text:
Determine the velocity of the O fragment after the dissociation

A $\ce{SO3}$ molecule is radiated with light with a wavelength of $\pu{193 nm}$, which results in photodissociation.
$\ce{SO3} + \text{photon}(\lambda=\pu{193nm})\ce{-> SO2 + O}$
It is measured that the $\ce{SO2}$ fragment is in the electronic ground state, but in the vibrational excited state where bending is in $v=3$.

Calculate the velocity of the $\ce{O}$ fragment after the dissociation.

I know that:

The $\ce{SO3}$ molecule is at rest before the radiation, and we don't have to look at the rotation.

The dissociation energy is $\pu{5.0 eV}$

The normal vibration frequencies for $\ce{SO2}$ is $\omega_1= \pu{1151 cm^{-1}}$ (sym stretch), $\omega_2= \pu{518 cm^{-1}}$ (bending) and $\omega_3= \pu{1362 cm^{-1}}$ (asym stretch)

My first thought was to approach it like the photoelectric effect where you can determine the velocity of the ejected electron. But I can't seem to figure out how to do it using the dissociation energy and vibrational modes. I hope someone can help me with this?

Your instinct is correct. Convert all units into a common energy unit. Then subtract the vibrational energy remaining in $\ce{SO2}$ from the incoming photon energy. The result is the kinetic energy of $\ce{O}$. Convert that into a velocity.

What about the dissociation energy? Do I have to subtract that as well?

If I subtract it as well, I get the velocity to around 3510 m/s

Yes, of course, you also have to subtract the dissociation energy.


text:
Detecting trash (plastic) burning smoke with sensors
I’m not a chemist, but a software engineer. I’m trying to build a device that could detect if someone is burning trash (mostly plastics and similar materials) vs wood or diesel. At my disposal I have the following sensors: https://www.mysensors.org/build/gas
What sensor or combination of them would be most reliable to detect smoke resulting from burning illegal waste?

This is not trivial (especially in the sense of what is illegal - is breakdown product A of a tire "legal" versus breakdown product B).  You'll find a lot of pertinent research under the airport/national security sector (John Kauer's research on the "electronic nose" might be a good jumping off point).

Even though this is a good question, I have a feeling that those gas sensors are too primitive and probably won't consistently detect "good" or "bad" smoke as it consists of numerous complex organic molecules (which are actually making it hazardous or add certain smell). Gas chromatography is what would give a definitive answer, but it requires a bulky complex device.

It doesn’t have to be 100% accurate. It only has to give enough probability to merit sending a physical patrol with more accurate equipment for suspicious smoke in chimneys.

The typical gas sensor marketed to hobbyists and enthusiasts (except for $\ce{CO2}$) are of the metal oxide semiconductor type, not to be confused with MOS as in MOSFET.
How do these work?
In these detectors, a semiconductor material made of the oxide of a metal, usually tin.
I'll quote from this question:

...sintered composite based on the semiconductor material $\ce{SnO2}$, which is the metal oxide. The different sensors have different admixtures and operating conditions to achieve different sensitivities to different gasses.
The principle of operation of MOS-type gas sensors is explained in Figaro sensors' Operating principle; MOS type.

The Figaro link shows the operating principle in the GIF below:
 GIF (Source: Figaro)
When the $\ce{SnO2}$ is heated by the (usually 5V but not always) heater supply, oxygen in the air will dissociate at the surface and oxygen ions will stick to the surface, depleting the semiconductor and cutting off the flow of current. When a carbon-containing gas (e.g. $\ce{CO}$ or $\ce{C_xH_y}$ (but not $\ce{CO2}$)) comes along, it will combine with the adsorbed oxygen and remove it from the surface. The more gas, the lower the equilibrium concentration of adsorbed oxygen and the higher resulting current flow through the semiconductor.
Not as gas-specific as you might think
By adjusting the amount and types of additives to the sintered $\ce{SnO2}$ and adding catalyzing metal clusters to its surface, the manufacturer can enhance sensitivity to some gases and reduce it to others, but there is no single-gas specificity with these sensors.
Since the carbon in $\ce{CO2}$ is already fully oxidized, it won't react with the adsorbed oxygen. Those sensors (e.g. MG-811) use a chemical cell sensor based on a solid electrolyte. They need very high impedance amplifiers that measure millivolts, a bit like pH sensors, in fact, a lot like pO2 sensors!
CO and VOCs
I would recommend you use a $\ce{CO}$ sensor (e.g. MQ-7, MQ-9, MQ-307A or MQ-309A) since burning garbage will likely be incomplete (not in an industrial incinerator) and a "Air Quality" VOC sensor like MQ-135.
You can experiment with others, some people like to build an "electronic nose" which is an array of a bunch of these, then they plot the data for different kinds of smoke and environments to see what happens, and try to use the pattern to identify the likely problem.
PM2.5
Inefficient combustion can also produce particles as soot, and there are more expensive "dust sensors" or "PM2.5 sensors" which measure small particles (2.5 microns or larger) in the air. Soot is a hallmark of incomplete combustion (think of the smoke from a badly tuned engine) and a huge problem from incomplete or low-temperature combustion from a garbage fire.
None of these will be conclusive, you'll have to look at your data and look at when there is a fire, and learn how the signals correlate to the events.
Ultraviolet light
There are UV light based fire detectors as well, but you'd need direct line-of sight access to the flames. If the sensor is designed to be solar-blind by adding a filter that removes all wavelengths from the Sun that make it to the ground, then it might be helpful here. However, they are usually designed to work indoors where indoor lighting may have less UV than outdoor sunlight. This would take some more investigation.
read the data sheets carefully
Finally, read each sensor's data sheet carefully and note carefully the specific voltages that each sensor needs. The "Air quality" sensors need you to run for a certain amount of time (about a minute) at one voltage, then a different amount of time at another voltage because the surface chemistry is a bit more complex.
Sensors can also be found here:

http://wiki.seeed.cc/search.html?q=gas+sensor
https://www.pololu.com/category/83/gas-sensors

source: Stack Exchange

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

sort: votes


text:
The effect of lactose monohydrate on acid-base extraction of tolterodine tartrate
I am doing an organic chemistry project for school in the form of an acid-base extraction of tolterodine tartrate (Detrol) in ammonium hydroxide (or sodium hydroxide, haven't decided yet) and dissolving the freebase in benzene.
I am planning on crushing up the tablets, and dissolving them in water. Almost all the inactive contents of the tablets are not soluble in water except for lactose monohydrate. Will the presence of lactose monohydrate effect the reaction of the acid base extraction when I add the hydroxide?

If your intention is to release free base and get it to organic layer, then do not use $\ce{NaOH}$. You may noy able to isolate tolterodine, because of its phenolic moiety. I'd used $\ce{NaHCO3}$ instead. Also, please use toluene in place of benzene.

I agree, NaOH is too strong a base and will deprotonate the phenolic -OH.Lactose is highly water-soluble and unlikely to extract into your organic phase. DO NOT USE BENZENE. You should not even have access to benzene

Thanks for the response. But my question is will the presence of lactose monohydrate in the disolved suspension (along with the detrol) effect the acid base extraction? It shouldnt right? My only question is how lactose monhydrate would potentially impede the free basing process. Im flexible with which solvent ill use.

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


text:
Why does chlorine gas have a higher boiling point than hydrogen iodide
Hydrogen iodide, $\ce{HI}$, is a dipolar molecule much larger than chlorine, $\ce{Cl2}$.
The melting point of $\ce{HI}$ $(222.35\ \mathrm K)$ is definitely higher than that of $\ce{Cl2}$ $(171.6\ \mathrm K)$. However, the boiling point of $\ce{HI}$ $(237.79\ \mathrm K)$ is less than that of $\ce{Cl2}$ $(239.11\ \mathrm K)$.
This is very confusing since $\ce{HI}$ has more London dispersion forces than $\ce{Cl2}$. It also has a dipole, which chlorine does not.
A similar question to this looked at how silicon tetrafluoride $(\ce{SiF4})$ had a higher melting point than sulfur tetrafluoride $(\ce{SF4})$, and the answer concluded that the exception happened because symmetrical molecules fit into lattices better than asymmetrical molecules. However, I am not sure if this applies to the situation between $\ce{Cl2}$ and $\ce{HI}$ because they both only contain two atoms. This situation is also in the opposite order, where the molecule with greater London dispersion forces and a dipole has a higher melting point but a lower boiling point.
With this in mind, can anyone explain why this exception between $\ce{Cl2}$ and $\ce{HI}$ exists?


text:
Difference between relaxation and resonance leading to an absorption spectral feature?
I need help understanding the physics behind this insightful comment below the question Does water really have strong EM absorption at 3 kHz in solid and 2 GHz in liquid? Why the huge shift?:

Please note: This is a relaxation spectrum. The transitions do not come from a resonance (absorption, like in NMR, IR, etc.), but an applied electric field has a lossy (at low frequencies) or elastic (at higher f.) effect on some mode of molecular orientation of electrical dipoles in your sample.

The following comment links to https://en.wikipedia.org/wiki/Dielectric_spectroscopy which confirms this.
Both mechanisms can to a peak in the imaginary part of the dielectric permittivity at a certain frequency or energy, and at this point the slope of real part is steepest.
But is there a simple yet intuitive way to understand the physical difference between a spectral absorption feature due to a resonance process versus a relaxation process?

Two things. Concerning the actual example, the link and comments should suffice. It is the nature or level of interaction that differs. The interaction between the applied em field and the sample results in a different orientation of the dipoles in the sample and it is not quantised at molecular level. Each same dipole would equally interact with the field as in a merely electrostatic interaction. The spectroscopic character is due to scanning the frequency at which the field is reversed, so that different samples adapt to it differently, promptly or not, or even not at all.

Concerning terminology, relaxation can be encountered also in a more molecular spectroscopy context, when indeed transitions due to resonance between the applied em field and quantised energetic levels (electronic, vibrational, rot. and their mixes) take places. Except that you won't scan for them, but parameters like the time needed for the sample to return back. A simple example: you excite a molecule at an energy you know it suffices to give luminescence emissions, than you follows the luminescence decay in time. Here you have a resonance process to pump a relaxation phenomenon.

@Alchimista What is the overall positive result of leaving compressed answers as comments? I really like your answers when you write them up!

For me nothing. The overall result is well discussed in the linked answer. For you: you are the only one knowing if my comments were useful or not. I did compress anything.  Just unable to formulate a nicely concise answer. If the comments weren't useful I can delete them. Let me know.

Didn't notice you weren't polemic so perhaps I have been so in last comment. But it answers well. I do have an intuitive answer, the problem is to convey that. I am confident I went the right direction. I can add one more thing: in the original Q you mention absorption peak. But while this might be done in a colloquial scenario, the peaks are in Permittivity, at the end. Not Absorbance. I think this note helps, too.

