bar pressure when total pressure is 20 bar , 0.5 moles of X gas will exert the same 10 bar pressure when total pressure is increased to 40 from 20 , thereby keeping the ratio same.

You don't even need to do all that ... Just looking at the fact that what is happening to the temperature/pressure/volume is enough to determine where the reaction will proceed. And as for calculation could you send me the Kp and moles intially or at equilibrium so that I can try as well?

@Abdullah I edited the answer


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Change in total gas pressure
The question reads:

For the part were they ask about the total gas pressure, would it be correct to say that the total gas pressure doesn't change? Instead of $0.8$ $atm$ of $N_2O$, it is now $0.8$ $atm$ of various partial pressures of $N_2O$, $N_2$ and $O_2$?

If you start with 0.8 atm, you won't necessarily end up with that many. You should write up the reaction and look at how many moles are gas are on either side of the equation. You can then interconvert from (pressure reactant) $\to$ (moles reactant) $\to$ (moles product) $\to$ (pressure product) by using the ideal gas law.

Thanks, I see the flaw in my logic now, since this reaction doesn't produce 1 mole of oxygen and 1 mole of nitrogen it definitely wouldn't be 0.8 atm after. Thank you

Process is probably isochoric - constant volume not pressure.


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Is palladium on carbon acting as an oxidising or reducing agent?
In the conversion of piperidine to pyridine, palladium on carbon (Pd/C) is used. But it is reducing agent.

Clear the confusion, please.

Palladium on carbon ($\ce{Pd/C}$) is neither an oxidant nor a reductant. The reaction is the dehydrogenation of piperidine: $\ce{C5H11N -> C5H5N + 3H2}$ and here, $\ce{Pd/C}$ is merely a catalyst.
Even in standard hydrogenation reactions using $\ce{H2(g) + Pd/C}$, the $\ce{Pd/C}$ is not a reducing agent. The $\ce{H2}$ is the reducing agent and Pd/C is again only a catalyst.

(The above answer brings up hydrogenation using Pd/C, be weary of the answer because they're not addressing the question.)
The above answer is partly true, but on the whole incorrect and would fail in a test scenario.
Yes, Pd/C is a catalyst therefore in a full reaction cycle the Pd/C should return to its original oxidation state.
However, Pd/C can only accept the N-H bond via oxidative addition onto Pd metal.  By this definition, because the Pd is oxidized while adding the N-H bond across Pd, then Pd must begin the cycle as a "reductant" (i.e. the Pd metal initially reduces the N-H bond). The organic compound must be the "oxidant" because it oxidizes the Pd (i.e. oxidative-addition to Pd).
Where the first answer fails to be correct is: if Pd/C has been previously oxidized, it cannot perform the first step in the catalytic cycle and cannot be catalytic.  It is more correct to say that Pd/C is a reducing agent (reductant, what have you).
Where the first answer fails more is: most Pd/C reductions in a lab setting are NOT catalytic and require a molar equivalent or even excess of Pd/C for a reduction, which goes against the definition of a catalyst.

I don't agree with some of your points: (1) I already addressed OP's dehydrogenation reaction in my very first sentence. (2) Even if the Pd is initially oxidised, I don't think I would say that it acts as a "reducing agent" in the grand scheme of things. There's plenty of Pd(0) catalysis out there with the whole ox. add. / ligand exchange / red. elim. procedure, which (at least in terms of the ox. states involved) is similar. Do people call Pd(0) a reducing agent in those? I would consider that to be extra detail about the catalytic cycle, not a definition of the fundamental role of Pd.

Am open to being convinced otherwise, though. I am also mildly surprised when you say that hydrogenation is often not catalytic – it's been a long time since I did anything, but I used to do Lindlars with 0.5 mol% Pd/CaCO3. Anecdotally, the amount required depended on the supplier.


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Geometries of SCl2(OCH3)2 and SF2(OCH3)2
I am not able to understand the difference between structure of $\ce{SCl_{2}(OCH_{3})_{2}}$ and $\ce{SF_{2}(OCH_{3})_{2}}$.
I am already well versed with Molecular Orbital and VSEPR theories. I found the geometries,
but don't understand why they are:

Why is the $\ce{F}$ at axial positions and $\ce{Cl}$ on the other hand in equatorial positions? (I am not able to understand Bent's Rule to be straight-forward, I have read Wikipedia already.)

Related: Why does F replace the axial bond in PCl5?

As a general reminder: The involvement of d-orbitals into the bonding is one of the many shortcomings of VSEPR theory. The sp³d hybridisation simply is not evident (and it is not necessary) in the bonding of these molecules. I can't stress that enough. Textbooks which still rely on this explanation are outdated and desperately need revision.

Ok, I think I understand it now.....  Things became more clearer by Ron's answer to the question suggested above by Orthocresol (Thanks!) and I got to know that my textbook is outdated, cheers to Martin.

@orthocresol I'm not sure that Bent's rule is straight forward to use here, and I am especially not sure, whether the given geometries are indeed correct. From my initial calculations, the lowest energy structure also has a co-linear arrangement of Cl-S-Cl.

Check the bond angles. The larger the angles around a bond, the more S character.
Electronegative elements prefer higher P character(Therefore less the bond angles).
Axial position has lesser bond angles.Hence higher electronegative element takes axial position.( F>O>Cl)
This concept however may not be applicable to every compound.

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Material for melting metal oxides
I am constructing an apparatus for high temperature, strong alkaline melts (metal melts), with extremely high oxygen rich environments.
Here what I have so far.
Iridium is a good candidate - is it better as an alloying agent or would it suffice in an almost pure Iridium composition?
Tantalum - has good temp but from what I understand is weak against alkaline reactions.
If nothing else gets answered, could this please.
If I have a tantalum oxide layer coating the tantalum will that protect it from further corrosion in a high temperature, oxygen rich environment?
Tungsten seems to also have potential, I remember reading a paper from the 50's stating that under high alkaline, temperature conditions the product contained an unsuitable level of impurity from the tungsten? with the properties of tungsten would this be the case.
What I am enquiring about here does anyone know what metals or alloys off-hand that are currently used in these conditions?
Basically I am only looking at refractory metals as the temperature range is (1900 deg C - 2900 deg C)
Also there can be no carbon in the composition of the alloy.

You seem to have three questions that are somewhat related.  Is there something that you are really curious about or trying to do, that you are asking questions to try and solve?  This feels like it could be could be a X-Y Problem

I am trying to create a crucible for melting corundum and other metal oxides at the temperatures specified. Need a vessel to handle the heat and oxygen without corrosion

Sometimes I think the XY "problem" should be defined as "when an arrogant person refuses to answer the question asked because they think the question is not worth answering." – Aaron McDaid May 1 '16 at 11:07
This is ment as a joke. :-)

It's not even arrogance, it's about asking about " asking about your attempted solution rather than your actual problem."  I don't know nearly enough to answer your question, but I'd imagine that if you asked directly about what you wanted to do, and outlined your ideas, you might get a solution, not necessarily what you envisioned.  Something along the lines of, "Crucible material for high temperature metal oxides" or something.  Then again, you may very well get the solution you are looking for here, I don't know.  Best of luck, either way!

I did do a brief search, and it's unfortunate about your carbon requirement, because apparently there are crucibles that would deal with the temperature and pH requirements. I'm guessing you already eliminated vitreous/glassy carbon crucibles.

my apologies about the comment I just found it funny I understand what you are saying and my comment centralised my question.

yes unfortunately I have looked at those. Most Insulators at this temp have carbon. its basically your refractory metals and their alloys. thank you for taking the time to respond, it is appreciated.

https://chemistry.stackexchange.com/questions/16640/how-do-you-melt-metals-with-super-high-melting-points

Why do you want to melt them?

I actually asked someone I know that worked with high temperature melts, but for crystal growing, and they said that iridium or platinum crucibles may work.  Either as a material would be very expensive, but it depends on what your budget is.  Both have the melting temperature that you need, and I believe that platinum may be the cheaper option of the two.

Thank you for that fyrepenguin. What industry is he in? What crystals do they grow sapphires per chance?

platinum has a melting point of around 1600 deg C but iridium or an alloy containing iridium may be best.


text:
How do you melt magnesium without causing its combustion?
How do you melt magnesium without causing its combustion?
I own a foundry capable of reaching temperatures around 2000 °F, and I recently bought some magnesium for melting before doing my research. I cannot find any information on how this process works or if it is even possible.

You'd have to do it with an inert atmosphere (no oxygen). It would require specialized equipment.

As MaxW noted, you will need an inert atmosphere within specialized equipment. Just as a note, magnesium auto ignition temperature is 746K (from Wikipedia) so you have an increased safety concern not only when you are melting but during any time the molten metal is above this temperature.

