text:
Why is the graph of fluorescence vs. wavelength continuous?
If the emission of a photon from an excited state to the ground state occurs with a precise energy, then why is it that wavelengths that differ from lambda max have values on a fluorescence vs. wavelength graph?

In a low pressure gaseous state atomic transitions would be a very particular wavelengths. However in liquids and solids there are all sorts of vibrational, rotational, and intermolecular attractions that broaden line spectra in "bands", very broad peaks.

There are two main reasons for what you're describing.
The first is the fundamental properties of the molecules involved, as mentioned in the comments. Even for isolated molecules, the transition is not always between the same vibrational states of the excited and ground state. Since the vibrational states are quantized, we would expect a series of very closely spaced emission lines, each representing a different combination of vibrational states.
If the molecules are in a solution or in liquid or solid phase, then the interactions of the emitting molecule with the molecules around it can also cause small changes in the energy of the emission as the electronic structure of the molecule is perturbed.
The general term for these phenomena is "line broadening," and most reference books about spectroscopy will have a discussion of the various mechanisms that lead to it.
The second reason for it is the error in measurements. Most spectrometers cannot filter wavelengths perfectly, so, for example, light of wavelength 590 nm will contribute to a signal at 589.5 and 591.5 nm, less as you get farther away. Sometimes the slit width is an adjustable parameter, but even at the narrowest setting, most instruments will not give extremely high resolution. As a result, a series of closely spaced lines will appear instead as a smooth curve.

There is a third reason involving Heisenberg uncertainty and lifetime of the excited state ...

You make good points about why "line spectra" wouldn't have monochromatic peaks. // In UV-Vis range one thinks of "band spectra" as opposed to line spectra. // In x-ray spectra it is rare to observe "line spectra" broadening due to anything other than instrumental errors.

@Fedoran, I would not add anything about the uncertainty principle as up today its widely accepted interpretation is debatable.

@Feodoran *      typo mistake

@user1420303 it's a fundamental part of quantum mechanics. the Interpretation of it is not the point here. There is debate about interpretation of whole of quantum mechanics, but that does mean it is not applied.

@Feodoran "It's a fundamental part of QM" applies for example under the Copenhagen interpretation, but doesn't under the ensemble interpretation. The time-energy relationship is more complicated as time has not an operator associated. What $Delta t$ means have been treated with little care in many sources, but it is not obvious. Few years ago (about 2012) there was debate. I do not know about the actual state. I remember have read some interesting works by Aharanov.

@user1420303 If I understand you right, then you are saying: The uncertainty principle might actually not exist, just because its interpretation is unclear?

@Feodoran, In such lines, I tried to say something like:The uncertainty principle exists and has very clear formal expressions. However, their implications about nature are under debate. In other words, there is no agreement about what they are describing (for example, many authors denies that the principle has  anything to do with simultaneous measurements). Because of this, I would try to avoid arguing in terms of this principle (except when it consists in applying the inequality to obtain another expression).
In my view, it is used mostly because the popularity of the Copenhagen Interpret.

@user1420303 I just don't see the how the interpretation is relevant here. It exists and it broadens the lines (somewhat), it is therefore relevant to the question. whether that is on a purely empirical basis or has some further implications is not the point here.


text:
How do I identify the compound(s) that formed from a failed synthesis of another compound?
A general answer to the question is great, but specified to my problem is very much welcomed.
My team had a lab project — to synthesize chrome alum crystals.
We first dissolved $0.3 \, \rm g$ of potassium dichromate on $4 \,\rm mL$ of $1 \,\rm M$ sulfuric acid, then added $2 \,\rm mL$ of $99\%$ ethanol and let it rest for a day. There were some crystals formed, but the color was orange (it should've been purple).
From the instruction that we followed, we should heat up the solution, so we did that, and carelessly let the temperature go up past $70 \,^{\circ} \rm C$. It become greenish in color and doesn't form any crystal after a week.
We resorted to evaporating the liquid part and obtained some leftover greenish yellow powder at the bottom which is very much far from the desired chrome alum.

My best bet is you have made chromium(II) acetate instead. Look here and here.

Hold on, did you just add 4 mL of 1 M H2SO4 to the powder of K2Cr2O7 without dissolving the dichromate in water first? In other words, what was the total volume you were trying to crystallize the alum from? Also, have you controlled the temperature when you were adding ethanol (it should've been between 15-35 °C), and have you noticed any smell at that point (was it more like apples or vinegar)?

I added the dichromate to the acid and it dissolved completely. No water. Yes, the temperature was controlled, but I didn't pay attention to the smell as it was done inside the fume hood.

Theoretical part
Chrome alum $\ce{KCr(SO4)2 * 12 H2O}$ is synthesized according to the following reaction (average oxidation numbers are shown above the elements undergoing redox reaction):
$$\ce{K2\overset{+6}{Cr}_2O7 + 3 \overset{-2}{C}_2H5OH + 4 H2SO4 →[15-40~°C] K2SO4 + \overset{+3}{Cr}_2(SO4)3 + 3 \overset{-1}{C}_2H4O + 7 H2O}$$
after joint crystallization of both $\ce{K2SO4}$ and $\ce{Cr2(SO4)3}$ with water forming a complex which is better described as aquacomplex $\ce{[K(H2O)6][Cr(H2O)6](SO4)2}$ with no sulfate(VI) anions in neither of the first coordination spheres.
Note:

It's absolutely crucial to control the temperature during the addition of alcohol to acidified dichromate. If the temperature below 15 °C, the reaction is too slow, but if it's above 40 °C, ligand exchange starts to take place: $\ce{H2O}$ ligands in $\ce{[Cr(H2O)6]^3+}$ are partially substituted by $\ce{SO4^2-}$, which is reflected in color change from blue-violet to green. Also, evaporation of alcohol also happens quicker at elevated temperatures, depleting the reducing agent at higher rate.
Whereas you need stoichiometric amounts of dichromate and sulfuric acid, alcohol is usually taken in excess to ensure complete reduction of $\ce{Cr^6+}$ to $\ce{Cr^3+}$ and to salt out crystals of chrome alum from the solution.
Working with such small quantities (few mL) of pretty abundant and cheap chemicals seems unreasonable to me as it increases the error of measurement, increases the negative impact of impure chemicals or erroneous concentrations and makes the experiment less demonstrative.
I'd advise to work with larger quantities (e.g. 1-2 g of dichromate) in the future.

Experimental results and discussion

There were some crystals formed, but the color was orange …

It's extremely likely that the orange crystals are unreacted potassium dichromate(VI), so it looks like the reduction was incomplete.
That's why I doubt that alcohol was ever oxidated further to acetic acid, hence I see the formation of chromium(II) acetate as suggested in the comments as improbable.
Also, chromium(II) acetate is very unstable and would've decomposed under these conditions.
There is also a rudimentary method of checking the oxidized product by its smell.
If it smells like apples, then it's acetaldehyde $\ce{C2H4O}$; if it smells like vinegar, then the oxidation took further and alcohol has been oxidized to acetic acid $\ce{CH3COOH}$.

… (it should've been purple).

Since you haven't dissolved potassium dichromate in water and just added sulfuric acid to the powder, the concentration of sulfate(VI) was too high.
Plus, you've added all alcohol at once, which might've resulted in temperature spike. Both factors effectively prevent formation of chromium(III) aquacomplex, leaving no chance for alum to crystallize. (see Addendum for the reference)

From the instruction that we followed, we should heat up the solution, so we did that, and carelessly let the temperature go up past 70 °C. It become greenish in color and doesn't form any crystal after a week.