I think that they will be helpful to someone who is in the process of formulating an answer so please leave them here...

"Both mechanisms can to a peak in the imaginary part of the dielectric permittivity at a certain frequency or energy, and at this point the slope of real part is steepest."  No. At a resonant QM absorption, you would get a peak in the real part, and a wiggle in the imaginary, but they return to the  same value on the left and right side.

@Karl what I wrote seems to agree with this plot from the linked Wikipedia page in the question, which shows the peak in the imaginary part of the dielectric permittivity, and the wiggle (if it occurs) in the real part.

That plot is a pure relaxation spectrum, the peaks come from rotation of dipoles, shifting of charges agains their equillibrium position, and shifting of the whole electron cloud. No quantised transition. Look at the abscissa, the peaks have a width of nearly an order of magnitude. Not peak in IR, UV-Vis or NMR can ever be so broad.

One more comment: Your question title says "absorption", but that's not what's on the ordinate of a relaxation spectrum. Permittivity is certainly related to absorptivity (and refractivity), but it's not the same.

@Karl distinction without a difference. If you have a sample of ordinary matter and  you have a non-zero imaginary part of the permittivity you have absorption, and vice-versa. Use $(n+ik)^2=\epsilon'+i\epsilon"$ to get $k$ then Beer–Lambert to get absorption of power per unit depth.

@uhoh That's what I meant to say. Sorry for my sub-complex comment. ;-) You can calculate one from the other (if you have the full complex value, not just a magnitude).

The difference can be viewed in two ways, I guess.
Firstly, for relaxation spectroscopy, you apply a forced oscillation on your sample. That can be mechanical (dynamic rheology, mHz to .1kHz, several kHz with very special, not commerically available equipment) or electrical (dielectric spectroscopy, low mHz to GHz range). The mathematical concepts behind both methods are very similar. For vibrational, MR, UV, etc. spectroscopy, the sample does just not interact with frequencies outside of the linewidth of some QM transition. The width of a resonant peak is given by the energy needed for the transition, and the lifetime of the excited state. The width of a feature in a relaxation spectrum is typically at least half an order of magnitude, from "sample can follow the external stimulus and stay close to a (dynamic) equillibrium" to "external stimulus is much too fast for the sample to follow at all".
The other point is that in relaxation spectroscopy, energy is transferred and dissipated into the sample below a specific frequency (invariably as heat), and reflected (elastically) at higher frequencies. The energy is not quantised, because the induced change in the sample is translatory. The peak frequency itself has no real significance, it's inverse is called a "relaxation time" of the sample. A statistical quantity, mostly. For QM transitions, energy is only transmitted at the given frequency/energy, and nothing happens otherwise, except Rayleigh scattering.
Basically, resonant spectroscopy deals with individual quantum mechanical species, and relaxation spectroscopy deals with the classical, unquantised newtonian properties of the whole sample. NMR is a bit of a hybrid: The induced transitions are purest QM, but you can only observe the classic induction of the ensemble, never individual photons.
Technically, relaxation spectra are usually taken by applying the stimulus at discrete frequencies, stepwise, and you want to have a phase-sensitive detector (voltage - current, deformation - torque).

Thank you for your post, I will now give this some thought. It would be great if you could add a supporting link or two to provide an opportunity for further reading for me and future readers. In both cases you apply an electromagnetic stimulus using an antenna or pair of plates or something similar; why would it be called a forced oscillation only for relaxation spectroscopy and not in the case of resonant spectroscopy?

Well, outside of resonance, oscillation can only be forced. The difference is that you are actually interested in the nonresonant reaction of the sample. In e.g. optical spectroscopy, the sample also interacts with off frequencies, by elastic (Raleigh) scattering.


text:
Why is triatomic hydrogen relatively short-lasting?
The first chapter of Fleming's "Molecular  Orbitals and Organic Interactions," describes the formation of triatomic hydrogen. In summary, the sigma MO of diatomic hydrogen interacts in phase with the third hydrogen's AO. In consequence, a new bonding and antibonding MO are formed. The antibonding MO of diatomic hydrogen does not interact due to antisymmetry.
Since two electrons will fill the bonding MO and one will fill the antibonding MO, there is a overall bonding interaction. This raises my question as to why triatomic hydrogen is relatively short-lived,  a duration of one millionth of a second prior to decomposition.

According to Wikipedia, the molecule exists only in the excited state. If I get it right, it means that, as shown here: https://upload.wikimedia.org/wikipedia/commons/thumb/0/07/Trihydrogen-cation-MO-diagram.svg/1920px-Trihydrogen-cation-MO-diagram.svg.png The energy required for the molecule to exist (SOMO energy level) is higher than the energy of the "generating reactants": H2 and H (H3+ is shown in the image, but the energy levels involved are the same). By releasing an electron, the molecule can become the more stable H3+ cation, with a low energy HOMO.

@The_Vinz Wouldn't combining the sigma MO with the third hydrogen's AO lead to an overall reduction in energy? The antibonding MO should be as "antibonding" as diatomic hydrogen's antibonding MO.

Possibly, but the antibonding MO of H2 is LUMO, unoccupied, right? That means that it doesn't represent the ground state energy of H2. After combination with another H, that LUMO or the new one (which I don't know if they are isoenergetic, actually) is occupied, and the molecule is destabilized

Is the short life due to high reactivity (2 H3 --> 3 H2), which should happen at EVERY collision, or to a unimolecular decomposition?


text:
What is the electronic configuration of La⁺?

Second IP of La is most likely to second IP of which element:
(1) Be
  (2) Ba
  (3) Ca
  (4) Zn

The given answer is calcium. I don't understand if the 6s electron or the 5d electron leaves on ionization. Any help is appreciated.

During ionization only the electrons in last shell leaves.So its 6s


text:
Platinum-carbon calayst in autoclave
Is it safe to treat platinum-carbon catalyst in an autoclave at 200 °C for 24 h in inert (nitrogen) atmosphere? If no, can you suggest safety measures?

Do you mean "safe" for the humans doing the process, or "safe" for the catalyst?  I'm not aware of any significant dangers from heating the catalyst, assuming it is not freshly pulled from an anaerobic atmosphere containing hydrogen.  However I don't know if it will degrade the catalyst in any way.


text:
For a gas having molar mass M, express specific heat at constant pressure in terms of heat capacity ratio

For a gas having molar mass $M$, express specific heat at constant pressure in terms of heat capacity ratio.

My attempt:
using $ C_p  - C_V = R $ and
$ \dfrac{C_p}{C_V} = \gamma $
$ C_p = \dfrac{R \gamma }{ \gamma - 1} $
But answer is
$ C_p = \dfrac{R \gamma }{ M(\gamma - 1)} $
But from where this $M$ is coming from? Isn't specific heat independent of $M$?

I agree with your answer.

The answer you've got $\left(C_p=\dfrac{\gamma R}{\gamma-1}\right)$ has the SI units $\dfrac{J}{mol\cdot K}$. The provided answer is in $\dfrac{J}{g\cdot K}$.


text:
Is the Standard hydrogen electrode potential at T other than 298 K also zero?

The standard hydrogen electrode potential by conventional at 298 K is taken to be 0.00 volts.

This is what I have been taught. It talks about SHE at 298 K, so is the hydrogen electrode potential at a temperature other than 298 kelvin also defined to be zero ? If not, how do we calculate at any other temperature?

According to the definition used by IUPAC, the standard electrode potential $E^\circ$ of the standard hydrogen electrode is zero at all temperatures.

For solutions in protic solvents, the universal reference electrode for which, under standard conditions, the standard electrode potential ($\ce{H+}/\ce{H2}$) is zero at all temperatures.

The absolute electrode potential $E^\circ(\mathrm{abs})$, however, depends on temperature. Its value can be calculated from thermodynamic quantities (e.g. $\Delta G^\circ$). The recommended value for $T=298.15\ \mathrm K$ is $E^\circ(\mathrm{abs})=4.44(2)\ \mathrm V$.

For those (like me) who have never heard of absolute electrode potential: It uses the free electron in a vacuum as reference point, see e.g. https://en.wikipedia.org/wiki/Absolute_electrode_potential

Wait, if the standard electrode potential is zero at all temperatures, what would the electrical potential be of a cell where each half-cell was at a different temperature?


text:
Heating of cyclic dicarboxylic acid
How do we predict the products formed when we heat cyclic dicarboxylic acids? I know that in case of aliphatic dicarboxylic acids we apply Blanc's rule and predict the product formed based on the number of carbons between the acid groups.
In case of aliphatic dicarboxylic acids:
$$
\begin{array}{cc}
\hline
n & \text{Product} \\
\hline
0 & \text{Oxalic acid} \\
1 & \text{Malonic acid} \\
2 & \text{Succinic acid} \\
3 & \text{Glutaric acid} \\
\ldots & \ldots \\
\hline
\end{array}
$$
Here $n$ is a number of carbons between $\ce{-COOH}$ groups.
I tried searching on the net but could not find a convincing answer. Please correct me if I am wrong somewhere

Blanc's rule applies to heating dicarboxylic acids in acetic anhydride NOT just heating them.


text:
Dissolving sand (or something similar) in a bottle of water?
I would like to create an illustration to demonstrate something during a talk. The idea goes that there is sediment in a bottle of water that sits at the bottom until shaken up then adding another substance to the water makes the sediment dissolve or appear to vanish.
My question is; What would make this possible? What materials or chemicals would work?

Dissolving sand would require hydrofluoric acid or molten alkali. Unfortunately, neither is a good fit for an auditorium demo, so you are probably looking for "something similar".

From the description, I think you can be quite flexible as to what there is at the bottom of your bottle. Maybe it would be a good idea to more specifically define what you would like it to look like: What is the most desired color of the undissolved sediment? What is the most desired color of the liquid once dissolved? How fast do you want it to dissolve (some things take hours or days to dissolve)?

It would be best for the illustration if the liquid ended up clear and the reaction was fast, seconds if possible!

To clarify: it sounds like you want a layer of insoluble sediment, so that if you shake the bottle, the water looks cloudy. Maybe it settles back out quickly, or maybe the water stays cloudy.
Then, you add something and shake the bottle again. This time, the sediment disappears.
There are lots of possibilities here. One that comes immediately to mind is lime (calcium hydroxide), which is only slightly soluble in water. Add acid (vinegar if you're timid and don't mind adding a large quantity, hydrochloric acid if you want fast results), and it will be converted to a highly soluble salt.
Chalk (calcium carbonate) would work, too, but it would fizz (produce bubbles of gaseous carbon dioxide) when you added the acid.
If you can say a little more about the context of your demonstration, we might be able to provide better ideas. Is this actually about solubility, or is it a metaphor for something else?

