text:
How to make an aluminumoxide coating on an electrode?
The Questions is above.
I am not a chemist but that is what I know:

Aluminum has an Oxide Layer normally
Aluminum-Oxide Layer can be broken down with NaOH
Zink coatings are done by galvanic coating using a certain voltage

Another way to ask: Can I do this with Aluminum?

Are you asking how to electroplate aluminum onto a substrate?

yes essentially, And if thats possible without any further difficulties/ what i should consider when doing so

It may be possible to electroplate , but I have only seen vapor deposition in vacuum or "flame"spray ( actually done like electric welding)  which makes a relatively  coating ( used for cathodic protection. ). and Alonizing , done at very high temperature.

Alonizing , a proprietary process, deposits aluminum vapor on steel at about 1800 F. It is used to coat high temperature equipment ( eg. furnace tubes) for oxidation and sulfidation resistance. Not decorative.

Does this answer your question? Can I anodize an aluminum coating on steel to have better corrosion inhibition for the steel?

@cngzz1 That linked article doesn't, I intended to use the passivated layer as a way to isolate the electrode electrically.

Yes, it is definitely possible, but also relatively difficult as compared to electroplating zinc onto another metal.
It is a significantly important industrial process. Although aluminium itself is reactive, it develops an impervious oxide layer that quite much renders it unreactive.

However, comparatively larger negative standard electrode potential of Al/Al(III) couple (− 1.67 V vs. SHE) precludes its deposition from the aqueous baths.

So it is usually considered impossible in aqueous (water-based) solution so is conducted in either organic liquids or fused (molten) state.
Fused: Usually a bath of an eutectic mixture of $\ce{LiBr, KBr}$ & $\ce{CsBr}$ is taken along with molten $\ce{AlBr3}$ (melts at about 94-98°C). A concentration of $80\%$ (by wt.) of $\ce{AlBr3}$ is reported to give a shiny finish.
Source $1$ reports that a satisfactory result was obtained with constant current of $1-3$ $\pu{A}$ passed for $1$ hour.
Non-aqueous solution:
From Source $2$ (emphasis mine)

..the commercial processes for electroplating of Al are based on organic solvents such as SIGAL (Siemens-Galvano-Aluminium) and REAL (Room-temperature Electroplated Aluminium) processes. In the SIGAL process which was developed in Siemens laboratories Al is electrodeposited at 100 °C from alkylaluminium compounds in toluene. In the REAL process which was developed in the laboratories of Philips $\ce{Al}$ is deposited at room temperature from $\ce{AlCl3}$ and $\ce{LiAlH4}$ in tetrahydrofurane.

The process must be conducted avoiding any traces of water, or the water would be electolysed first, leading to observable evolution of gas.
TL;DR: Electroplating aluminium is not something I would recommend trying at home.

References:

Aluminum electroplating on steel from a fused bromide electrolyte, P. K. Tripathy et al, Surf. Coat. Tech., Vol. 258, 15 Nov. 2014, 652-663. DOI:10.1016/j.surfcoat.2014.08.021
Electroplating of mild steel by aluminium in a first generation ionic liquid: A green alternative to commercial Al-plating in organic solvents, S. Zein El Abedin et al, Surf. Coat. Tech., Vol. 201, Issues 3–4, 5 Oct. 2006, 1352-1356. DOI:10.1016/j.surfcoat.2006.01.065

Thank you very much. Very informative. I think i will then try some other material though. Thats too much of a hassle to homebrew. But I have one question left: Why do some people recommend using NaOH to destroy the oxide Layer and why isn't it possible form Al+1 or Al+2 to go into solution and then using a voltage to redirect the ions to flow to the other electrode?

You're welcome. $\ce{NaOH}$ forms an aluminate complex, and hence dissolves $\ce{Al2O3}$ under moderate heat (read more here). And no, forming and stabilizing +1 or +2 states would be really challenging (and maybe impossible altogether) at room temperature. +1 state is rare but existent, while I don't believe the +2 state exists. If you've studied basic electronic configuration, you might know that Al prefers a +3 state.

Yeah I had studied that, but its a while ago and i never considered one impossible. Thank you very much for the answers


text:
How to determine toxicity of a substance?
I often read description of various compounds or elements as being poisonous. Now I know that through experience we have discovered the toxicity of various substances like lead, chlorine or phosgene etc. We know nitrogen isn't poisonous as its present in the atmosphere etc.
What I wanted to ask is whether there is a perfectly analytical or experimental way through which chemists determine whether a newly discovered substance is or isn't poisonous? Or is it always found out only by exposure to humans (which I don't think is always possible)? Or perhaps whether such an information can't always be deduced?
I think animal testing is a possible (but immoral) method but again some substances could be harmful to them but harmless to us.

For $\mathrm{LD}_{50}$ you get some rats and dose them. When the dose is large enough to kill 50% of them you have your answer.

@MaxW But this clearly states a lack of  validity in $LD_{50}$.

The only way is to do some real tests. Drugs for people have to be rigorously tested before people are exposed. Since Thalidomide, we test animals for teratogenicity, for example, before use in people. (thalidomide is harmless to adults but causes terrible deformations in embryonic development). There is no simple short cut.

@MaxW There is not killing like killing.Killing for food is fast. Killing by poison can be very nasty. Testing of drug candidates is inevitable, but not all testing is justified.

I think that this is broad, at least if we want to keep some chemistry/biochemistry in it. I suspect it varies depending on why / for what a molecule came to the attention (new dtug, new material for processes, etc.). Fortunately for rats, not all the thousands of new molecules synthesised each day in the labs have to be tested. Which, en passant, points to the necessity of being scrupulous on the bench, and avoid dangerous habits such as touching, smelling around, drinking coffee and so on.

@Poutnik — And this is why ethical committees exist in many countries and that is a necessary thing IMHO.

@SteffX I do agree.

No, there is no way to determine toxicity to humans with a "magical method".
Most of our knowledge about toxicity is extrapolated from animal testings and from reported accidental exposures in human.
Also, Nazis carried out a large number of human testing on Jews, and this is not limited to toxicity.

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text:
Potassium carbonate vs. Potassium hydroxide in Wood Ash
I'm trying to get more concrete information about the chemical process of creating soap from wood ash.
What I’m confused about: Many soap-making materials say that you create potassium hydroxide directly from filtering water through wood ash. As far as I can tell this is false? You are actually getting potassium carbonate (and some small amount of other residue.)
To create potassium hydroxide from this, historically you would add a solution of calcium hydroxide (calcium oxide/burnt limestone dissolved in water). Calcium carbonate would precipitate out of the solution, leaving behind potassium hydroxide. From my understanding, a lot of modern soap-makers are simply misunderstanding the chemistry, or not realizing that the lime has to be added before they actually have potassium hydroxide.
However, some sources say that if the wood is heated hotter and given a more complete burn, thus combusting more of the organic material, then more hydroxides will be dissolved instead of carbonates, thus negating the need for lime. Is this true?
I have also read that if you take the potassium carbonate, dehydrated, and heat it to an (uncertain) degree, it will burn off carbon dioxide and create potassium oxide, which added to water will become potassium hydroxide. Is this true? How hot do you need to burn it, if so?
However, other sources say heating potassium carbonate only purifies it, creating pearl ash, which is still potassium carbonate.
Basically, I just need more information about potassium carbonate vs. potassium hydroxide from wood ash sources.

According to this reference https://www.researchgate.net/publication/239372324_Thermal_Stability_of_Potassium_Carbonate_Near_Its_Melting_Point  Potassium carbonate has a melting point of 900C under a nitrogen atmosphere but some volatilisation occurs below that temperature so I guess it depends on how hot the wood fire got but you are probably correct that you mostly get potassium carbonate

You can never get $\ce{KOH}$ directly out of wood ash. One can recall that potash $\ce{=K2CO3}$ is indeed pot-ash! So you are right that historically calcium hydroxide was needed to obtained $\ce{KOH}$ out of potash. The rest of the stories are all false. $\ce{KOH}$ is unstable in air with respect to carbonate formation. It is hygroscopic as well. Wood ash remains bone dry forever. I don't think there is even a trace of potassium hydroxide in wood ash.
Until and unless you don't have a furnace, which can generate temperatures above 1300○ Celsius chances of getting $\ce{KOH}$ in useful amounts are almost negligible.
Secondly, a soap made with tons of calcium salts will be nothing but literally scum (=insoluble calcium soaps). Calcium should not be a part of soap.
Here is a nice abstract on the composition of wood ash as a function of temperature. [1]

The elemental and molecular composition of mineral matter in five wood types and two barks was investigated as a function of temperature using thermal gravimetric analysis, differential thermal analysis, inductively coupled plasma emission spectroscopy, and X-ray diffraction. Low temperature ash was prepared at 500°C, and samples were heated in a tube furnace at temperature increments to 1400°C. The dissociation of carbonates and the volatilization of potassium, sulfur, and trace amounts of copper and boron were investigated as a function of temperature. Overall mass loss of the mineral ash ranged from 23–48% depending on wood type. The mass of K, S, B, Na, and Cu decreased, whereas Mg, P, Mn, Al, Fe, and Si did not change with temperature relative to Ca which was assumed to be constant. Sintering of the ash occurred, but fusion of the ash did not occur. In the 600°C ash CaCO3 and K2Ca(CO3)2 were identified, whereas in 1300°C ash CaO and MgO were the main compounds. The implications for ash deposition in furnaces is discussed.

References:

Misra, M. K.; Ragland, K. W.; Baker, A. J. Wood ash composition as a function of furnace temperature. Biomass Bioenergy 1993, 4 (2), 103–116 DOI: 10.1016/0961-9534(93)90032-Y.

Great information, thank you. About the comment "Calcium should not be part of soaps," I believe you mean the act of adding CaCO3? If you use that method to get from potassium carbonate to hydroxide, after precipitating, I was under the impression you would have 2 KOH + CaCO3, and all the CaCO3 would settle out of the solution. Is this incorrect? Where does the insoluble soap scum come from?

The scum would originate from remaining calcium hydroxide. Calcium hydroxide needs to be added in excess to convert K2CO3 to KOH. Calcium hydroxide has some solubility in water. I assume the scum during soap making can be removed. I have tried this experiment, but at one time I was also very curious about wood ashes for feeding to plants.

Thanks for the clarification. Looking at the solubility of calcium hydroxide (0.066 g/mL) vs. potassium hydroxide (162.9 g/100 mL) --This is for a solution at boiling 100Celcius -- I imagine it would be possible to simply boil off the water from the solution. Basically, using recrystalization as a purification technique. The Calcium hydroxide would be forced to precipitate, and then the solution containing the potassium hydroxide could be poured off. Does that sound reasonable?

Sounds reasonable on paper. KOH cannot be crystallized as far as I know. I have never tried this experiment (sorry for the mistake in the previous comment). As I said my only interest in wood ashes was its use as a plant food but never for soap making. Are you making an organic soap for some reasons?