At this stage you pushed the reduction further and probably evaporated both ethanol and acetaldehyde completely, and the solution was consisting of very concentrated chromium(III) sulfate, still no chance for alum to crystallize due to above mentioned reasons. (see Addendum)

We resorted to evaporating the liquid part and obtained some leftover greenish yellow powder at the bottom

Green powder is likely a mixture of potassium chromium(III) sulfate $\ce{KCr(SO4)2}$ and also minor amounts of chromium(III) oxide $\ce{Cr2O3}$ formed along the way.
Now yellow powder is with a very high confidence just potassium chromate(VI) $\ce{K2CrO4}$ leftovers from the dichromate(VI) once the acid was used up:
$$\ce{Cr2O7^2- + H2O <=> 2 HCrO4^2- <=> 2 CrO4^2- + 2 H+}$$
One might also claim that it's elemental sulfur formed by reduction of sulfate(VI) after chromium(VI) was completely converted to chromium(III), but I find it rather implausible.
Addendum
From Ullmann’s Encyclopedia of Industrial Chemistry [1, p. 4441]

Potassium chromium(III) sulfate [10141-00-1], [10279-63-7], [7788-99-0],
  potassium chrome alum, $\ce{KCr(SO4)2 · 12 H2O}$, $M_\mathrm{r}~499.11$, $\pu{1.813 g/cm3}$, crystallizes in the cubic system forming violet regular octahedra
  which decay in air and melt at 89 °C, the color changing to green; the enthalpy of formation is − 5788 kJ/mol. The solubility in water at 25 °C is 11.1 wt %. The solution is violet when cold, but becomes green above 50 °C, this change being accompanied by a decrease in the molar conductance. For 0.125 M solution at 50 °C, the molar conductance is $\pu{221 Ω−1 cm2 mol−1}$ for the violet form and $\pu{202 Ω−1 cm2 mol−1}$ for the green form. This change is reversible and its rate is increased by acids. The green form always occurs as an amorphous solid and crystals
  are unknown.
[…]
For the preparation of potassium chromium alum, a saturated potassium dichromate solution is reduced with sulfur dioxide in the presence of sulfuric acid.
  During the reaction, the temperature must be kept below 40 °C by cooling to prevent the green modification being produced. Apart from sulfur dioxide, such organic compounds as formaldehyde, methanol, or starch are suitable as reducing agents. Crystallization starts after sulfuric acid has been added and the temperature is kept
  further below 40 °C. If the solution is allowed to settle in vats, large crystals are produced, whereas ﬁne ones result if the solution is stirred.

References

Ullmann’s Encyclopedia of Industrial Chemistry, 40 Volume Set, 7th ed.; Wiley-VCH, Ed.; Wiley-VCH: Weinheim, 2011. ISBN 978-3-527-32943-4.

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text:
Thickening and/or slowing evaporation of IPA-based ink
The conundrum
I have a security printing ink that I make by adding an IR-absorbing dye to a 50/50 mix of Isopropyl Alcohol and distilled water. The ink solution is pale green but dries virtually invisible to the human eye.
The problem is that when the ink is applied to the stamp the ink evaporates within seconds but I need it to remain viable for ~10 minutes. Even if I shortcut the 10 minutes to just a few seconds, the ink evaporates enough to cause an uneven distribution (i.e., it is faded in some areas.)
Requirements

Provides an even distribution of ink (to me, this means viscous)
As little bleeding as possible (to me, thin coat on stamp, also viscous)
Stays viable on a stamp for ~10 minutes
Allows the final inked product to be virtually invisible after ~24 hours

A solution that works (almost)
I have solved this problem before with a different ink. It was a UV-reactive ink which uses an identical IPA/water solvent. I was able to meet all of my requirements by adding 14% Glycerin to the solution. The increased viscosity really helped with consistent and clean (no bleed) marks. The slower evaporation gave me plenty of time to stamp. There was much rejoicing.
However, the Glycerin has an unwanted side effect with the IR ink in that it remains visible on the substrate. I'm guessing this is because the Glycerin prevents the ink from fully drying. I tried heating final product to see if that dried out the Glycerin but it didn't help.
You may ask why the UV ink ended up being invisible (requirement #4). Simple - that ink is clear to begin with.
My questions

I think what I need is a viscous fluid (maybe not as viscous as Glycerin) that evaporates slowly but fully in ~24 hours. Does such a wonder exist?
Glycerine solves all of my problems and creates one (visible marks.) Is there a way to use Glycerin, while still getting an invisible mark? (I've tried varying the amount of glycerin, but it's a trade-off of usability and visibility.)
Is there another chemical (hopefully easily obtainable by an individual) that could work as well as Glycerine while also allowing invisible marks?
I would love to understand this problem better so I can troubleshoot it myself (I'm a software engineer, not a chemist.) Any random insights would be welcome.
It seems to me that viscosity is inversely proportional to evaporation rate. Is that true?

Background info
I am trying to apply this ink to long thin strips (~7mm each) of a porous paper substrate (wrapped around plastic.) This needs to be done hundreds of times. To simplify the alignment process, I have a large rubber stamp with a series of ~100 duplicate strips embossed into it. This stamp is placed face up into a rig with alignment grooves on either side. From here, I apply the ink to the stamp and then slide each strip, one by one, into a pair of adjacent grooves and press down to stamp. This is much faster than re-inking a single thin stamp for every strip. This process takes about 10-15 minutes.
UPDATE
When I originally wrote this post, I was unable to openly discuss the underlying application, which hindered my ability to describe the problem clearly. I have since open-sourced the application, which is a technology for card magic that uses a small computer to scan bar codes marked on the edges of cards.
If this stimulates your interest in answering this question, you will find the GitHub project here. In the documentation, I explain the process of marking the edges of cards using a stamp and a variety of inks.
The section on IR Absorbing Inks covers the challenges with working with this ink. Specifically, this section outlines these challenges and details the extensive process that I use to mitigate them.
As describe in the original question, I was able to mitigate these same issues for UV reactive inks by using Glycerine, which is outlined here.
The intent of this question was to find a way to simplify the mitigation process of working with the IR absorbing inks, similar to working with the UV reactive ink.

(1) Am I correct to assume that you do not want to change your inking stamp setup which you already have designed? (2) To process a batch of strips takes 10-15 minutes, so am I correct to assume that most of that time is fiddling with loading the paper strips? (3) How long between inking the stamp and stamping the strips? (ie can you load strips, then ink stamp, then stamp to reduce cycle time between inking and stamping?)

(1) Before arriving at this solution, I tried many things, including hacked up inkjet printer. I'd prefer to keep using it if possible. (2) Time is relative to the number of items. Just a few seconds each: pick one up, line it up in the proper adjacent grooves, stamp evenly, set it aside, grab another. (3) Once I ink the stamp, I start to stamp immediately. First one lands on the stamp ~4 seconds after the stamp is fully inked. The problem is, by the time I've finished inking the entire stamp, the first part I inked is nearly dry with pure IPA because it's a thin coat of ink to avoid bleeding.

A little late, but: Unfortunately glycerin does have a pretty high boiling point. However, perhaps you could apply a coat of glycerin w/o dye to the strips, allow that to dry, and then apply the +IR dye stamp. Perhaps that will remove any contrast against the background.

This is more of comment than a solution.
It has often been noted here that chemistry is an experimental science so often you just need to try things.
There are a number of considerations here. It seems like it takes on the order of 10 minutes to stamp a batch of "strips" (whatever that is...) and that:

You need a stamp with multiple "stamp faces" to stay wet that long.
You also want the ink solvent to "dry" so that the ink solvent doesn't leave a noticeable "watermark" on the paper.
You want to use a minimal amount of solvent so that the dye stamp doesn't bleed into the paper.

(1) The dye must be fairly polar to dissolve in a solution of 50/50 mix of isopropyl Alcohol (2-propanol)  and distilled water. You can look at a table of solvents with their polarities listed at this webpage. You can try other solvents that you have, or mixtures that you can make, to get a rough idea of how polar the solvent (mixture?) must be.

Reasonable to assume about 1 gram/ml for any of the solvents listed. Estimate polarity of a mixture by assuming ideal solution, ie that total solvent polarity is equal to molarity of a solvent times its mole fraction. (ie 50 wt % is not the same as 50 mole %.)
You can check to see if the ink dissolved by seeing if the solution is "clear" - in other words shine a bright light through it. If you see scattering then the dye didn't dissolve.
You can probably estimate from your work with glycerine how concentrated the dye must be to be invisible when dried, but visible under fluorescence. (I'm guess that the dry dye itself is green. So you don't want to "paint" the surface with particles of the dye. Rather you want the dry to be carried into the fibers of the paper as the solvent carrying the dye is adsorbed.)

(2) Look at other solvents that have been used for inks, for example on this webpage. Choose some solvents to test which have a polarity that works from step (1). Now look at solvents vs vapor pressure. The lower the vapor pressure the slower the solvent will be to evaporate.

Note that a mixture won't evaporate homogeneously. So with an acetone/glycerin mixture the acetone evaporates preferentially to the glycerin.
There is also an effect here in that the ink solvent is adsorbed into the paper not just sitting on the surface of the paper like a drop of such solvent on glass. So the correlation between vapor pressures and drying time would be "good" but certainly not perfect.

