text:
How is chlorocarbene synthetised?
The wikipedia page on cyclononatetraene says that the aromatic cyclononatetraenyl anion can be synthetised by a rearangement of 9-chlorobicyclo[6.1.0]nona-2,4,6-triene.
My first idea on how to synthetise this compound is to react cyclooctatetraene with chlorocarbene $(\ce{:\!CHCl})$. However, in contrast to dichlorocarbene $(\ce{:\!CCl2})$, chlorocarbene $(\ce{:\!CHCl})$ is much less stable. Can it be synthetised by an action of a strong base on dichloromethane?
It seems, from this pdf, that $t$-$\ce{BuOK}$ will make $\ce{:\!CPhCl}$ from $\ce{CHPhCl2}$. Is it, however, a strong enough base to deprotonate $\ce{CH2Cl2}$? Or must some stronger one, e.g. LDA, be used?

https://pubs.acs.org/doi/abs/10.1021/ja01506a041 uses nBuLi to generate chlorocarbene from DCM, and this article is mentioned in https://pubs.acs.org/doi/abs/10.1021/ja01077a032, which mentions that the chlorocarbene addition gives the wrong diastereomer of the target bicyclic compound (it is instead made by reaction of $\ce{C8H8^2-}$ with chloroform).

The formation of dihalocarbenes (at least cloro- and bromo-counterparts) in reactions of haloforms with bases has been well established (e.g., Ref.1,2). In 1959, Closs and Closs introduced a novel reaction to prepare chlorocarbene as a reaction intermediate in situ in order to synthesize chlorocyclopropanes derivatives (Ref.3). They have reported that when n-butyllithium is added to methylene chloride in the presence of olefins at $\pu{—25 °C}$, chlorocyclopropanes are formed in fair to good yields depending on the structure of the olefin. They further claimed that the novel procedure strongly suggests the existence of chlorocarbene as a reaction intermediate.
They further developed the method and published a full paper as a general method (Ref.4) where abstract states:

Alkyllithium compounds react with methylene chloride in the presence of olefins to form chlorocyclopropanes. This reaction is believed to proceed through the intermediate chlorocarbene. Seven alkyl-substituted chlorocyclopropanes have been prepared by this method. Configurations are assigned to the products by considerations of predictible non-bonded interactions in the transition states of the addition of chlorocarbene to the olefins. The configurations are confirmed by n.m.r. spectra.

Later, Billups and Haley have used the same method to synthesize spiropentadiene (Ref.5).
References:

W. von E. Doering, A. Kentaro Hoffmann, “The Addition of Dichlorocarbene to Olefins,” J. Am. Chem. Soc. 1954, 76(23), 6162–6165 (https://doi.org/10.1021/ja01652a087).
W. von E. Doering, Wm. A. Henderson, Jr., “The Electron-seeking Demands of Dichlorocarbene in its Addition to Olefins,” J. Am. Chem. Soc. 1958, 80(19), 5274–5277 (https://doi.org/10.1021/ja01552a065).
Gerhard L. Closs, Lisellote E. Closs, “Synthesis of Chlorocyclopropanes from Methylene Chloride and Olefins,” J. Am. Chem. Soc. 1959, 81(18), 4996–4997 (https://doi.org/10.1021/ja01527a057).
Gerhard L. Closs, Lisellote E. Closs, “Carbenes from Alkyl Halides and Organolithium Compounds. I. Synthesis of Chlorocyclopropanes,” J. Am. Chem. Soc. 1960, 82(21), 5723–5728 (https://doi.org/10.1021/ja01506a041).
W. E. Billups, Michael M. Haley, “Spiropentadiene,” J. Am. Chem. Soc. 1991, 113(13), 5084–5085 (https://doi.org/10.1021/ja00013a067).


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What is the meaning of moment in Dipole moment
Today I read about dipole moment in my standard textbook. It said something like it is the measure of charge separation, and then proceeded to the formula , ie. Delta'charge'×bond length.
But I can't intuitively imagine what exactly dipole moment is. I can't imagine whose moment is this concept talking about. Please help me with this.

https://chemistry.stackexchange.com/questions/48444/what-are-dipole-moments-in-a-molecule-supposed-to-act-upon

https://chemistry.stackexchange.com/questions/74683/is-dipole-moment-a-vector

and here https://chemistry.stackexchange.com/questions/116554/is-there-a-difference-between-dipole-and-dipole-moment#comment220232_116554

Think of it as pushing on a lever - the moment depends on distance to pivot and strength of push. The moment of pushing with 1 N on a 2 m lever is the same as pushing with 2 N on a 1 m lever.

source: Stack Exchange

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/

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How do blue shifts in Raman spectra indicate a co-crystalline compound?
The short conference paper The Acetylene-Butane Co-Crystal: A Potentially Abundant Molecular Mineral on Titan (mentioned in EurekaAlert's 'Bathtub rings' around Titan's lakes might be made of alien crystals) says:

Experimental: Acetylene (purified to remove acetone stabilizer) and butane (99% purity) were condensed into a liquid nitrogen-cooled cryostage (Linkam Scientific Instruments Ltd.) at 90 K or 130 K. Raman spectra within the cryostage were obtained using a high-resolution confocal dispersive micro- Raman spectrometer (Horiba Jobin Yvon LabRam HR) equipped with a 50 mW Nd:YAG laser (frequency- doubled 532 nm) as the excitation source. Thermal stability studies were performed by warming in 5 or 10 K increments and obtaining Raman spectra following a 5-minute equilibration time.
Results: Blue shifts (12–29 cm-1) of the C≡C and C-H stretching modes indicate the formation of a co-crystalline compound. We observed similar shifts for co-crystals comprised of benzene and ethane (2–12 cm- 1 red shifts; Vu et al. 2014) and of acetylene and ammonia (7–16 cm-1 red shifts; 42–66 cm-1 blue shifts; Cable et al. 2018). These shifts indicate a change in the chemical environment of the molecular species, typically modification of a host crystal lattice to accommodate a guest molecule.

Question: Is it common to conclude that such blue shifts in Raman spectra "...indicate the formation of a co-crystalline compound"? There is a solid with both molecules in close contact so they interact, but how does this point specifically to there being  a o-crystalline compound?

No. Those are the results of a specific case. They could have said "Red shift (12-29 cm^-1....) and I would have kept reading without wondering. Indeed, this is the case for the benzene-ethane sample. Also, confocal microraman is a microscope, allowing to spatially resolve the spectroscopic features. That is what probably points to co-crystallisation, i.e.  no or little segregation of phases.

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Why might litmus paper and universal indicator give conflicting pH readings in an acid-carbonate reaction?
The resulting neutralised (which wasn't actually neutralised due to not realising CaCO3's insolubility) solution between Hydrochloric Acid and Calcium Carbonate, as per balanced equation
CaCO₃ + 2HCl → CaCl₂ + CO₂ + H₂O
turned red litmus paper blue, and universal indicator solution orange (repeated twice).
I performed another reaction between Citric Acid and Calcium Carbonate and I received consistently acidic results. There is a chance I might've mixed up these results though, and the Citric Acid reaction WAS the conflicting one.
I'm wondering if anyone could shed some light on why this might've happened? I'm guessing it can't be due to the insoluble calcium carbonate, as my other reaction didn't have issues.

It seems there is several subjective effects involved, affecting results. I suggest to repeat experiments in defined manners, allow establishing of equilibrium, that do measurements before and after filtration of excess carbonate.

Make sure you  mix the contents well before pH measurements. I would also assume that calcium citrate will be a basic solution (salt of a weak acid with very strong base).


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What will be the product ratio if I do chlorination of p-toluic acid?
If I do chlorination of p-toluic acid using Lewis catalyst, e.g. $\ce{FeCl3},$ what will be the ratio of major product to side product? Main product shall be 3-chloro-p-toluic acid and other product may be 2-chloro-p-toluic acid for monochlorination.
So, what will be approx ratio of these two?

I couldn't find any literature evidence for the nuclear chlorination reaction of 4-methylbenzoic acid (p-toluic acid). Instead, I found a patent in Japanese for the nuclear chlorination of 4-methylbenzoic acid chloride, of which, I believe close enough to your question (See below the translation given in Google).
According to the patent, when mixture of 99.7% pure 4-methylbenzoic acid chloride ($\pu{927.6g},\; \pu{6.0mol}$) and ferric(III) chloride ($\pu{2.9g}, 0.3 \:wt\%$) was heated at $\pu{50\!-\! 55 ^\circ C}$ for $\pu{22 h}$ in the presence of $\ce{Cl2}$ gas. After completion of the reaction, 84.9% of 3-chloro-4-methyl benzoic acid chloride was isolated. In addition, 0.1% of 2-chloro-4-methyl benzoic acid chloride and 2.2% of dichloro-4-methyl benzoic acid chloride were also isolated (dichloro-4-methyl benzoic acid chloride was not identified, but it should be 3,5-dichloro-4-methyl benzoic acid chloride, judging from the ratio of 3-chloro-4-methyl benzoic acid chloride to 2-chloro-4-methyl benzoic acid chloride).

Reference:
Inventor: 智規 加藤亮太 甲斐幸基 小林, "Isopropyl 3-chloro-4-methylbenzoate and method for producing same." Patent application # (2013): WO2013015203A1 (https://patents.google.com/patent/WO2013015203A1/en).


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Are there any exceptions to Le Chatelier’s principle?
Peter Atkins says

In introductory chemistry, we meet the empirical rule of thumb known as  Le Chatelier’s principle:
When a system at equilibrium is subjected to a disturbance, the composition of the system adjusts so as to tend to minimize the effect of the disturbance.
For instance, if a system is compressed, then the equilibrium position can be expected to shift in the direction that leads to a reduction in the number of molecules in the gas phase, for that tends to minimize the effect of compression.Le Chatelier’s principle, though, is only a rule of thumb, and to understand why reactions respond as they do, and to calculate the new equilibrium composition, we need to use thermodynamics.

That Le Chatelier’s principle, is only a rule of thumb, and thermodynamic calculations need to be done to determine new equilibria.
So are there any exceptions to this rule of thumb ??

Metastable systems, where the 'disturbance' pushes it over the edge towards a global minimum energy, won't obey it particularly well.

I'll say if you find an exception to Le Chatelier, a closer look will show that you've had a wrong idea (i.e. not the whole picture) about what's actually happening. Very strong hint that you should have a closer look.


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Why doesn't beryllium, an elemental solid, obey Dulong and Petit's law?
Beryllium has a molar heat capacity of $\pu{17.7 J mol-1 K-1},$ which is quite far from the $\pu{25 J mol-1 K-1}$ as predicted by Dulong and Petit's law. What is the reason for this discrepancy?

Boron and graphite also deviate significantly. Now, what is Dulong and Petit's "law" based on?

No elementvreally obeys the"Law" which is just a high temperature approximation for the solid state.  Some elements have ambient temperatures high enough, others don't.

First, a reminder that most scientific "laws" are only convenient mathematical approximations. They aren't guaranteed to work for every case, and they aren't guaranteed to have underlying physical meaning. As such, don't think of exceptions as "not obeying the law"; it's more correct to say that the law isn't obeying reality,
In the case of the Dulong-Petit Law:

Despite its simplicity, Dulong–Petit law offers fairly good prediction for the specific heat capacity of solids with relatively simple crystal structure at high temperatures. It fails, however, at room temperatures for light atoms bonded strongly to each other, such as in metallic beryllium, and in carbon as diamond. In the very low (cryogenic) temperature region, where the quantum mechanical nature of energy storage in all solids manifests itself with larger and larger effect, the law fails for all substances.