Thank you—this sounds helpful. The demonstration is a metaphor for hidden problems in life that resurface when we are shaken and cloud our judgment! I will give the chalk a go. Perhaps with a hole in the lid for the carbon dioxide to escape!

You can likely use big grains table salts, with some timing and preparation. No need for an extra ingredient except water and salt. Just experiments before to find out the amounts that work well.

Start with sugar in ethanol or isopropanol (rubbing alcohol), then add water to dissolve. For fast dissolution, use  confectioners sugar.

Another trick is to use refractive index matching of the solid and the solvent. There are several organic solvents which match the refractive index of silica. Chlorinated solvents match silica RI very closely. However the best/simplest trick is by Karsten.
Search refractive index matching.

You need (1) Any solid in powdered form (2) Recommended: Any other product that would change the nature of the solid to make it soluble in the solvent (let's inaccurately call it "solubility enhancer"), (3) Any solvent in which this solid would dissolve without producing gases or unsoluble co-products.
You would pick the solid and solvent based on:
A) How much solid you would like to dissolve in the liquid. The Ksp (Solubility Product Constant) would be a good indicator of this, telling you how many molecules you can dissolve in a given volume of solvent at the given temperature you'd like to do your demo. If you want to have a lot of sediment that would disappear completely, you would need a compound that is highly soluble in that solvent and at that temperature (in the form resulting from the reaction with the solubility enhancer if you chose to use one).
B) How quickly you want it to dissolve (i.e. "dissapear"). You can influence on this independently of the Ksp in at least to ways: The first is how finely ground your solid is (as it influences its contact surface), you could go from pellets down to "nano-powders", the second is how the reaction itself will influence the conditions determining the speed of the reaction: the solubility of many compounds increases with temperature for instance, and some reactions give out energy (called "exothermic" reactions), which means that you could chose a reaction that increases it's own rate of reaction as it proceeds (carefully within the limits necessary not to increase the container's temperature or internal pressure beyond what would be safe for your demo).
I would recommend one setup whereby:
(1) The solid to be dissolved is first set at the bottom of your container
(2) The solid to be dissolved is covered by a very thin layer of another solid densely packed in such a way as to insulate the first layer as much as possible from the solvent. This solid won't dissolve but the relative quantity is so small that no one will notice.
(3) The solvent is then very carefully added as not to mess up the two layers of powder
(4) You keep the recipient untouched, preferably right were you need it for the demo
(5) Right before your demo you carefully add to the solvent the chosen "solubility enhancer"
As you start shaking the bottle, the insulating solid layer will disperse, allowing for the underlying solid powder/pellet? mass to contact (1) the "solubility enhancer" it will react with to create a more soluble product (2) the solvent it will dissolve in. Now based on the solvent, the solid, and the product you make it react with, you could be shaking for a dozen seconds, or for a couple hours. :)

Thank you for your answer—it's sounds a little too involved for my needs. I was also hoping for the solid to cloud the liquid before adding a substance to dissolve it.

When you say "cloud the solution" do you mean, clouding it completely, or just enough that people see that there is tiny particles suspended in the entire volume of solvent in the bottle? (the latter case would happen for most reactions which will take at least a few dozen seconds to proceed)


text:
Flame Test: Why does the flame colour of calcium disappear quickly?
While performing flame tests for calcium fluoride and strontium fluoride, I observed strontium gives a long-lasting rose-red flame color but the brick red color for calcium quickly disappears. I also found some books mention the flame color for calcium as transient red.
Why does the flame color of calcium is transient?

A flame is a constantly upward "moving" chemical reaction. The velocity is dependent on the flow rate of the gases entering the burner. There is an extensive upward motion (on Earth). In a gravity-less flame, the flame is round, still there is an upward movement. Now another important thing to note is that the Group II flame emission is not due their atoms but their hydroxides, such as CaOH. In an ordinary Bunsen burner flame you can expect "weird" molecules, which cannot exist in an ordinary bottle. The typical flame temperature is not hot enough to cause atomic excitation and completely break calcium compounds into calcium atoms. If this were true you would see a violet flame (422 nm) after introducing calcium. We see reddish color which is always contaminated with sodium's yellow color. The flame color is then dependent on the thermal stability and volatility of the emitting compound being formed in the flame.

Not sure this answers the q, because no difference between Ca and Sr is given.


text:
Gibbs free energy calculation for redox reaction
I was wondering if it is correct to calculate the Gibbs free energy of reaction at $P = \pu{1 atm}$ and $T = \pu{1100 K}$ using the NIST-JANAF thermochemical tables. For example, for the reaction of ammonia and nitrous oxide:
$$\ce{NH3(g) + 1.5 N2O(g) -> 2 N2(g) + 1.5 H2O(g)}$$
Can I later use this result to calculate the electric potential knowing the amount of electrons transferred?
I'm no electrochemist, I'm just trying to estimate the open circuit voltage for a fuel cell.

You should always be able to compute open-circuit potentials from free energies.  I gather you know this and are asking about the appropriateness of your approach for computing free energies.  Is that what your question really is?

Yes, I am familiar with the formula and learnt how to use it. I would like to know if the way I am computing the free energies is correct.


text:
Why isn't aluminium involved in biological processes?
There are so many biological processes which are dependent upon ions of lighter metals (upper part of periodic table) such as $\ce{K+}$, $\ce{Na+}$, $\ce{Mg^{2+}}$ and even early transition elements ($\ce{Fe}$, $\ce{Mn}$, $\ce{Cu}$, $\ce{Ni}$), but I haven't yet come across dependence of biological phenomena on aluminium. Is this because there is less use of trivalence or is there some other reason?

The metal’s trivalency is certainly not an issue. Iron and cobalt form trivalent compounds (in the $\mathrm{+III}$ oxidation state) in many of their biologically relevant complexes. In fact, in metalloprotein surroundings the actual ‘valency’ (oxidation state) is not so much of a deciding factor; what matters is the number of ligands and the required rigidity of the resulting complex. While some metals prefer octahedral complexes and others tetrahedral ones, many can take either shape and switching between these is inherent to catalytic activity of many metalloproteins.  Aluminium has an all-empty d shell and in its compounds is spherically symmetric (only core shells filled) so it should allow for a flexible coordination environment—like zinc.
One thing that aluminium certainly cannot become is a redox catalyst: in compounds it is practically always $\ce{Al^3+}$. But zinc is likewise practically always $\ce{Zn^2+}$ and yet used in a number of (non-redox) metalloproteins. So that’s not the issue. In fact, some enzymes’ catalytic activities might (as a thought experiment) be increased when zinc is replaced by aluminium because the latter has a higher charge and thus might be able to better catalyse hydroxide generation. On the other hand, aluminium typically displays slower ligand exchange rates than other comparable metal ions (exemplified by the fact that it is precipitated as a hydroxide when its solutions are alkalised with ammonia) reducing its catalytic utility.
Reading that, one might well be inspired to think of aluminium as a structural metal, assisting in the correct folding of proteins. Again, there is nothing intrinsic to aluminium stopping it from doing this and it should fare comparable to other redox-innocent metals.
Obviously, the enzymes living organisms use nowadays are optimised towards using non-aluminium metals, so replacing one with aluminium will usually result in a less active catalyst or less-stabilised structure. But if aluminium had been used, enzyme structures would have evolved accordingly to allow for the effective use of aluminium in active cores and/or as structural metals.
As already hinted at by the other answer, aluminium has a low bioavailability. In fact, at $\mathrm{pH\,7}$ aluminium is predominantly precipitated as $\ce{Al(OH)3}$ but partially redissolving as $\ce{[Al(OH)4]-}$. At slightly lower pH values—typical of the surroundings of most living beings—it is almost entirely precipitated as $\ce{Al(OH)3}$ so practically biounavailable. Of course, specialised ligands can bring it into solution but these need to be evolved first. The problem of silicates further complicates bioavailability.
In conclusiong, I will agree with the already posted answer that bioavailability, especially low solubility at typical pH values are the most likely answer; when compared to e.g. iron, the latter’s redox abilities make it much more attractive for use in enzymes.

Thanks Jan.  This should probably be the accepted answer.

@CurtF. Gosh, too much honour D=

Glad to see you got the checkmark here; it was a better answer.  (Also I just discovered that because of that, I now have a shiny new Gold "Populist" badge.  What a weird badge.)

One argument put forward has been that aluminum is very poorly bioavailable, moreso than many other elements.  Aluminum oxide is very insoluble in water.  In addition, any dissolved aluminum that does form in seawater is likely to be precipitated by silicic acid, forming hydroxyaluminosilicates.
From Chris Exeter's 2009 article in Trends in Biochemical Sciences:

But how has the by far most abundant metal in the Earth's crust remained hidden from biochemical evolution? There are powerful arguments, many of which influenced Darwin's own thinking [15], which identify natural selection as acting upon geochemistry as it acts upon biochemistry. I have argued previously that the lithospheric cycling of aluminium, from the rain-fuelled dissolution of mountains through to the subduction of sedimentary aluminium and its re-emergence in mountain building, depends upon the ‘natural selection’ of increasingly insoluble mineral phases of the metal [7]. The success of this abiotic cycle is reflected in the observation that less than 0.001% of cycled aluminium enters and passes through the biotic cycle. In addition, only an insignificant fraction of the aluminium entering the biotic cycle, living things, is biologically reactive. However, my own understanding of such an explanation of how life on Earth evolved in the absence of biologically available aluminium was arrived at by a somewhat serendipitous route! In studying the acute toxicity of aluminium in Atlantic salmon I discovered that the aqueous form of silicon, silicic acid, protected against the toxicity of aluminium [16]. Subsequent work showed that protection was afforded through the formation of hydroxyaluminosilicates (HAS) [17] which, intriguingly, are one of the sparingly soluble secondary mineral phases of the abiotic cycling of aluminium! The discovery that silicic acid was a geochemical control of the biological availability of aluminium, though now seemingly obvious in hindsight, was a seminal moment in my understanding of the bioinorganic chemistry of aluminium, and although it helped me to understand the non-selection of aluminium in biochemical evolution, it also provided me with a missing link in the wider understanding of the biological essentiality of silicon.

Dr. Exeter is one of the few scholars who appears to have written in depth about this issue.  Thus, perhaps it is fair to say that (a) your question doesn't have a definitive answer, but (b) the poorly accessible nature of aluminum over geological time due to its interaction with and precipitation by silicic acid is the leading hypothesis.
It's worth noting that when aluminum is artificially introduced into metalloenzymes in place of naturally occuring metals, the resulting alumino-enzymes can retain activity, as a 1999 article in JACS by Merkx & Averill shows.

Do you know what is the average solubility of the most common hydroxyaluminosilicates? Would you also happen to know how it is handled in the body of mammals?