I mix it in the IPA first (91%) then add water. It takes ~1.5 hours until the particles are no longer visible in the IPA. It does not dissolve in water. (I discovered the use of a flashlight almost right away.) I have tried many things (my kitchen table is full of chemicals) including Hypromellose, xanthan gum, and Acrylates C10-30 alkyl acrylate cross polymer. .

This is good info for things to look into and learn from. Thank you. Side question: the flashlight works for solvency, but how do you know if two miscible liquids are fully mixed?

If there are two phases (two liquids that don't mix) you should be able to see a line between the too. Think of mixing a salad dressing with oil and water. The oil layer floats on top.

OK, just wanted to make sure. I have been waiting for the dye to dissolve, but just shaking the bottle for a few seconds for liquids until the solution looked homogenous.

You said, "The higher the vapor pressure the slower the solvent will be to evaporate." The murov table seems to indicate the opposite. Am I missing some basic understanding? My understanding: A substance with a low vapor pressure will have less volatility, reaching equilibrium at lower pressures (due to less volatility pushing against the pressure.) This adds up to lower evaporation rates.

MaxW Now that I've had a chance to pour through your reply, I wanted to add another thank you. Your post has sent me down some really cool educational rabbit holes. And that table of solvents (murov) is priceless! I've been looking for exactly that for weeks, but didn't know the terms (ex: vapor pressure is a new concept to me.)

LOL - You comment made me look at what I wrote. Of course the lower the vapor pressure of the liquid, then the  slower that liquid will evaporate.

The chemical manufacturer recommends ethanol for a solvent (relative polarity of 0.654) so I'm guessing the dye will also dissolve in benzyl alcohol (polarity of 0.608). This would be good, considering it has a higher viscosity than water (about 6x) and much lower vapor pressure (0.094).
Now if only there were a place locally I could pick some up so I wouldn't have to wait for a shipment.

The vapor pressure of IPA at 2.4 degrees C is 1/4 that of its room temperature value (ref).
The viscosity of IPA also increases dramatically as temperature cools to ~0 (ref).
Could you simply conduct the evaporation with the same solution but in a colder environment? A normal household fridge should be in the right range for what you're looking for.
The viscosity will also likely increase as evaporation happens since the concentration of the dye will increase, so you may be able to start the evaporation process in the fridge and then let warm to room temperature (or higher if needed) to fully drive off the solvent within your time constraints.
A possible process could be to pre-cool your solution to below 0 degrees Celcius, conduct the stamping at room temperature (within your 10 minute window), place the stamped substrates in the fridge to allow for slow evaporation over 8-18 hours, then warm to room temperature to drive off remaining solvent.

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text:
Deposition of Nb on SiO2 wafer (Z-Cut)
I am trying to sputter $\ce{Nb}$ on $\ce{SiO2}$ Z-Cut wafers (quartz crystal).  I was wondering if $\ce{Nb}$ will stick well to the surface of the wafer.
I've routinely deposited $\ce{Nb}$ on $\ce{Si}$ wafers but have never done so on quartz crystal wafers.

Worried about Nb "sticking" to SiO2 or "reacting" with SiO2?


text:
ICE table with changing volume without Kc
Just having trouble understanding how to approach an ICE table question with no Kc and a volume change
Given

If you had 4.4 mol of a solid, A, placed in a 1L container which decomposed in to B and C (gas), B would slowly increase until 1.20M
reaction: $\ce{A(s) <=> B(g) + C(g)}$

Question

If the volume doubled and equilibrium was re-established, how many mol of A would there be?

My attempt:
Kc = 1.2M + 1.2M (not including solids)
Kc = 2.4M
for new ice table
2.4 =  ((1.2 + x)(1.2 + x)) / 2 (because now litres is doubled)
0 = x^2 + 2.4x - 3.36
x = 0.9908 (which is wrong apparently)

Welcome to StackExchange Chemistry. Do you know what the equilibrium constant expression is for this reaction. You could use the first equilibrium to calculate Kc and then apply it to the second equilibrium. Could you edit your question to include how far you got, and where specifically you got stuck?

David, were you given the reaction or did you propose it?

This reaction was given to me

How you calculated the Kc is wrong. The Kc is the equilibrium constant using concentrations. // Frankly using an ICE table is overkill and confusing as far as I am concerned.

Why are you adding values to get $K_{\mathrm{c}}$? I'm hard pressed to come up with even a contrived situation where that would be the correct approach.

For the numerical result, see the answer by MaxW.
Problems with your attempt

Kc = 1.2M + 1.2M (not including solids)

The concentrations of B and C should be multiplied, not added. So Kc is 1.2 squared.

2.4 = ((1.2 + x)(1.2 + x)) / 2 (because now litres is doubled)

Your ICE table is about amounts, and the equilibrium concentration is about concentrations. The equation should read:
$$K_c = \frac{1.2 + x}{2} \times \frac{1.2 + x}{2}$$
Why you don't need an ICE table
This problem is similar to an ionic solid dissolving:
$$\ce{NaCl(s) -> Na+(aq) + Cl-(aq)}$$
This reaches equilibrium when the solution is saturated. If you add more water (dilute it), more will dissolve until the solution is saturated again or you run out of solid. If you don't run out of solid, the concentration before dilution and after dilution will be the same (that concentration, based on the amount of NaCl dissolved, is called solubility, and the equilibrium constant for the process is called solubility product).

You should be able to do this in your head...
You start with 4.4 moles of A. A decomposes into B and C which are both 1.20 molar in a 1 liter container, so there are 1.2 moles of B and 1.2 moles of C in the gas phase. Since both B and C are 1:1 with loss of A, there are 4.4 - 1.2 moles = 3.2 moles of A left.
If the container volume is doubled the concentrations of B and C stay the same when the new equilibrium is established, 1.2 molar. However the volume doubles to 2 liters so there are now 2.4 moles of B and C in the gas phase. Thus there are 4.4 - 2.4 = 2.0 moles of A left.

Let’s look at this at different approach. Suppose your initial volume is $V$.
$$
\begin{array}{ccc}
\text{Reaction:} & \ce{A (s) & <=> & B (g) &  +  & C (g)}\\ \hline
\text{initial amount:} & \pu{4.4 mol} &  & \pu{0 mol} &   & \pu{0 mol}  \\
 \text{amount at equilibrium:} & \text{some solid} & & \pu{1.20 mol} &  & \pu{1.20 mol} \\
\text{equilibrium concentration:} & \text{some solid} &  & \pu{\frac{1.20}{V} mol\: L^{-1}} &   & \pu{\frac{1.20}{V} mol\: L^{-1}} \\
\text{amount at equilibrium at 2V:} & \text{some solid} & & \pu{x mol} &  & \pu{x mol} \\ \text{equilibrium concentration at 2V:} & \text{some solid} &  & \pu{\frac{x}{2V} mol\: L^{-1}} &   & \pu{\frac{x}{2V} mol\: L^{-1}} \\ \hline
\end{array}
$$
Thus, $K_\mathrm{C}$ when volume is $V$ can be derived as:
$$K_\mathrm{C} = \pu{\frac{1.20}{V} mol\: L^{-1}} \times \pu{\frac{1.20}{V} mol\: L^{-1}} = \pu{\left(\frac{1.20}{V}\right)^2 mol^2 L^{-2}} $$
Similarly, because it is the same reaction, $K_\mathrm{C}$ can also be derived when volume is $2V$ as:
$$K_\mathrm{C} = \pu{\frac{x}{2V} mol\: L^{-1}} \times \pu{\frac{x}{2V} mol\: L^{-1}} = \pu{\left(\frac{x}{2V}\right)^2 mol^2 L^{-2}} $$
As a result,
$$K_\mathrm{C} = \pu{\left(\frac{1.20}{V}\right)^2 mol^2 L^{-2}} = \pu{\left(\frac{x}{2V}\right)^2 mol^2 L^{-2}} $$
Solving this equation, you may calculate values for $x$:
$$ x = \pu{1.20 \times 2  mol}= \pu{2.40 mol}$$


text:
Reaction of ketones with sodium amide

Find the product of the reaction of dimethyl ketone i.e. $\ce{CH3C(O)CH3}$ with $\ce{NaNH2}.$

Should I do nucleophilic addition and form the product in the picture?

