text:
What's the reason for the instability of ethylene dione?
Ethylene dione, with a chemical formula $\ce{C2O2}$ and an IUPAC name of ethene-1,2-dione, is an oxocarbon.

(source: wikimedia.org)
The Wikipedia page says:

The existence of $\ce{OCCO}$ was first suggested in 1913. However, despite its deceptively "simple" structure, for over a century the compound had eluded all attempts to synthesize and observe it. Such elusive nature had earned $\ce{OCCO}$ the reputation of a hypothetical compound and a mysterious, "exceedingly coy molecule".

I cannot understand what the rest of the article is telling me. So, why is ethylene dione unstable?

Well, being a dimer of CO doesn't do it any good.

Expanding what Tan Yong Boon said, there is a full scientific article detailing why ethylene dione is unstable Chem. Eur. J., 1998, 4, 2550-2557.
Long story short: Two CO molecules are far more stable than OCCO. The ground state of ethylene dione is a "diradical". This is, OCCO has two unpaired electrons instead of forming the intuitive chain of double bonds (O=C=C=O). However, if you add one or two additional electrons, the resulting anion becomes stable and adopts the shape of glyoxal (but without hydrogen atoms).

The strength of a bond can be determined by the amount of electron density accumulated in the bonding region. A general rule is that the more electron density, the stronger the bond. This is due to a stronger attraction of the nuclei to the bonding electrons. (Of course, if there is too much, it may in turn become overall repulsive, like in the fluorine molecule.)
With an oxygen atom attached to each carbon, there is a strong inductive effect pulling electrons away from the carbon-carbon bonding region of the ethylene dione molecule. Thus, the strength of the carbon-carbon double bond in ethylene dione is greatly reduced, resulting in its instability. This is one plausible explanation.

If we count $\pi$ electrons, we discover that $\ce{C2O2}$ has an unfavorable count for forming a stable molecule whereas both $\ce{CO2}$ and $\ce{C3O2}$ are much more favorable. This effect is discussed in the context of whether there can be aromaticity without rings.
When you have a linear chain of $sp$ hybridized atoms, as in $\ce{C_mO2}$ oxides, the $\pi$ orbitals are all doubly degenerate, with no nondegenerate low-energy orbital like conventional aromatic rings. Therefore the favored $\pi$ electron count for maximum stabilization is $4n$, not $4n+2$. Well, in $\ce{C_mO2}$ we find there are $2m+6$ $\pi$ electrons, so with both $m=0$ and $m=2$ we get an unfavorable count. With the double degeneracy we then get diradical species, and when we draw out molecular orbitals we find that the unpaired electrons are in $\pi$ antibonding orbitals effectively weakening the bonds holding the molecule together.
With $m=0$ ($\ce{O2}$) the only alternative is to break up a lot of bonding orbitals leaving the individual atoms, but with $m=2$ the poorly stabilized $\ce{C2O2}$ molecule can and does break into two relatively stable $\ce{CO}$ monomers.
Comparing the $\pi$ electron structures for $m=0,1,2,3$:
$m=0 (\ce{O2}): \color{red}{6}$ $\pi$ electrons, diradical, but too small to decompose into alternative stable species.
$m=1 (\ce{CO2}): \color{blue}{8}$ $\pi$ electrons, nonradical species with strong resonance stabilization.
$m=2 (\ce{C2O2}): \color{red}{10}$ $\pi$ electrons, diradical; can and will decompose to stable smaller molecules.
$m=3 (\ce{C3O2}): \color{blue}{12}$ $\pi$ electrons, nonradical, resonance stabilized; can undergo nucleophilic/electrophilic addition (polymerization) but not decompose; storable under controlled conditions.


text:
Could An Equivalence Point Be Calculated or Deduced Based on Limited Data with Accuracy?
I have recently been tackling a problem from one of my A-Level chemistry exercises on the topic of acids and bases, particularly with strong base and weak acid titration curves:

A titration is carried out in which $\pu{25 cm3}$ of $\pu{0.08 mol dm-3}$ propanoic acid $\ce{CH3CH2COOH}$ of pH 2.5 is put in a conical flask. Sodium hydroxide solution of concentration $\pu{0.1 mol dm-3}$ is added from a burette until $\pu{50 cm3}$ has been added.

One of the sub-questions asked to state the equivalence point of the pH, but I do not know whether drawing a rough sketch (as the exercise had suggested) would be validly sufficient to deduce this. From my reading of theoretically calculating such a point, a $\mathrm{K}_b$ constant is required such as that used in a solution in a similar conundrum: https://socratic.org/questions/calculate-the-ph-at-the-equivalence-point-for-the-titration-of-0-20-m-hcl-with-0 But the exercise does not provide such a value.
While the volume of the strong base, sodium hydroxide, required to reach the end-point is easy to calculate, as...
$\pu{0.1 mol dm-3} = \frac{\pu{2 * 10^-3 mol}}{v}$
$v = \pu{0.02 dm3} = \pu{20 cm3}$
...I am still unsure what would be the best way to tackle such a problem in which a roughly accurate answer could be obtained.

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text:
Which of the following compounds were involved in the titration?
We have a sample that weighs $\pu{0.3148 g}$.This sample may contain one,two or even three of the following compounds: $\ce{NaOH, Na2CO3, NaHCO3}$. We have as a clue, that the sample was titrated with $\pu{20 mL}$ $\ce{HCl}$ $\pu{0.09 M}$ using phenolphthalein as an indicator. Later, the sample was titrated with $\pu{50 mL}$ of the same $\ce{HCl}$ using orange methyl as an indicator.
The question is how do we understand which of the compounds were involved? Should we set for example mass of $\ce{NaOH} = x$, mass of $\ce{Na2CO3} = y$ and mass of $\ce{Na2CO3} = z$, then write down the reactions between all of the compounds with $\ce{HCl}$ and then extract some equations from the equivalent point? One equation comes from the total mass.

Yeah,  you just gotta do that.
Assume the variables $x$, $y$, $z$. You need three equations to get the values.
First equation: Add all three and equate to $\pu{0.3148g}$.
Second equation: In the titration involving methyl orange, due to it's pH range being 3.1 to 4.4, changes color as soon as carbonates and bicarbonates are neutralized. That will give you the second equation.
Third equation: Consider all three bases for phenolphthalein titration, since phenolphthalein changes color at 8.3 to 10.0.This gives the third equation.
Solve all the three, and enjoy the efforts of your hard work.

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text:
Oleum concentration description
What is the meaning of "free" $\ce{SO3}$ and "combined" $\ce{SO3}$ in the oleum sample concentration description?
Also what are these maximum and minimum values of % labelling in oleum sample?

There is virtually no SO3 molecules in oleum, as they are strongly Lewis acidic.

Guys, I don't understand how this question is unclear. The question makes sense. And to consider, the asker can only ask about the terms that they have learnt.

sigh I guess this "combined" is SO3 + H2O and free the rest as if there was just H2SO4 and SO3 unreacted which isn't true.

The English wiki (https://en.wikipedia.org/wiki/Oleum) is not discerning "free" and "combined" $\ce{SO3}$, still working with fuming $\ce{H2SO4}$ with 30% and  >60% $\ce{SO3}$ (http://www.sigmaaldrich.com/catalog/product/aldrich/778990?lang=en&region=US) keeps a mark.

I also think this should be two separate questions.

I think the term combined "$\ce{SO3}$" is nothing but $\ce{SO3}$ of $\ce{(SO3 + H2O)}$ and free $\ce{SO3}$ means that is $\ce{SO3}$ in oleum sample.
And value of % leveling of oleum is from (100% to 122.5%).

Why is the upper limit 122.5%?


text:
Is there a difference between parchment paper for writing on and the type you use in cooking?
I have tried searching this question many different ways but to no avail. My instinct and common sense say the processing is probably exactly the same but I would like to be sure.

Parchment paper for cooking is coated with silicone.  Therefore, most traditional writing implements would not even be able to leave a permanent mark on it.
In addition to the coating, at least one brand says it uses additional manufacturing steps to strengthen its cooking parchment paper:

A spokesperson for Reynolds Consumer Products, says that the company’s
  parchment paper is pressed into a sheet, then dipped into an acid
  bath, washed, and “passed over a series of hot rotating drums that
  realign the fibers and give the paper its strength,” before the
  silicone coating is applied.


text:
How do metal's density affects the corrosion rate?
Based on different formula that can be used to compute the corrosion rate of metals, all of them involves weight loss, time, surface area, and density. I thought that the weight loss, surface are, and time should definitely be considered. But, I got confused about the relevance of the metal's density. How does that affects metal's corrosion?

The reason you need density is to determine the mass lost by the system, assuming that the equations you are talking about look something like:
$$
m_{loss} = k\rho S_{area} t
$$
Where $k$ is a constant defining how fast the corrosion is occurring in your system (in this case k would be in units of [length]/[time]), $\rho$ is the density of the material, $S_{area}$ is the surface area, and $t$ is time: the equation outputs the total mass of material lost.
In more comprehensive corrosion equations $k$ would be substituted with equations describing the system causing the corrosion and may be dependent on (not a complete list): pH, temperature, surface activity, liquid flow, concentration of certain chemicals, passive layer growth, ect.


text:
Ultrasonic bath operation time
I use an ultrasonic bath in my lab for some chemical treatment of samples. It usually takes several hours with short breaks or even without it. I have a timer at the bath with a limit 90 min. So I have to run it every time again after it ends. My colleague told me that if I'm not giving a rest to the bath after every 90 min run, then the ultrasonic generator will be broken very soon. So, in general, how much time should be between the runs? Can external chilling enhance the safe treatment duration? It is a 3 liter, 100 W, 40 kHz Finnsonic bath with a heater.

Read the user manual. If the manufacturer is concerned about overheating the element or the driving circuit, it will say. Or call their service rep or applications engineer.

You'd be amazed how often outdated, incorrect advice gets passed down as fact in just about any field. Ask your colleague if they have a source for their statement aside from "XX told me."

Keep an eye on the temperature of the water in the bath. I have found significant temperature rises over hours of operation.

That advice from your colleague might be nonsense (see @chipbuster 's comment), but do YOU have aquired any evidence that five hours of ultrasonication is actually helpful for your problem? (Sorry if you have, but you wouldn't be the first who wasted many hours on a similar, useless procedure. It's called Cargo Cult Science. )

Yes, I do. It's not for cleaning of some parts. It is for chemical etching of glass. According to some article, it can take up to several days depending on the etching volume.

Beside Jon Custer's comment, to be taken with a grain of salt (as the op mentions a heating unit implemented in the ultrasonic bath), if the manual allows, it depends.
You will recognise that by simple use (even if nothing but water is in the trough), that the temperature of the water will increase simply bu use.  Hence, to keep the results reproducible from one batch, to the next, you power on the bath until this temperature stabilises.  (It is better to briefly switch off the sonification to record the temperature with a thermometer.)
In terms of duration of sonification: while one aims to keep the reaction time under sonification short(er) than by conventional heating (alone), like

(source, doi 10.1016/S0008-6215(00)90155-1)
there are examples, where the sonification reportely has to last much longer than $\pu{90 min}$, for example

(source)
I've seen colleagues run reactions from morning till noon, for the sake of selectivity of the reaction intentionally cooling the bath by intermittent addition of water ice into the trough.
To attenuate the cavatational noise the bath may generate, esp. if your flat bottom vessel hovers with some distance just above the resonators (the hot spots), either consider a different room (or at least, keep the sash closed), or have a "noise-proof shoe box" like

(source)
that you may design by yourself, too, covering your bath durin the operation.  Instead of the foam-lining, a wooden box lined with polysterene is even cheaper, but offers less "relief". It is even better to place the bath (if weight permits) at the same time on a sheet of expanded polysterene that you get "for a smile" at at your departemental chemical shop, or appliance shop.
But do not forget the ultrasound bath underneath the box / in the other room, especially if your heating unit is used.