@Veritas no I don't know offhand.  I have seen many studies where folks have shown that intentional ingestion of silicic acid lowers aluminum excretion in the the urine, though.  See e.g. https://www.sciencedirect.com/science/article/pii/S2352396417304280

I have seen them as well (in the context of Alzheimer's studies for instance), but I'm wondering in this context of this (very interesting and plausible) explanation as many silicic acid salts aren't much soluble either (Ca2SiO4, Mg2SiO4, Fe2SiO4, Mn2SiO4 for instance), which makes me wonder about the source of the purported relative impact on Aluminum solubility.

For the explanation to hold true, which it very much might be, that Al did not take a major role in biological processes because it is rendered insoluble by the action of Si(OH)4, one needs to show not only that hydroxyaluminosilicates aren't much soluble, but that this contrasts with the case of other elements that are involved in biological processes (Ca, Mg, Zn, Cu, Fe, Mn, Se, Mo, ...). Demonstrating the relative uniqueness of Al in that aspect is the very first step in demonstrating that this is the reason why it isn't involved in biological processes.

Agreed.  That's why I said "leading hypothesis" in my answer, but it is definitely an active research area.   I did find this paper https://agupubs.onlinelibrary.wiley.com/doi/full/10.1002/2013JC009202 , which says that dissolved aluminum is single-digit nanomolar in sea water, but also seems to argue against the hypothesis here.

The mentioned hypothesis, though convincing, doesn't argue at all about trivalence. Does this mean it isn't a factor to worry about?

Bioavailability is just one factor. See this paper. Fe³⁺ also has a low bioavailability but it is  accumulated in the body via specialized mechanisms. In the paper and also a book by the same name, it was mentioned that association and dissociation rates of Al³⁺ with ligands is too slow for biological processes. Also, it is not a good redox species (my guess). I was expecting some answers in this direction in [chemistry.se]. That's why I flagged it for migration.


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What's the mechanism behind mineral oil loosening cyanoacrylate bonds?
Per industry literature I found that mineral oil is used to loosen out "super glue" and aid in removal from skin. I am exploring the chemistry of this loosening and I am intrigued. Please help me find which option applies here:

Does the mineral oil disturb the cohesive forces of the cured cyanoacrylate?
Does the mineral oil interact with the oil on the skin to loosen the cured adhesive and thus break the bond between skin and cyanoacrylate?
Does the mineral oil break the bond between the skin and cyanoacrylate by another mechanism?

Based on this answer, I am trying to find out if mineral oil will loosen the cured cyanoacrylate on a different substrate as well, namely PVC plastic. Please help!


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Why do chiral biological molecules only exist as one enantiomer? Does it have any advantage?
Why is it that chiral biological molecules are enantiomerically pure? The other enantiomer would have the same reactivity, and the only difference is their angle of rotation of plane polarized light. Why, then, is one enantiomer preferred over the other?
Is it that the enantiomer found in our bodies has some advantage over the other enantiomer, or is it random and luck that the structure we see naturally was chosen?
Are all the biochemicals that our body uses enantiomerically pure or are racemic mixtures too?

There are no advantages. The lucky isomers were lucky enough to be the first which organized in structures capable of replicating themselves

Enantiomers are preferred because they work. Most of the things they interact with are also enantiomers and they won't interact equally with their mirror images (if a left handed person uses their left hand to try to shake the right hand of someone else, it doesn't go the same way as two right hands shaking). Enzymes made of random enantiomers would have a very different structure than their enantiomerically pure real versions.

Are all the biochemicals that our body uses enantiomerically pure or are racemic mixtures too?

Many molecules exist in both forms in nature. One fun example are the enantiomeric terpenoids R-(–)-carvone and S-(+)-carvone. The R-form smells like spearmint while the S-form smells like caraway. The difference in smell shows that properties other than the optical activity are different for two enantiomers.

Why, then, is one enantiomer preferred over the other?

Nucleic acids
The_Vinz stated correctly in the comments that for replicating structures like RNA, choice of the enatiomer was random; once established, one chiral form prevailed. DNA building blocks have the same chirality as those of RNA because they are made by the same biochemical pathway.
Amino acids
One amino acid, glycine, is not chiral. Many amino acids are found in both forms. L-amino acids are used to make proteins, but D-amino acids are made in bacteria and used in the context of cell walls and natural antibiotics.
Proteins
Proteins made by ribosomes (from amino acids attached to tRNA by tRNA-synthetases) use L-amino acids exclusively. The tRNA-synthetases are highly specific (including stereo specific), and they don't link tRNA to D-amino acids. It helps that there are very little D-amino acids made in a typical cell. Why one form was chosen over the other is probably luck again. Why all amino acids have the same chirality at the alpha carbon is more intriguing. Some are made from the same precursor, so that will contribute. Right-handed alpha helices require that the amino acids in them be L-amino acids. If proteins had a mixture of L- and D-amino acids (e.g. all alanines are D-alanines but all aspartates are L-aspartate), alpha helices would be more constrained in the possible sequences, and some of them would be left-handed.
All other molecules
Most steps in the synthesis of biomolecules are catalyzed by enzymes. Enzymes, as chiral catalysts that have lots of interactions with reactants, are often highly stereospecific. So the presence or absence of enzymes catalyzing certain reactions largely determines which products are made, and there is no additional cost of making a chiral product from non-chiral precursors (very different from a typical lab synthesis).

Enzymes are very specific for the reaction to enantiomers. For certain products you have only one enantiomer in nature.
It may be the case that there is just one enantiomers which can react with an enzyme because of steric hindrance. But it is also possible that a racemic mixture can occur in fermentation products. So there is no general answer.
The biochemical reaction of two enantiomers can be very different in the human body. For example the contergan / thalidomid can be indicing sleep in the one form and in the other form can cause birth defects.

Other than that the answer is a bit short, it is a good answer for saying that enzymes are stereo-specific catalysts, and that in context of the human body, enantiomers can have very different properties (because proteins bind with stereospecificity). As an aside, thalidomide unfortunately racemizes in the body, so administering a single enantiomer instead of the mixture still causes birth defects.


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Organic Chemistry - Reaction scheme and structures of molecules
Compound P and compound Q are straight-chain constitutional (structural) isomers with the
molecular formula C5H12O. Compound P can form optical isomers, whereas compound Q
cannot.
When reacted with concentrated sulfuric acid, compound P forms two products, compounds
R and S; compound Q forms only one product, compound S.

How come compound Q is pentan-1-ol and not pentan-3-ol?

@The_Vinz There is no much place for so- called geometrical isomerism here nor the text is suggesting that. You somehow mixed up description and nomenclature. OP should show some reasoning on h. side otherwise I'm afraid it is a homework Q

Woops you are right, I lost the point and misunderstood the question

Q is pentan-1-ol and not pentan-3-ol as on dehydration, of pental-1-ol, only one product (alkene) is formed. Whereas on dehydration of pentan-3-ol, we get two alkenes, as the alkene can show geometrical isomerism.

Why the downvote?

Probably because  you have attacked the dehydration / geometrical isomerism issue,  too, without considering the optical isomerism.


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Why does the hydride ion in LAH act as a nucleophile rather than a base in the reduction of carbonyl compounds?
Why does the $\ce{H-}$ ion in $\ce{LiAlH4}$ act as a nucleophile rather than a base in the reduction of carbonyl compounds ??
Because $\ce{H-}$ being a small ion would act as a better base than a nucleophile and I have been taught that it mostly acts as a base and it is very rarely that it is a nucleophile.
Also $\ce{H-}$ after extracting $\ce{H+}$ should become $\ce{H2}$ and leave the system, and also favour this reaction and move it forward.

Heavily related, same answer and hence possible duplicate of Why does a hydride like NaH act as a base, but a hydride like NaBH4 act as a nucleophile?

Because there is not actually ionic separation i.e H- is not formed.


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Why enthalpy changes in phase transition when there is no change of temperature?
Why there is a change in enthalpy in phase transition?
I'm asking because enthalpy depends only on temperature:
$$\mathrm dH = C_p\,\mathrm dT,$$
and the change in temperature is $\mathrm dT = 0.$

It depends on n, P, V. too. dH = Q isn't general as in your statement. Your book or notes should have all.

For a single phase, enthalpy is a function of temperature and pressure, and, at constant pressure, $\mathrm dH = C_p\,\mathrm dT.$
However, at a change of phase, enthalpy (per unit mass) undergoes a jump change, even at constant temperature and pressure.  If the phase change is from liquid to vapor, for example, this jump change in enthalpy is called the heat of vaporization.


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Low or high enthalpy of combustion in relation to stability
I know that this is a very basic question, but I am new to chemistry and I cannot figure it out on my own:
A combustion is always exothermic, so ΔH is negative. It says in my textbook that more stable molecules have a lower heat of combustion (I am assuming here that this is basically identical to stating that it has a lower enthalpy of combustion).
What I do not understand is the following: more stable molecules have a lower enthalpy of formation than less stable molecules. When undergoing combustion, stable molecules should be expelling less heat than less stable ones. So shouldn't the enthalpy of combustion of more stable molecules be higher (i.e. less negative)?
N.B. The question is related to the following:
I was asked to determine which of the following two compounds has a lower enthalpy of combustion. I understand that B is more stable than A, but I am not able to translate that to "a lower enthalpy of combustion" because of my lack of understanding of the above…

Many thanks!

I have redrawn your two structures A and B in their most stable conformations. Note that A has all equatorial methyl groups while B has two equatorial groups and one that is axial. The estimated difference in the heats of formation and combustion is 1.8 kcal/mol, the same as the difference in energy between the equatorial and axial conformations of methyl cyclohexane. Because A and B produce the same amount of CO2 and H2O upon combustion, the more stable A's heat of combustion is 1.8 kcal/mol less than that for B. Compare your A and B with their enantiomers S and T at this ChemSE site at which time I addressed a similar question.

Thank you! So A is more stable, and therefore its enthalpy of combustion is lower, correct? Just summarising to see if I fully understood it...

Correct. The sum of the heats of formation and combustion in this example are the same. If you have a more negative heat of formation, i.e., more stable, you have a smaller heat of combustion.

The following schematic (not meant to represent the actual changes in $H$) illustrates the problem:

The first step (from leftmost reagents to middle products) represents the formation reaction (not balanced), the second step (from middle to oxidation products on the right) the combustion reaction.
It pays to be careful using words such as "greater" or "less" when comparing signed quantities. For instance, stating "$H$ for B is less than for A" might not make it clear whether one means "more negative" or "smaller in magnitude".
In this schematic it is clear that $\Delta \Delta _\text{f} H(A-B)>0$, that is, A has greater (more positive) enthalpy of formation than B. On the other hand $\Delta \Delta _{\text{comb}} H(A-B)<0$, that is, the heat of combustion is more negative for A than for B. Note that where we place A and B is not important for the purposes of this illustration, where we compare the relative enthalpies qualitatively (not quantitatively). What's important is that $H_A>H_B$.