I'm not sure what your foundry is made of, but if it forms intermetallic compounds or alloys with the magnesium (which I strongly suspect it does), you have even more problems than just combustion.  Do some research and as other comments imply, get the right stuff.

Also note that water can't extinguish magnesium fires.  In fact just the opposite. Water causes magnesium fires to flare.

https://chemistry.stackexchange.com/questions/16640/how-do-you-melt-metals-with-super-high-melting-points

The choice of the "inert" gas will matter with magnesium. carbon dioxide won't work as magnesium will burn. Nitrogen and magnesium will also react if the magnesium is hot enough (burning magnesium will keep burning in pure nitrogen). You might need argon.

I have seen molten magnesium; it was covered with charcoal and other fluxes and in a furnace that had a gas shroud. (I think methane that burned any oxygen that got into the shroud atmosphere.) When an immersion thermocouple was put into the magnesium, the open spot immediately burst into flames. I recommend that you not try to melt magnesium; you certainly must get more information than you can get on the internet.

This is not Plausible. Magnesium reacts with carbon dioxide and carbon monoxide to form magnesium oxide and carbon.  Further Magnesium forms carbides (specifically acetelyides). Charcoal and related gases cannot effectively shield magnesium

Magnesium actually does not form carbides with elemental carbon. Or so I have not seen. Usually we get magnesium carbide from magnesium plus acetylene. So charcoal could be used for the covering.

Mg does not form carbides; It is added to cast-iron ( 2 + % C) to make malleable cast iron . Mg helps develop graphite nodules instead of flakes; does not make carbide.

Magnesium is best melted under a flux – these fluxes are normally based on magnesium chloride, with additions of barium-halogen compounds. Obtaining the flux could be difficult due to the toxicity of the barium compounds – it is even difficult to obtain when you are in the industry like myself.
Another option is a reactive cover gas like $\ce{SF6}$ or R134a, however these are expensive, and are extremely bad greenhouse gases.

As with any science, advances are made by doing something stupid or accidental - the key is being able to do something intentionally stupid safely enough to do it again. That being said, wear proper safety gear - flame retardant clothes, blacksmithing gloves, a full face shield (i.e welding mask) not just safety goggles from a home improvement store. Once you're pretty sure you're not going to get hurt, then you can proceed.
Before you begin, you should know that this is not the cheapest way to do this, nor the most efficient, but it is effective, and will produce results you're looking for.
You will need:

A crucible (an old/empty steel fire extinguisher with the top cut off will work great)
Inert gas (you can almost always find some locally, but do your research first to find pricing and proper storage of it)

Find an extra-tall, nonporous crucible (steel should be just fine), and place it in the cold foundry, making sure it has a sturdy base. Add the magnesium into the crucible, and fill to the top with argon. Begin heating up the foundry and  crucible. Argon is heavier than air, so it will push the air out as it fills the container, which if you're using a charcoal foundry, will be evident by a ring of charcoal around the crucible momentarily going cold as the argon replaces the oxygen. At this point, sit back, and go about the melt like you would anything else.
If you're casting it, the trick is getting the mold to be encased in an argon bath as well, for that you will need a much larger container and a lot more argon.
Look into how they melt titanium – it has all the same issues.


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Will this reaction form magnesium carbonate?
I have magnesium oxide pills I take. I know if I dissolve them they'll form magnesium hydroxide, and I'm trying to make pure magnesium from my pills. I've come up with a double replacement reaction, will it do what I think it will?
$$\ce{Na2CO3(aq) + Mg(OH)2(aq) -> MgCO3(s) + 2NaOH(aq)}$$
Will this precipitate out magnesium carbonate?
Another thing: I was planning on using electrolysis to separate the magnesium, but if it's insoluble then I'm not sure what to do. How can I separate the magnesium from the carbonate?

There's basicly no way for you to make pure Mg nor should you do that as a private person if you somehow could.

If you really want (close-to-)pure magnesium, just order some off Amazon.

If such a reaction takes place at all it is not very efficient because magnesium hydroxide is not very soluble.
And if you want to get the metallic magnesium neither the carbonate nor the hydroxide will really work.  Both decompose at a relatively low temperature to the oxide, which is then solid up to $\pu{2800\!^\circ C}$.  Convert the magnesium to the chloride and melt that, preferably as a eutectic with alkali metal chlorides.  Then electrolyze the molten salt.  An aqueous solution of magnesium chloride  (or anything else containing magnesium) does not work.

The reaction of $\ce{Na2CO3}$ with $\ce{Mg(OH)2}$ will not take place to any significant degree due to solubility issues.
The problem is that $\ce{Mg(OH)2}$ is about 20 times less soluble than $\ce{MgCO3}$, the latter of which is still only about $\pu{140 mg/L}$ at room temperature. So, even if you were very patient, only a tiny fraction of the $\ce{Mg(OH)2}$ would dissolve then precipitate out as $\ce{MgCO3}$.
The "good" news is that the $\ce{MgCO3}$ synthesis problem doesn't really matter, because it is basically impossible to convert magnesium carbonate (or hydroxide, or oxide) to magnesium metal, outside of a properly equipped laboratory anyway! That is, assuming your home lab doesn't include a $\pu{2300^{o}C}$ furnace. If you had all the appropriate equipment, all you would need is some carbon and you could do the following (from this Wikipedia article):
$$\ce{MgO(s) + C(s) -> Mg(g) + CO(g)}$$
That is correct, Mg would actually be formed as a gas at $\pu{2300^{o}C}$.
Sorry to have such negative news for your idea. Please don't hesitate to ask for any clarification in the comments below.

"Magnesium would be formed as a gas".  So would carbon monoxide.  Our intrepid chemist might get a visit or two from local and Federal authorities when the air becomes a bit toxic, no?

You can easily make magnesium carbonate from magnesium oxide by bubbling it in $\ce{CO2}$ gas.
$$\ce{MgO + CO2 → MgCO3}$$
But the problem is that it is impossible to make pure magnesium from magnesium carbonate(@airhuff). Magnesium carbonate decomposes on heating to form magnesium oxide which is very stable.(@Oscar). Instead, you can reduce magnesium oxide into magnesium using reducing agent such as carbon(@airhuff), beryllium or calcium.
$$\ce{MgO + C -> Mg + CO ; ΔH^{0}_{298}= 491.3 kJ / mol}$$

Reaction between magnesium oxide and carbon to form magnesium and
  carbon monoxide. The reaction proceeds at a temperature above
  2000 ° C.(Source)

$$\ce{MgO + Be ->[\Delta] BeO + Mg}$$

Reaction interaction beryllium and magnesium oxide to form magnesium
  and beryllium oxide. The reaction proceeds at a temperature of about
  1075 ° C.(Source)

$$\ce{Ca + MgO ->[\Delta] CaO + Mg}$$

Reaction between magnesium oxide and calcium to form calcium and
  magnesium oxide. The reaction proceeds at a temperature of about 1300
  ° C.(Source)

Carbon is best for this purpose although it takes much more heat than calcium and beryllium. Beryllium takes the lowest amount of heat but I would recommend against it as beryllium is poisonous and expensive. For this reason, calcium can also be used as an alternative.
Also for these reactions, you need a expensive furnace that conducts such type of reaction and can be found only in metallurgical industry. For this reason, these reaction are not part of simple home experiments and cannot be conducted at home. (@airhuff).

Carbothermic reduction of magnesia sounds like more fun than making explosives. You need a strong vacuum, a way to not condense soot in your superheated magnesium gas, and then condense the magnesium.

Calcium is also expensive and difficult to handle as a metal.  It is prone to oxidation, and it does not form a protective layer like beryllium or magnesium itself.  Magnesium is sometimes made by a combination of processes, whereby the oxide is reacted with both carbon and chlorine to make the chloride which is then electrolyzed.


text:
CO stretching frequency band in the IR spectrum for tetracarbonylnickel and tetracarbonylferrate(II)
A past exam question says:

With reference to the main bonding interactions between a transition
  metal and a terminal $\ce{CO}$ ligand, explain whether you would expect to
  observe a lower $\ce{CO}$ stretching frequency band in the infrared
  spectrum for $\ce{[Ni(CO)4]}$ or for $\ce{[Fe(CO)4]^{2-}}$

My attempt at an answer is:

There are essentially two bonding interactions at play in the carbonyl
  ligand: a ligand-to-metal $\ce{n → d\sigma}$ interaction and a
  metal-to-ligand ${d\pi → \pi^*}$ interaction. The latter interaction is
  called backbonding, because the metal donates electron density back to
  the ligand.