(source)
Beside this, you may interested in periodicals like Ultrasonics and Ultrasonics Sonochemistry, or Luche and Capelo-Martínez as secondary references, too.

Just to add to my comment to the OP - if the heating unit was part of the OEM equipment, their manual should give the appropriate limits. If you added it yourself, well, you really should call their application engineer. Looking at one manufacturer (not the one mentioned by the OP), they note their heated baths allow for continuous operation, just like their normal ones.

This is an important aspect I agree with if there were the aim to install post factum a heat unit not foreseen by the manufacturer should be done only in agreement with the corporation.  In addition, for safety / insurance and OSHA (and perhaps I forgot one point, too), my university would allow such a modification only if both announced in advance and agreed, and eventually performed by a trained & authorized personnel (including a new safety-check sticker on the back + entry in the log).

In "Eco-efficient ultrasonic responsive synthesis of pyrimidines/pyridines" by Ramakanth Pagadala, Suresh Maddila, Sreekantha B. Jonnalagadda (Ultrasonics Sonochemistry 21 (2014) 472–477, doi 10.1016/j.ultsonch.2013.08.024), several sonifications last 2 (two), some even 4 (four) hours.


text:
What happens to the nitrous oxide in the whipped cream when it is on your food?
Nitrous oxide is used to create whipped cream but there is no information available on what happens to that gas when you eat it in the cream.

Here's a related post: https://chemistry.stackexchange.com/questions/76505/why-does-whipped-cream-use-nitrous-oxide-instead-of-nitrogen-gas

From a personal medical point of view? Or what your body might do with it (more bio than chem)? What, specifically, are you asking about?


text:
How do we know the number of s and p orbitals in a shell?
How do we know how many s & p orbitals will be in a shell?

There is one 2s orbital and three 2p orbitals in shell #2.
There is one 3s orbital, three 3p orbitals and five 3d orbitals in
shell #3.

Is there a reason for this? Wouldn't four 2s orbitals for shell #2 be a configuration of lower energy?

The solutions to the Schrödinger equation provides this result.

The number of orbitals in a shell is restricted by the possible values for the quantum numbers. Those come directly from the solution of the Schrödinger Equation for the Hydrogen atom (except for the spin, but let's keep it simple here), and are applied to all other atoms as well.
The quantum numbers and their possible values are:

the principal quantum number $n=1,2,3,4,5 \dots$, which relates to the period
the angular momentum quantum number $l=0,1,\dots, (n-1)$ which defines the shape of the orbitals $s$, $p$, $d$ and so on
the magnetic quantum number $m_l=-l,-l+1,\dots,+l$ further define the shape, e.g. $p_x$ or $d_{z^2}$
the total spin for electrons is always $s=\frac{1}{2}$ (this may be different for other elementary particles)
the relative spin of an electron is $m_s=-\frac{1}{2},+\frac{1}{2}$, and represents $\alpha$ and $\beta$ or up and down spin

Each electron needs to have its own unique set of quantum numbers (within one atom). This is know as Pauli Principle.
Thus for $s$ orbitals ($l=0$) there is only $m_l=0$ possible, hence there can only be one for each shell ($n$). Having four $2s$ orbitals would violate the Pauli Principle.
Additionally having a $2d$ orbital would also not be allowed as $l=2$ would be larger than $(n-1)=1$. So the $n=2$ shell is restricted to one $2s$ orbital ($m_l=0$) and three $2p$ orbitals ($m_l=-1,0,+1$).

I guess, technically, having four empty 2s orbitals wouldn't violate the Pauli principle. Having four 2s orbitals that are all populated would. IMO the Schrodinger equation argument is more fundamental than an argument based on the Pauli principle.

Technically, an orbital is defined as a one electron wave function, in other words the quantum mechanical description of an electron. Thus the presence of an orbital would already imply the presence of an electron. An unoccupied (or virtual) would just be nothing (the vacuum state of the electron).

Fair enough, then.


text:
Chelation-controlled aldol: why is the α substituent pseudo-axial?
In this Mukaiyama aldol reaction, taken from a 1999 paper by D.A. Evans,1 when $\ce{TiCl4}$ was used as the Lewis acid catalyst, the aldehyde 4a formed a chelated intermediate whereas 4b did not (due to the bulk of the TBS group). This was demonstrated by the stereochemistry of the product. Aldehyde 4a produced a syn:anti ratio of 3:97, whereas aldehyde 4b led to a ratio of 93:7:

The chelated intermediate which aldehyde 4a goes through is a standard six-membered ring with a half-chair conformation. Perplexingly, the α-methyl group has to be placed pseudo-axial instead of pseudo-equatorial (or else the syn aldol product will be obtained):

Ordinarily, substituents would adopt a pseudo-equatorial position in the dominant conformer. Why is this an exception?
Reference

Evans, D. A.; Allison, B. D.; Yang, M. G. Chelate-controlled carbonyl addition reactions. The exceptional chelating ability of dimethylaluminum chloride and methylaluminum dichloride. Tetrahedron Lett. 1999, 40 (24), 4457–4460. DOI: 10.1016/S0040-4039(99)00739-X.

Having the methyl group psuedo-axial means that when the nucleophile approaches, the transitiation state is chair-like. The energy penalty for having the methyl group initially in the less favourable position is compensated for the low energy TS.
The ring-flipped version, in which the methyl group is psuedo-equatorial, would open via a high energy twisted TS (the opposite of the argument presented above, i.e. the low energy conformation of the starting material doesn't compensate for the high energy of the twisted TS).

Looking at the Evans' paper specifically, the TS he draws has an addition isopropyl group on the beta-carbon, which is equatorial (this would further stabilise the starting material, since the isopropyl would be far higher in energy if axial compared to the methyl).

In your second TS, you can have the nucleophile attack from below, no? And that would give the syn aldol product, but still via a chair TS. [I saw the TS with the iPr group, but he repeated it with a different substrate lacking the iPr group (the one I drew above), and the result was similar.]

I suppose theres a steric argument, that the nucleophile (in this case a bulky silyl enol ether) doesn't want to approach across the methyl group

Yeah, I think that was my guess, but I just wasn't super sure about it.


text:
What factors other than Gibbs free energy determine formation of a compound?
The $\Delta G_{f}^{\circ}$ value for $\ce{Fe2O3}$ is less negative than that of $\ce{Fe3O4}$ in JANAF table at room temperature but still, $\ce{Fe2O3}$ is what's formed. How does one explain this given that it's not the thermodynamically favored one?

Thermodynamics only dictates whether or not the process is favorable. It gives no indication as to whether the process will actually take place or how quickly if it does.
You're asking about chemical kinetics which measures how quickly reactions proceed. In this case, the formation of the more stable product under ambient conditions is slow because the barrier to reaction is high.

I understood what you said and thanks for sharing this information but the thing is I am analysing a GdFe2 sample and XRD shows formation of Fe2O3 on it's surface rather than Fe3O4 (which has more negative"delta G") when the sample is left at room temperature for a long time (a few days). My question is why is Fe3O4 not forming instead based on Gibbs energy value and if I have to explain the formation of Fe2O3 what other factors or energies should I be looking at. As of now I read surface energy, interface energy and orientation of planes can be considered but what else should I consider?


text:
Structure of AsF3Cl2
Since the molecule has 5 bonding pairs and zero lone pairs, the shape should be trigonal bipyramidal. But one thing i am confused about is which atom(s) will occupy the axial and the planar positions, respectively, and why?
Thanks

Related (duplicate?): Why does F replace the axial bond in PCl5? (permeakra's comment on that question sums it up quite nicely.) So, I'd surmise the first two fluorines preferentially go axial. The third fluorine goes equatorial because there are no more axial positions left.

But why axial ligands have more significant negative charge on them ? Thanks

I'm not entirely sure about the background of this question, but $\ce{AsF3Cl2}$ before it decomposes above $-75 ^{\circ}$C in liquid phase (1, p. 100) has the following trigonal bipyramidal structure with two axial F atoms:

Crystal structure of solid compound though relates to the mixed halide of $\ce{AsCl4[AsF6]}$ formula (2) with discrete $\ce{AsCl4}$ and $\ce{AsF6}$ units:

(1) Chemistry of arsenic, antimony, and bismuth, 1st ed.; Norman, N. C., Ed.; Blackie Academic & Professional: London ; New York, 1998.
(2) Preiss, H. Z. anorg. allg. Chem. 1971, 380 (1), 45–50.
DOI 10.1002/zaac.19713800109

The axial positions see 3 substituents at 90 degrees, while the equatorial positions only see 2.
To first order, it is generally favorable to place the smaller substituents in the axial positions. Unfortunately, when you have fluorine, higher order effects (like minimizing dipole moment) may overcome steric considerations, but I would still prefer to place the fluorines axially as my first guess.


text:
Regarding rate dependence on redox potential
I am stuck with an equation wherein chloroauric acid reacts with oxalic acid to form Au(s). Even though one of both are enough for the reaction:

How does the rate change with increased oxalic acid?
Does it have anything to do with low reduction potential of oxalic acid?


text:
Color Visible in Large Quantities of Ethanol
I am currently writing a sci-fi novel and have come upon a rather ponderous question that I have thus been incapable of discerning an answer for. If a planet were to have oceans composed not of water, but rather of ethanol or a similar alcohol, what would they look like from space, assuming an atmosphere similar to earth's? What I've found thus far implies that ethanol and other pure alcohols are simply clear, however, in small quantities (such as in a glass), water also has an apparently clear color. Knowing that water, in larger quantities, absorbs red, yellow, and green light more readily than blue, making it appear blue, I was assuming that other liquids might perform similarly, making large quantities of them a far different color than in isolated quantities.

If somebody has access to this database, this paper may have the info you are looking for:
http://pubs.rsc.org/en/Content/ArticleLanding/1968/TF/tf9686402081#!divAbstract


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What does "René" mean in René alloys?
René alloys are nickel-based super alloys (René 41, René 88, René N5 etc. etc.). But how did it get the name René? Is it an acronym or contraction of other terms? It was created by Earl Ross and the General Electric Materials Science group, so it doesn't seem to be the name of the discoverer.

