Electronegativity of Potassium (Pauling)

It depends which electronegativity scale you choose to use.  There are a lot of different scales for comparing electronegativities of the elements.  I've included a list of 11 of them below if you're interested.  In any case here are the electronegativities reported for potassium using 5 different electronegativity scales.  All values are in "Pauling units" so that the numbers can be compared.  The Pauling scale is the most commonly used, so 0.82 would be a reasonable estimate for the electronegativity of potassium.

Pauling 0.82
Sanderson 0.45
Allred Rochow 0.91
Mulliken-Jaffe 0.73
Allen 0.734

On the Pauling scale, electronegativities range from around 0.7 to 3.98 (fluorine), so as expected, potassium is at the electropositive end.  None of these scales report a value of 1.

List of 11 Different Electronegativity Scales

Pauling Scale: (1932)  Obtains values by thermochemical methods
Mulliken Relation: (1934)  Defines a relation that depends upon the
orbital characteristics of an atom in a molecule. Mulliken
electronegativity is the numerical average of the ionisation
potential and electron affinity
Gordy Scale: (1946)  Defines electronegativity in terms of the
effective nuclear charge and the covalent radius. (Zeff)e/r
Walsh Scale: (1951)  Relates electronegativity to stretching force
constants of the bonds of an atom to a hydrogen atom.
Huggins Scale: (1953) Alternative to Pauling's thermochemical
procedure
Sanderson Scale: (1955)  The ratio of the average electron density of
an atom to that of a hypothetical "inert" atom having the same number of electrons. This ratio is a measure of the relative compactness of the atom
Allred-Rochow Scale: (1958)  Defines electronegativity in terms of the effective nuclear charge and covalent radius. Like the Gordy scale but uses (Zeff)e/r2
Jaffe Scale: (1962)  Uses the electronegativity of orbitals rather
than atoms to develop group electronegativities for molecular fragments (e.g. $\ce{CH3}$ vs $\ce{CF3}$) that take into account the charge of a group, the effects of substituents, and the hybridization of the bonding orbital
Phillips Scale: (1968)  Defines electronegativity in terms of the
dielectric properties of atoms in a given valence state
Martynov & Batsanov Scale: (1980) Obtained by averaging the
successive ionisation energies of an element's valence electrons
Allen CE Scale: (1992) Configuration energy (CE), the average
one-electron valence shell energy of the ground-state free atom, is
used to quantify metal-covalent-ionic bonding


When, how and why does sodium metal react with cycloalkenes, cycloalkadienes, cycloalkatrienes?

The negatively charged particles are conjugate bases of respective compounds, which (reactants) acts due to presence of active methylene group. Depends on the stabitily of conjugate base the reaction goes forward.


Does there exist a black coloured metal?

Highly and degenerately doped semiconductors and semi-metals can be at least dark in visible light in many cases yet have good electrical conductivity. However it would be a stretch to call them either "metals" or "black" without qualifying those labels.
They appear dark in visible light for (superficially at least) the same reason that obsidian or black marble appears dark; they mostly transmit light entering the surface rather than reflect it, then slowly absorb it internally over microns or millimeters. The same is true for many semiconductors, but since their index of refraction in visible wavelengths is in the 2.5 to 5 ballpark their specular Fresnel reflection can be 20 to 50%! Compare to glass with an index of 1.4 or 1.5 and Fresnel reflection of ~4%.
See all the excellent answers to Why are most metals gray/silver? in Physics SE.
There's a short discussion of plasma frequency and semiconductors in this answer which begins:

Metals have a high density of free carriers that can be though of as a plasma. The plasma frequency of metals is usually in the ultraviolet (except for Born & Wolf; Alkali metals transparent to UV? Cesium transparent to blue?), meaning any electromagnetic wave below the plasma frequency would be rapidly absorbed and re-emitted (fancy words for "reflected" by the metal). Just like Earth's ionosphere reflecting HF frequencies and lower but transmitting most VHF and higher, a metallized film that reflects visible light and infrared light (heat) would also reflect all RF radiation; anything below the plasma frequency.
$$\omega_{pe}=\sqrt{\frac{n_e e^2}{m^* \epsilon_0}}$$

A polished surface could look different depending on what is being reflected, but a roughened or ground surface will look grayish.
This piece looks fairly smooth and so it has a shiny appearance due to Fresnel reflection, but it only reflects about 40% of the light, whereas a polished metal can be 95% or higher in some cases.


What is the difference between λ and Λ with regard to molar conductivity?

Look in the Wikipedia article Ion transport number.

Lower case lambda, $\lambda$, is the molar ionic conductivity

Upper case lambda, $\Lambda$, is the molar conductivity.

$\lambda = t \dfrac{\Lambda}{\nu}$

Where $\nu$ is the number of ions for the cation or anion in the molecule. So $\ce{Na3PO4}$ would have 3 cations per molecule, but only 1 anion per molecule.


Is the Hartree-Fock determinant always the minimum energy single Slater determinant solution to a molecular Hamiltonian?

By construction, any determinant $|\chi_{i_1}\ldots \chi_{i_N}\rangle$ where the $\chi_{i_j}$ are distinct solutions of the Hartree-Fock equation $f(x) \chi(x) = \epsilon_i \chi(x)$ will have stationary energy $\langle \chi_{i_1}\ldots \chi_{i_N} | H |\chi_{i_1}\ldots \chi_{i_N}\rangle$ (where stationarity is with respect to perturbing the orbitals).

This is not true: only the converged Hartree-Fock determinant $|\Phi_0\rangle = |\chi_{i_1}\ldots \chi_{i_N}\rangle$ has stationary energy with respect to orbital perturbations.
Stationarity of energy requires all orbitals in the determinant to fulfill the Hartree-Fock equations
$$
\hat f \chi_i = \varepsilon_i \chi_i
$$
where the Fock operator
$$\hat{f} = \hat h + \sum_j^N(\hat{J_j} - \hat{K_j})$$ depends on the $N$ occupied orbitals $\{\chi_j\}$ included in the determinant.
The Hartree-Fock determinant $|\Phi_0\rangle$ fulfills this condition and is therefore stationary with respect to orbital perturbation, because the constituting orbitals are converged (in SCF iterations) solutions to these equations, where the Fock operator is calculated from the $N$ orbitals with lowest energies.
Any other (excited) determinant $|\Psi\rangle$ is generally not stationary with respect to perturbing the orbitals. Since other orbitals are occupied in this determinant, it corresponds to a different Fock operator $\hat {f'}$ where the sum is over the orbitals in the determinant $|\Psi\rangle$ which differs from the set of orbitals included in $|\Phi_0\rangle$.
Using this Fock operator $\hat{f'}$ in subsequent SCF cycles, the total energy of the excited determinant is again minimised by tweaking the orbitals. In this course, I suppose, it is quite likely that the excited determinant relaxes back to the ground state Hartree-Fock determinant  $|\Phi_0\rangle$. The iterations might also converge to a state with higher energy, but one-determinant representations of excited states are rather inaccurate anyway.
In principle, in should be possible for an excited determinant to converge to a determinant with lower energy if and only if the initial converged solution is merely a local minimum. However, this is very unlikely if using a reasonable initial guess.


What is the signal at around 2090 wavenumbers in the nitroguanidine FTIR spectrum?

Its common Nujol, a hydrocarbon oil used to prepare paste-samples for IR from powdered materials. Hence the C-H valence around 2090 cm-1


Is there any selectivity between the C-2 and C-4 positions of pyridine in SNAr reactions?

True, the para (C4) and ortho (C2) positions give similar resonance structures but consider the fact that:

For an $\ce{S_{N}Ar}$ reaction the electron rich nucleophile must attack. Nitrogen, being highly electronegative, does draw electrons away from the ortho and para positions but because the electron clouds would be larger around nitrogen, it is harder for the nucleophile to approach the ortho than the para. You could think of it as the lone pair on Nitrogen being a steric shield of some sort.
An attack on the para position gives a symmetric intermediate. This would be lower in energy compared to the intermediate produced by an attack on the ortho position.

As such, the para product would be the kinetic product of the reaction whereas the ortho product would be the thermodynamic product. Statistically ortho should be attacked more often as well, as there are two positions whereas there is only one para position.


Does oxygenating tap water remove the tap water flavor?

Many public water systems spray water into the air or trickle over rocks (aeration) to let some dissolved gases bubble out (e.g. odorous $\ce{H2S}$). This also may make the water less acid and thereby precipitate dissolved manganese and iron.

When tap water treated with chlorine, bromine, ozone or other disinfectant is allowed to stand, open, then those chemicals also can escape.

However, I doubt anyone actually tastes dissolved oxygen (the human body can sense $\ce{CO2}$, but not oxygen, which is why miners kept canaries. In addition, one can taste the acid of $\ce{CO2}$ in soda water, and feel the bubbles bursting.)

So the practice of letting water stand and refrigerating is useful in improving flavor, but $\ce{O2}$ has little to do with it. Chilling water does increase the solubility of gases, though.


What advantages does the mole have over counting large numbers using SI prefixes?

Of course, it’s convention and has been so for a good century. And there is no real reason why this convention (and not a different one) happened in the first place — it is entirely conceiveable to define an ‘imperial mole’ so that the atomic mass of an element is equivalent to the same mass in ounces. 1 imperial mole of carbon atoms would then be twelve ounces of carbon atoms.

So there are basically two questions in this:

Why replace a large number with a unit just meaning ‘this certain large number’?
Why is the presently defined mole a good choice for this unit?

The thing about large numbers is that they are large. Almost everybody who grew up in a metric country can name at least three SI-prefixes: kilo, centi and milli. Thanks to IT, many people now also know mega and giga (and maybe tera), even if they don’t use metric units at home. But tera only gets you to $10^{12}$. We need some $10^{21}$ for moles.

I often work with milli- or micromoles of substances in my research. In plain numbers, that’s $10^{20}$ or $10^{17}$ — I don’t know those prefixes and thus would have to learn an entirely new subset. With the mole, everything one uses in the lab will nicely fall into something between nano and kilo.

It also helps to have a single unit there. Molar mass is expressed in grams per mole, concentrations in moles per litre and many more. If there were no unit, it would be simple grams per 1 or 1 per litre — precisely the reason why some people prefer to use rad or some other way to show radians rather than just writing the number. If the unit is there, you’re unlikely to forget it, you know if your calculations are good and more. If there was no name for this unit, it would have to be invented.

So if the mole didn’t exist, it should be invented for simplicity.

The good thing about the size definition of the mole is, as noted above, that it brings everything into one general range. Whether it’s mass, volume, concentration or amount, every unit is going to be prefixed by only a small subset of the SI-prefixes: kilo, milli, micro, maybe nano. Thankfully, those are the ones that are most used in everyday life, too (excluding nano and maybe micro).

It doesn’t really matter where one ends up. If the mole had originally been defined imperially,[1] that would be fine, I wouldn’t have memorised $12.01\,\frac{\mathrm{g}}{\mathrm{mol}}$ for carbon but $340.48\,\frac{\mathrm{g}}{\mathrm{mol}}$. That would create significantly larger molar masses, but that shouldn’t be a problem it should only mean that nano is more prevalent.

