text:
Why don't we produce water from combustion?
My friend who is a biology major proposed that we can produce water by burning any hydrocarbure or wood or ... to  obtain water.
$\color{red}{f(x,y,z,\Lambda)}\ce{C}_x\ce{H}_y\Lambda_z+\color{red}{g(x,y,z,\Lambda)}\ce{O_2->}\color{red}{x}\ce{CO_2}+\color{red}{\frac{y}{2}}\ce{H_2O}+\color{red}{\text{other substances}+\text{heat}}$
I know this may be a stupid question but why isn't this method used to obtain pure water? Is it maybe because it releases a lot of heat and a lot of carbon dioxyde or is it because the fuel is more expensive than water?

In semiconductor fabs, the water for steam oxidation of silicon is made by burning hydrogen - it is 'easy' to get very pure $H_{2}$ and $O_{2}$ so this way no impurities (like Na) are introduced into the equipment.

He proposed this solution for the water scarcity problem that is to produce large quantities of water.

You need to capture and separate your water from your $\color{red}{\text{other substances}}$ as well as any unburnt fuel.

Like you say, fuel is more expensive than water. You need to burn one liter of fuel for two-three liters of water, and then you still have to condense it out of the hot exhaust gas.

@Karl You could burn any cheap substance but I understand this is impractical.

@BPP Nothing that burns is much cheaper than oil or wood, or people would use it instead.

The answer is in the formula.  The 'other substances', which are either volatile matter or ash byproducts, are formed along with the water.  The water is not pure because some of volatile substances can condense with the water which will require further treatment for removal.  Removal of the volatile matter and the disposal of the rest of the 'other substances' will cost too much money for this solution to be economically feasible.

Since water and \color{red}{the other substances} have different degree of condensation we can easily separate them. Even better we can use only hydrocarbures or substances that don't produce nocive substances.

Distillation (which is a type of treatment process involving separation by differing condensation temperatures) is conceptually easy; however, relative to current technologies the energy requirement for distillation make implementation expensive.   ... If alternative hydrocarbons are available then it is likely a source of water is available as well.  Why would it not make more sense to distill impure water, which likely has much fewer contaminants, to obtain pure water.

Maybe we could do some kind of isothermal combustion at a temperature where water is liquid and all other substances are gazes.

The problem about drinkable water is that we need more "clean and unsalted" water, and less CO2; we don't need more water at the moment. With the climate unbalance, less snow (clean and unsalted) is accumulated in high mountains in winter so fewer water is released in rivers when the weather becomes warmer. It is a big problem between some middle east countries which are along the river because some built barrages on the river to get energy and other countries get less water because of that.
The question is much larger than just getting water.

This is a good point.  Geo-politics surrounding clean water is an ancient problem.  People have been squabbling over clean water and burning things for a long time.  If there were an economical way to produce clean water from burning stuff we would have likely found out how to do it many years ago.

I know this may be a stupid question but why isn't this method used to
  obtain pure water? Is it maybe because it releases a lot of heat and a
  lot of carbon dioxyde or is it because the fuel is more expensive than
  water?

Cost. Fuel cost a lot more than water. And in most cases, it just does not make sense to burn fuel of any type to obtain pure water. It is cheaper to just use that energy from burning said fuel to power machinese to purify water from a local body of water (ie lake, river, ocean)
However, in space missions, hydrogen fuel is oxidized (electrochemically, never burnt as fires in space is an absolute nightmare) in a fuel cell to obtain both energy and drinking water.

Why would we? Pure water is widely available and, where it isn't, there are cheaper and easier ways to get it.
Sure, you can make water from combustion of hydrocarbons. Better still by burning hydrogen (where it is the major product). But why do that if there are better ways to get potable water?
The majority of countries have widespread sources of clean (or easy to clean) water. There are lakes and rivers. extra water can be obtained by arranging to collect rain (often in artificial lakes called reservoirs).
Even in countries short of rain, rivers or lakes, there are usually vast quantities of salty water available. You can't drink it directly, but there are plenty of relatively cheap ways to make it potable (distillation or reverse osmosis use far less energy than generating water from the fuels that would provide power for those processes).
Maybe, were you a stranded astronaut on Mars with large supplies of oxygen and hydrogen, the best way to generate water would be to burn the hydrogen (this is part of the plot of a well-known movie). But why use this technique even in a desert country where you could generate far more useable water by running a desalination plant with the same fuel?
To summarise: if you need water the best way to get it will be the cheapest and easiest way to get it. Unless you are stuck on Mars, burning fuel will never be the cheapest or easiest way to do it.

There is an attractive solution particularly well suited for the desert areas near the ocean, like California or Mexico.  Use an old oil rig (or a retired aircraft carrier) for a floating desalination/hydrolysis plant (solar or nuclear powered).
H2 gas can be piped ashore at a few psi for many miles, over mountain ranges, where ever it is needed, far easier than pumping water$^1$.
At its destination, every pound of H2 burned will yield around 9 pounds (1 gallon) of water, plus energy from H2 combustion (in a turbine) can be used to for mechanical energy or to make electricity.
Burning wood or coal unfortunately yields a lot of CO2. Even natural gas, CH4, would yield 44 pounds of CO2 for every 36 pounds of water.
$^1$ the pure O2 by product of water hydrolysis would also be saleable.


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How can weaker nucleophiles kick out stronger ones in the reaction of chloroalkanes with sodium iodide in acetone?
I have the following SN2 reaction, in an acetone:
1,4-dichlorohexane (1 mol) reacting with NaI (1 mol) to form 4-chloro-1-iodohexane and NaCl.
However, chloride ions are stronger nucleophiles than iodide ions in an aprotic solvent and iodide ions are weaker bases than chloride ions. Reactions generally form weaker bases, why does this reaction occur? My text shows it does, but I don't understand why. What am I missing?

While everybody makes mistakes, it would be preferable that when you copied the question falsely and an answer has already been given, you would ask a new question instead. In the very least you can inform the author of the answer about the changes.

Acetone is the key here. The solubility of NaI in acetone is about 50 g in 100 mL while NaCl is virtually insoluble in acetone, thus the reaction is driven towards the alkyliodide due to Le Châtelier’s principle, since NaCl will be precipitating from the reaction. This process is called a Finkestein Reaction. Also see this.

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Calculating the theoretical percent purity of a recrystallization
The sample contains some percent A and some percent B with A being the majority. The solubility for both is given at two different temperatures. One where both are very soluble and one where both are much less soluble. From these five values I'm supposed to determine the theoretical purity of the recrystallized compound.
I'm surprised that this is possible and I have no idea how to even begin approaching it. I always thought that crystallization was structurally depended as well. Shouldn't the lattice favor the dominant compound? What if the impurity is less soluble at the low temperature than the majority compound? How can we mathematically represent such a (seemingly) complex phenomena with such little information?
Note: I don't mean percent yield, the question specifically stated purity.

I can give you a hint, imagine you have a mixture containing one consitutant $A$ only in your solution with a solvant.
Doing the balanced equations about the mass of $A$ in each phases you have :
$$\begin{cases}
m_A^\mathrm{t}=m_A^\mathrm{c}+m_A^\mathrm{liq} \\
m_A^\mathrm{i}=x_A^\mathrm{c}\cdot m^\mathrm{c}+x_A^\mathrm{liq} \cdot m^\mathrm{liq}
\end{cases}$$
Where $m^\mathrm{t}$, $m^\mathrm{c}$, $m^\mathrm{liq}$ and $m^\mathrm{i}$ denote respectively the mass of $A$ total, in the cristal form, dissolute in the solvant (liquid), and at the begining (initial) and $x$ the corespondant mass-fractions.
Then you must obtain, $$m^\mathrm{c}=m^\mathrm{t}\left(\frac{\frac{m_A^\mathrm{i}}{m^\mathrm{t}}-x_A^\mathrm{c}}{x_A^\mathrm{liq}-x_A^\mathrm{c}}\right)$$
Notice that $\frac{m_A^\mathrm{i}}{m^\mathrm{t}}=x_{A,\mathrm{i}}^\mathrm{liq}$ which is the mass-fraction at the begining in the liquid and $x_A^\mathrm{liq}$ is the mass-fraction in the liquid at the end.
If you know how what you put at the begining, you can determine the yield. Now if you want to know the purity, you'll need to do the same with both of your component to see how much of them have cristalised at the end and then determine the purity of your marjoritory compound.
I hope it can help, have a good day ! :)

Be careful when you do the balanced equations if there is also a gas phase which can happend sometime.

For such seemingly complex problem, you are usually expected to use an ideal case as an approximation. Here, I suppose you should use "ideal crystallization" so that A and B will form pure crystals. So the contamination of A by impurity B will only depend on their relative solubility in the medium.
HTH

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Determining absolute configuration when the lowest priority group is not facing backwards
I have learnt that the absolute configuration of a chiral centre can be labelled as (R) or (S) according to the Cahn–Ingold–Prelog rules, when the hydrogen (or other lowest priority group) is facing backwards, i.e. going into into the page, and the remaining groups decrease in priority clockwise or anticlockwise respectively. On the other hand, if the hydrogen is coming out of the page (i.e. as a wedge) we swap our answer from (R) to (S) or vice versa (to get the hydrogen going into the page).
But what do we do when the hydrogen is flat and not a wedge or dash, as is the case here?

The priorities are

Chlorine atom
Carbon chain on left (propyl group)
Carbon chain on right (methyl group)

This gives an anticlockwise or (S) configuration. But the answer is given as clockwise or (R) configuration.
Why is this the case? The hydrogen isn't coming out of the page. Is there something else that has to be done when the hydrogen is flat or in the plane of the page?  If I visualise it rotating to bring the H to the back, I still end up with an (S) configuration.

One of my favorite ways to figure out the absolute configuration is to imagine the tetrahedral chiral center as a steering wheel. Make the lowest priority atom the "axle" and the three other atoms the "wheel". Mentally visualize which way the wheel would have to rotate to achieve 1-2-3. Right is R and left is S.

Here is my favorite method: https://chemistry.stackexchange.com/questions/99721/twisting-stereoisomers-with-rings-to-determine-r-s/99750#99750

You had the right idea when you said "if the hydrogen is coming out of the page (i.e. as a wedge) we swap our answer from R to S or vice versa (to get the hydrogen going into the page)".
The difficulty seems to have came with your manipulation of the drawing. The diagram you have implies that the hydrogen is coming out of the page towards us (i.e. anticlockwise = (R)).
You could also flip the entire molecule through by 180° to give you the chlorine pointing out of the page towards you.

Thanks for the answer @NotWoodard
Why does my drawing "imply the hydrogen is coming out of the page"?
Is this some sort of convention? I know the Cl was going into the page due to the dashed line, but the hydrogen isn't shown. How did you know it was out of the page? I assumed it was just flat in the plane of the page (like the carbon chains on either side) and not inward or outward.
Also, do you mind telling me which drawing software you used for the diagrams please?