Similar to inox for inoxydable (inoxidable), it equally is a French coinage.  To tune mechanical properties of iron, mixing additives leads to numerous alloys of iron.  In case of rene or rené as contraction of résistant (durable) and on base of Nickel, and is example of the superalloys.
This source technologie des métaux et alliages - particulièrement en aéronautique (technology of metals and alloys, especially in aviation industry) by Dominique Ottello (2003), section 4.6.1.1 "Les alliages de nickel: / the alloys of nickel" on page 15 states:

Ils  conviennent pour les trois types de sollicitations : MÉCANIQUES, THERMIQUES et CHIMIQUES (très grande résistance à la fatigue, au fluage et à  la corrosion à chaud). Il existe plusieurs familles d'alliages de NICKEL :

Les alliages NICKEL + CUIVRE : alliages "MONEL" qui présentent une bonne résistance à la corrosion chimique, utilisables jusqu'à 500 °C.
Les alliages NICKEL + CHROME + FER : alliages "INCONEL" qui sont très tenaces et utilisables jusqu'à 700 à 800 °C.
Les alliages NICKEL + CHROME + COBALT : alliages "NIMONIC" et alliages
  "RENE". Ils sont très tenaces et présentent une bonne résistance au FLUAGE, utilisables jusqu'à 950 °C.

NOTA : Dans les alliages de NICKEL à haute température, on rencontre :

Des alliages plus spécialement réservés à la FONDERIE.
Des alliages pour CORROYAGE.
Des alliages à durcissement structural par traitements thermiques.

For our non-Francophon readers:

These [Nickel-derived alloys] are suitable for three fields of application / strain: mechanical, thermic, and chemical (very high endurance against fatigue, compression, and corrosion at high temperatures).  There are several groups of nickel-based alloys

alloys based on nickel and copper:  "monel" alloy, with a good resistance against chemical corrosion, serviceable up to $\pu{500 ^\circ{}C}$ [i.e., $\pu{932 F}$]
alloys based on nickel, chrome, iron:  "inconel"alloy, tenacious and deployable up to $700$ or $\pu{800 ^\circ{}C}$ [i.e., $1292$ to $\pu{1472 F}$].
alloys based on nickel, chrome, and cobalt: "nimonic" and "rene" alloys. These are very tenacious and exhibit a good resistance against traction, deployable up to $\pu{950 ^\circ{}C}$ [i.e., $\pu{1742 F}$].

Note: One finds, among the nickel-based alloys with high melting point:

alloys foreseen / dedicated for deployment in foundries / casting houses
alloys for wrought (e.g., forging, rolling, etc.)
alloys for hardening by thermal treatment

Voilà !

conjunction seems like you meant portmanteau?

@cat Actually, "contraction".

I'm confused. If it's a contraction of "résistant" and "nickel", why is it "rené" rather than "réni"?

@DavidRicherby I speculate "rene" is easier to use internationally.  Not only in terms of trade, but in terms of comparing qualities and establishing standards.  There was a time where German Krupps and French-Alsatian De Dietrich were competing each other, not to forget other manufactures, too. In addition, a French "réni" sounds maybe not identical, yet ressembles an UK English "rene/é".  Eventually, the French source does not claim who invented rene alloys, too.

@cat Yes, portmanteau is exactly the word intended.  "Contraction" seems to fit, too.

Very interesting.  I had wrongly assumed someone at INCO had just dreamed up the names.

René means “re-born” in French, it is related to one of the alloys which was not weldable when invented, but after modification it was re-born and used in production.

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text:
At what point does decreasing solvent temperature cause a decrease in gas solubility?
Example: At constant pressure, carbon dioxide becomes less soluble in water as temperature increases. We also know that carbon dioxide becomes more soluble in water as temperature decreases.
Does introduction of order and loss of kinetic energy of solvent molecules eventually lead to a decrease in a solvent's ability to accommodate gas molecules? It is hard for me to believe that ice cubes can hold gas more effectively than the same volume of a glass of water.

Intuitively I would expect ice to hold gas MORE than liquid water could, as ice is less dense. Perhaps for the majority of substances which get MORE dense as they freeze, your assumption would make sense to me...
Of course, I'm going off intuition, not data, so I could be wrong.

@iammax: You are reasoning by analogy. Any gas becomes practically insoluble in anything at the phase transition into the solid solvent. There is no space for a random gas molecule in a crystal lattice of similarly sized host molecules. Exceptions: Chladathes and hydrogen, at elevated pressure.

Rules such as "solubility of A in B rises with lower temperature" are only meant to be used if there is no phase transition of the participating substances.


text:
How do/did chemists observer/infer the structure of organic molecules and reactions?
As I understand organic chemistry there is no simple generalization that allows us to predict reaction outcomes- rather we transform molecules using a sequence of known reactions.
This has me wondering how these known reactions came to be known. My first guess would be that the structure was empirically observed before and after the reaction.  If this is the case what instrumentation would be used to observe this?
If not, how else were these reactions determined and if they were not all developed by trial and error, what is the name of the branch of chemistry that deals with hypothesizing them?

"How do" and "How did" are two completely different, albeit both very good, questions. Voting to close as too broad. Pick one for this question, and if need be ask a new question for the other.

Spectroscopy is how they do it.

My understanding of spectroscopy is that it gave you information about composition, not structure, but I'm still a student so maybe that's not quite right.

Nope, spectroscopy can give lots of direct structural information. The first thing that comes to mind is the different 2D NMR techniques.

Because other people can describe them better, here's a list to get started, in the order I recall them: NMR, UV-vis, IR, XRC if possible, HPLC-MS, radiography, chemical testing, physical testing.

With a mix of techniques and chemistry. The question is closed but I wonder nobody mentioned chemistry at is best. Chemical structures well pre-dates spectroscopy especially the most advanced one. Fisher established the entire sugars using just chemistry and a polarimeter. Benzene was known to be a regular exagon by looking at various derivatives.  An so on. Of course nowadays is way much easier.

Various crystallographic techniques exist, the lab technique I have used most is x-ray crystallography
https://www.ncbi.nlm.nih.gov/pmc/articles/PMC1186895/pdf/mp53000008.pdf
Currently I specialize in computational chemistry, which can estimate (with very high accuracy) the geometry of a molecule which gives the lowest energy (aka the equilibrium geoemetry) through a process called geoemetry optimization
https://www.shodor.org/chemviz/optimization/students/background.html


text:
Do atoms react to fill outer shell or 8 valence electrons?
I recall in middleshool chemistry we simply said electron shell configurations were 2,8,8,many which I would have no problems with.
But now I'm learning that shells contain $2n^2$ electrons where n is the shell number.
Then why is "full outer shell" and "8 valence electrons" still used interchangeably when describing stability? i.e we are taught that atoms react in ways that achieve an outer shell of 8 valence electrons. But at the same time, they react to form full outer shells.

Excellent question I must say.
To middleschoolers they really don't teach advanced concepts regarding orbitals. But since this question came to your mind, I guess you're capable of handling what I have to say next.
Unlike what's taught to you, shells don't exist just like that, they exist as a collection of subshells. Subshells are given names such as $\mathrm{s}$, $\mathrm{p}$, $\mathrm{d}$, and $\mathrm{f}$. The number of subshells a shell has is given by it's principal quantum number, or simply the shell number. In this way, the 1st shell has only one subshell, which is $\mathrm{1s}$. The second shell has two subshells; $\mathrm{2s}$ and $\mathrm{2p}$. The third has three subshells; $\mathrm{3s}$, $\mathrm{3p}$, and $\mathrm{3d}$.
Every subshell is further composed of orbitals. Orbitals are different from orbits, orbits have electrons in a precisely defined path, while orbitals show a cloud of probability of where the electrons could exist. Each orbital holds two electrons.The orbitals also have exciting shapes:
The $\mathrm{s}$ subshell consists of just one orbital, the $\mathrm{s}$ orbital. It's a spherical cloud of electron probability:

The $\mathrm{p}$ subshell consists of three orbitals, $\mathrm{p}_x$, $\mathrm{p}_y$, and $\mathrm{p}_z$. They are dumb-bell shaped, oriented along the $x$, $y$, and $z$ axes respectively:

The $\mathrm{d}$ subshell has five orbitals, $\mathrm{d}_{xy}$, $\mathrm{d}_{yz}$, $\mathrm{d}_{zx}$, $\mathrm{d}_{x^2+y^2}$, and $\mathrm{d}_{z^2}$. They have interesting shapes.

The $\mathrm{s}$ subshell can hold two electrons, the $\mathrm{p}$ subshell can hold six electrons,the $\mathrm{d}$ subshell can hold ten electrons, and the $\mathrm{f}$ subshell can hold fourteen electrons.
In middleschool, they teach you to fill the lower electrons first, and move to the top, if you want to find the electronic configuration of a given element. Unfortunately, that's not entirely correct. Infact, considering the subshell concepts, you must use the Aufbau Principle. The rule tells us to fill the electrons in the order given in the picture below:

(source: boundless-cdn.com)
If the picture wasn't clear, the order is $\mathrm{1s}$, $\mathrm{2s}$, $\mathrm{2p}$, $\mathrm{3s}$, $\mathrm{3p}$, $\mathrm{4s}$, $\mathrm{3d}$, $\mathrm{4p}$, $\mathrm{5s}$, $\mathrm{4d}$, $\mathrm{5p}$, $\mathrm{6s}$,...
Do you notice that $\mathrm{4s}$ comes before $\mathrm{3d}$? Thats why you don't get those ten extra electrons of $\mathrm{3d}$, since the $\mathrm{4s}$ subshell is being filled for the elements potassium and calcium.


text:
Spectra of Methylene Blue
Does the spectrum of Methylene Blue change on prolonged exposure to light?
I've studied about the photochemical reactions and about energy transfer reactions. Methylene Blue is a triplet sensitizer used for the production of singlet oxygen through energy transfer reaction, can this formed singlet oxygen attack methylene blue? Why is leuco(colorless) form of Methylene Blue is formed?
One more thing is still confusing me, while the energy transfer reaction we say that energy is transferred from sensitizer to another molecule, but which form of energy is transferred is the light or heat energy, if it is light energy which is transferred then the light of same wavelength is transferred which caused the excitation of the sensitizer?
If anyone can please help me through these confusions...

That MB "attacks itself" via the production of singlet oxygen makes much sense. It should be what is done in a large number of studies in conjunction with (photo)catalysts in the field of photocatalytic cleaning, or TiO2 solar cells. Type MB photodegradation and you will find a multitude of references. Unfortunately, at a first glance I did not see anything really accessible.


text:
Calculation of Bond/Angle/Dihedral force constant
I am a big fan of Mopac (Though I have a very little knowledge about the tool). I have some queries regarding some calculations in Mopac2016:
Problem:
How to calculate Bond or Angle or Dihedral force constant for Vanadium (V) (with other elements: Oxygen and Chlorine)?
My compound (smiliar to): CN(OV(=O)ON(C)C(C)=O)C(C)=O
Structure:

I am not sure I understand your question correctly: Do you wish to calculate a force constant using a method in Mopac, such as PM7?

yes. I need the method to calculate the force constants using Mopac2016.

It should be possible to perform a "FORCE" calculation, visualize the resulting vibrational modes and extract the force constant from the appropriate frequency. Note that you need a well-converged geometry for this, preferably without negative frequencies.

1 and 2 provide you with two CIF files (all distances and angles are present) of chloro-oxo-bis(N-phenylbenzylhydroxamato-O,O')-vanadium(V), replace phenyl groups with protons and you have perfect starting geometry.