No worries, I understood your previous comment. It's fine that you haven't tried it, but your understanding of chemistry is probably better than my own. I will have to try it myself, when my accommodations allow. I am a science fiction author and am writing a new series where authenticity is important. The least I can do is take some time to research the material I'm writing about so it's not blatantly wrong.

$\ce{CaCO3}$ is easily decarboxylated to $\ce{CaO}$ in the heat of the fire. By dumping the ash into water you get a solution of $\ce{Ca(OH)2}$ and $\ce{K2CO3}$, and since $\ce{CaCO3}$ is less soluble than $\ce{Ca(OH)2}$ and $\ce{K2CO3}$ you end up with $\ce{KOH + Ca(OH)2}$ solution in one step and $\ce{CO2}$ dissolving into this solution will deplete calcium ions before it will start depleting $\ce{KOH}$

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text:
Oxidation of Aromatic Methyl to Aldehyde by CAN
I read in my text that Ceric Ammonium Nitrate is used for oxidation of primary alcohols. But the surprising part was that, according to the book, it could oxidise aromatic methyl group to a -CHO group.
Is this actually true?

This review should tell you.  https://pubs.acs.org/doi/full/10.1021/cr068408n?src=recsys

I would imagine that it proceeds via a single electron transfer, and the first step would be to create the benzyl alcohol.


text:
Does ultrasonic bath cleaning damage laboratory volumetric glassware calibration?
As an analytical chemist, I'm always ensuring my glassware is thoroughly clean before I start an analysis. However, my supervisor always states that using ultrasonic bath to clean volumetric glassware damages its calibration, rendering it imprecise for analytical usage. Does that hold true, or is it okay to use an ultrasonic cleaner?

Unless it fractured the glassware I don’t see the issue.

I sometimes sonicate volumetrics to help dissolution for a short period. However, I am actually curious about the following test, which I vaguely recall (and therefore could be wrong). If we make up the volume exactly to mark, and then sonicate, what happens to meniscus? It goes down slightly below the mark temporarily. This could be from local heating of the glass. Glass and solutions do heat up during sonication.

A temporary effect due to temperature is sure. But volumetric glassware is supposed to be used at about room T.

According to technical information for volumetric measurement provided by Brand, it is acceptable to use an ultrasonic bath.

Both glass and plastic labware may be cleaned in an ultrasonic bath. However, direct contact with the sonic membranes must be avoided.

Nevertheless,

For gentle treatment of labware, clean immediately after use – at low temperatures, with brief soaking times, and at low alkalinity. Glass volumetric instruments should not be exposed to prolonged immersion times in alkaline media above 70 °C, as such treatment causes volume changes through glass corrosion, and destruction of graduations.

If your supervisor is unconvinced by Brand's guidance on this (as supplied by Loong in his answer), you could always test the effect of ultrasonic cleaning directly:
Take, say, six volumetric flasks, and weigh the amount of water each holds when filled to the line.  Then subject three to several cycles of ultrasonic cleaning (the other three can be left to sit in the ultrasonic cleaning solution, but without turning the machine on), dry all six, and repeat the weighing measurement. Determine if the before-and-after difference for the ultrasonically cleaned flasks is significantly different from that seen in the control flasks.
While the control flasks do control for temperature differences between the before-and-after measurements, it would be cleaner if you could do the before-and-after measurements at the same temperature.  That way the only difference you see should be random experimental error plus any systematic differences due to cleaning.
[Or more simply, if your boss is unconvinced by Brand's guidance, you could ask Brand if they could supply their test data on the effect of sonication.]
EDIT, BASED ON COMMENTS: If cumulative damage is the concern, then one could always employ an accelerated wear technique.  In this case, one would increase the sonication time until it matched the total sonication time anticipated over the lifetime of a volumetric flask.  E.g., if  standard cleaning time is 20 minutes, and glassware is typically cleaned 600 times, then one could leave it in the sonicator continuously for 8 days (assuming the sonicator wouldn't be damaged by this).  If you don't have a spare sonicator, you could do this over four successive weekends.  Or perhaps one weekend would be sufficient to satisfy your manager.
Note that this accelerated wear technique is a standard method used in product testing -- e.g., if you have a part that over its, say, 10-year lifetime is expected to be flexed no more than 10,000 times, then you put it in a device that does those 10,000 flexes over a very short period of time (or does sufficient flexes on several parts to determine a mean time to failure).

I suspect his adviser's objection is to repeated cleanings, not a single (or even short cycle) event. For the experiment to hold true, you'd need to perform potentially hundreds of cleanings.

@Valorum No. Put it in the ultrasonicator over the weekend and be done with it. Of course the result will be the same, glass does not yield to ultrasound. Repeating the same experiment a thousand times just in the hope it will one day give a different result is not good scientific practice.

@Karl - A beaker in the lab might get re-used thousands of times. Unless you can adequately model what happens after x thousand cleanings, your results will be worthless. Cleaning a table with a too-harsh cleaner once won't usually harm it in any measurable way. Cleaning it hundreds of times will result in the patina coming off.

@Karl It's not "repeating the same experiment a thousand times." It's possible that one ultrasonic wash doesn't make a large enough difference to measure, but that, over the life of a flask, the damage will add up to something significant. How many washes would a weekend correspond to?If it's something like a lifetime of washing, fine; if it's not, then it's not enough.

Can anyone propose a mechanism that could impair a flask's calibration with repeated sonication? Does the advisor think cavitation will erode the flask's interior?


text:
Beer's law with a polychromatic light source
As is well known, Beer’s law (aka the Beer-Lambert law, among others) is predicated on a number of conditions, one of which is that the light source be monochromatic or, as a practical approximation, that the spectral bandwidth of the light source be much smaller than the spectral bandwidths of whatever absorbers are being investigated. For present purposes, a polychromatic light source may be considered to have spectral bandwidth, $\ce S_{source}$, that is not negligible relative to that of the relevant absorbers. The question posed below will deal with what happens when the light source is decisively polychromatic.
Assume a cell path length, b, of 1 cm, and assume a single absorber, having triangular molar absorptivity spectral profile (Fig. 1):

The peak molar absorptivity, $\ce \epsilon_{450 nm}$, is 100 L/mole cm at 450 nm. Assume the absorber concentration, c, ranges from 0 to 0.01 M. In the figure, the absorber’s molar absorptivity spectral profile is modeled as an isosceles triangle. In reality, absorber profiles are often modeled [1, p. 164] as Gaussian, Lorentzian, Voigt, or whatever the underlying physics requires. Here, however, an isosceles triangle is used for mathematical simplicity.
The figure also shows a narrower red isosceles triangle, centered (and having unity transmittance) at 425 nm and having full width at half maximum height (FWHM) equal to 20 nm. This red triangle is the transmittance spectral profile [2, equation 3-60, p. 70] of the light that exits a monochromator, assuming white light of uniform (unity) intensity is incident upon the monochromator and also assuming both 10 nm/mm reciprocal linear dispersion (RLD) and 2 mm slits ($\ce W_{slits}$). With white light incident on the entrance slits of the monochromator, the exiting light has $\ce S_{source}$ = RLD x $\ce W_{slits}$ = 20 nm. Note that $\ce S_{source}$
 is defined here as the FWHM of the spectral profile: this is not the only possible such definition, but it is satisfactory for present purposes. Before proceeding, note that a 0.1 M monochromator typically has something like RLD = 4 nm/mm and 2 mm slits are relatively huge, as slit widths go, so the light source’s profile is polychromatic (and not good at all). Indeed, the FWHM is 20 nm, while that of the absorber is 50 nm.
As always, $\ce T(\lambda) = I_{transmitted}(\lambda)/I_{incident}(\lambda)$ and $\ce A(\lambda) = -log[T(\lambda)]$. If $\ce c = c_{max} = 0.01 M$ and b = 1 cm, then the next figure shows the absorbance (black) and transmittance (red) spectra from 380 to 520 nm (Fig. 2):

If the light source was monochromatic, at 425 nm, a Beer’s law plot would be a straight line with slope = $\ce \epsilon_{425 nm}$ x b = 50 L/mole cm x 1 cm = $50 M^{-1}$.
For absorber concentrations from 0 to 0.01 M, $\ce A_{max} (425 nm)$ = 0.5 at $\ce c_{max} = 0.01 M$.
However, for the light source profile in Fig. 1, the Beer’s law plot will curve downward, as undergrad textbooks like to show in cartoon form, and the absorbance at $\ce c_{max}$ will be less than 0.5.
Question: For the above scenario, with the specified polychromatic source profile in Fig. 1, what is the absorbance at $\ce c_{max} = 0.01 M$?
Hints: 1) Forget about undergrad textbooks for this. 2) There is no added noise or error. 3) The light source cannot be modeled as a pair of monochromatic lines.
Assume the photodetector responsivity is flat from 200 to 700 nm and dark response is zero. There is no detector or preamplifier noise, no digitization error, and no source shot noise. This is idealized in every reasonable way except for the polychromatic source, which, of course, no one would deliberately use.
References:

T.C. O’Haver, “Effect of the Source/Absorber Width Ratio on the Signal-to-Noise Ratio of Dispersive Absorption Spectrometry”, Analytical Chemistry, 63 (1991) 164-169.
J. D. Ingle, Jr. and S. R. Crouch, Spectrochemical Analysis, Prentice Hall, Englewood Cliffs, NJ, ©1988, pp. 70-74.

Generally this sort of extended mathematical analysis is useless. As an analytical chemist I'd create a calibration curve.  If it is non-linear then you need more standards to get a good fit.

Sure! Using a polychromatic source obviously hurts sensitivity, peak resolution and linearity, but how much?

Well, to solve the problem you're missing the detector sensitivity as a function of wavelength.

Assume the responsivity is flat from 200 to 700 nm and dark response is zero. No detector or preamplifier noises, no digitization error, and no source shot noise. Idealized in every reasonable way except for the polychromatic source, which, of course, no one would deliberately use. By the way, having read many thousands of questions and many answers and comments, I thought it would be fun to have my first question not involve propane, napalm, or other such things. So, I don't answer those and no one has to answer this one. Seems fair.

@EdV is this question something related to a problem you have actually encountered or a sort of "challenge problem" that you came up with? Either way it's an interesting question, but I think you should specify if you already know the answer and are just looking to see other responses/self answer.

Dear Ed, this is a very interesting question...is this a exam question you used to write for graduate students? Write the answer yourself. The quality of discussions is decreasing esp. in chemistry. This happens in every forum with time.

@Tyberius I already know the answer (I think!) and thought it would be fun to see what would happen if I asked it here. In real research, it often happens that what is very well understood is right next door to something that was never even suspected. So it is just a little puzzle for anyone who might want to play with it and maybe someone will gain a small pleasure from a bit of unorthodox thinking about our old friend Beer's law.