The carbon of interest is tetrahedral. As you drew it, you had the methyl and the propyl group in the plane of the paper, with the chlorine going behind the paper therefore the hydrogen must be coming towards you in order for the carbon to indeed be tetrahedral. The hydrogen is implicit.       The software is called ChemDraw, but unless you're at a university it's prohibitively expensive— ChemDoodle is a much cheaper alternative that will do you well throughout your high school/college studies :)

I see. I didn't even think of the tetrahedral shape. Cheers :)

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Can you differentiate between Kc and Kp in this problem?

I know how to do this problem. We must find the value of K to compare to the value of Q, which can easily be calculated.
I know that the equation we use to get a K out of delta G is as follows:

However, does this equation give us a Kp or a Kc value? What would it give me in this problem if I plugged in the provided delta G standard value?
I calculated the value of this K (whatever it is) to be 1.00, and I also calculated the Qc and Qp for this problem to be 0.00005 and 0.36, respectively. So clearly, Qc and Qp are both less than K. But how can we determine of Qc is really < Kc in this problem?

Just out of learning curiosity, the system isn't at equilibrium at 25'C? and reaction begins with both reactants and products present? Can we say for any reaction Qc < Kc initially and Qc increases as more products are created towards equilibrium?

The equation yields Kp.
$$\Delta G^{o}=-RTln(K_p)$$
There is, however, a relationship betweek Kp and Kc.
$$K_p = K_c(RT)^{\Delta n} $$
From the balanced equation there are 3 moles of gas on the right (2+1) and 1 mole of gas on the left. In chem, delta means final minus initial and we arrive at $$\Delta n = 3-1=2$$
and you now can find Kc. Note that you should use 0.08206 L•atm/mol•K for R.

Sources:

Zumdhal example 13.5
https://www.quora.com/In-a-gas-equilibrium-which-K-constant-is-used-in-the-equation-%CE%94G-RTlnK-Kp-or-Kc


text:
Why does magnesium form covalent bonds with carbon?
According to my professor, magnesium and lithium form covalent bonds with carbon but calcium does not. He did not elaborate.
Why is this so? The electronegativities are:
Mg — 1.31
Li — 0.98
Ca — 1
C — 2.55
So a $\ce{C-Mg}$ bond would have an electronegativity difference of 1.24 and a $\ce{C-Li}$ bond would have an electronegativity difference of 1.57. Both are considered covalent. So how come a $\ce{C-Ca}$ bond wouldn't be covalent? There exists an electronegativity difference of 1.55 — comparable to the difference in the $\ce{C-Li}$ bond.
I know that Mg likes to be tetra-coordinated (Grignards) but I don't think this has much to do with this discussion. So, what makes Mg and Li special compared to Ca?
Summary of possible explanations others have come up with

$\ce{Mg^2+}$ has an ionic radius 72 pm, $\ce{Ca^2+}$ has an ionic radius of 100 pm, and for $\ce{Li+}$, it's 76 pm. So perhaps the greater charge density of $\ce{Mg^2+}$ and $\ce{Li+}$ allow them to pull electrons toward themselves a bit more, thus making the bond more covalent in character? That being said, $\ce{Li+}$ has a single positive charge while $\ce{Ca^2+}$ has a double positive charge, and I haven't calculated the actual charge densities for any of these ions so I don't think we can come to a definitive conclusion here.

http://www.wiredchemist.com/chemistry/data/metallic-radii

related https://en.wikipedia.org/wiki/Group_2_organometallic_chemistry Covalent that, ionic this, is sort of silly. On the other hand article I linked does not mention calcium Griniard analogues at all - organocalcium chemistry looks like dwarf in comparison with organomagnesium.

Magnesium forms a rather smaller cation, which would likely be better at polarising the fluffy electron cloud of the would-be carbanion and forming a more covalent bond. Magnesium has a diagonal relationship with lithium, too, which also forms a very small cation and has a significant amount of covalent bonding to carbon in organolithiums.

@NicolauSakerNeto - are elements diagonal from each other on the periodic table always related in this way (cation size)?

As far as I'm aware diagonal relationships are only really invoked for elements in the second and third periods, and I think it has to do in part with the anomalously small size of the 2p orbital.

The crystal structure of calcium carbide gives a subtle hint of covalent bonding.  It looks like a rock salt structure, but note that the carbon atoms in the anions have their lone pairs directed towards the calcium atoms and so are favorably oriented for covalent overlap. The structure might be considered either a rock salt structure or a series of parallel $\ce{–Ca–C#C –}$ polymeric chains.

Source

Sorry ! The question was not related with $\ce{CaC2}$. It was about the nature of the bond between Magnesium and Carbon

I mentioned calcium carbide to illustrate that the bonding does not go from covalent to ionic in one blow just because we go oe period down in the group. MgC2 has those chains too, albeit stacked differeny.

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text:
How do I calculate the isoelectric point of amino acids, each of which has more than two values of pKa?
For most amino acids, the $\mathrm{pI}$ is simply the arithmetic mean of the amino and carboxyl $\mathrm pK_\mathrm a$s. However, for tyrosine and cysteine, which have more than one $\mathrm pK_\mathrm a$ value, this rule of thumb doesn't apply.
I see that for tyrosine, it's the $\mathrm pK_\mathrm a$s of the carboxyl and amino groups that are averaged, but for cysteine it's those of the carboxyl group and the side chain.
I haven't been able to find an explanation of why this is the case, or what the reasoning behind the calculations is?

Since the $\mathrm{pI}$ is the $\mathrm{pH}$ at which the amino acid has no overall net charge, you need to average the $\mathrm pK_\mathrm a$ values relevant to the protonation/deprotonation of the form with no net charge. Here are the acid-base equilibria for tyrosine:

The form with no net charge is in red (+1 and -1 cancel out to give no net charge). It is the $\mathrm pK_\mathrm a$ values on either side of this form (in blue) that matter, hence the $\mathrm{pI}$ of tyrosine is $5.66$ (the average of $2.20$ and $9.11$).
It just so happens that $2.20$ is the carboxyl $\mathrm pK_\mathrm a$ and $9.11$ is the amino $\mathrm pK_\mathrm a$. If the side chain $\mathrm pK_\mathrm a$ were lower than $9.11$, then you should average the carboxyl and side chain $\mathrm pK_\mathrm a$'s instead.
The same logic applies to cysteine (look up the $\mathrm pK_\mathrm a$ values and draw out the differently protonated forms). You'll find that since the side chain has a lower $\mathrm pK_\mathrm a$ than the amino group, you average the carboxyl and the side chain $\mathrm pK_\mathrm a$'s.
This procedure can of course be extended to the amino acids with acidic side chains (aspartic acid; glutamic acid) and those with basic side chains (lysine; arginine; histidine).


text:
E/Z stereoisomers : assigning priorities with a cycloalkane group
For the following compound

Right hand side of double bond:
$\ce{CO2CH3}$  is highest priority due to the extra carbon (as we work along both chains)
But for the left hand side oft he double bond I am stuck.
I thought maybe we take the blue path, and as there is 3 carbons as opposed to 2 on the orange path, the blue path is the highest priority :

But my book gives the opposite answer :

There is a similar question here How to assign E/Z configuration according to the Cahn-Ingold-Prelog rules when subsituents differ only by stereochemistry   but in the case of my question the carbons on the cyclo which are directly after the double bond do not have groups attached.

The reasoning you have used is clearly faulty as there is no meaning in assigning more number of carbon to one end in a cyclic system.
Just look at the both the branches you have marked using orange and blue. The logic followed is in accordance to Cahn-Ingold-Prelog(CIP) rules.

The first atom in branches, marked as blue and as orange , have two hydrogen and one carbon each.So, there is nothing to clearly distinguish between the two branches.
So, we look at the only distinguishing feature between the two branches , which is the substituents on the carbon atom(marked as 2) attached to the carbon atom(marked as 1) on both sides.
We can see that the carbon atom marked as 2 on orange marked side has two carbon atoms and one hydrogen atom attached to it. This is of higher priority than the carbon on the blue marked side which has one carbon atom and two hydrogen atom attached to it.

Hence the Z-nomenclature as answered in the book.

That is fantastic. I understand now. Before you explained it I didn't realise the orange C2 carbon connects to the blue C2 carbon again (and for some reason thought we don't go back around the cycle when assigning priorities). Thanks!

The simple way to solve it is : look at the carbon number 1 from each side they are the same , then look at the carbon number 2 above there are three bonds but carbon number 2 down there is only 2 bonds
So above take the highest priority ..
The highest priority doesn't depend on the number of carbon it's depend on the same carbon in each side
I hope my answer is clear for you
I tried my best to explain it .
Good luck


text:
Word for substance being diluted
The liquid we use to dilute is the diluent, what is the substance that is being diluted called? I considered "dilutant" but it seems ambiguous. I can't seem to find a common term in literature either.
A possible analogy is solute and solvent. You'd think there'd be a counterpart for diluent too.

I'd go with stock solution, probably.

Maybe diluate?

I like diluand ☺

If you think about it, both liquids are being diluted, no? So maybe they are both diluents...


text:
Theoretical Drying capacity of CaSO4
What is the Theoretical Drying capacity of CaSO4 when used to dry solvents? e.g. How much water can 1 gm of CaSO4 remove?
Now since CaSO4 itself comes in several forms (anhydrous, hemihydrate, dihydrate) I am wondering which is the form best suited for water absorption?
If I assume that the drying transition is from CaSO4 (anhydr.) to CaSO4.2H2O (dihydrate) then I get a drying capacity of approx. 0.26 gm H2O per gm CaSO4 used.
Is this the right capacity? Or am I using an incorrect reaction?

The actual drying agent is calcium sulfate semihydrate. It has low drying capacity, which decreases already after the uptake of $6.6\%$ water (theoretical maximum is $15.7\%$). Within these limits, however, it is possible, for example, to dry air at room temperature down to $0.0004\ mg$ of water per liter of air. It is suitable for drying almost all liquids and gases. It can be regenerated by heating to $190-230^{\circ}C$. Above $300^{\circ}C$ it loses its drying properties. The data comes from an old Merck's handbook on drying agents.

Thanks! Is the "semihydrate" the same as the "hemihydrate"? i.e. CaSO4.0.5H2O?

@curious_cat Yes, those are the same.


text:
Is the compound CH2(OH)(ONa) possible?
I was just pondering about the reaction mechanism of Cannizzaro reaction.
$\ce{H2C^{+}-O- + OH- -> H2C(OH)O-}$
$\ce{HCHO + H2C(OH)O- -> H3CO- + HCOOH}$
$\ce{H3CO- + HCOOH -> HCOO- + H3COH}$
I wondered why in 2nd step instead of receiving $\ce{H-}$ from $\ce{H2C(OH)O-}$; $\ce{H2C^{+}-O-}$ wouldn't react with NaOH to form $\ce{H2C(OH)ONa}$.
I don't think this compound will be able to exist. But don't know why! Everything seems right about it.

It's not clear what compounds you are talking about. The species you seem to be implying $\ce{H2C+O-}$ does not exist - the central carbon only has 6 electrons! Please clarifying what steps in the reaction you mean.