The thing about the definition of the unit is that it doesn’t bother $99.9\,\%$ of the scientists working with the unit. Seconds are defined (by SI) according to a number of transitions occuring in some weird isotope that I wouldn’t even know how to measure. I would explain a second by saying it’s the 86400th fraction of a day if someone asked me. Same meaning, different exactness. If the moles are soon defined by mere counting rather than weighing atoms then so be it. Nothing will change for me in practice. (Maybe the fourth digit of a molar mass but I don’t really count those.) So as long as a new definition doesn’t break anything, we’re good to carry on.

Notes:

[1]: With imperially, I meant to assume the following definition using imperial units:

  One mole is the number of atoms in exactly $12.00~\mathrm{oz}$ of atoms of the carbon isotope $\ce{^{12}C}$.


Does the distillation of alcohol cause the fractioning of isotopes and carbon-12 enrichment?

Does the distillation of alcohol cause the fractioning of isotopes[...]?

The short answer is it depends.

Distillation of ethanol from water can be done nearly to completion.  There are published methods for determining isotopic composition of sugars in fruit juices which involve fermentation of the juice to yield ethanol, followed by nearly quantitative distillation of the ethanol from the resulting brew.  Then, the 13C content of the ethanol is measured.  If the ethanol distillation step (or fermentation step) substantially altered the isotopic composition of the sugars, this method would not be suitable.  The main thing which determines the isotopic composition of the ethanol in this case is the type of photosynthetic metabolism used by the plants from which the juices or syrups are derived.  Quoting from the linked paper:

  The fruit juices or maple syrups are completely fermented with yeast, and the alcohol is distilled with a quantitative yield (>96%). The carbon-13 deviation (delta13C) of ethanol is then determined by IRMS. This parameter becomes less negative when exogenous sugar derived from plants exhibiting a C4 metabolism (e.g., corn or cane) is added to a juice obtained from plants exhibiting a C3 metabolism (most common fruits except pineapple) or to maple syrup.

If distillation is not done to completion, there can be a fractionation.  In fact this has been measured fairly accurately in 1990 for distillation of pure ethanol.  Here is a figure from that paper.

It plots the ratio $R_D$ of 13C to 12C of the distillate ($D$) normalized to the same ratio $R_Q$ for the starting material $Q$, as a function of the fractional yield or progress $\rho$ of the distillation (so $\rho = 0$ corresponds to no distillation and $\rho = 1$ corresponds to complete distillation).  It also plots the same curve for the remaining liquid $L$.

As you can see, at $\rho = 0$, when only a few molecules of ethanol have distilled, the liquid (unsurprisingly) has an unchanged isotopic composition relative to the starting material, and the tiny amount of vapor that does come over has an isotopic composition that is enriched in $\ce{^{13}C}$.  That's right, for pure ethanol at least, distillation is a process with an inverse isotope effect.
As $\rho \rightarrow 1$, the isotopic content of the distillate tends to 1.0,  consistent with the idea that for quantitative distillation, isotopic composition doesn't change.
At 50% completion, the distillate has about 0.2% more $\ce{^{13}C}$ than the starting material, and the remaining liquid has about 0.2% less $\ce{^{13}C}$ than the starting material.  So even though the composition does vary, the effect isn't very big.

  Does the distillation of alcohol cause [...] carbon-12 enrichment?

No, contrary to what I would have guessed before I read the literature, distillation of alcohol can cause carbon-13 enrichment.


You add dissicant until it suspends easily in the solvent when you swirl the flask. Some people say: "Until it snows".

When you add a bit of magnesium sulfate or other fine powdered dissicant, it forms a lump and goes to the bottom.
Many times it just sticks to the bottom when you swirl the flask.
You add more solid and swirl until it suspends and then slowly falls to the bottom, like it is snowing. At that point it is taking no more water and it has no sense to add more.


Relation of temperature and intensive property

After looking at the list in the Wikipedia article linked in the comment by M.Farooq, I suspect there are few properties that are guaranteed to be independent of temperature, an exception being molality and mole fraction in a closed system and absence of reactions. Other concentrations that are functions of volume are bound to be functions of temperature. Even in the presence of reactions, the atomic mole fraction is bound to be a constant (absent nuclear reactions, that is).

Standard molar quantities and some other properties are defined at fixed temperatures and are therefore strictly temperature independent, but that makes for a circular argument, and presumably this is not the type of property the question is about.


Experimental calculation of solubility product constant

It's probably an experimental error. The 'barely' may be important here.

For one, you may not be sure how accurate your initial molarities are. A mistake in one of them ruins it all. Titrate them properly and check. Titration usually gives two significant digits.

Also, since its a cube overall, a small mistake can get much larger. You may want more accuracy in your readings (one significant digit isn't good enough), and you need to improve on the 'barely'. Though a factor of 11 needs a variation by a factor of 2.3 in both readings. So I don't know.

Have you redone the experiment? Doing an experiment more than once is integral to getting good values.


I'm not sure what your question is. The reaction from NI3 is largely exothermic (Wikipedia gives –290 kJ/mol as the reaction enthalpy), which is consistent with its explosiveness. The reaction from NI3 · NH3 is also exothermic: you quote the heat of formation of the NI3 · NH3 as being +146 kJ/mol, and the heat of formation of product NH4I is –202 kJ/mol. Both are largely favored entropically because they release gas.


https://physics.stackexchange.com/questions/45838/how-to-deduce-e-3-2kt

http://hyperphysics.phy-astr.gsu.edu/HBASE/kinetic/molke.html#c1

http://en.wikipedia.org/wiki/Kinetic_theory#Properties

These contain proof at various level . Choose according to your mathematical level


Is there a relationship between Free Radicals and Beta Particles?

There is high-energy in the sense of chemical reactivity, and then there is high energy in the sense of nuclear reactivity. The energy associated with the latter is many orders of magnitude greater than the energy associated with the former, so that there is generally no direct interaction between the two (where the free radical is chemically reactive, and the beta particle arises from nuclear reactivity).

A free radical is a chemical species, whose electron configuration is such that it is highly reactive towards other chemical species. However, this energy is far too low to affect any nuclear reactivity.

A beta particle is a high-energy electron or positron emitted by the radioactive decay of an atomic nucleus. As these particles are emitted from the nucleus of the atom, the electron configuration (i.e. a free radical) is irrelevant; only the stability/reactivity of the nucleus determines whether an atom will emit a beta particle, not the chemical species.

If any of this seems incorrect or unclear, please do not hesitate to ask for clarifications in the comments.


Can carboxylic acids be used in the aldol reaction?

Carboxylic acids form enolates

I think I am right in saying that carboxylic acids do not form enolates because the carboxyl proton is much more acidic and so this reaction could not take place under basic conditions...

They do, if you beat them to it:

Dissolve the carboxylic acid in a polar, aprotic solvent, such as 1,2-dimethoxyethane, 1,4-dioxane, tetrahydrofuran, etc.

Add one quivalent of sodium hydride to deprotonate and form the carboxylate.

Add one equivalent of a strong base, such as LDA (lithium diisopropylamide) to deprotonate the carboxylate at the $\alpha$-position.

Add your aldehyde

What about acyl chlorides?
Adding a base, such as Hünig's base (plus catalytic amounts of a Lewis acid) to an enolizable acyl chloride under anhydrous conditions gives rise to a different pathway: a ketene can be formed!
How do ketenes react with aldehydes?
In the presence of a Lewis acid, they form $\beta$-lactones!


What is the major product of the reaction of 2-ethoxypentane with hydrogen iodide?

Strictly speaking, none of those answers are correct, provided your stem is right.
The reaction will start with protonation of the ether to give you the following species:

And the iodide nucleophile can attack either side via $S_{N}2$ mechanism. Since the left side is more accessible, it'll probably attack there first giving you:

Then, since we're still in the presence of HI and it is a strong acid, it protonates the alcohol and converts it into a corresponding alkyl iodide via either $S_N2$ like in the example above or $S_N1$:

A general rule of thumb here for the future reference is that in a hydrolysis of an ether with a strong hydrohalogen acid (HI, HBr), you get a mix of two corresponding alkyl halides with halogen where the oxygen used to be.


It might not be the only factor, but I think that the steric effect of the huge phenyl groups in Wittig ylides might contribute in making them less nucleophile.
If you give a fast look at a "space filling" model of the two molecules, you see that the negatively charged carbon in a Wittig Ylide is much more masked by the phenyl groups

Note that in order for a nucleophile-electrophile interaction to occur, an electrophile should approach to the carbanion's HOMO orbital (approximatively orthogonal to the screen in both the pictures).


The one solvent I see referenced the most for polyurethane foams is N-methyl-2-pyrrolidone or as the the cool kids call it: NMP. You don't have to just take my word for it a certain supplier that I have no affiliation with sells it for this purpose: Spray Foam Systems NMP.

As far as the purpose of ethanolamine, it must either be reacting to reduce chain length.  If you look at the urethane group below the Carbon bonded to two oxygens and a nitrogen is susceptible to attack by the basic ethanolamine. This reaction would break the molecular bonds making the polymer chains shorter and thus more mobile and more easily dissolved, thus saving time. This explination is a little hand-wavy but I didn't want to go deep into organic reaction mechanisms.


Why are hydrocarbon flames yellow or blue?

It depends on the amount of oxygen available and on the ratio of hydrogen to carbon in the fuel.

Hydrogen burns with an almost invisible flame no matter how much oxygen is present, but solid carbon burns with a yellowish (black-body) flame. If the fuel combusts in a multi-step process where all reactants are gaseous, you get a bluish flame. For example, the combustion of methane produces methanol, then formaldehyde, formic acid, carbon monoxide etc.

The combustion of acetylene at lower temperature, with little oxygen, however, starts with cracking of the C-C triple bond, releasing carbon particles that have a luminous flame.

So the greater the hydrogen:carbon ratio, the easier it is to get a blue flame.


The Combining of Carbon Dioxide and Hydrogen

I believe you are referring to the Sabatier process. $$\ce{CO2 + 4H2 -> CH4 + 2H2O}$$

The reaction uses a $\ce{Ni}$ catalyst and is done at $300-400\,^{\circ}{\rm C}$ and high pressures. The reaction is exothermic, $\Delta H = −165.0~\mathrm{kJ\,mol}^{-1}$, but requires initial heating in order to start the reaction off due to the activation energy.

Read more on Wikipedia.


Internal and external pressure during a gas expansion

If the frictionless massless piston is not moving, then internal and external pressures must be equal.  If the piston has mass, the internal pressure must be greater when it is stationary, to support the weight.  If there is static friction, all bets are off.

Suppose the piston were a right cone rather than a right cylinder.  Would the excess surface area change the outcome, all other things being equal?