Nickel is further right in the periodic table than Iron, so $\ce{[Ni(CO)4]}$ has a greater stretching frequency (a greater $\ce{C–O}$ bond order). This is due to the periodic trend in orbital energy. As we move left to right across the periodic table, the $\ce{d}$ orbital energies decrease and the energies of the $\ce{d\pi}$ and ${d\pi^*}$ orbitals separate. As a result, the backbonding orbital interaction is worse in  $\ce{[Fe(CO)4]^{2-}}$ (as strong orbital interactions require well-matched orbital energies)." This means that the ${\pi^*}$-orbitals of the $\ce{CO}$ ligands are significantly more populated in $\ce{[Fe(CO)4]^{2-}}$, so the $\ce{CO}$ bond is accordingly weakened and $\ce{v(CO)}$ is reduced.
Is this correct?

I can't explain but for a quick answer I'd use the fact that as the charge on the central metal atom increases, more electrons are donated into the anti-bonding MO of $\ce{CO}$. As a result the $\ce{CO}$ bond length increases and its bond strength decreases.

This reads ok to me. The CO  molecule frequency is $\approx 2143~\pu{cm^{-1}}$ and is $\approx 2060$ in the Ni carbonyl and $\approx 1790$ in the isoelectronic Fe carbonyl. As the extent of M-C back bonding increases the CO frequencies are lowered, as you suggest, because the CO must accept more $d\pi$ electrons from the metal to prevent accumulation of charge.

This is correct, it would be a good idea to take the answer out of the question and put it into an answer.


text:
What is the OCNH2 group called?
what is the product of reaction of Traut's reagent with aliphatic hydroxyl called? I am reacting Traut's reagent with Fmoc protected Doxorubicin and I just need to know the name of the product after Traut's regent reacts with the hydroxyl group. What is $\ce{OCNH2}$ called? I am looking for a name for $\ce{OCNH2}$ circled in red.

The group itself unprotonated is called a carboximidate as noted by the Wiki page.  The acid that the group is derived from is carboximidic acid.  As is common practice with imines, the nomenclature for the corresponding protonated halide is done by removing -es, adding -ium, and the halide name.  If this is done using the acid name, the group will be called a carboximinium chloride.

Maybe carboximiDium chloride?

There is a paper (1) on imino-ether salts with rather similar structure including $\ce{-C(NH2+)O\! -}$ group:

This compound comes with two defined names, which OP can use as an example:

Hydrogenethylpropylimidate tetrachloroaurate(III) (paper)
1-(Ethoxy)propylimidium tetrachloro-gold(III) (generated by CCDC)

Subjectively, carboximidate fits perfectly if the group wouldn't carry a positive charge; as long as this is the case, probably carboximidium would fit better.
(1) Potts, R. A.; Gaj, D. L.; Schneider, W. F.; Dean, N. S.; Kampf, J. W.; Oliver, J. P. Polyhedron 1991, 10 (14), 1631–1637. DOI: 10.1016/S0277-5387(00)83774-2


text:
Purpose of an Iodine Flask
Is there any specific reason why Iodine Flasks are used instead of Erlenmeyer Flasks when doing Iodine Determination tests?
Iodine Flasks are quite expensive when compared to Erlenmeyer Flasks.
Does the funnel-shaped top form serve a specific function?

Iodine flasks differ from normal conical (Erlenmeyer) flasks in three main ways:

The ground glass joint.
The modified stopper.
The funnel shaped top.

Iodine itself, and iodine related molecules produced during the analysis are, to a certain extent, volatile (you can see this if you just put a bead of iodine in a sealed vial and watch the vial fill with vapour). The ground glass joint of the iodine flask prevents these vapours from escaping, improving the accuracy of your analysis.
As for the funnel, its simply there to help get your stuff into the flask in the first place, preventing loss through spillage and preventing the need for a separate glass funnel. Anything that ends up on the tapered neck is simply rinsed directly into the flask.

The funnel-shaped top form serves a specific function. In the standardization of sodium thiosulfate with potassium iodate for example, there is a waiting time for liberation of iodine. That is when the analyst fills the tapered neck with purified water to "seal" the flask and prevent escape of iodine. When the lid is removed for the titration step, let the purified water go through the flask and act as a rinse.


text:
Is there any connection between "static correlation" and the Born-Oppenheimer approximation?
Beware: High probability for that I am mixing two concepts that should not be mixed, just because the same term is used when explaining the two concepts!
For some reason I get really confused when I think about static, or nondynamical, electron correlation. I somehow want to make a connection to the Born-Oppenheimer (BO) approximation.
As we know, the BO approximation is no longer accurate when two electronic states come close in energy, due to the increase in the nonadiabatic coupling terms. These terms couple the electronic motion to the nuclear motion, and arise because the electronic and nuclear motion is not fully separable in the Schrödinger equation.
Static correlation, sometimes called a "near-degeneracy effect", describes the correlated movement of electrons belonging in different electronic configurations. Certain systems are not described properly by just a single-determinant wave function. One example is the H-H molecule at long bond distances, for which the bonding and anti-bonding $\sigma$-orbitals become degenerate. (this was probably a bad explanation, as I struggle with this concept!)
So, as we have just seen, the BO approximation breaks down when two electronic states come close in energy (near-degeneracy), and the static correlation becomes extremely important in molecules with near-degenerate electronic states. Is there a connection here?
Imagine we are in the ground state at a point on some potential energy surface, and the first excited state is very close in energy. At this point, there should be a high coupling between the electronic and nuclear positions (which means what, exactly??). Also, at this nuclear geometry, the electronic wave function should be described as a linear combination of these two near-degenerate electronic states.
It is the BO approximation that introduces the idea of a potential energy surface (PES). The PES itself can be thought of as an approximation, as the nuclei should ideally be fully described by quantum mechanics. The nuclear positions should be described probabilistically by a wave function, and the very idea of a molecular geometry then becomes fuzzy.
What is the connection, if there is any?

There exist methods that calculate coupling terms between different PES. They are called sth like "nonadiabatic dynamics"..

I'm talking about routine, "static" calculations. If two electronic states are almost degenerate, then the Born-Oppenheimer approximation breaks down. Do we take care of this when doing calculations on molecules with near-degenerate states?

afaik the BO doesn't really break down when two PES "come near each other", it just breaks down when you try to calculate the change from one to the other. I' not really an expert in BO, but I'll show this someone who is and maybe he can really help you ;)

I just edited the question to make my own confusions clearer. The question appears to be a different one now, but I think it explains my confusions better.

okay it's a lot clearer now what you want to know, if there's no satisfying answer on monday, I'll get to it...

Is there a yes/no answer? :P

Probably. I'm inclined to say no, but I don't want to say something wrong

So after I confered with my colleague we came to the conclusion that the answer would be "No". If two PES come near, it's not a near-degeneracy. It's as if you would try to say "The banana grows on trees because paper is white".
Static correlation is important in degenerate electronic configurations, which has not much to do with two PES coming near each other.
That's my two cents. If somebody knows better, please feel free to bash me.

Perhaps you could write up an answer?

Okay, I'll see what I can come up with.

Logically they must be independent, because the concept of "static correlation" occurs "only" (which is in practise almost always) when one builds multielectron wave functions from one-electron-wave functions. Purely mathematically this is not neccessary, so stat. corr. is a quasi-artefact. While the BO approximation is completely independent on the strategy you tackle to solve the Schrödinger eq. However a famous case comes to my mind where both (BO breakdown + stat. corr.) come togehter: NO3^. see here https://molspect.chemistry.ohio-state.edu/institute/2007/stanton/no3.pdf .

I think static correlation only becomes important once the nuclear configuration reaches a state where the energies of the two PES's become (nearly) degenerate. Regardless, I believe the static correlation in that configuration should still be entirely independent of the application of the BO approximation.

Credit to Rudi_Birnbaum for coming up in a comment to the question with what I think is the key here.
Static correlation is a difficult to grasp concept of which even a clear definition is beyond me. However, it seems clear that a full Configuration Interaction (FCI) calculation yields all correlation energy, thus, must correctly handle static correlation. From the discussion of the Born-Oppenheimer approximation (BO) by Jensen,$^{[1]}$ I gather that the BO approximation is essentially a modification of the Hamilton operator. This shortcoming cannot be healed by a FCI calculation that only considers the electronic problem. So, in this sense, the two concepts are unrelated.
(And now for the speculative part of my answer.)
However, I cannot speak as to whether they frequently show up together in chemical systems. Though I cannot find a reference for it, I recall that the beryllium atom requires a multi-reference treatment. Being a single atom that is rather heavy, the BO approximation could be acceptable for it. Thus, this would be a case where the two are not related.


text:
In the mechanism of Cannizzaro reaction why does the hydride migrate rather than the benzene ring?