Not a  'Law' a such it is more like a 'rule of thumb' based on a limited set of data measure at high (i.e. room) temperatures. The heat capacity is not constant but varies with temperature and for metals reaches a limiting value of approximately $3R$ at high temperatures. (The limiting value is higher for molecules due to the many translational, vibrational and rotational energy levels).
The heat capacity is the rate of change of the energy with temperature, i.e. the slope  of a plot of energy vs temperature.  At low temperatures only the lowest few energy levels are populated, so the rate of change vs temperature is almost zero. As the temperature is increased, more energy is added, more levels become populated and $C_V$ rapidly increases because the rate of change of energy is large. At high temperature, very many levels are now populated and the rate of increase of energy with temperature is constant, and therefore $C_V$ levels off becoming almost constant.
In some metals, those with large energy gaps between levels, room temperature is insufficient for them to reach the point where the slope of internal energy increase with temperature becomes constant and so the heat capacity is smaller than $3R$. The value at higher temperatures (say 1000 C) should reach the limiting value.

The Dulong–Petit law applies in the classical limit, i.e. when temperature is high enough that the quantisation of energy levels (as prescribed by quantum mechanics) is not readily  apparent. The larger the gaps between energy levels, the higher the temperature needs to be such that you no longer "see" these gaps. The comparison is fundamentally between thermal energy $k_\mathrm BT$ and the energy gap $\Delta E$: the Dulong–Petit law applies when $k_\mathrm BT \gg \Delta E$.
There are three kinds of energy gaps which are under consideration here, namely translational, rotational, and vibrational. It turns out that vibrational energy levels have the largest gaps between them, and hence offer the strongest "challenge" to the requirement for $k_\mathrm BT \gg \Delta E$. From the simple harmonic oscillator we know that $\Delta E = \hbar\omega = \hbar\sqrt{k/\mu}$, so we can identify several cases where we expect the Dulong–Petit law to break down:

when temperature is small;
when the force constant $k$ is large, i.e. very strongly bonded atoms;
when the reduced mass $\mu$ is small, i.e. light atoms.


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Reduction of conjugated double bond by sodium borohydride
Some of my teachers were debating about whether a double bond in conjugation with a ketone is reduced or not by $\ce{NaBH4}.$ One of my teachers presented the evidence from Clayden which had an example where a cyclic ketone was reduced alongside the conjugated double bond.
However, there was no example for a straight chain compound. Also, the book said that steric hindrance may affect the reduction process and only favour the reduction of ketone.
I want to know that is there any authentic source which has an example of reduction of a conjugated double bond and ketone via $\ce{NaBH4}.$

Scifinder shows lots of reactions such as you describe. Often (but not always) copper is used to help favour the 1,4-addition (alkene reduction) process.

The simplest example I can see is the reduction of 3-penten-2-one[[2]] - when treated with NaBH4 the majority (89%) is reduced at the ketone only, while 11% of the doubly-reduced product is formed:

Obviously different substrates would give different product ratios, but certainly this is a plausible (and precedented) reaction.

Thanks a lot. But why does it have such a low yield in the example you gave. I don't see much steric hindrance. Do you know the reason?

Related question: https://chemistry.stackexchange.com/questions/87138/why-does-lialh4-reduce-the-double-bond-of-cinnamaldehyde-whereas-nabh4-does-not/87195#87195 As a rule, NaBH4 reduces 1,2.


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Reaction rate of Mg with aryl halides
I have been thinking of the rate of reaction of aryl bromide and aryl chloride with  $\ce{Mg}$ to form Grignard.
When considering the electronegativities the carbon in aryl chloride is more positively polarized. But when considering the bond energies carbon chlorine bond is stronger.
I want to know which factor is more significant when we consider the reaction rates of aryl chloride and aryl bromide with $\ce{Mg}$ to form Grignard.

Bond energy is key.  Magnesium is easily electropositive enough to barrel its way in unless facing a high activatiin energy.

So that means aryl bromide forms grignard faster than aryl chloride right?

On target.  In fact bromides generally react faster than chlorides.


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What drives vibrational cooling in an excited stated?
As we excite a molecule from its ground state, $S_{0,v=0}$, to some excited state in a higher vibrational state, i.e. $S_{1,v'=3}$, what drives vibrational cooling within that excited state manifold (such as $S_{1,v'=0} \leftarrow S_{1,v'=3} $?) This is a fast process, typically on the order of hundreds of femtoseconds, and it occurs prior to any photorelaxation (as per Kasha's rule). Due to the vibrational density of states being quite high near the first excited state, is it the anharmonic coupling to these states that drives vibrational cooling? Furthermore, is there a quantum mechanistic/mathematical way of depicting the probability/rate of this process (as in, what would the relevant perturbation term $\hat{H}{'}$ for dealing with this problem in the context of Fermi's golden rule)?

Where are you, in a condensed phase or in a dilute gas? A compact lattice has no space for local vibrations.

Good point - I guess the form of vibrational cooling depends on the nature of the state. I'm more interested in the condensed phase, where solvent/bath interactions or possible. My interest mainly lies in the crystalline phase.

Normal modes are technically orthogonal so no energy could flow from one to another, but of course this is just a mathematical construct to make life simple for us, and in reality  'anharmonicity' (physical and electronic) will allow the energy to move about. This is to say that these modes are able to couple to one another.  Fermi golden rule is the usual starting point although the density of states may not be that high at low vibrational levels. The calculation involves calculating the Franck-Condon weighted density of states from the initial to final level. The process is called IVR or intramolecular vibrational redistribution.
In many types  molecules, when isolated in the vapour phase, the initially excited level can still remain excited for many nanoseconds and the fluorescence lifetime and spectrum  from these levels can easily be measured, e.g. naphthalene, pyrene, benzanthracene etc. This is particularly the case when the FC excited mode has a distinctly different frequency and displacement to other modes so that the FC factors to other vibrational levels is small. An interesting case is that of many types planar aromatic dyes which show persistent wavepacket behaviour of the FC excited mode even in solution at room temperature.
The theory of radiationless transitions has been thoroughly worked out mainly for intersystem crossing but the theory is essentially the same with different operators. Look for papers from the 1970's by S. H. Lin et al, W. Siebrand et al,  J. Jortner et al. and K. Freed et al.


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What kind of bonds (σ or π) are formed by px, py and pz orbitals?
Our chemistry teacher said that two  $\mathrm{p}_z$ orbital only form sigma bond whereas  two $\mathrm{p}_x$ and two $\mathrm{p}_y$ orbitals form only pi bond.
He also stated that only $\mathrm{p}_z$ orbital forms a sigma bond with the $\mathrm{s}$  orbital.
But in our book it is written that $\mathrm{p}_x$, $\mathrm{p}_y$, and $\mathrm{p}_z$ orbitals can form both sigma and pi bonds.
Please give me the correct explanation.

See, the three p orbitals look exactly the same. How do you know which of them is px and which is pz?

@Ivan Neretin Ok so does that mean that any p orbital can form a sigma or pi bond.

Yes and no. Some people (including you teacher) use a different convention.

It's a question of
(1) labeling conventions (what you call x, y and z in the coordinate system defining the position of two bonded atoms)
and
(2) the orientation of the two overlapping orbitals from the bonding atoms
If an overlap maximum occurs on the internuclear axis between the atoms, you call the resulting bonding orbital a "sigma" orbital. If maximum overlap occurs on an axis parallel to the internuclear axis, you call this type of bonding "pi". Either s or p orbitals can form sigma bonding orbitals, but s orbitals generally form sigma orbitals, whereas of the 3 conventional p orbitals, only one has lobes oriented along the internuclear axis (this orbital is usually called $p_z$ because its principal symmetry axis coincides with the internuclear axis, which is also labeled the z axis) and forms sigma bonding orbitals. The remaining two (x and y) form pi orbitals. This is illustrated schematically in the following figure:

(note the AOs don't overlap significantly, focus is on showing their relative orientation and labels)

I appreciate to explain the following paragraph by using a picture "only one has lobes oriented along the internuclear axis (usually called $p_z$ because its principal symmetry axis coincides with the internuclear axis which is also labeled the z axis) and forms pi bonding orbitals. "

@AdnanAL-Amleh Thanks for bringing the garbled sentence to my attention. Also, I added a figure.

Thank you sir for the answer. I wanted to clarify a point. If we had taken the x axis(vertical direction ) as the internuclear axis then we would say that only the px orbital forms sigma bond and pz and py orbitals form pi bonds. It depends on which axis we want to place the nucleus of the atoms . Please correct me if I have misunderstood. Thank you

@AshokSharma Yes that is correct. But beware conventions, as I mention in my answer. If you have two atoms, some people (maybe your teacher) might always use the convention that the internuclear axis is the "z" axis and call the orbital on that axis pz.

Is it by convention that $p_z$ orbitals lies on the plane?

@Kaushik There is a lot of freedom in deciding what you call the "z" coordinate, but usually the convention is that x and y are related by a rotation symmetry whereas z is unique and not related to x or y by a rotation.

If $p_z$ is considered to be internuclear axis, then which orbital is filled first? Is it $p_x, \ p_y \ or \ p_z$? In this convention $p_z$ is forms sigma bond. Isn't? And if the order of filling $e^-$ is $p_x, \ p_y \ or \ p_z$ , how is $\sigma$ bond formed before $\pi$ bond (which is true) ? Because $p_x$ orbital is filled first which forms $\pi$ bond by the convention we chose? What have I misunderstood?

Sigma bonds usually form first (lower in energy), so in this case pz would be filled first. But I'd hate to generalize without knowing what particular problem you are referring to.

@BuckThorn please check the edited comment. please let me know if you don't understand my comment

@Kaushik The orientation of the orbitals between the atoms is more important then what you call them (x,y or z). If x lies in the internuclear axis, then px is involved in sigma bonds.

So orbital along internuclear axis is always filled first.

@Kaushik Please see my previous comments.


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Why protons combined with electrons will release heat?
During water electrolysis, the electrochemical reaction on the cathode side can be expressed as protons combined with electrons to form hydrogen. I believe this electrochemical reaction is a heat release process, but I don't know the reason.
Why protons combined with electrons will release heat?
It will be better if you could provide some references.

Well, there is the binding energy to consider...

If the process of uniting protons and electrons is assumed to occur isothermically in a vessel of constant volume, then the heat released will be equal to the change in internal energy of the system:
$$q_v=(\Delta U)_v$$
Now if the mixture were to behave as an ideal gas, the change in internal energy would be zero and therefore so would the heat. But we can expect a mixture of oppositely charged species such as protons and electrons to behave as an extremely non-ideal mixture, consistently with the large Coulombic potential between the particles. Since $\Delta U$ for the process of uniting protons and electrons (and thereby also minimizing repulsions) is negative, the process is then expected to be exothermic ($q<0$).

Thank you for the reply. Do you have references for me to read in order to have a better understand of the mechanism?