How to oxidise iron in spinach to iron(III)

Use ceric ammonium sulfate.[1,2]  Quoting from reference 2:

2 Reagents
(i) 0.25 per cent ceric ammonium sulphate in 8.8 N sulphuric acid.
(ii) 40 per cent potassium thiocyanate solution.
(iii) (a) Stock standard ferrous iron solution - 100 μg iron per ml.
(b) Working standard ferrous iron solution - 0.3 to 10 μg iron per ml
2.2 Procedure
To five ml of clear water sample in a test tube 0.5 ml of ceric ammonium sulphate solution is added and mixed thoroughly by shaking. Finally, 0.5 ml of potassium thiocyanate solution is added and again mixed properly. Exactly 5 minutes after addition of thiocyanate solution, the absorbance of ferric thiocyanate complex is measured at 480 nm in a spectrophotometer (Spekol, Carl-Zeiss, Jena).
A series of freshly prepared working standard ferrous iron solutions (upto 10 ppm iron) were treated similarly and the absorbances measured at 480 nm. From the calibration curve, plotted with concentrations of different standard iron solutions against absorbances, the iron content of the unknown sample is evaluated.

References

Willard, H. H.; Young, P. Ceric Sulfate as a Volumetric Oxidizing Agent. Ii. Determination of Iron. J. Am. Chem. Soc. 1928, 50 (5), 1334–1338. DOI: 10.1021/ja01392a013.

Goswami, D. C.; Kalita, H. Rapid Determination of Iron in Water by Modified Thiocyanate Method. DSJ 1988, 38 (2), 177–182. DOI: 10.14429/dsj.38.4835.
Also available via CiteSeerx. (The pdf icon has a cached version of the article.)


Why are tertiary alkyl halides obtained in higher yield than secondary or primary during free radical halogenation of alkanes?

The reason for the ordering is that tertiary radicals have a lower energy (and are thus easier to form) than secondary radicals, which are in turn easier to form than primary radicals. This is because of a phenomenon called sigma conjugation - the weak overlap between the half filled orbital and the adjacent side group orbitals gives a slight lowering of energy.
This means that it is easier to remove a hydrogen to give a tertiary radical intermediate than to give a secondary radical intermediate and so on.
This is similar to Markovnikov's rule.

The reason a tertiary product is not formed is that for butane there is no tertiary centre to begin with - nowhere that a Cl radical can react to give a tertiary butyl radical. Forming a tertiary product would require a rearrangement of the carbon chain.


Moment of inertia for a triatomic molecule

The moment of inertia is defined for a molecule as it would be for any other solid: you consider the nuclei as points with mass, ignore the mass of the electrons, and you can then calculate either the scalar moment of inertia or the tensor of the same name.

The moment of inertia of a molecule is involved in the quantification of its rotational movements. Of the top of my head, the two areas of chemistry where this plays a big role are:

rotational spectroscopy: the rotation of a molecule is quantized, and the value that the angular momentum can take are related to the moment of inertia of the molecule.
the rotational entropy of a molecule is linked through statistical mechanics to its inertia tensor.


What are modern approaches to describe and/or quantify aromaticity?

I'm more familiar with the Electron Localization Function approach to aromaticity commented by Quantum AMERICCINO, it's been proposed that the average value of the $\text{ELF}_{\sigma}$ and $\text{ELF}_{\pi}$ bifurcations would be used for a general scale of aromaticity of a molecule or molecular cluster. The points of bifurcation can be interpreted as a measure of the interaction among the different basins or, simply, electron delocalization. Although the ELF depends on the density (i.e. no separate information of $\pi$ or $\sigma$ bond), a topological analysis on a separated ELF formed only by the $\pi$ orbitals and other formed by the $\sigma$ orbitals can be done (also you can do a separation for the spin $\alpha$ and $\beta$ contributions to the ELF). [J. Chem. Theory Comput. 2005, 1 (1), 83–86.]

There is another approach based on how a bond responds when its electron density is perturbed by a finite amount. Specifically, they used the interaction coordinate (IC) to see this. Quoting

  If the delocalization in the aromatic system is more extensive, for constraining a bond by unit displacement from the equilibrium geometry, it requires more energy. So the response of the other internal coordinates will be greater for constrained optimization and is proportional to the effective delocalization and hence related to the aromatic stabilization. The IC is a measure of the response of a bond (its electron density) for constrained optimization when another bond or angle is stretched by one unit (its electron density is perturbed by a measured amount).

Instead of taking the full molecule, they take only the aromatic skeleton to account the electron density perturbation on the system. Also, it says that this is very general yet, so the IC approach can take new directions.
[J. Phys. Chem. A 2016, 120 (18), 2894-2901.]

Another approach is making a relationship between the amount of the electron density (due to ring size, number of $\pi$ electrons, etc) and the non-nuclear magnetic shielding (cf. NICS).
[Phys. Chem. Chem. Phys. 2010, 12 (39), 12630–12637.]


Can we force reactions with positive delta G?

The change of the Free / Gibbs enthalpy $\Delta_{\text{R}}{}G$ consists of a contribution by reaction enthalphy $\Delta_{\text{R}}H$, and a contribution by reaction entropy, $\Delta_{\text{R}}S$:

$$ \Delta_{\text{R}}{}G = \Delta_{\text{R}}H - T \cdot \Delta_{\text{R}}S $$

Hence, the answer to your question is no.  By variation of temperature $T$, for example, a reaction previously endergonic ($\Delta_{\text{R}}{}G > 0$) may become exergonic ($\Delta_{\text{R}}{}G < 0$), as summarized here.


Nucleophilic attack on cyclic intermediates in dibromination and oxymercuration reactions of alkenes

Just to clear your doubts these are electrophilic addition reactions

Case 1: oxymercuation-demercuration

The reaction is initiated by electrophilic addition of $\ce{Hg^2+}$(mercuric) ion to the alkene to give an intermediate mercurinium ion, whose structure resembles that of a bromonium ion. Nucleophilic addition of water as in halohydrin formation, followed by loss of a proton, then yields a stable organomercury product.
The final step, demercuration of the organomercury compound by reaction with sodium borohydride, is complex and involves radicals.

The regiochemistry of the reaction corresponds to Markovnikov
addition of water; that is, the $\ce{-OH}$ group attaches to the more highly substituted carbon atom, and the  $\ce{-H}$ attaches to the less highly substituted carbon.

  When the π electrons of the alkene attack the mercury cation, a
  three-membered intermediate forms in which both carbons share a single
  bond with the mercury and a partial positive charge. It turns out that
  rearrangements are unlikely to occur from this intermediate (as
  opposed to the discreet carbocation intermediate in the acidic
  hydration reaction).

A water molecule then attacks the more substituted carbon, which bears a greater share of the positive charge. Cleavage of the carbon-Hg bond is accomplished with a reducing agent called sodium borohydride $\ce{NaBH4}$.

See images:

So where is the carbocation present?

The end result of an oxymercuration-reduction reaction is the Markovnikov addition of water to an alkene without rearrangement

The regioselectivity on this reaction is partly based on inductive effects resulting from the shifting of electrons in a s bond in response to the electronegativity of nearby atoms. I believe the other reason is based on hydride shift which converts the carbonation into a tertiary one and this is thermodynamically favourable. (Not entirely sure on second reason though)

Case 2: Dibromination

Halogenation of Alkenes: Addition of $\ce{Br2}$

Follows the reaction sequence (proposed by George Kimball and Irving Roberts)

  that the reaction intermediate is not a carbocation but is instead a
  bromonium ion, $\ce{R2Br+}$, formed by electrophilic addition of $\ce{Br+}$ to the
  alkene.

The bromonium ion is formed in a single step by interaction of the alkene with $\ce{Br2}$ and simultaneous
loss of $\ce{Br-}$.

See image:

I have updated to include an even more detailed mechanism:


Gallium-induced structural failure of aluminium

As you state, it forms an amalgam, just as mercury does. @Jon Custer's explanation is correct, and furthermore, the alloy does not form a protective oxide, so the aluminum oxidizes to $\ce{Al2O3}$ and quickly crumbles. This happens even faster with an Hg-Ga alloy!

BTW, gallium should be safe enough, though expensive, but I would not like to work with mercury without a hood, and without careful containment.

Thanks for the question; this encouraged me to view that amazing demo on Vimeo!


Mechanically speaking, it is not possible to separate $\ce{SiO_2}$ from these minerals. The only way to do it is carrying out a chemical transformation. Here is an example of such a treatment.
Mix a sample of one of these minerals with $\ce{Na_2CO_3}$ in excess, and heat it in a platinum crucible with a Bunsen burner. When the substance is liquid, wait $10$ minutes. Let it cool down. The mineral is now transformed into sodium silicate $\ce{Na2SiO3}$ plus some oxides and carbonates. If I choose the specially simple olivine $\ce{MgFeSiO_4}$ as an example, the following reaction may happen : $$\ce{MgFeSiO_4  +  Na_2CO_3 -> Na_2SiO_3 + MgCO_3 + FeO}$$ The reaction is somewhat more complicated, because usually $\ce{FeO}$ is oxidized in air into $\ce{Fe_2O_3}$, and $\ce{MgCO_3}$ is partially decomposed into $\ce{MgO}$ and $\ce{CO_2}$. Furthermore, the sodium silicate $\ce{Na2SiO3}$is partly transformed into $\ce{Na4SiO4}$. Fortunately this side reactions do not really modify the rest of the process.
Anyway, put the crucible in a small beaker. Add $20$ mL water. Let it boil for a couple of minutes. When the substance is pulverized, add an excess of diluted $\ce{HCl}$. Heat on a water bath until the substance is dry. Add $2$ mL concentrated $\ce{HCl}$. The following reactions happen : $$\ce{Na_2SiO_3  +  2 HCl ->  H_2SiO_3 + 2 NaCl}$$ $$\ce{MgCO_3 + 2 HCl-> MgCl_2 + CO_2 + H_2O}$$ $$\ce{Fe_2O_3 + 6 HCl-> 2 FeCl_3 + 3 H_2O}$$ After $10$ minutes, add some hot water and filtrate. $\ce{H_2SiO_3}$ remains on the filter. Calcium, magnesium and iron chlorides are soluble into water and are eliminated.  Then $\ce{H_2SiO_3}$ can later be heated to obtain $\ce{SiO2}$ $$\ce{H2SiO3 -> SiO2 + H2O}$$  This is one way for obtaining $\ce{SiO2}$ from the proposed silicates. Other processes are known, using for example fluorhydric acid $\ce{HF}$. But, of course, using this acid is not recommended, because of its toxicity.


Is an iron (II) sulphate solution supposed to turn cloudy when silver nitrate is added?

According to this resource for solubility rules, $\ce{BaSO4}$ is insoluble and $\ce{Ag2SO4}$ is slightly insoluble, which would explain the the cloudiness of each solution, as well as why the $\ce{BaSO4}$ solution is cloudier than the $\ce{Ag2SO4}$ solution.

According to this table, $\ce{BaSO4}$ has a $K_{\mathrm{sp}}$ of $1.1 \times 10^{-10}$. Comparably, $\ce{Ag2SO4}$ has a $K_\mathrm{sp}$ of $1.2 \times 10^{-5}$, which is considerably higher than $\ce{BaSO4}$.