The hydride ion is not very stable. On the contrary a negative charge on a $\ce{sp^2}$ on the benzene ring would be much more stable. So why doesn't the benzene ring migrate instead of the hydride, as the benzene ring with a negative charge would be a better leaving group (as we know that less basic groups are better leaving groups)?
I've seen some examples where the benzene ring does migrate instead of the hydride for example when two nitro groups are attached at ortho positions. But I think the benzene migration should always happen instead of the hydride migration for reasons which I stated above.
P.S: Someone suggested me that the "real" mechanism is the transition mechanism. But even that does not explain why the transition state involving o-dinitro benzene would be more stable than the one involving $\ce{H}$ but the one involving just a benzene would be less stable than $\ce{H}$. How do we logically explain the discrepancy?

One possibility:  hydrogen has better overlap for forming the delocalized bond in the transition state.  You have to heavily withdraw electrons from the benzene ring (e.g. nitro groups) to make the hydrogen non-hydridic and force the benzene ring to migrate instead.

Actually I dont know why you think phenyl anion is very stable( not saying incorrect, just asking why?) . Its not resonance stabilized. If it had been Phenylacetaldehyde then maybe it would be stable anion but in that case the reactant may give Aldol condensation reaction. What am I missing?

@AlphaRomeo Because it has a $sp^2$ carbon which stabilizes the negative charge.

How? Pka of benzene is in 40s .. and I dont see much stabilization by inductive effect as resonance wont occur. Anyway, I will brush up my concepts :P

@AlphaRomeo Who's talking about inductive effect here? I'm talking about $sp^2$ hybridization.

AFAIK Inductive effect is also caused by $sp^2$ hybridisation.

@Mesentery Not in this case.

Has it got anything to do with the bond energies?

Why are we even talking about $\ce{H^-}$ and $\ce{Ph^-}$? Do they really form in solution? I don't think so.


text:
How can I predict whether a reaction will occur?
I'd like to know how can I predict if two substances will react. I already know that there's a reaction if one of the products is a weak electrolyte or a precipitate but that leads me to the question:

If we have two substances such as salts and they react with one another forming two other salts, when is the reaction double replacement and when there is no reaction?

I'll try giving two examples from the problems I came across:
Which one of the following substances will react with one another in a solution –

$\ce{(NH4)2CO3, CuCl2, K2SO4, Ba(NO3)2, HCl}$
$\ce{Na2CO3, Ba(NO3)2, Na2SO4, AgNO3, CaCl2}$

Relevant: https://chemistry.stackexchange.com/a/73860/43942

well thank you for that,so the way i understand it is that whenever a weaker acid is formed,there is a reaction. And it does apply to salts as well: whenever a salt thats from a weak acid is formed, there will be a reaction?

It would work if the constituent base of the salt obtained as the product and the constituent base of the reactant salt is the same.

could you perhaps illustrate that with an example using the salts i mentioned in the post. I apologize about my inability to understand these things and i thank you for that wonderful edit!

For the first example, I definitely wouldn't want to react $\ce{CuCl2}$ with $\ce{HCl}$, because then I would end up getting the same products again. I'm primarily driven by the motive to create something new, albeit the reaction must be a feasible one. Many factors influence feasibility, and you've mentioned the formation of a precipitate is one of them.
Reactions involving neutralization:
$\ce{HCl}$ would react with $\ce{(NH4)2CO3}$, $\ce{K2SO4}$ and $\ce{Ba(NO3)2}$, as the reactions give out $\ce{H2CO3}$, $\ce{H2SO4}$ and $\ce{HNO3}$ respectively which are weaker acids.
Reactions involving precipitation:
I see a very tempting reaction here:
$$\ce{K2SO4 + Ba(NO3)2 -> 2KNO3 + BaSO4\downarrow}$$
$\ce{BaSO4}$ is a very good precipitate, and I highly emphasize this fact. Just as a fun fact: It can be partially soluble if you try to dissolve it in boiling concentrated $\ce{HCl}$ (or boiling concentrated $\ce{H2SO4}$).
On similar grounds, $\ce{Ba(NO3)2}$ would react with $\ce{(NH4)2CO3}$ to give out the precipitate of $\ce{BaCO3}$. How do I know this? Partly because of the fact that while testing for the presence of $\ce{CO3^2-}$ ions in a laboratory we use the lime water test. Instead of using $\ce{Ca(OH)2}$, baryta water or $\ce{Ba(OH)2}$ is also used at times yielding strikingly similar results.
Also, I know this partly because $\ce{CO3^2-}$ ion is notorious for forming precipitates. It forms a precipitate with a large number of ions, some exceptions (they're the only exceptions I know) being $\ce{Na+, K+, Rb+, Cs+}$ and $\ce{NH4+}$.

However, what happens to the reactions like these?

$$\ce{(NH4)2CO3 + K2SO4 <=> K2CO3 + (NH4)2SO4}$$

Neither of the product is a precipitate. Hey that's easy? $\ce{K2SO4}$ is made up of $\ce{KOH}$ and $\ce{H2SO4}$. And look the base is the same for $\ce{K2CO3}$, that means I've made carbonic acid from sulfuric acid and thus the reaction takes place. Right? Wrong.
In your haste you forgot to take note that I could've used the same logic if I were to deal with $\ce{(NH4)2CO3}$ and $\ce{(NH4)2SO4}$. Wait! This reverses my previous reasoning! Yes it does. Or maybe we get a little confused.
Perhaps these cations are spectators, shall I strip them off? Yeah, let's give it a try too. We're now (sadly) left with:
$$\ce{CO3^2- + SO4^2- <=> CO3^2- + SO4^2-}$$
That's bizarre. What's happening? Nothing. And that's what it is in its true form, the ions basically are just combining with each other, there are four ions, two cations and two anions, four possible combinations. Does that make it a reaction? Perhaps not. It's just an equilibrium, there's no going forward or backward. When we say there's a reaction taking place, we are implying that the equilibrium sways to the the product side, which apparently isn't the case here.


text:
How does the position of an electron donating group affect acidity in benzoic acid?
I  know that the closer the electron withdrawing group to the $\ce{-COOH}$, the stronger is the acid. I am interested as to how this relationship works out in the case of an electron donating group.

This is the exact goal of Hammett plots and analysis of linear free energy relationships (LFERs).

@Zhe, judging  by the question I guess your attempt here is to further confuse the person asking the question.. :P

@AlphaRomeo Yeah, that wasn't super helpful... Let's actually read the question and not just the title and try again: Electron donation groups increase the negative charge density on the conjugate base, which destabilizes the conjugate base. The less stable the conjugate base, the less acidic the corresponding acid. The effect is exactly opposite with electron withdrawing groups. The closer the substituent is on the ring, the stronger the inductive effects. Just be careful if you're working with substituents that can participate heavily via resonance.

Keep in mind, resonance effects do not matter much, since the $\ce{-COOH}$ isn't in conjugation with the ring.

https://chemistry.stackexchange.com/questions/12309/inductive-effect

https://chemistry.stackexchange.com/questions/7683/ortho-effect-in-substituted-aromatic-acids-and-bases

The strength of the acid is determined by strength of the positive charge that is on the carbonyl carbon so in case of the presence of an electron donating group you have also to check what is its inductive effect whether it has an electronegative part which would attract electrons towards it or not ....the best way is to draw the resonating structure for every case and you will reach how is the activity affected by the EDG in different positions..one last factor you have to bear in mind is that the presence of an OH group in ortho position to the COOH will form a hydrogen bond which makes it even more stable than other positions ..if we take salicylic acid as an example then the strength is ortho>meta>para


text:
Why does the electrolyte used in Lechlanché cell turn blue?
In my school there are jars which are turned into Lechlanché cells. Basically the cell has aqueous ammonium chloride solution as the electrolyte which is obviously transparent in the beginning. The electrodes used are zinc and the second electrode is a porous pot which has a carbon rod and crushed manganese(IV) Oxide. The electrolytes in cells which have been used for a long time have turned blue. I want to know where this colour is coming from. Because ammonium salts are colourless, zinc complexes cannot show colour due to d-d transition and I have never seen a blue coloured manganese compound. So what is it that is forming due to which the solution is turning blue?

How pure is the zinc?  How much copper (-> deep blue ammonia complex) is in it?

I think this is due to the colour ammoniated electrons (blue)

Solvated electrons in water solvent?  Not sure they are blue in water or would last long enough to be seen  without special equipment.

Not sure. We use zinc rods so I don't think there will be much impurities in the rod which could turn the solution deep blue.