@TAR86 : Thank you. And force did work , please write as an answer and I will close this question. But it gives back result in a weird way? How to interpret it? example: http://openmopac.net/manual/Formaldehyde_output.htmlstion

Finally found it! SEMINAR10, JORGE M. "Calculation of intramolecular force fields from second-derivative tensors." (1996). http://compbio.chemistry.uq.edu.au/mediawiki/upload/2/2a/Seminario_Hess2FF.pdf

@DevashishDas You can also answer your own question, and this can potentially help others in the future.


text:
Why does tryptophan absorb UV light?
Why is that tyrosine, phenyalanine, and  tryptophan absorb UV light while other amino acids don't even absorb visible light? Does it have something to do with aromatocity?

Your "even" is misplaced. Tryptophan doesn't absorb visible light either. True, it does absorb UV, so the whole question is valid. And yes, it has something to do with aromaticity, or rather, with a system of conjugated $\pi$-bonds in general.

Consider benzene, phenol and indole, which all absorb UV and are each parts of (in turn) phenylalanine, tyrosine and tryptophan.

Doesn't practically everything absorb some sort of UV light?

I have another problem I realized that why then histidine also absorb UV light it also has aromaticity ?

Any molecule can absorb UV light. What the question is probably going for is why those three molecules absorb at longer wavelengths than other amino acids. This has to do with the conjugated pi bonds from aromaticity. Looking at a list of amino acids, we can see that most of them don't have conjugated pi ystems.
amino acids (source)

Conjugated pi bonds decrease the HOMO-LUMO energy gap the more they are conjugated. This allows longer wavelength (lower energy) light to excite the molecule by promoting an electron from its HOMO to LUMO.
UV light is in the range of about 10-400 nm. A single bond (sigma bond) would absorb around 130 nm. Ethene's single pi bond is at 165 nm. A benzene ring's conjugated double bonds peak primarily at 180 and 200 nm. Further conjugation can absorb longer wavelengths and, like anthracene, begins edging into visible light, which as a result has a yellow color and more transitions in the UV-VIS spectrum. NIST has UV-VIS spectra for both benzene and anthracene which you can compare.
Benzene

Anthracene

The more conjugated molecule has more transitions and absorbs at longer and longer wavelengths. This principle is why carrots are orange for example, because they contain a highly conjugated molecule, with 11 pi-bonds, that absorbs well into the visible light range. This molecule is beta-carotene:

All that said, the other amino acids do have double bonds and lone pair electrons (carbonyl groups) that can absorb light in the UV range (remember any bond, including single bonds, will absorb UV light if you go low enough). The lone pairs on a carbonyl group can absorb at higher wavelengths than benzene. These involve electronic transitions between a non-bonding electron to the LUMO; the transition energy is so low because a non-bonding electron does not have its energy lowered like a bonding orbital. The gap between a non-bonding orbital and anti-bonding orbital is therefore much smaller than usual.
There was a good post on pi-conjugation elsewhere on the site, here.
You can read more about carbonyl excitations here. For general excitation values, this page was useful.

The carrot color nice point , It helps me feel that point  ;)

As far as the molecule is concerned there is no distinction between visible and uv light. In each case the transition is from a ground state to an electronically excited state. This is caused by a photon being absorbed and its energy taken up by the molecule. In doing so an electron is promoted from a lower orbital to a higher one. The lowest energy transitions are from the LUMO to HOMO.
The conjugation in a molecule lowers the orbital energy because, in effect, an electron is able to 'spread out' and is not confined to be associated with just two atoms. Thus the greater the extent of conjugation the lower the orbital energy and the smaller the energy gap between the ground state and lowest excited state. You can think of this rather like a particle in a box, the longer the box the lower and closer the energy of the various levels becomes.
A conjugated molecule has transitions associated with electrons in pi orbitals; $\pi$ to $\pi^*$ transitions. Molecules with non-bonding electrons can also have $n \rightarrow\pi^*$ transitions and when there are only sigma bonds the high energy transitions are $\sigma \rightarrow \sigma^*$ in nature.
In addition to the lowest electronic transitions there are transitions to higher electronic states, where an electron is promoted to a higher anti-bonding orbital than the LUMO. Each electronic state so produced also has its own set of rotational and vibrational levels in the same way to the the ground state. Thus the spectrum can become congested even when molecules are studied in the low pressure vapour phase where the influence of collisions with other molecules and solvent is absent.


text:
Can cis and trans isomers have widely different flash points?
Sigma seems to have two products listed for sale as "Cinnamic Aldehyde". One compound clearly seems the trans isomer but the other one seems like an unspecified mixture. (They even have two different CAS nos.: 14371-10-9 vs 104-55-2 )

What surprises me is that the corresponding SDS lists the Flash Point of one as 71°C and the other as 125°C.
Is this really possible or is this just an error? The boiling points listed seem fairly close: 250°C vs 248°C.
P.S. Even if one accounts for methodological differences in Flash Point Determination e.g. closed cup vs open cup those are typically at worst 10°C apart. Here the flash points seem a good 55°C apart.
What gives?

Which has the lower flash point? Trans is more stable. Runaway isomerisation of the cis component? Also the cis isomer is rather unstable in air, anway.

First glance suggests that the first compound is not cis-, but some kind of mixture of isomers.

It is not totally surprising giving that they are stereoisomers. Think about maleic and fumaric acid, which due to their very different properties were given different names.
Moreover, but perhaps you are aware of, there is not in general a relationship between boiling point and flash point. In the first case factors which depends on the actual configuration - as molecular symmetry, polarity - should be considered for how they affect intermolecular forces. For the flash point, their influence to the intrinsic stability of the molecule and to the overall shape and accessibility for oxygen, should be considered instead.
For the less experienced ones: in few words volatility is required to reach the concentration (in air) at which the vapour can be ignite at a given temperature (flash point). More a liquid is volatile, easily that concentration is reached. But this does not imply a lower flash point, which depends on the thermodynamics and the kinetics of the reaction between the vapour molecules and oxygen in air.


text:
Forming the resonance structures for the nitrite anion
Yesterday I learned about resonance structures and came across this problem.
The resonance structures are formed when there is a delocalized π electron cloud. I understand that for benzene, there are two resonance structures. For making π bonds I know that there are three main reasons:

The two orbitals should overlap each other.
There should only be one electron for each orbital.
The two orbitals should be in one level.

In benzene there are six $\mathrm{p}_z$ unhybridized orbitals, therefore they can share their electrons with each other (delocalize the π bond electrons). So benzene has 2 resonance structures.
But I don't understand how this works for the nitrite anion, $\ce{NO2-}$.

In $\ce{NO2-}$ there is also π electron delocalization, so there are 2 resonance structures. Now I am going to say how I think the delocalization is happening.
When the first structure changes into the second structure, nitrogen has to make a π bond with the right side oxygen with each of the $\mathrm{p}_z$ orbitals before, but now that π bond's two electrons take to the right side oxygen, and then there are 2 electrons in the right side $\mathrm{p}_z$ orbital and no electrons in the nitrogen $\mathrm{p}_z$ orbital. But the left side oxygen (in the first structure, $\ce{O-}$) has 2 electrons in its $\mathrm{p}_z$ orbital. Then the nitrogen $\mathrm{p}_z$ and oxygen $\mathrm{p}_z$ overlap each other and make a π bond.
I have said before "There should only be one electron for each orbital." But, according to my understanding above, the π bond is made by $\mathrm{p}_z$ with 2 electrons (oxygen) and $\mathrm{p}_z$ with no electrons (nitrogen) overlapping each other.
So, is there any mistake in my imagination?

If you want to speak in terms of orbitals, then there is no such thing as resonance structures.

@Osal Edit your question to correct punctuation and orthography errors.

According to that, is there not a connection between orbitals and resonance. When there are resonance structures then the pi electron cloud is also change every time. Electons are in orbitals. Therefore  there is a connection between orbitals and resonance.

Resonance structures are a convenience to help make Lewis structures more applicable in a general sense. However, since Lewis structures are still quite limited in terms of expressive power, this is just a stop gap. Don't try to read too much into the physical interpretation of resonance structures: they don't exist.

Sure enough, there is a connection between orbitals and resonance, and it goes as follows: once you learn the former, you should forget the latter.

Yes, I can understand your theory. But my problem is how a pi bond is form when there are 2 electrons in one orbital and none in other, according to above reason.

It is formed just as well as when each orbital brings one electron. See, all electrons are identical. They don't "know" where they came from.

There are 3 centers and 4 electrons. It's a 3-center-4-electron bond.

Another way to state what Zhe posted is that the true structure of nitrite ion is halfway between the two "resonance structures". The two N-O bond lengths are equal and the two oxygens have equal partial negative charges.

Just curious, does my response answer your query? If it doesn't, where did I fall short?

Really, I found that there is a mistake in my imagination. I said that 'There should only be one electron for each orbital' when forming a  π bond. That sentence is incorrect and the  π bond can make as a dative bond also. My above imagination is correct but that sentence is wrong. Thank you very much for everybody , help me in this problem.

Yes. I think that there is a mistake in your imagination. Nitrogen is evidently $sp^2$-hybridised in this case. Nitrogen has 5 valence electrons. Two electrons are used to make 2 $\sigma$ bonds with two hybrid orbitals. Another two electrons become the lone pair and reside also in a hybrid orbital. This leaves us with one electron in one unhybridised $p$ orbital. This $p$ orbital is then used to make the $\pi$ bond to either oxygen atom, contributing one electron to the $\pi$ bond while the oxygen atom also contributes one electron from its $p$ orbital.


text:
What do s, m, w, m-w, and w-s mean in the context of infrared spectroscopy?
I am referring to Donald Pavia's Introduction to Spectroscopy, and I found this table that tells about the intensities of the peaks of vibrations of certain bonds.

I can't understand what s, m, w, w-s, etc. are supposed to mean. My guess is that s could mean small, m could mean medium and w could mean weak. But that's crazy, since small and weak mean the same thing? So what do they mean actually?

I'm quite sure this is described in the book itself.

@Mithoron Unfortunately it isn't.

@Mithoron how is ft-ir an appropriate tag for this? My question has nothing to do with Fourier transform spectroscopic technique, it's just regular infrared.

This is the tag for IR spectroscopy, ft is standard procedure.

@hBy2Py is correct. I have found a similar table on another page which is a bit more explicit in what its abbreviations mean and every thing lines up with the the notation of  strong/ medium/ weak. https://www2.chemistry.msu.edu/faculty/reusch/virttxtjml/Spectrpy/InfraRed/infrared.htm

It is described in the book, p 28 (5th ed) "Often, when reading the literature of organic chemistry, you will find absorptions referred to as strong (s), medium (m), weak (w), broad, or sharp."

Given the general agreement surrounding my comment on the question and the similar abbreviations and descriptions in the chart on the page linked by Tyberius in his comment, the meanings appear to be:

strong
moderate (or medium)
weak

Notations such as m-w and w-s presumably indicate a peak intensity that is variable in the indicated range, depending on the properties of the particular molecule of interest.


text:
What is the hybridization of chlorine in vinyl chloride?
Is chlorine in vinyl chloride sp2 or sp3 hybridized?
Calculating using steric number it is found to be sp3 hybridized. But according to I. L. Finar* it is sp2.
*I.L. Finar: Organic chemistry Vol.1 Fundamental principles. Sixth Edition. Orient Longmans: 1973. Page 328.

Let's put it this way: what observable difference would that make?

Terminal, electronegative atoms are in 99% of the cases sp hybridised. (There are no bunny ears.)