If you are not aware the $K_{\mathrm{sp}}$ is a mathematical measurement describing the solubility of a salt at a certain temperature (usually $25^{\mathrm{o}} \mathrm{C}$), and is found by multiplying the concentration of the salt's dissolved ions, each raised to the power of their respective stoichiometric coefficients. The higher the $K_{\mathrm{sp}}$, the more soluble a salt is. For example, the $K_{\mathrm{sp}}$ for $\ce{Ag2SO4}$ would be given by the expression: $K_{\mathrm{sp}}=[\ce{Ag+}]^2\cdot[\ce{SO4^{2-}}]$.


The BHT is only present to prevent small amounts of potential radicals polymerising the monomer. One option is to simply consume it i. e.  Put in one extra per cent of your initiator.
Otherwise,  similarly to basic alumina, use a basic resin. Once you consume the BHT,  your monomer is prone to polymerisation, so you may need to run it into your reactor under your reaction conditions e.g. flushed with nitrogen, pre-cooled, etc


Predict the products of the controlled oxidation of 2-methylpentan-2-ol

Tertiary alcohols are resistant to oxidation in neutral or alkaline solution, but are readily oxidized by acidified oxidizing agents.

In your case two products will be acetone and propionic acid.

http://chemistry.tutorcircle.com/organic-chemistry/tertiary-alcohol.html


According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), the preferred name (PIN) for the parent ring component consisting of three benzene rings fused in a linear manner is the retained name anthracene.

Subsection P-25.1.1, Table 2.7, Entry (11) shows that anthracene has a special numbering:

According to Subsection P-14.4 (a), this fixed numbering must be used in PINs and in general nomenclature.

The prefixes ‘hydro’ and ‘dehydro’ are used to indicate addition and subtraction, respectively, of hydrogen atoms (see Subsection P-31.2.1).  Hence, the PIN for the ring component of the structure given in the question is 9,10-dihydroanthracene:

The corresponding monovalent substituent groups derived from such ring systems by the removal of one hydrogen atom are named by replacing the final ‘e’ of the name of the hydrocarbon by ‘yl’ (see Section P-29). Thus, the name for the substituent group derived from 9,10-dihydroanthracene is (9,10-dihydroanthracen-2-yl):

(Note that locants are placed immediately before that part of the name to which they relate (see Subsection P-14.3.2); i.e. ‘…anthracen-2-yl’. Thus, the name ‘…-2-anthracenyl’ given in the question is not correct.)

The substitutive prefixes for the remaining groups are
‘methyl’ $(\ce{-CH3})$,
‘hydroxy’ $(\ce{-OH})$,
‘oxo’ $(\ce{=O})$, and
‘carboxy’ $(\ce{-COOH})$.

The name for the complete substituent group is (7-carboxy-1,3,4,6-tetrahydroxy-8-methyl-9,10-dioxo-9,10-dihydroanthracen-2-yl):

The bases for carbohydrate names are the structures of the parent monosaccharides in their acyclic form (see Subsection P-102.2).
Alditols are named by changing the ending ‘ose’ in the name of the corresponding aldose into ‘itol’ (see Subsection P-102.5.6.5):

The prefix ‘anhydro’ expresses the corresponding intramolecular ether, which is formally obtained by elimination of water from two hydroxy groups of a single molecule of a monosaccharide or monosaccharide derivative (hence the structure is commonly called an intramolecular anhydride). The prefix ‘anhydro’ is preceded by a pair of locants identifying the two hydroxy groups (see Subsection P-102.5.6.7.1).


All electrochemistry is relative.  And any time you have a reduction or oxidation reaction, you are dealing with an electrochemistry problem.

NADH is "more energetic" than NAD+ because the hidden assumption is that energy levels are relative to reaction with the atmosphere.  NADH can react with oxygen in the atmosphere to release energy.  But on a planet with no oxygen and lots of hydrogen in the atmosphere, NAD+ would be the "more energetic" molecule.

Another less exotic example is anaerobic fermentation in yeast.  NADH builds up in glycolysis, and must be converted back to NAD+ through formation of ethanol + CO2.  That process consumes energy; thus, in anaerobic environments, often times NAD+ can be thought of as a higher-energy molecule than NADH.


Is perovskite symmetric under exchange of A and B atoms?

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


Why can’t lanthanum through lutetium and actinium through lawrencium all be in group 3?

Since I am neither clairvoyant nor able to travel back and forth in time, I can only guess what IUPAC was thinking.  But there is no guessing that the inner transition elements do not completely fit in with Group 3 chemically.
One major chemical difference is in the oxidation states the elements may achieve, and thus the stoichiometry of compounds with nonmetaks that have strong ionic character.  Whereas the Group 3 elements overwhelmingly form compounds in the +3 oxidation state, other oxidation states are more likely for at least some inner transition elements.  While in some other groups there is more variance linked oxidation states, this is little seen in Group 3 whereas, below, variance in oxidation states is more common with inner transition metals.  Especially note the higher oxidation states in the examples that follow, which would not be expected in a group generally defined by elements normally having only three valence electrons.
Among lanthanides, cerium achieves a +4 oxidation state commonly enough to be called "ceric" and be available as a reagent for redox reactions, as in ceric ammonium nitrate and cerium(IV) sulfate.  Cerium can also form +4 as well as +3 ion cores in the metal by either including or excluding its $4f$ electron in the metallic band structure, leading to a transition between two FCC phases [1].  At the opposite end, europium accesses the +2 oxidation state much more readily than a true Group 3 element, leaving half-filled valence shell, and thus overlaps the heavier alkaline earth metals in its chemical properties.  Wikipedia reports that

In anaerobic, and particularly geothermal conditions, the divalent form is sufficiently stable that it tends to be incorporated into minerals of calcium and the other alkaline earths.

With actinides, oxidation states departing from the +3 predominance of Group 3 are more common.  Well-known examples include thorium forming a highly refractory oxide in the +4 oxidation state and uranium achieving oxidation states as high as +6 when combined with fluorine.
Reference
1.
Johansson, B., Luo, W., Li, S. et al. Cerium; Crystal Structure and Position in The Periodic Table. Sci Rep 4, 6398 (2014). https://doi.org/10.1038/srep06398


According to this Wikipedia page:

  Martensite, named after the German metallurgist Adolf Martens
  (1850–1914), most commonly refers to a very hard form of steel
  crystalline structure, but it can also refer to any crystal structure
  that is formed by diffusionless transformation.1 It includes a class
  of hard minerals occurring as lath- [emphasis mine] or
  plate-shaped crystal grains.

Lath basically means the shape of a woven screen when viewed close up, or like a grid of alternating appearance.


You need to factor in the electrons that are being moved around

Chromium is doing this:
$\ce {Cr^{2+}_{(aq)} + 2 e- ->  Cr_{(s)}}$

Aluminum is doing the reverse of this:

$\ce {Al^{3+}_{(aq)} + 3 e- -> Al_{(s)}}$

Now, like a math equation flip around the aluminum reaction.
And balance out the electrons

$\ce {[Cr^{2+} + 2 e- ->   Cr_{(s)}] \times 3}$

$\ce {[Al_{(s)} -> Al^{3+} + 3 e- ] \times 2}$

$\ce {3Cr^{2+} + 6 e- + 2 Al_{(s)} -> 3Cr_{(s)} + 2 Al^{3+} + 6 e-}$

Hope this helps


How can I specify pyridine as a substituent?

The issue, I think, is to specify how the pyridine is connected. This means you have to identify which carbon on the pyridine ring is connected. This picture shows the numbering:

So, using your example the structure and names of the various possibilities would be (using connections on the 3-position of the benzonitrile ring as an example):

I'm sure you can generalise from this example.


What is the convention on color of molecular orbitals in publication figures?

The colours represent the signal/phase of the molecular orbital, i.e. positive or negative.
Below is the HOMO (made with ChemCraft) of butadiene (PBE-D3BJ/def2-SVP with ORCA) as an example.

Figure 1: HOMO of butadiene (cutoff 0.03)

ChemCraft shows by default negative phase as blue and positive as red.
VMD seems to follow the same convention.

You can certainly change the colours if you want.
Below is how the controls for surfaces in ChemCraft look like:

Figure 2: surface controls in ChemCraft

Other programs may use different conventions, such as NBOView (yellow and blue).


How can I calculate the energy of a solution by knowing all of it's elements?

Since you ask specifically for internal energy, would it be that you already have some ideas?

So what is internal energy? It is the energy due to the interaction among the species plus the energy due to the interaction between the species and any external field. That is, the energy it takes to "assemble" the system.

Then for gaseous species, it would be mainly the bonding energies of the molecules. Those are usually inferred from experiments. And you can look up tables in textbooks and IUPAC website.
For ionic crystals, you would reason this way:  first ionize the species and them find the Columb interaction among them in crystal. Ionization energies are measured. Columb interaction is probably calculated and is a nightmare to calculate.
The further problem is from NaCl(s) to (aq) where the solvation involves quite a bit of interactions between the ions and water. For the case of Na+ Cl-, water molecules do not form regular cages around them, even though we CAN study the ideal case of regular cages. I think there has been some minimal models which can possibly guide a simulation study for the solvation energy. Though I don't think there has been any simulation study or any theoretical study that is sufficient to predict solvation energy.  I can be wrong. On the other hand, direct measurement should at least be able to measure the heat change during solvation.

So in digression, I don't think you can find the complete energy too well.


Is there any replacement for ethyl mercaptan in LPG?

Why would one want to replace it?

The choice of ethanethiol is a reasonable compromise between its toxicity and the extremely low detection limit around 1 ppb (1mg/1000 kg).


In accord with this forum's policy on not providing health advice, I  will just comment on the chemistry surrounding any created Al3+ interacting with the superoxide radical anion (which is incidentally, present in the human body). As a source, here is a 2014 article by Tushar Kanti Das, et al, published in Archives of Neuroscience, available here.
The authors explicitly references, in Figure 4, “Formation of Aluminum Superoxide Semi reduced Radical Ion and Aluminum Superoxide Complex (43)”, with described reactions proceeding as follows (also adopting notation and water complexing from yet another article):
$$\ce{[Al(H2O)4](3+) + O2•− <-> [Al(O2•−)(H2O)4](2+) }$$
$$\ce{[Al(O2•−)(H2O)4](2+) + Fe(3+) --> O2 + [Al(H2O)4](3+) + Fe(2+)}$$
And, in acidic conditions:
$$\ce{[Al(H2O)4](3+) + O2•− + H+ <-> [Al(O2•−)(H+)H2O)4](3+)}$$
$$\ce{[Al(O2•−)(H+)H2O)4](3+) + [Al(O2•−)(H+)H2O)4](3+) --> 2 [Al(H2O)4](3+) + H2O2 + O2}$$
where the above chemistry suggests a role for Al3+ in possibly assisting with the subsequent production of miscellaneous powerful radicals associated with a Fenton reaction  as sourced from an iron ion presence (as in blood).
Now, as to the possible originating source of Al3+ itself, in the particular context of baking with Aluminum pans, and further washing said aluminum ware in a dishwasher also employing Baking Soda, is that the latter NaHCO3 on warming in aqueous solutions, apparently will attack Al metal, commencing with the reaction:
$$\ce{ NaHCO3 + H2O + Heat --> NaOH + H2O + CO2 }$$
And further, to quote Wikipedia:

Sodium aluminate is also formed by the action of sodium hydroxide on elemental aluminium which is an amphoteric metal. The reaction is highly exothermic once established and is accompanied by the rapid evolution of hydrogen gas. The reaction is sometimes written as:
$$\ce{ 2Al + 2NaOH + 2H2O → 2NaAlO2 + 3H2 }$$
however, the species produced in solution is likely to contain the [Al(OH)4]− ion or perhaps the [Al(H2O)2(OH)4]− ion.[6]

This problem with Aluminum ware is well known (see warnings, for example, here, and also here).
I hope this, in part, advanced chemistry presentation nevertheless is of assistance in answering your question "Is using oxidized aluminium baking forms safe?".