@M. Farooq It is not an exam question for the students I used to teach in the grad Spectroanalytical Chemistry course. I will post my answer in due course, but I am genuinely curious to see if an answer from someone else gets posted. And my reputation is now so large (!) that maybe I should put a bounty on the question! Anyway, your comment is on target and there does seem to be quite a few questions with either no answer or no accepted answer. I have no idea why that is the case.

@EdV Cool! Just wanted to make sure people understood that when they came across the question.

I do not understand what is in figure 2, if compared to the text. And besides integration, isn't just matter of how areas overlap / incoming "area"?

@Alchimista Figure 2 is just a possible convenience: it simply shows the absorbance and transmittance when c is 0.01 M. As for integration and overlap, I thought it might be helpful to remember that the nice triangular molar absorptivity profile does not mean that transmittances have a triangular spectral profile. So Fig. 2 helps to keep that in mind.

I see it now. Lambda max refer to a monochromatic ideal csse. The picture doesn't. I was confused.

There are several ways to answer the question, so here are my three. The first answer involves using a simple integer wavelengths approximation, i.e., noticing that Figs. 1 and 2 have 1 nm point spacing, and then using a spreadsheet to perform a summation. The answer explained in this part also serves to define the terms used in the second answer, which involves analytic calculation of the relevant overlap integrals. Then the third answer, not discussed in detail, uses a Mueller calculus-based method. Before proceeding to the explanations, the numerical answer to the question is $A_{maximum} = 0.46951302113867$ at $C_{max} = 0.01 M$ and Fig. 4 shows the absorbance as a function of concentration:

The absorbance is 93.9% of the ideal value (i.e., 0.5) at $C_{max}$, despite the source being significantly polychromatic, and the Beer’s law plot is convex, but perhaps less than might have been expected intuitively.
All three of the answers and explanations that follow are based on the model shown in Fig. 4 below:

As per the question’s assumed conditions, the light intensity incident at the monochromator is unity, i.e., 1, with fictive units of Watts. Note that units may be ignored since they will divide out when transmittance is computed later. The 50:50 beam splitter sends half of the light it receives to the sample (upper) beam path and the other half of the light it receives goes to the reference (lower) beam path. Let $I_{incident}(\lambda )$ denote the light intensity incident upon the absorber and $I_{sample}(\lambda )$ denote the attenuated light intensity that gets through the absorber. Then $I_{sample}(\lambda )$ is the incident light intensity on the upper photodetector and transimpedance amplifier (PD-TIA).
From Fig. 4, it is clear that $I_{reference}(\lambda )$ = $I_{incident}(\lambda )$, assuming the mirror is ideal. This is simply the unattenuated incident light intensity on the lower PD-TIA. The photodiodes were stated to have flat responsivities, i.e., spectrally constant responsivities, so their photocurrent responses are independent of wavelength. Hence, the responsivities, in this idealized scenario, cancel out when transmittance is computed later. Likewise for the constant transimpedances, which are assumed equal. Hence, it suffices to simply ‘add up’ the light intensities, for each color of light, incident at each PD-TIA. Now to the answers and explanations.
Answer 1: As per Figs. 1 and 2, assume a wavelength spacing of 1 nm. There is no light outside the region 405 to 445 nm and wavelengths are effectively “discretized”, i.e., are treated as discrete integer values. Therefore, the number of discrete wavelengths, where light intensity is non-zero, is just 39, i.e., from 406 to 444 nm, in 1 nm increments. At 405 nm, $I_{incident}(\lambda )$ is 0 and it increases linearly to 0.5 (at 425 nm). It then decreases linearly to 0 (at 445 nm). The absorbance, $A_{absorber}(\lambda)$, increases linearly from 0.1 (at 405 nm) to 0.9 (at 445 nm). Hence, the transmittance through the absorber, $T_{absorber}(\lambda)$, decreases non-linearly from about 0.7943 (at 405 nm) to about 0.1259 (at 445 nm). As always, $T = 10^{-A}$, so there is no difficulty computing $T_{absorber}(\lambda )$ at any of the relevant wavelengths.
Accordingly, $I_{sample}(\lambda )$ = $I_{incident}(\lambda )$ x $T_{absorber}(\lambda )$ and $I_{reference}(\lambda )$ = $I_{incident}(\lambda )$ x $T_{reference}(\lambda )$ = $I_{incident}(\lambda )$, since $T_{reference}(\lambda )$ = 1. All of the necessary calculations are shown in the spreadsheet screenshot in Fig. 5 below:

In Fig. 5, column A contains the wavelengths and column B is a unitless “x” variable that is relative to 405 nm, i.e., at 405 nm, x = 0 and so forth. Column C contains the incident light intensities, column D has the associated absorbances and column E has the corresponding transmittances through the absorber. Column F has the unity transmittances for the reference beam path, column G has the $I_{sample}(\lambda )$ values and column H has the $I_{reference}(\lambda )$ values. Adding up the 41 values in columns G and H yields total intensities: 3.39164 and 10.000, respectively. Then T is simply 0.339164 and $A_{maximum} = 0.469590$. This is close to the true value, as computed in the second answer, despite the rather coarse “discretization” of the wavelengths.
Answer 2: From the first answer, it is clear how the relevant overlap integrals should be calculated. First, x is the independent (unitless) wavelength variable and it is continuous from 0 to 40. As x increases from 0 to 20, $I_{incident}(x)$ increases from 0 to 0.5, so $I_{incident}(x)$ = x/40. As x increases from 20 to 40, $I_{incident}(x)$ decreases from 0.5 to 0, so $I_{incident}(x)$ = 1- (x/40). For the absorbances, $A_{absorber}(x)$ = 0.1 + 0.02x for x = 0 to 40. Hence, $T_{absorber}(x)$ = $10^{0.1 + 0.02x}$ and $I_{sample}(x)$ = $I_{incident}(x)$ x $10^{0.1 + 0.02x}$.
For the sample beam path, the total intensity, $I_{sample}^{total}$, is the following overlap integral: $$I_{sample}^{total} = \int\limits_0^{40} {I_{sample}(x)\ \mathrm dx} $$
But this is simply the sum of two definite integrals: $$\int\limits_0^{20} {(x/40)\ 10^{0.1 + 0.02x}\ \mathrm dx} + \int\limits_{20}^{40} {[1-(x/40)] 10^{0.1 + 0.02x}\ \mathrm dx}$$
Using the fact that $10^{-x} = e^{-xln(10)}$, and defining $a \equiv -0.02 ln(10)$, then results in
 $$ \frac{e^{5a}}{40} \int\limits_{0}^{20} {xe^{ax}  \mathrm dx} + e^{5a} \int\limits_{20}^{40} {e^{ax}  \mathrm dx} - \frac{e^{5a}}{40} \int\limits_{20}^{40} {xe^{ax}  \mathrm dx}$$
The integrals are easily evaluated, resulting in $$ \frac{e^{5a}}{40a^2} {[(20a - 1) e^{20a} + 1]} + \frac{ e^{5a}}{a}{[e^{40a} – e^{20a}]} - \frac{e^{5a}}{40a^2} {[(40a - 1) e^{40a}] - [(20a - 1) e^{20a}]} $$
With $a \equiv -0.02 ln(10)$, $I_{sample}^{total}$ = 3.39224318670537.
For the reference beam path, the integrals are easier because $T_{reference}(x)$ = 1. Hence $$ I_{reference}^{total} = \int\limits_0^{20} {(x/40) \ \mathrm dx} + \int\limits_{20}^{40} {[1-(x/40)] \ \mathrm dx} = 5 + 5 = 10$$ Alternatively, this is simply the area of the triangle: 0.5 x 40 x 0.5 = 10. Therefore, $T =  I_{sample}^{total}\ / \ I_{reference}^{total} = 3.39224318670537\ /\ 10 = 0.339224318670537$. Then $A_{maximum} = 0.46951302113867$.
The analytical method worked, as expected, but the system was so heavily idealized that it was essentially a “spherical cow” and they don’t milk. Any realistic complications, e.g., even simply using a tungsten bulb light source, make the analytic method harder and it does not take much to make the equations all but analytically intractable. Then numerical methods may be employed, but the difficulty then mostly lies in getting the right model equations: correctly evaluating equations is not necessarily the same as evaluating correct equations.
What about convolution? Convolution, per se, is unnecessary in this heavily idealized problem, for the reasons discussed below. First, note that the spectral profile of $I_{incident}(\lambda )$ is the same as the monochromator’s transmittance profile (shown in Fig. 1), except for having a maximum value of 0.5 rather than unity. The spectral profile is symmetric. Second, the absorber’s transmittance profile is as shown in Fig. 2 and it is also symmetric. Third, the convolution of two real functions involves reversing one of the functions and then computing the overlap integrals as a function of a shift parameter, i.e., a relative displacement of the two functions. Since the two functions to be convolved are both symmetric, nothing needs to be reversed, so the convolution reduces to computing the overlap integrals as a function of a shift parameter, $\tau $. Zero displacement, i.e., $\tau = 0$, can be conveniently defined as when the two functions are perfectly aligned.
The next step is to compute the convolution function either analytically, if possible or desired, or numerically, if necessary or convenient. Then the convolution function would be evaluated at $\tau = 25$ (nm), since this corresponds to the 25 nm shift between the two spectral profiles and $\tau $ is integer valued, i.e., data values are available every nm, as shown in Fig. 5. With finer wavelength spacing, the shift would scale up accordingly. If the convolution is computed analytically, then $\tau $ is a real number.
For the data in Fig. 5, the numerically computed convolution is shown in Fig. 6:

From Fig. 6, the value of the convolution at 425 nm is 3.39164, as in Fig. 5. If the two spectral profiles do not overlap, then the value of the convolution is 10, as expected. But note that all that was needed was the single overlap integral at the known $\tau = 25$ (nm), and that value is 3.39164, as shown above. For the reference beam path, the overlap integral was 10, as shown in the trivially simple calculation above.
This problem is so heavily idealized that there is no necessary reason to compute the convolution function. In a more realistic situation, the computation would be harder and even numerical evaluation of the convolution function might be fruitless.
Answer 3: Figure 4 appears to just be a schematic block diagram, but it is actually a screenshot of a Mueller optical calculus-based computer simulation program. Although discussion of the fundamental principles of the several optical calculuses is beyond the scope of this answer, both Mueller calculus and Jones calculus were invented in the early 1940’s and there is a considerable body of information about them in the literature, e.g., [1-4].
In the simulation depicted in Fig. 4, stacks of Stokes vectors (one per discrete color) are passed from one optical component to the next optical component(s), e.g., from the lamp block to the monochromator block. Each component has a resident stack of matrices, with components as appropriate. When an optical component receives an incoming stack of optical vectors, each vector is multiplied by the corresponding matrix and the resulting stack of processed optical vectors is sent out to the next optical component (or components) “downstream”. All the stacks of optical vectors are received, processed and then sent on in a single computer simulation step.
The plot shown in Fig. 3 was obtained by using 1,001 concentrations: 0 to 0.01 M, in increments of 10 µM. The wavelength range was from 405 to 445 nm, in 1 pm increments. Thus, with every one of the 1,001 simulation steps, stacks of 40,001 Stokes vectors (used in Mueller calculus) were passed, starting with the lamp block and ending with the sample and reference PD-TIA blocks. Both of those blocks, for every step, computed 40,001 intensities, multiplied them by responsivities, summed the resulting photocurrents and then multiplied by the transimpedances.
In the present case, the responsivities were simply constant default values (0.1 A/W) and the transimpedances were $1 k\Omega$. For every step of the simulation, T was computed as the quotient of the real number outputs of the two PD-TIA blocks and $A_{maximum} = -log(T) = 0.46951302122$. Compared with the exact answer, this differs in the ninth decimal place. As well, there was no need to solve any equations or use a spreadsheet: no wheel had to be re-invented.
Another of the advantages of using optical calculus-based spectrometric simulations is that they can easily accommodate non-idealities, such as realistic lamp spectral profiles and photodetector responsivities, imperfect beam splitters and mirrors, polarized light, optical activity, circular dichroism, stray light, light source modulation and so on. In fact, any linear optical effect can be modeled with an optical calculus. In addition, noise effects are easily included, e.g., detector/preamplifier noises and light source noises, including low frequency noises.
Summary: Answer 2 provided an exact solution to the posed question, but it was a bit tedious to obtain and would not be useful if the system was altered in any significant way, e.g., if the photodiode responsivities were those of a real red-sensitive photodiode. Answer 1 was adequately accurate, despite its simplicity, but not particularly amenable to generalizations. Answer 3 gave an accurate answer and made it simple to get the plot shown in Fig. 3. There is no significant difficulty in generalizing to more realistic scenarios.
Lastly, it is clear that using the decisively polychromatic source in Fig. 1 caused a relatively small reduction in $A_{maximum}$ and the Beer’s law plot is convex, but not badly so. The deleterious effect on spectral resolution was not investigated here, but could easily be done, perhaps as another question.
References:

W.A. Shurcliff, Polarized Light, Harvard U. Press, Cambridge, MA ©1962 & 1966.
D.S. Kliger, J.W. Lewis, C.E. Randall, Polarized Light in Optics and Spectroscopy, Academic Press, Boston ©1990.
Mueller calculus article in wikipedia and references therein.
Jones calculus article in wikipedia and references therein.

Why use Mueller/Jones matrices when you are not overtly dealing with a polarisation problem ? The response you measure $I_{sample}$ will be the convolution of your instrument slit width (monochromator transmission in fig 1 of the question) with the molecular transmission profile as you scan the former over the latter.

Your answer 2 seems to be a straight integration not convolution.

There is no need it is a well understood process.


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List of primary, secondary and tertiary alcohol
Is there a list of primary, secondary and tertiary alcohols examples anywhere?

There can be infinitely many alcohols of each kind. Why would you want a list anyway? It's like asking for a list of instances you went to the restroom.

I would like to see some examples of them.

Is Google blocked in your country?

I googled it and kept getting results that had nothing to do with Alcohols.

Try this.

If you know, what primary, secondary, tertiary alcohol is, you need not a list. If you do not know it, learn it first.

If there are 2 hydrogen atoms connected to the carbinol carbon(the one with a hydroxyl group) its a primary alcohol.
If there is only one hydrogen atom connected to the carbinol carbon it's a secondary alcohol.
The one with no hydrogen atoms is a tertiary alcohol.
Now you can make the list yourself.

Unless the carbon belongs to aromatic  system, in such a case we consider it rather as a phenolic than alcoholic group.


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What are orbitals actually?
If s, p, d and f subshells have these kinds of boundary surface diagram, then why don't they include the probability of finding electrons in their preceding shells?
As in, if d subshells have boundary surface diagrams as stated by the great scientists, then it must include the nodes of the preceding subshells, like that of s and p shells.

May be add a picture of your question to clarify it. Make a diagram of your question in power point and share a picture. Orbitals are solutions of a differential equation proposed by Schrodinger. When this equation is solved for a single electron bound to a nucleus of positive charge. Under various mathematical restrictions, you get various "shapes" of orbitals.

10 is greater than 9; how come it doesn't have the figure 9 in it? Well, just like that. Same thing here.

Is it me or the title is a totally different question than the actual question?

There are no requirements for orbitals as wave functions  to respect geometrical details of other orbitals.
There are 3 meanings of the term orbital:

A complex wave function $\Psi$ being a particular solution of the quantum wave equation
A quantum state of an electron, following 1., described by particular quantum numbers $n$, $l$ and $m$, representing discrete values of electron energy, orbital angular momentum and one of component of angular momentum, respectively.
3D shape, following 1. and 2., representing statistical  probability of the electron presence at given point by$|\Psi|^2$, determined by the conventional value by residual external probability and by the isoprobability 3D surface.

In one electron atomic system, like $\ce{H}$,   $\ce{He+}$, orbitals are independent on each other.
In a multielectron system, an electron occurence probability depends on quantum states of other electrons, due kernel shielding and electron repulsion.
The typical example is the relative energy of ns and (n-1)d orbitals.
For quantum chemistry calculations, there are done simplifications like electron independence and orbital shielding factors ( s > p > d > f ).
Electrons in f orbitals do not shield much, what is the major contribution factor of lanthanoid contraction.


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What is the mechanism of this pyrylium synthesis?
In this preparation, http://www.orgsyn.org/Content/pdfs/procedures/CV5P1106.pdf, 2,4,6-trimethylpyrylium perchlorate is prepared from 4 molecules of acetic anhydride and 1 molecule of t-butanol.
However, I cannot find a mechanism of this reaction, nor can I think of one. Could someone please provide the mechanism? Thanks.

From some quick research it seems the first step involves forming isobutylene from the t-butanol. Both ac. anhydride and t-butanol have basicity potential (oxygen lone pairs), but the t-butanol conjugated acid can undergo [Dn] and [De] steps to form isobutylene, which can then act as a nucleophile on the reaction (the piece that seems to be missing in the reactants). I didn’t go further than that yet, but I’ll try to figure it out later if no one does it by then.

Look at Method II in your link. They use mesityl oxide (4-methylpent-3-en-2-one) and 2 equiv. of Ac2O. Mesityl oxide is formed from isobutylene and Ac2O. Form the extended enol of mesityl oxide, acylate it and cyclize.

I tried to work out a mechanism from the information I could gather about the reaction. If someone notices something odd/wrong please comment and I'll fix it.
The steps are either showing the protonation or skipping directly to the protonated forms of the reactants, them being in acidic medium.
Formation of isobutylene

Not very complicated, in acidic conditions we remove the hydroxyl group from the t-butanol to form isobutylene. The latter behaves as a nucleophile in the next steps of the reaction. This step is acid catalysed, so the acid is not consumed.
Formation of Mesityl Oxide

Also relatively simple step. Nucleophile addition to the protonated C=O bond [Adn] followed by elimination of Acetic Acid.
Tautomerism

Shows the Mesityl Oxide tautomerism, important because the tautomer that will react to form the final product is the one in the right (with the double bond in the terminal carbon).
Enol Formation

Here a Enol is formed by a nucleophilic attack of the Mesityl Oxide over another Acetic Anhydride molecule (protonated by the acidic medium). In the last step I added the deprotonation of the more acidic oxygen to respect Pauling's rule of charges by keeping an overall +1 charge in the molecule. The left oxygen is more acidic since the other oxygen can better stabilize the charge with delocalization. I wrote the molecule with wavy bonds because I'm unsure about the stereospecificity of the step, but maybe the Z conformation is favored due to the charges on the oxygens repelling themselves?
Cyclization

Final step is cyclization of the intermediate, by nucleophilic attack of the carbon by the deprotonated oxygen. The intermediate is dehydrated and rearranged to form the final product (releasing a proton to reach aromatic status). This step had an intermediate with +2 charge, which I'm a little skeptical about due to the Pauling's rule of charges, so I also draw the mechanism as a concerted step:

If the rule of charges isn't that strict, maybe the non-concerted path is a closer description of what's happening, but I'm keeping both here for now. If anyone thinks one of the two is the right one, comment and I'll update the answer.

I would enolize mesityl oxide by protonating the oxygen to form the dienol and then protonate acetic anhydride to the nucleophilic addition. You don't want a negative charge on oxygen in acidic medium. The structure above acetic acid needs to be a Z-double bond in order to cyclize.

@user55119 I omitted the protonation trying to make the mechanism less messy, but I'll try to draw the mechanism without omitting them later. By the way, regarding the Z double-bond, is the reaction of the mesityl oxide with acetic anhydride stereoselective? Or will I have both the Z and E forms? And do you have any idea why the method claims more acetic anhydride equivalents are required?

The yield is only 55% so some alpha acylation of mesityl oxide (MO) may occur. As to E an Z isomers, they interconvert via the enol of the diketone. The stoichiometry is correct. Two equivalents of Ac2O are required to form the diketone. Another Ac2O reacts with water from t-BuOH and the fourth equivalent of Ac2O reacts with the water formed in the last step. If you start from MO, then only two equivalents of Ac2O are required. The idea is to sequester all water as acetic acid.

One other thing, in your seventh structure change the -OH to =O. –

Under remaining steps: reverse the directions of your arrows on mesityl oxide. You can create the dienol this way without invoking hydride loss (as the system is under acidic conditions).

@user55119 I updated the answer adding the protonated forms and tried to organize it better. Still think I'm using less Acetic Anhydride in my mechanism than it seems to be indicated by the method on the link provided by the OP..

@Beerhunter Thanks, I fixed it in the updated anwer!

@IanC maybe use some enol instead of carbonyl during your mesityl oxide final stages/ ring closing steps. You don't need to have all the protonation you've drawn and it consistently  keeps your charges down to +1

@Beerhunter Wouldn't the enol be protonated on acidic conditions? The idea of showing the protonated forms was to have a closer picture of the reaction intermediates. By the way, I'm not sure it's worth it to keep working on this answer, spent quite some time already and not a single vote, even the OP seems to have either forgotten about the question or moved on..

If there weren't double positives in the mechanism I'd have definitely upvoted

@Beerhunter I tried to account for that by deprotonating the enol in the "Enol Formation" step (I could have it right before the nucleophilic attack in order to keep the double charge of ever happening though) and by adding the possibility of the cyclization being concerted (no double charges would happen then). If the cyclization had an enol instead of carbonyl, like you proposed, wouldn't the negative oxygen be readily protonated in the next step anyways? Another way of avoiding the double charge in the cyclization would be having $\ce{OH-}$ as a leaving group instead of $\ce{H2O}$

Though the $\ce{OH-}$ would quickly be consumed in acidic conditions to form $\ce{H2O}$ anyways, only difference is if it that would happen while it's still attached to the molecule or after it being released, or right between both (concerted option)


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Why is Polonium the only metal to crystallize forming Simple Cubic Cell?
I have read that Polonium is the only metal that crystallizes forming SCC. I searched for a concrete explanation but all they were saying is the stability of Polonium at STP. I can't really comprehend their explanation.