Geez, it's imply mesomeric structure written so it should be more clear. One molecule accepts hydroxide and becomes H- donor, other functions as acceptor.

Of course, you shouldn’t start from $\ce{H2C+-O-}$ unless you note that it is a mesomeric form of $\ce{H2C=O}$ — the more relevant mesmeric structure.
The mechanism as you wrote it and as it is typically written is given without any counterion, i.e. using just $\ce{OH-}$ and not $\ce{NaOH}$. You could instead write the mechanism using a counterion for $\ce{OH-}$ — $\ce{Na+}$, which would result in the first step being:
$$\ce{H2C=O + NaOH -> H2C(ONa)(OH)}\tag{1}$$
This is essentially the compound you are suggesting. Hence it exists, but it is a minor constituent of the overall solution. It’s not part of the second step, it is the first step. For the second step to happen, you need another molecule of $\ce{H2C=O}$, which has not yet reacted to the geminal diolate $\ce{H2C(ONa)(OH)}$.
See this scheme for the transition state of the Cannizzaro reaction and where $\ce{H2C(ONa)(OH)}$ comes into play:

While the mechanism that you refer to is not correct (as Jordan mentioned in his comment), the question is why this compound - $\ce{H2C(ONa)(OH)}$ - doesn't exist. As you say yourself, you see nothing wrong with this compound. Neither do I! However, I see that this compound can undergo simple reactions in either acid or base to become simple formaldehyde. (If you don't see them yourself, I can explain further, but it may be better to just try figuring out yourself.)
There is a small note on this matter that I think is worth mentioning. While in most cases, compounds with carbonyls are much more stable than the corresponding geminal diols, formaldehyde is an exception to this rule (one of the driving forces for the equilibrium towards the carbonyls is the steric interactions in the geminal diol which are greatly reduced in $\ce{H2C(OH)2}$). Therefore, the compound that you drew up is even more likely to occur than if you had an $\ce{R}$ group instead of one or both of the hydrogens.

Mechanism is ok, simply OP has problem with mesomeric structures. Gem-diolate is present as reactive species - in small amount which answers another question.


text:
Why is the Still-Gennari reaction Z-selective?
Why is the Still-Gennari reaction Z-selective?

I guess dipole interactions have the biggest influence here. They somehow destabilize the reversion of the oxaposhpetane to the betaine and prevents formation of the more stable E isomer but I am not sure.

As you hopefully realise, the stereochemical outcome of the HWE reaction (and variations) is due to the relative orientation of the R-groups in the phosphaoxetane intermediate.
If the R groups are syn as in 5A then the (Z)-olefin is formed, with an anti arrangement 5B favouring the (E)-olefin as the product.

Source: Modern Carbonyl Olefination, Wiley, 2004

The pertinent question when looking at the selectivity is why do we get dominant elimination from the syn phosphaoxetane (leading to the correct olefin), rather than dominant elimination from what is arguably the sterically most favourable anti phosphaoxetane (in which the R groups are pointing away from one another).
As it turns out, the initial 'aldol' addition into the aldehyde is rate determining, with the anti addition being favoured over the syn addition on steric grounds.

Source: Stereoselective alkene synthesis, Springer-Verlag, 2012   (a really useful book, FYI)

Once the aldol addition has taken place, the subsequent steps are rapid due to the high electrophilicity of the Still-Gennari phosphonate (with the additional electron withdrawing groups).
This means that the entire process is essentially controlled by the initial addition to give a syn or an anti adduct. Once formed, the adducts are sufficiently short lived that despite the cis phosphaoxetane being less favoured due to the clash between R-groups, the (Z) olefin is formed via quick and irreversible elimination.
The HWE therefore is (E)-selective, in part, due to the fact that elimination is less rapid and as such there is equilibration to the more favourable phosphaoxetane from which elimination may occur.

Why is the (Z) olefin faster obtained?

The other thing I learnt is that, Still-Gennari conditions usually have something to sequester the metal ion - typically 18-crown-6 for K+. In a normal HWE the metal ion would stabilise the starting materials via chelation (as your diagram shows) and this would make the oxaphosphetane formation "more reversible" i.e. lower activation energy for reverse reaction.

The sequestering thing is pretty interesting. There are some really nice papers where they probe the effects of using various crown ethers etc.

In the first image you defined syn and anti as the relationship between the two R groups, in the second image it is between R and phosphonate group. Could you define in both cases syn and anti in the same way for clarity?

No.  This would be incorrect notation.  Though admittedly you would be understood

I am not convinced by the transitions states given in "Stereoselective Alkene Synthesis" because they rely on chelation to restrict the conformation of the phosphonate anion AND the orientation of the aldehyde.
And yet, the reaction conditions are KHMDS/18-crown-6 which are clearly designed to eliminate the possibility of coordination to the potassium cation and maximise the rate of addition of the "naked" alkoxide anion to the electron deficient P.
My own suggestion for the intrinsic "anti" (in this example) selectivity of the fluorophosphonate anion is that it simply mirrors the intrinsic selectivity of the unstabilised Wittig ylid and Horner-Wittig (Ph2PO) stabilised anions both of which produce Z-alkenes stereoselectively.


text:
Internal (Z-Matrix) to Cartesian coordinates conversion
Is there any way to convert from Internal to Cartesian coordinates using simple formulas?
I have studied and implemented: INTERNAL TO CARTESIAN by Jack D. Kunitz.
But somehow the x, y and z values are not correct. I am not implementing the transformation back to the initial frame as it says, because I do not understand that step. Is that the reason for not being able to get correct answers or is this not a good tutorial or solution to follow.

The link you have provided returns 403 (forbidden) - it cannot be accessed. You might consider quoting the relevant information from the material you linked in your question to avoid problems relating to linking content in the future.

I added the original DOI to the link. It's a BASIC implementation of Internal to Cartesian from 1995.

Sure, it's just trigonometry. There are a variety of open source implementations.
For example:

InternalToCartesian from Open Babel
ChemKit
QCL Python

There are dozens in a search on GitHub if you don't like C++ or Python.
There's also a web app from Shodor for z-matrix conversion.

I have just one more question: If I have (x y z) coordinates at first. Then I change them to Internal coordinates. And then I change them back to (x y z) coordinates, what is the possibility that the new (x y z) coordinate points are EXACTLY the same as the 1st ones up to 4 floating points.
Or in other words... Is it okay if they are not that similar. e.g. if I had a point in x=3.1455  and the new point is at x=3.4398. Is that acceptable?

It's highly unlikely for the Cartesian coordinates to be identical. For example, the first atom will usually be placed at 0.0, 0.0, 0.0, regardless of the initial position. What matters is not the Cartesian frame of reference, but the relative positions between the atoms.

Lets just say I can move the first three atoms of the new molecule to be at the exact same place as the old one. Still is it unlikely that they will be different? Or lets assume that the first atom (coıncidentally) was placed at (0, 0, 0). the second atom at (x, 0, 0) and the third one also perfectly and coincidentally placed at the right spot. Assuming this, will the molecule be EXACTLY identical to the previous one or still it might have some floating point errors?

OR in other words, will the relative positioning be EXACTLY the same upto... lets say... three or four floating point numbers. for example: old position = (3.14556, 6.3434, 9.9594) and new position = (3.1459, 6.3439, 9.9560)

No there will often be floating point issues even if the orientation is exactly consistent.

Now I am having another problem: If I convert from Cartesian to Internal and then from Internal to Cartesian a few times (lets say 10), then my whole molecule is messed up. I noticed a few things: First is the the angles and lengths are changed after every time I perform the conversions back and forth, but the torsions remain exactly the same. Is this normal?

Also I believe that small decimal point errors are acceptable in this case but after 10 conversion back and forth, my molecule seems to converge to  a straight line in 3D space with all the atoms in a row. It looks like a long stick with atoms on it. Is this one also normal?

I also just noticed that the lengths are decreasing, the angles are increasing by some degrees and the torsions/dihedral angles remain exactly the same throughout the conversions back and forth.


text:
Is perovskite symmetric under exchange of A and B atoms?
A perovskite is any crystal with the general formula $\ce{ABX3}$ where if the $\ce{B}$ atoms form a cube, an $\ce{A}$ atom sits in its center and the $\ce{X}$ atoms sit between neighboring $\ce{B}$ atoms.
Is the perovskite structure symmetric under exchange of $\ce{A}$ and $\ce{B}$ atoms? That is, if I replace every $\ce{A}$ atom with a $\ce{B}$ atom and vice versa, will I end up with the same exact structure as before?

No.  In the perovskite structure, the "A" atoms are larger than the "B" atoms.  The B atoms are 6-fold coordinated, while the "A" atoms are in 12-fold cuboctohedral coordination.

I'd say the size is not the reason - the coordination/symmetry difference does the job to explain why the sites are not equivalent. The larger volume available in that center certainly helps explain why a larger atom likes it, but if it can't hack the coordination requirements it won't want to be a perovskite.


text:
What is the mechanism for the oxidation of Cu(I) by ammonium nitrate?
I can't seem to find any resource online that describes the arrow pushing mechanism behind the oxidation of Cu(I) by ammonium nitrate in acid.
Does a radical oxygen form? Or does it require both copper atoms simultaneously? I've drawn out a few possible mechanisms, but all seem plausible...
Here is the mechanistic formula:
$$\ce{2Cu(I) + 2H+ + NH4NO3 -> 2Cu(II) + H2O + NH4NO2 -> 2Cu(II) + N2 + 3H2O}$$


text:
A little help understanding some basic algebra(I think) in conversion pathways
Attached is a picture from my textbook.

Here is my question:
Must the 3.18cm(1.25in X 2.54cm) have to be cubed(^3) in order to cancel the cm in the 1cm^3 that is beneath the 22.59g Osmium?
Is that a standard rule in algebra? And if it is, under what section might I find it.
I've looked in Blitzer's 6th edition of Introductory Algebra for a clue but didn't find anything there. I've checked out Principles of General Chemistry by Martin S. Silberberg for a similar conversion but didn't find anything similar; all there was in that chemistry text book were much easier conversions.
I've even gone onto youtube and searched "multiplying fractions with unit and exponents" but couldn't find anything.
I find it strange that the entire cm^3 doesn't cancel out and just the cm does.But cubing the 3.18cm is necessary to arriving at the given answer. I could assume that the 1cm^3 should be cancelled, but we cannot cancel the cm^3 on one side and not cancel the cm^3 on the other side, but if we did that, we would be removing the ^3 which is necessary to arriving at the given answer.
Thank you all

The image is drawn poorly, the entire cm³ should be canceled.

Welcome to Chemistry.SE! Take the [tour] to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax. A screenshot or picture of an exercise is not searchable. Please consider rewriting it, so that it can be of help for future visitors.

On the matter itself. You have a cube with a length of $l=1.25~\mathrm{in}$ (Please don't make it a habit to use non-SI units.) In order to obtain it's volume, you need to multiply the lengths of all edges $V=l_1\cdot l_2\cdot l_3$ which are all equal in a cube, hence $V=l^3$. Then you have the density given. In order to obtain the mass of you cube you have to multiply the density by the volume, $m=\rho\cdot V$. It is not actually a conversion as it is more a relation of properties of an object. The only unit conversion you do is from inch to centimetres.