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Viscosity and concentration of solutions
I will need a fluid around a specific viscosity, using a mixture of water and something like sugar.
Is there a way to calculate viscosity based on concentration?


text:
What is prefix L for amino acids?
I know the "L" at the beginning of amino acids pointed out to show the substance is digestive.  The thing I don't know is that what is the "L" stands for and how it works.

The stereochemistry at the chiral centre of each individual amino acid can be D or L or racemic. L is the naturally occuring form so most digestive enzyme are set up to metabolise polypeptides made of L amino acids.

Originating from right and left, though due to nomenclature stratification d and l labels do not always correspond to the actual verse, identified by (+) or minus instead.

From Chemgapedia:
Most of the aminoacids in nature have different spatial structures. These properties are due to a chiral center (on the C-alpha atom), which has four different substituents. One differentiates a D-form (from Latin dextro = right) in which the amino group of the C-alpha atom in the Fischer projection to the right, and an L-shape (from levo = left), in which it is to the left.  Since they are the same molecules that differ only in their spatial structure, they also have identical chemical and physical properties. With one exception: enantiomers turn the plane of oscillation of polarized light either to the left (-) or to the right (+).  But attention: One can not conclude from the D / L-form on the rotation characteristics +/-, since the division D / L refers only to the arrangement of the substituents at the C-alpha-atom!
See the link above for pictures to support the explanation.

BTW, dextro is the root of the word dexterous, implying the person is skilled ("right" handed). Levo is a synonym for sinister, or left handed.


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Trans vs. cis, or Z vs. E
I am new to organic chemistry and am confused regarding the following question:
Does the geometrical isomerism of 3-methyl-2-heptene fall into the category of trans vs. cis, or the category of Z vs. E? And what is the rule that determines whether it falls into the first or the second category?
My textbook is not clear on this.
Many thanks!

Thank you! The thing is that it says somewhere in my notes that cis vs. trans requires there to be two H atoms, obviously on opposite sides of the double bond. This is of course not the case here. Yet I cannot find any external support for this "rule" that miraculously made it into my notes…

That rule in this context is correct. The corresponding wording in the current IUPAC recommendations is: "… 'cis' and 'trans' are used … only to describe double bonds having to hydrogen atoms, one on each carbon atom. …"

(Z) is the same like cis
(E) is the same like trans
BUT cis and trans is usually only for double bonds with 2 different substituents. Confusion can arise when, for instance, you have 2-halobut-2-enes, because some take the highest priority when determining if this is cis or trans whereas others may take the identical groups.
You can use (Z) and (E) for more than 2 substituents, for those with higher priority. Like in your example.
In general it is better to use (Z) and (E) nomenclature to avoid these ambiguities.


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The time-dependent pH value in a fermentation
I want to know if someone has supporting tips or solutions to my problem:
In biotechnology, fermentations are used to grow microorganisms. These need a specific media/broth in which temperature, pH, substrate and many other factors conclude to an optimum. Only if the surrounding broth is in optimal condition for the individual microorganism, the microorganism will grow as best as possible.
In my media the broth is buffered with $[\ce{HA}] = \ce{H2PO4^-}$ and $[\ce{A-}] = \ce{HPO4^2-}$.
While the fermentation is running the pH is corrected by $2\ M\ \ce{NaOH}$ which is pumped into the broth. Via balances, the weight is recorded and the input can be calculated.
I know the initial concentrations of my buffer solutions and the concentration of the base that is dropped into at every minute of record.
How can I calcute the pH for an individual timepoint in my aqueous fermentation? My tries via the Henderson-Hasselbach-formula did not work so far...
(Neglecting other surroundings such as stirring, the other media components, the biomass' metabolites, temperature influence on the pKs, etc.),
Thanks in advance!

I guess you are aiming to control the $p^H$. Are pH probes + a feedback/feedforward control system not possible?

I need to calculate the pH at different times troughout the process to be able to implement a pH-model on a calculation/formula. It should be used to compare it to the probes and control system.Furthermore to the model made from probes + feedback/feedforward control system.

The pH is not corrected mathematically, it is stabilized chemically by addition of NaOH when the pH sensors get out of range. Your pH is controlled to be constant over the fermentation period (+/- a little bit).
The acid produced by the fermentation is neutralized by addition of NaOH to keep the pH constant. What you can calculate from the data you get is the rate of acid production. The moles of acid produced will equal the number of moles of NaOH added, to a good approximation. The pKa of H2PO4- is 7.2 and the pKa of acetic acid (a typical carboxylic acid) is 4.76, so the acid produced will be "strong" - i.e., fairly acidic in your medium, which appears to be aimed for about pH = 7 - and the NaOH is strongly basic, so they will balance out (approximately).
Interestingly, sodium acetate has a pH range of 8.5 - 9.9 in real life, according to SDS. Calculation is so much more pleasant, where you can calculate pH to three decimals and feel good about it (until you test it in real life and have to calibrate your meter, etc, etc.)

Could you recommend literature on this specific topic?
Such as implementing a control system for fermentations on basis of a pH probe.
If the pH is corresponding to the acid concentration of the microorganism, it is responing to biomass itself and can therefore be used for prediction models.

Google brings up a lot of fermentation control system literature - but it seems to be mostly for controlling beer and wine. Major industries, of course! But maybe the info on one of these systems (before you buy!) would give you some ideas. In addition to the pH measurement, you could consider measuring temperature, O2 uptake (a little more difficult, but very interesting), CO2 production, conductivity, viscosity. Some of these parameters would just be measured for your model, and some could be controlled to eliminate their variability.


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Is there an easy way to detect methanol in spirits (which may have been adulterated)?
Is there an easy way to detect methanol in spirits, which, you suspect, may have been adulterated?

Related: How to test for methanol?; How to find trace methanol content in fermented fruit/distillate?.

Thanks, I should've checked for similar questions first. I'm sorry.

Well that depens on your possibilities.
a) Destillation / Boiling Point (easy, unsafe)
b) Refractive index (depends on impurities)
c) GC (best, very expensive)
d) 1H NMR


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Theoretical p-Phenylenediamine synthesis
I am thinking on an alternative synthesis of pPDA, in which I need counsel. I was thinking about these easy reactions, and I want to ask if it can be done, or maybe you can hint me some reference. I am trough the reading of nitration and reduction of aniline, or from 4-nitrochlorobenzene or the DuPont synthesis.

The first route smells a bit fishy, but I think with enough heat, and molar ratios it can be done.
The second reaction is more promising for me since it is made from economic reagents, I'm fairly sure the ascorbic acid reduces the imine to the amine form. The only problem that could occur is with the reaction of 1,4-benzoquinone with ammonia. Can you validate my thoughts? I see no reason why it should not work.
Please if you have any suggestions or help, I gladly welcome it.
EDIT 1:
The comment from Waylander in wich he/she states the step from 1,4-benzoquinone to the imine form is probably not achievable. Howered i found also an another route from benzoquinone, wich is to react it with hydroxil-amine forming the oxime. *1 Then the question left behind is the step from oxime to amine, maybe the ascorbic acid still can do the job, however i dont find any source for it, but my assumption is that it still works.

For the top route - I don't think it is a runner. Firstly treatment of phenols with PCl3 does not result in chlorobenzenes, it gives triphenylphosphites. Secondly treating chlorobenzene with ammonia, if one NH2 group goes in, it makes the ring more electron rich and less likely to undergo SNAr. I can find no reports of the bottom route though I do not have full database acces. Since it is a fairly obvious transformation I can only conclude that it does not work.

I'd say the suggested reaction of benzene-1,4-diol (hydroquinone) with $\ce{PCl3}$ to get 1,4-dichlorobenzene would not work (see  Waylander's comment). However, you can buy 1,4-dichlorobenzene very cheap (e.g., USD 21.50 for $\pu{100 g}$ bottle from Sigma-Aldrich) so why you need this trouble?
The suggested reaction of 1,4-dichlorobenzene with $\ce{NH3}$ to get benzene-1,4-diamine (p-phenylenediamine) would work, however, because there is a literature evidence for that (Ref.1). To do so, you need to use copper(I) chloride catalyst ($\ce{CuCl}$) and use 28% $\ce{NH4OH}$ solution as $\ce{NH3}$ source. Also need to heat the reaction mixture above $\pu{200 ^\circ C}$ in a pressure bomb (you may need to read this reference carefully before you start the reaction).
References:

Armand J. Quick, "The Preparation of p-Phenylenediamine and Aniline from their Corresponding Chlorobenzenes," J. Am. Chem. Soc. 1920, 42(5), 1033-1042 (https://doi.org/10.1021/ja01450a018).

Thank you for your time and answer. The reference is welcome.
"why you need this trouble?" It just a theoretic approach sketched crude.


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Finding EMF of a galvanic cell without standard potentials

For the galvanic cell
$$\ce{Ag | AgCl(s), KCl (\pu{0.2 M}) || KBr (\pu{0.001 M}), AgBr(s) | Ag}$$
find the EMF generated given $K_\mathrm{sp}(\ce{AgCl}) = \pu{2.8e-10},$ $K_\mathrm{sp}(\ce{AgBr}) = \pu{3.3e-13}.$

This is the question from JEE exam (1992).
How to start solving the problem since the $E^°$ of individual half reactions is not given? How to write the $E^°$ for the cell without it? Or is it not needed?

I am upvoting this because several of us learned something today! Thanks for the nice question!

How to start solving the problem since the $E^\circ$ of individual half reactions is not given?

This is a concentration cell, i.e. the half reactions at the anode and at the cathode are the same (except for the direction).
AgCl(s) electrode
$$\ce{AgCl(s) <=> Ag+(aq) + Cl-(aq)}$$
$$\ce{Ag+(aq) + e- -> Ag(s)} $$
AgBr(s) electrode
$$\ce{Ag(s) -> Ag+(aq) + e-}$$
$$\ce{Ag+(aq) + Br-(aq) <=> AgBr(s)}$$
The standard reduction potentials will cancel out, i.e. $E^\circ (\mathrm{cell}) = 0$.
Further thoughts

[Comment by EdV] The Ag|AgCl electrodes I have are commercial, but making them in the lab is just a matter of oxidizing Ag wire in a chloride solution, so the electrode is Ag wire with an adherent coating of AgCl... I have never seen one of these made by just sticking an Ag wire in the Ag halide, but I guess it would work.

I added more to the answer prompted by that thoughtful comment.