There is no simple explanation. If you think you have one, you are wrong.

(nit-pick: research is not a fancier version of "search", but means a carefully planned investigation, in which you expect to find something that is not yet known by others.)

The inorganic crystal structure database (ICSD), by specifying that
      the structure contains only a single element, you can reduce the
      number of hits to 55. Structure examples are Cr, Po, Mn, Sb, Se, P,
      Ca, Li, As. Some are specific phases occurring at different
      temperature or pressures.

So Po is not the ONLY metal to have a simple cubic structure, but
the other cases require special conditions.

The origin of the stabilized simple-cubic (SC) structure in Po is
  explored by using the first-principles band calculations. We have
  found that the prime origin is the inherent strong spin-orbit (SO)
  interaction in Po, which suppresses the Peierls-type structural
  instability, as usually occurs in p-bonded systems. Based on the
  systematic analysis of electronic structures, charge densities, Fermi
  surfaces, and susceptibilities of Se, Te, and Po, we have proven that
  the stable crystal structure in VIA elements is determined by the
  competition between the SO splitting and the crystal-field splitting
  induced by the low-symmetry structural transition. Our study suggests
  that the large SO interaction would suppress the Peierls instability
  which is generally expected to occur in one-dimensional conductors.

In fact, Po becomes airborne with ease: if a sample is heated in air
  to 55 °C (131 °F), 50% of it is vaporized in 45 hours to form diatomic
  Po2 molecules. (The melting point of polonium is 254 °C and its
  boiling point is 962 °C.)

Putting it all together, Po is in the column with Se and Te, and just to the right of Bi, none of which are typical of "normal" metals. The formation of $\ce{Po2}$ molecules at near ambient temperatures suggests that the outer electrons of Po form bonds of a highly covalent nature.
But it is indeed unusual and fascinating, and if Po were not 5000 times more radioactive than radium, I might be tempted to do a little chemistry.
References

Quora:
"What are examples of simple cubic crystal structure?" Gervais Chapuis, Nov 3, 2016
B. I. Min, J. H. Shim, Min Sik Park, Kyoo Kim, S. K. Kwon, and S. J. Youn, Origin of the stabilized simple-cubic structure in polonium: Spin-orbit interaction versus Peierls instability, Phys. Rev. B,  73, 132102 – Published 7 April 2006
https://journals.aps.org/prb/abstract/10.1103/PhysRevB.73.132102
https://en.wikipedia.org/wiki/Polonium


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Would dissolving Mg in NH3 help me purify Mg from all elements other than Li, Na, K, Rb, Cs, Fr, Ca, Sr, Ba, Eu and Yb?
According to Wikipedia all of the alkali metals, as well as $\ce{Ca, Sr, Ba, Eu,}$ and $\ce{Yb}$ (also $\ce{Mg}$ using an electrolytic process[4]), dissolve in ammonia to give the characteristic blue solutions.
Could I just dissolve $\ce{Mg}$ in $\ce{NH3}$ and then filter it to remove traces of any elements other than those substances (supposing my aim is to get any other element to get $<\pu{0.1ppm}$)?

Blue solutions are temporary, until solvated electrons ( metals are already in form of ions ) reduce ammonia to amide and hydrogen. When electrolysis is involved, I doubt about selectivity.

What other elements do you mean, alkalis and earth-alkalis? Why and in what form would they precipitate? And for what purpose? If you condense NH3 gas into a clean flask, it's already pretty ion-free. Sorry, totally unclear what you ask.

@Karl All other elements. They wouldn't precipitate, but would simply not dissolve to start with. My goal is to purify Mg, not NH3.

You are not likely to find much francium in your metal.  There is only an ounce or so in all of Earth's crust.

@OscarLanzi: You are right. I just listed all the elements that theoretically dissolve in NH3.

I can only give you a qualitative answer based on my experience. But I'm doing a lot of Europium based chemistry where I use Europium metal as reagent, often together with sodium as a fluxing agent. The sodium is then removed using liquid ammonia by constantly dissolving it, decanting it off and warming it up.
Just to give you an idea, I'm using about 90 - 100 mg of Sodium metal usually. My container is filled with, I'd say about 10-15 ml ammonia per run. It will take me about 20-25 repetitions until I have most of the Sodium removed, which is about 1-2h of work. And that is just for the Sodium. If we use Lithium it will take much longer. And I usually find metals like Sr, Ba, Sm or Yb still in my washed sample.
If you have traces of other metals present, say Eu or Sr the removed sodium will often adopt a golden color. So it's usually a mixture and not well separated, just as you assumed it. But I imagine you could perhaps dissolve the Magnesium and separate it from other metals that way. But using standard pressure would most likely take forever. I never tried Magnesium but I can tell about Sodium, Lithium, Strontium, and Europium. And the latter ones are extremely slow.
This could perhaps be increased if you left the metal for a while in the ammonia or if the pressure or temperature is increased.
Our setup works by cooling down the vessel to - 80°C so the ammonia will liquify. And then immediately pour it off to save time. So there isn't much time for the metals to dissolve and the pressure is then often at around 350 - 400 mbars.
But as I said this is just from my experience, a specially designed setup may work in your case.

Thanks a lot for your answer! Would you perhaps have an idea of another way to purify magnesium? I'm most concerned with Al & As which I'd like to bring down under 0.1ppm or so. For As I guess I could oxidate my Mg sample (by calcination?) and then dissolve & filter it "rinsing out soluble As oxides", and repeat a few times if necessary. When it comes to Aluminum, I don't have the slightest clue how to bring it down that much in a convenient way... Maybe I should ask a new question for that specifically?

I would guess magnesium takes so long to dissolve that it decomposes before a large concentration is built up. That's why they use an electrolytic process.


text:
Understanding the unit cell of Chalcopyrite vs its formula
According to the Wikipedia article for Chalcopyrite it has a formula of: $\ce{CuFeS2}$ and a unit cell of:

But from the unit cell it has 13 Copper, 10 Iron and 8 Sulphur.  This doesn't match with the formula.

Some of the Copper and Iron lie on the faces of the unit cell.
Copper atoms
The 8 Copper atoms at the corners all represent the same Copper atom so that means only 13 - 7 = 6 Copper atoms.
Then there are an additional 2 repeats on the top and bottom faces and front and back faces bringing it to 4 Copper atoms.
Iron atoms
The Iron atoms similarly have a pair repeat on the left and right faces, the top and bottom faces and the front and back (10 - 3 = 7).
Additionally the 4 Iron atoms in the centre line (from left to right) represent the same atom so thats 7 - 3 = 4 Iron atoms.
This is how the formula and the unit cell are consistent with each other.


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Can we tell if a compound is explosive by just looking at its chemical structure?
If I showed you a compound's chemical structure will you be able to tell if it can be used as an explosive or we can only know that by conducting chemical experiments?

Some materials you wouldn't think of as explosive (depending on your background) are very explosive under the "right" (or very wrong) circumstances.

As @BuckThorn pointed, it depends on the conditions in which the compound has been put and for unknown or new ones, you will have to experiment. There is no known (none to me atleast) way of telling whether something's explosive.

https://chemistry.stackexchange.com/questions/7587/what-makes-substances-explosive?noredirect=1&lq=1

Looks like more precise dupe of https://chemistry.stackexchange.com/questions/117195/can-we-tell-the-chemical-properties-of-a-compound-with-its-chemical-structure

It's not like "no way"; no, there is a way, but it is far from absolute.

Surely looking at cubanes or tnt at least one suspects something.  Experiments are nevertheless necessary, also because something that might be explosive enough to be potentially dangerous doesn't necessarily qualify as an explosive in a technical sense (mining, weaponry, demolition).

A compound, a mixture may explode under different conditions.  As an example, for some of them it is heat, other require electric discharge or mecanical impact to decompose.  There are classes of compounds with known examples of explosives (azides, picrates, fulminates) so one may infer new compounds within these classes equally might be explosives, too.
Among modern explosives, increasing the number of nitrogroups or / and increase of nitrogen content over other elements present is a theme, like octanitrocubanes.  See the Klapötke group in Munich / Germany working on 5,5’-azotetrazolate salts, too.

(dihydrazinium 5,5’-azotetrazolate dihydrazinate, source)

In short, things are explosive if they contain lots of nitrogen. For example, C2N14 (1-diazidocarbamoyl 5-azidotetrazole) is highly explosive because of its high nitrogen content. This is because nitrogen gas is very stable, so when compounds release it, they also release a lot of energy. The nitro group is often found in explosives because of this and because it contains oxygen, which can react and oxidize the rest of the molecule.
We can also look at bond strength. For example, in peroxides the O-O bond is weak and can easily break, releasing reactive radicals. Many peroxides, such as ascaridole, are explosive.
Also, there are steric effects, which occur when parts of a molecule are close together. Nitrogen triiodide is explosive because its three iodine atoms are very close together and are under lots of strain.
There are many exceptions to these rules. For example, urea, despite containing nitrogen, is not explosive.

Being explosive is ill defined as context matters
Part of the issue with what is explosive is the context of the situation where the substance explodes. Flour (the sort used for making bread) is not normally thought of as an explosive, but serious explosions in four mills are known and easy to recreate in the lab (finely divided flour dispersed in air is an explosive mixture).
This illustrates that "being explosive" is not an entirely simple idea.
So, even if we stick to the narrower question of whether a single pure compound is explosive, it isn't always entirely clear what we mean.
Some classes of compounds are usually explosive. Azides and fulminates often are. Highly nitrated compounds often are, but then again, circumstances matter. Guncotton, for example, is highly nitrated cellulose. It has been used as a propellant in firearms (where it deflagrates–which is an explosion travelling slower than the speed of sound in the substance). early billiard balls made from it occasionally exploded with little provocation during normal play. But as a loosely aggregated ball of cotton or a thin film of paper, it doesn't explode, it just burns quickly.
The military explosive C-4 clearly is an explosive: armies use it as a demolition charge. But they have also been known to just burn it in the field to heat rations. It will explode but needs a good detonator to trigger that behaviour.
The point it that the situation and context matters for what will explode.
So, while there are many classes of compounds that are likely to be explosive in some contexts, it is a little hard to generalise. Highly nitrated compounds are often explosive but some will explode at the slightest provocation or when there is a full moon (like nitroglycerin); others need the right context (like TNT).
Many highly nitrogen rich compounds are explosive and highly nitrated, nitrogen rich compounds are in a class of their own And Derek Lowe explains in his "Things I Won't Work With" series talking about Hexanitrohexaazaisowurtzitane:

Hexanitro? Say what? I'd call for all the chemists who've ever worked with a hexanitro compound to raise their hands, but that might be assuming too much about the limb-to-chemist ratio. Nitro groups, as even people who've never taken a chemistry class know, can lead to firey booms, and putting six of them on one molecule can only lead to such.