I recommend an introductory physical chemistry or thermodynamics textbook that explains basic electrochemistry. Perhaps if your background in basic electrostatic theory is not so strong, also reading a introductory physics textbook (undergraduate level).

I checked the Wikipeida, it said the electron affinity is the heat released (negative energy term) when a free electron combines with a neutral atom to make a negatively charged ion, and this process could release heat. In my case, protons (not neutral atom) combined with electrons to form hydrogen (not a negatively charged ion). Can I still use electron affinity to explain my problem?

The thing about your questions is that you attempt to take leaps between different scenarios. Take my answer for instance: I answered a very narrow (specific) question, what is the heat released when you combine proton and electron in an isothermal chamber. I think however you are more interested in what happens at an electrode when hydrogen is reduced: $\ce{2H+ + 2e- -> H2}$. There is at least one important additional step between this reaction and the case I present and that is the formation of an H-H bond.

So the affinity of a proton for an electron is definitely an important step of the process, so I'd have to answer that you should use the electron affinity, but that is not enough.

One way to think of it is the electrical potential energy of the separated charges is released (as heat energy, eventually) when the charges unite. First the energy will be released as photons, and these photons will be absorbed by nearby molecules, causing them to increase in kinetic energy, which on the macro scale, is heat.

If an isolated proton and electron combine when initially widely separated, energy is conserved and they will just separate again as total energy will be above dissociation energy. No photons emitted as initial state is a continuum. In a medium, say a liquid or gas, energy will be taken away from pair primarily by collisions and ultimately end up as thermal energy, ie translation, vibrational and rotational motion. Some emission is possible from excited levels but is slow process compared to collisions so not a major pathway

A proton combining with an electron doesnt necessarly release energy.All the posts forget that there is the weak and strong interaction.For example let's take hydrogen atom.If the electron from the 1s orbital is combined with the proton(core) it will leave a neutrally charged neutron.But neutron is unstable when it isnt in the core of elements , so it will decay into a proton , an electron and an electron anti-neutrino . So you see not all cases are what you describe.

This is chem.SE. We're not in the business of synthesising elements.

Hmm?Why synthesise elements?I didnt say anything wrong.It doesnt always release energy.

Honestly? You cannot produce neutrons via chemistry.

Yes there is always the possibility of something like this happening.In Hydrogen atom the electron is at the 1s subshell so it is possible for it to be fused with the proton of the core.

Ah, the famous inverse neutron decay. A very powerful and versatile cooling device. Only it doesn't work, because it can only take heat in chunks of 0.8 MeV.


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Calculating amount of base needed to achieve a specific pH in a mixed ethanol-water solvent
I'm looking for some way to calculate how much base I will need to add to an unbuffered solvent consisting of a mixture of ethanol and water in order to basify it to a specific reading (10) on an a pH probe calibrated with an aqueous standard.

I have several mixtures of ethanol and water, ranging from about 85% ethanol to nearly 100% ethanol, and I plan to basify these solutions.  Using the method defined by ASTM D6423, one of these solutions was measured to have a pHe of 5.2 due to the presence of some acidic species, but I don't expect these species to act as buffers.
Now, I'm aiming to get these solutions to a pHe of approximately 10, and in practice I'm just going to add the base and check the pHe as I go.  However, motivated by the fact that I'll potentially need to do this on quite a large scale at some point, I'm curious if it's possible to put a mathematical backing behind this.
If this were a pure water solvent, I would calculate $[\ce{OH^-}]=10^{-8.8}$, since the self ionization constant of water is $K_W = 10^{-14}$ and $K_W \approx [\ce{OH^-}][\ce{H^+}]$.  From there it's relatively simple to calculate the amount of base I should need to add to reach $[\ce{OH^-}] = 10^{-4}$, resulting in a pH of 10.
If this were a pure ethanol solvent, I would do something similar using the self ionization constant for ethanol, except nobody has published one as far as I can tell.  This answer also failed to find one.  The Wikipedia page on the leveling effect depicts ethanol having a wider acid-base discrimination window than water, so I would expect $K_{EtOH}$ to be somewhat higher than $K_W$, but no numbers are given.  I checked the source of the graphic as well, and they also give no numbers.
Finding $K_\ce{EtOH}$ wouldn't even fully solve the problem though, as there is a non-negligible amount of water in the solvent.  The only source I was able to find discussing pH in mixed-solvent system was this one, which discusses using a correction factor ($\delta$) to allow the measurement of something analogous to pH in mixed solvent systems using an electrode calibrated in an aqueous standard.  However they only determined this correction factor in solutions of up to about 72% ethanol, and the relationship is decidedly non-linear, as shown by this plot:

We know that when our probe reads 10 we'll get good results with the rest of the process, so presumably this means in the math I would want to target a pH of $10+\delta$, were $\delta$ known.
This is as far as I've been able to get with my own literature search.  If anyone here has any insight into this problem I would greatly appreciate it!

EDIT: Additional detail.  I just had some 200 proof ethanol measured by the ASTM D6423 method, and the pHe was measured as 6.5.  This is counter to what I expected.  I expected it to have a pHe greater than 7 since pure ethanol should be a somewhat weaker acid than water based on everything I've read.  I'm not certain if this is helpful information at all, but it's another data point.

I'm looking for some way to calculate how much base I will need to add to an unbuffered solvent consisting of a mixture of ethanol and water in order to basify it to a specific reading (10) on an a pH probe calibrated with an aqueous standard.

I think you have summarized the main issue yourself. The reading of a pH probe in a basic alcoholic solution is just a positive real number. It has no significance whatsoever. The value of 10 means nothing. The pH probe has no idea that it is being dipped in an aqueous organic solution. I am sure the probe will take a very long time to equilibrate to give a stable "real number" output. Since those numbers such as pHe =10 have no theoretical basis it is hard to calculate anything. This is confirmed by a quick look at the standard method.
The ASTM D6423 method says " Readings are taken at exactly 30 s because the meter reading will change throughout the analysis due to glass-electrode voltage drift caused by dehydration of the glass-electrode membrane. Because alcohol dehydrates the glass-electrode membrane, time spent soaking in alcohol solution should be minimized.
4.3 The electrode is soaked in water-based pH 7 buffer between readings to prepare it for the next sample. This rehydrates the glass electrode, a necessary step to preserve the electrode's response characteristics.
4.4 The electrodes are cleaned/rehydrated at least every ten samples by alternately soaking several times in 1 mol/L NaOH solution and 1 mol/L H2SO4 (or 1 mol/L HCl) for about 30 s each. New electrodes are treated by this procedure before first use."
The practical way to "calculate" perhaps is to do a acid base titration. Calculate the titer for each batch with a suitable indicator and add the amount of base accordingly.

Well, the reading on the probe has some significance at least to our process outcomes, and it is at least analogous to pH in that adding basic compounds moves the value upward while adding acidic compounds moves it down.  But I take your point that using the same scale as pH is basically an arbitrary choice and that those numbers no longer correspond directly with concentration.  If we scale this process up I guess I'll just need to generate a range of calibration curves for the process.

Calculations of pH might not take all the factors into consideration, and the high and variable ethanol content is an unusual restriction.
If the unknown acidic species is a contaminant or simply unwanted, and you desire a reproducible constant reference pH, you might consider treating your ethanol solutions with magnesium hydroxide. In water, it will give about pH 9.5. It should absorb/adsorb the acid. If the acid salt becomes an undesired solute, the ethanol solutions could be filtered thru a column of Mg(OH)2; this would more likely adsorb and remove the magnesium salt.
MgO is less reactive than Mg(OH)2. MgO comes in 2 forms, light burned and dead burnt, and either of these might be better because of particle size (settle faster) or column operation (faster thruput).


text:
Titration of barium(II) hydroxide by hydrochloric acid

$\pu{25 ml}$ of $\pu{0.024 M}$ $\ce{HCl}$ is titrated into a volumetric flask which initially contains $\pu{45 ml}$ of $\pu{0.034 M}$ $\ce{Ba(OH)2}$. Calculate the $\mathrm{pH}$.

This is how I calculated.
$$\ce{Ba(OH)2 + 2HCl -> BaCl2 + 2 H2O}$$
Moles leftover / excess moles will determine the $\mathrm{pH}$ because the others that have reacted will neutralise each other.
So calculating moles of $\ce{Ba(OH)2}$ and $\ce{HCl}$ in their respective volume
$C=\frac{n}{v}$, hence $n= c \times v $.
Moles of $\ce{Ba(OH)2} = \pu{0.034M} \times \frac{45}{1000} = 1.53 \times 10^{-3}$
Moles of $\ce{HCl} = 0.024 \times \frac{25}{1000} = 6 \times 10^{-4}$ but $2\times6\times10^{-4}$ since it’s $\ce{2HCl}$ which is $1.2 \times 10^{-3}$.
Now to find leftover moles, you’ll have to subtract as such
${1.53} \times 10^{-3} - 1.2 \times 10^{-3} = 3.3 \times 10^{-4}$, this is the excess moles of $\ce{Ba(OH)2}$.
Looking at solution of excess $\ce{Ba(OH)2}$
$\ce{Ba(OH)2  -> Ba^2+ + 2 OH-}$
Since $\ce{Ba(OH)2}$ : $\ce{2OH-}$, moles of $\ce{OH-} = 3.3 \times 10^{-4} \times 2 = 6.6 \times 10^{-4}$ moles of $\ce{OH-}$
So, concentration $= \frac{6.6 \times 10^{-4}}{\frac{25+45}{1000}} = \frac{6.6 \times 10^{-4}}{0.07} = 0.009..$
$\mathrm{pOH} = -\log \ce{[OH]} = -\log[0.009] = 2.026$
Since $\mathrm{pH} + \mathrm{pOH} = 14$, $\mathrm{pH} = 14 - \mathrm{pOH}$
So $\mathrm{pH} = 14 - 2.026 = 11.97$
HOWEVER, the book is saying $\mathrm{pH} = 12.55$. I have done this problem multiple times and cannot figure it out.

"Moles of $\ce{HCl} = 0.024 \times \frac{25}{1000} = 6 \times 10^{-4}$ but $2\times6\times10^{-4}$ since it’s $\ce{2HCl}$ which is $1.2 \times 10^{-3}$".