[...my own comment] I pictured the silver electrode submerged in the solution, with the solid halide on the bottom. I am puzzled now too. Does it make a difference if the electrode touches the solid, the liquid, or both?

What are the actual reduction potentials?
$$\ce{AgCl(s) + e− <=> Ag(s) + Cl−}\ \ \ \ E^\circ_\mathrm{red} = \pu{+0.22233 V}\tag{1}$$
$$\ce{AgBr(s) + e− <=> Ag(s) + Br−}\ \ \ \ E^\circ_\mathrm{red} = \pu{  +0.07133 V}\tag{2}$$
$$\ce{Ag+ +  e−  <=> Ag(s)}\ \ \ \ E^\circ_\mathrm{red} = \pu{  +0.7996 V}\tag{3}$$
Are they related?
If you subtract (3) from (1), you get the dissolution reaction of AgCl, if you subtract (3) from (2), you get the dissolution reaction of AgBr. Thus, standard reduction potentials for (1) and (3) should be different by
$$ -\frac{RT}{zF} \ln K_\mathrm{sp}(\ce{AgCl})$$
and standard reduction potentials of (2) and (3) should be different by
$$ -\frac{RT}{zF} \ln K_\mathrm{sp}(\ce{AgBr})$$
Finally, standard reduction potentials (1) and (2) should be different by
$$ -\frac{RT}{zF} \ln \frac{K_\mathrm{sp}(\ce{AgBr})}{K_\mathrm{sp}(\ce{AgCl})}$$
Numerical answer using half reaction (1) and (2)
$$\ce{AgCl(s) + Br-(aq) <=> AgBr(s) + Cl-(aq)}$$
$$Q = \frac{[\ce{Cl-}]}{[\ce{Br-}]} = 200$$
$$E_\mathrm{cell} = E^\circ_\mathrm{cell} - \frac{R T}{z F} \ln(Q)$$
$$= \pu{(0.22233 V− 0.07133 V) - 0.13612 V = 0.0149 V}$$
Numerical answer using half reaction (3) twice
$$\ce{Ag+(c) + Ag(b) <=> Ag(c) + Ag+(b)}$$
"c" stands for chloride side, and "b" stands for bromide side. For consistency, I am using the following values for the solubility products (derived from difference of standard reduction potentials of half reactions (1), (2) and (3)).
$$K_\mathrm{sp}(\ce{AgCl}) = \pu{1.74e−10}$$
$$K_\mathrm{sp}(\ce{AgBr}) = \pu{4.89e−13}$$
$$[\ce{Ag+}]_c = K_\mathrm{sp}(\ce{AgCl}) / [\ce{Cl-}] = \pu{8,27e−10}  $$
$$[\ce{Ag+}]_b = K_\mathrm{sp}(\ce{AgBr}) / [\ce{Br-}] = \pu{4.89e−10}  $$
$$ Q = \frac{[\ce{Ag+}]_c}{[\ce{Ag+}]_b} = 0.560 $$
$$E_\mathrm{cell} = E^\circ_\mathrm{cell} - \frac{R T}{z F} \ln(Q)$$
$$\pu{= 0 - (-0.0149 V) = 0.0149 V}$$

@EdV I wrote up the numerical answer calculated both ways. I gave myself some extra significant figures for the calculation of the solubility products. When you say "your expression", I think there was some misunderstanding. With a standard potential of 0 V and silver ion concentrations similar in both cells, I'm not sure how you got -0.173 V. In any case, you can see that the chloride and bromide concentrations are used both times because the reaction quotient contains the silver ion concentrations in one case, and the bromide and chloride ion concentrations in the other case.

Excellent, we got the same answer! The -0.173 V was using the OP's solubility product constants. Thanks for the clarification!

@KarstenTheis Sir, you just said that E° = 0 but now by putting values you are saying that E°≠0, also this was an exam type situations so the E° was not given. Also, how did you write the equation 3 directly ie.$$\ce{Ag+(c) + Ag(b) <=> Ag(c) + Ag+(b)}$$ why did you ignore the Chloride/Bromide radicals from the equation ?? What I tried making was the equation from reaction 1 and 2 but since E° was not given I couldn't proceed. Please explain the equation 3. :)

@DivMit $$\ce{Ag+(aq) + e- <=> Ag(s)}$$ is a balanced half reaction, and if it describes both half-cells, you get the chemical equation you have in your comment. The concentration of silver cations in the two half cells indirectly depends on the concentration of chloride or bromide (as you can see in the calculation). Writing it this way, however, chloride and bromide are not part of the electrochemical equation and so they don't appear in the reaction quotient Q of the Nernst equation directly.

Even though this question 1) has an answer with multiple upvotes (and I was the first upvote), 2) the OP has accepted the answer and 3) I have great respect for @Karsten Theis, having co-taught a quant class with him back in 2008 and knowing, first hand, that he is an excellent scientist and teacher, nonetheless, I have several problems with this trick exam question.
First, the $K_{sp}(AgCl)$ is about $1.8 x 10^{-10}$. This is the trivial problem, nothing more than a typo. More importantly, the $K_{sp}$ solution does not tell the whole story. So suppose T = 298.15K, i.e., standard temperature, n = 1 equivalent/mole, and all activity coefficients are assumed to be unity, so molar concentrations can be used in place of (tossing the molarity units) the unitless activities. Then, under standard state conditions, we have the following voltaic cell:

Thus AgCl(s) will be reduced to Ag(s) plus $Cl^-$ ions, on the right hand side (RHS) of fig. 1, and Ag(s) will be oxidized to AgBr(s), on the left hand side (LHS) of fig. 1.
But what happens if [$Cl^-$] = 0.2 M and [$Br^-$] = 0.001 M? Then both reduction potentials increase (become more positive), but the new Ag|AgCl reduction potential is still the most positive, so that is where reduction is spontaneous wrt (with respect to) the other half cell reaction reduction potential. This is shown is fig. 2 below:

In this figure, n = 1 equivalent/mole, T = 298.15K, R = 8.314472 J/(mole K), F = 96485.3383 C/equivalent, so RT/nF = 25.6926 mV and (ln10)•RT/nF = 59.1594 mV. Plugging in the numbers, as per the $E_{cell}$ equation in fig. 2, yields +0.0149V. If [$Br^-$] = 0.0001 M, then the cell potential would be -0.044V. This means the Ag|AgCl electrode would be the anode and the Ag|AgBr electrode would be the cathode. Note that the OP’s question actually showed the Ag|AgCl electrode as the anode, since this is the standard convention. But it wasn’t! So this was a rather nasty trick question, in my professional judgment.

Thanks for finding all the problems in my answer. I think we have converged on a single result. Of course, because the OPs question has slightly different numbers, the numerical result to that question is a bit different. I learned a lot today.

I learned a lot as well! The way you did this problem is very sweet! Many thanks!


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How Would A Vast Mixture (Like an Ocean) of Ammonia-Water Behave in Terms of Evaporation and Freezing?
First, I'll go ahead and say I'm not very versed in chemistry. My question comes from my interest in planetary-sciences, specifically the speculation that there are likely exoplanets and moons with surface oceans of water-ammonia. However I can't find much hard info on how a vast body (i.e. ocean) of water-ammonia would behave differently than just water.
In short my question is how might a water-ammonia ocean differ from a "pure" water ocean in terms of:

Ice(s) formation
Evaporation
Changes in water/ammonia ratio based on the above two factors (ice formation and evaporation).

This seems largely like a chemistry question to me. If I'm wrong please let me know which Stack Exchange you think I should post in. :)
Parameters:

5% - 40% ammonia.
Temperature range: -80C to 15C
Atmospheric Pressure: 0.5 bar to 30 bar.

More Info:
I've heard that the freezing and boiling points of the mixed liquid would exist on some continuum between the freezing and boiling points of its constituent liquids (ammonia and water), varying depending on the exact mix ratio and of course pressure. Is that correct?
What I'd like to most wrap my head around is whether the water-ammonia mix would behave as a single uniform liquid, freezing and evaporating at a rate determined by the average of the mix's freezing/boiling point, or whether the ammonia molecules would start to evaporate before the water molecules, and the water molecules start to form ice before the ammonia molecules.
If the later behavior is correct I could imagine an ocean of water-ammonia varying substantially in its water/ammonia ratio from pole to equator thanks to the greater heat near the equator and the ammonia molecules evaporating.
Thanks!

Water/ammonia oceans tend to be subterranean.

They are in the Galilean moons but there is no reason water/ammonia surface oceans can't exists. In fact they are predicted to exist, possibly in some abundance.

The concentrations will definitely vary tremendously over the climate zones.

https://astronomy.stackexchange.com/questions/32676/how-would-water-ammonia-oceans-behave

https://en.wikipedia.org/wiki/Uranus#Internal_structure BTW I can't say I'm fully OK with this one, too.

Freezing

I've heard that the freezing and boiling points of the mixed liquid would exist on some continuum between the freezing and boiling points of its constituent liquids (ammonia and water), varying depending on the exact mix ratio and of course pressure. Is that correct?

For freezing points, that is incorrect. Pure water freezes at 273 K and pure NaCl (table salt) "freezes" at 1074 K. However, if I dissolve NaCl in water, the freezing point of water will be lowered. In you water/ammonia scenario, water will freeze as pure water, but the freezing point will be lower. Ice formation would increase the concentration of ammonia in solution.
Evaporation
Water and ammonia mix in the liquid state, and everything mixes in the gas state. Evaporation (and condensation) depend on the ratio of components in the liquid state, the partial pressure of components in the gas state (or ratio of components and total pressure), and the temperature. At equilibrium at earth-like temperatures, there would be more ammonia than water in the atmosphere (the boiling point of ammonia is much lower than that of water, so the vapor pressure of ammonia is higher than that of water). A sudden drop in temperature would make it rain ammonia, mixed with a bit of water. So condensation and precipitation would increase the concentration of ammonia in the liquid state, and a sunny day would lower it.

If the later behavior is correct I could imagine an ocean of water-ammonia varying substantially in its water/ammonia ratio from pole to equator thanks to the greater heat near the equator and the ammonia molecules evaporating.

You are making a lot of assumptions here: that the exoplanet rotates fairly quickly around its own axis, that its axis is roughly perpendicular to the star, and that the star is close enough to heat up the planet.

What I'd like to most wrap my head around is whether the water-ammonia mix would behave as a single uniform liquid, freezing and evaporating at a rate determined by the average of the mix's freezing/boiling point, or whether the ammonia molecules would start to evaporate before the water molecules, and the water molecules start to form ice before the ammonia molecules.