I'm not sure what you are thinking would be substituted. There are no good leaving groups on dimethyl ketone. Given that sodium amide is a strong base, you should ask yourself how dimethyl ketone can act as an acid, and what reactions could occur from there. I'll give you a hint: there should be a nucleophilic addition step.

Sodamide is a strong base and so I  believe it will deprotonate any organic compound and form a sodium salt of that organic compound.

try thinking on 1) $\ce{2*CH3COCH3}$ in presence of soda amide and 2) ofcourse what @ringo said is also true.

https://www.chegg.com/homework-help/questions-and-answers/acetone-ch3-coch3-undergoes-following-acid-base-reaction-sodium-amide-nanh2-please-note-pk-q9919579

Addition is irrelevant here.

@ayush not that nucleophilic addition step. Another base will be formed before any addition occurs

It's just aldol condensation.


text:
Meaning of polarization pulse in microwave (rotational) spectroscopy
In the recent rotational spectroscopy literature, "microwave polarization pulse" is often used in conjuction with pulsed Fourier transform microwave spectroscopy (Google Scholar).
Sometimes the pulse angle is also mentioned such as $π/2$. I gather this term is similar to FT-NMR where the net magnetization vector is tipped at a certain angle. When we talk about microwave rotational spectroscopy, what is being polarized?
In other words, why do we call the microwave pulse as a polarizing pulse? I wanted to confirm that this is related to the net electric dipole moment vector of the molecules which is being oriented after excitation?

In NMR the sum of the magnetic dipole vectors is called the magnetisation, and this is induced in the sample by the large fixed external magnetic field acting on the nuclear spins. As you mention it is influenced, i.e. moved about,  by the magnetic field of the rf pulses applied to the sample and so ultimately cause the nmr signal. In optics the similar thing  can occur where an intense optical field can align the electric dipoles and cause a polarisation, i.e. macroscopic dipole in the sample, this is the optical equivalent of magnetisation. The experiment is far more difficult to do in optics, however, as the dephasing time (T2 time in nmr) is several orders of magnitude faster than in nmr as is the lifetime of any level optically excited.
The field is generally called optical resonance spectroscopy and/or coherent transient spectroscopy with effects such as transient nutation, photon echos and self-induced transparency.

Thanks. The technique I as specifically referring to is called FT-MRR, FT-Molecular Rotational Resonance Spectroscopy. You give a short chirped pulse and then monitor microwave emission by FID. I guess that also explains the word resonance. Do you recall a reference which talks about macroscopic dipole formation with reference to microwaves pulses? I could not find a definition per se that polarization = macroscopic dipole formation by an intense electric field?

I'm not familiar with the particular technique you mention, but have a look at J. Steinfeld ' Molecules & Radiation' and W. Demtroder 'Laser Spectroscopy' for the background to all these types of experiment which is basically the same for all types of transitions. Conventional spectroscopy books do not generally cover the coherent optical processes you need to understand.

Thanks. I think the ideas must be similar to FT-NMR.

Yes, thats right


text:
What is the mechanism of the reaction of a ketone with sodium nitrite/ hydrogen chloride?
What is the mechanism of reaction of a ketone with $\ce{NaNO2/HCl}$?
Can $\ce{NaNO2/HCl}$ be considered as freshly prepared $\ce{HONO}$ here? I am confused about how the reaction will proceed.

Consider also production of nitrosyl chloride

I know how Sodium nitrite reacts with amines. I was just curious about how it would react with a carbonyl group.

$\ce{NaNO2/HCl}$ is as you suggest a source of $\ce{HONO}$. This will be protonated by $\ce{HCl}$ to give the nitrosonium ion $\ce{NO+}$ (in equilibrium with nitrosyl chloride). This strongly electrophilic species will react with the enol of the ketone to nitrosate the alpha position. This nitroso compound then rearranges to give the alpha oxime.

Image from: Emilia Iglesias and Isabel Brandariz, A further study of acetylacetone nitrosation. Org. Biomol. Chem. 2013, 11, 1059-1064. DOI: 10.1039/C2OB26073E; available as pdf from pdfs.semanticscholar.org.

Slightly weird diagram ... two methyl groups disappeared

I noticed that, think they were meant to be hydrogens

See also: https://chemistry.stackexchange.com/questions/86061/what-is-the-mechanism-of-the-reaction-between-hexane-2-5-dione-and-nitrous-acid/86090#86090

@Waylander can this same reaction take place with aldehydes too ?

Multiple literature sources state that this reaction works for carbonyl compounds, but I can find no examples with aldehydes.


text:
Synthetic ruby - damaged by heat less than its melting point?
A bit of a jewellery question but chemistry related. I want to encase a synthetic ruby (corundum, Al2O3) in silver by melting the silver around it. Either in a furnace or by casting. The MP of silver is 960 °C.
The ruby will be facteded and polished. Now, in theory the ruby should be unharmed given that its MP is double that of the silver and is much harder. However, would there be any damage to its optical properties?

There could be risk of fracture or minor defects due fast and uneven temperature changes, related to the material thermal dilation.

However, would there be any damage to its optical properties?

It is almost guaranteed. Corundum is incredibly sensitive to thermal shock and unless you can do this in a furnace that heats very slowly, it will fracture/crack.
As far as any other properties, your ruby should not see any change in other optical properties for such a short duration, but for longer durations (hours at high temperature) metals can diffuse into the ruby and cause color changes, but I doubt this is an issue for what you describe.

So the only method would be slow heating and cooling. I shall have to try it and see what happens. Experiment rules...


text:
Resonating structures of (pentan‐2‐ylidene)oxidanium

Draw resonating strictures of the following compound:

In this molecule there are two π-components, but they are not conjugated.
According to me, as there is no conjugation, it should not have a resonating structure.
But it has. Why?

@SoumikDas It does have two mesomeric structures not tautomers. In case of cations structures with broken bonds are important contributors.

Resonance is not limited to pi systems; you probably have heard of hyperconjugation, too. The structure has at least two more significant contributing structures. Draw the structures placing the charge at the adjacent atoms. || I personally do not agree with the close votes on this question; I find it clear enough, providing enough context of the thought process, and not trivial.


text:
Find the increasing order of formation of ortho product

Find the increasing order of ortho product formation when Y is (i) F, (ii) Cl, (iii) Br, (iv) I.

We know that halides are ortho, para directing for EAS reactions and we consider the electronic stabilization effect as the reason for it. According to me, the order should be 1 > 2 > 3 > 4 because fluorine will give better electronic stabilization due to P(pi) – P(pi) bonding, and steric effect of iodine will reduce the amount of ortho product
But the answer given is 4 > 3 > 2 > 1.


text:
Oxidation reaction of dimethylselenide / Selenourea with hydrogen peroxide
For my paper,  I need to find the reaction between $\ce{(CH3)2Se2}$ and $\ce{H2O2}$ and between $\ce{SeC(NH2)2}$ and $\ce{H2O2}$. The mechanism below shows how I think it goes. The reaction should give an anion, an inorganic form of the selenide or selenourea but I'm unsure when the forms become anionic.

And
$$\ce{\overset{+IV}{Se}O3^2- + H2O2 -> \overset{+VI}{Se}O4^2- + H2O}$$
If anyone could give me a push in the right direction or can confirm if the reactions in the picture are correct, I'd appreciate it very much.
Thanks in advance!

I) I see no reason for Se-O-Se group formation II) I would be weird for H2O2 to cause SET, so unless there's catalyst, I don't see something like this happening to thiourea.

For the sulfur analogue search thiolsulfinate. I see no reason why selenium should not react in a similar manner.

I'm not familiar with selenium chemistry but am trying to help with the oxidation reaction of selenourea with hydrogen peroxide part. Here what I found on that reaction in literature (Ref.1):

Abstract: Reactions of biological oxidizing agents, such as hydroxyl radicals ($\ce{^•OH}$), singlet oxygen ($\ce{^1O2}$), hydrogen peroxide ($\ce{H2O2}$), and peroxynitrite ($\ce{ONOO-}$) with selenourea were studied. The kinetics of the reactions was followed using time-resolved techniques, and the bimolecular rate constants were determined. In all these reactions, under aerated conditions, elemental red selenium was produced as one of the reaction products. The average size of the selenium particles could be controlled and stabilized in the range of $\pu{20−100 nm}$ with the addition of bovine serum albumin (BSA) or sodium dodecyl sulfate (SDS). The particles were characterized by dynamic light scattering studies (DLS), which revealed that the size and distribution of the particles depended mainly on the amount of selenourea undergoing oxidation. Other factors such as the nature of the oxidant and the concentration of the stabilizer also are important in stabilizing the particles. Nanoselenium-reduced $\ce{ABTS^{•-}}$ to colorless $\ce{ABTS^2-}$ ($\ce{ABTS} =$ 2,2‘-azinobis(3-ethylbenzothiazoline-6-sulfonate) and oxidized dichlorodihydrofluoresecein diacetate (DCFA) to fluorescent dichlorofluorescein (DCF) indicating its ability to participate in redox and free radical reactions. The reactivity of selenium nanoparticles with these systems varied linearly with the surface area of the particles. The studies demonstrate that selenourea undergoes oxidation with both one-electron and two-electron oxidants to produce elemental selenium, which, on stabilization to nanometer size, exhibits size-dependent redox activity.