The amount of reacted $\ce{Ba(OH)2} =\frac{\text{the amount of}~\ce{HCl}}{2}=\frac{6\times{10^{-4}}}{2}=\pu{3\times{10^{-4}
mol}}$

Your solution is correct up to the point you assumed that you can double the concentration of hydrochloric acid.
Unfortunately, this is wrong and not at all how stoichiometry works.
Let's focus on what's important.
In the nutshell, we are dealing with a typical neutralization reaction:
$$\ce{H3O+ + OH- <=> 2 H2O}$$
and note that $\ce{BaCl2}$ as a salt of a strong acid and a strong base won't undergo hydrolysis and, as a consequence, won't have a noticeable impact on $\mathrm{pH}$.
As you [semi-]correctly calculated, there is an excess of a base ($n_0$ values refer to the initial amounts):
$$
\begin{align}
n_0(\ce{H3O+}) &= n(\ce{HCl}) \\
  &= c(\ce{HCl})\cdot V(\ce{HCl}) \\
  &= \pu{2.4e-2 mol L-1}\cdot\pu{2.5e-2 L} \\
  &= \pu{6e-4 mol}
\end{align}
$$
$$
\begin{align}
n_0(\ce{OH-}) &= 2\cdot n(\ce{Ba(OH)2}) \\
  &= 2\cdot c(\ce{Ba(OH)2})\cdot V(\ce{Ba(OH)2}) \\
  &= 2\cdot\pu{3.4e-2 mol L-1}\cdot\pu{4.5e-2 L} \\
  &= \pu{3.06e-3 mol}
\end{align}
$$
One can see that there is a tremendous excess of hydroxide of the following amount:
$$
\begin{align}
n(\ce{OH-}) &= n_0(\ce{OH-}) - n_0(\ce{H3O+}) \\
  &= \pu{3.06e-3 mol} - \pu{6e-4 mol} \\
  &= \pu{2.46e-3 mol}
\end{align}
$$
and the leftover concentration is
$$
\begin{align}
c(\ce{OH-}) &= \frac{n(\ce{OH-})}{V(\ce{HCl}) + V(\ce{Ba(OH)2})} \\
  &= \frac{\pu{2.46e-3 mol}}{\pu{2.5e-2 L} + \pu{4.5e-2 L}} \\
  &= \pu{3.51e-2 mol L-1}
\end{align}
$$
Finally, as you rightfully suggested (use the correct value for $c(\ce{OH-})$, however):
$$
\begin{align}
\mathrm{pH} &= 14 - \mathrm{pOH} \\
  &= 14 + \log(\pu{3.51e-2}) \\
  &= 12.55
\end{align}
$$
Notes

Using the term "moles of X" is incorrect.
The proper terminology here is the amount of substance.
Never try to substitute a physical quantity with its unit.
Watch out for the proper capitalization, which is actually a big deal in natural sciences.
$\ce{HCl}$ is a formula for hydrogen chloride, whereas $\ce{HCL}$ is not and may denote "hydrogen-carbon-ligand" or be an abbreviation.
Same goes for math operators, including $\mathrm{pH}$ and $\mathrm{pOH}$: $\ce{Ph}$ is phenyl and $\ce{PH}$ can be read as "phosphor-hydrogen", or be yet another abbreviation.

Thank you. I do have some questions. 1) why is it "H3O+ OH -> 2H2O" and not "2H + 2OH -> 2H2O" or are these equivalent? 2) Why did you multiply the moles of OH by 2? Is it because of Ba(OH)2 having 2*OH? 3) From where I went wrong and looking into your 1st bullet point in the notes, what topic should I look into? I don't exactly know what I don't know if that makes any sense.

@soho 1. Yes, both are pretty much equal (check out Balancing redox equations in acidic media: to use H+ or H3O+?); 2. As it's mentioned, $\ce{Ba(OH)2}$ is a strong base: $$\ce{Ba(OH)2 <=>> Ba^2+ + 2 OH-}$$ 3. See the most recent Meta post Buzzwords Season 9 - Number of Moles (reboot) and have a look at the links inside.


text:
Is fluorine also hybridised in carbon tetrafluoride?
I had read that when an atom forms multiple bonds, it hybridises it's orbitals to minimise the repulsion and decrease the overall energy. But do the terminal atoms also exist in hybridised form?
E.g. in $\ce{CF4}$, are fluorine orbitals also $sp^3$ hybridised, or maybe has some other type of hybridization, or no hybridization at all?
Also in a molecules like say $\ce{HF}$, in this case is $\ce{F}$ hybridised or not?

Hybridisation is a mathematical concept, you can use it or not, it does not make a difference. It cannot be measured. Please also note that atoms cannot be hybridised, only orbitals can. It is more than sloppy to say that 'F is hybridised'.

@Martin-マーチン In other words, you mark «F is hybridised» with a F.

@Martin-マーチン Hybridised orbitals do exist in NBO calculations, which is essentially VB theory but more perfect.

I see that you keep commenting things like "not to use hybridisation for TM complexes" which is only useful for properties depending on the canonical MO's, e.g. the energies(ionisation, transition etc). Hybridisation is perfectly OK for intuitive things- go calculate some chromium carbonyl's NBO's.

@KanghunKim While indeed, NBO localises orbitals to be interpreted as hybrid orbitals, it's in no way as complete as VB theory; and especially it is not more perfect. It is an interpretation of a wave function, nothing more. It's fairly obvious that it has a lot of benefits, but it cannot provide a complete picture. Also, if you take the time to read the comments you refer to on the TM complexes, they always refer to the central atom. Go ahead and run the calculations, I very much doubt you'll find hybrid NBO on chromium. Except of course you're using a really old version.

No Fluorine is not hybridised because every atom of Fluorine in CF4 make one bond which is the typical according to the octet rule.

No, Flourine does not undergo hybridisation generally. In Carbon tetraflouride, there is a sigma bond formed by sp³- 2p orbital. Same applies for carbon tetrachloride and bromomethane.(atleast from exam's point of view) lone pair electrons of Flourine is in s and p orbitals.


text:
Is there a reliable chemical theory that predicts pKa based on structure?
Obviously, there are general stability arguments that can be made to estimate relative pKas, such as evaluating the stability of the conjugate base of an acid, or thinking about how polarized the bond to hydrogen is.
However, is there a theory that can be absolute and account for the many exceptions to the above guidelines? What are the factors that are preventing a rigorous theory from being proposed?

Unfortunately acid/base strengths are still major weak points in quantum theory; there are no rigorous and affordable theories or approximations available. Most approaches have to be calibrated against highly accurate experimental measurements.

This 2015 review article is behind a paywall, but its abstract seems to provide a good summary of the state of the field, namely that it continues to be an active area of research, and there are two basic approaches researchers are pursing:  https://onlinelibrary.wiley.com/doi/abs/10.1002/wcms.1218

@theorist the article you mention is available to the public, either on request on research gate (https://www.researchgate.net/publication/274141867_Computational_estimation_of_pKa_values) or directly and freely (no paywall), e.g., http://mercuryconsortium.org/furman/pubs/SeyboldsS2015.pdf

thanks for the responses, I talked to Seybold, and it seems that using rigorous quantum chemical calculations is too computationally expensive. It is the solvent effects and the corresponding entropy changes that make things complicated really fast.

@NathanMatthieuTang It would be great if you could summarize the comments and your conversation with Seybold in an answer to your own question.

@KarstenTheis, Seybold wrote: "The central problem in predicting pKas comes from the difficulty in accounting for the interactions between the acid compounds and the surrounding solvent.  Gas phase acidities, which are described in terms of delta Gs, can be rather accurately estimated if one applies sufficiently advanced quantum techniques. But solution acidity computations, which usually rely on continuum approximations for the solvent, aren't really up to the challenge.  Remember that one now must account for thermodynamic alterations in both the solute compound and the surrounding solvent."

@NathanMatthieuTang I would again encourage you to summarize comments and the quote in an answer. Eventually, someone else will probably write one, but you should get credit for your own research.

Somewhat surprising that this hasn't been linked here: Calculating pKa's using Gaussian


text:
Why are there only 6 series (Lymen, Balmer, Paschen, Bracket, Pfund and H.Davy) in the hydrogen atomic spectrum? Why can't there be more?
The Balmer series, for example, is formed when the electron jumps from $n_2 = 3, 4, 5, \ldots, ∞$ to $n_1 = 2.$
The Humphry Davy series (The last series is formed when electron jumps from $n_2 = 7, 8, 9, \ldots, ∞$ to $n_1 = 6.$
Can't there be another series formed when electron jumps from $n_2 = 8,9,10... \ldots, ∞$ to $n_1 = 7?$

There can be and surely is another series, and another after that, and another after that, and another after that. It is just that we ran out of names. After all, infinity is infinite, and the number of guys who ever lived is finite (that is, a bit smaller).

I agree with Ivan. There are certainly other series that are theoretically known but are yet practically observed by scientists. Hence there are no name for the series yet.

@NilayGhosh They are observed, it's rather a matter of notability.

Note also that for a large enough value of $n$. The differences between energy level $n$ and $n+1$ is small enough that they are very similar. As an exercise, you can try to figure out the critical value of $N$ such that for $n>N$, the difference in energy between levels $n$ and $n+1$ is less than threshold $E$. That's just a long way to say, as $n$ increases, the differences between series get small.

@Zhe 1.What does threshold E mean though?                                                       2. I understood the part where you said that the energy difference gets smaller as the radius increases.. Does that mean that if I have a REALLY SMALL energy difference it won't form a spectral line, or is there some other explanation?

It'll still be a line, but it might blend a bit depending on how sharp the line is. If you don't understand the comment about thresholds, don't worry. It was just an exercise if you have that kind of training.

As Ivan commented, there are actually an infinite number of possible series of this type. So your question is really why there are only six named series.  The reason is part of the culture of science.  Typically, a result is named if it is, to use Mithoron's term, sufficiently notable.  Thus we have the Diels-Alder reaction, the Hartree–Fock method, the Wentzel–Kramers–Brillouin (WKB) approximation, the Southern blot, and so on.
Hence the reason why the initial series of hydrogen spectra were named is because they were notable scientific results. [In addition—and perhaps someone more expert in the history of science can comment on this—it is possible that identifying the, say, fifth and sixth series weren't that scientifically notable, but a tradition had become established of naming successive spectra of hydrogen.] But beyond the sixth, finding more series was essentially deemed routine. The last published series I know of is the seventh, identified by Hansen and Strong, but not named after them [Peter Hansen and John Strong, "Seventh Series of Atomic Hydrogen," Appl. Opt. 12, 429-430 (1973)].
That's not to say it wouldn't be experimentally difficult to see higher-numbered series, since they become increasingly faint. Rather, it's that, at this point, finding them isn't interesting science.


text:
The Trend Between Boiling Point and Solubility in Organic Chemistry
Is it true in assuming that a higher boiling/melting point means that an organic compound will be more soluble in water?
I'm trying to distinguish between the solubility of aldehydes vs. ketones based on boiling point. I received a question in a test today about it, comparing propanal with propanone, based solely on boiling points given, and my structural understanding of the two. I thought propanone had lower solubility of the two, as the ketone group is surrounded by carbons, while the propanal is exposed. Scouring through the internet, I'm not so sure my answer was correct!

You mean propanone as in acetone? It is miscible in water, whereas propanal has limited solubility. Both compounds have similar bp and mp (particularly bp), so solubility would be very difficult (imho) to decide on the basis of this alone.