I don't think that this is the best choice of SE for such a question... Not that we can't help you with it.

Yes, 3.18cm has to be raised to power 3.
$$Density=Mass/Volume$$ or upon rearrangement
$$Mass=Density*Volume$$
The information is provided about edge of the cube made of Osmium and its density. So in order to obtain mass of the cube, it is necessary to find the volume of the cube(since the information about density is supplied). The volume of the cube is edge raised to the power 3,i.e.,
$$Volume of cube=edge^3$$
Since the information regarding edge is provided in inches it has to be converted into cm because the units of density is provided in $g/cm^{3}$.
Once the the volume is also obtained in $cm^{3}$, the mass can be obtained in g.
Follow these steps:

Convert the length into same unit as the the length unit in density(here it is $cm^{3}$). So, 1.25 in=3.18cm
Now to obtain volume use the formula for volume,i.e., to cube the length. So, volume=$$(3.18cm)^{3}\approx 32.16cm^{3}$$
Multiply volume with density to obtain mass. $$Mass=32.16cm^{3}*22.59g/cm^{3}\approx723g$$


text:
Which thermodynamic ensemble does Gaussian's BOMD functionality sample?
Someone sent me a Born-Oppenheimer molecular dynamics output with the following settings line:
# b3lyp/sto-3g BOMD=(MaxPoints=35, Ntraj=1, Update=15, RTemp=298) guess=read geom=allcheck

Based on my understanding of the BOMD keyword documentation, it sounds like the temperature is only used for putting energy into the vibrational modes, which are then used to initialize the nuclear velocities. Since there is no mention of a thermostat in the output or the documentation, this implies that calculations can only be run in the NVE ensemble.
Are my assumptions correct?

@pentavalentcarbon: As of now (Gaussian16), I believe that BOMD calculations can only be run in the NVE ensemble.
However, you can do ADMP (also available with Gaussian and uses an extended Lagrangian scheme), and offers equivalent functionality to BOMD. Thermostatting is available here, but the caveat is it implements only the simple velocity rescaling algorithm. On performing simple tests this seems to be "nearly" canonical on some systems I have worked on. i.e small ion-water clusters.


text:
Failing to optimize geometry of tricalcium bis(phospate) with GAMESS
I'm trying to learn how to use GAMESS. I used Avogadro to download the structure of tricalcium bis(phosphate), then I accidentally ran a runtyp=energy on the structure and it succeeded, so I assumed the resulting structure is better than what avogadro got and used that one instead.
I don't even know if geometry optimization is the first thing I should do. Maybe I should use another runtyp first and then optimize the result?
Maybe the fact that it's a salt is what's getting in the way? Maybe I have to specify the charges of the cations and anions somewhere.
Maybe I have to preoptimize it in Avogadro, because I don't think the straight line of 3 Ca followed by 2 $PO_4$ perfectly aligned remotely resembles what $Ca_3(PO_4)_2$ looks like. (save the pastebin as .log and open it in avogadro so you'll see how it is structured)
Here's a pastebin of the result. The input is printed in the begining, I took off the extended credits section on the begining.
http://pastebin.com/wivjv9WY

Why would you want to optimize a non-existent molecule in the first place?

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

You seem to think that all existing compounds are made of molecules. This is very much not so. The compound $\ce{Ca3(PO4)2}$ does exist, of course, but there is no such molecule.

Can you clarify what do you mean by that? Geometry optimization is only supposed to be done on molecules, is that it? Like I said, I'm learning. I'm not a chemist, My knowledge of chemistry is bellow amateur level.

You really need a more thorough understanding of chemistry first, then you need some decent understanding in quantum mechanics, because you need to combine that to run meaningful computations. Try optimising simple organic molecules first. Use something like water, ammonia, hydrogen chloride, methane, ethane, etc. These have already quite a lot of pitfalls to start with. Obviously you can optimise the geometry of $\ce{Ca3(PO4)2}$, but how meaningful is that as a molecule in the gas phase?

On a technical note. Please include the relevant parts of the input directly in the post so that it is persistent over time. Also note that this is a site about chemistry in the first place. Questions about running the calculations are not really on topic here. A more suitable place is [scicomp.se].

This pastebin will persist over time. It is set for never to expire.

I'm trying to learn how to use GAMESS.
I'm not a chemist, my knowledge of chemistry is below amateur level.

There is a serious contradiction out there: quantum chemistry is way above amateur level. Unlike playing computer games, quantum chemistry can not be done in a just-get-in-and-drive way. You need a serious theoretical background to make meaningful calculations.
Thus, for instance, you have to understand that usually by default all the calculations in quantum chemical programs are done for an isolated molecule in gas phase. Doing geometry optimization for ionic compound in such settings is, well, let's say, it is not so meaningful. Very likely, you want to do something totally different, say, establish the crystal structure of solid $\ce{Ca3(PO4)2}$, but that is a totally different task. In fact, you do even need a special program (like CRYSTAL) to achieve that.

What I really want is to build a "good enough" computational approximation of reaction rate, equilibrium constant and activation energy for any given reaction. I'm making a Minecraft mod, and I wanted it to be realistic and educational. So after building a good enough model that would work for at least big set of reactions then I would come up with more practical questions, like how to separate this product frome this solution, etc.

@FinnTheHuman, well, then I strongly advice you to stop wasting your time, since building such an approximation is currently beyond the dreams of chemists.

Thank you, I didn't know that. Can you give me advise on one or more really large databases where this kind of data was obtained experimentally? NIST is really disappointing on that matter. Do you know about any other?


text:
Reaction of Iron with Concentrated and dilute hydrochloric acid
Does Iron react differently with dilute and concentrated hydrochloric acid?
A. I know that with dilute hydrochloric acid Iron (II) chloride is formed.
B. I read on the internet that with concentrated hydrochloric acid Iron reacts to form Iron (III) Chloride.
Could someone tell me:
1. Is FeCl3 formed as a product of reaction between HCl (conc.) and iron?
2. If yes, then what is the complete chemical reaction?
Thank you.

Related: Formation of ferrous chloride or ferric chloride

No, $\ce{FeCl3}$ is not formed upon reaction with Conc. $\ce{HCl}$. Both dilute and concentrated $\ce{HCl}$ give $\ce{FeCl2}$ as product.
$$\ce{Fe + 2HCl_{dil} -> Fe^2+ +  2Cl^- + H2 ^}$$
$$\ce{Fe + 2HCl_{conc} -> Fe^2+ +  2Cl^- + H2 ^}$$
Source: Vogel's Qualitative Inorganic Analysis

You want to add this statement to your answer:-"Liberation of hydrogen prevents the formation of ferric chloride".

@NilayGhosh  why does hydrogen oppose the reaction? Can you elaborate?

@DatBoi Hydrogen prevents oxidation from $\ce{Fe^2+}$ to $\ce{Fe^3+}$. Refer to the previous question: "Acids can oxidize iron since the redox potentials E for pH=0 show that $\ce{H+}$ can oxidize iron to $\ce{Fe^2+}$ but non-oxidizing acids (in this case HCl) cannot further oxidize $\ce{Fe^2+}$ to $\ce{Fe^3+}$." If there is presence of oxygen, it can oxidize to $\ce{Fe^3+}$.


text:
Looking for a common solid which is very soluble in water, insoluble in a volatile solvent, and doesn't corrode ferrous metals
I'm looking for a way to make a shiny object 'matte', but temporarily. I'd like to be able to spray on a coating which is a uniform colour and wash it off easily, even from small 'nooks and crannies' in the surface.
I've looked into chalk paint ( https://www.montana-cans.com/en/spray-cans/spray-paint/chalk-400ml/montana-chalk-400ml ) but these use suspensions of eg. Carbon black or Titanium Dioxide, which are insoluble in water and difficult to wash off.
I'm wondering if there is a readily available compound which forms a suspension in a volatile solvent (for when it's sprayed on), which is soluble in water (so it's easy to wash off) and doesn't corrode ferrous metals (because the object I'm trying to coat is made from chrome steel with small rust pits already).
Any ideas?

You could try a dye such as rhodamine 6G its highly soluble in water but sparingly soluble in acetone and probably less so in less polar solvents such as hexane etc.  Rhodamine is used to dye bait in fishing so its very cheap.


text:
Why Silica and normal alumina can damage or destroy many metal complexes?
in case of organic ligand there r several techniques for purification,but in case of metal complex(Cr,Mn,Fe,Co) with schiff base ,  i have not seen any literature for the purification of them except recrystallization. give some suggestion

There are a lot of reasons why silica and alumina can damage a compound: they are not neutral, they contain 12-15% water, they have a huge specific area (typically 550 m2/g) which greatly enhances interactions between the compound and the stationary phase, they can sort of "chelate" metals... Furthermore, in your case, imines (Schiff bases) are usually not stable to chromatography.
Although your metal complexes will likely not be stable enough, you can give it a try on 2D TLC to check for decomposition. Maybe adding 2-5% TEA in the eluent can limit damaging your compounds. If oxygen is a problem, you can manage to remove it from the solvents and carry out your column chromatography under argon.
HTH


text:
Skunk Water Formula
I was reading about Skunk Water (a "good"  measure from a policing perspective as a method for crowd dispersion). I would like to find out if anyone has any idea as to the chemical composition of Skunk Water.
The official line is that it's composed from water, yeast, bicarbonate soda and other ingredients, however I have a suspicion that the soda is a red herring since it's also used to counteract the smell of actual skunk spray.

Fermenting any organic matter, especially the one rich in nitrogen and sulfur, should give you a cheap and effective malodorant. The original Skunk Water composition is a trade secret.

The sodium bicarbonate is to maintain the pH at a level where the yeast can produce the odor causing substances http://www.skunk-skunk.com/image/users/121755/ftp/my_files/MSDS_Skunk.pdf?id=3225191

Could just use this conveniently canned food: https://en.m.wikipedia.org/wiki/Surstr%C3%B6mming.

Here's a good looking list: The chemistry of Skunk Spray
However, many stink bombs are commonly made of a very simple molecule called ammonium sulfide.
But if it were up me (in a war), I might be tempted to use cadaverine and putrescine, for that long-term, never-come-back-to-the-area kind of smell.
Now, I know that this is a chemistry forum, but I think it's important to note that there are some bacteria (like Bacteroides thetaiotaomicron, Hafnia alvei, and Serratia liquefaciens, to name a few examples) which produce many of these stinky molecules, and I beleive that proper inoculation of a drum of eggs, would be a very simple, low-tech way to produce a putrefied product (that would certainly make me run away).
However, I would very much advise against any sort of assult as you may find yourself in prison or worse for manufacturing or using chemical or biological weapons.
And there's always tear gas (chloroacetone) and pepper spray which  are commonly used by police and in warfare.
Other notable compounds include skatole, hydrogen sulphide, mercaptans/thiols, and other organic sulfoxides; dimethyl sulfoxide is comparatively weak but has a certain savory robustness that really adds a natural tinge to whatever smells bad. A bit of garlic essential oil goes a long way.