The reaction mechanism is depicted in following Scheme, which is directly from Ref.1:

Hope this help.
References:

B. Mishra, P. A. Hassan, K. I. Priyadarsini, Hari Mohan, “Reactions of Biological Oxidants with Selenourea:  Formation of Redox Active Nanoselenium,” J. Phys. Chem. B 2005, 109(26), 12718–12723 (DOI: 10.1021/jp051328n).


text:
What kind of system is required for dynamic equilibrium to be set up?
Most websites and books mention a closed system, but technically isn't an isolated system required? If energy can be transferred, then the position of of the equilibrium will continuously change and no equilibrium will be set up, right?

Who cares about closedness? An equilibrium is possible in any system.

Do you know what dynamic means? You're talking about static not dynamic situation.

@Mithoron he used the term correctly.

If energy can be transferred, it does not mean it is transferred.
If there is e.g. zero net thermal energy transfer, a system is in thermal equilibrium with the neighborhood, even if it is not isolated.
As Ivan says, an equilibrium is possible even in an open system, if net transfer of particular matter kinds is zero.
The thermodynamic equilibrium consists of thermal, radiative, chemical and mechanical equilibrium.
Note that thermodynamic equilibrium happens at following conditions:

For a completely isolated system, the entropy $S$ is maximum at thermodynamic equilibrium.
For a system with controlled constant temperature and volume, the Helmholtz free energy $A$ is minimum at thermodynamic equilibrium.
For a system with controlled constant temperature and pressure, the Gibbs free energy $G$ is minimum at thermodynamic equilibrium.

$$\begin{align}
G&=H-TS \\
\Delta G&=\Delta H-T\Delta S \\
\end{align}$$
Total entropy of the not isolated system and it's surrounding increases in time.
Entropy of the system alone may decrease with time, if the Gibbs energy of system decreases, as this latter decrease also means
$$\Delta S_{\rm surrounding} +  \Delta S_{\rm system}  \gt 0$$

What I found after discussing this with my friend is that Gibbs free energy = Enthalpy change - (Temperature * Entropy change) has to equal to 0 for equilibrium to be established. In a closed system this is possible, provided your temperature is kept constant (as entropy increases with time). Is that what your answer is saying as well or is it different?

Gibbs free energy change = ..  G=H-TS DeltaG=DeltaH-T.DeltaS

See the update.

If energy can be transferred, then the position of of the equilibrium will continuously change and no equilibrium will be set up, right?

When a system is at equilibrium, its state does not change. So there can't be a net transfer of energy into the system or out of the system, that would change its state. There also can't be a net transfer of matter into the system or out of the system, that would also change its state.
An open system allows energy and matter exchange with the surrounding. However, once the system and the surrounding together have reached equilibrium, the net transfer of matter will be zero (and the net transfer of radiation), and there will be no heat transferred between system and surrounding, and not work done by the system on the surrounding or the other way around.
So it is not that you have to prevent energy transfer, but that the processes at equilibrium will not show a net transfer of energy even though it is possible and probably happened as the processes approached equilibrium.

What kind of system is required for dynamic equilibrium to be set up?

There are no requirements, any kind of system is fine, including the universe (we're not at equilibrium yet, though).

I don't know German, but my perception is that German readily forms new words by making compound words out of what would phrases in English. So in English adjectives get thrown around. You have to learn the lingo.
A closed system is one in which matter isn't being exchanged with the outside environment, but energy can be.
An isolated system is one in which neither matter or energy is being exchanged with the outside environment.
An open system allows both the exchange of matter and energy.

A closed system is required for the dynamic equilibrium to set up.


text:
How should we express concentrations and partial pressures in ΔG=-RTlnK?
I understand that $K$ will be dimensionless. Let's say an equilibrium involving $K_p$ and the reaction is $\ce{A -> B}$, $K_p$ will be dimensionless. If the reaction $\ce{A + B -> C}$, then based on the formula of $K_p$ will be:
$$ \frac{(\text{partial pressure of C})}{[(\text{partial pressure B})\times(\text{partial pressure of A})]}$$
Since partial pressure can be in $\mathrm{Pa}$, or $\mathrm{bar/atm}$, with the difference between them will be roughly 100000 times. So the $K_p$s will have 100000 difference. Putting the $K_p$ into $\Delta G=-RT\ln K$, there will be a huge difference of $\Delta G$. So what units of $K_p$ should we use in this case?

Related: What is the correct unit for the equilibrium constant? In short: In $\Delta G = -RT\ln K$ the standard equilibrium constant is used, which is not $K_p$.

An argument of logarithmic function must be unitless by principle.

What units of K should we use in ΔG = -RT lnK?

The full formula is:
$$\Delta_r G^\circ = - R T \ln K$$
The circle after the G refers to a standard state, where reactants and products are present at standard concentrations/pressures. The equilibrium constant has to be given with reference to the same standard state. The subscript r stands for reaction. The balanced chemical reaction has to match between the Gibbs energy of reaction and the equilibrium constant expression. Let's say our balanced reaction is:
$$\ce{ \nu_A A + \nu_B B + \nu_C C + ... <=> \nu_Y Y + \nu_Z Z + ...}$$
Here, the different $\nu$ are the stoichiometric coefficients.
The correct equilibrium expression for this reaction uses activities of species raised to their respective stoichiometric coefficient. Activities are dimensionless, and are equal to one at standard state. For dilute solutions, these can be estimated as the ratio of a concentration (or partial pressure) divided by the standard concentration (or partial pressure):
$$ a_\text{species} \approx \frac{c_\text{species}}{c^\circ_\text{species}} = \frac{p_\text{species}}{p^\circ_\text{species}}$$
As a consequence, the equilibrium constant expression will always be dimensionless, and will not depend on the concentration measure used, as long as the standard state concentration is expressed using the same measure.

Since partial pressure can be in Pa, or bar/atm, with the difference between them will be roughly 100000 times.

No, because you are dividing by the standard state, using the same concentration measure (and for ease of cancelling units, in the same units). For example, let's say the standard state of a species in the gas phase is defined as one molar. At ambient conditions, this would correspond to a partial pressure of about 0.04 atm. If the equilibrium concentration is - say - 2 M, that would correspond to a partial pressure of about 0.08 atm. Either measure of concentration gives the same result: The species is present at twice the concentration (partial pressure) as the standard state. No matter what concentration measure I choose, I get the same value for the equilibrium constant.
The same goes for switching pressure units. As you are always dividing by the same standard pressure, you always get the same result (either by converting the units of the partial pressure, or the units of the standard partial pressure, and then cancelling).
Switching to a different standard state
In biology and biochemistry, a different standard state is used. This affects both the values of the standard Gibbs energy of reaction, and the values of equilibrium constants. However, these changes are self-consistent, i.e. you still can calculate K from the Gibbs energy and vice versa, and you can convert easily from one convention of the standard state to the other.


text:
Molarity Numerical on Water Gas Shift Reaction

For the water gas shift reaction below, $K_c = 3.491$ at a certain temperature. What are the equilibrium concentrations of all the components of the reaction if $\pu{0.3815 mol}$ of $\ce{CO}$ and $\ce{H2O}$ are initially mixed in a $\pu{250 mL}$ flask?
$$\ce{CO(g) + H2O(g) <=> CO2(g) + H2(g)}$$

I was able to calculate the molarity of $\ce{CO}$ and $\ce{H2O}$ and set up an ICE table, but my answer seemed inaccurate. My work so far:
$$3.491 = \frac{[\ce{CO2}][\ce{H2}]}{[\ce{CO}][\ce{H2O}]}$$
$$3.491 = \frac{x^2}{(1.526-x)^2}$$
Because $x$ should be much smaller than $1.5$, it can be ignored when squaring $1.526$. Therefore, $3.491 = x^2/2.32$
$$3.491\cdot 2.32 = x^2$$
The square root of that value is $2.85$, which is impossible because there weren't $\pu{2.85 mol}$ of $\ce{CO}$ or $\ce{H2O}$ to begin with.