No, they would do different things. Pure water at terrestial atmospheric pressure freezes at 0 $^\circ$C, while ammonia already boils at -33 $^\circ$C. Mixing them will lower the freezing point of water and increase the boiling point of ammonia, so there is no temperature where you would separate them completed by freezing all water and boiling off all ammonia. On the other hand, they don't freeze or boil in lock step. If you lower the temperature, water will form ice and the ammonia concentration in the liquid will increase. If you increase the temperature, more ammonia than water will turn into gas (evaporating or boiling), so the ammonia concentration in the liquid will decrease.

Thanks for the reply. Your the first person to go over in some detail what you think the behavior would be. I appreciate that! May I ask where your deriving some of this information? Is it a synthesis of info you're already aware of, or are you referencing something? If you do have a reference I'd love to see it so that I could study it myself. Not trying to dispute your points, just hoping to learn more from primary sources.
And yes, I was making a bunch of assumptions about the planet, but assumptions that serve my purpose.

– Oh, I see you teach biochemistry. Great! Would it be alright if I asked you a few follow-up questions? I'm really trying to understand how these mixes work, and as I mentioned I'm not terribly versed in chemistry.

@n_bandit Sure. If they questions are substantial, just ask a new question referencing the current one. If they are short and refer directly to my answer, ask them in the comments here, and I or someone else will try to respond. (And I'll incorporate it in the answer if it makes sense.)

Can anyone tell me why my question is "too broad" and therefore on hold? How would you recommend rephrasing it without diluting the substance of the inquiry?

@n_bandit RE:references: Phase transitions, freezing point depression and boiling point elevation are concepts you encounter in physical chemistry or general chemistry. Once you know the concepts, you can look up the specific data (boiling point etc) in textbooks, the literature or online databases.

@n_bandit Concerning "too broad". You can comment below the question, and ask for clarification (by using the mechanism with the "at" sign).
The two most critical parameters you did not specify are temperature and pressure, and maybe gravitational acceleration (the smaller the planet or moon, the harder to hold on to light molecules).

I've added some constraints. Hope it helps!

@n_bandit Are you saying the temperature varies daily or seasonally within that range on the surface, or you want different planets with a set temperature? I guess the pressure would be for different planets, although if all the components in the atmosphere are close to their boiling point (not like for us, where the main components N2 and O2 are always in the gas phase), it could vary a lot. Also, are there effects that you are hoping for (like springs of pure water or something like that)?

@n_bandit I added a bit to my answer to specifically address one of your questions. If you want to get the question re-opened, you probably either ask for parameters where certain effects happen, or stick with one scenario (i.e. as completely as you could describe earths ocean and atmosphere) and then ask your questions. Alternatively, you move on an ask a new question, referencing this one. Mind you, it would have to be a substantially different question.


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How to predict whether an ionic compound will act as acid or base when dissolved in water?
How would we know whether an ionic compounds such as $\ce{NH4Cl}$ reacts acidic or basic in water?
The explanation I have seen is that you have to figure out what acid and base could combine to form the ionic compound. In the case of $\ce{NH4Cl}$, you could combine hydrochloric acid and ammonia:
$$\ce{HCl(aq) + NH3(aq) -> NH4Cl(aq)}$$
It would be a strong acid and a weak base, making the solution acidic. If we don't know the strength of both the acid and the base, this method doesn't help.

On adding $\ce{CO2}$ in water $\ce{H2CO3}$ is formed which is a weak acid thus the solution will be acidic.

Generally the oxides of non metals are basic in nature like $\ce{SO3 , SO2, NO2 , NO3, CO2}$ thus on hydrolysis the give acidic solution.

Similarly oxides and hydroxides are basic in nature

More advanced theory won't help you with fewer data. Quite the opposite, in fact.

You mentioned the strengths of the acids and bases can be used to determine whether the salt is acidic or basic. This is largely correct because a strong acid has a very weak conjugate base, while a weak acid has a relatively strong conjugate base; the same applies to bases and their conjugate acids. There are a few other rules of thumb though that can be used to predict whether a salt is acidic or basic.
Oxides
In general, metallic oxides (base anhydrides) are basic in aqueous solution. For example, calcium oxide, $\ce{CaO}$, produces a basic solution when dissolved in water:
$$\ce{CaO + H2O -> Ca(OH)2}$$
Nonmetal oxides (acid anhydrides) are acidic in aqueous solution. Since carbon dioxide, $\ce{CO2}$, is a nonmetal oxide, it is acidic in solution:
$$\ce{CO2 + H2O <=> H2CO3}$$
Cations
The conjugate acids of weak bases will act as proton donors. For example, $\ce{NH4+}$ can donate a proton to water:
$$\ce{NH4+ + H2O <=> NH3 + H3O+}$$
Additionally, small, highly charged metal cations are also acidic because they can act as Lewis acids, with water acting as a Lewis base. For example, $\ce{Fe^3+}$ ions can form the $\ce{[Fe(H2O)6]^3+}$ ion. The electrons in the water molecules are pulled toward the metal ion, making the $\ce{O-H}$ bonds more polar. Thus, water molecules are more likely to become deprotonated, donating a hydrogen ion to nearby water molecules. For example:
$$\ce{[Fe(H2O)6]^3+ + H2O -> [Fe(H2O)5(OH)]^2+ + H3O+}$$
Cations of group 1 and 2, as well as those with a charge of $+1$, are neutral in aqueous solution. They are too large or have too little charge to polarize the nearby water molecules to any significant extent.
There are no basic cations because protons are repelled by the positive charge.
Anions
There are very few acidic anions, because protons are attraced to the negative charge. However, hydrogen sulfate ($\ce{HSO4^2-}$) and dihydrogen phosphate ($\ce{H2PO4-}$) are among the few exceptions.
The conjugate bases of strong acids are neutral. Because the acids are so strong, their conjugates are very weak. Examples include $\ce{Cl-}$ and $\ce{NO3-}$.
The conjugate bases of weak acids are basic. Because the acids are weak, their conjugates will be stronger. Examples here include $\ce{F-}$ and $\ce{NO2-}$.
The Acidity or Alkalinity of the Salt
To predict whether a salt will produce an acidic or basic solution when dissolved in water, we have to consider both the cation and the anion. In some cases, these rules of thumb are insufficient, such as with ammonium nitrite, $\ce{NH4NO2}$; the cation is acidic while the anion is basic. In this case, it's necessary to just carry out the equilibrium calculations.

Source: Atkins, Peter and Loretta Jones. Chemical Principles: The Quest for Insight, 5th edition.


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Strange batch of potassium nitrate with lower than expected melting point
I'm part of a rocket building team and I'm working with propellants. We use saltpeter as the oxidizer, bought as fertilizer and purified by recrystallization. Recently we bought it from another source, that identified the chemical as potassium nitrate, as usual, but when we heated it to make the propellant it simply melted at a temperature of $231~^\circ\rm{C}$, $100$ degrees lower than expected.
It works as an oxidizer, as making the propellant with this chemical gives good results, so it must be something similar to $\ce{KNO_3}$. Maybe there is something working as flux? It tastes like saltpeter (I eat a lot of materials we have in the workshop) and has that characteristic high temperature-dependent solubility.
Do you have any idea of what could it be, and what can I do to find out? Have in mind that our workshop doesn't have much in terms of lab equipment, only thermometers, various things to heat stuff and some solvents for a variety of applications. Also, note that I'm a telecommunications engineering student, so I don't have too much of deep knowledge of chemistry.

Flame test? What colour does it give off when put in a flame?

@Waylander  a bit orange

Perhaps it is a mixture of nitrates. Also, please do not eat chemicals in he workshop! Adulteration of substances is increasingly a problem, and some adulterants can be lethal!

Did you operate as usual? Did you check the melting point of the recrystallised salt alone? When we melted it to make the propellant is unclear to those not familiar with the hobby.

STOP EATING CHEMICALS !!!

It might contain $\ce{NaNO3}$ as well as $\ce{KNO3}$. This paper by Foong et. al. cites some thermal measurements, such as melting point and heat capacity, that you could make with a thermometer and heater. Note that there are inflections in the heat capacity curve as the different components ($\ce{NaNO3}$, $\ce{KNO3}$ and the eutectic of the two) melt.
If you have access to a spectroscope, or even a hand-held diffraction grating, you could compare the spectrum with that from samples of known substances, such as $\ce{NaNO3}$, $\ce{KNO3}$, $\ce{NH4NO3}$ and $\ce{Ca(NO3)2}$.

makes a lot of sense, as the eutectic melts pretty much at 100K lower than pure KNO3

Excellent answer, but sodium nitrate should give a pretty bright yellow emission in a flame test. Curious it is only a bit orange.

@EdV, agreed... but one persons "yellow" could be another's "orange", therefore the grating was suggested.

Pure sodium nitrate gives yellow, but a mixture of sodium and potassium could easily look orange.


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What ratio of bicarb to sugarcane juice with a pH level of 4.5-5.5 to neutralize pH level?
I am thinking of adding baking soda to sugarcane juice to eliminate the fruity acid that is present. The pH of sugarcane juice ranges from 4.5-5.5. I wonder how much baking soda should I incorporate every 1 liter of sugarcane juice. Thank you so much. This would really help me in my thesis.

Is experimentation at smaller scales possible? That would be the best answer, probably.

You need to know which acids are present in the sample, or determine the required amount experimentally.
Without any knowledge of the pKa values of the acid or acids present, there could be very different scenarios. A pH of about 5 could be due to a high concentration of a weaker acid or a low concentration of a stronger acid. For example, 0.01 mM of HCl would bring the pH to 5, or 1 M of a weak acid with pKa of 10. Or it could be a buffer composed of an acid and its conjugate base present at similar concentrations, with the acid having a pKa of around 5.
This paper states that the main acidic component is aconitic acid (derived from citric acid), at 3 g per 6 liter (i.e. a concentration of about 3 mM). The pKa value for the first ionization is 2.8, and for the second 4.46. So if all of these assumptions are correct, the extract is buffered with a buffer that has a pKa of 4.46 and a concentration of 3 mM.
If possible, the best thing would be to add a buffer at higher concentration (say, 10 mM) with a pKa in the range of the desired pH. Alternatively, you could add any solution (or solid) that increases the pH, and determine through an experiment how much you need.
It is possible that the composition of the extract will vary from sample to sample as you are starting with biological material that shows some variation in its composition.