Did you mean cadaverine and putrescine? I worked with those compounds, they are not so bad.

Butanoic acid smells like skunk (both the animal and cannabis).  Foxes have a similar smell.  Butanoic acid used to be made by the action of the bacteria in cheese on sucrose.  It wouldn't have to be too concentrated to make you smell awful.  Sodium carbonate is used to neutralise the smell (sodium butanoate).  When I worked in a school chemistry department, we used it - very carefully - to make ethyl butanoate, which smells like pineapple.


text:
Meaning of "ad" in a pharmaceutical patent: "Acetic acid ad pH 5.0"
Can the abbreviation "ad" mean "up to" in pharmacology? I saw a translator translate the Russian line "до pH 5.0" ("pH level up to 5.0") in a table describing the composition of a drug product as "ad pH 5.0". The meaning of the Russian record was "the content of the excipient in the product may be as high as to bring the product's pH to 5.0, but not higher".
I looked up a list of pharma abbreviations, and "ad" there does not have the meaning "up to".
I googled and found this on Google Patents:

Ammonium acetate 0.77 mg Sodium chloride 7.21 mg Benzyl alcohol 10.0 mg Polysorbate 80 0.2 mg Acetic acid ad pH 5.0 0.1 q,s, NaOH 0.1 N ad pH 5.0 0.1 q,s, Water for injection ad 1.0 ml

But this is unlikely to mean "up to", yes? I wonder what the meaning is there.
To be on the safe side, I used "up to" in my translation, but now I'm curious.

I suggest not to ask about things easily found in dictionaries, like http://www.dictionary.com/browse/ad

@Mithoron - I was not sure that  it was used as "up to" in this particular situation.

Most likely yes. In latin,(1) ad means "to, towards, until", like in "ad libitum" (aka ad lib) which means "keep going until I am satisfied", or "ad nauseam".
(1) Pharmaceuticals very often use latin.

Ad hoc... for this

Nice! I did not even think about that one :-)


text:
How were silver and gold separated using the salt cementation process?
Gold and silver are often extracted from the same ores and are difficult to separate due to their chemical similarity, which prevents the use of techniques such as cupellation. Acid-based methods were discovered during the medieval period but were too expensive for use, so the main method of gold parting (the separation of gold from other metals) until the C16th was by salt cementation, in which sheets or granules of low-purity gold were heated with salt (NaCl) and/or other other compounds.
I would like to understand how this process was actually used. Wikipedia says the silver chloride (AgCl) produced is 'removed' but does not describe how. It also mentions that AgCl is volatile and the vessel used for the process is sealed to prevent its escape, which seems to contradict the previous statement. At the same time, it mentions the process is kept below 1000degC to stop the gold from melting, which is a long way below the boiling point of AgCl at atmopsheric pressure (although I don't know what pressure would be reached inside the vessel or how that would affect the volatility of AgCl). Furthermore, I thought that upon heating AgCl would normally decompose to silver (and chlorine) anyway, which in a simple, sealed vessel would presumably just remix with the gold. Some sort of alembic arrangement, maybe? I'm pretty sure I've misunderstood something fairly basic here.
I'm then further confused by this 1974 paper by Notton (open access) which seems to describe a recreation of what sounds like salt cementation process but then compares it to cementation (meaning it must differ significantly in some way I'm missing), and also sheds no light on the mechanism for recovering silver, although notes that the alumina plug of the sealed container ends up discoloured by silver salts after heating. Were the silver salts absorbed by the walls of the container, as in lead/silver cupellation?
I am interested in historical (pre-C16th) application of this process, rather than modern solutions.

I've read this work before and always found it confusing.  I can see that an equilibrium can be formed whereby silver chloride is formed.  The silver chloride subsequently sublimes from the gold but that would mean sodium cation remains.  I think that instead, sodium carbonate is formed from carbon dioxide and water vapour within the fire; however, this isn't mentioned in any of the literature I've seen.  After the process gold could be washed with water, whilst the sublimed silver chloride could be scraped from the sides of the vessel for further purification.

@user1945827 glad it's not just me - many thanks for the comment. From the sound of it the vessel is sealed so I'm not sure there would be enough water present, although Theophilus mentions adding urine.

This is an old process
M. Berthelot gives it from Papyrus V from the Leyden Papyrus X, in his Introduction à l'étude de le Chimie, des anciens et du moyen âge (Introduction to the study of Chemistry, of forefathers and the middle age).
The papyrus is at least from the III° century, the process very probably older. From page 14 (my translation):

"Take piquant vinegar, thicken, take some [lacuna], 8 drachmas of common salt, 2 drachmas of lamellar alum (schist), 4 drachmas of litharge, crush with vinegar for 3 days, separate by decantation and employ. Then add to the vinegar 1 drachma of copperas, half an obole(1) of [lacuna], 3 oboles of chalcite(2), one and a half oboles of sory(3), one silique(4) of common salt, two siliques of Cappadoce salt(5). Make a lamella having two quarters (of an obole?) Submit it to the acion of fire... until the lamella breaks, then take the pieces and view them as refined gold.

1 drachma = 6 oboles
copper ore, such as pyrite
product of alteration of pyrite, which can include both copper sulfate and basic iron sulfate.
silique = third of an obole, a measure of weight
a variety of rock salt"

What about the chemistry?
Berthelot cites the process of départ par cémentation (departure by cemetation) from Macquer's Dictionnaire de la Chimie (1778). From the latter, page 312 (still my translation):

"First prepare a cement composed of 4 parts crushed bricks passed on a sieve, one part green vitriol calcined to red, and one part common salt: mixe the whole very-exactly, and make a firm paste, moistening it with a little water or urine. This cement is called cément royal, because it is used to purify gold, which chemists view as the king of metals.
On the other hand, reduce the gold one wishes to prepare, to lamellas about as thin as billon pieces: at the bottom of a crucible or cementation pot, place a layer of the cement of thickness the width of a finger: stratify the gold lamellas on this layer: place on top a new layer of cement: thus fill the pot, always placing the gold between two layers of cement; and cover it with a lid sealed with sand and clay. Place the pot in a furnace or oven ; heat progressively, until the pot is a dull red ; maintain this level of heat for about twenty four hours: it is very-essential that the heat not be able to melt the gold. After that, let the pot cool down, and open it to remove the gold, which must be separated thouroughly from the surrounding cement: it also has to be boiled in large quantities pure water several times. One can try this gold on a touchstone or otherwise ; and if it is found insufficiently pure, submit it a second time to the process.
The vitriolic acid of the brick and of the calcined vitriol, releases the acid from the common salt during this cementation ; and this last one dissolves the silver alloyed to the gold, and seperates it by this mean.

Berthelot has this to say about the chemistry occuring here, (page 16):

By proceeding thus, the silver and other metals dissolve in the sodium chloride, with the help of the oxidative -- and then chlorinating -- action exerted by the iron oxide derived from the vitriol ; while the gold remains unchanged.

And elsewhere (page 15):

[...] At a stretch, it could be that sodium chloride, in the presence of basic salts of [ferric oxide] [...], releases chlorine [...].

So what I think happens is: the silver reacts with $\ce{Cl2}$ and maybe $\ce{FeCl2}$ (speculation?, b.p.= 1023°C, vapor pressure at 800°C = 0.1bar) to give $\ce{AgCl}$. Then the $\ce{AgCl}$ evaporates (b.p.= 1547°C, vapor pressure at 800°C = 0.1mbar), condensates on the cement, and dissolves in it. It is unclear whether the silver stays in oxidized form. I would bet yes, but I am no inorganic chemist.

Recovering the silver
Macquer also says (page 313):

We can see it is essential to wash thouroughly the gold, after the process is completed, to remove all the disolved silver parts, which would otherwise stay mixed with it.
Concerning this silver, it can subsequently be separated from the cement, by heating it with a sufficient quantity of lead and litharge, and taking the resulting silver containing lead slag and cupelating it to remove the lead."

So apparently you can redissolve the silver in lead, and then remove the lead by normal cupellation.

Food for (even more) thought
Macquer (page 313) also mentions that the process works just as well if you replace common salt by saltpeter. Which is weird because $\ce{AgNO3}$ decomposes at 440°C.
He then adds that you can add both common salt and saltpeter, and not only does the silver still get dissolved, the gold doesn't. Which is also weird because you have all the components of aqua regia, but it doesn't attack the gold. Macquer does acknowledge his low confidence in his sources for this last process.

Thank you very much for this, particularly the second source. I can follow a fair bit of it but would you be able to clarify the actual chemistry occuring here? Green vitriol is (I believe) iron (II) sulphate, but if it's calcinated first the 'red' referred sounds like it would be iron (III) oxide, so I don't understand where the 'vitriolic acid' referred to later would come from - is the author just guessing at a mechanism because green vitriol is involved? From this it sounds like the silver chloride possibly evaporates and condenses on the cement, possibly also decomposing to silver?

I assume conditions are close to atmopsheric pressure - how highly could a clay-sealed crucible be pressurised?) and no higher than about 1000degC, since this source reiterates that the gold must not melt.

I don't really understand why adding lead would help with recovering the silver either - as I understand it cupellation was primarily to remove lead (I may need to ask a similar question about the process of cupellation at some point).

Edited to address your very constructive comments, thank you for that. I retranslated the last part for better clarity, and translated additional bits from Berthelot. Also added my guess at a mechanism.

Not at all - thank you very much for your effort!

Actually, I wanted to thank you. I had read about this process years ago in Berthelot's book, and found it confusing. Your question made me elucidate a bit of that confusion, and that's pretty cool. Bon vent !


text:
Calculating singlet-triplet gap and open-shell singlet character

I am trying to calculate the singlet-triplet energy gap ($S_0-T_1$) of a conjugated system (18 atoms). For the singlet $S_0$ state, assumed to be closed shell, the Gaussian options were
opt=tight b3lyp/6-311g(d,p) integral(ultrafinegrid)
...
0 1
...

and for the energy of the triplet $T_1$ state
opt=tight rob3lyp/6-311g(d,p) integral(ultrafinegrid)
...
0 3
...

Does the energy difference of those two calculations have any physical significance? (i.e. should have I used  ...td-ub3lyp...0 3 ... instead?)
I believe a system I am currently working on may have some open-shell character for the singlet ground state so I performed the following with the optimized geometry I got from a TD calculation of a triplet state:
#p ub3lyp/6-311+g(d,p) scf=(tight) guess=(mix,always) gfinput gfprint iop(6/7=3)
...
0 1
...

For the stability check, do I have to use the guess=(mix,always) on they keyword route as well?

1.a you need to use same level of theory (open shell or unrestricted) for both S0 and T1.
1.b you should consider using non-dft, as spin-polarized dft has somewhat questionably fundation. It is OK for big systems when there is no choice, but for systems of 18 atoms something like ump2/6-311++g**  should be used, at least for final energies.