@OscarLanzi    My change in molarity was greater than the actual moles of reactants involved.

@KarstenTheis       H2O and CO both start at 1.526 M and have x subtracted from them. CO2 and H2 start at 0 M and have x added to them.

Taking the square root of both sides gives
$$\frac{x}{1.526-x}=1.868$$
So
$$x=1.526\frac{1.868}{2.868}=0.994$$

That's one beautiful elephant in the room I failed to notice when I wrote my answer solving quadratic equation:( Nicely done!

Your reasoning looks correct up to point you assumed $x\ll 1.5$.
There is no evidence to support this claim and eventually you just have to solve the equation:
$$K_c = \frac{x^2}{(c_0 - x)^2}\label{eqn:1}\tag{1}$$
where $c_0$ is the initial concentration of the reactants:
$$c_0 = \frac{n_0}{V} = \frac{\pu{0.3815 mol}}{\pu{0.250 L}} = \pu{1.526 M} \tag{2}$$
Equation \eqref{eqn:1} can be rewritten as a typical quadratic equation of $ax^2 + bx + c = 0$ form:
$$(K_c - 1)x^2 - 2K_cc_0x + K_cc_0^2 = 0 \tag{3}$$
with the following set of roots:
$$
\begin{align}
x_{1,2} &= \frac{2K_cc_0 ± \sqrt{4K_c^2c_0^2 - 4K_cc_0(K_c - 1)}}{2(K_c - 1)} \\
  &= \frac{K_cc_0 ± \sqrt{K_cc_0(K_cc_0 - c_0(K_c - 1))}}{K_c - 1} \\
  &= \frac{5.327 ± 2.851}{2.491}\tag{4}
\end{align}
$$
$$x_1 = 3.283;\quad x_2 = 0.994$$
Since $x < c_0$, only $x_2 = 0.994$ is physically meaningful, resulting in the final answer:
$$[\ce{CO2}] = [\ce{H2}] = x = \pu{0.994 M}$$
$$[\ce{CO}] = [\ce{H2O}] = c_0 - x = \pu{1.526 M} - \pu{0.994 M} = \pu{0.532 M}$$


text:
Software for conformational search
I need a free software to perform a conformational search on molecules with 10 to 16 rotational bonds. What software would you recommend me? Where can I download it? Where can I find a manual of this software?

It is easy to generate the conformations, and hard to evaluate them by some measure. What property of the conformation are you trying to optimize?

While not free Spartan is very good and I think worth the price

Avogadro, Openbabel, RDKit


text:
Comparing solutbility of acetanilide and diethyl ether in water
Would acetanilide be more soluble in water at 25 degrees Celsius because it has an $\ce{N-H}$ and an oxygen that can participate in hydrogen bonding, whereas diethyl ether only has an oxygen?

Your argument would be good if you'd  talk about acetamide, but that's N-phenylacetamide. Phenyl group changes things quite a bit if attached to a smaller molecule.

What are you comparing? Acetanilide in water with acetanilide in diethyl ether? Or acetanilide in water with diethyl ether in water?

I have a question on an organic chemistry assignment asking me to choose the least and most soluble molecules in water at 25 degrees celsius. Acetanilide and diethyl ether are the choices for the least soluble and i don't know how to choose which one is more soluble in water. I realize that the phenyl group weakens the H-bonding capabilities of acetanilide, but does that make it less soluble than diethyl ether?

Acetanilide is approximately 10x more soluble in diethyl ether (1g dissolves in18ml) than in water (1g dissolves in 185ml).
Source here
As @Mithoron said in the comments, that phenyl group has a big effect.

I believe you are asking solubility of diethyl ether and acetanilide in water. If that the case, then diethyl ether is is more soluble in water than acetanilide at $\pu{25 ^{\circ}C}$:
Solubility of diethyl ether in water at $\pu{25 ^{\circ}C}$: $6.05\% (w/v)$ (Wikipedia)
Solubility of acetanilide in water at $\pu{25 ^{\circ}C}$: $\lt 0.56\% (w/v)$ (Wikipedia)
Here, as a novice to organic chemistry, you may concern mainly on number of carbons in each molecule (aside other factors such as presence of $\ce{N}$ and $\ce{O}$) to decide the solubility. But in future, you may learn the importance of that single $\ce{O}$ atom in ether molecule when you learn subjects like Grignard reagents.


text:
Free induction decay in NMR (as stated in Pavia)
There is a statement in a very popular textbook of spectroscopy by Pavia (Chapter 3, Intro. to Spectroscopy), it says "The observed FID is actually an interference signal between the radiofrequency source (300 MHz in this case) and the frequency emitted by the excited nucleus"
As per my understanding this statement is not fully correct, as the free induction decay signal is acquired after the excitation pulse is turned off. Does anyone know if there is a overlap in timings of pulse excitation and signal acquisition?

AFAIK, the signal from the nuclei must be mixed with a signal at the reference frequency (300 MHz) before it can be digitised. So the FID no longer contains the frequency of the nucleus (300 MHz +/- a bit), but rather the offset frequency (0 +/- a bit). The reference frequency signal would come from the rf source. See eg Levitt Spin Dynamics 2ed §4.5.

Thanks for that book. Nobody ever talks in that level of detail. So I gather that is "down conversion" of raw data. The interference pattern has a fundamental original from the de-phasing of the precessing nuclei. Do you think this is correct? This down conversion protocol is a signal processing approach. Levitt writes "A similar procedure takes place in an ordinary radio receiver. In that case, the modulated r.f. waves, travelling through space, are detected and down- converted to the audible frequency range, where they are transformed into mechanical oscillations to generate sound. "

As it is not really practicable to digitise 300MHz or higher frequencies as @orthocresol states the 300MHz is mixed with the signal from the experiment (which is at almost the same frequency) and the difference frequency, now in the KHz region, is digitised instead.

Thanks @porphyrin. It makes sense what Pavia is trying to say, but do you agree that FID is not an interference pattern formed by mixing this reference frequency and the frequency emitted by precessing nuclei?

Not really interference as the two signals are multiplied together not added. For example if they are sine waves then the product is  two cosine waves, one of the sum and one of difference frequency. The math identity is $2\sin(v_1)\sin(v_2)=\cos(v_1-v_2)-\cos(v_1+v_2)$. The high frequency is filtered out leaving low frequency which is easily digitised . Electrical/communication engineers call this heterodyning.


text:
How do I calculate the total volume when i mix 1000 kg of glucose with 5000 L of water
I want to know how to calculate the resulting volume when I mix 1000 kg of glucose into 5000 L of water.
How does the calculation change if I'm mixing NaCl into the water, instead of glucose? Or how does the calculation change if im mixing several things into water (CaCl, Nacl, ect.)?
Sorry if this is a basic question, I have searched alot for a way to calculate this, but could not find answers that clearly explained how to calculate.

This is something that is usually far easier to just measure than to calculate. There will be tabulated results for common compounds, but, even so, the fastest way to find out is to do the measurement which could easily take less time than the online search required to find a reliable answer.

You could calculate it only if

you know the density of the substance being dissolved and of the solvent
there is no interaction between the 2 compounds, causing the deviation from the Raoult law and the shift of the final density.

If both assumptions are met, the final volume is
$$V_{\rm mix}=\frac{m_{\rm 1}}{\rho_{\rm 1}}+\frac{m_{\rm 1}}{\rho_{\rm 1}}$$
where m and $\rho$ are respective masses and densities.
On the other hand, if the mentioned deviation is in place, the calculation will be wrong.
E.g. if you mix $\pu{1 L}$ of ethanol with $\pu{1 L}$ of water, then after cooling down ( as the mixture would heat itself), the result will not be $\pu{2 L}$, but $\pu{1.9 L}$
In such cases, it can be determined experimentally, or from tabulated data of solution densities.
Or, the calculation above can be usually used as a good estimation, especially if there are no significant temperature changes during dissolution.