Disclaimer: An important point to remember is, that hybridisation always follows the molecular structure, it is never the cause for a certain structure. As such, hybridisation is an interpretation tool, nothing more.

As a rule of thumb, terminal atoms (excluding hydrogen) are almost always (maximally) approximately sp hybridised. While other hybridisation schemes can be applied, they are usually not a good representation.*
One popular example is water. In many texts the central oxygen is described as having (approximately) four sp3 orbitals, which essentially makes the lone pairs equivalent. That view, however, is not in agreement with the photoelectron spectrum, which clearly shows that the lone pairs are not equivalent. At this point I'd like to refer you to Michael Laing's article: "No rabbit ears on water. The structure of the water molecule: What should we tell the students?" (J. Chem. Educ. 1987, 64 (2), 124.).
The same principle described there applies to terminal atoms. Because of the local $C_\mathrm{\infty}$ symmetry the notion of three equivalent lone pairs is likely to be false. The most likely (main resonance contributor) structure for these atoms are (if hybridisation is at all feasible, i.e. mostly in the second period) one sp orbital forming the bond, one sp lone pair, two (perpendicular) p lone pairs.
In this particular example, I have calculated the molecule on the DF-BP86/def2-SVP level of theory and analysed it with the natural bond orbital theory (NBO6). Below you find the localised orbitals which best fit with the common Lewis structure of the molecule.
I have ordered the orbitals starting with the carbon-carbon σ- and carbon-chlorine σ-bonds on the bottom. Following that is the carbon-carbon π-bond. Continuing with the three carbon-hydrogen σ-bonds, and finally ending with the three chlorine lone pairs. The occupied orbitals are in blue and orange, while the corresponding virtual orbitals are next to it in red and yellow. The detailed analysis in numbers is at the end of the post.

In ast's answer there is the claim that chlorine must be sp2 hybridised because of a resonance contributor. While this explanation is extremely tempting because it easy to understand, you cannot judge from the possibility of resonance to the electronic structure. As another general rule of thumb you can remember: the lesser the orbitals are hybridised, the more likely the structure is.
The calculation also provides an analysis in terms of natural resonance theory. On the DF-BP86/def2-SVP level the contribution of the second configuration to the total electronic structure is only 6%. The main configuration contributes with about 90%, other configurations (ionic) make up the remaining 4%.
In summary, I'm afraid, but your book is incorrect, or at least incomplete. The description with two equivalent sp2 lone pairs is unnecessary complicated and does not accurately reproduce the electronic structure.

Truncated results of the NBO analysis (skipping core orbitals)

     (Occupancy)   Bond orbital / Hybrids
 ------------------ Lewis ------------------------------------------------------
   8. (1.99378) LP ( 1)Cl  6            s( 81.55%)p 0.23( 18.44%)d 0.00(  0.01%)
   9. (1.97199) LP ( 2)Cl  6            s(  0.18%)p99.99( 99.79%)d 0.18(  0.03%)
  10. (1.90237) LP ( 3)Cl  6            s(  0.00%)p 1.00( 99.95%)d 0.00(  0.05%)
  11. (1.99642) BD ( 1) C  1- C  2
               ( 48.61%)   0.6972* C  1 s(  0.00%)p 1.00( 99.94%)d 0.00(  0.06%)
               ( 51.39%)   0.7169* C  2 s(  0.00%)p 1.00( 99.95%)d 0.00(  0.05%)
  12. (1.99535) BD ( 2) C  1- C  2
               ( 49.24%)   0.7017* C  1 s( 40.16%)p 1.49( 59.76%)d 0.00(  0.08%)
               ( 50.76%)   0.7124* C  2 s( 44.47%)p 1.25( 55.47%)d 0.00(  0.06%)
  13. (1.96941) BD ( 1) C  1- H  3
               ( 61.15%)   0.7820* C  1 s( 29.81%)p 2.35( 70.15%)d 0.00(  0.04%)
               ( 38.85%)   0.6233* H  3 s( 99.91%)p 0.00(  0.09%)
  14. (1.98105) BD ( 1) C  1- H  4
               ( 61.33%)   0.7832* C  1 s( 30.09%)p 2.32( 69.87%)d 0.00(  0.04%)
               ( 38.67%)   0.6218* H  4 s( 99.91%)p 0.00(  0.09%)
  15. (1.98492) BD ( 1) C  2- H  5
               ( 62.36%)   0.7897* C  2 s( 32.79%)p 2.05( 67.18%)d 0.00(  0.03%)
               ( 37.64%)   0.6135* H  5 s( 99.89%)p 0.00(  0.11%)
  16. (1.99056) BD ( 1) C  2-Cl  6
               ( 43.14%)   0.6568* C  2 s( 22.75%)p 3.39( 77.00%)d 0.01(  0.25%)
               ( 56.86%)   0.7541*Cl  6 s( 18.31%)p 4.43( 81.12%)d 0.03(  0.57%)
 ---------------- non-Lewis ----------------------------------------------------
  17. (0.09516) BD*( 1) C  1- C  2
               ( 51.39%)   0.7169* C  1 s(  0.00%)p 1.00( 99.94%)d 0.00(  0.06%)
               ( 48.61%)  -0.6972* C  2 s(  0.00%)p 1.00( 99.95%)d 0.00(  0.05%)
  18. (0.01232) BD*( 2) C  1- C  2
               ( 50.76%)   0.7124* C  1 s( 40.16%)p 1.49( 59.76%)d 0.00(  0.08%)
               ( 49.24%)  -0.7017* C  2 s( 44.47%)p 1.25( 55.47%)d 0.00(  0.06%)
  19. (0.00909) BD*( 1) C  1- H  3
               ( 38.85%)   0.6233* C  1 s( 29.81%)p 2.35( 70.15%)d 0.00(  0.04%)
               ( 61.15%)  -0.7820* H  3 s( 99.91%)p 0.00(  0.09%)
  20. (0.01032) BD*( 1) C  1- H  4
               ( 38.67%)   0.6218* C  1 s( 30.09%)p 2.32( 69.87%)d 0.00(  0.04%)
               ( 61.33%)  -0.7832* H  4 s( 99.91%)p 0.00(  0.09%)
  21. (0.02582) BD*( 1) C  2- H  5
               ( 37.64%)   0.6135* C  2 s( 32.79%)p 2.05( 67.18%)d 0.00(  0.03%)
               ( 62.36%)  -0.7897* H  5 s( 99.89%)p 0.00(  0.11%)
  22. (0.03245) BD*( 1) C  2-Cl  6
               ( 56.86%)   0.7541* C  2 s( 22.75%)p 3.39( 77.00%)d 0.01(  0.25%)
               ( 43.14%)  -0.6568*Cl  6 s( 18.31%)p 4.43( 81.12%)d 0.03(  0.57%)

Summary of Natural Population Analysis

                                     Natural Population
             Natural    ---------------------------------------------
  Atom No    Charge        Core      Valence    Rydberg      Total
 --------------------------------------------------------------------
    C  1   -0.43992      1.99995     4.42868    0.01129     6.43992
    C  2   -0.22801      1.99996     4.20978    0.01827     6.22801
    H  3    0.22758      0.00000     0.76996    0.00246     0.77242
    H  4    0.22498      0.00000     0.77165    0.00337     0.77502
    H  5    0.23463      0.00000     0.76237    0.00300     0.76537
   Cl  6   -0.01926      9.99998     7.00353    0.01575    17.01926
 ====================================================================
 * Total *  0.00000     13.99989    17.94598    0.05414    32.00000

Optimised Geometry DF-BP86/def2-SVP

6
symmetry cs ( E = -538.027155015 A.U. )
C        1.309538928      1.048660302      0.000000000
C        0.000000000      0.759346920      0.000000000
H        1.630332304      2.102388803      0.000000000
H        2.083175437      0.265507033      0.000000000
H       -0.793639761      1.523771533      0.000000000
Cl      -0.633947150     -0.867041806      0.000000000

* It is also important to understand, that atoms are never hybridised, only orbitals are. The presence of an sp hybrid orbital does not exclude the presence of a sp3 hybrid orbital.

Would u elaborate on what's on the second column of the picture you added?

@Mockingbird These are the corresponding (anti-bonding) virtual orbitals that belong to the bonding orbitals. Note that they all have a nodal plane perpendicular to the bonding axis.

Why do lone pairs don't have corresponding antibonding orbital? Or, you just didn't mention?

@Mockingbird They don't form a bond. There is only one combination, that's why lone pairs are usually considered as non-bonding.

But is it possible to rationalise this using some simply chemistry explanation, instead of just basing this theoretically on a computational calculation? For example, I was thinking, it is likely terminal atoms are sp-hybridised because they often only make one sigma bond to the central atom and so to maximise the strength of this bond, it would be reasonable that the terminal atom uses an sp-hybrid for bonding since the sp-hybrid orbital is directional and has high s character. Would that be a reasonable simple chemistry rationalisation of your computational result?

@TanYongBoon Not really. The concept of hybridisation is only useful to explain certain geometries. The scheme can only be applied after the fact, i.e. you already know the bonding situation reasonably well. Terminal atoms will always need a p-orbital to form a bond, the extent to which the s-orbital mixes depends on the sp-gap of the element to be considered. Consider the local coordination of a terminal atom, which is always linear. Given that hybridisation is applicable in the first place, the only reasonable choice for a linear coordination is sp.

Can I just clarify? Why do you say that terminal atoms will always need a p orbital to form a bond?

@TanYongBoon Well, not always, but in almost all cases where hybridisation can reasonably be applied. That is because they are directional and higher lying in energy than s-orbitals.


text:
Assigning labels and answering questions using given DSC graph
I'm currently working through past exam papers in preparation for my upcoming Inorganic Chemistry exam and I got to this question. We really haven't looked at DSC graphs, so I'm not completely sure how to answer this. I've shown my attempt, but am looking for someone to confirm if I'm right, if not, where I went wrong.
Question:
The DSC thermogram depicted below was run on a sample of a new compound suspected of possessing mesogenic properties. The thermogram was obtained at a scan rate of 10 °C min−1 and the solid line represents the heating scan, while the dashed line represents the cooling scan.

Interpret the DSC thermogram, naming and explaining the processes occurring at peaks (a) – (e), using the correct terminology.
Does the DSC thermogram confirm that the new compound is a mesogen? If so, what can you say about the liquid crystal phase(s), and what name is given to this class of mesogen? What word describes this particular kind of thermal behaviour?
My answer: e = Crystallization, a = melting point (crystalline -> smectic), b = transition point (smectic -> nematic), d = (re)crystallization (nematic -> crystalline), c = melting point (crystalline -> nematic). Since a nematic phase is a subset of liquid crystal (synonym for mesogens), it confirms that the new compound is a mesogen. Molecules in the nematic phase are (somewhat) orientationally ordered (in contrast to the smectic phase which is (somewhat) orientationally and positionally ordered). I don't have an answer to "What word describes this particular kind of thermal behaviour?".