As andselsik pointed out, definetley no in the general sense. Just think of some practical examples like PTFE (Teflon) - which melts at about 330 °C but is definetly not water soluble.
However there is one connection I'd like to point out that allows to relate the enthalpy of melting to the solubility:
Consider phase equilibrium between a soluble compound C and water. At equilibrium, we have equality of chemical potentials:
$$\mu^{solid}_c(p,T,x_s)=\mu^{liquid}_c(p,T,x_l)$$
Assume that $x_s$ at equilibrium is 1 (there would be a tiny fraction of water in the solid, but its fair to neglect that). Then the LHS is just the chemical potential of the pure solid:
$$\mu^{*,\:solid}_c(p,T)=\mu^{liquid}_c(p,T,x_l)$$
Now using the solvent convention for the chemical potential of the liquid mixture:
$$\mu^{*,\:solid}_c(p,T)=\mu^{*,\: liquid}_c(p,T)+RT\mathrm{ln}(x_c^l \: \gamma_c)$$
Rearranging this and recognising that a chemical potential difference of a pure substance across a phase boundary is the gibbs energy of that phase transition gives
$$\Delta g_{c;\: melt} = \Delta g_{c;\: S \rightarrow L}=\Delta \mu_{c;\: S \rightarrow L} = RT\mathrm{ln}(x_c^{l} \gamma_c)$$
and
$$\Delta g = \Delta h + T \Delta s$$
using the limiting behaviour of the activity we find that for $x_c , \gamma_c \rightarrow  1$, the RHS tends to zero; we can reexpress the entropy of the phase change by setting $\Delta g$ in the above equation zero, and substitute the result into the equation obtained from equality of chemical potential:
$$\Delta h_c^{*,\:melt} \Big(\frac{T}{T^{melt}}-1\Big) = RT\mathrm{ln}(x_c^l \gamma_c)$$
...
$$x_c^l\gamma_c=\mathrm{exp} \Bigg( \frac{\Delta h_c^{*,melt}}{R} \Big(\frac{1}{T^{*,melt}}-\frac{1}{T}\Big)\Bigg)$$
The activity coefficient itself is obviously also a function of the composition, so this does not give us a straightforward way to calculate the solubility of compound C (it does however provide us with a route to $\gamma$, which is normally harder to measure than a solubility) but it shows that phase change properties have all sorts of implications!

What do you mean with " phase equilibrium between a soluble compound C and water"?

Between solid solute and solvated solute. There is mechanical, thermal and chemical equilibrium and two distinct phases, so this is also a phase equilibrium

Ok, maybe you want to edit for clarity.

By the way, the OP asks about the connection between the boiling point and solubility, not the melting point.

I interpreted "higher boiling/melting point" as referring to both boiling and melting point

This is a bit too complex for my understanding (only a high school student) but what would you say for the solubility of comparable ketones and aldehydes? Is the boiling point not concrete evidence for higher solubility? Thanks!

I think its hard / next to impossible to deduct a very general trend. However a high bp can point to hydrogen bonding, which would mean favourable interactions with water. For propanon / propanal your approach was right but I'd argue that the keton is so miscible because the oxygen is in a symmetrical central position, whereas the aldehyde has a hydrophilic and a hydrophobic end - maybe there is more, surprised it would have such a strong effect with a 3C chain length (that two-tail effect is used in tensides / soaps, where the chain length is around 12-18 Cs)

$\gamma_{c}\to 1$ would apply only for sparingly soluble solutes. The assumption would deviate further away from reality at higher $x_{c}$ values.

The boiling point is directly dependent on the various forces of attraction that the chemical species can exert. However, for solubility, it is the relative strength of these forces of attraction.
Note that higher molecular weight species, in general, boil at higher temperatures due to greater magnitudes of van der Waals interactions.
Boiling Points
The boiling point of organic compounds of comparable molecular weight increases as the types of forces of attraction increase. As an example, methane exhibits only van der Waals interactions. Formaldehyde shows polar interactions as well. Ethanol can also hydrogen-bond. Note that this is with respect to a pure component, i.e. no solvent has been introduced yet.
Propanal and propanone (acetone) show van der Waals interactions and polar interactions.
Solubility
Imagine methane in water. Methane molecules can have only van der Waals interactions with water. Moreover, by just being present inside the bulk of water, methane molecules disrupt the hydrogen bonding between surrounding water molecules. Water is "unhappy" with this loss of stabilization and methane, as expected, has a very low solubility in water. In an aqueous solution of formaldehyde, some hydrogen bonding between water molecules is disrupted. But, formaldehyde can hydrogen-bond with water due to its $\ce{O}$. In this case, the overall energy of the system is better reduced when formaldehyde is completely mixed with water. The case for ethanol is similar.
In a propanal-water and acetone-water systems, there are van der Waals interactions, polar interactions and hydrogen bonding (as with formaldehyde). Their magnitudes, however, differ, resulting in different solubilities.
Conclusion
Boiling point and solubility depend on similar properties. Solubility also depends heavily on the solvent properties. There is no direct correlation between boiling point and solubility (for non-homologoues)


text:
Balancing chemical reactions, issues of Gauss's elimination method
I am trying to balance some chemical reactions by using the Gauss' elimination method and I noticed that this method does not work if the creation of a matrix with exactly one degree of freedom is not possible.
In this example
$$\ce{H2O2 + NO3- + H+ → O2 + NO + H2O}$$
I succeeded in creating just 4 equations
$$\ce{\alpha H2O2 + \beta NO3- + \gamma H+ → \delta O_2 + \epsilon NO + \zeta H2O}$$
$$
\left\{
\begin{aligned}
2\alpha+\gamma&=2\zeta &\qquad &(\ce{H})\\
2\alpha+3\beta&=2\delta+\epsilon+\zeta &\qquad &(\ce{O})\\
\beta&=\epsilon &\qquad &(\ce{N})\\
-\beta+\gamma&=0 &\qquad &(\text{electrons})
\end{aligned}
\right.
$$
obtaining
$$(\alpha, \beta, \gamma, \delta, \epsilon, \zeta)=\left(\zeta-\frac{1}{2}\epsilon, \epsilon,\epsilon,\frac{\zeta+\epsilon}{2}, \epsilon, \zeta\right)$$
I know that the solution should be $(3,2,2,3,2,4)$, but, in having two degrees of freedom, I am not able to choose the two free variables in order to obtain what I expect, i.e. $\epsilon=2$ and $\zeta=4$.
Is there any (just one more) equation that I missed?
Is there anything else that I have to add further?
Or do I have to consider that Gauss' elimination fails with problems with a number of free variables $\neq 1$?

Also see https://chemistry.stackexchange.com/questions/111375/finding-all-chemical-reactions-given-products-and-reactants/111718#111718

Your "should-be" solution,
$$\ce{3H2O2 + 2NO3- + 2H+ → 3O2 + 2NO + 4H2O}$$
is a linear combination of
$$\ce{H2O2 + 2NO3- + 2H+ → 2O2 + 2NO + 2H2O}$$
and
$$\ce{2H2O2  → O2 +  2H2O}$$
Any other linear combination is valid as well.

I did first write this answer and then voted to close the question, which might be viewed as a mixed message. I'm happy to delete this answer if there are more down votes (1 up 1 down as I write this).

There's no prob. with answer on duplicate.


text:
What makes determining entropy changes theoretically so hard compared to enthalpy?
It is really hard to determine the entropy change of a reaction accurately because of subtle solvent effects and a lack of standard entropies. On the other hand, enthalpy is relatively easy and through a thermodynamic cycle, it can usually be calculated. But what about our knowledge of entropy limits our ability to measure it? Without using straight empirical data, why is using force field and electronic interactions not very reliable? Many reactions face this problem, like the mixing of liquids.

Generally speaking, what you say is not true, tables of standard entropy (and enthalpy ) can be found easily, such as in the data section of Atkins' phys chem textbook. Any problems of accuracy will exist for both entropy and enthalpy.

Since G = H-TS, any inaccuracies in H or S reflect on each other directly.

Porphyrin, that is simply not true when it comes to solvent effects. You can't just use standard entropies when you are trying to solve for the solvation entropy change. Also there are many compounds that do not have standard entropies in tables when it comes to acids and bases.


text:
Is any symmetry element conserved during a supra-antarafacial cyclo addition of two cis-butadiene units
I tried to construct a correlation diagram for the [4+4] cyclo addition of two cis-butadiene units. But I failed to construct any meaning full orbitals for the "educt" side where the two butadiene molecules still have their pi-systems intact.
I thought that a C2 rotation axis is conserved during a supra-antarafacial reaction pathway but aligning the pi system of both units leads to a conflict. The backbone of the supra-facial part cannot be combined with the antarafacial unit to sustain C2 symmetry. This would only be possible for a combination of a cis-butadiene with a trans-butadiene. But without any conserved symmetry element correlation is impossible.
So I'm wondering if the Woodward-Hoffmann rules are applicable at all in this case or if I am being too strict with symmetry. Is it legit to neglect the backbone when thinking about the symmetry?


text:
Is cyclic ether oxidised by periodic acid?
Is the three member ring of cyclic ether oxidised by periodic acid? As this also has oxygen attached to adjacent carbon so I think that cyclic ether of three members should be oxidised by periodic acid.
I know the mechanism of oxidation of 1,2 diols by periodic acid but ain't able to relate to this question. Please help me out here.

Well if you manage to hydrolyse the oxirane first I suppose it will oxidize it.

Cyclic ethers usually do not get oxidized by periodic acid. However, epoxides would do only if they were acidified first or use acidified periodic acid in the reaction (Ref.1). In general, acidified periodic acid has been used as a qualitative test for the presence of epoxide moiety in organic compounds (Ref.1). The procedure is a modification of the qualitative test for  glycols (Ref.2). Hence, epoxide must be converted to 1,2-diols first by acid and oxidation by periodic acid must be followed. The test is described as follows:

Exactly 2 drops of concentrated nitric acid is added to 2 ml. of a 0.5% solution of periodic acid, and 1 or 2 drops of the unknown is added. Water-insoluble unknowns should first be dissolved in 2 ml. of dioxane or acetic acid. The mixture is shaken and 1 or 2 drops of 5% silver nitrate is added at room temperature. A positive test is the appearance of a white precipitate of silver iodate which usually forms immediately, but may require up to 5 minutes for complete precipitation. In all cases a blank should be run for comparison. The test is negative for simple aldehydes, ketones, and alcohols.

References:

R. Fuchs, R. C. Waters, C. A. Vanderwerf, “General Qualitative Test for Epoxides,” Anal. Chem. 1952, 24(9), 1514 (https://doi.org/10.1021/ac60069a047).
R. L. Shriner, R. C. Fuson, Identification of Organic Compounds; 3rd ed., John Wiley & Sons: New York, NY, 1948, pp. 115-116.

This quote is from Comprehensive Organic Name Reactions and Reagents by Zerong Wang.

The oxidation of adjacent diols with periodic acid or its salt in aqueous solution is generally known as the Malaprade reaction. Several solvents have been used to increase the solubility of organic substrates and the reaction proceeds faster under acidic conditions. This reaction has been further extended to the cleavage of α‐hydroxy carbonyl compounds, 1,2‐dicarbonyl compounds, α‐amino alcohols, α‐amino acids, and polyhydroxy alcohols. .

The mechanism of Malaprade reaction reaction is as follows:

As per the requirements of mechanism , the three membred ether should be converted to a adjacent diols (viscinal) diol for it to be oxidised by periodic acid.

The epoxide in acidic hydrolysis undergos ring opening to give 1,2 diol .This would undergo oxidation with periodic acid.


text:
What are the limitations of the Theorem of Corresponding states?
I recently came across the theorem of corresponding states, which states that all gases exhibit the same compressibility factor, under same reduced pressure and temperature (and also the critical compressibility factor). Thus:
$$Z=f(P_r,T_r,Zc)$$
where
$$Z=\frac{PV}{RT}$$
$$P_r=\frac{P}{P_c}$$
$$T_r=\frac{T}{T_c}$$
My question is, what are the limitations of this theorem? Can I consider it as an absolute truth, or is it just like any other gas law, being valid only within certain ranges and for certain acccuracies?