How is nomenclature decided in this case (formyl vs oxy)?

According to the new recommendations of IUPAC nomenclature rules, when choosing the senior parent structure, you have to first consider the senior parent structure, which has the maximum number of substituents corresponding to the principal characteristic group or senior parent hydride in accord with the seniority of classes (functional groups). Then, if there is a choice, choose the principal chain, which has the greater number of skeletal atoms. Next in order is the chain, which has the greater number of multiple bonds, and then has the greater number of double bonds (This is according to Blue Book in 2013, which was displayed in 5 steps you have mentioned from Wikipedia).
Accordingly, your most senior function is carboxylic group, which would have the highest priority:

Then, there is a choice to choose the principal chain, which has the greater number of skeletal atoms. Both choice have six carbons, but the numbered chain indicated in the image has additional function, thus it is the principle chain. You also need to consider the stereochemistry if they are indicated. Here, there is a double bond in trans-orientation. That should be included in the name. Although, $\ce{C}$-5 is a chiral carbon, its stereochemistry is not indicated so you can leave it alone. Therefore the correct IUPAC name is: (2E)-2,5-dimethyl-6-oxohex-2-enoic acid.


Is iron(III)nitrate changed and if so how?

If a stoichiometric amount of $\ce{SCN-}$ is added then the reaction is mostly

$$\ce{Fe^3+(aq) + SCN-(aq) <=>Fe(SCN)^2+(aq)}$$

If an excess of of $\ce{SCN-}$ is added then the reaction is mostly

$$\ce{Fe^3+(aq) + 2SCN-(aq) <=>[excess SCN-] Fe(SCN)2^+(aq)}$$


Methanation is a process whereby synthesis gas is converted to methane.

$\ce{CO + 3H2 -> CH4 + H2O}$

Methanol, like nearly all organic molecules, can be gasified to form synthesis gas at elevated temperatures.

$$\ce{CH3OH -> CO + 2 H2}$$

(Note: the gasification reaction is actually a complex balance of many types of reactions.  Often times mixtures of $\ce{CO}$, $\ce{H2}$, $\ce{H2O}$, and $\ce{CO2}$ are obtained.  The reaction above is the ideal case where complete conversion is achieved.)

Thus a process for converting methanol to methane could proceed via gasification followed by methanation.

However, since methanol is made commercially from methane by essentially the reverse process, this route is unlikely to be economically viable!


Is there an actual distinction between through-space charge transfer and excimer emission?

The excimer / exciplex ground state is wholly repulsive, molecules $\ce{M  and M^*}$ diffuse together to form excimer $\ce{MM^*}$ the excimer or exciplex if formed from different species. The complex does not exist in the ground state. Emission is observed from both $M^*$ and $MM^*$. ($MM^*$ can also diffuse apart before fluorescence) The reason the excimer/exciplex emission is broad and structureless is because the ground state is repulsive. The exciplex is formed from different species so there has to be some charge transfer character, i.e. potential energy is sensitive to solvent polarity. Intramolecular excimers and exciplexes were first synthesised and studied in the late 1970's.
CT complexes are intimate in the ground state (via $\pi$ stacking etc.) and the CT absorption is generated upon absorption $(DA) + hv\to (D^+A^-)$. Any emission is then from this CT state plus normal fluorescence from either D* or A* should they be isolated in solution (as DA is in equilibrium with D and A) and then excited.
For details of kinetic see J.B. Birks, 'Photophysics of Organic Molecules' 1970, publ Wiley, and Organic Molecular Photophysics' 1975, publ Wiley. M. Gordon & W. Ware 'The Exciplex' 1975, Academic Press.


Should the relatively low solubility of a base be considered in the calculation of the pH?

The $K_\mathrm{sp}$ of  $\ce{Ba(OH)2}$ is $0.005$.

Therefore, the solubility is great enough that 0.055M  $\ce{Ba(OH)2}$ can all be in solution.

Secondly, you are diluting the $0.055~\mathrm{M}$ solution, so the concentration will be even less.

Finally, even if there was too much  $\ce{Ba(OH)2}$ to dissolve in water, the $\ce{HCl}$ would react with the $\ce{OH-}$ that is in solution, causing more  $\ce{Ba(OH)2}$ to dissolve.


Titration of sulfuric acid with sodium hydroxide

That is one way of doing it.

In reality, this is a stoichiometric question where you are trying to find the molarity of an unknown polyprotic acid by titrating it with a strong base (woo-hoo!) of known molarity. I prefer using this method as it is easier to see the relationship between the units as well as use chain-ink conversion which most people are familiar with.

We know that

$$ \text{M} = \frac{\text{mol}}{\text{L}} $$

Therefore,

$$ 55 \ \text{mL} \ \ce{ NaOH} * \frac{250 \ \text{mmol} \ \ce{NaOH}}{1000 \ \text{mL} \ \ce{NaOH}} * \frac{1 \ \text{mmol} \ \ce{H2SO4}}{2 \ \text{mmol} \ \ce{NaOH}} * \frac{1}{ 35 \ \text{mL} \ \ce{H2SO4}} = $$

Which equals

$$ 0.19643 \ \frac{\text{mol}}{\text{L}} $$


Are the odours added to natural gas completely flammable?

On looking it up on Google and a few more places, it turns out that natural gas is primarily methane, along with a few other hydrocarbons.
Well, I have found out through some searches online that in the case of propane based fuel, like LPG (Liquid Petroleum Gas) in India, ETHYL MERCAPTAN is a popular option. It's smell is described as a

  strongly disagreeable odor that humans can detect in minute
  concentrations

And that it's smell resembles that of

  leeks, onions, durian or cooked cabbage, but is quite distinct.

(source :Wikipedia)

And also, most important point of all, it is highly explosive.

This is what the New Jersey Department of health and safety has to say:

(Source : This document)

So yeah, You're right. Based on this example, I think it logically follows that these "additives" are flamable and result in products that are ODOURLESS.


What is the exact meaning of GAMESS (US) SCF density convergence criteria?

The difference is calculated in the subroutine DDIFF (scflib.src) and corresponds to the largest absolute change in the density matrix


Must a desiccant be made anhydrous before it can be used for drying?

In the oil and gas industry, glycols are used in a scrubber to remove water from natural gas [methane] for a number or reasons: water is very soluble in glycols, water is extracted into “dry” glycols due to equilibrium forces, and there is efficient contact between the gases and the liquid.  Another huge part of the stribber is the regeneration portion of the system, the stripper – getting to the point of your question.  Heat is applied to the water/glycol mixture after having been in contact with the “wet” methane, driving off the water and regenerating the sorbent, the glycol.  The glycol can then be recycled.  Although I have described this as a batch process, the actual process is continuous.
In an industrial setting, “throwing something away” is used as a last resort.
Below is a schematic of a glycol dehydrator and regeneration unit.

https://www.sciencedirect.com/science/article/pii/S2352484719301039#fig1


What exactly happens when vegetable oil is partially hydrogenated?

You actually don't need a polyunsaturated fat to be partially hydrogenated. The "partial" refers to the macroscopic view of the fat, rather than a per-molecule basis. That is, if you have, say, one mole of monounsaturated fats and hydrogenate the double bond in half a mole of them, then you've partially hydrogenated the original mole of fat. Taken individually, each molecule has either been hydrogenated or it's still monounsaturated - no single molecule has been "partially hydrogenated". However, the mole taken as a whole has only been partially hydrogenated - 50% of the double bonds that were originally present have been hydrogenated.

Regarding trans fats, the important thing to realize is that (parts of) the hydrogenation reaction is reversible. The addition of molecular hydrogen is not concerted. Each hydrogen is added in a step-wise fashion (roughly). It's in that one-hydrogen added, one-to-go stage where the reaction can reverse course and go back to a double bond. But prior to reforming, the catalyst has broken the double bond and you have free rotation around it. Which way it's put back together is a matter of chance.

So you start with a fat which has all cis bonds. (Because the stereo and regio-selective enzymes that make them only make cis fats.) You then start the hydrogenation process. Each time you add hydrogens some goes to completion, and you end up with saturated bonds. But a fraction of the reactions get part way and then reverse. But as the hydrogenation catalyst is not stereo- and regio-selective, it's just as happy reversing to a trans bond as it is to a cis bond. So it's those bonds - the ones which go part way and then reverse - which end up as the trans fat in the final product.

The reason why fully hydrogenated fats don't have this issue is because they're produced with an excess of hydrogen. Yes, some of the double bonds react part way and then reverse to trans fats, but the catalyst is not stereo- and regio-selective, and it's able to convert the trans bonds to saturated bonds just as easily as the cis bonds. All the trans fats that get formed in the reaction are subsequently destroyed by the complete hydrogenation.


In Williamson ether synthesis, the reactive species is an alkoxide, which can act as a strong nucleophile or base. When reacted with an alkyl halide, if the alkoxide acts as a nucleophile in an SN2 reaction, the ether product is formed. If the alkoxide acts as a base in an E2 reaction, then an alkene is formed.

So when considering a the reactivity of alkoxides, there is a competition between SN2 and E2. Which pathway will dominate is determined by several factors, and the steric nature of the nucleophile and electrophile are critical.

With respect to the alkyl halide, we know that SN2 reaction is favored by sterically unhindered electrophiles. So the trend in rates is methyl > primary > secondary >> tertiary (to the point where this is generally impossible). Steric hindrance of the nucleophile will also slow the SN2 reaction. The E2 reaction is relatively insensitive to sterics, so if there is a competition between SN2 and E2, a bulky nucleophile/base will favor E2. That is why often potassium tert-butoxide is used to perform an E2 reaction.

Let's analyze each of the reactions presented.

1) A primary alkoxide (sterically unhindered) is reacted with a tertiary bromide (very hindered). SN2 reaction is extremely slow for tertiary halides, and nine beta-hydrogens are available to deprotonate from the halide, so E2 reaction dominates, and we would expect the alkene is formed.

2) A tertiary alkoxide (very sterically hindered) is reacted with a tertiary bromide (very hindered). Again, SN2 reaction is extremely slow for tertiary halides, and nine beta-hydrogens are available to deprotonate from the halide, so E2 reaction dominates, we would expect that the alkene is formed.