As example, Skoog / Leary's principles in instrumental analysis contained a section about DSC in the chapter of thermoanalysis.  Your university library likely has some of them or other books about the topic, too.
Then, beside monitoring glas transitions, melting, decomposition (while heating) or solidification / crystallisation (while cooling) I suggest to attribute the other phase transitions with (in the best case: simultaneous) visual observation of the sample (heat stage microscopy, like here) at these critical temperatures.  As these transitions take time, from a practical perspective, a heating rate of $\pu{10 K/min}$ is maybe rather quick.
After reading "liquid crystals" in the title, I thought a cholestic phase would be mentioned, too, but if your data support its absence, ok.  On the other hand, one of PerkinElmer's brochures includes this representation

(source)
which might be useful to complement the structural description of your liquid crystals.

I appreciate your answer but I checked the textbook, specifically that chapter, and it essentially just describes how DSC works and wasn't able to help me answer the questions. It wasn't even able to guide me to the right answer. Thanks anyways.

I went to my professor to discuss the answer with him and this is the solution I obtained:
As the sample is heated from 0 °C nothing significant happens until around 90 °C when at peak (a) the melting point [crystal → liquid crystal (1)] occurs as signified by the large endotherm.
Heating continues until the tiny peak (b) indicates a transition point [liquid crystal (1) → liquid crystal (2).]
Further heating results in the medium-sized endotherm (c) at the clearing point [liquid crystal (2) → isotropic melt].
Upon cooling there is a super-cooling effect and an exotherm at (d) [isotropic melt → liquid crystal] and then the liquid crystal phase remains until the crystalline phase is formed again at (e) [liquid crystal → crystal].
The DSC confirms two mesogenic phases upon heating [between (a) and (b) and another between (b) and (c), and a single mesogenic phase on cooling [between (d) and (e)].
This is a thermotropic mesogen and as liquid crystal phases are observed on both heating and cooling it is enantiotropic.


text:
Phosphoric Acid Problem

What is the $\ce{[HPO4^2-]}$ of a solution labeled "$\pu{0.10 M}$ Phosphoric Acid"? Given the $K_\mathrm{a}$ values of $7.1 \times 10^{-3}$, $6.3 \times 10^{-8}$ and $4.2 \times 10^{-13}$ for the first, second and third ionizations, respectively.

Some sources have shown that the answer is $6.3 \times 10^{-8}$, but following through the long calculation through each ionization to get the equilibrium concentration of $\ce{HPO4^{2-}}$ I get an answer of $3.8 \times 10^{-5}$. Hoping someone can enlighten me on this matter.
Attempt at solving:
So first what I did was to individually find the concentrations using the given $K_\mathrm{a}$ values. Working from the top, I calculated for the concentration of $\ce{H2PO4-}$ from the given concentration of $\ce{H3PO4}$. Had to go quadratic with this one since the first ionization constant is a relatively big number. After getting the concentration of $\ce{H2PO4-}$, went on to get the concentration of $\ce{HPO4^2-}$.
After this one, used the final ionization constant to get the concentration of $\ce{PO4^3-}$ produced, which I found to be so small that the $\ce{HPO4^2-}$ from earlier isn't all that much affected. And after all that I get the answer of $\pu{3.8e-5 M}$ for $\ce{HPO4^2-}$.

It would be very beneficial if you could include your whole approach. That way we can better follow your train of thought and may point out where you went wrong.

do you think the current answer is sufficient to accept?

Let's list what's going on here. Orthophosphoric acid dissociates stepwise:
\begin{align}
\ce{H3PO4 + H2O & <=>[$K_\mathrm{a1}$] H3O+ + H2PO4–} \tag{R1}\\
\ce{H2PO4- + H2O & <=>[$K_\mathrm{a2}$] H3O+ + HPO4^2–} \tag{R2}\\
\ce{HPO4^2- + H2O & <=>[$K_\mathrm{a3}$] H3O+ + PO4^3–} \tag{R3}
\end{align}
Mass balances for protons and phosporous, respectively:
\begin{align}
C_\ce{H} & = [\ce{H3O+}] + [\ce{HPO4^2–}] + [\ce{H2PO4–}] + [\ce{H3PO4}] = [\ce{H3O+}] + \sum_{n=1}^3 [\ce{H_$n$PO4^$(3-n)-$}] \tag{1}\\
C_\ce{P} & = [\ce{PO4^3-}] + [\ce{HPO4^2–}] + [\ce{H2PO4–}] + \ce{H3PO4} \tag{2}
\end{align}
Now we compare the equilibrium constants:
\begin{align}
K_\mathrm{a1} & = \frac{[\ce{H3O+}][\ce{H2PO4-}]}{[\ce{H3PO4}]} = 7.1 \times 10^{−3} \tag{3}\\
K_\mathrm{a2} & = \frac{[\ce{H3O+}][\ce{HPO4^2-}]}{[\ce{H2PO4-}]} = 6.3 \times 10^{−8} \tag{4}\\
K_\mathrm{a3} & = \frac{[\ce{H3O+}][\ce{PO4^3-}]}{[\ce{HPO4^2-}]} = 4.2 \times 10^{−13} \tag{5}
\end{align}
Notice that $K_\mathrm{a1} >> K_\mathrm{a2}$ (more that $10^5$), therefore by the first dissociation step phosphoric acid behaves as an acid of medium strength, producing equal amounts of $\ce{H3O+}$ and $\ce{H2PO4-}$ (considering charge balance):
$$[\ce{H3O+}] \approx [\ce{H2PO4-}] \tag{6}$$
which directly leads us to the answer. Using expression for $K_\mathrm{a2}$:
$$[\ce{HPO4^{2-}}] = K_\mathrm{a2} \times \frac{[\ce{H2PO4-}]}{[\ce{H3O+}]} = K_\mathrm{a2} \tag{7}$$
That's it, $[\ce{HPO4^{2-}}] = \pu{6.3 \times 10^{−8} (mol/L)}$

Now for the fun part, if you really want to solve quadratic equation, we can also find solution pH and $[\ce{PO4^{3-}}]$.
Based on the values of equilibrium constants, it is safe to assume that pretty much only first dissociation step defines solution's pH. We can rewrite equation for $K_\mathrm{a1}$ as follows:
$$K_\mathrm{a1} = \frac{[\ce{H3O+}]^2}{C_0 - [\ce{H3O+}]},$$
where $C_0$ is the concentration of the solution ($\pu{0.10 M}$). Using the given numbers we have
$$7.1 \times 10^{-3} = \frac{[\ce{H3O+}]^2}{0.10 - [\ce{H3O+}]}$$
$$[\ce{H3O+}]^2 + 7.1 \times 10^{-3} [\ce{H3O+}] - 7.1 \times 10^{-4} = 0 $$
$$[\ce{H3O+}] = \frac{-7.1 \times 10^{-3} + \sqrt{(7.1 \times 10^{-3})^2 + 4 \times 7.1 \times 10^{-4}}}{2} = \pu{2.33 \times 10^{-2} (mol/L)}$$
$$\mathrm{pH} = - \log{[\ce{H3O+}]} = 1.63$$
WolframAlpha thinks the same.
Now we know $[\ce{H2PO4-}] = \pu{2.33 \times 10^{-2} (mol/L)}$ and we can finally find $[\ce{PO4^3-}]$ from the expression for $K_\mathrm{a3}$:
$$[\ce{PO4^{3-}}] = K_\mathrm{a3} \times \frac{[\ce{HPO4^2-}]}{[\ce{H3O+}]} = 4.2 \times 10^{−13} \times \frac{6.3 \times 10^{−8}}{2.33 \times 10^{-2}} = \pu{1.1 \times 10^{-18} (mol/L)}$$
This demonstrates that in pretty concentrated solution of phosphoric acid anions $\ce{PO4^3-}$ are practically non-existing.


text:
How do manufacturer remove oxygen from metal during melting?
How do manufacturers remove oxygen from metal during melting? I melt aluminium with nichrome wires and during melting it, it forms lots of slag (aluminium oxide) and I have to remove it manually.

If the metal reacts with oxygen at high temperatures, then the processing needs to be done under an inert atmosphere (i.e. nitrogen, argon, etc.).

Putting a lid on the crucible might already help a lot. Also: Thicker wires, larger batch size. ;-) And yes, the oxide swims on top, and you remove it mechanically, if necessary.

They use flux, that's how. That is, an inert substance which melts easier than the metal and floats on top, preventing it from oxidation in the air. Common fluxes for aluminum seem to be based on eutectic mixtures of various chlorides (say, $\ce{NaCl + KCl}$).
Covering your crucible with a lid does not offer complete protection, and working in inert atmosphere is much more costly (though sometimes they do that, too).

what does "lid does not offer complete protection" mean? It implies a lid offers partial protection.

Well, yes, that's right.

Ok, cool. I assume you've noticed that a lid helps some. Can say a bit about how much it helps? Thanks!

I don't think we can say anything more specific in the general case. There are different molten metals and different lids.

Can we say a tight lid will seal out more oxygen than a leaky one?

I guess so.$!$

@IvanNeretin this partially answers my question for soda can recycling. Can you see if you can add your answer there as I am looking for something before the melting starts because soda cans are thin sheets so they must be dynamically insulated from oxygen prior to its melting point.

There are two aspects to this, firstly is actually reducing the pure metal from its oxides, usually called smelting, most metal ores are metal oxides so this is clearly an important part of primary metal production For example in iron production iron oxides are reduced by carbon, usually from coke which is also the fuel which provides the energy for the reaction.
In the case of remelting metals for casting it is usually a case of limiting oxide formation rather than reversing it. For decent quantities it is usually self limiting to and extent as the oxides tend to float to the surface of the crucible, limiting and further oxidation. Cover fluxes can also be used which sit on the surface and provide a soft seal to the crucible.
This is aided by use of fluxes which are able to trap the oxides which do form and trapping them in a semi-fluid glass mass which is easier to remove. In iron casting limestone is used, other additives (such as sawdust) can be used as slag coagulants to make the resulting slag less sticky and easier to handle.
In the case of aluminium especially thin foil and wires surface oxide layers can make of a significant volume of the material before you even start melting it and there is not a lot you can do about that. Aluminium has a particularly high affinity for oxygen and the only practical way to reduce it is by electrolysis.
If you are melting a range of sizes of material (eg a mixture of scrap and ingots) it may help to melt the larger pieces first as they have a lower surface area to volume ratio and then adding the smaller bits so they melt while submerged in an established pool of molten metal.
Strictly speaking the oxides you get on the surface of molten aluminium are called dross, this term applies to metal oxides produced during melting while slag is a glassy material which is either  a mixture of metal oxides and flux or silicates from the ore.
In industrial production there are various approaches to slag removal. It can either be skimmed off the surface of the ladle/crucible or you can tap from the bottom of the bottom (as in a cupola furnace). Some casting applications also use ceramic slag filters to prevent any stray bits from getting into the mould.

awesome. You listed about 6 different methods, where other answers only offer 1.

Re the self-limiting layer: isn't this a good reason NOT to STIR the pot? I've noticed some youtubers continually stir the pot, as if it's soup :D

@johnywhy maybe they stir the pot to ensure no bubbles are trapped or something.