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Nature of the bonding in hexaaquaaluminium complex
$\ce {[Al(H2O)6]^3+}$ is a well-known complex ion formed by the $\ce {Al^3+}$. I have always had the understanding that only the aqua complexes of transition metal ions, such as $\ce {[Fe(H2O)6]^3+}$, would legitimately possess dative covalent bonding between the water ligands and the central metal ion. I had always thought that the primary reason for this is that ions such as $\ce {Mg^2+}$ and $\ce {Al^3+}$ lack vacant low-lying orbitals for them to be able to accept electrons donated from the ligands.
However, after further thinking on the issue, it seems quite ridiculous to say that there are no low-lying vacant orbitals on $\ce {Al^3+}$. For example, $\ce {AlCl3}$ is a fairly good Lewis acid and is able to accept dative covalent bonds from various Lewis bases. If the $\ce {3p}$ orbital in the $\ce {Al}$ atom in $\ce {AlCl3}$ is of sufficiently low energy to accept electron donation, then clearly, the $\ce {3p}$ orbital in $\ce {Al^3+}$ would certainly be energetically accessible as well since the large magnitude of positive charge on the ion would lower the oribtal energies greatly.
The answer in this post posits that the interaction between $\ce {Na+}$ and water ligands is mostly of the electrostatic nature while there is much more covalent involvement in the complexes of the transition metal ions. However, no reasons were given as to why this was so. Perhaps, someone with expert knowledge in this area of chemical bonding would like to enlighten me on this issue.
I have also read up some literature on the aqua complexes of alkali metal ions, which have been studied in great detail due to their biochemical significance. Talekar (1975) informs that the bonding between alkali metal ions and water ligands is approximately $\ce {20}$% covalent and $\ce {80}$% ionic and he further adds that the covalent component is due to donation of electrons into vacant orbitals on the alkali metal ions. However, the computational methods used at the time to compute these data may have been quite limited and thus, it may be unreliable.
Reference
Talekar, S. V. The nature of metal-ligand bonding in the primary hydration shell and in ionophores with alkali metal ions. Biochimi. Biophys. Acta 1975, 375, 157-164.

So what would you thought the bonding was? You start with imaginary Al 3+ and 6 H2O -  s and 3  p orbitals of Al and 6 lone pair and they somehow mix - you can write some 3-c-4-e bonds or whatever.

@Mithoron I had always thought that the bonding in this aluminium complex was mostly ionic, between the ion and the water molecules. However, I have seen some sources describing a high degree of covalency in the bonds.


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Radical monobromination of cycloalkane
I'm trying to figure out the isomeric products of the radical monobromination of bicyclo[2.2.1] heptane. I'm not sure at which site the bromination would happen and what makes one site better than the other (how can one determine this without given the reaction enthalpy ?). I was also given the hint that there are stereoisomers but I can't think of one.

Would help the discussion if you supplied a drawing with atom numbering

The radical monobromination of bicyclo[2.2.1] heptane could give $\ce{1}$ ,$\ce{2}$ and $\ce{3}$.
However , $\ce{2}$ and $\ce{3}$ , are unstable .On observing $\ce{2}$ , planar radical at bridgehead is unstable due to angular strain introduced by planarity of radical at bridgehead .
Similarly in $\ce{3}$ , due to angular strain introduced in to the neighboring bonds,this  radical also is unstable.
The situation leads to product from ${1}$ , that gives  stereoisomers as shown below.


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Anodization Titanium current vs voltage
I am fairly new to anodization of Titanium. It concerns anodization using oxalic acid with the sole purpose of coloring Titanium grade 2 parts (= type III?). After some research online, I found a lot of information on the required voltage levels per color. However, to my understanding, the "purpose" of the voltage is only to induce a certain current. I did find a few articles and discussions on other forums about anodization based on the current density, but I mainly found information related to a constant voltage. Considering that the current generated by a certain voltage also depends on the resistance of the anode, cathode, electrolyte, etc., I'm wondering why there is this little information on the required current density instead of the required voltage.
Can anyone give me (a link to) the information that I'm looking for or explain me why my reasoning is not correct?
If any information is missing, please let me know!


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What does "reaction rank" in Reaxys stand for?
By exporting some search results from Reaxys, I found "Reaction rank" to be one of the columns in the Excel file. In my file, values descend from 107 to 52 and skip some values in-between too. Does anyone know what reaction rank stands for here?

Depending on how you exported this data, could it just be the sequential numbering of the reactions in your search results, after selecting the reactions you wanted? For example, you may have gotten 500 search results, but with some particular restrictions you exported only a few dozen results, specifically between hits 52 and 107.

For the Elsevier's Reaxys data base, there is a support page you may consult without subscription, nor log-in to the product.  Actually, this is more like a searchable directory, too, because you search for keywords, like «rank», too.
Doing so here -- time stamped 25/04/2019 -- offers insight about «How do I view my results?».  About half-way through this document, in subsection «Reactions. Order, limit or export your search results», the documentation states:

"[Reaxys ranking] Defines the method of determining 'relevance'. To
  change the sort type, click 'Reaxys Ranking' and define another sort
  option: 'No of References', 'Reactant Availability', ‘Product
  Availability'', ‘MW of product’, ‘Yield’ or ‘Publication Year’."

And because the content of columns of numeric content (like molar mass, number of reagents, year of publication) may be sorted, this may affect the presentation of the export, too.  It equally may explain the occurrence of gaps in the values you mention.


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How to make aluminium monostearate?
Can I make aluminium monostearate $(\ce{C18H37AlO4};$ $\pu{344.472 g mol-1})$ by reaction between stearic acid $(\ce{C18H36O2};$ $\pu{284.484 g mol-1})$ and aluminium hydroxide $(\ce{Al(OH)3};$ $\pu{78.00 g mol-1})$ or by adding more aluminium hydroxide in aluminium tristearate?

Aluminium hydroxide isn't a partilarly reactive or soluble compound, so you probably don't want to directly mix it with stearic acid; also, you'd likely obtain a tristearate this way.
According to a procedure described in US3056819A patent, a synthetic path would be to obtain aluminum dihydroxychloride $\ce{Al(OH)2Cl}$ first, and then add stearic acid and sodium hydroxide (please disregard full stops after units, this isn't a standard notation these days):

(1) 2.7 g. (0.1 mole) of metallic aluminium is dissolved in 50 cc. of 2N-HCl (0.1 mole) under heating to obtain a solution of a basic aluminium chloride whose formula is $\ce{Al(OH)2Cl}.$
(2) To an aqueous solution of 47.58 g (0.1 mole) of potassium alum $(\ce{AlK(SO4)·12H2O})$ is added a solution of sodium bicarbonate to form aluminium hydroxide.
  The aluminium hydroxide is dissolved in 50 cc. of 2N-HCl with stirring and the solution is heated for 1 hour at 100 °C. to give a solution of a basic aluminium chloride whose formula is $\ce{Al(OH)2Cl}.$
(3) To a solution of 24.2 g. (0.1 mole) of hydrous aluminium chloride in 100 cc. of water is added gradually 16.8 g. (0.2 mole) of sodium bicarbonate with stirring and the solution is heated for 1 hour at 100° C. on a water-bath.
Any of the solutions of the basic aluminium chloride produced by the above three methods is diluted with water to 200 cc. and added to a solution of 28.45 g. (0.1 mole) of stearic acid and 4 g. of sodium hydroxide (0.1 mole) in 1000 cc. of water, and the mixture is heated on a water bath until the product precipitates.
  After washing with water to remove chlorine ion, the product is extracted with acetone in Soxhlet's extractor to remove free stearic acid, obtaining 33 g. of aluminium monostearate whose formula is $\ce{Al(OH)2(C17H35COO)}.$

References

Hideo, T. Aluminium Soaps and Their Production. US3056819A, October 2, 1962. (PDF)


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Where do the electrons come from to make the carbon stable during bombardment of alpha particles on beryllium
When beryllium is bombarded with alpha particles, where do the electrons come from to make the carbon stable?

Source: https://en.wikipedia.org/wiki/Triple-alpha_process

A carbon nucleus with no electrons around it is stable. Charged, yes. Happy to grab electrons, yes. But it is stable and given time will likely find electrons to become neutral.

Physical stability and chemical stability are totally different and unrelated things, much like crown and crow.

When beryllium is bombarded with alpha particles, where do the electrons come from to make the carbon stable?

Nowhere.  This is a nuclear reaction, not a chemical one.  Charge is conserved and no extra electrons were supplied.
The carbon nucleus is stable with respect to nuclear decay.
Chemical stability relates only to the behavior of electrons with respect to atoms.  It has no place in this nuclear  reaction.
Eventually the carbon nucleus may attract additional electrons from some external source, but that's nothing to do with stability of the nucleus.  The nucleus does not require any electrons to be stable.

The electrons are there, all around and intermingled with the nuclei.  They maintain the requirement of charge neutrality but, at the temperatures in stellar cores, they don't do much else because there is too much thermal energy and entropy to stabilize complete atoms.  So in studying reactions, we ignore them as we would a "spectator species" in an ordinary chemical reaction.
Until, that is, the star dies.  Then the quantum mechanical characteristics of those electrons come to the fore in a big way, enabling the formation of white dwarfs.  The electrons are ultimately important after all.


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Last steps of mechanism of Swern Oxidation
My question is regarding Swern oxidation. Why is the methyl group of the DMSO deprotonated and not directly the red hydrogen atom of the R2-CH-O bond deprotonated? Shouldn't it be that oxygen more electronegative (stronger -I effect) than sulfur and shouldn't that make that corresponding hydrogen atom more acidic? Or is it perhaps that sulfur has a positive charge and therefore acts as a stronger electrophile that attracts the electrons from carbon and therefore its hydrogen groups are more acidic?

Related: ylide stability.
Also, mechanisms usually show the shortest path. Try going your way and see what ends up happening.

If I go my way... I guess a carboanion is built, which is very unstable? But then the electronegative oxygen will take the electrons anyway right? Wouldnt the net effect be the same? Just that there doesnt have to be a second attack?

The oxygen is electronegative, but is it electron-deficient?

The hydrogen alpha to an oxygen is not generally very acidic at all - consider diethyl ether. The +ve charge on the sulfur makes the proton on the neighbouring carbon pretty acidic

S increases acidity more than O to begin with, compare sulfides and ethers; and the positive charge makes the proton next to S even more acidic.


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How can spike recovery be over 100%?
I am confident enough about my knowledge about spiking and the process of standard addition but I came across this process which claimed to have a percent recovery of 115% and it was considered as a good recovery process. How it is considered a good recovery? Is it because you add too much of the spike and recovered most of them?

A spike recovery value of >100% does not mean that you added too much or the recovery process is too good. All it means is that the analytical method which is being followed "in letter and spirit" gives a positive error. There may be a bias in the method or the precision of the method is poor. We cannot say just by looking at one number. At this stage it is assumed that there was no operator error; only the analytical method is to be blamed. A single number like 115% in analytical chemistry has no meaning. Whoever called it a good recovery process needs to revisit that statement.
Whenever we develop a new method, recovery is done is several matrices and even different types of samples. As a result one gets a range of recovery values in different matrices or for different samples. It is better to say that a method shows gives a recovery between 95% to 115%. This statement is far more meaningful than just stating that a recovery of 115% is good. Certainly, that method must be pretty good.