Not "how valid" but ask "what are the limitations" from the start.

The van-der-Waals and Redlich-Kwong equations (and perhaps others)  can both be recast in terms of reduced quantities therefore, in so far as an actual gas is described by  these equations, the the law of corresponding states should apply. Certainly gases/vapours as diverse water, CO2, N2 and isopentane, among many others, follow this law very well. Your question is thus about the applicability of, for example, the vdW equation.

Corresponding states very nearly causes the curves describing the state behavior of all gases to collapse to a single set of curves.  In 1955, Pitzer added the ascentric factor as an additional parameter, which improves the correlation even more.  The ascentric factor of each individual gas accounts for differences in polarity of molecules.  In practice, corresponding states does a superbly accurate job.

@ChetMiller thanks for this info, so does this mean that I can write $Z$ as $$Z=f(P_r,T_r,Z_c,\omega)$$ where $\omega$ is the acentric factor?

The usual form is $$Z=Z^{0}(P_r,T_r)+\omega Z^{1}(P_r,T_r)$$See Smith and van Ness, Introduction to Chemical Engineering Thermodynamics.  They provide the plots of $Z^0$ and $Z^{1}$


text:
Is there any bad side effects for using calcium chloride to replace a running dehumidifier?
My dehumidifier costs to much to run, so I’m thinking about using calcium chloride to keep my basement dry(er) ALSO, this may keep my mouse problem to a minimum... my questions are -Will this be harmful to me breathing this when I’m in the basement?
-Will any gases be omitted to me in my main floor?
-Will this remove to much moisture that my block foundation will start to degrade?
-Are there any other side effects I’m not aware of?
The only person that goes in basement is me so I’m not concerned about anyone consuming it.

It is only harmful when eaten (and even then, not much), won't release any gases, and I don't believe you can remove too much moisture this way or any other way. That said, why do you think it is cheaper?

@IvanNeretin - perhaps they are ignoring the fact they will need to dry the material over and over again in an oven? Of course, filling the basement with calcium chloride is a good way to stop worrying about the humidity levels there...

...for a while.

Actually, some commercial products intended for room drying are made of calcium chloride (scented, in some cases), so no worries. I guess it works when humidity is relatively contained, anyway: it is cheap, but you can't expect the performance of an electric dehumidifier without buying a lot of it

As for safety, calcium chloride, $\ce{CaCl2}$, is in sea water and added to foods, such as beans and pickles. It's about as dangerous as table salt, sodium chloride, $\ce{NaCl}$ (actually, $\ce{CaCl2}$ is safer for concrete).
However, it absorbs only so much water until it dissolves and stops being effective. What would you do with the puddle of $\ce{CaCl2}$ solution? It can be dried with heat and reused, but then your just moving energy usage from a continually running electric dehumidifier to a batch process using $\ce{CaCl2}$... though if you can get heat for free, it would be cost effective! If you don't have a solar concentrator or access to geothermal energy, there would probably be no energy savings, though.

Concrete samples have been rapidly destroyed when
stored in CaCl2 solutions under certain conditions (Peterson, 1991). The most critical
concentration and temperature seems to be 30 % and 5 °C, respectively. [Peterson, C.: The chemical effects on cement mortar of solutions of calcium
Magnesium acetate and other deicing salts, Report TVBM-3045, Lund Institute of
Technology, Sweden.]

@user482745, thank you for that concrete info. How does that compare with immersion in NaCl solution?


text:
Will silicon dioxide react with hydrochloric acid?
I have the following equation, based on the reaction of $\ce{SiO2}$ with $\ce{HF}:$
$$\ce{SiO2 + HCl -> SiCl4 + 2 H2O}$$
I want to know if this reaction is indeed feasible, what conditions may be needed to induce it, or if it isn't possible why that's the case.

No, HCl does not etch SiO2 appreciably.

@JonCuster You say 'appreciably', what do you mean by that?

You can store HCl in a glass bottle. Please don't do that with HF.

Well, Cl is very much not like F. For this reason, the reaction is totally impossible. Or rather, it will go quite readily, only backwards.

@IvanNeretin What is it about Cl that makes it unsuitable compared to F? Is it simply proton count or something more?

@JonCuster Rest assured I'm well aware of the risks of dealing with HF, I'm trying to develop a process that avoids it.

@faissaloo - in grad school somebody really messed up a hood by being lazy and pouring HF waste into the (glass) waste bottle sitting there (for aluminum etchant). What a mess to clean up!

Well, they are different chemical elements; isn't that enough of a justification for any of their properties to be different? In particular, size matters. F is small, which makes the formation of $\ce{SiF6^2-}$ possible and even desirable. Not so with Cl.

HCl reacts with oxides of Ge, Sn and Pb but not Si. See this answer: https://chemistry.stackexchange.com/questions/60757/which-hydrogen-halides-react-with-group-14-oxides?r=SearchResults

The combination of both those comments answers my question.

The reaction works with a high efficiency in the opposite sense : $\ce{SiCl_4}$ reacts quite easily with the slightest amount of water to produce huge fumes of silica + HCl. This reaction has been used in war times for hiding tanks advancing in the direction of the enemy.

According to the paper by Chung and Sturm [1], $\ce{HCl}$ will do a very small amount of etching at high enough temperatures (525 °C will give you about 1 nm/min), but the presence of hydrogen limits the reaction because the formation of $\ce{HCl}$ is preferred to the formation of $\ce{SiCl4}.$
References

Chung, K. H.; Sturm, J. C. Chlorine Etching for In-Situ Low-Temperature Silicon Surface Cleaning for Epitaxy Applications. In ECS Transactions; ECS: Chicago, Illinois, 2007; Vol. 6, pp 401–407. DOI: 10.1149/1.2727426.


text:
Why are atomic radii calculations all over the map?
If you look up an element on Wikipedia, it will list covalent, ionic, Van Der Waals, high-and-low -spin, and finally 'empirical atomic radius'.
Then, on the page for atomic radius, the chart of 'empirically-determined atomic radii' shows numbers that are all higher than on the individual elements' pages.
Then, farther down on the Atomic radius page, it lists 'calculated radii' which are even higher!
(Almost always, in all these cases, these numbers are supposedly referring to covalent states).
The latest measurements seem to be getting further and further away, smaller and smaller, from the calculated radii.  What gives?
The 'high-spin covalent' number refers to electron spin, not nuclear, right?  But when is a covalent bond in a high spin state?  I thought most molecules were in a singlet state?

"Atomic radius" is an umbrella term for a great many values, all vastly different, barely related at all.

Cross-posted to Physics.

I'm voting to close this question as off-topic because it's cross-posted here.


text:
How do scientists study liquid tungsten?
Inspired by this What If article: https://what-if.xkcd.com/50/
In the above article, Randall Munroe mentions that liquid tungsten is difficult to study because of its extremely high melting point. Because of this property, containers for the tungsten tend to melt before the metal itself. This might be a fairly elementary question, but given this difficulty, what methods do modern chemists use to study it, or other materials with extreme melting points?

One way is to use electristatic or electromagnetic forces to hold the liquud in place.  This abstract refers to a "non-contact method" and an "electrostatic levitate", which us enough to reveal the basic strategy.  Unfortunately the article is behind a paywall, so you have to put up to get details on the good stuff.
Industrially, metals like tungsten are not processed as liquids.  They are ... see here.


text:
Prove that idempotency is a necessary and sufficient condition for a densiy matrix to be N-representable
As I understand, a necessary and sufficient condition for a density matrix $P$ to be represented by a wavefunction $|\Psi\rangle$ is that it is idempotent, i.e. $P^2=P$. It is easy to see that if $P=|\Psi\rangle\langle \Psi|$ then $P$ is idempotent:
$$
P^2=|\Psi\rangle\langle \Psi|\Psi\rangle\langle \Psi|=|\Psi\rangle\langle \Psi|
$$
since by normalization $\langle \Psi|\Psi\rangle=1$. However, I have had trouble with the other direction - how can I show that if $P^2=P$ then $P=|\Psi\rangle\langle\Psi|$ for some $\Psi$?
I have been able to demonstrate that if $P^2=P$ then
$$
P=\sum_{j} |\Psi_j\rangle\langle\Psi_j|
$$
where the sum over $j$ has no more terms than the dimension of the density matrix $P$. My argument was analogous to the one found here. I suspect that I'm almost there - I think it's just a change of basis that I haven't seen yet. How can I complete this proof?

A density matrix $D$ is positive semidefinite, hermitian, and has trace one.
Because of hermiticity we may assume that it is diagonal.
Let's denote the eigenvalues with $\lambda_i$ .
Because it is positive semidefinite, we have $0 \leq \lambda_i \leq 1$.
The matrix $D^2$ has eigenvalues $\lambda_i^2$. Since it is idempotent ($D = D^2$) and has trace one, we may write:
$$\sum_i \lambda_i = \sum_i \lambda_i^2 = 1$$
The only possibility for this to be true is, if there is one and only one $j$
for which $\lambda_j = 1$.
So $D$ has one $1$ in the diagonal at the $j$-th column and all other values in the matrix are zero.
You can easily represent this matrix with:
$ e_j e_j^T$. ($e_j$ being the $j$-th unit vector.)
If you transform back to you original nondiagonal form, you obtain your $\Psi$.

Thank you! I see what I was missing - the key had been in the defining properties of the density matrix.

I think that what you have described is a pure state, a stationary state (eigenstate of the time-independent Schr. Eq.). However if you have a superposition or coherence then I don't think D would be reducible to the form you describe, or?

If you have no pure state, then D would not be idempotent. Because lambda would not equal lambda^2.


text:
Is it possible to dissolve concrete with acid
So I have this question. Can you dissolve concrete with acid? I know that there are many types of concrete, but the common and dominant component, as far as I know, is $\ce{Ca3SiO5.H2O}$ (hydrated tricalcium silicate). Will the acid react with it? What kind of acids are best for the job? Specifically I'm interested in nitric acid ($\ce{HNO3}$). If the reaction is possible, I also want to know, will it generate heat and how much?

From experience, Nitric Acid will happily react with concrete producing NOx. Cannot give the exact reaction and the generated heat, so will leave it to others to answer.

There is a thing called chlorine trifluoride which is able to ignite sand and concrete.

Well, chlorine trifluoride would ignite most of things, including human body. Not much suitable for home experiments.

@Jeppe Nielsen  I suppose HNO3 will not dissolve concrete, but at the best will convert it to some silicate / hydrated silica based residue.

Concrete is frequently treated with dilute hydrochloric acid to etch the surface prior to painting or coating with epoxy. After treatment with acid, good rinsing is suggested. This etch treatment will react with CaCO3 at the surface (Ca(OH)2 produced as a product of the hydration of cement reacts with CO2 in the air to form CaCO3 over a period of months), and it will likely bubble up a little. Then the acid will react with Ca(OH)2 which is deeper in the concrete, and some sand and stones will be loosened. If you brush these away, the acid can go deeper.
Nitric acid offers no better results.
Silica is generally formed after acid treatment, but is so finely divided that it can be rinsed off. If you want to dissolve the silica, you would use hydrofluoric acid, carefully, because of its human health hazard.
The sand and stone will not be affected much at all.
If you have lots of time, you could use mild reagents, like sugar dissolved in a little water. The concrete beneath tanks where sugar is purified suffers degradation as this very weak acid just complexes the calcium into solution. Months, years, maybe, depending on how much sugar and how much rinsing away there is.
Heat generated is fairly small because the reaction is slow with bulk concrete, especially if it is strong, but can get larger if you grind up the concrete first.