Yes , I also think that the colour couldn't be due to ammoniated electrons because there is water in excess. But then again I am not completely sure.


text:
How to create the total reaction out of the two steps behind denitrification?
I can see that denitrification consists of processes involving mainly two steps:
Step 1 (nitrate to nitrite):
$$ \ce{3NO3- + CH3OH -> 3NO2- + CO2 + 2H2O} $$
Step 2 (nitrite to molecular nitrogen):
$ \ce{2NO2- + CH3OH -> N2 + CO2 + H2O + 2OH-} $$
Total reaction is supposed to be:

$$ \ce{6NO3- + 5CH3OH -> 3N2 + 5CO2 + 7H2O + 6 OH-} $$

Now I'm a bit rusty.
As far as I can remember, the right hand side on the first step of the reaction will be the starting point for the left hand side of step 2 in the overall reaction process.
In other words, what is present on the right hand side of step 1 and on the left hand side of step 2 can be subtracted from each other (which is logical). Also, the start of the total reaction would be represented by the left hand side of step 1 and the end of the total reaction would be represented by the right hand side of step 2.
The coefficients 5 and 7 seem very strange, but they balance everything so that makes sense. However the balancing of the Os seem to be the challenging part.

The attempt yields:
$$ \ce{3 NO3- + CH3OH -> N2 + CO2 + H2O + 2OH-} $$
Balancing N:
$$ \ce{6NO3- + CH3OH -> 3N2 + CO2 + H2O + 2OH-} $$
C and H atoms are already balanced.
However balancing O is more complicated because changing any of the Os affects the balance of the others (like C, H). Are there any recommendations for easing out that process or methods that can be applied in general?

After balancing carbon and nitrogen, I'd recommend you go for balancing the charge in the equation:
$$ \ce{6NO3- + CH3OH -> 3N2 + CO2 + H2O + 6OH-} $$
Now, we've to balance oxygen and hydrogen without disturbing the already balanced species. It would be interesting to note that here the stoichiometric coefficients of methanol and carbon dioxide should be the same or the number of carbon atoms wouldn't be equal on both sides of the equation.
So, let's assume the stoichiometric coefficient for both these two species to be $x$. Also, let's assume the stoichiometric coefficient of water on the right hand side to be $y$. The equation now looks like this:
$$ \ce{6NO3- + xCH3OH -> 3N2 + xCO2 + yH2O + 6OH-} $$
We now balance the number of oxygen atoms:
$$ 18 + x = 2x + y + 6 $$
And the number of hydrogen atoms:
$$ 4x = 2y + 6 $$
Solving these two equations, you get $x=5$ and $y=7$. And therefore your overall equation finally is:
$$ \ce{6NO3- + 5CH3OH -> 3N2 + 5CO2 + 7H2O + 6OH-} $$

That definitely eases the process to a high degree, thanks!
However regarding the assumptions for the stoichiometric coefficients that you use where you define the stoichiometric coefficient of water to be equal to y:
is it normal to treat water in this way when balancing chemical reactions? Or is it rather so that it is not a rule of thumb, but something one finds by trial and error and/or experience? So perhaps this doesn't only apply to water.

@jibo It just struck me that this was the quickest way I could get to the answer, I'm aware that this is really a vague comment, but now you're interested I'll write a canonical post about balancing equations. It would then make your concepts crystal clear. Please allow me some time.


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Why does light and heat yield different stereochemistry in electrocyclic reactions?
In my organic chemistry II class, we started learning about electrocyclic reactions. We learned heat and light yield different products. So my question is, how do they work differently to yield those different products?

The explanation may be a bit too long here. You should look up how the symmetry of the HOMO and LUMO impact how pericyclic reactions unfold.

Have a look at Woodward-Hoffman rules.


text:
What's the relationship between supersaturation and temperature?
In gravimetric analysis, what does supersaturation mean? And, how is it achieved by lowering the temperature, even though increasing the temperature makes a higher number of solute particles more soluble?

Supersaturation means simply that more of the substance is dissolved than should be. Supersaturation is an unstable condition. Given some "trigger" precipitation will occur and equilibrium conditions will then be restored.
The trigger could be a seed crystal, scratching a glass rod on a side wall or a piece of dust which acts as a nucleation point.

So can this trigger be lowering its temperature ?

Lowering the temperature is used to reach supersaturation in the first place.

yeah i don't understand how ..isn't it supposed that increasing the temp is needed to reach supersaturation??

Nope. Increasing the temperature would increase the solubility. You want to decrease it.

in order to precipitate?

In order to make the solution supersaturated, and ultimately, yes, to precipitate.

@SalmaM.Hanafy Whether temperature is increased or decreased, depends on the situation at hand. A good number of solutes will dissolve less if temperature is increased in some range; but decreasing temperature is still more common. Precipitation is usually a slow process, so before you optain an insoluble solid, there is a time period when the concentration exceeds maximum solubility at the particular temperature. Mathematically, relative supersaturation turns positive.

@SalmaMHanafy Glad we could help!


text:
first-order binding kinetics with multiple identical binding sites
I have beads in solution at concentration $B$ which have $N$ independent binding sites for a single target molecule at concentration $T$. $T$ is small compared to $B$, so most beads will probably not have anything bound at all. The on and off rates for a single target binding a single site on the bead are $k_{on}$ and $k_{off}$.
I want to determine the equilibrium distribution of the number of bound targets per bead. I expect it to be a Poisson distribution (right?) with an average binding count that depends on the parameters above.
Define $B_n$ to be the concentration of beads with $n$ bound targets. First order kinetics suggest we should have the coupled system of equations:
$\frac{dB_n}{dt}=k_{on}T(B_{n-1}-B_n)+k_{off}(B_{n+1}-B_n)$,
$\frac{dT}{dt}=\sum_{n=0}^{N-1}\left(-k_{on}TB_n+k_{off}B_{n+1}\right)$
with $B_{-1}\equiv 0\equiv B_{N+1}$
First question: is there an analytical solution for this system of equations? I am perfectly happy to simulate it, but if there is a way to avoid that, it would be nice.
Second question: the wikipedia page for Dissociation Constant has some ambiguous wording suggesting that even if the receptors are all identical, the rate constants will be dependent on $n$ as well, via
$K_d(n) =K_d \frac{n}{N-n+1}$
where $K_d=\frac{k_{off}}{k_{on}}$
How does that play here? Are my equations in need of modification, to something like:
$\frac{1}{k_{on}}\frac{dB_n}{dt}=T(B_{n-1}-B_n)+K_d \frac{n}{N-n+1}(B_{n+1}-B_n)$.
If so, could someone explain where that modification comes from?
EDIT: it seems that this is probably due to the fact that there is a higher change of dissociation if a bead has many targets bound since there is an equal probability for each target to unbind. However, it seems that the fraction presented is 1 for $n=N$, which suggests that the dissociation constant is defined in terms of a fully bound bead and not on a per-site basis. Could someone show me the derivation?

The scheme you need seems to be of the form
$$\ce{S + M_i  <=>[k_2][k_{-1}] M_{i+1}    }$$
where S is the molecule in solution and M the bead with i sites filled.
If there is no restriction on how many S may be associated with any bead and as the  S are indistinguishable molecules the number of ways of arranging the S among a total of M beads is
$$ W=\frac{M!}{\prod M_i!}$$
and the constrains are $\Sigma_0^N M_i = M$ and $\Sigma iM_i = N$ where N is the total number of S molecules.
Solving with these constraints is done using Lagrange multipliers, in the same manner as the Statistical Mechanics calculation of Boltzmann distribution, and leads to the distribution of finding i molecules of type S in a bead as,
$$p_i = \frac{\bar n ^i}{(1+\bar n)^{i+1}}$$
where $\bar n$ is the average occupancy of a bead. This distribution, which is clearly not Poissonian, has the form that it decreases as i increases even for moderate $\bar n$.
If the kinetic scheme is changed slightly to
$$\ce{S + M_i  <=>[k_2][(i + 1)k_{-1}] M_{i+1}    }$$
where the ‘off’ rate constant now depends on how many S are on a bead then the probability becomes Poissonian. This change/trick effectively makes the molecules distinguishable and now
$$p_i = \frac{\bar n^i}{i!}e^{-\bar n}$$
with $\bar n = k_2\ce{[S]}/k_{-1}$.
Which of these or other models applies will have to be determined by experiment.
If you want to solve the equations numerically then it is best to try a Master equation approach.
The set of equations for the population of M is written as
$$d\vec M/dt = \vec k \vec M$$
and has the formal solution $\vec M=\vec M_0e^{\vec kt}$ where $\vec M$ is a vector of concentrations of species i and $\vec k$ a matrix of rate constants such as
$$ \bar k=\left [\begin {matrix}
-Sk_2 & 2k_{-1} & 0 & 0 & 0 & \cdots\\
Sk_2 & -Sk_2 -2k_{-1} & 3k_{-1} & 0 & 0 & \cdots \\
\vdots& \vdots&\vdots & \vdots& \vdots& \\
\end {matrix}\, \right ]$$
where S is the concentration of S and is assumed to be in excess so that its value is constant. (If you want S to vary then you will need to do a numerical integration on all N equations instead of this method).
To solve, the secular determinant is set up and the eigenvalues $\lambda$ obtained. The populations at time t are given by
$$M(t)= x[e^{\lambda t}]x^{-1}M_0$$
where x is the matrix of eigenvectors corresponding to eigenvalues $\lambda$ and $[e^{\lambda t}] $ is a diagonal matrix of $e^{\lambda_1t}, e^{\lambda_2 t} \dots $ etc.