Generally speaking, manufacturers remove oxygen from metal by using deoxidizers.  For instance, a manufacturer will melt steel, and pour liquid aluminum (or various other viable substances that act as reducing agents) into the liquid steel.  This aluminum will react with the oxygen in the steel to form aluminum oxide (Al2O3).  Because the oxide is lighter than the steel, it flows to the top, and is collected from there.  There is more satisfying detail here.
There are more ways one would go about the deoxidation of the metal, like blowing the oxygen out with noble gases like Argon (since having a nonreactive gas in the metal won't be a problem).  There are more methods like the vacuum method which is described extensively online.  Regarding the deoxidation of aluminum as it pertains to your specific question, there is great stuff to read about on this site.  Some of the information on that website is about copper, but the concepts are the same.
I just realized that I misunderstood your question, and that Ivan's answer is the correct one.  However, I will still leave this explanation here.  Hopefully it serves as good context for the process of metallurgy.

Surely the oxygen is the oxidiser. Wouldn't the alumunium would be a reducer in this senario?

Yes, my mistake.  When Aluminum reacts with oxygen, oxygen gets reduced and aluminum gets oxidized.  Therefore, aluminum is the reducing agent and oxygen is the oxidizing agent.

Although the original question is How to deoxidize oxidized melt?, in practice, deoxidizing may be more difficult than prevention or removal.
This answers the separate questions:

How to prevent oxidation in melt?
How to deoxidize oxidized melt?
How to remove oxidation slag (dross) from the melt?

This is a collection of options, from numerous sources and commenters. Although it overlaps some answers already given, hopefully it's a comprehensive answer. And hopefully, more clearly organizes these three separate questions. (In some cases, it's unclear whether the method will prevent, deoxidize, or remove oxidized metal.)
Be aware that some methods, such as fluxes, may vary depending on which metals you're working with.
How to prevent/reduce oxidation?

Not Oxidation: Your slag might not be oxidized metal at all-- it might be impurities, or contamination by other metals.
Self-Limiting: As mentioned in another answer, it is self-limiting to an extent, as the oxides float to the surface, limiting further oxidation.
Electrolysis: Aluminium has a particularly high affinity for oxygen and the only practical way to reduce it is by electrolysis.
Boric Acid: materials like borax fluorite, silica, sodium carbonate are used. As mentioned in another answer, these sometimes form undesirable "Spinels" in the liquid metal. Boric acid should melt and then (at least partially) decompose into boric oxide and water. Water isn't as dangerous as you'd might think as long as it's at the surface, so as long as you give it ample time to decompose before thrusting anything through the layer it shouldn't be a problem.
Flux: When the metal fully melts into a liquid, carefully sprinkle a small amount of flux onto the molten metal. 'Marvelux flux' is an example (not an endorsement). It will sit on the surface and provide a soft seal to the crucible. After pouring, allow the piece and the mold to cool completely. Then remove the piece, and pickle it to clean off any leftover flux. Then thoroughly rinse it. Alcohol and evaporation seems a better choice than vinegar and water, as vinegar can cause corrosion.
Glass: On brass, you'd use broken glass bottles.
Charcoal: With low temps, crunched charcoal is the usual remedy.
Kitty Litter: some bullet casters use it.
Remove the Oxygen: possibly replacing it with an inert gas. Nitrogen is commonly used, and it's non-toxic (it's already in the air we breathe). Watch that you don't use silicone tubing to connect your N2 cylinder - it is quite O2 permeable. Ditto silicone gloves. There are several ways to do this-- all require an airtight environment. Oxygen must be removed before the melting metals are exposed to the atmosphere. Some methods listed are used in manufacture, some might work for small-scale production or DIY, and some are speculative:

Burn off the oxygen with a flame. Depending on volume, this may cause a pressure difference between the volume and the outside environment, which in turn may require replacing the oxygen with an inert gas.
Pump out all the air, and operate in a vacuum.
Adsorb it with an oxygen "getter". A getter is a deposit of reactive material that is placed inside a vacuum system. When oxygen molecules strike the getter material, they combine with it chemically or by absorption, thus reducing the amount of oxygen in the space.

Commercial gloveboxes use 10% H2 in N2 (formier gas) over heated Pd catalyst to convert O2 to water, which is then removed by drying train (molecular sieve). This may produce CO2, which may need to be removed.
Pass the air through a heated column of copper powder. Ideally, 63,55 g copper powder should remove 24,5 L of oxygen from the air - which leaves us with >80 L of nitrogen left. Now that's not a whole lot of copper for a 30-40 L glovebox! However, depends on column temperature, flow rate, mesh size etc. The atmosphere in the glovebox circulates all the time, and the catalyst can go from 20% O2 to 0ppm in like 15 minutes.
Food-safe deoxidizers: dry iron powder, sodium and sometimes activated carbon. Two 500cc of such oxygen absorbing agents can absorb up to 500cc of oxygen. Normal dry air contains 20.95% oxygen, so you'd need a volume of oxygen absorbing agent equal to 21% of your insulated workspace. More numbers here.
Pass it through a heated tube (circa 500 degreesC) stuffed with steel wool. Unsure if this can work with air, or just water.
Other recipes comprise a porous reducing support such as activated carbon, carbon black, coal, petroleum coke and titania which is impregnated with about 10% to about 90% by weight of an alkali metal oxide or an alkaline earth metal oxide.
A simple tray of vinegar and steel wool. The vinegar dissolves the protective layer on the steel wool, accelerating the rusting process, which sequesters oxygen from the local atmosphere.
Get some of those chemical hand warmers and cut them open and pour the powder in a dish in your glove box. That stuff is reduced iron combined with catalysts that make it more prone to oxidation.
A Zirconium-based getter.
An open petri dish of sodium/potassium eutectic, maybe with mag stirrer.
Get a hold of this stuff. Researchers from the University of Southern Denmark have synthesized crystalline materials that can bind and store oxygen in high concentrations. Just one spoon of the substance is enough to absorb all the oxygen in a room. Wow!

Lid: as mentioned in another answer, a crucible lid may reduce oxidation.
Pipes: Connect your crucible directly to your mold via a pipe, so the molten metal is transferred without exposure to the air.
Reduce area: A skinny crucible, or one with a narrowed upper section, will reduce the surface area exposed to the air. This will be impossible if your solids are thicker than the neck of your crucible. Consider starting with metal powder, instead of ingots.
Eliminate the crucible: Put dry metal powder directly into the mold, pack it down, seal the mold, and put it in an oven. Known as "Compacting and sintering".
Eliminate the crucible, 2: Put a thin layer of metal powder on a flat work surface. With a CNC laser, fuse the powder in the shape of the bottom of your pattern. Add another layer of metal powder, and fuse it. Repeat until the whole part is built up in layers. Known as Direct Metal Laser Sintering.

How to deoxidize oxidized metal?

Heat: turn the pot up really hot and it dissolves back in for the most part.
Charcoal: To deoxidize the layer of dross/oxide sprinkle with powdered charcoal.
Deoxidizers. As mentioned in another answer, a manufacturer may pour liquid aluminum (or other viable reducing agents) into liquid steel. The aluminum will react with the oxygen in the steel to form aluminum oxide (Al2O3). Because aluminum is lighter than the steel, it flows to the top, and is collected from there.
Apply electric current. In 1906, English engineer A. G. Bloxam used direct current in a vacuum, to produce filaments for incandescent lamps. The applied current was particularly effective in reducing surface oxides. (they might mean "prevent", not "reduce"). This has applications in powder-metal sintering.

How to remove oxidized metal?

Sawdust: Stir it and scrape the sides of the pot with a wooden stick. Pick out the trash and leave a layer of ash on top. As mentioned in another answer, sawdust may add hydrogen to the liquid Al causing a new set of problems.
Limestone: Used in iron casting, as mentioned in another answer.
Skim: It can be skimmed off the surface of the crucible.
Filters: use ceramic slag filters to prevent any stray bits from getting into the mould.
Bottom Tap: Extract from bottom of pot, instead of top.

https://www.practicalmachinist.com/vb/general/preventing-oxidation-molten-lead-280491/

Charcoal is added to the flux. Things like sawdust can add hydrogen to the liquid Al causing a new set of problems. Better check references for Al flux ;  materials like borax fluorite, silica, sodium carbonate are used but sometimes form "Spinels"  in the liquid metal = more problems.

what's the role of the charcoal? thx


text:
Oxidation of an aldehyde - how is this a loss of electrons?
I've been reviewing redox reactions and have been using the OIL RIG acronym to classify oxidations vs. reductions. Generally this makes sense, and many resources I've looked at repeat the "Oxidation Is Loss - Reduction Is Gain of electrons" statement.
But, I also know that aldehydes can be oxidized to carboxylic acids:
$$
\ce{RCHO -> RCO2H}
$$
The compound $\ce{RCHO}$ does not lose electrons when oxidized to $\ce{RCO2H}$ -- in fact, it gains a whole new oxygen atom. So, there appears to be a net gain of electrons here. Does OIL RIG not always hold, or am I misunderstanding something?

Take a look at the functional group carbon. Upon the reaction taking place, it loses a $\ce{C-H}$ bond and gets a $\ce{C-O}$ bond. Oxygen is more electronegative than hydrogen is relative to carbon, so in a way, oxygen pulls electrons away from carbon, so there's an oxidation.

My previous comment didn't mention oxidation numbers, so have a read about it here: https://en.m.wikipedia.org/wiki/Oxidation_state

Roughly speaking, compounds don't get oxidized. Elements do. Think of it this way for a while. Then turn back to the usual way.

OILRIG always holds, but you need to be careful at what you are actually looking at. When we say a compound gets oxidised, that is technically not true. In this framework, only atoms get oxidised or reduced.
In the reaction
$$\ce{R-CHO + [O] -> R-COOH}$$
we usually omit the reduction part, because it's always the same: the oxygen.
To identify the whole redox process, use oxidation numbers. Keep in mind that these are simply a bookkeeping tool and the partial charges of the atoms involved might be quite different. Setting $\ce{R {=} CH3}$ or the above reaction it boils down to this:
$$\ce{
\overset{+1}{H}_3\overset{-3}{C}-\overset{\color{\red}{+1}}{C}\overset{+1}{H}\overset{-2}{O}
+ [\overset{\color{\green}{0}}{O}]
->
\overset{+1}{H}_3\overset{-3}{C}-\overset{\color{\red}{+3}}{C}\overset{-2}{O}\overset{\color{\green}{-2}}{O}\overset{+1}{H}
}$$
Now you can see that the carbonyl carbon increases its oxidation state from $\color{\red}{+1}$ to $\color{\red}{+3}$, so formally it lost $\pu{\color{\red}{2} e^-}$. The oxidising agaent, indicated with $\ce{[O]}$, at the same time decreases its oxidation state from $\color{\green}{0}$ to $\color{\green}{-2}$, therefore formally gaining $\pu{\color{\green}{2} e^-}$. The redox reaction is complete.
Let's look at a real world example, the Baeyer-Villiger oxidation:

As you can see, the ketone gets oxidised, more precisely the carbonyl carbon gets oxidised as increases its oxidation state. On the other side, the peroxide oxygens get reduced as they decrease their oxidation state.

I actually got confused with this before as well. But when you start to think of things based on the perspective you have to look at them from, it becomes easier to understand. It's all a matter of separating the meaning of oxidation and reduction when it comes to organic chemistry. Simply put, in organic chemistry what is meant by "oxidation" is simply an increase in an atom's number of bonds to oxygen, while the opposite implies reduction. For example, when you oxidize an alcohol, the C-O single bond becomes a C=O double bond and so in essence carbon's bonds to oxygen increased by one and thus the carbon is oxidized.
In relation to electrons, I think it's not a loss of electrons in the singular sense like how we could look at it in general chemistry, but more of a shift of electron densities when the bonds change. Here it says that when atoms lose bonds to hydrogen and gain bonds to other heteroatoms, it is generally an oxidative process, considering that H is the least electronegative. So when an alcohol gets converted to a ketone there is a net loss of electron "density" on the carbon atom - synonymous to the loss of electrons.