Density of liquid solutions usually cannot be calculated in a simple way. You need density tables (or a Thermodynamic Activity Coefficient Model like UNIFAC or similar), e.g.:
The Engineering ToolBox: https://www.engineeringtoolbox.com/density-aqueous-solution-organic-sugar-alcohol-concentration-d_1954.html
Bettin, H.; Emmerich, A.; Spieweck, F.; Toth, H.: Density data for aqueous solutions of glucose, fructose and invert sugar. Zuckerindustrie 123 (1998) (5) 341-348
Comesaña, J. F.; Otero, J. J.; García, E.; Correa, A.: Densities and Viscosities of Ternary Systems of Water + Glucose + Sodium Chloride at Several Temperatures. J. Chem. Eng. Data 48 (2003) (2) 362–366
https://www.internetchemie.info/chemie-lexikon/daten/n/natriumchloridloesung-dichtetabelle.php


text:
Mechanism of the formation of peracetic acid
Wikipedia says that the equilibrium $$\ce{H2O2 + CH3COOH ⇌ CH3COOOH + H2O}$$ occurs. What is its mechanism?
The following is my speculation.
The first possibility is that $\ce{CH3COOH}$ is protonated into $\ce{CH3CO(OH2)+}$ because of the strong acid condition and then turns into $\ce{CH3C+O}$. Because the oxygen atom in $\ce{H2O2}$ is electron rich, it will bond with the carbon atom with positive charge to form $\ce{CH3C(=O)O(OH+)H}$ and then peracetic acid is formed by deprotonation.
The second one is that the oxygen atom in $\ce{H2O2}$ attacks the carbon atom in $\ce{MeCOOH}$, then the $\ce{OH}$ in $\ce{COOH}$ and
one of the $\ce H$ in $\ce {H2O2}$ leave.
Is the mechanism above right or not? If it is not, what's the correct one?
P.S. (this question does not answer my question)

You are not too far off. It is somewhat of a mixture of the two mechanisms you proposed. The carbonyl oxygen is much more basic than the non-carbonyl oxygen and will be protonated preferentially. Then hydrogen peroxide can attack, and once the tetrahedral intermediate collapses, deprotonation yields peracetic acid.


text:
What does it mean by 'the residue was taken up in solvent' in synthesis protocol?
The below is a synthetic protocol I need to perform:

N-trifluoro acetoxy succinimide Synthesis
Trifluoroacetic acid anhydride was added to N-hydroxysuccinimide at 0 °C. The reaction was allowed to warm up to r.t. over 24 h. Then toluene (50 mL) was added and the volatile was removed in vacuo. The residue was taken up in DCM (15 mL) and the remaining solvent was removed in vacuo.

I don't quite get the last sentence. What does it exactly mean by 'taking up a residue in solvent'? Does it mean I just need to add 15 mL DCM to the reacting mixture?
Also, the second bit also gets me confused — if I put my mixture under vacuum, wouldn't all the solvents get sucked up? what would be the point of adding DCM if I am placing my mixture under vacuum immediately after adding DCM?

Looked at one database on a website and toluene and DCM don't seem to form an azetrope. // Wouldn't trifluoroacetic acid be a side product? Would that evaporate better with DCM?

You add the DCM to the material left after removing the toluene and then evaporate it off under vacuum. The reason is to ensure that all the trifluoroacetic acid is removed as this would adversely affect the next reaction, and TFA is readily soluble in DCM.

Thank you! but why do you need both Toluene and DCM to remove trifluoroacetic acid? Is it just to doubly make sure that TFA is all gone?

Exactly so. TFA will really mess us the next stage (which I presume is adding the N-Trifluoroacetoxysuccinimide to a nucleophile)


text:
What, if any, are the differences between radical scavenging and radical quenching?

Hindered amine light stabilizers (HALS), introduced in the 1970s, function mainly as free radical scavengers, although they also may act as quenchers or peroxide decomposers. (Polypropylene: The Definitive User's Guide and Databook)
Anthocyanins, as other flavonoids, have been shown to act as scavengers of various oxidizing species, that is, superoxide anion, hydroxyl radical, or peroxy radicals. They may act as quenchers of singlet oxygen or they may react with metal ions and thereby indirectly decrease hydroxyl radical production.(Comprehensive Natural Products II Volume 3, Chemistry of Flavonoid-Based Colors in Plants)

Sources like these (all emphasis mine own) have led me to question whether radical quenching and scavenging are indeed the same thing.
The article "Radical Scavenging and Singlet Oxygen Quenching Activity of Marine Carotenoid Fucoxanthin and Its Metabolites" appears to use different measurements for scavenging and quenching, the effective concentration for 50% scavenging (EC50) and quenching rate constants ($k_\mathrm{Q}$, $\pu{×10^10 M−1 s−1}$). While it is clear that one quantifies rate while the other measures concentration, it is unclear why both quantities couldn't be termed similarly.
From what I can see, it appears that "quenching" is generally more commonly used for singlet oxygen, an example being the second quoted passage. One other example would be: "Lidocaine: a hydroxyl radical scavenger and singlet oxygen quencher".
However, in a number of examples, antioxidants that react with singlet oxygen are also termed as scavengers, such as in: "Tocopherol as singlet oxygen scavenger in photosystem 2" and "Ascorbic acid as a scavenger of singlet oxygen".
Are "scavenger" and "quencher" used differently because of a significant difference in how singlet oxygen and "other" ROS (such as superoxide anions, hydroxyl radicals, or peroxy radicals) are neutralised, which makes one more suitable than the other in either case? Is termination of all radicals not fundamentally similar? If so, elaboration on the reaction mechanisms in both cases would be appreciated.

I do comment because I am not sure, and in particular my answer comes from a semantic pov, and considering that there is not a quencher in my language. Scavenger emphasises the property of removal of something (in our case free radicals) while quencher emphasises that a process is put to a stop. A subtle nuance is that quenching  should result in a stable enough specie, while generic scavenging can be even capturing a free radical and forming another still reactive one.

To clarify a clean use of the terms would be "adding a radical scavenger such as X quenches the reaction.  .".


text:
Oxidation of 2-Methyl, 2-Phenyl Propane
Considering oxidation mechanism of side-chains in arenes, it is known that side-chains get oxidized to carboxylic group ($-COOH$).
Below is a picture of 2-Methyl, 2-Phenyl Propane. When it gets refluxed with acidified $KMnO_4$ or acidified $K_2Cr_2O_7$, it should be oxidized to give benzoic acid.
But according to the provided source, it seems 2-Methyl, 2-Phenyl Propane doesn't react with the above oxidizing agents.
My question is,
Why does this compound disobey the general oxidation of benzyl derivatives?
Note, the source has a OH on position 4 of the ring, that makes it difficult for me to name it.
Source: https://papers.gceguide.com/A%20Levels/Chemistry%20(9701)/9701_s13_qp_41.pdf
Qno:5 c) i)
Or is the OH that makes the difference in this question?
2-Methyl,2-Phenyl Propane:

For oxidation of aromatic side chains, you must have at least one Benzylic hydrogen. tert-butyl Benzene (which is actually 2-methyl,2-phenyl propane) doesn't have any benzylic hydrogen..

@SoumikDas Oh, hadn't considered that? Should I remove this thread?

@Amar30657 No. I rather suggest either comment or answer your own question to let others with the doubt have it answered on ChemSE; you may close the question later on. (The question has been answered by Soumik, tho)

For oxidation of aromatic side chains, you must have at least one Benzylic hydrogen. tert-butyl Benzene (which is actually 2-methyl,2-phenyl propane) doesn't have any benzylic hydrogen.
-By @Soumik Das


text:
Why does energy need to be "compensated" in order for solvation to occur?
I am currently learning about the physical properties of alcohols.
I understand that the main intermolecular forces between alcohols and water are hydrogen bonds. When these two are mixed, the intermolecular hydrogen bonds between alcohol and water are broken and hydrogen bonds between alcohol and water molecules are formed. I also read that as an alcohol's hydrocarbon chain gets longer, it is less soluble, as the hydrocarbon part is hydrophobic and that is becoming more significant (while the OH becomes less significant).
There's one part I'm confused about. I read a website which stated that:

The energy released when these new hydrogen bonds are made more or less compensates for that needed to break the original ones.

I understand what the author is trying to say here, but I'm confused about why this needs to happen? Why does the energy need to be "more or less compensated"? Why can't long alcohols just break their hydrogen bonds and form weaker van der Waals forces with the water (the hydrocarbon part cannot form hydrogen bonds).

Such mixing would be too endothermic for the entropy increase to cover it.