Thanks. Not quite what I had in mind, but there was enough information here for me to get the answer I needed. I do need S to vary, so I will have to do this numerically, sadly.

I was able to figure out the answer to my second question using the information provided by @porphyrin in the accepted answer, but I am posting my own solution here for the sake of completeness. Specifically, the reasoning behind the effective rate constants changing when there are multiple bound targets per bead discussed cryptically on the wikipedia page here.
Let $k_{on}^{(0)}$ and $k_{off}^{(1)}$ be the rate constants for binding and unbinging of a target molecule to a single site on the bead. Because we are assuming linear kinetics, if there are multiple binding sites free on the bead, the rate of binding to that bead will simply be multiplied by the number of free binding sites. Similarly, if there are multiple bound targets, the unbinding rate for that bead will be scaled accordingly.
Consider the reaction $$\ce{T + B_{n-1}<=>[k_{on}^{(n-1)}][k_{off}^{(n)}]B_n}$$
Because there are $(N-(n-1))$ free sites on the bead on the left hand side, the effective on rate is $$k_{on}^{(n-1)}=k_{on}^{(0)}(N-(n-1))$$
On the right hand side, because there are $n$ bound targets, the effective off rate for this bead is $$k_{off}^{(n)}=nk_{off}^{(1)}$$.
Therefore, the dissociation constant for this particular reaction is $$K_d^{(n)}=\frac{k_{off}^{(n)}}{k_{on}^{(n-1)}}=\frac{k_{off}^{(1)}}{k_{on}^{(0)}}\frac{n}{N-n+1}$$ as required.


text:
Schottky Defect in AB2 type crystals
Are there any examples of Schottky defect in ${\ce{AB_2}} $ type crystals.
Does Schottky defect require that the cation and anion be of equal size?
According to me it is not necessary that the two ions are of equal size as even ${\ce{NaCl}} $ shows Schottky defect but ${\ce{Na+}} $ has radius is 95 pm and ${\ce{Cl-}}$ has radius of 181 pm which is a bit less than double of that of ${\ce{Na+}} $

I suppose schottkey or other defects are based on thermodynamics that is if  energy required to pair a cation and anion is higher that the energy required for the defect then that defect is more probable to be seen. I think it is just an observation that cation and anion of similar sizes show  schottkey defect

Schottky defect occurs when equal number of positive and negative ions are missing from the lattice thereby maintaining net electrical neautrality and leading to a decreased density in the lattice (since there are now less number of ions) .
It is observed to be occurring in cations and anions of similar sizes.
NOW CONSIDER THAT AN ION IS AT THE LATTICE POSITION AND A DEFECT IS TO BE OCCURRED.
There are 2 ways stoichiometric conditions can be satisfied. - either by loosing this ion as well as another of opposite charge in equal ratio(SCHOTTKY) ,
OR not losing this ion at all , but just shifting it's position from lattice position to interstitial one.(FRENKEL)----
If we think logically ,
Since it's a stoichiometric defect it should maintain neautrality as well as loose cations and anions in a certain ratio.
If the sizes were very different , there are chances the smaller ion would occupy an interstitial space causing it to create a hole and still maintain neutrality and ratio without losing any ions. (Which is what happens in Frenkel defect) . Since however ions are of similar sizes , one can't occupy an interstitial space .. and therefore losing equal no. Of anions and cations is the only way to go.
You can think of these as 2 cases of defect... Yes it's not necessary the ions are of similar sizes for a defect to occur BUT IF THEY ARE...They are most likely to cause the Schottky defect and if not , then Frenkel Defect.
As for your NaCl doubt , I would suggest try finding the critical radius ratio for one ion to occupy the interstitial space . This way you can verify whether it can perform Frenkel defect or not ..Or why it's performing Schottky.
Hope this helps.
EDIT:- I can't find any examples of ${\ce{AB2}} $ type as such , but you can again try finding critical ratio of ions in ${\ce{AB2}} $ lattices to decide what type of defect they are likely to occur.

Please consider editing your answer to make it more readable.

@Mr.Momo Yeah,I will check.

I didn't say it was you f^_^; sorry. Just wanted to know why it was downvoted.


text:
Are lustre and conductivity directly related?
In a question that I had asked before, Is chromium the shiniest metal?, I received the answer that silver was, in fact the most lustrous metal for visible wavelengths. At the same time, silver is also the best metallic conductor, having a resistivity of $1.59×10^{-8}$. I know that lustre of metals was explained by considering the oscillations of free electrons in them, so, I could hypothesize that silver allows the electrons to move around better than other metals, allowing to shine and conduct more. As a comparison, the pictures of polished silver, aluminum and iron are shown below.
Silver: ($\rho =1.59×10^{-8}$)

Aluminum: ($\rho =2.65×10^{-8}$)

Iron: ($\rho =9.71×10^{-8}$)

Can I safely conclude that conductivity and lustre are directly related?

There aren't related, see https://en.wikipedia.org/wiki/Pyrite for example.

As interesting as the question is, posting pictures of different metals polished to different extents doesn't really help justify the "Lustre-Conductivity dependence" ;)

@Mithoron I don't see how that's related. Please explain.

@paracetamol I made sure the pictures were of maximum polished.

What @Mith meant: Some samples of (iron)pyrites are more lustrous than some common metals (when polished to the same degree), but pyrites have poor electrical conductivity... so the Lustre-Conductivity "relationship" isn't something you should hinge upon ._.

Those pictures look like they're straight off Google Images... I seriously doubt that they're polished to the greatest degree ;P

@I see, thanks for explaining. And yes, they're from Google Images. I typed "polished iron" and I got that.


text:
Does adding slices of lemons to water reduce its chlorine content?
Often restaurants add slices of lemon to water. They seem to make the water taste better and perhaps reduce or at least cover up the chlorine taste.
I've heard that Vitamin C tablets can be used to remove chlorine from tap water.
Q: Does anyone know if lemons, being rich in vit C, will encourage the removal of chlorine from tap water?

See https://inspiredliving.com/chloramine-filters/vitamin-c-removes-chloramines.htm

My only guess would be that adding an organic compound will allow the residual Cl to react with and oxidize the compound, thus reducing the Cl content.

Yes , a quick and cheap way to reduce chlorine residual is to add ascorbic acid
 (VITAMIN C), which if added in an equal molar amount to free chlorine would completely eliminate chlorine residual within 1 min.
I am quoting from ABSTRACT from a paper from, Int. Journal of Engineering Research and Applications , ISSN : 2248-9622, Vol. 3, Issue 5, Sep-Oct 2013, pp.1647-1651.
"Society’s preference for bottled water mainly comes from a dislike of the mildly pungent taste and odor of tap water. Such odor and taste may be caused by chlorine residual (in the form of free chlorine or chloramines), which is maintained in a water distribution system to prevent regrowth of microorganisms and is thus present in tap water.
"Two methods of reducing chlorine residual, boiling tap water and adding ascorbic acid to
tap water, were tested. The results showed that, for free chlorine, a quick and cheap way to reduce chlorine residual is to add ascorbic acid, which if added in an equal molar amount to free chlorine would complete eliminate chlorine residual within 1 min "

Reduction of Chlorine Residual in Tap Water with Temperature

source : https://www.ijera.com/papers/Vol3_issue5/JN3516471651.pdf
As shown aboce, the chlorine residual decreased faster at higher temperatures.