3) A primary alkoxide (sterically unhindered) is reacted with methyl bromide. Not only is methyl bromide very unhindered, but there are no beta-hydrogens available to participate in an E2 reaction. SN2 is the only viable pathway, so the ether will be the only product formed.

4) A tertiary alkoxide (very sterically hindered) is reacted with methyl bromide. Again, not only is methyl bromide very unhindered, but there are no beta-hydrogens available to participate in an E2 reaction. SN2 is the only viable pathway, so the ether will be the only product formed.


Can powdered beta-tin be made from raising the temperature of alpha-tin?

$\alpha$Sn and $\beta$Sn are the two solid allotropes of Sn.  As you not, below 13C the stable phase is $\alpha$Sn, which has a diamond cubic crystal structure (like diamond, Si, and Ge) and is a semi-metal.  Above 13C, the thermodynamically stable phase is $\beta$Sn, a body-centered tetragonal crystal. So, cycling back and forth around 13C varies the thermodynamically stable phase back and forth, with the only question being the kinetics of the transformation.  The video in the comment by @JasonPatterson shows that the transformation indeed takes place reasonably quickly (unlike diamond to graphite for carbon).

As for melting, if you rapidly heated $\alpha$Sn and avoided the phase transformation, you would find that the melting temperature would be lower. Using the SGTE thermodynamic data (A.T. Dinsdale, CALPHAD 15(4) 317-425 (1991)), one finds that the $\alpha$Sn -> liquid phase transition would occur at about 430.7K, ~75K lower than the standard $\beta$Sn -> liquid melting point at 505.06K.


Chromium from electrode not oxidizing in electrolysis

Some of the ways you aren't at standard condition.
Metal solution so the activity of the metals are at their concentration in the alloys
The solutions have far below 1mol/L of their respective metal ions
The oxygen that you are reducing in basic conditions is probably about 258 micromol per litre-
whereas it should be 1 mol/L. The potential might be too high- its hard to tell the solution resistance is an effect.

If you were at standard condition with low voltage then the first experiment makes perfect sense-
From wikipedia full list of electropotentials
V2+ + 2 e−
    V(s)     −1.13

Cr3+ + 3 e−
    Cr(s)    −0.74
Vanadium is preferred.


I need to find the Ka of a weak acid in titration with a strong base

I’m not sure if I’m missing your question, but I thought that perhaps you could use the relation $\mathrm{pH}=\mathrm{p}K_\text{a}$ at the half equivalence point? The answer would be still incorrect but closer?

Otherwise, you could probably use stoichiometry to solve for $K_\text{a}$ at the beginning or the end of the titration by constructing an ICE table or otherwise. By considering the deprotonation at the start, so the concentration of protons would be equivalent to concentration of conjugate base. Then, instead of inserting the pH at the half equivalence point, insert the pH before the titration.

If you need help doing the algebra, I could try...

But I think what you did wrong was inserting the concentration at the half equivalence point, rather than before the titration.


Why are paramagnetic materials attracted to ferrimagnetic materials(such as iron, nickel, and cobalt)?

First of all, in your question i feel a confusion about the nature of magnetism and paramagnetism. These magnetic interactions have nothing to do with attraction/repulsion of electric charges, and magnets do not attract or repel Coulomb charges (when we discuss the magnetic part of the interaction). In short static, magnetic and electric properties are two different stories.

Second, iron is ferrOmagnetic and not ferrImagnetic (capitalisation is by me). Forget ferrimagnetic materials for a second, and let's just talk about ferromagnetic, paramagnetic and diamagnetic materials:

Diamagnetic materials have no free electronic spin in the system, therefore they do not interact with magnetic field as a first approximation. There is a very small repulsive interaction coming from the polarisation of paired spins, but it is very small compared to any other interactions. Most organic compounds are in this category.
Paramagnetic materials have free electronic spins, which can flip / rotate. When put the sample in magnetic field, this field polarises the spins which turns into an attractive interaction. This is analogous with the electric field induced dipole moments, but this is a magnetic interaction, not an electronic interaction. The alignment of spins are constantly changing, as simple thermal excitation can rotate them. Therefore paramagnetic materials has no permanent magnetic moments like magnets. Typical examples are molecular oxygen, transition metal complexes, etc.
In ferromagnetic materials the free electronic spins are having a strong interaction with each other, and aligned parallel in strict order. This alignment is so stable that the magnetic moment of spins adds up, and there can be a macroscopical magnetic field around the sample (i.e. it is a magnet).

When you see the attraction between paramagnetic and magnetic materials, you see the polarisation of the paramagnetic materials in the magnetic field of ferromagnetic materials, which is an attractive itneraction.


Nomenclature of complexes: Cyano vs Cyanido ; Chloro vs Chlorido

The IUPAC Red Book is the official reference for recommended inorganic nomenclature. The full change to the additive nomenclature system was made in the 2005 update:

  Some new recommendations represent breaks with tradition, in the interest of increased clarity and consistency. For example, the application of the ending ‘ido’ to all anionic ligands with ‘ide’ names in additive nomenclature (e.g. chlorido and cyanido instead of chloro and cyano, and hydrido throughout, i.e. no exception in boron nomenclature) is part of a general move to a more systematic approach.

This is discussed in more detail in section IR-7.1.3 on page 112. Basically, anionic ligand names used as prefixes get changed from '-ide', '-ate', and '-ite' to '-ido', '-ato', and 'ito', respectively. Neutral and cationic ligands are usually used unchanged except for things like aqua, ammine, carbonyl, etc.


As pointed out in the comments above, the expected product would be acetaldehyde, not acetic acid.

If you stop and think about the mechanism of any thermal decarboxylation, you immediately stumble upon the issue. The decarboxylation would lead to a d1 type synthon, which from organic chemistry you should realise is the wrong polarity for a carbonyl group (the natural polarity of the C=O bond is to have a partial positive charge on carbon and partial negative charge on oxygen).


net charge nucleobases at alkaline conditions

The conjugate acids of the adenine and guanine have pKa's between 9-10. Guanine also has a nitrogen with a pKa of 12.3, which would be mostly protonated at pH 12. The pKa for the phosphoric acid proton of a nucleotide is significantly lower than 7.

Therefore, in a pH 8 buffer, the basic nitrogens of each adenine and guanine are protonated, while the phosphate is deprotonated. This gives an overall net charge closer to zero than you would see at pH 12.

In a pH 12 buffer, where the pH is above the pKa of both the phosphoric acid protons and the nitrogen conjugate acid protons (except the last one on each guanine), all of these positions would be deprotonated. Therefore, the formal charge on the nitrogens would be zero (again, except the most basic nitrogen on guanine), but the phosphates would still have a charge of -1 each. This would lead to a net charge of significantly higher charge magnitude because there would be very few positive charges to balance out the negative charges.

At pH 12, the net charge should be approximately: $$(n_N +n_u) - (n_c + n_g).$$
Approximate net charge at pH 8: $$n_N - (n_a + n_c + 2n_g).$$

nN= number of nucleotides

nu= number of uracils/thymines

nc= number of cytosines

ng= number of guanosines

na= number of adenosines


Short answer? Possibly, but these ones aren't.

So, the other thing mentioned in the video is that the specific material he's using has a molecular weight of approximately four million. If we assume it's ordinary poly-ethylene oxide, we can get the rough number of $\ce{(CH2)2O}$ subunits by just dividing by the molecular weight of the subunit:

$(12 \times 2) + (1 \times 2) + 16 = 42$
$\frac{4000000}{42} \approx 95000$

So, if we assume that each $\ce{C-O-C}$ back-bone unit is stretched out like a di-methyl ether molecule, a quick check on a fairly standard Ghemical force field gives me a subunit length of approximately 3.7 Å.

$3.7 \times 10^{-10} \times 95000 = 0.00003515 \text{m}$

So, these molecules should be about 0.04 millimetres long. Not exactly astronomical distances, but impressively long for a single molecule.

In theory, I don't think there's anything stopping you making a molecule as long as they claim, but these ones certainly aren't. I think also that the longest (synthetic) polymers in common use are of similar molecular weights and thus similar lengths: ultra-high molecular weight polyethene seems to top out around 4.5 million for commercial applications, from a quick search.

Biological molecules frequently get longer than this, though, especially DNA: the average human chromosome apparently contains a single DNA molecule approximately 5 centimetres long.

The man explaining in the video may have erroneously given the figure for the total length of all the molecules in that beaker.


What is the mechanism of the reaction of ethyl prop-2-ynoate to ethyl (2Z)-2,3-dibromoprop-2-enoate?

A search of SciFinder will also produce the formation of the (E)-dibromide 3. The conditions described for the formation of the (Z)-dibromide 2 are taken from a paper by Myers1 as reported earlier by Tippett.2 Rossi3 has reported the formation of the (E)-dibromide 3 using pyridinium hydrobromide perbromide.
Given the more vigorous conditions for the formation of the (Z)-isomer 2 compared to the milder conditions for the formation of the (E)-isomer 3 suggests that the former reaction is under thermodynamic control while the latter is kinetically controlled.  There is the distinct possibility that bromine radicals may effect the isomerization via species 4 as may surreptitious acid.

1) A. G. Myers, M. M. Alauddin, M. A. M. Fuhry, P. S. Dragovich, N. S. Finney, and P. M. Harrington, Tetrahedron Letters, 1989, 30, 6997.
2) R. G. Hall and S. Trippett, Tetrahedron Letters, 1982, 23, 2603.
3) F. Bellina, A. Carpita, M. De Santis and R. Rossi, Tetrahedron Letters, 1994, 35, 6913.


Can fungi metabolize pluronic polyol f127?

It is known that some species of fungi can use hydrocarbons as their carbon source. Polyethers like pluronic are less chemically inert than hydrocarbons. Indeed, degradation of polyethylene glycols by microorganisms is a well studied process (see DOI: 10.1007/s00253-001-0850-2 for example). Gloeophyllum trabeum is known to degrade poly(ethylene oxide) by a non-enzymatic, oxidative process (Proc. Natl. Acad. Sci. USA. 1998, 95(18):10373-7). Given those facts, there is a very good chance that some species of fungi can metabolize pluronics and similar compounds.


By what mechanisms can molecules with the same empirical formula be so different?

What makes both of those molecules so stable?

This is a fundamental question. Let's start by looking at the following diagram; a chemist would call it a "potential energy diagram." The y-axis shows the energy of a molecule as it travels along the reaction coordinate (x-axis; transforms, reacts) from the starting material on the left to the product on the right.

Notice that there are 3 "potential energy wells" (valleys) in the diagram, these reflect molecules that are stable enough to have a finite lifetime.  Notice too that to move from one well to another, an energy barrier must be surmounted.

So to return to your question, a molecule would be "kinetically stable" if it was protected on all sides by high energy barriers.  A molecule would be "thermodynamically stable" if its potential energy well was located at an energy that had a very low value on the y-axis.  A molecule that is very stable in a thermodynamic sense, could still be kinetically unstable (and therefore very reactive) if it was surrounded by low energy barriers.  By the same token, a molecule that is very unstable in a thermodynamic sense (exists at a high energy value on the y-axis) could be kinetically stable if it were surrounded by high energy barriers to reduce the likelihood of further reaction.