Hmm, I would think that Nitric Acid would perform significantly better.

In commercial practice, concentrated muriatic acid (HCl) is diluted to acid wash concrete or mortar. Nitric acid is more expensive, more hazardous when concentrated and turns your skin yellow if it doesn't burn it off. (The rinse-off would be a better fertilizer if it were neutralized enough so as not to kill vegetation!) You might think faster is better, but not only would nitric acid be no faster, a slower etch is often desired for better control. Hydroxyacetic acid is used for slower etch and complexation of multivalent metals to prevent discoloration of concrete.


text:
Electronic configuration in [Ni(CN)4]2-
$\ce{[Ni(CN)4]^2-}$ is a low spin , inner orbital complex having $\mathrm{dsp^2}$ hybridisation. According to Crystal Field Theory, $\ce{CN^-}$ is a strong field ligand that causes pairing of electrons i.e electrons occupy the orbitals of the lower $t_{2g}$ level completely before they occupy the higher $e_{g}$ level. If I am not wrong, Hund's rule of maximum multiplicity is followed while filling the orbitals of $t_{2g}$ or $e_{g}$.
$\ce{Ni^2+}$ has a configuration of $[Ar]3d^8$. If we write down the electronic configuration of $\ce{Ni^2+}$ in this complex, 6 of the d-electrons would occupy the $t_{2g}$ level and the remaining 2 would occupy then $e_{g}$ level.
If we search for diagrams of electronic configuration of $\ce{[Ni(CN)4]^2-}$, many of them show the last 2 occupying the same orbital in $e_{g}$ which seems to contradict Hund's rule in my view. But I am pretty sure that the the diagrams on the Internet are correct according to experimental data of magnetic moments and other features.
My question is whether strong field ligands can pair up electrons within orbitals of $t_{2g}$ and $e_{g}$ like in this case where the 2 $e_{g}$ electrons are paired up within one orbital instead of letting them occupy two separate ones. Or is there some other factor at play?

Please see: Why is it wrong to use the concept of hybridization for transition metal complexes?

There is no such thing as $\mathrm{t_{2g}}$ and $\mathrm e_g$ in $\ce{[Ni(CN)4]^2-}$. Those labels only apply to octahedral complexes. So the question does not really make much sense.

Your thinking is correct. Within a set of degenerate orbitals (i.e. equal in energy), the orbitals will fill one electron at a time, and only after all have one electron will they start to pair up.
The situation with $\ce{[Ni(CN)4]^{2-}}$ is that it has square planar geometry, so the two orbitals that are $e_g$ in an octahedral complex are separated in energy. This diagram from Chemistry LibreTexts shows it nicely:

Because of the separation in energy, the $\mathrm{d}_{xy}$ fills completely (two electrons) before any electrons fill the $\mathrm{}_{x^2-y^2}$.


text:
Which molecule has maximum bond angle amongst BF₃, BCl₃ and BBr₃?
$\ce{Br}$ is the least electronegative from the given compounds. So, in $\ce{BBr3}$ the electrons will be the closest towards boron and thus due to electron-electron repulsions $\ce{BBr3}$ will have maximum bond angle.
Steric hindrance due to large size of bromine atom can be a factor too.
But the bond angles are same according to most of the websites.

I think for judging bond angles you should take a look at VSEPR structures for them. For all practical purposes, the changes due to electronegativity are negligible in most cases.

Unfortunately, I still don't get the motive of this question.

How would the bond angles be different?  If you have a ball and stick model, how do you move the bonds around so that you don't have an equilateral triangle (due to symmetry)?

@EricTowers I think the angles they're comparing between are the X-A-X vs X'-A-X' angles. Each of the X-A-X angles will necessarily be identical, but for different identities of A and X, you can change the angles (e.g. ammonia being pyramidal).

You might find this related answer interesting https://chemistry.stackexchange.com/questions/14981/why-does-bond-angle-decrease-in-the-order-h2o-h2s-h2se/57973

@Eric Towers Yes! It is now so obvious. Because the shape is trigonal planar for all of them, hence it must be 120*.

With VSEPR theory you can predict the structures of $\ce{BX3}$. Since boron only has three valence electrons, all of which are used for σ-bonding to the halogen atoms, they arrange in a trigonal planar fashion. This means that all molecules $\ce{BF3}$, $\ce{BCl3}$, and $\ce{BBr3}$ have the same point group $D_\mathrm{3h}$. From this symmetry you can deduce that the bond angle is $\angle(\ce{X-B-X}) = 120^\circ$ for all of them.
Bent's rule won't have any effect on that.
There is also considerable π-backdonation in these molecules, but that only has an effect on bond lengths, not bond angles.


text:
Protein purification pressure limits confusion, about the X-MPa limit for FPLC?
I have always been a bit confused about column pressure limits. I have to elaborate a bit on the problem to get to my question:
In protein purification we use the äkta purification machines (e.g. äkta explorer), which always come with a high pressure and low pressure alarm setting for the column in their unicorn software. The most common column is the IMAC 5-mL HisTrap column to capture proteins that contain e.g. a 6xhistidine tag. When I check the column specifications I can see that the column has a flow limit of 0.3MPa which should be rounded up to 0.5MPa for the alarm (as the flow restrictor gives an additional 0.2 MPa pressure).  The second type of column is the SEC column, one example is the analytical "superdex 200, 10/300 GL" for separation of protein between 10 and 300 kDa. The specification of this column is that the  $\Delta$p is 1.5 MPa, however, its typical that most people are sceptical to go above 0.5 MPa for any type of column.
So my questions regarding pressure are as following:

The 0,5MPa limit for a HisTrap column is what is called the $\Delta$p pressure limit?
It always the specified  $\Delta$p for each individual column that I have to care about? (not the pre or post column pressure limits)?
Can I then assume that it is always safe to run the column all the way up to the specified $\Delta$p limit to e.g. equilibrate or wash the column? So for the mentioned SEC column, I could run in at a  $\Delta$p value of 1,5MPa and assume that the column and protein are just fine?

Proteins are fine at high pressure. Beads don’t like high pressure difference, and columns crack if total pressure is too high. So look up column limits ( different for plastic, glass and steel) and chromatography medium (material, bead size matters).

yes i know, thats why i specified the delta-p for the column, but your answer is then - yes i can be 100% sure that its okay to run the column up to the specified delta-p limit ?

edited and formulated my questions a bit better


text:
Why does the earth have more radon in some locations?
I am not a chemist so I'm not going to give specific amounts of radon; different countries use different units. My question: why do some areas have more radon? For instance, Australia has a spreadsheet of all locations and their radon levels. I've looked it up and where I used to live, nearby, has half the listed Radon levels of where I live now. Wikipedia was a bit general to me for American examples:

Iowa has the highest average radon concentrations in the United States due to significant glaciation that ground the granitic rocks from the Canadian Shield and deposited it as soils making up the rich Iowa farmland.

So it is rock or soil types or weather or something else that causes the different amounts?

Well, there are parts of the earth that have significantly more water than others.

To me that is a bit unsound. Water is complex: ice, snow, moisture. I know where your angle is: radon often comes from groundwater.

While radon and groundwater may have some association (I truly don't know), Radon is part of the U238 decay chain, so what you are looking for is types of rocks, particularly those that contain uranium. Granite happens to be one.

My point was actually why would you expect uniform radon levels around the earth? Yes, the comment was snarky, and I apologise for that, but the point remains: the earth is not uniform, so there will be differences in abundance in certain elements. So the why part of your question can't really be answered (at least not on chemistry.se), but the part of where it comes from might be answerable.

Your comment is not to me scientific; in my OP quote from Wikipedia there are characteristics of the earth that lead to higher or lower Radon levels. Also, human influence has to be factored in such as uranium mining.

@athornton To me this seems more like a question of "What are the different natural or artificial events that lead to radon production?"; obviously radon is more likely to be found where such events are abundant. Their distribution is uneven, hence, so is the radon distribution. Could you please clarify if this is not your question?

So, you need uranium-bearing rocks (granite is good). In the Canadian shield, that means huge pieces of solid, continuous granite. So, any radon generated inside a huge chunk of granite isn't getting out anytime soon, and will decay, ultimately making lead at the end of the decay chain. Now, pulverize the rock and all of a sudden much of the radon being produced gets out of the time bit of rock and into the soil, diffusing up to the surface where humans are and can breath it in and potentially get cancer from the radon decay occurring in their lungs.

https://certmapper.cr.usgs.gov/data/PubArchives/radon/georadon/3.html would be an excellent place to start.

BTW, the moon got its minimal argon "atmosphere" the same way - decay of potassium, in this case.

Radon nuclides are short-lived members of the natural radioactive decay chains of thorium and uranium. On a time scale that would be relevant for geology, you should better ask why some locations have more uranium or thorium. The distribution of radon concentrations in the ground air is a result of the distribution of uranium or thorium in the ground.
The main reason for the inhomogeneous distribution is igneous differentiation.  The original liquid magma might have been rather homogenous. But when the magma cooled down, fractional crystallization started. This removal of mineral precipitates also changed the composition of the remaining melt. Uranium and thorium ions are incompatible, i.e. they do not fit well in the crystal lattices of the silicate minerals that precipitate first. Therefore, they were accumulated in the remaining melt. Primary deposits of uranium and thorium minerals (such as pitchblende) were formed where the last parts of the cooled magma solidified.
Furthermore, secondary deposits were formed when uranium or thorium minerals were dissolved, transported and precipitated in ground waters. (This includes the nicely coloured and fluorescent uranyl ($\ce{UO2^2+}$) minerals.


text:
Lead nitrate and sodium iodide
As we know that if you mix lead nitrate and sodium iodide solutions together, a yellow precipitate forms. Can you maybe add nitrate in water that may or may not have lead in it and then mix it with a sodium iodide solution and get the same results if lead was present in the water?

First, the iodide precipitates with lead ions, $\ce{Pb+}$, called cations because of the positive charge. It doesn't matter what the anions are in solution, whether nitrate, acetate or other, so there's no need to add nitrates or nitric acid for this test.
Second, yes, this is a test for lead in solution, but it is not at all sensitive. At room temperature, about 0.75 g of the product, lead iodide or $\ce{PbI2}$, could dissolve in a liter of water. That means that there could be a fatal amount of lead in water, yet that wouldn't be enough to form a yellow precipitate. One approved test for lead in drinking water is an atomic absorption (AA) spectrophotometer.
This reaction is not sensitive enough to be a useful test for lead, and the amount of lead needed to show a positive reaction is dangerous, so this should not be done by unwary students (unless the teacher is trying to reduce class size). Finally, safe disposal of the chemicals would be an issue.


text:
Determining geometry of coordination compund without experimental data
How can I determine if a coordination compound has a tetrahedral or square planar geometry without any experimental data (such as magnetic spin) by just looking at its formula?
For example, what is the geometry of the following complexes:

$\ce{[CoCl4]^2-}$
$\ce{[Ni(CN)4]^2-}$
$\ce{[CuCl4]^2-}$

Do they have a Jahn-Teller distortion?