@permeakra 1.a not necessarily. If either formalism is a good description it does not matter. If not then both are inappropriate. See also http://chemistry.stackexchange.com/q/49174/4945 1.b UMP2 will be worse than any DFT (and that is not considering that there could be a transitionmetal in there). Since OP did not specify what kind of atoms these 18 are, it is hard to judge if a higher method is suitable.

@Martin-マーチン Hey,
They are pi-conjugated systems containing phosphorus, nitrogen, sulfur, carbon, and hydrogen. I have been reading about some  methods such as broken symmetry DFT, and spin-flip TD-DFT, which help in cases of potential open-shell singlet ground states.

There is a related discussion on which method to use for a singlet triplet guess, which I encourage you to also read:
U- or RO-method for Singlet-Triplet Gap?
This post is going to be a little longer, so I'll present the key points upfront. It is reasonable to calculate (and/or optimise) the wave function with a restricted method, then test it for instabilities. The same applies to to triplet (etc.) wave functions, where I would also try to start with a restricted open shell approach. To answer 1, it is a reasonable first approach for a singlet-triplet gap. If you are more interested in the electronic excitations a time-dependent treatment is necessary.
As a general recommendation I would advise to split the calculations up and read the respective wave functions from the checkpoint file to perform additional calculations and checks, so that you always have a fallback.
Hence the answer to 2, once you have created and converged a solution to the wave function you can read it in again. Do not generate a new initial guess.
You should also always keep an eye on spin contamination, because this is essentially telling you if the method you are using is suitable in the first place. When it comes to broken symmetry solutions, you should always consider to run multi-reference calculations, too.
(Note that guess(always) applies only to geometry optimisations; see below.)

I'm going to exercise a couple of calculations on dioxygen, because it's a nice study case. I will be using the BP86 functional with the def2-SVP basis set and density fitting. I have done the calculations with Gaussian 09 rev. D.01.
It is somewhat an expansion on the example in Exploring Chemistry with Electronic Structure Methods, 3rd Edition (p.71).
I am going to present the full input of the calculations and at the end the absolute energy to compare to, in case you'd like to run the calculations yourself. Please note that I have not run any geometry optimisations and assumed a equilibrium bond distance of 120 pm. I think that is a reasonable choice since this is just an example and the optimisations can be considered as additional exercise.
I labelled the the used methods RHF, ROHF, and UHF even though I am using density functional approximations.
Because the main focus is the use of the stable keyword, we start with a wave function that is already know to be not stable. The RHF solution of dioxygen, filename bp86svp.rhf.sing.gjf:

%chk=bp86svp.rhf.sing.chk
%nproc=2
%mem=10000MB
#p RBP86/Def2SVP/W06
DenFit
gfinput gfoldprint iop(6/7=3)
symmetry(loose)
pop=full pop=nbo
scf(xqc)

title card unused

0 1
O  0.6 0.0 0.0
O -0.6 0.0 0.0

This input uses a few keywords, that are not strictly necessary to complete the calculations. For example the route section #p requests detailed output. DenFit is already implied with the choice of W06 as an additional basis set. I like to write it in nevertheless, to exactly know what I am doing. The same goes for symmetry(loose); occasionally I like to turn symmetry of, so I know this time I really wanted it. The line gfinput gfoldprint iop(6/7=3) is included so that molden (❤) can read (and display) the orbitals. There is no shame in always requesting population analyses.
In this very case even scf(MaxCycle=9999) will not lead to convergence; with scf(xqc) we are allowing the algorithm to switch to quadratic convergence if conventionally the convergence cannot be achieved. Note that this may converge to a different solution than originally attempted.
For publication type calculations I would also include int(ultrafine) and scf(tight); but I forgot it here.
From here I went on to calculate the stability of the wave function, i.e. bp86svp.rhf.sing.stab.gjf:

%chk=bp86svp.rhf.sing.stab.chk
%nproc=2
%mem=10000MB
%oldchk=bp86svp.rhf.sing.chk
#p RBP86/Def2SVP/W06
DenFit
geom=check guess=read
gfinput gfoldprint iop(6/7=3)
scf(xqc)
stable

title card unused

0 1

Note that in this case the SCF will converge within the conventional cycles and hence the energy is different. You can additionally check this by reading in the checkpoint file and perform a new SCF cycle.
However, in the stability analysis you will find the following output:

 ***********************************************************************
 Stability analysis using  singles matrix:
 ***********************************************************************

 Eigenvectors of the stability matrix:

 Excited state symmetry could not be determined.
 Eigenvector   1:      Triplet-?Sym  Eigenvalue=-0.0560505  =2.000
       8 ->  9         0.70645

 Excited state symmetry could not be determined.
 Eigenvector   2:      Singlet-?Sym  Eigenvalue= 0.0372059  =0.000
       8 ->  9         0.70689

 Excited state symmetry could not be determined.
 Eigenvector   3:      Triplet-?Sym  Eigenvalue= 0.1233031  =2.000
       7 ->  9         0.70493

 Excited state symmetry could not be determined.
 Eigenvector   4:      Triplet-?Sym  Eigenvalue= 0.2035093  =2.000
       5 ->  9         0.70373

 Excited state symmetry could not be determined.
 Eigenvector   5:      Singlet-?Sym  Eigenvalue= 0.2902800  =0.000
       6 ->  9         0.70685

 Excited state symmetry could not be determined.
 Eigenvector   6:      Singlet-?Sym  Eigenvalue= 0.3353297  =0.000
       5 ->  9         0.70254
 SavETr:  write IOETrn=   770 NScale= 10 NData=  16 NLR=1 NState=    6 LETran=     118.
 The wavefunction has an RHF -> UHF instability.

As expected we see a couple of instabilities. At this point we can run another calculation with stable=opt to find such a wave function. For reasons I don't quite understand why it finds the the broken symmetry solution instead of the triplet solution.
For exercise purposes we'll manually generate the broken symmetry solution; i.e. bp86svp.uhf.sing.gjf:

%chk=bp86svp.uhf.sing.chk
%nproc=2
%mem=10000MB
%oldchk=bp86svp.rhf.sing.chk
#p UBP86/Def2SVP/W06
DenFit
scf(xqc)
gfinput gfoldprint iop(6/7=3)
symmetry(loose)
pop=full pop=nbo
geom=check guess=(read,mix)

title card unused

0 1

The keyword guess(mix) generates a UHF guess by mixing the HOMO with the LUMO. You only need to do this once. Note that the keyword guess(always) only applies to optimisations and will perform a new initial guess at every point along the optimisation. I personally have no idea why this option exists or what it is used for.
Now we can go ahead and test for stability; i.e. bp86svp.uhf.sing.stab:

%chk=bp86svp.uhf.sing.stab.chk
%nproc=2
%mem=10000MB
%oldchk=bp86svp.uhf.sing.chk
#p UBP86/Def2SVP/W06
DenFit
scf(xqc)
gfinput gfoldprint iop(6/7=3)
symmetry(loose)
pop=full pop=nbo
geom=check guess=read
stable

title card unused

0 1

Note that we do not need to mix the orbitals in the wave function again. The stability section will now read like this:

 ***********************************************************************
 Stability analysis using  singles matrix:
 ***********************************************************************

 Eigenvectors of the stability matrix:

 Excited state symmetry could not be determined.
 Eigenvector   1:  1.012-?Sym  Eigenvalue=-0.0000016  =0.006
      8A ->  9A        0.70640
      8B ->  9B       -0.70640

 Excited state symmetry could not be determined.
 Eigenvector   2:  1.006-?Sym  Eigenvalue= 0.0930085  =0.003
      8A ->  9A        0.70688
      8B ->  9B        0.70688

 Excited state symmetry could not be determined.
 Eigenvector   3:  2.245-?Sym  Eigenvalue= 0.2031898  =1.010
      7A ->  9A       -0.70381
      7B ->  9B        0.70381
 SavETr:  write IOETrn=   770 NScale= 10 NData=  16 NLR=1 NState=    3 LETran=      64.
 The wavefunction is stable under the perturbations considered.

We already know that the triplet state should be more stable. As I hinted, I like to start with a ROHF calculation. In principle these calculations are always instable with respect to an UHF solution. However, usually this leads to faster results, especially if you have already a converged RHF solution. In this case we could readily work from the broken symmetry solution and directly calculate the triplet from there. Choose whatever is more convenient for you.
For this illustrative purpose I chose to generate a new initial guess; i.e. bp86svp.rohf.trip.com:

%chk=bp86svp.rohf.trip.chk
%nproc=2
%mem=10000MB
#p ROBP86/Def2SVP/W06
DenFit
gfinput gfoldprint iop(6/7=3)
symmetry(loose)
pop=full pop=nbo

title card unused

0 3
O  0.6 0.0 0.0
O -0.6 0.0 0.0

And get the UHF triplet from there; i.e. bp86svp.uhf.trip.gjf:

%chk=bp86svp.uhf.trip.chk
%nproc=2
%mem=10000MB
%oldchk=bp86svp.rohf.trip.chk
#p UBP86/Def2SVP/W06
DenFit
gfinput gfoldprint iop(6/7=3)
symmetry(loose)
pop=full pop=nbo
scf(xqc)
geom=check guess=read

title card unused

0 3

And then obviously check for instabilities; i.e. bp86svp.uhf.trip.stab.gjf:

%chk=bp86svp.uhf.trip.stab.chk
%nproc=2
%mem=10000MB
%oldchk=bp86svp.uhf.trip.chk
#p UBP86/Def2SVP/W06
DenFit
gfinput gfoldprint iop(6/7=3)
symmetry(loose)
pop=full pop=nbo
scf(xqc)
geom=check guess=read
stable

title card unused

0 3

We will find the following stability section:

 ***********************************************************************
 Stability analysis using  singles matrix:
 ***********************************************************************

 Eigenvectors of the stability matrix:

 Excited state symmetry could not be determined.
 Eigenvector   1:  3.003-?Sym  Eigenvalue= 0.2313384  =2.004
      6B ->  8B        0.70567
      7B ->  9B        0.70567

 Excited state symmetry could not be determined.
 Eigenvector   2:  3.003-?Sym  Eigenvalue= 0.2313396  =2.004
      6B ->  9B       -0.70568
      7B ->  8B        0.70568

 LASSW sees vector with all elements less than  0.100000000000E-03
 LASSW sees vector with all elements less than  0.100000000000E-03
 Excited state symmetry could not be determined.
 Eigenvector   3:  3.002-?Sym  Eigenvalue= 0.2704717  =2.003
      6B ->  8B        0.70711
      7B ->  9B       -0.70711
 SavETr:  write IOETrn=   770 NScale= 10 NData=  16 NLR=1 NState=    3 LETran=      64.
 The wavefunction is stable under the perturbations considered.