Millions of molecules are stable at room temperature, but if you supply enough energy to the system (heat them for example) they can all transform into other compounds.  At room temperature there are high enough energy barriers separating fructose and formaldehyde and they will both be stable in a jar, but under more energetic conditions such transformations might be feasible.


derivation of enthalpy internal energy relationship

First a brief note on the mathematics of Legendre Transforms

Consider a function in two independent variables: $ f(x,y)$

We right the differential of this function as follows:

$df(x,y) = \Big( \frac{\partial f}{\partial x}\Big)_ydx + \Big( \frac{\partial f}{\partial y}\Big)_xdy$

Now, I shall define $u := \Big( \frac{\partial f}{\partial x}\Big) $ and $w := \Big( \frac{\partial f}{\partial y}\Big)$

so, $df = udx + wdy$ [equation 1]

Here, u and x (and similarly w and y) are called conjugate variables. (i.e $ux$ and $wy$ will have the same units as $f$)

[Think of it this way: In thermodynamics, PV, TS have the same units as H, U,G,A]

Now, our goal is transform our original function $f(x,y)$ to a function $g(x,w)$. We wish to replace y, with it's conjugate variable and the function $f $ and $g$ will have the same units.

so, we take the differential of the product $wy$ : $d(wy) = ydw + wdy$

We subtract this from [equation 1] and get $d(f-wy) = udx - ydw $
and can deduce that our Legendre Transformed function $ g(x,w) := f-wy$

Now, we can easily apply this method to Internal Energy (U) to obtain Enthalpy (H)

We have $U(S,V)$ and $dU(S,V) = \Big( \frac{\partial U}{\partial S}\Big)_VdS + \Big( \frac{\partial U}{\partial V}\Big)_SdV $

Now, here we seek $H(S,P) \equiv g$

a)  $f \equiv U$

b) $S \equiv x$ is the variable we hold fixed

c) $V \equiv y$ is the variable we switch

d)  $\Big( \frac{\partial U}{\partial S}\Big)_V := T \equiv u $(the conjugate of the variable we are holding in place

e) $\Big( \frac{\partial U}{\partial V}\Big)_S := -P \equiv w$ (the conjugate of the variable we change

$ g =f-wy $

so, $H(S,P) = U-(V)(-P)$

Or, $H(S,P) = U + PV$

From here, you can proceed to write down: $dH = dU + PdV + Vdp$

and for macroscopic changes: $\Delta H = \Delta U + P\Delta V + V\Delta P $


Does Gallium (liquid or solid) corrode all forms of brass?

From here

  Gallium is corrosive to all metals except tungsten and tantalum, which have a high resistance to corrosion.

Beta-Brass is embrittled by Gallium

Due to its highly reducing character it is highly corrosive material.


Determining order of a reaction without isolation

Although this is indeed a bit more a math problem than chemistry, I think it is useful to answer it, because this is a commonly encountered problem in reaction kinetics.

First of all, Yes it is possible to get the rates with respect to each reactant separately, as long as you have sufficient experimental data. With sufficient I mean: $1+\text{number of reactants}$. This is because you don't only need the order for each reactant, but also the rate constant for the total reaction

The rate equation

In general a rate equation will have the following structure:
$$ r = k [A]^\alpha [B]^\beta[C]^\gamma...\text{etc}$$
where with etc I mean that you can continue for as many reactants that you have. In your example you have 2 reactants so your rate equation in general will look like this:
$$ r = k [A]^\alpha [B]^\beta $$
To find the reaction order with respect to each reactant we need experimental data on $r$ for given $[A]%$ and $[B]$ like you provided.

The fitting procedure

You will have to write a short program that takes an initial guess for $k$, $\alpha$ and $\beta$ and checks how good or bad that guess is by comparing the calculated rates $R$ with the experimental rates $r$ for each set of $[A]$ and $[B]$. The normal way to make this comparison is based on least-squares:
$$S = \sum_{i=1}^N (r_i-R_i)^2 $$
So you sum up the squared differences between the reaction rate you found experimentally and the calculated rates for all $N$ experimental datasets.

To get the reaction orders and rate constant you want $S$ to be minimal, preferably 0 (although that normally doesn't happen with experimental data due to small measurement errors). Normally $S$ is not immediately optimal on your first guess for $k$, $\alpha$ and $\beta$, so you have to 'guess again'. You could indeed really guess again, but there are much smarter ways of determining the new guess, which will make sure your new guess is a better one than the old one. A common technique for this is called Newton's method. With the new guess you again check the value of $S$. You then repeat the cycle until you cannot find a lower value for $S$, which means that you have found the values of $k$, $\alpha$ and $\beta$ that are optimal for this experimental dataset and thus should be the correct reaction orders for the reaction.


The solubility of different compounds relative to each other in water

First of all, let's temporarily make a distinction between ionic and covalent compounds. Generally speaking, covalent compounds are more soluble in nonpolar solvents while ionic compounds are more soluble in polar solvents. Of course ionic and covalent bonding exist on a spectrum and some compounds you would expect to be ionic are indeed more covalent in nature. This concept is often visualized using the Van-Arkel Ketelaar triangle of bonding

As described by the Fajans' Rules, highly polarizing cations will distort polarizable anions, pulling their electrons into the inter-atomic space to the extent that they essentially form ionic compounds. This is one of the reasons solubility decreases with increased polarity. Solubility does not increase with increased covalent nature. In water, it would decrease. The best example of this is beryllium. Almost all $\ce{Be^{2+}}$ compounds actually exhibit covalent chemistry, as we can see with  beryllium peroxide in the Van-Arkel Ketelaar triangle above. This is because beryllium has extremely high charge density.

In order to properly understand solubility trends of ionic compounds, we should examine the enthalpy of dissolution using a Born-Haber cycle.

The first step is vaporization of the crystalline lattice, which has a large positive enthalpy. This is affected by a few factors, but it basically represents the stability of the crystal. Generally, dipositive anions/cations will have much greater lattice energy because of stronger electrostatic interactions, and lattice match should also be considered. A significant mismatch in size will lead to lower lattice energy. In the Born-Haber cycle, the ion-ion interactions are broken when the crystal is vaporized, then ion-dipole interactions are formed when water molecules surround the gaseous ions, so the other  factor to consider in the Born-Haber cycle of dissolution is hydration enthalpy, which is negative. Generally, this is increased with increasing charge density. Sometimes net enthalpy is positive, and in these cases, entropy may also come into play. For the most part, you can think of ionic solubility as a balance between lattice energy and hydration enthalpy.

  Solubility increases with decreased lattice enthalpy increased hydration enthalpy

and you're right that

  Solubility decreases with increased polarizability

Sodium's ionic radius is 227 pm, while ammonium's is 175. Often ammonium is a good substitute for sodium, as a large, low charge density cation (for stabilizing highly polarizable anions like bicarbonate). In fact, ammonium bicarbonate is one of the only soluble non-group 1 bicarbonates. Still, ammonium is still a little smaller than sodium. As a solid, ammonium will stabilize chloride better, but bicarbonate worse.

Since bicarbonate is much much more polarizable than chlorine, the two bicarbonate salts will probably be less soluble than the two chlorides. Ammonium will stabilize chloride better by size match, so ammonium chloride will be a little less soluble than sodium chloride. Due to it's large size and low CD, though, sodium will stabilize bicarbonate better, so sodium bicarbonate will be a little less soluble than ammonium carbonate.

The way I see it, the order should be:

  $\ce{NaCl}$ > $\ce{NH4Cl}$ > $\ce{NH4HCO3}$ > $\ce{NaHCO3}$ (decreasing solubility)

Of course solubility is a pretty complex concept and these are only generalizations. To get a better handle on it, I would try looking at the enthalpy and entropy cycles of different types of salts in each of the main groups. Trends exist, but the way they work is a little different depending on the group. I think, for example, that lattice match is mostly important in group 1 salts because of their lattice types. Anion charge becomes an important factor in group 2. Trends can be established but a lot of the time they are only consistent within a given group.


Why is Charcoal such an excellent adsorbent?

Activated charcoal is a great adsorbent because of it's huge surface area. While it doesn't bind very many ions/atoms/molecules per surface area (which is the characteristic of a 'good' adsorbent), due to very big surface area per unit of mass it can adsorb a lot of particles. Actually, process of 'activating' charcoal is designed to maximize the surface area to mass ratio.


There are many endothermic physical processes such as melting ice or sublimation of solid carbon dioxide $\ce{CO2}$. One should also mention thermoelectric cooling effect (Peltier device), which is is rarely effective in practice, yet still an option.

A suitable endothermic chemical reaction would probably be an electrolytic dissociation of certain salts, among which dissolving ammonium nitrate $\ce{NH4NO3}$ in water is probably one of the most well-known and effective examples.


Why does bromine increase the viscosity of olive oil?

Olive oil is a triglyceride, and the main fatty acid is oleic acid below.

Now the bromine does not add to the double bond between oxygen and carbon; instead the bromine reacts with carbon-carbon double bond and add itself to it. If you use e.g. coconut oil which has minimal unsaturated (meaning having carbon-carbon double bond) fatty acids (and assume you don't shine light onto it, heating it or otherwise forcing the reaction to start) there should theoretically be no reaction.

  But why would bromine want to break the bonds?

Simply put, bromine loves electron. The carbon-carbon double bond happens to be rich in electrons, so the bromine binds to it. While the carbon-oxygen double bond also has electrons, oxygen loves electron more than bromine and tends to keep the electron for itself, so bromine cannot add there.

  Does this increase the viscosity of olive oil?

The individual molecules that make up olive oil is held together by weak forces, that arise from van der Waals' forces, which is attraction due to small, unevenness in electron distribution (which leads to very small positive and negative charges on the molecules that subsequently attract other molecules).
This is overly simplified concept though so it's not entirely accurate.

Bromine atom is very large (hence bromine, which molecules consists of only two bromine atom, exists as a liquid in room temperature). The more the electron, the more likely there are some unevenness in electron distribution, and hence stronger attraction between molecules. Also, bromine loves electron so it tends to pull electron closer to itself, creating a small but rather permanent positive and negative charge on the molecule. This also strengthen the force between brominated oil molecules, hence the viscosity goes up.

  Viscosity goes up because less double bondings makes the fatty acids more “packable”

This is rather about melting point of the fat / oil, as when freezing, the molecules in liquid need to "pack" into a solid structure. In liquid however, molecules tends not wanting to maintain the "packing", or ordered structure. Adding bromine does increase the melting point but that's another story.


Generating neat orbitals/surfaces from molden/wfn-files

Often different quantum chemical programs require different analysing tools. Unfortunately there is no universal standard (program), that can do it all. If there were, someone would come along and just invent something new.

I personally prefer ChemCraft for producing publication ready images. It is a commercially distributed program. It mainly supports any Gaussian and GAMESS file formats (and Firefly by extension). It can visualize quite a few more formats now. What I especially like about the program is that it can read the files generated by the NBO programs. It makes visualisation of all kinds of localised and canonical orbitals much simpler.
There are some more features of the program I really got attached to, like symmetrizing molecules or building molecules from scratch. It has the big advantage of writing the symmetry adapted data group for GAMESS for custom basis sets, which I have not been able to find anywhere else. But this is certainly not the target of this post.
 Unfortunately it is not able to read molden or wfn[1] files.