So, while there are classes of compounds that are very likely to be dangerous and very likely to explode given their structure, there is a big grey area where compounds can be explosive in some contexts but not others and some where it is almost inevitable that they will be dangerous.
Thank goodness there are some people who are willing to do the experiments on this rather than you.


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What are the system and surroundings when dissolving sodium nitrate in water?

When sodium nitrate ($\ce{NaNO3}$) dissolves in water at room temperature, the
  solution cools down.  $\Delta H$ is positive because heat flows into the system during the process.

I thought though, that if the water cools down, the water would be the surroundings as heat is being taken from the water in solution to dissolve the $\ce{NaNO3}$ however my professor said:

The system is the calorimeter and its contents, including all of the
  water/ $\ce{NaNO3}$. The surroundings are everything outside of the calorimeter (room etc.)

I'm slightly confused here.
Why does heat flow into the system if water forms part of the system in the first place?

$\Delta H$ represents the amount of heat that would have to flow into the system of water and NaNO3 (from the surroundings) if the final temperature was to be held constant at the initial temperature.  If that amount of heat is not supplied to the system from the surroundings, the system will cool down.

So the solution initially cools down and then there is heat flow from the surroundings  in order for the solution and the surroundings to reach the same temperature again, correct? @ChetMiller

If the solution is insulated from the surroundings, no heat will flow and the solution will cool down and just stay there.  If it is contact with surroundings that are at the initial temperature, heat will be flow from the surroundings to the solution, and, when the final state is reached, the system and surroundings will be in thermodynamic equilibrium and back at the original temperature.

"System" and "surroundings" are labels for regions in space, and do not define the direction of heat flow. In general you define the system to be the part of interest (where a reaction occurs for instance) and the surroundings are everything else.

see for instance https://chemistry.stackexchange.com/questions/117115/regarding-enthalpy-of-a-reaction/117120#117120

This is the basic principle of cooling a system. for example when water cools in an earthen pitcher, the heat is taken from the water, pitcher and the atmosphere to convert the surface molecules into vapors. Yet water is considered as a system there and not surroundings.So in your question,during solvation of sodium nitrate, energy is absorbed even from water to dissociate the sodium nitrate into ions (as solvation energy).

Which one's the system and surroundings again? Please explicitly state that.

Yes please do define the system and surroundings just for clarity!

The system and surroundings are designations to help simplify the universe. There are things to focus on and details to ignore. Those, respectively, are very rough definitions of what the system and surroundings are.
From a chemical standpoint, the system is everything of interest where the chemistry is happening. Dissolution is a process that occurs because a solute and solvent interact. Therefore, it would a mistake to look at the solute or solvent only when examining a dissolution process. Both of those components at least must be in the system.
The surroundings are generally understood to be large enough that they are more or less fixed or unchanging. Nevertheless, if there is heat transferred to or from the surroundings, even though this cannot change the temperature of the surroundings, it will change the entropy.
For calorimetry, you need to limit heat transfer so that you can measure accurately the enthalpy change of the system. A calorimeter helps to do this, but it is now a heat sink and will need to be accounted for in your calculations. Since its temperature is changing significantly, it should not be included in the surroundings, but rather as part of the system.


text:
How many dipeptides are possible by the reaction of glycine and alanine?
How many dipeptides are possible by the reaction of glycine and alanine?
Glycine has no chiral centre. Alanine has one chiral centre.
The products are alanylglycine and glycinylalanine (I not sure whether this is the correct name or not. Please excuse), and I got a total of 4 isomers for each, (using D,L notation and optical isomers) so I got a total of $4+4=8$ products.
But the answer is given as $9$.
Where is my mistake?

The number of isomers are ${9}$ as shown below.

AlaAla has ${4}$ isomers.
AlaGly has ${2}$ isomers.
GlyAla has ${2}$ isomers.
GlyGly has ${1}$ isomer.


text:
Calculating the pH of Strong Acid Solutions
Calculate the $\mathrm{pH}$ of a $\pu{1.5e–11 M}$ solution of $\ce{HCl}.$ How is the $\mathrm{pH}~7?$

Whats wrong with the pH being 7? The pH would actually be slightly lower than 7 but I'm assuming it will round to 7.00 as it is standard to give pH to 2 dp.

What pH would you expect, considering water self ionisation ?

Pure water dissociate according to the equation:
$$\ce{2H2O(l) <=> H3O+(aq) + OH- (aq)}$$
Assume that $K_\mathrm{w} = 1.00 \times 10^{-14}$ at room temperature. Thus, $\ce{[H3O+]}$ of solution is $\pu{1.00E-7 M }$. Now, if you add a trace amount of strong acid, this equilibrium would be disturbed and backward reaction occur to reduced some of added acid, according to the Le Chatelier's principle.
Suppose you added $\pu{1.5E−11 mol}$ of $\ce{HCl}$ to $\pu{1.0 L}$ of pure water (this means you made $\pu{1.5E−11 M}$ $\ce{HCl}$ solution). Since $\ce{HCl}$ strong acid, it would completely dissociate to $\ce{H3O+}$ and $\ce{Cl-}$. As stated above, the dissociation of water is vary depending on $\ce{[H3O+]}$  concentration of the solution.  If $\alpha$ amount of water dissociates due to addition of $\ce{HCl}$, the total concentration of the $\ce{H3O+}$ in the solution will be $1.5\times 10^{−11} + \alpha$  and the concentration of the $\ce{OH-}$ in the solution will be $\alpha$.
$$\therefore \; K_\mathrm{w} = [\ce{OH-}][\ce{H3O+}]= (1.5\times 10^{−11} + \alpha)\alpha = \alpha^2 + 1.5\times 10^{−11}\alpha = 1.00 \times 10^{-14}$$
Using the solution for quadratic equation,
$$\alpha= \frac{-1.5\times10^{-11} \pm \sqrt{(-1.5\times10^{-11})^2+ 4.00\times 10^{-14}}}{2} \approx \frac{-1.5\times10^{-11} \pm 2.00\times 10^{-7}}{2} \\\approx 1.00\times 10^{-7}$$
Thus, the total concentration of the $\ce{H3O+}$ in the solution is $(1.5\times 10^{−11} + 1.00\times 10^{-7}) \approx 1.00\times 10^{-7}$.
Therefore pH of the solution is 7.00.


text:
Why is there a z² label for d orbitals, but no x² and y² labels with corresponding shapes?
I know that the $\mathrm{d}_{z^2}$ orbitals are named as a result from the conversion of the angular component of the hydrogen-like wave function in terms of Cartesian coordinates, but how come we lack analogous orbitals along other axes, such as $\mathrm{d}_{x^2}$ or $\mathrm{d}_{y^2}$? My naive assumption would be due to the fact that $z$ axis is considered the principle axis in regards to chemical bonds, but I don't feel that this is the best answer, nor can I seem to come across anything in textbooks or articles online.

See: Is there a sixth type of d-orbitals with index 2z²-x²-y²? Actually, I'm not sure that it's a duplicate question, but my answer to it should still address your problem.

TL;DR The $\mathrm{d}_{z^2}$ orbital is a result of solving the Schrödinger equation for the hydrogen atom in the most mathematically convenient way.

To properly understand this, it is necessary to go back to the fundamentals. The complex d-orbitals are obtained by solution of the Schrödinger equation. In general, these d-orbitals are made up of a radial part $R(r)$ and an angular part $Y(\theta,\phi)$. The radial part is not relevant to the question, because its value does not depend on the direction in which you look, so it cannot explain why there is a $\mathrm{d}_{z^2}$, but not a $\mathrm{d}_{x^2}$. So, it is instructive to look at the angular component (the spherical harmonics), which are tabulated below. The normalisation factors are ignored here:
$$\begin{align}
\mathrm{d}_{+2} &\propto R(r) \cdot \sin^2\theta \cdot \mathrm{e}^{\mathrm 2i \phi} \\
\mathrm{d}_{+1} &\propto R(r) \cdot (-\sin\theta \cos\theta) \cdot \mathrm{e}^{\mathrm i \phi} \\
\mathrm{d}_0 &\propto R(r) \cdot (3\cos^2\theta - 1) \\
\mathrm{d}_{-1} &\propto R(r) \cdot \sin\theta \cos\theta \cdot \mathrm{e}^{-\mathrm i \phi} \\
\mathrm{d}_{-2} &\propto R(r) \cdot \sin^2\theta \cdot \mathrm{e}^{\mathrm -2i \phi}
\end{align}$$
These are combined in an appropriate manner to generate the real d-orbitals. It turns out that the $\mathrm{d}_{z^2}$ orbital is simply the same as the original $\mathrm{d}_0$ orbital, which is already real (the radial component $R(r)$ is real). Since $\cos\theta = z/r$ and $r^2 = x^2 + y^2 + z^2$ (spherical coordinates), the angular component can be rewritten as
$$\begin{align}
3\cos^2\theta - 1 &= 3\left(\frac{z}{r}\right)^2 - 1 \\
&= \frac{3z^2 - r^2}{r^2} \\
&= \frac{2z^2 - x^2 - y^2}{r^2}
\end{align}$$
and conventionally we denote this as $\mathrm{d}_{z^2}$. There is no analogous $\mathrm{d}_{x^2}$ orbital because the conventional way of deriving real d-orbitals does not yield any orbital with an angular form of $(2x^2 - y^2 - z^2)/r^2$.
But that still does not properly answer the question. To do so, we should look more closely at how the complex d-orbitals are obtained. The Schrödinger equation, $\hat{H}\psi = E\psi$, only prescribes that we need to find eigenstates of $\hat{H}$, but does not say how. It turns out that the easiest way to accomplish this is to look for solutions which are simultaneous eigenstates of $\hat{H}$ (energy), $\hat{L}^2$ (total orbital angular momentum), and $\hat{L}_z$ (projection of orbital angular momentum onto the z-axis). The eigenvalues of the solutions under these operators are directly tied to the quantum numbers $n$, $l$, and $m_l$, respectively.
Fundamentally, the $\mathrm{d}_{0}$ orbital (i.e. the $\mathrm{d}_{z^2}$ orbital) arises because we have chosen it to be an eigenstate of the operator $\hat{L}_z$. If we were to solve the Schrödinger equation in a different way, by searching for simultaneous eigenstates of $\hat{H}$, $\hat{L}^2$, and $\hat{L}_x$ (instead of $\hat{L}_z$), we would indeed get a $\mathrm{d}_{x^2}$ orbital. This is an equally valid solution of the Schrödinger equation (as it is an eigenstate of $\hat{H}$), but it is simply not the conventional solution.
So, why choose $\hat{L}_z$ and not $\hat{L}_x$ when solving the Schrödinger equation? The answer is simply mathematical convenience: the Schrödinger equation is most easily solved in spherical coordinates, and in spherical coordinates, $\hat{L}_z$ has a particularly nice form of being equal to $-\mathrm{i}\hbar(\partial/\partial \phi)$. The eigenstates of this are none other than $\mathrm{e}^{\mathrm{i}m\phi}$, which is exactly what we observe in the complex d-orbitals at the beginning of this post ($m$ is, of course, quantised, so there are only five valid solutions with $l = 2$, i.e. d-orbitals).
But why does $\hat{L}_z$ have a nice form and not $\hat{L}_x$? Ultimately, this mathematical convenience is a result of the definition of spherical coordinates. Because $\theta$ is chosen to be the angle made by the vector with the z-axis, the z-axis has a special status in spherical coordinates:
$$\begin{align}
z &= r\cos\theta \\
x &= r\sin\theta\cos\phi \\
y &= r\sin\theta\sin\phi
\end{align}$$
So, at the end of the day, it is a completely arbitrary convention. And it should be, too, because no axis should have an inherently special status in a spherically symmetric system like the hydrogen atom. As mentioned previously, you could obtain a $\mathrm{d}_{x^2}$ orbital by solving the Schrödinger equation in a different way. You could even choose to redefine the axes in such a way which gave you a $\mathrm{d}_{x^2}$ orbital (it's quite easy; simply rename $(x,y,z)$ to $(y,z,x)$).
But there is no way you can have both at the same time. Because $\hat{L}_x$ and $\hat{L}_z$ do not commute with one another, it is not possible to "solve" the Schrödinger equation in a way which gives you both $\mathrm{d}_{x^2}$ and $\mathrm{d}_{z^2}$ orbitals at the same time.