A time-dependent calculation will look for excitations based on the reference wave function. Therefore the solution will not be different. If you don't need to perform this calculation because you are interested in for example the UV/Vis spectrum, then there is no reason to do it.
I guess we have run the necessary scenarios now, so here are the results:

Command file              Functional            Energy / Hartree ( cycles )
bp86svp.rhf.sing          E(RB-P86)       =       -150.155033509 (      3 )
bp86svp.rhf.sing.maxc     No energy statement found.
bp86svp.rhf.sing.scf      E(RB-P86)       =       -150.154053425 (     16 )
bp86svp.rhf.sing.stab     E(RB-P86)       =       -150.154053425 (     16 )
bp86svp.rhf.sing.stab.opt E(UB-P86)       =       -150.201668045 (      4 )
bp86svp.rohf.trip         E(ROB-P86)      =       -150.213780181 (      9 )
bp86svp.uhf.quin.stable   E(UB-P86)       =       -149.651924155 (      8 )
bp86svp.uhf.sing          E(UB-P86)       =       -150.201668044 (      8 )
bp86svp.uhf.sing.stab     E(UB-P86)       =       -150.201668044 (      1 )
bp86svp.uhf.trip          E(UB-P86)       =       -150.215687576 (      8 )
bp86svp.uhf.trip.stab     E(UB-P86)       =       -150.215687574 (      1 )
bp86svp.uhf.trip.td       E(UB-P86)       =       -150.215687575 (      9 )

Lastly and maybe most important: with DFA it is never enough to just use one method. For more on this: DFT Functional Selection Criteria. Winter is coming!

Martin!! . Thank you for such a detailed and illustrative explanation!!! I am still confused as of why you specify scf(xqc) together with guess=read in the stability calcs. 1) I naturally understand you exemplified a typical HF procedure in which the SCF calculates/creates/guesses a wavefunction until the energy and density (orbitals!) criteria are met, plus the fact that it is a very simple diatomic molecule (thus no optimization); if I was doing a DFT on the pi-system I would naturally need to first optimize then,can I use that optimized  geometry for all the wavefunction stab calculations?

To be clear, the opt keyword on the stable=opt portion of a stability input job does not refer to a geometrical optimization, right? Thanks a lot.

@kimdazar I think there was no particular reason to specify xqc after reading in the orbitals other than laziness, or maybe the stability analysis needed it. I don't remember; it shouldn't do any harm though. You should use the optimised geometry for stability calculations. If you find instabilities you need to optimise again with the new wave function. Stable=opt is only for single point calculations; yes.


text:
EI-MS: M-28 peak in aldehydes
I've noticed a phenomenon in the EI-MS spectra of aliphatic aldehydes. Many of them show a significant peak (sometime it's even the base peak) at $M-28$.
To my best understanding, the only thing that can explain this – and is common to all aldehydes – is removal of the carbonyl $\ce{C=O}$ group.
However, I fail to understand the chemistry behind this. How can the $\ce{C=O}$ disconnect, while the aldehyde hydrogen stays – somehow replacing the entire group (forming a new $\ce{C-H}$ bond)? Or am I missing something and this is not actually removal of the $\ce{CO}$?
For example, here is the EI-MS spectrum of (2Z)-3-methylpent-2-en-4-ynal:

Molecule drawing from www.chemspider.com:

M-28 can also represent loss of ethylene (e.g. via a McLafferty rearrangement although I don't see how that would happen with this particular molecule). As for fragmentation at the carbonyl carbon, I would expect a M-29 fragment (loss of the formyl group) - consistent with the fact that even MW compounds often give odd MW fragments and vice versa and the rationale is as you explained in the question

but Silverstein's Spectrometric Identification of Organic Compounds (7ed) writes: "Observation of a fragment ion mass different by 1 unit from that expected for a fragment resulting from simple cleavage (e.g., an even-numbered fragment mass from an even-numbered molecular ion mass) indicates rearrangement of hydrogen has accompanied fragmentation."

@orthocresol That exactly was my thought. I cannot find any possible rearrangement that would cause the loss of ethylene, and I'm trying to figure out whether there is some kind of rearrangement that would move the aldehyde hydrogen to some location on the "main" structure (triple and double bond) while cleaving only the remaining C=O.

There's probably 1,5 hydride shift before decarbonylation.

This is indeed a quite intriguing question. Thus I am afraid that my answer will be mostly incomplete and quite speculative on some aspects.
First of all,  book and litterature searches lead to find that this M-28 loss for aldehdyes is mostly attributed as a loss of ethene and not CO. This can be traced back to the 1957 article by Gilpin and McLafferty which describes the EI fragmentation of straight chain aliphatic aldehydes. In the article, they demonstrate, through $\ce{^{18}O}$ labelling of the aldehyde, that it is a neutral ethene loss and not a $\ce{CO}$ loss. I did not find a clear proposal for a mechanism leading to this ethene loss, but as it occurs on long straight chain aldehydes, I would tend to favor a mechanism occuring through multiple $\ce{H^{\bullet}}$ migration within a series of isomeric distonic ions following the scheme, taking hexanal as an example:
But of course this is not sufficient in the case of the example that is the initial question, as there is no terminal ethene loss possible. I wondered if it would apply for other aldehydes, and I was not long to find in the NIST database terminal branched aldehydes (such as for example 5-methylhexanal, for which the EI spectrum is available here) that also lead to a strong M-28 peak, which obviously cannot be obtained through the above mechanism: one would expect a M-42 peak which happens to be present, but with a similar intensity to that of the M-28 peak.
And now we enter the speculative part. Someone with access to a high resolution mass spectrometer and a collection of aldehydes could easily identify if the M-28 is loss of $\ce{CO}$ or of ethene. I will assume that in the case of the question, as well as in the case of terminally branched aldehydes it is a loss of $\ce{CO}$, but this should be experimentally checked. (If someone also finds a reference to high resolution spectra showing loss of CO, please edit this answer accordingly.)
In the case of the (2Z)-3-methylpent-2-en-4-ynal, as well as other aklene aldehydes, one could probably explain the loss of CO as a $\ce{H^{-}}$ (hydride) migration from the carbonyl carbon towards an ionized double (or triple bond), followed by a loss of $\ce{CO}$, following the mechanism:

This mechanism leads to the formation of the methyl-methylenecyclopropene ion. According to Shay et.al, the ionized methylenecyclopropene ion is the most stable structure for these types of ions, but it could also produce any of the isomers, so I chose to start with a 1,5 hydride shift, although other hydride transfers could work out fine.
Although satisfying for the initial question, this does not address the case of the branched aliphatic aldehydes, for which no hydride transfer can lead to the formation of a carbonyl group before cleaving the carbonyl - carbon bond. Considering that the final product is in this case an ionized alcane, the mechanism for the rearrangement cannot proceed through a final bond cleavage which would lead to a bond deficient (unsaturated) system. Somehow, the system has to go through a partially bonded state at one stage. The following proposal could work out:

Initially, the system proceeds though standard α bond cleavage which would usually lead to formation of the m/z 29 $\ce{HCO^{+}}$ ion (present in small abundance in the original question as well as in most aldehyde spectra). An ion-neutral intermediate complex is formed in some instances (although eventually the two steps might be not absolutely separated), which can lead to a proton transfer between the $\ce{HCO^{+}}$ ion and the alkyl radical. This transfer is thermochemically favorable and would lead to loss of $\ce{CO}$.
Point in favor of this hypothesis are:

The thermochemistry for final proton transfer is favorable. Not many radicals have thermochemical data, but one takes the propyl radical (Enthalpy of formation for the propyl radical 100 kJ/mol; Ionization energy of propane 10.94 eV; Enthalpy of formation for propane -104.7 kJ/mol; Enthalpy of formation for $\ce{H^{+}}$ 1530 kJ/mole) one can evaluate its proton affinity at 680 kJ/mol. In comparison, proton affinity of $\ce{CO}$ (at C) is of 594 kJ/mol, thus the proton transfer is exothermic by more than 80 kJ/mol. For ketones, α-cleavages lead to the formation of $\ce{RCO^{+}}$ ions, with R an alkyl group, which cannot as easily transfer a proton (proton affinity of the neutral ketenes on the order of 825 kJ/mol) which also explains why similar rearrangements would be specific of aldehydes.
Fragmentation of the ionized alcane should occur and be present in the mass spectrum. Indeed, as shown in the figure below, some ions (notably m/z 71, 43 and 27) are present in the fragmentation of ionized 2-methylpentane and also in that of 5-methylhexanal, and can harldy be explained by α-cleavage of the aldehyde or by a McLafferty rearrangement alone. Nevertheless, this is a rather weak point: loss of water is also typical of aldehyde and leads to ions which could further fragment into these ions also.

A concern with this hypothesis is that a proton transfer to a radical is not a well established process. Indeed, formation of an ionized alcane leads to an ion which cannot be directly represented using standard Lewis notation, which makes writing mechanisms involving this species a bit worrisome.
The second problem is that although ion-induced dipole interactions are not negligible, the binding between $\ce{HCO^{+}}$ and the alkyl radical is likely to be weak compared to other ion neutral complexes. Thus its lifetime might be too short to allow a proton transfer step.
Further work on this subject should, in my opinion, try to focus on, (i) experimental validation of the loss of CO, and ideally, using D-labelled compounds validate the hydrogen transfer hypothesis. MS/MS fragmentation of the M-28 ion would also allow validation that an ionized alcane is indeed formed in the process. Finally, the mechanism should be validated through some theoretical calculations, at least for the formation of an ion neutral complex and for the barrier (and ideally RRKM kinetics) for the proton transfer within the ion neutral complex.

Wow, this is great, I learned so much just from your answer; thanks!


text:
Multiplet shape in proton NMR of morpholines
I recently carried out a Buchwald–Hartwig reaction to attach a morpholin-4-yl group to an aromatic ring:

In the proton NMR of the (columned) product, I found the peaks corresponding to the morpholine protons. The multiplet at 3.86 ppm comes from the protons next to oxygen, and the protons next to nitrogen give rise to the multiplet at 3.04 ppm:

The spectrum was obtained on a 400 MHz spectrometer and the values of the "coupling constants" are shown in the second multiplet. (They are identical in the first one.)
Why do the peaks have the fine structure that they do? Based on a simplistic analysis (the "n+1 rule"), I would expect a 1:2:1 triplet for both of these peaks.
In my crude NMR (pre-column) there was some excess morpholine and those peaks had exactly the same fine structure.
Does it have something to do with the chair conformation of the six-membered ring and the differentiation between axial and equatorial protons?

I did notice that type of ‘crown’ structure before in NMRs of THF, but I am at a loss what actually causes it …

I once naively gave morpholine as a spectroscopy unknown to students, assuming it would give a nice clean pair of triplets.  After the student came to ask about the spectra, they got a new unknown and morpholine went on the "do not assign" list.

It looks as if the NMR of morpholine is an AA′XX′ spectrum (the chemical shift difference is 0.80 ppm, or 320 Hz on your spectrometer, two orders of magnitude larger than the coupling constant).
Unlike linear AA′XX′ systems where the bonds can rotate, in the morpholine it very much has a fixed conformation.