The program comes with a large number of visualisation schemes, which can easily be modified to produce a personalised style. Below is a small demonstration.

Most of the times I would use one of the above, depending on whom I have to impress. Often it is far more beneficial to have monochrome images, like the ones below.

A quite recent addition came with styles that allow you to generate simple 2D like structure. It is still possible to overlay these structures with orbital plots. (Unfortunately the transparency is hard-coded and cannot be set explicitly.)

The caveat is obviously, that it cannot read molden or wfn[1] files, not even Turbomole output.

A feature that I am still missing is the batch production of orbital pictures. You still always have to save them by hand, which can be incredibly tedious for large molecules (and large active spaces). (But I have written a feature request for that.)
Support for the program is quite good and the available help files are even helpful. Most of the features can be understood intuitively, however, as always, it takes time to get used to working with it.
Another caveat is, that the windows version outperforms its little linux brother by a lot. If you already run windows somewhere, that is probably fine for you. If you run linux almost everywhere, then you have to deal with some missing features (or recently it can be ported via wine, making it slow though).

I do use molden (Xwindow) quite a lot for intermediate visualisation, since it is one of the fastest interfaces available. It is very robust and has low graphics; you can easily use it whit a rather slow internet connection. It is free for private and non-commercial usage. Since my first days as a computational chemist involved the use of 56k-modem for checkup routines, this was one of the only go to programs. It will always be the first program I install on a work-dedicated computer. One simply does not forget his or her oldest friend.
A very big plus is the very well structured and written postscript files produced by molden. Some of the purists might even like it. Here are a few converted samples, but you can download the source here (via tinyupload.com).

Such files can become extremely handy, when preparing posters (especially when using LaTeX). I have on occasion seen molecular plots generated by molden in publications; I am even guilty of that. However, nowadays I believe there are more elegant methods.

Visualising Turbomole output has been, is and will probably always be a painful experience. For some of the deeper features you should probably be an expert in gnuplot, too. For the basics of structure and orbitals, the current recommendation is to use TmoleX, the GUI to turbomole. If it is done by the same people who designed define, then it's will be quite painfully to use. When I was a heavy user of turbomole, this did not exist, so I should probably shut up.
Judging from the current (version 7.0) manual the workflow I used is still somewhat recommended. It is to let the native modules create the binary *.plt files, which are gOpenMol input. Unfortunately, what was once a very aspiring program is nowadays not developed any more. You can still obtain an unsupported copy. The source code is available, but the current license state of the program is somewhat unclear. Since VMD (see Brians post) and TmoleX do now support gOpenMol files, the necessity of learning a deprecated program is probably not useful any more. One of the prime features of this program however was that it has a tcl interface and hence was fully scriptable - apparently something that VMD now also has.
It is well capable of producing nice orbital pictures, you might have seen them on this page already. (I tweaked the Gaussian output a bit, since I do not have Turbomole available.)

Although it supports quite a few file formats, molden input and wfn[1] is unsupported.

[1] Let's come to the last part: wfn files. However you produce them, they are probably not up to date. The main (and probably only) purpose for those files is the analysis with AIMPAC -  a program that is no longer developed. They have one very big disadvantage: they cannot deal with functions higher than of f type. If you are using a triple zeta basis set, you probably have those functions. Another disadvantage is, that they don not include information about possible effective core potentials.
For the purpose of analysing the electron density with the quantum theory of atoms in molecules (QTAIM) the much superior version wfn extended (*.wfnx or *.wfx) should be used. I am currently only aware of Gaussian to create such files correctly. The main program which uses these files is AIMAll. It is also a commercially available program. You can learn more about these files on the specification page there.
The only valuable point of creating molecular orbitals with this program is, if you want to overlay it with the bond paths instead of just sticks. Just like in this example:

Apart from this usage, the process is very tiring and needs many steps to be completely performed. Working the program non-local is very, very slow. And to be perfectly honest, what is the purpose of the above other than impressing a few organic chemists?
Another negative aspect of the program is the very high demand of memory and storage, especially for large molecules.

In conclusion: If you are not doing QTAIM analysis you should probably refrain from *.wfx files for visualisation purposes. For every QC program there usually comes a workable interface/visualisation program. With any program there is the activation barrier of learning how to do things to overcome. In any case, I suggest having a few ready to work with, you never know where you go and you might have to learn a new program any time.
I think the biggest loss of any visualisation tool is the neglect of writing good tutorials or even a basic documentation.


What meaning do multiple real solutions of a truncated virial equation have, if any?

The virial equation is set up for the gas phase only.  The virial coefficients $B$ and $C$ are found from gas phase measurements.

If you get three solutions, I suspect that it is the largest one that you want.  I'd do a trial solution with something like water vapor to see.

Higher order virial equations are problematic since it is exceptionally difficult to determine higher virial coefficients from experimental data.  Higher order virial equations do have theoretical interest, but that's another story.


Clemmensen reduction of α-, β-keto acids

The likely reason for this is a steric hindrance. The Clemmensen reduction is very vulnerable to hindrance. See the image below:
There was also another reaction conducted that had a 67% yield for a gamma keto acid with resonance. See the image below:

Therefore, the main factor that makes α- or β-keto acids not undergoing Clemmensen reduction is likely not because of resonance, but primarily of steric hinderance. But this also may raise another question, why doesn't the carboxylic acid group get reduced? The reason for this is that the carbon in a carboxylic acid is not quite partially positive because OH is an electron-donating group. The mechanism involves a transfer of an electron from zinc to the partially positive carbon atom, which, as mentioned, carboxylic acid doesn't have.
However, even for diketones, the Clemmensen reduction doesn't often yield what one would expect. See the image below:

The reason is because of the mechanism. In the mechanism, radicals are formed on the carbonyl carbon. However, if you have a diradical, they can come together and form a bond if they are close enough. This is just some extra information.
The source I took the images from: https://www.sciencedirect.com/topics/chemistry/clemmensen-reduction#
Extra papers I found that may be helpful:
https://d.lib.msu.edu/etd/37284/datastream/OBJ/view
https://pubs.acs.org/doi/abs/10.1021/jo01017a512


Is pK_{In} = pH at equivalence point still true for weak basic organic indicators?

It is true that acid-base indicators ($\mathrm{pH}$-indicators) are either weak acid or weak base. What I understood reading your question is that you have wrong impression about bases. To clear your view, not all bases contain $\ce{OH-}$ ions. Specially, most organic bases are weak and contain electronegative ion (e.g., $\ce{N}$ or $\ce{O}$) with at least one lone pair. This fact is well cleared by Poutnik's answer, thus, I'm not going to elaborate it more. Instead, I'd like to explain the action of a $\mathrm{pH}$-indicator (e.g., $\ce{HIn}$ in acidic form) to some depth.

Since Indicators ($\ce{HIn}$) are weak acids (or weak bases), in aqueous medium, it's in an equilibrium:

$$\ce{HIn + H2O <=> H3O+ + In-} \tag1$$
$$\therefore \ K_\mathrm{aIn} = \frac{[\ce{H3O+}][\ce{In-}]}{[\ce{HIn}]} \tag2$$
Taking log on both side followed by simplification gives:
$$-\log [\ce{H3O+}] = -\log K_\mathrm{aIn} + \log \left(\frac{[\ce{In-}]}{[\ce{HIn}]}\right) $$
$$\text{Hence}, \ \mathrm{pH} = \mathrm{p}K_\mathrm{aIn} + \log \left(\frac{[\ce{In-}]}{[\ce{HIn}]}\right) \tag3$$

This is a resemblance of Henderson-Hasselbalch equation. The equation $(3)$ shows that when $[\ce{HIn}] = [\ce{In-}]$, $\mathrm{pH} = \mathrm{p}K_\mathrm{aIn}$.

Let's consider the indicator, Litmus, which is we commonly use in labs. Litmus is a weak organic acid (see the structure in bottom box of the diagram; $\mathrm{p}K_\mathrm{aIn} = 6.5$). The $\color{green}{\text{green circle}}$ shows the $\ce{H}$ of $\ce{HIn}$ molecule. The un-ionized litmus ($\ce{HIn}$) is $\color{red}{\text{red}}$ (below $\mathrm{pH}$ 4.5), whereas the ion part $(\ce{In-})$ is $\color{blue}{\text{blue}}$ (above $\mathrm{pH}$ 8.2):

Usually color change happens within a range of $\mathrm{pH}$. It is a rule of thumb that this range falls between the $\mathrm{p}K_\mathrm{aIn}\pm 1$. That means, acid color persists if $[\ce{HIn}] = 10 \times [\ce{In-}]$ and base color dominates when $10 \times [\ce{HIn}] = [\ce{In-}]$. For example, apply these values in the equation $(3)$:

$$\mathrm{pH} = \mathrm{p}K_\mathrm{aIn} + \log \left(\frac{[\ce{In-}]}{10 \times [\ce{In-}]}\right) = \mathrm{p}K_\mathrm{aIn} + \log \frac{1}{10} = \mathrm{p}K_\mathrm{aIn} -1 $$

Similarly,

$$\mathrm{pH} = \mathrm{p}K_\mathrm{aIn} + \log \left(\frac{10 \times [\ce{HIn}]}{[\ce{HIn}]}\right) = \mathrm{p}K_\mathrm{aIn} + \log 10 = \mathrm{p}K_\mathrm{aIn} +1 $$

Hence, where $\mathrm{pH} = \mathrm{p}K_\mathrm{aIn}$ is the most intense point during color change. For litmus, color change happens between the range, $\mathrm{pH} = 5.5-7.5$ since $\mathrm{p}K_\mathrm{aIn} = 6.5$ for litmus.

Let's see litmus in basic solution where $\ce{In-}$ ions are predominant ($\color{blue}{\text{blue}}$ solution). Now you start adding acid to the solution and you can see what would happen to the solution (equation $(1)$) by using Le Chatelier's Principle. Since excess hydronium ions disturb the equilibrium, it'd shift to left hand side of equilibrium to reduce excess hydronium ions by reacting $\ce{In-}$ ions. Eventually, solution changes the color to $\color{red}{\text{red}}$. Opposite would happens if you added hydroxide ions to the $\color{red}{\text{red solution}}$ in equilibrium.

The scaning UV spectra in the upper box of the diagram shows how color change happens in solution with $\mathrm{pH}$. The indicator here is bromothymol blue, which is yellow in the acidic medium absorbing light at about $\pu{430 nm}$. When $\mathrm{pH}$ has changed systematically with time (by adding known aliquot of $\ce{OH-}$ solution with in fixed time), the solution changed the color accordingly, and bromothymol blue has changed to its basic color blue, absorbing light at about $\pu{620 nm}$. Note that the peak at $\pu{430 nm}$ decreases while the peak at $\pu{620 nm}$ increases while $\mathrm{pH}$ increases.