Does the nomenclature of d-orbitals change by changing the inter-nuclear axis from $z$ to $x$ or $y$?

@Apurvium That depends on two things. Firstly, do you want the d-orbital axis coordinate system to line up with the internuclear coordinate system? (Probably yes, but you don't have to; you can keep on calling the d orbitals whatever you like. After all, it's not as if they are humans which you have to name correctly. They don't care what you call them.) Secondly, recall that the d orbitals are a set of five orthogonal functions, but any linear combination of them is also a valid set of d-orbitals. So do you want to stick with the original linear combination ...

... which means that the dz2 is now no longer pointing along the z-axis, but rather the x-axis, in which case you should probably call it the dx2 orbital? Or (more likely) do you want to reorient the set of d-orbitals you have chosen, i.e. choose a new linear combination, to match the new coordinate system? In this case, the labels remain the same, but the d-orbitals rotate. So the old dz2 orbital no longer exists. By taking a linear combination of the old d-orbitals to reflect the new coordinate system, it has been replaced with a new dz2 orbital, which has a different orientation.

TL;DR (1) If you want the orbitals themselves to move along with the new coordinate system, then the labels don't change, but the orbitals will be rotated. (2a) If you don't want to rotate the orbitals, then you need to decide whether the orbitals are going to be named according to the new coordinate system (in which case the labels do change), or according to the old coordinate system (in which case the labels don't change, but you have to mentally keep track of two different coordinate systems).

@orthocresol Do you know if anyone derived shapes of 5 d orbitals using for example quaternions instead of spherical coordinates so that the result are 5 "identical" shapes, just rotated?

@KamilSzot Not that I know of, no. In a sense, the complex orbitals are related by rotations in that $\hat{L}\pm|\mathrm{d}m\rangle = |\mathrm{d}{m \pm 1}\rangle$ (where $\hat{L}\pm = \hat{L}_x \pm \mathrm{i}\hat{L}_y$); but I don't know if it's possible to take linear combinations of the five orbitals to form five things with the same shape.

@orthocresol I found alternative solutions for d orbitals that are identical: https://sci-hub.se/https://doi.org/10.1021/ed045p45 What's kind of surprising to me that this still creates a special axis, so any atom that has electrons energetic enough to be in d orbitals is never spherically symmetrical. This approach naturally generalizes to f orbitals and further.

It’s just the naming scheme for the hydrogen atom $l=2$ wavefunctions in real form.
When you solve the Schrodinger equation you do it in spherical coordinates and the solution you get out involve complex numbers.
Thanks to the properties of the Schrodinger equation you can transform these 5 complex valued orbitals in spherical coordinates into 5 new real values orbitals, also in spherical coordinates.
They are named because of their shapes.
First, the $d_{z^2}$ is actually short for $d_{2z^2-x^2-y^2}$ and and represents a wavefunction which is proportional to $\frac{3z^2-r^2}{r^2}$. Similarly, $d_{xy}$ is proportional to $\frac{xy}{r^2}$.
We do this because our new real d orbitals are both real and orthogonal, which are not only properties we like but back when quantum mechanics was new these properties made the math a lot easier.

Some citations needed for the last paragraph: (1) there's nothing inherently difficult about using complex d-orbitals. The real d-orbitals are useful for discussing chemical bonding because they have convenient directional properties, i.e. it's easier to describe & visualise their overlap with other orbitals. (2) What does "converting orbitals back into Cartesian form" even mean? And how do you end up with 6 orbitals?

No you don't end up with six orbitals.  You do have six functions $x^2, y^2, z^2, xy, xz, yz$.  But the first three actually are shorthand for  (functions proportional to) $2x^2-y^2-z^2, 2y^2-z^2-x^2, 2z^2-x^2-y^2$ and only two of those are linearly independent, the sum of all three being zero.  So you can make only five orbitals no matter how you slice and dice space into coordinate surfaces.

Maybe you want to elaborate and address the comments that have been brought up. Generally speaking if you have n functions that represent an orthonormal basis spanning a space, then any unitary transformation into another orthonormal set spanning that space will generate n functions.

@OscarLanzi I get it now. I didn't realize that only two of those will actually count. I have made the edit.

It is also worth noting that doing this conversion to purely real orbitals is entirely for convenience in visualizing them, but this does not come without a cost. In fact, the transformation to all real orbitals makes each of the orbitals that had an imaginary component no longer an eigenfunction of $l_z$.


text:
Reduction of ester in presence of ketone group

As there is no ring expansion I'm assuming that a carbocation should not form at ester carbon as it leads to ring expansion. But is there any an exception to it?
Also the hint provided along with this question says "use ethylene glycol". How does ethylene glycol help here?

Actually you just have to reduce the ether. The reason you use ethylene glycol is to block the carbonyl by forming an acetal as the ketone can get reduced too.

okay we can now use lialh4 followed by h+ addition to yield alcohol but how does the acetal revert back to ketone?

Just hydrolyze it again.

@SameerThakur Sorry, my page was not updated when I looked at it. Never mind.

Example 1

Ethylene glycol is often used to make cyclic acetals; its acetals are called ethylene acetals (or ethylene ketals).This interconversion makes acetals attractive as protecting groups to prevent ketones and aldehydes from reacting with strong bases and nucleophiles

Example 2

Reference
Page 861, Organic Chemistry , Eight edition , L.G.Wade

The hint is to form the dioxolane (cyclic acetal) of the ketone with ethylene glycol then reduce the ester (LiAlH4 will do this, the dioxolane will not react) and finally deprotect the ketone. The reduction can also be accomplished by hydrolysing the ester and reducing the resulting acid with borane.


text:
Why isn't the base reformed in base-catalyzed hydrolysis of esters?
In acid-catalyzed hydrolysis, the hydronium ion is clearly reformed; however, in the base-catalyzed hydrolysis, an alcohol is formed as a by-product, but the strong base isn't reformed. Is it a misnomer to call it base "catalyzed"?

The initial products of saponification are a carboxylic acid and alkoxide. These two react to form a carboxylate and an alcohol, the leveling effect. The reaction is stoichiometric in base. In this sense you are correct, it is a misnomer.


text:
Soluble compounds to absorb radio-frequency electromagnetic radiation
Are there any soluble compounds capable of absorbing radio-frequency or long microwave  radio-frequency electromagnetic radiation? I have noted that metals are often used to absorb these signals, but I cannot find any absorption spectra for radio/long microwave wavelengths.

Of course! Salt water is effective, because it is an ionic solution. Submarines and SCUBA divers cannot use radio-frequency (RF) communication over any distance because of that issue. Very high power, very-low-frequency (VLF) land-based transmitters can penetrate to perhaps 20 meters depth to subs, but salt water is essentially opaque to RF and microwaves.


text:
Acid/Base Indicator
I'm not sure how the concept and math work on this problem. Can someone show me how you can get the answer, both conceptually and using math?

You can look at the question this way: when we have a strong acid/strong base titration, at the equivalence point we would end up with a neutral solution (the $\mathrm{pH}$ would be ~$7$).
If we had a weak acid/strong base titration, at the equivalence point we would end up with $\mathrm{pH}$ above 7 (basic solution has $\mathrm{pH}$ between $8-10$).
If we had a weak base/strong acid titration, we would end up with an acidic solution with $\mathrm{pH}$ $4-6$)
Coming back to your question: since it is asked about the methyl red indicator, we are going to need something which at the equivalent point would have acidic solution which means you need weak base/strong acid combinations,only option (C) fulfils this.


text:
What do GGN, AAP, TCN, CAP, CCN, TGQ, etc. mean in DNA analysis?
What do the letters Y, Q, P and N mean in three-letter combinations involving DNA and/or RNA? I know, of course, about the bases A, C, G, T and U but especially in articles (and forensic tv shows) I see, they are connected to three-letter amino acid codes, letter combos like GTN, GGN, etc. What do they mean?

It is not always possible to obtain an absolutely precise result from automated DNA sequencers, mostly due to too much background noise or two very close peaks which the computer has difficulty reading.
An example of a very noisy four-color chromatogram from a sequencer (sanger):

As you would notice, there are several ambiguous spots and the computer algorithm can be confused or mis-call the wrong nucleotide. Often humans are called in to resolve such ambiguities. It's a fairly common practice.
Sometimes the ambiguity goes beyond a solid conclusion, like when two or more peaks have almost the same intensity.
In that case it is better to explicity label those ambiguous nucleotides as either, say A or B, rather than report a possibly incorrect sequence. Thus was born the incompletely specified nucleic acids.
Here's where IUPAC comes in: the single letter code recommendations.

The "N" in GGN, TCN and CCN would mean the third codon was completely ambiguous and could be any one of tbe four bases. Y would mean pyrimidine (C, U or T).
This would satisfy your query with the exception of the letters P and Q, which I assume were wrongly picked up, or the code was for proteins (amino acid) sequence single letter code.


text:
What happens to the carbon in carbon steel when it rusts?
Carbon steel is iron containing some carbon in various forms of iron carbide.
When it rusts (oxidises), the iron becomes part of all kinds of crystalline and amorphous iron oxides and hydroxides, mostly as Fe3+.
Where does the carbon go in this process? Does it also oxidise to carbon dioxide or monoxide and leaves the system as a gas? Does it combine with the iron to form siderite (iron carbonate)? Does it remain as graphite or some other form of elemental carbon? Something else?