While this is all true and the IUPAC even states that oxidation is the gain of oxygens or loss of hydrogens it doesn't explain the actual question: How is it a loss of electrons?

I think it's not a loss of electrons in the singular sense like how we could look at it in general chemistry, more of a shift of electron densities when the bonds change.

Here  it says that when atoms lose bonds to hydrogen and gain bonds to other heteroatoms, it is generally an oxidative process, considering that H is the least electronegative. So when an alcohol gets converted to a ketone there is a net loss of electron "density" on the carbon atom - synonymous to the loss of electrons.

You make an excellent point and I encourage you to include it into your answer, which would make it a lot stronger and would offer another insight then just my oxidation number approach.


text:
Calculating the UV/Vis spectrum of perylene
I recently started to do computational chemistry and am now trying to write some basic inputs for important calculations. One, which I guess is quite important to me would be to predict UV/Vis spectra of molecules. So I looked for a basic input file on the net, and although my tutor warned me about using DFT for absorption spectra (it's not a topic of the lecture and therefore he wasn't sure what to use either) the first few simple molecules, mostly long conjugated systems and a few azo-compounds gave really good solutions in comparison to literature values.
Now I wanted to predict a perylene in benzene as solvent as well and gave the structure optimization (HF/VDZ) six cycles, which took forever to calculate before I stopped it. Then I used the geometry at each of the six steps to predict the spectrum and compared the results. What surprises me is that the optimization, while becoming better, moves the first absorption peak at ~435 nm further from experiment. From its position and the direction, it seems to better describe the second peak (~410 nm) instead of the first one.
I am also only getting one peak only but for the other molecules, it had been pretty much the same as well, where one peak was quite big and the rest, although being at the right positions were too small to really see.
Now I know about the quantum mechanics and the theoretical backgrounds of UV/Vis spectroscopy but we never really discussed the effects in real spectra. I know from NMR that peaks can split and multiply in experiment, while your theoretical model only contains one signal.
Is there a reason, perhaps due to the big delocalized system why perhaps the bands shift or multiply in some way or why I am not able to see the first band? Or perhaps my calculation shows the first band and it's really far off?
All UV/Vis calculations were done with Orca using RI-BP86/SV(P) and COSMO for the solven.
Here is the experimental spectrum I am referring to:

Please note: Those are my very first calculations and I am still trying to understand when the simple approaches yield good results and when I need to switch to different approaches.

I won't comment on the accuracy or otherwise of you calculation but that does not seem to a problem just now.
In the spectrum you show and in similar ones, the electronic transition energy is at approximately the mid-point of the first absorption and emission peaks, at approx 22800 cm$^{-1}$ in your figure. The different bands you see are due to vibrational transitions from various normal modes 'on top', as it were, of the electronic 0-0 transition. The energy gaps between peaks in the absorption spectrum measure the vibrational frequency in the excited state and vice versa for the fluorescence spectrum. Look up a Jablonski diagram to see generic sketches vibrational levels and transitions.
The number and magnitude of the vibrational transitions depends on the shape of the potential energy surface and is governed by the Franck-Condon factors which are $|\int \psi_e\psi_g|^2$ between the ground (g) and excited state(e).
The width of the lines in solution spectra (such as the one you show) is partly also due to rotational lines  but mainly due to collisions with solvent shifting and broadening energy levels. Consequently if you look up gas/vapour phase spectra at high resolution, such as for benzene as there a many published spectra, you will see many narrow lines corresponding to different vibrational normal modes and if resolution is high enough, even rotational lines.


text:
Silicone with good thermal conductivity
I am looking for a material with the same properties of silicone (flexible, easy to manipulate, soft, etc.) but without being a thermal isolate.
I need this material to do the junction between a temperature sensor and the human skin (this is why I need it to be soft).
So far I've used the ELASTOSIL® E43 but without good results.

For microelectronic packaging people use 'loaded' epoxies to increase thermal conductivity. Silver is a common loading and can be used if you want conductivity (ground contact). If an insulating bond is desired, the often use alumina powder. The trick is to find a balance of loading and other properties of interest.

Beryllium oxide is used to improve the thermal conductivity of silicone pastes used on heat sinks...  but BeO is toxic. See https://en.wikipedia.org/wiki/Beryllium_oxide


text:
Which oxygen will form a better hydrogen bond?

Which oxygen will form a stronger hydrogen bond , the one in water or the one in hydrogen peroxide ??

According to me , the one in peroxide will form the stronger H bond as the O here being less electronegative will be more willing to donate lone pair to an incoming dipole .

Electronegativity is a property of the element. All oxygens are equally electronegative, so your line of reasoning is not correct.
A better argument would probably involve looking at the $\alpha$ effect and how lone pairs on vicinal atoms destabilize each other (generally used to justify why the peroxide oxygen-oxygen bond is weak). One might imagine then that each oxygen could form a stronger hydrogen bond, since this would help to stabilize the repulsion.


text:
Neutral and positively-charged polyoxometalates - proper name for the class of compounds
Would you still call neutral and positively-charged molecular metal oxides polyoxometalates, or do you know a better/proper way to name this class of compounds (Polyoxometalloids/Polyoxometallic species/Polyoxometallides/...)?
This question occurred on the premise of an interesting paragraph in the paper (1) regarding the nomenclature of newly-synthesized polynuclear oxovanadium-lanthanide species:

Given that polyoxovanadates are increasingly being used as diamagnetic
  ‘‘ligands’’ for lanthanide ions, the cage structure in 3 could be
  described as a tubular $\ce{[V^V12O32]^4-}$ ligand coordinating to two
  $\ce{\{Dy(OH2)4\}^3+}$ centres, with the latter plugging the ends of
  the tube and trapping the interstitial chloride. However, it is surely
  preferable in this case to consider the dysprosium units as part of
  the cage, and this to be a heteropolyoxovanadium complex. One must
  also note that, very unusually, 3 is a cationic cage system, and
  that ‘‘heteropolyoxovanadate’’, with the -ate suffix denoting a
  negative charge is here incorrect according to normal IUPAC
  nomenclature rules. On the other hand, there are examples of
  positively-charged oxygen-bridged metal cages which correspond to well
  known motifs from polyoxometalate chemistry such as the
  $\ce{\{Al^{III}13\}}$ $\varepsilon$-Keggin-ion structure
  $\ce{[AlO4Al12(OH)24(H2O)12]^7+}$.

Compound $\ce{[AlO4Al12(OH)24(H2O)12]^7+}$ (2) was investigated in 1960, though it has not been explicitly denoted as polyoxoaluminate at that time.
Here are the structures of these polynuclear complexes:

It also seems like IUPAC does not really care, always referring to polyoxometalate term. So please feel free to share your opinion!
(1) Šimuneková, M.; Prodius, D.; Mereacre, V.; Schwendt, P.; Turta, C.; Bettinelli, M.; Speghini, A.; Lan, Y.; Anson, C. E.; Powell, A. K. RSC Advances 2013, 3 (18), 6299. DOI 10.1039/c3ra40385h
(2) Johansson, G.; Gullman, L.-O.; Kjekshus, A.; Söderquist, R. Acta Chemica Scandinavica 1960, 14, 771–773. DOI 10.3891/acta.chem.scand.14-0771

Many years ago ago I dealt with [PMo$_{12}$O$_{40}$(VO)$_2$]$^{q−}$, which is basically a Keggin polyoxomolybdate (-3) capped with two vanadyl groups (2*(+2)). It was in a highly reduced form, but depending on how many blue electrons were residing in the Keggin cage, it might conceivably present neutral or positive charge. My work was theoretical and actually revolved around the control of this reduction state.
While I did not find any canonical source to support it, when privately thinking about the possibility of a positively charged polyoxometalate, I always called the q>0 version of the molecule a polyoxometalonium. I did this in analogy with the well-known positively charged polyatomic ions ammonium, hydronium... but also other similar species, I guess, such as sulfonium, phosphonium. [Folk linguistics warning: no IUPAC was involved in this nomenclature].
Back then I did not think about the neutral (q=0) version, but I'd say polyoxometallic complex sounds a reasonable approximation to me. To me it does not inherently say "no charge", but it sure does not imply negative (nor positive) charge. To make the meaning clear, I'd go for neutral polyoxometallic complex.

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How right is defining elements with the number of protons?
Atomic number is the number of protons a certain atom has. It's the defining attribute of a certain element. But, How right is that?
From the viewpoint of chemistry, the definition is right. Because, for a certain chemical reaction, isotopes of a certain element will react in the same way.
For example, $\ce{H2 + 1/2O2 -> H2O}$
$\ce{ ^2H2 + 1/2O2-> ^2H2O}$
So, it's useful to define elements in terms of their number of protons.
But, other than reactivity in chemical reactions, many properties of an atom differ wildly due to difference of  number of neutrons and electrons eg. half life of radioactive decay, type of radioactive decay, magnetic moment, stability, nuclear quantum states etc. So, even having the same number of protons can't ensure much similarity in some aspects.
I think to a rational perspective, the similarity of reactivity of isotopes in chemical reactions should weigh more than the other nuclear difference between isotopes. So, in that logic, the definition is right.
But, difference in nuclear properties should do matter. So, how right is defining elements with atomic number i.e. number of protons?

But aren't elements already defined via $^A_Z\ce{E}$ notation when it is necessary, i.e. in nuclear reactions?

@andselisk That is true. But, it's defined mainly by atomic number. And my question is why is that?

Look up isotope and nuclide.  Isotope is primarily a chemical term and treats the proton number as more significant than the neutron number.  Nuclide is a physicist's term and treats both equally.  There are also terms isobar and isotone.

You just rehearsed the definition of element and isotope, and why we do have these two different, and why they are useful. Good. So what is the question? Why isotopes are not called elements and vice versa?

You are trying to split hair. The terms are defined as they are, because those definitions are useful for the applications people had in mind when they made them. Chemistry, in this case.

Most of chemistry doesn't care a whit about all of the properties you named: they're in the realm of nuclear physics, as noted by user badjohn.  Even radioactive decay is something of a black sheep, included in physical chemistry as a convenient example of a first-order kinetic process. To my knowledge, it's not substantially used in conjunction with any other sub-discipline of chemistry, as it'd be more of an annoyance than anything ("Gah, the cobalt center of my complex keeps changing into nickel!").  It does of course have plenty of applications (medical imaging, nuclear power, etc.), but they're arguably in the realm of physics, with any attendant chemistry being a secondary consideration.
Magnetic moment, neutron count (isotopes), and nuclear quantum states do have some slight influence over certain spectroscopic properties (primarily NMR, as far as I know), but other than the somewhat special cases of the spin isomers of hydrogen and the kinetic isotope effect of deuterium vs. protium I'm not aware of any appreciable effect of these on chemical reactivity.
For the chemist, the identity of a nucleus is sufficiently defined (within an experimentally measurable perturbation, at least) in virtually all cases by its atomic number.