Reduction of Free Chlorine and Monochloramine in Tap Water through Boiling

source : https://www.ijera.com/papers/Vol3_issue5/JN3516471651.pdf
As shown in Figure 2, both residuals of free chlorine and monochloramine decreased with boiling time, and the residual of free chlorine decreased slightly faster than that of monochloramine.
The boiling time required to completely remove free chlorine was about 33 min.$\ce{^1}$

Effect of Ascorbic Acid on Free Chlorine and Monochloramine

Ascorbic acid added in a molar ratio of 1:1 (free chlorine to ascorbic acid) completely eliminated free chlorine. The DPD indicator(A DPD solution can also be used for the measurement of chlorine residual) did not produce a red color when added within 1 min,indicating complete elemination chlorine residual within 1 min.$\ce{^2}$
The reduction of free chlorine by ascorbic acid can be expressed as follows:
$\ce{C6H8O6}$ + $\ce{HOCl}$  →$\ce{C6H6O6}$ +$\ce{HCl}$ + $\ce{H2O}$
Ascorbic acid $\ce{(C6H8O6)}$ is oxidized to dehydroascorbic acid $\ce{(C6H6O6)}$ . Simultaneously, free chlorine $\ce{(HOCl)}$ is reduced to chloride ions (HCl).
In the experiment conducted it was determined that 10 mg of ascorbic acid is required to treat 1 L of tap water. and the quickest way to reduce the concentration of free chlorine in tap water is to add powdered vitamin C (ascorbic acid) tablets in less than 1 min.$\ce{^3}$

Thanks to N4v i will be using this cost effective method to remove chlorine from water.(wow).
Reference

https://www.ijera.com/papers/Vol3_issue5/JN3516471651.pdf
https://www.ijera.com/papers/Vol3_issue5/JN3516471651.pdf
https://www.ijera.com/papers/Vol3_issue5/JN3516471651.pdf

Yes, vitamin C can be used to remove dissolved $\ce{Cl_2}$ from water. The same applies for dissolved $\ce{Br_2}$ and $\ce{I_2}$. Whether a lemon slice (or several) has enough vitamin C to quantitatively react with all the dissolved $\ce{Cl_2}$ in a glass of water is another matter and also a simple hypothesis that can be tested easily in an undergraduate quantitative analysis or intrumental analysis lab. The next two paragraphs are from my old pre-lab lab write-up where students used in situ generated iodine for the quantitative determination of vitamin C in commercial 1 g vitamin C tablets. Bromine, generated in situ by oxidizing bromide ions, also works well. We never tried generating $\ce{Cl_2}$ by in situ oxidation of chloride ions because of the interfering side oxidation reaction.
Commercial vitamin C tablets are not pure vitamin C: it is not suitable for formulation in pure form.  Rather, commercial vitamin C tablets are comprised of pure vitamin C (ascorbic acid, denoted as HAsc) plus various binders, processing additives and excipients.  To determine how much HAsc is in a tablet, we determine the number of moles of HAsc in the tablet and use the molar mass of HAsc.  From the measured mass of the tablet, the % purity is then computed.  Alternatively, we may determine the % purity of the tablet and then use its measured mass to compute the mass of pure HAsc the tablet contained.  Either way, it is necessary to determine the number of moles of HAsc in the tablet or in a measured fraction of the tablet.  In practice, the latter is done since commercial 1 g vitamin C tablets contain inconveniently large amounts of HAsc.
The technique of coulometric titration is used to determine the number of moles of HAsc in the titrated aliquoted of solution prepared from the tablet.  Iodine is capable of rapidly and irreversibly oxidizing HAsc to dehydroacsorbic acid, with iodide ion being another product.  Molecular iodine is poorly soluble in aqueous solution, unless iodide ion is present, in which case the solubility is much higher due to formation of the tri-iodide ion.  The tri-iodide ion rapidly and readily releases iodine molecules for reaction with HAsc in aqueous solution.  Thus, since one HAsc molecule reacts with one diatomic iodine molecule, the iodine molecule can be generated electrolytically in solution by oxidizing a pair of iodide ions, which are deliberately present in excess in the titration vessel.  Therefore, removing 2 electrons fron 2 iodide ions yields 1 iodine diatomic molecule which then reacts quantitatively with one HAsc molecule.  The 2 iodide ions are thus regenerated and keep cycling around.  The stoichiometry is 2 equivalents per mole of HAsc.  (One equivalent is the number of moles of electrons required to reduce one mole of hydrogen ions to hydrogen atoms.)


text:
Why does Nitrogen gas appear in the equation for making producer gas?
My book gives the equation for synthesis of producer gas as follows:
$$ \ce{2C{(s)} + O2{(g)} + N2{(g)} -> 2CO{(g)} + 4N2{(g)}} $$
And it says that the stuff on the right-hand side is called producer gas.
My question is why does $\ce{N2}$ even appear in the equation? It isn't reacting is it? It's just a mixture.

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

In the wikipedia page, the nitrogen has a (seemingly) empirically derived stoichiometric coefficient, so I would also be interested in knowing why that level of detail is presented (since I believe it is not facilitating/participating in the reaction). I would hazard a guess and say that if you had a process that made producer gas, you would track $\ce{N2}$ per unit volume because that would effect the caloric value of the mixture to be used as fuel.

I can tell you why the coefficient is four, or rather four times that of oxygen: precisely because in air, they're found in roughly a 4:1 ratio. Yes, the reaction could definitely be written without the presence of N2. But as it stands, the N2 has been included to emphasise the presence of N2 in the reaction mixture (which, technically, is open air) and to indicate that our focus is on the collective mixture of carbon monoxide and nitrogen.

As the comments allude, the coefficient is 4 to represent the approximate 4:1 ratio of oxygen to nitrogen in air (more precisely 21:78 or 7:26).  This added nitrogen must be heated along with the oxygen and carbon.  From a stoichiometric perspective, nitrogen is just a spectator and could be excluded, but from a thermal/total system perspective as nitrogen increases the specific heat of the system requiring more heat to be inputted for heating the mixture which subsequently requires more heat dumping to cool the system, reducing the efficiency of the system compared to a system of pure reactants.


text:
What is the best way to convert SMILES strings to file names?
I would like to convert SMILES strings into file names. There are many characters to avoid in file names, i.e. \/:*?"<>| or maybe even []()^ #%&!@:+={}'~ (see discussion) for convenience.
I would like to escape such characters that appear in SMILES. However, I could not see a list of allowed characters for SMILES to decide on how to escape them. There might be also some existing convention to do this for databases, so I am all ears for any help.
An alternative is using InChIkeys but since they are not human readable, I want to avoid that option.
Edit: I am mainly interested in small organic molecules with less than 20 atoms.

Are the SMILES strings not going to make awfully long file names for anything but the most trivial of molecules? Long file names might be allowed now, but can cause issues with some software / systems

I am mainly interested in small organic molecules, (less than 20 atoms) so length won't be  a problem. Thanks for pointing it out, I'll edit the question to specify it.

So you like to filter out, char wise, all but 0-9 and A-Z, and collapse this string --  with the examples on the site given by you and on https://en.wikipedia.org/wiki/Simplified_molecular-input_line-entry_system, it sounds like a suitable question for the code golfers (https://codegolf.stackexchange.com/).  Awk, grep ...

I am not after the code to play with the string. I think that part is trivial. My question is more on finding convenient rules to escape such characters. Probably, one can live with  []()^ #%&!@:+={}'~ but I think others (\/:*?"<>|) should be replaced.

InChI keys are more appropriate, since they are in fact hashes. But there is always many ways how to repesent single structure in SMILES string, and there are also multiple ways how to normalize those strings.

Also useful is: https://en.wikipedia.org/wiki/Filename#Comparison_of_filename_limitations  My main concern is that if you're going to escape names. How will do map the characters onto something else? Most modern file systems have name lengths of up to 255 characters, which is a lot until you start encoding your special characters...

Maybe a better question to ask is: why are you doing this?

I personally find SMILES quite unreadable, even for only small compounds. I think they are even less readable when performing substitutions. Given the vast amount of special characters they use, I do not think they are a good way to create a naming scheme.

An alternative is using InChIkeys but since they are not human readable, I want to avoid that option.

While that is true, InChI strings are human readable, especially for small molecules. In principle only the layers need to be separated by something other than / (underscores would do that).
Other problematic characters are *;?+() which do not fit in [0-9a-zA-Z-.,_], and you would have to find ways around them when using in scripts.

I am mainly interested in small organic molecules with less than 20 atoms.

If you are really only interested in molecules with less than 20 atoms, general nomenclature should be the easiest to comprehend, and should yield reasonably short file names. For these you probably only have to worry about spaces.

The question is about SMILES.

@NoelO'Boyle Exactly, that's why I am writing that SMILES is a bad way to generate filenames and then offer alternatives.

I've used base64 encoding in the past for escaping SMILES strings for use as filenames.

I might be wrong here, but that is equally unreadable as an InChIkey, and it doesn't even have a fixed length. Otherwise I don't understand your answer (which is more like a comment to begin with).

I read the original question as the OP looking for a way of escaping the SMILES string for use as a filename. For sure, base64 encoding is not human readable.

From my experience, the only characters that have been a problem are '/', '\', and '#'. I replace them with 'z', 'x', '^' respectively. It's not a convention but just something I've come up with. If you were to use the file name in a command, '(', ')', '$' would also need to be escaped.

Use the PubChem Compound ID (CID) to name your files. You'll be able to name your files with numbers containing, at most, 9 digits. With over 100 million compounds in PubChem, your compounds are likely to be present.
Download the CID-SMILES.gz file from the PubChem FTP Server. It contains 2 columns: CID and the isomeric SMILES, separated by a tab.


text:
Noyori hydrogenation and the Curtin–Hammett principle
My question is relating to how the Curtin–Hammett principle applies to homogeneous hydrogenation using chiral ligands.