@DrMoishePippik not sure. Black things in rust could be magnetite, specular hematite and a bunch of other things that are unrelated to carbon.

@DrMoishePippik but the carbon is part of the iron carbide in the beginning. Do we know that when iron oxidises the carbon is left behind as elemental carbon? This is what my question is about.

apologies - my previous (now deleted) comments referred to cast iron, in which most of the carbon is graphite inclusions in the iron matrix. See answer below.

Cast iron has considerable free carbon as graphite, but carbon steel has iron carbide in cementite, $\ce{Fe3C}$ in a ferrite ($\ce{α-F}$e) matrix, along with other phases and additives. In a corrosive environment, the ferrite and cementite form a galvanic couple, causing increased corrosion of the $\ce{α-F}$e while effectively protecting the $\ce{Fe3C}$. According to Acta Metallurgica Sina,, "So, the ferrite phase diminishes gradually, and the accumulated cementite multiplies on the surface of the specimens along with the development of corrosion."
That said, what now happens to the remaining $\ce{Fe3C}$? Wikipedia states, "While cementite is thermodynamically unstable, eventually being converted to austenite [$\ce{γ-Fe}$] and graphite at higher temperatures, it does not decompose on heating at temperatures below... 723 °C..."
So, at room temperature, the remaining carbon would be specks of cementite amid the rust.


text:
Can air conditioning cause urine smell?
On rainy days there is a strong urin smell on trains in Tokyo.
I suspect this smell comes from ammonia.
On Wikipedia I read:

Ammonia and ammonium salts are also found in small quantities in rainwater […] Ammonium salts are found distributed through fertile soil and in seawater.

Both is valid for Tokyo. In rainy season especially there is a lot of rain. And Tokyo is located at the sea.
Air conditioners work by drying the air.
Tokyo trains make extensive use of air conditioning.
Could this process intensify or make the ammonia to fall out and thus create the strong urine smell?
What other possibility is there for the strong smell?

I'll provide an alternative and obvious hypothesis: in other cities with subway systems I've been in, some people are known to take the liberty of relieving themselves in the subways (and urine is the least offensive smell). This is particularly true when homelessness is rampant, public restrooms are a rare commodity, and/or cold or rainy weather is not uncommon.

You might buttress your hypothesis by determining what airborne concentration of the offensive substance would be required and see if an oceanic source could account for it. You might want to check on geology SE if necessary.

Oh, in addition to people in need for a restroom there are other (perhaps more obvuously) possible sources: animals, including pets (dogs), rats and (maybe?) birds.

Ammonia is in the air, from sources Buck mentioned above, and when the condensed water+ammonia gets warm again in the sewage, it releases the ammonia again. It's trapped.

We can rule out animals or passengers. First this is Japan, there's free toilets on any station. Secondly this smell is general in any car on most lines. Even if there would be people who pee in the train. They'll hardly sprinkle all cars in Tokyo

I can't clearly point to specific chemical but a pungent odor was typical of german ICE wagons. Consider that clean people can smell, if confined in a "close" environment. Moreover, what we refer to smell, depends also on T and humidity. Conditioning (to my sense) often gives rise to a pungent odour, a bit similar to the "smell of (prior of) the snow". https://newengland.com/today/living/new-england-nostalgia/does-snow-have-a-smell/ in my and a friend case, hydrogen peroxide / ozone like.

Ammonia is also used as a refrigerant in soma AC systems, so it’s possible the subway system uses it and very tiny amounts (but enough to smell) are escaping from somewhere.

First, human sweat contains urea, so it doesn't require incontinent commuters. Despite constant cleaning, this effluvia accumulates in the rail cars.
Second, air conditioners precipitate water, entrained dust and volatile organic compounds, making an ideal environment for bacteria, fungi and protists to swim around, excreting their wastes. Microbial volatile organic compounds are none too pleasant, and may have the smell of urine or ammonia (though the waste products of yeast in baking bread can be quite appealing).
Third, there are probably birds in and around the train stations. Birds produce uric acid, which is rapidly converted to ammonia by fecal flora.
Fourth, in older drainage systems, rain water and waste water may be mixed. Given a high water level, wastes may be closer to the surface... or overflow.

Thank you for your answer. The problem I have with it is. That one day prior when it hot and people sweat much more. There is no such smell. It seems somehow related to the rainy weather condition. It changes rapidly from one day to another.

The production of ammonia need water for hydrolysis. Water is more available at higher humidity due hygroscopicity. Additionally, at higher humidity, more water gets adsorbed ( and ammonia dissolved in it ) on aerial dust microparticles, leading to higher content of ammonia on nasal sensors.


text:
I need to identify the peaks in this boron mass spectrum
As part of a bigger project we were given this image. Its a mass spectrum produced in an accelerator with a Boron cathode. It was implanted in diamond. I assume the first peak is just $B^-$ but I have no clue what the rest is. Any help would be appreciated

Given that boron has two isotopes (10 and 11), and your low mass peak looks more like it is at mass 9, you indeed have some issues. Further, why a negative ion (unless you have a tandem accelerator and are just looking at the SNICS source)?

It is indeed a SNICS source

You said the spectrum taken with a boron cathode. It is seemingly, $\ce{^{10}B}$ enriched cathode and I clearly see the presence of $\ce{^{10}B3}$ (very weak), $\ce{^{10}B4}$ (weak), and $\ce{^{10}B5}$ (strongest) peaks. To assign peaks at 38 and 49, I need to know how was the cathode prepared (such as with $\ce{Ag}$ or $\ce{C}$ or etc.).

The supervisors for the experiment did not tell us anything about the preparation, it is a side task for the main experiment, which is implantation of ions in diamond to create nitrogen vacancy centers.

@MathewMahindaratne - But I see no indication of a $B_{2}^{-}$ peak, which shows up clearly in the scans in the Negative Ion Cookbook. But, yes, the OP should clarify what cathode was used.

i will ask tomorrow and come back, thanks for the help everybody!


text:
Occurence of double displacement reaction
$\ce{AgNO3 + NaCl}$ gives $\ce{NaNO3 + AgCl}$ but why it's vice versa is not possible. My question is how to predict if a double displacement reaction will occur naturally or not.

Strictly speaking, vice versa is possible, it's just way far from completion. In general, one remembers which compounds posses good solubility and which don't. If in doubt, have a look a their solubility products. Note, however, that solubility (and reactivity) may change drastically by introducing third-party component such as complexing agent or another solvent.

@andselisk couldn't you possibly write an answer based on this?

On a microscopic level, you still have «a forward» reaction described by
$\ce{AgNO3 + NaCl -> NaNO3 + AgCl}$
and a «backward reaction», which may be described by
$\ce{AgCl + NaNO3 -> AgNO3 + NaCl}$.
However, reaching the thermodynamic equilibrium of these two reactions is heavily influenced by the low solubility product of $\ce{AgCl}$ at about $\sqrt{1.1 \times 10^{-7}}\,\mathrm{mol}$ (or $1.9\,\mathrm{mg}/\mathrm{L}$) in pure water at room temperature (reference).
Consequently, you may either replace the «balanced» double arrow about chemical equilibria
$\ce{AgNO3 + NaCl <=> NaNO3 + AgCl}$
by one tilted in favour to one side
$\ce{AgNO3 + NaCl <=>> NaNO3 + AgCl v}$
where $\ce{v}$ is used to highlight the preciptation of solid $\ce{AgCl}$ shifting the equilibrium.
Or write the reaction equation as if there were no microscopic reversibility at all:
$\ce{AgNO3 + NaCl -> NaNO3 + AgCl v}$

Depending on school and what you want to put emphasis on, the microscopic reversibility often is «omitted» (i.e. the double arrow is dropped in favour of the single arrow) for reactions with a $K > 10^4$ or $K < 10^{-4}$.  On the other hand, you see in many textbooks
$\ce{2 H2O <=> H3O^+ + OH^-}$
despite a $K_w \approx 10^{-14}$, too.


text:
Ionization energy and charge localization: π-bonds/systems and ionization energy
Page 35 of my textbook, Mass Spectrometry by Jürgen H. Gross, says the following in a section on ionization energy (IE) and charge localization:

Molecules with π-bonds have lower IEs than those without, causing the IE of ethene to be lower than that of ethane. Again the IE is reduced further with increasing size of the alkene. Aromatic hydrocarbons can stabalize a single charge even better and expanding π-systems also help making ionization easier.

My questions are as follows:

Why do molecules with π-bonds have lower IEs than those without?
Why does expanding π-systems help making ionization easier?

I would greatly appreciate it if people could please take the time to clarify this.

$\pi$ MOs are higher in energy than $\sigma$ MOs, so they are somewhat 'easier' to remove. Furthermore $\pi$ MOs are more delocalized (spread out in space), thus helping to stabilize a charged species by lowering the overall charge density.

@LoschmidtsSchnitzel Thanks for the comment. Just to clarify, so the answer to number 2 is that expanding $\pi$-systems are further delocalized, thus lowering the overall charge density and making ionization easier?

The delocalization helps to increase the stability of the charged species and makes it thus more likely to actually remain in the ionized state. Think of an allyl carbenium ion vs a secondary carbenium ion

@LoschmidtsSchnitzel So molecules sometimes lose their ionization after becoming ionized? And so a challenge is ionizing them and keeping them ionized? Is this the phenomenon that number 2 is referring to?

Have you heard of flame colouring? If you dip a sodium solution into a flame, the flame will turn bright yellow. The highest electron of sodium is raised to a higher energy level by the thermal input, then relaxes back into the lower orbital - releasing a characteristic amount of energy which happens to be a photon in VIS. (this is the basis of a lot spectroscopy techniques). Its not a textbook style example of ionization, but quite intuitive. Try any introductory organic chemistry book and look at reaction mechanisms involving charged species (SN1, E1 vs SN2/E2)

To get a feel for what chemists mean with charged species - some might disagree but the term 'charged' is hella undefined imo (or at least rather arbitrarily defined), which comes of the mix between classical / QM approach

@LoschmidtsSchnitzel I think this is what you are referring to: In chemistry, a conjugated system is a system of connected p orbitals with delocalized electrons in a molecule, which in general lowers the overall energy of the molecule and increases stability. https://en.wikipedia.org/wiki/Conjugated_system


text:
What was the process that created a plastic-looking material from the jam leftovers?
I was making apricot jam using fresh apricots and gelling sugar. All went well except I burnt it. Some of the solid parts sank to the bottom and stuck there. After putting the good stuff in jars, I then soaked the pot.
About five hours later I went to scrub it and the whole bottom layer came free in a perfect circle. The under side is shiny — looks like plastic. The upper side is matt and almost furry. As it dries out it is curling up and is brittle.
What happened?

Classic sol-gel process. The overheating part dehydrated and became somewhat solid. After you soaked it in water afterwards, the soft and hard parts dried (and shrunk) differently, hence the curling up.

You've created a sucrose Frisbee.