Your question is like asking "how do I tell whether an object weighs more than a kilogram without measuring its mass". For some objects, it's obvious: a feather is definitely under 1 kg, an elephant definitely over. For some other things, your eyes aren't good enough and the only way to know is to put it on a scale. A more nuanced question, and probably better, would be to ask what factors tend to favour tetrahedral geometries vs square planar geometries. Some of them are touched upon here: https://chemistry.stackexchange.com/q/40880/16683

Thank you for your time.
I know that question is a bit odd but this is a question from my last inorganic chemistry exam. Well, let me write that question exactly,
Wich one of those ligands you can observe Jahn-Teller Distortion? and why?
[CoCl4]2−
[Ni(CN)4]2−
[CuCl4]2−

As speculation: if it was an examination question, perhaps especially an oral one, see it as a check about «how would you answer this later in professional life by  yourself»: display both the pathway (your reasoning «because of chlorine ligands ...») as well as your answer («so x and y were square panar, z however tetrahedral») plus a probe («which could be experimentally confirmed by method A»).  For your comment, put forward your new arguments -- even if they do not lead yet to an answer, it will 1) show us where to help you and 2) show you to yourself what you already understood.

For example [Ni(CN)4]2− I've drawn energy diagram for that compound both tetrahedral and square planar, put electrons with respect to strong field ligand and as you did say and I've made consumptions about Jahn-Teller distortion for both geometries.
What i couldn't get is, if this the right way to do it? Note that i don't have any extra data

@orthocresol So for coordination compounds, there's  nothing similar to VSEPR,  hybridization or VBT that helps predict the shape of complex without experimental data?


text:
Metering Device decreases pressure yet produces saturated liquid?
This question pertains to the change of state of liquid using a Metering Device as the refrigerant enters an Evaporator Coil of an HVAC system:

The Metering Device takes liquid and then reduces the pressure and then spits out 75 percent liquid and 25 percent vapor. I don't understand why decreasing pressure, thus decreasing temperature will create a saturation point where liquid is both water and vapor. Shouldn't it be the other way around? Decreasing pressure should change saturation point so the liquid is both water and ice.
This is my rationale behind what I say:
In terms of changing of state of a liquid, as you add more heat content (molecules in motion or BTU), such as 340 BTU all the way up to 1310 BTU, all the liquid is converting into gas. This is latent heat. So adding heat changes liquid to gas. Conversely, when we remove heat it changes state of liquid to ice. At 0 degrees F and 0 BTU, you have ice.
In terms of gas laws, if you compress a gas to reduce its volume, both the temperature and pressure will increase. If you expand a gas to increase its volume, both the temperature and pressure will decrease.
We can use PSI or PSIA (absolute) to measure pressure per square inch of area. If you look at a Pressure Temperature Chart for Water, 212 Degrees F and 14.7 PSIA is at saturation where there is both liquid and vapor. If we add any heat, it will change state from liquid to vapor.
If we reduce the atmospheric pressure (e.g. from 14.7 PSIA to 0.33 PSIA) and if we put the water into a vacuum (e.g. from 29.6 inHg to 0.68 inHg), then we can change the state of the water at 70 degrees F. Why? Because decrease in pressure means decrease in temperature.
Consequently, how is reducing pressure creating vapor in the case of the Metering Device?

I'm a bit confused by your question -- your diagram shows the path of refrigerant through the cooling system, but you seem to be discussing water's state changes.
Liquid refrigerant under pressure enters the metering device. As it leaves the device at a lower pressure, it boils, producing some vapor, but not evaporating completely. The rest of the liquid evaporates as it passes through the evaporator coil, absorbing heat as it does so.

The refrigerant in the HVAC system is not water. In this case, apparently, it is R22 (chlorodifluoromethane).
(Note that use of the units °F, psi, inHg, BTU is deprecated. You might want to consider using proper SI units. Furthermore, you should not use gauge pressures when making thermodynamic calculations since such values depend on the ambient pressure; you should use absolute pressures instead.)
The R22 enters the metering device at a temperature of $T_0=105\ \mathrm{^\circ F}=40.6\ \mathrm{^\circ C}$ and a pressure of $p_0=278\ \mathrm{psi(g)}=20.2\ \mathrm{bar}$, which corresponds to a specific enthalpy of $h_0=250.26\ \mathrm{kJ/kg}$ (IIR reference state). Since the equilibrium pressure of R22 at this temperature is $p=211\ \mathrm{psi(g)}=15.5\ \mathrm{bar}$, the refrigerant is liquid at this point.
When passing through the so-called metering device (i.e. through an orifice), the pressure of the refrigerant is relaxed to $p_1=69\ \mathrm{psi(g)}=5.8\ \mathrm{bar}$. The equilibrium temperature at this pressure is $T_1=40.3\ \mathrm{^\circ F}=4.6\ \mathrm{^\circ C}$. The corresponding specific enthalpy is $h_{1,\mathrm l}=205.44\ \mathrm{kJ/kg}$ for liquid R22 and $h_{1,\mathrm g}=406.71\ \mathrm{kJ/kg}$ for gaseous R22.
Assuming a fast and adiabatic process, the evaporated fraction $x$ of R22 can be estimated using a simple enthalpy balance as follows:
$$\begin{align}
h_0&=x\cdot h_{1,\mathrm g}+(1-x) \cdot h_{1,\mathrm l}\\[6pt]
x&= \frac{h_0-h_{1,\mathrm l}}{h_{1,\mathrm g}-h_{1,\mathrm l}}\\[6 pt]
&= \frac{250.26\ \mathrm{kJ/kg}-205.44\ \mathrm{kJ/kg}}{406.71\ \mathrm{kJ/kg}-205.44\ \mathrm{kJ/kg}}\\[6pt]
&=0.223\\[6pt]
&=22.3\ \%
\end{align}$$
which is quite close to the given $x=25\ \%$.

As you stated: If we reduce the atmospheric pressure (e.g. from 14.7 PSIA to 0.33 PSIA) and if we put the water into a vacuum (e.g. from 29.6 inHg to 0.68 inHg), then we can change the state of the water at 70 degrees F. Why? Because decrease in pressure means decrease in temperature.
One minor correction is that a decrease in pressure means a decrease in saturation temperature, not necessarily a decrease in temperature. What causes a decrease in temperature is when a gas' pressure is reduced which causes the gas to expand.
If you lower the pressure of water then you can change the state of it into a vapor at a lower temperature. The same is happening with the refrigerant when it goes through the metering device. The refrigerant pressure is reduced, so it starts changing phase into a vapor. When that liquid starts changing phase into a vapor, it expands which causes the drop in temperature. That's my simple explanation.


text:
Prime in substituted benzene
I was sent data for a molecule referred to as 1,2'-bis(4-aminophenoxy)benzene. I haven't been able to figure out if that is a real molecule.
If I drop the prime, I know 1,2-bis(4-aminophenoxy)benzene should look like that:

and Google agrees. But with the prime, I have no idea what the molecule could be. Is there such a molecule, or is it a typo?

Your picture doesn't show 1,2-bis(4-aminophenoxy)benzene.

I believe the prime in the name, 1,2'-bis(4-aminophenoxy)benzene, is a typo as you originally suspected. However, the compound in your picture you thought is the correct one is neither 1,2-bis(4-aminophenoxy)benzene. The compound in the picture is 1,4-bis(4-aminophenoxy)benzene (CAS#: 3491-12-1).
Nonetheless, 1,2-bis(4-aminophenoxy)benzene exists (CAS#: 49860-97-1; Ref.1). It was synthesized and used to make ortho‐linked aromatic polyimides with aromatic dicarboxylic acids (Ref.1).

Reference:

Chin‐Ping Yang, Jeng‐Jieh Cherng, "Synthesis and properties of aromatic polyamides derived from 1,2‐bis(4‐aminophenoxy)benzene and aromatic dicarboxylic acids," Polymer Chemistry 1995, 33(13), 2209-2220 (https://doi.org/10.1002/pola.1995.080331315).

This is a typo. The prime 2' would refer to position 2 in a separate ring, but you are only numbering the central ring.
Note that you have drawn the structure of 1,4-bis(4-aminophenoxy)benzene.
You can test the names of chemicals and their structures using a tool such as OPSIN.


text:
Product of this Michael addition
I'm having problems when trying to know the outcome of this reaction.

The first step must be a Michael addition of the 1,3-dicarbonyl compound to the  α,β-unsaturated carbonyl compound, followed by a acid hydrolisis and a decarboxylation of the ester.

LDA must deprotone the least hindered hydrogen, that must be the $\alpha$-$\ce{CH3}$ to the non cyclic ketone. But at this point, I don't know how this reaction can continue to give one of the above products.

Intramolecular aldol reaction?

@user55119 So it was as easy as a Robinson annulation, but the LDA step was driving me crazy, it makes sense now, thanks.

I don't think you need LDA to finish an aldol condensation. The first step probably effect the Robinson annulation with the ester group intact. Acid hydrolysis would lead to CO2, methanol and compound C because the precursor is a vinylogous beta-ketoester. Probably a made-up problem. –

Although the work of Wilds and Werth1 does not align perfectly with the conditions of the OP's question, it does illustrate that Robinson annulation and decarbomethoxylation can occur under alkaline conditions. Catalytic base was found to be optimal to effect the Michael addition with methyl vinyl ketone (MVK). The aldol condensation was conducted in aqueous KOH at reflux.
Whether decarbomethoxylation occurs prior to aldolization (2-->3-->5) or subsequently (2-->4-->5) was not addressed. α'α-Dialkylated β-ketoesters generally undergo attack at the keto group rather the carboxylate by base. This generality may be tempered by the aromatic nature of the keto group in 2. Initial aldolization of 2 to 4 affords a structure that is a vinylogous β-ketoester; it is susceptible to base initiated decarboxylation.

1) A. L. Wilds and R. G. Werth, J. Org. Chem. 1952, 17, 1154.

The OP's original question has some flaws. Since he  is asking the most possible product among the given four, I'd give reasonable mechanism to achieve that product, which would be C. Following is the suggested mechanism:

Late addition: Evidently, lithium diisopropylamide (LDA) has been used to selectively abstract proton from the least steric position among three available acidic hydrogens (irreversible generation of the kinetic enolate). Such a non-nucleophilic, sterically-demanding, strong base will always abstract a proton from the least hindered side. The resultant lithium enolates will avoide Proton transfer at low temperatures in ethereal solvents, so that addition of a second carbonyl group will produce the desired aldol product (Organic Chemistry Portal). For an example for use of lithium dialkyl amide in aldol condensation, see the given reference:
Atsushi Seki, Fusae Ishiwata, Youichi Takizawa, Masatoshi Asami, "Crossed aldol reaction using cross-linked polymer-bound lithium dialkylamide," Tetrahedron 2004, 60(23), 5001-5011 (https://doi.org/10.1016/j.tet.2004.04.026).

Mathew Mahindaratne: Do you have a reference for the use of LDA in the context of your example? If so, could you post it in your answer?  Conceptually, it works! Practically, does it? I'm curious. PS: In your fourth structure, the keto form is undoubtedly better than the enol form.

@user55119: I have added a paragraph and a reference for your request.