In this fixed conformation, the protons are predominantly in a 'gauche' type arrangement due to the shape of the chair, this is where the appearance of the 'triplet' comes from in the NMR spectrum.
The image below is from Hans Reich's NMR notes, and shows some simulated NMR spectra depending on the gauche/anti ratio (Karpus equation), alone with a true spectrum of cyanomorpholine (which you can see looks basically like your spectrum above).

Source: Hans Reich (wisc.edu)

As a complete side-note. In TopSpin (which is what it looks like you used to process your NMR data) you can change the window function to use a Gaussian function, for strong spectra (which it looks like you have), this can help get a bit more resolution (at the expense of S:N- peaks on the baseline will disappear) such that what looks like a broad slightly messy spectrum actually gives resolved multiplets where the J values can be calculated.
In TopSpin, this can be done by entering the wm command; select "gaussian" from the drop-down list, and enter the two required parameters LB and GB (a decent starting choice is -2 and 0.2). The modified FID will now appear; this can be Fourier transformed using ft and automatically phase-corrected using apk to give the modified spectrum. [Even shorter is to sequentially type the commands lb -2, gb 0.2, and gfp into the command line.]
In MestreNova, the apodisation interface may be brought up from the Processing menu (or directly with the hotkey W) and the relevant parameters set in the same fashion.


text:
Difference between atomic mass, mass number, atomic weight, and relative atomic mass
I think this question is common, but i want a little help.
I'm little confused.
What I concluded.
Relative atomic mass/Atomic weight (both are exactly the same) - It is the mass of atom calculated as taking the sum of all the masses of the subatomic particles and dividing it with mass of (the same technique calculated) 1/12th of C-12 atom.
Atomic mass- It is the mass of the atom calculated as taking the mean according to their abundances in nature, of atomic weight of the different isotopes present in that sample.
Mass Number- It is the mass calculated as the sum of the nucleons , as there is only a slight difference in the mass of the proton and neutron, and the mass of electron is negligible as compared with the mass of the proton and neutron
But this is conflicting with definitions on wikipedia.
Also the relative atomic of carbon should be exactly 12, but its 12.011

Well, the atomic mass of carbon deviates from 12 because not all carbon in nature is C-12. As for the rest, your understanding is mostly right; what exactly contradicts it?

wikipedia says,  atomic mass and the relative mass are different but, it also says both are calculated as relative to 1/12th of the C-12 atom. Then where is the difference.

Related: Units of mass on the atomic scale, Quick and simple explanation of molar mass, molecular mass and atomic mass, What is the difference between “molecular mass”, “average atomic mass” and “molar mass”?, and Mass number, (relative) atomic mass, average mass

I can't even decide which is the correct duplicate. grumbles

Please correct me if I am wrong but in my understanding, the following are the definitions for the aforementioned terms:
Mass number: The total number of nucleons present in the nucleus of an atom(Dimensionless).
Atomic mass: The combined mass of all the nucleons present in the atom of any given element(Unit: amu/u).
Relative isotopic mass: Ratio of atomic mass to one amu(Dimensionless)
Average atomic mass: Weighted average of all the isotopes for the given element(amu/u).
Relative atomic mass/atomic weight: Ratio of atomic weight to one amu(Dimensionless).
Here is the Wikipedia page, explaining that "atomic weight" is simply the older term used for "relative atomic mass":
Relative atomic mass


text:
Why is difluoride a stronger oxidizing agent than dichloride?
I am preparing for a chemistry olympiad and I am stuck with the question that why $\ce{F_{2}}$ is stronger oxidizing agent than $\ce{Cl_{2}}$.

Fluorine has a more positive redox potential (probably the most positive) at +2.87 V  thus it spontaneously oxidises most everything else as they have a lower potential. Cl has a potential 1.35 V. Note also that Oxidation is loss of electrons, Reduction gain, thus  F is reduced in oxidising Cl.

Related: Why is fluorine an oxidising agent?, Why is fluorine the best oxidising agent, but also the most basic?, Are there stronger oxidizing agents than fluorine gas? and the links therein.

Also see: What is the role of hydration enthalpy in oxidation?

If you think about it, strong oxidizing agents must have a high "desire" or propensity to be reduced themselves (as the nature of redox reactions is such). We'll build up on the explanation from the ground up.
Comparing both fluorine and chlorine gas, both are halogens and both only need two electrons each to be reduced to fluoride and chloride ions respectively:
$$\ce{X2 + 2e- <=> 2 X-}$$
Since it appears that the two are similar in most respects, we should look at the clearest differences between Fluorine and Chlorine instead: the number of protons and the number of electrons.
Fluorine is a row 2 element with 9 protons an electronic configuration 1s2 2s2 2p5, whereas chlorine is a row 3 element with 17 protons and electronic configuration 1s2 2s2 2p6 3s2 3p5. Although chlorine has more protons to attract electrons (and therefore be reduced / act as an oxidizing agent), it also has two inner shells worth of electrons that create a shielding effect to reduce the attraction. Fluorine, however, only has one inner shell to do this.
From a simplistic physics point of view, electronic attraction is governed by the charge and the inverse square of the distance. You can approximate that the shielding effect will make the effective charge of the atomic nucleus the same in both halogens, so distance becomes the key deciding factor. Since chlorine is larger, and therefore the attracted electron further than it would be if it was instead bound to fluorine, the attraction (and therefore oxidizing power) is greater in fluorine than in chlorine.
However, looking at the electron affinity (EA) values you'd find that fluorine doesn't follow the group trend of decreasing EA and there's a very good discussion on that over at ChemLibre. Long story short, it's because fluorine is so small that you have to account for the electronic repulsion as well which reduces the bond strength of fluorine vs chlorine (36.6 vs 58 kcal/mol)
There is another measurement which seems to be consistent with this whole fluorine being more oxidizing than chlorine though - and that is reduction potential. The reason that there is a difference between the values for reduction potential and electron affinity even though they are both supposed to be measuring the same thing (that is the energy change related to accepting an electron), the key is the environment in which they are measured. Electron affinities are measured in the gas phase where it is purely the energy related to the electron transfer whereas reduction potentials are measured in aqueous solution and so factors such as hydration energy start to play a role.
As your textbook correctly points out, the high hydration energy is due to fluorine's higher charge density compared to that of chlorine.


text:
why bimolecular elimination reactions are stereospecific?
I know the mechanism of this reaction but it is stated that these are stereospecific in nature . why do we call these stereospecific explain with example.

The hydrogen being abstracted by the base, and the leaving group must be anti-periplanar. This requirement determines the stereochemistry unambiguously , and thus the reaction is stereospecific.


text:
How can stable O2 react with other metals on earth
$O_2$ is a molecule which has achieved a stable configuration, why then can spontaneous reactions occur between O2 and metals on earth ?  Is it that stealing of electrons is preferred above sharing of electrons ?

One view would be that the free energy of the reacted system is lower than the unreacted metal + oxygen system. Many metal oxides are highly favored.

Your question could apply to almost any reaction! All molecules have a stable electron configuration, N$_2$, benzene, chlorophyll, proteins, DNA etc. We, humans (and all life) are unstable with respect to oxidation, but the fact that a reaction is exothermic (spontaneous) does not mean that it has to react quickly. Thermodynamics does not inform us about time-scales just starting and ending energies. A reaction's activation energy can be huge compared to thermal energy and in many cases it often is so reactions proceed exceedingly slowly, e.g man + O$_2$ = CO$_2$+H$_2$O is, thankfully, very, very slow :)  Note also that the heats of formation of molecules are different, so different types of molecules possess different energy and hence have the ability to react.


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Concentrating Oxalic Acid by Evaporation
I bought Oxalic Acid in an aqueous solution from a hardware store years ago. I need to turn it to powder form for easier storage and other things.
Can I simply evaporate the solution? I seem to notice that evaporating it releases fumes that are similar when one is boiling Acetic Acid. Are there more Oxalic Acid being lost just as it is when boiling Acetic Acid (not distillation)

Depends on how hot the plate is I'd say. Oxalic acid is hygroscopic, so it will be hard to completely dry it by just subjecting it to heat as you might just cause it to break down.
There's a paper and procedure on Organic Synthesis on how to obtain the anhydrous form, so you should check that out to see if that meets your requirements.

I don't need the anhydrous form, I just need it to be crystallized. So I'm taking time with it and heat it until there is a bit of mist above the solution, careful not to boil it.

If you do not need the anhydrous acid, you can just leave the solution to evaporate for a few days. You do not even need to heat it. The dihydrate of oxalic acid is not hygroscopic. The white fumes are just mist created when the last remnants of water evaporate quickly taking some solid particles of oxalic acid with the vapor.


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The dissolution of calcium carbonates in water
Kindly help out. It is generally know that calcium carbonates has poor solubility in distilled water. Hence its ability to form slurry becomes difficult. Studies have shown that eggshells contain basically calcium carbonates.  I conducted an experiment in modification of eggshell powders with anionic surfactant in order to improve its solubility in water.  I carried out solubility and dissolution test on the modified eggshell powder. This was compared against the raw eggshell in order to confirm its solubility rate.  ICP was used to calculate the calcium  content in the solvent filtrate.  The reusults is showing that the calcium content in filtrate was higher than that of the modified eggshell.  Please who can assist in explaining this phenomenon. Thanks.

Eggshell also contains protein, which binds to the calcium carbonate and helps prevent  dissolution. The surfactant could have improved wetting of the carbonates, and might even have helped dissolve the protein. See http://antoine.frostburg.edu/chem/senese/101/consumer/faq/eggshell-composition.shtml.

What does ICP mean to be?

Regarding the slurry statement, you can only make slurries with insoluble compounds or they would dissolve and your slurry would go away leaving you with a clear solution!
Your attempt to alter the eggshell composition by extracting with the surfactant should have had little effect.  You didn't give any quantitative results from your ICP analysis, but I think one or both of two things happened:
1) There was residual Ca from the surfactant remaining on your eggshell
 and/or
2) You may have extracted the few percent of protein that makes up eggshells, leaving a more "concentrated" $\ce{CaCO3}$ residue as compared to the untreated shells.


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How can I obtain crystal-clear glycerin?
I'm trying to get perfect crystal clear glycerin for my experiments with decorative soap. Is there a special way to get it or should I buy a special formula of it? I have not succeeded in purifying the glycerin I have to more than a milky-translucent, semi-transparent soap.

You can buy high-purity glycerin from numerous vendors online. High-purity grades will be as crystal-clear as you can possibly get.
For instance, check out the premium grade glycerin product from Sigma-Aldrich which should be suitable for your needs but is quite expensive.
There are less expensive (but pure enough for your needs) options, so be sure to explore the other grades available from other vendors. For example, you can secure a gallon of Kosher USP food grade vegetable-derived glycerin for $14.99 from Amazon. This product (or similar) is adequate for your needs at a better price point that lab-grade reagents.

@orrachevsky - I'm amended my answer to include an option that is probably best for you. It pays to shop around, and you can get a gallon of glycerin (USP food grade, vegetable-derived and Kosher) for $14.99 from Amazon (see link in my answer, above). Good luck!