Figure 0: An example of a quadrant diagram used as a mnemonic to predict the facial selectivity of reduction (in this case of a 3-hydroxyketone)
We commonly draw mnemonics (i.e. quadrant diagrams) to 'predict' the most likely products, however this prediction surely works on the assumption that the energy of the intermediate complexes is responsible for the overall outcome of the reaction since in using these mnemonics we're assuming that steric clashes are what determines the 'lowest energy' binding mode (which is supposedly not the case in the example in figure 1).
Does the mnemonic take into account this kinetic preference, or are we only able to predict the product for hydrogenations not taking place via Curtin–Hammett kinetics? (and if so, can one generalise if the Curtin–Hammett principle is applicable for a given reduction? I seem to recall reading something along the lines of Ru(II) and Ru(I) systems operating by different kinetics, for example).

Ruthenium-BINAP complexes (and related complexes using modified ligand systems) are commonly employed for asymmetric homogeneous hydrogenation/reduction of a wide range of substrates (the general caveat is the need for 2 point binding in order to get suitable control).
An example is below in figure 1 using a diphosphine ligand to selectively reduce the enamine, which introduces a new stereocentre and essentially furnishes an amino acid derivative.

Figure 1: Example of a Ru-catalysed asymmetric hydrogenation (from Organotransition metal chemistry, from bonding to catalysis, Hartwig).

When thinking about these reactions, we often use 'quadrant diagrams' and other mnemonics in order to predict how the substrate will bind (based on minimising steric interactions), but essentially binding of the substrate to the complex yields two diastereomeric transition states which are able to interconvert (figure 2).

Figure 2: Proposed 'mechanism' of a Ru-catalysed asymmetric hydrogenation  (from Organotransition metal chemistry, from bonding to catalysis, Hartwig).

What is unusual in this reaction is that the energy (and hence population) of the diastereomeric intermediates does not have any influence on the ratio of products formed, with the higher energy/less populated intermediate being the one going on to form the product. This is an example of Curtin–Hammett kinetics, with the oxidative addition step, in this case, being the thing that determines which product forms.

Figure 3: Reaction profile (from Organotransition metal chemistry, from bonding to catalysis, Hartwig).

There's no way to predict this.
Mnemonics are just rules, and rules are just rules. There will be exceptions, and you won't know when you should apply them a priori. If you could, it would be part of the rule by definition.
The Curtin-Hammett conditions apply when the rate of interconversion between intermediates is significantly faster than the rate of reaction from the intermediates onto their respective products. If the rate of interconversion is slow, then the ratio of products is simply governed by the ratio of intermediates.
The problem comes about from determining if the interconversion is fast. If you're looking at proton transfers or some kinds of conformational changes, it's usually pretty straightforward. Otherwise, only intuition can guide you ahead of time. Once you have experimental results, you might be able to rationalize this fact and determine that the reaction is governed by Curtin-Hammett kinetics, but there's no way to know for sure. This is just due to the complexity of binding entropy, solvent effects, etc, coupled with the intrinsic barrier for the forwards reaction.
Chemistry is hard.

I agree with what you're saying, and as a general comment you're completely right about a priori prediction of kinetics. My question was more related to these reductions specifically. A mnemonic clearly does exist, and often works, even if the steric argument doesn't tally with the kinetics. Whilst we might not be able to 100% guarantee whats going on, there are surely some general principles at play

The last statement could be potentially dangerous depending on what you mean by "general principles." The science of chemistry is a set of such general principles, but the way nature works obeys those principles only as far as the principles are correct. We may find the that rules or principles apply, but it is indeed the limitation of science that we cannot ever be sure that the general rule is correct; we can only ever say that an hypothesis was refuted.

I'm not arguing against mnemonics. I just want to emphasize that they are always limited, and you should recognize that there are pitfalls to using them.

I don't disagree, your answer is just more philosophical than practical. We have models, which in many cases we're stuck with (rightly or wrongly, many predict the right thing for the wrong reasons). Since we do have these models, and we have a wealth of examples, we can and do rationalise and draw general conclusions (which is what I'm asking about, since many textbooks do indeed make these conclusions with little explanation).

Unfortunately, I do not have enough experience with Noyori hydrogenations to provide any insight or assistance with the rules. Sorry that this answer was too general to be helpful...

Not at all, its an interesting answer in its own right, and useful no doubt if someone searches Curtin Hammet generally!


text:
Equivalent weight for CaO

I was wondering how I'd go around calculating the equivalent weight
of CaO. From my knowledge, the EW of a compound is its (given mass
)/(valence factor, n). Since CaO has no charge and no change in
the oxidation number takes place, my valence factor comes out to be
0, which I know is wrong because the compound has to have some EW, which is 28 as I know that 28grams of CaO will be formed by 8g O

sigh Why are you even learning this outdated stuff? EW depends on specific reaction, in typical reactions of CaO it's 2.

@Mithoron Poor kid gets homework from outdated teachers.

Yes, Dev, you know the correct process of calculating EW, but the problem you are facing is how to calculate the valence factor?, right?
So, in this case you have to understand that EW for any compound is reaction-specific. So, you can't calculate the EW of $\ce{CaO}$ without any reaction reference.
Still, usually $\ce{CaO}$ reacts through 2 electron exchange mechanism, in other words, $\ce{CaO}$, on dissociation, gives total cationic charge +2 and total anionic charge -2.
$$\ce{CaO = Ca^2+ + O^2-}$$
Hence, in this case of calculating EW, you have to consider $n=2$ and so the EW of $\ce{CaO}$ is $(40+16)/2 = 28$.


text:
In this specific case is the proton considered product or catalyst?
Assuming we have a reaction of
$$\ce{CO2 + H2O ->  HCO3- + \color{red}{H+}}$$
then is the proton (in red) is considered a product or a catalyst?
I read the Wikipedia article titled Product (chemistry), but the issue is still not clear to me. The article gives the following illustration:

In my original example above, I assume that the proton is considered a product since the hydrogen was originally part of the water molecule.
But in this case (see here, at 6:45) the hydrogen proton of the product of the reaction goes to the hemoglobin and kicks off an oxygen molecule.

As you've illustrated with your figure from the Wikipedia article, a catalyst appears as the same compound on each side of a chemical equation. The short answer to this part of your question is that since the proton only appears on one side of the equation, it cannot be a catalyst. A hydrogen atom bound to a water molecule is not the same thing as a proton. Rather, it is the result, or product, of a chemical reaction. Also note that the reaction given in your question would be more correctly expressed as:
$$\ce{CO2 + H2O <=> H2CO3 <=> HCO3− + \color{red}{H+}}$$
In the second part of the question, that regarding the video excerpt, the proton from your reaction actually becomes a reactant in a second completely different reaction. Here, the proton reacts with hemoglobin resulting in two new products: the protonated hemoglobin an an oxygen molecule.
I hope that I have addressed the crux of your question. Please don't hesitate to ask for any clarifications in the comments below.

I guess the correct equation is:

$$\ce{CO2 + H2O <=> H2CO3 <=> H+ + HCO3-}$$

A catalyst is something that enters the chemical equation just to decrease energy needed for the reaction to begin and in this case the $\ce{H+}$ took part in the final product so I don't think it's a catalyst.


text:
Why aren't free radicals considered resonance forms?

In C, the answer states that electrons must come in pairs. And so C does not show resonance forms.
But here I read that if there is an unpaired electron, it is a free radical. Why can't free radicals be part of the resonance form?

(Di)radicals are legitimate resonance forms.

Thank you orthocresol. Just checking, are single radicals also resonance forms?

Yes, but you can't make a resonance form with only one unpaired electron from a resonance form with none - so in this case the usual depiction of O2 has no unpaired electrons, and any resonance form you draw can only have 0, 2, 4, ... unpaired electrons. If your original structure has one unpaired electron, then yes, certainly it can have other resonance forms that also have one unpaired electron.

Thanks @orthocresol, that makes sense.
Mithoron, You are right, I do no quite understand what you mean by "something usually is or isn't a radical".   Do you mean like a single hydrogen atom is a free radical but usually isn't a radical as it exists as H2? If not, do you mind giving me some search terms I can use to read more about what you are talking about (I don't really know what you mean so not sure what to google). Thank you.

In the case of dioxygen, I would not consider a diradical representation to be a resonance structure. The diradical representation is only attempting to account for the short comings of Lewis structures.

And in terms of free radicals, allyl radicals are the simplest example of a free radical that has resonance structures involving the radical. You should be using half-headed arrows, but otherwise, it's not unlike resonance structures with charges.