The compensation is required as the first step is unfavourable(breaking of alcohol-alcohol H bonds). The intrusion of water into alcohol's pre-existing network of H bonds would be "compensated" by water forming new H bonds with alcohol molecules.

Sorry, but what is meant by the 'intrusion' of water?


text:
Why is the reaction quotient Q always equal to one at standard state?
I understand that at equilibrium, $Q$ is equal to $K$. This should not depend how I express concentrations (as amount of substance concentrations $c$ or as partial pressures $p$ or as molality $b$ or as mole fraction $x$) and which units I use (e.g. partial pressure in Pa or in atm) as long as I am consistent.
Various expressions rely on $Q$ being dimensionless (because we take its logarithm) and on being equal to one when all species are at standard state. For example, the equation for calculating the change in Gibbs free energy for a reaction should result in the standard Gibbs free energy for the standard state:
$$\Delta_r G = \Delta_r G^\circ + R T \ln Q = \Delta_r G^\circ \text{(for standard state)}$$
This requires that $Q = 1$ at standard state. The Nernst equation is another (related) example.
What I am wondering about is why $Q$ is always equal to one at standard state, no matter how I express concentrations when calculating reaction quotients.
For example, if I have a reaction like $\ce{A(g) + B(g) -> C(g)}$, doesn't the value of $Q$ change if I switch from concentration $c$ to partial pressure $p$, or use units of atm instead of Pa when measuring partial pressures? Come to think of it, doesn't the value of the equilibrium constant $K$ change as well when I change measures or units of concentration?

This question is related, but not identical, to this question

$Q$ is derived from activities and/or fugacities. You cannot express concentrations exactly as you want. You need to use the proper activity (concentration in molar) or fugacity (pressure in bar). Of course, these approximations are generally only valid at lower concentrations/pressures.

As already said in the comments, you cannot use any unit you want in the reaction quotient; it has to be formulated in terms of activity (for solutes) or fugacity (for gases).
The "standard state" of a chemical species is a reference state in which the properties of the chemical species are defined. This standard state is a theoretical state, it does not need to be practically reachable by the chemical species. By definition, solutes in standard state have an activity of 1, and gas have a fugacity of 1. Therefore, whatever you reaction, if all the species are assumed to be in standard state, the reaction quotient will be a product of 1s, that is, 1.

Fugacity is not dimensionless, so it can't be 1 (it could be 100,000 Pa though).


text:
Is there a reason to use bases in electrolysis and acids (eg vinegar) for electroplating?
To me, electrolysis for rust removal and electroplating seem to be the same process, with different metals and goals.
However, in seeing electrolysis and electroplating videos, it seems that electrolysis (for rust removal) uses washing soda, while electroplating uses vinegar.
Similarly, anodizing titanium seems to use a washing/baking soda (base) solution.
I am an EE. It seems to me vinegar would be less conductive. But from a chemistry perspective, what difference does the pH make?
Thank you

This may help !


text:
IUPAC nomenclature of organic compound

I strongly feel that the answer should be option (1) as the ester functional group has a greater preference over the double bond and hence it should get a lower locant.
But the source of the question says that the answer is option (2). My question is why the double bond gets a lower locant than ester in this case?
As far as I know the lowest set out locants rule is applicable for substituents but when the comparison is between functional groups the preference order plays a greater role. Please let me know if I am wrong.

related: Doubts about IUPAC nomenclature of 4-ethyl-1-fluoro-2-nitrobenzene // Very related: IUPAC nomenclature: "Smallest sum of locants"?  // the lower set of locants is used, i.e. 2 is more correct than the others (but not the preferred IUPAC name either).

@Martin Lower set of locants is preferred when we are comparing only substituents but in my question preference order is between functional groups.

I considered 4-ethanoyloxycyclopen-2-ene as a prefix, I might be wrong; let's wait until the experts find this question. // Btw. You should [edit] this thought process into the question itself.

Related: https://chemistry.stackexchange.com/q/99458/7951

@Loong So that means leaving the principal functional group and unsaturations, rest all other groups which are used as prefixes are treated as substituents even if that is a functional group(not principal)? And in that case the unsaturation gets the immediate next preference after the principal functional group?

@suhridisen Yes, that's correct.

As already explained in a related answer, the most important simplified criteria for the numbering in such cases are:

lower locants for the principal characteristic group that is expressed as suffix
lower locants for multiple bonds
lower locants for prefixes
lower locants for substituents cited first as a prefix in the name

The corresponding actual wording in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows:

P-14.4 NUMBERING
When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority:
(…)
(c) principal characteristic groups and free valences (suffixes);
(…)
(e) saturation/unsaturation:
  (i) low locants are given to hydro/dehydro prefixes (…) and ‘ene’ and ‘yne’ endings;
  (ii) low locants are given first to multiple bonds as a set and then to double bonds (…);
(f) detachable alphabetized prefixes, all considered together in a series of increasing numerical order;
(g) lowest locants for the substituent cited first as a prefix in the name;
(…)

The principal characteristic group is the carboxylic acid. Therefore, the lowest locant is assigned to the $\ce{-COOH}$ group in accordance with Rule (c).
Next, a low locant is assgined to the double bond according to Rule (e). Therefore, the name of the structure without any further substituents is cyclopent-2-ene-1-carboxylic acid.
Finally, a low locant is assgined to the remaining substituent to Rule (f).
Thus, the correct name is 4-(acetyloxy)cyclopent-2-ene-1-carboxylic acid.


text:
Simplifying temperature profile
I am given the temperature profile below and the two equations:

$$Q = (2\pi R_0)LU_0(T_{1\infty}-T_{2\infty})\tag1\label1$$
$$\frac{1}{U_0R_0} = \frac{1}{h_1R_0} +\frac{\ln(R_\mathrm a/R_0)}{k}+ \frac{1}{h_2R_\mathrm a}\tag2\label2$$

and I am asked

How does \eqref{1} simplify for the temperature profile shown, i.e. the case where the temperature in the wall is very close to $T_{1\infty}$}?

I am not sure if this means that I am supposed to "remove" the middle term in \eqref{2} since it will more or less be the same as the first term and thus the overall heat transfer coefficient would look like this:
$${U_0} = \left(\frac{1}{h_1} + \frac{R_0}{h_2R_\mathrm a}\right)^{-1}$$
What I don't understand is that I just simplified $\ce{Eq(2)}$ and not \eqref{1} which the question asks for. Or is it that by simplifying \eqref{2} I in turn simplify \eqref{1}, or have I completely misunderstood the question.

Actually, the correct answer should be $$U_0=\frac{h_2R_a}{R_0}$$.  These are shown as two separate equations only for mathematical convenience.  You could eliminate Uo by combining them into a single equation.  What do you get if you substitute the approximation I wrote into Eqn. 1?

@ChetMiller Oh I forgot about the first term. Thank you! If I subtitute Your approximation into Eq.1 I get, $\ce{Q=(2πL)h_2R_a(T_{1∞}−T_{2∞})}$


text:
Magnetic moment from given CFSE value

Spin magnetic moment of a octahedral complex having CFSE = $– 0.8∆$ and surrounded by weak field ligands can be:
A. $\sqrt{15} \space \mu_B$
B. $\sqrt{8} \space \mu_B$
C. (A) & (B) both
D. None of the above

The CFSE value given in the above question corresponds to only $d^2$ configuration for weak field ligands. Hence, according to me, the value of spin magnetic moment should be only $\sqrt{8} \space \mu_B$.
For the answer to be $\sqrt{15} \space \mu_B$, the configuration would have to be $d^7$ requiring the CFSE value given in the question to have an additional 2P(twice of pairing energy) but that is not the case.
Hence answer should be option (B) but the source of the question says that the answer is option (C).
I am unable to understand the consideration of both $d^2$ and $d^7$ configurations.
Can anyone please explain?

The total spin for $d^2$ must be S=1 or 0 which using $M=\sqrt{4S(S+1)}$  gives  $\sqrt{8}$  or zero, and for $d^7$ the spin is $S=3/2$ or $1/2$ for low spin so $\sqrt{15} $ or $ \sqrt{3}$, so if the answer is (3) you must have an octahedral complex can be either S=1 ( two unpaired electrons) or S=3/2 (three unpaired) with little difference in energy between them.

The pairing energy is considered only when under the stong field of the ligand the electrons are forcefully paired and not when the electrons in the d orbital are filled according to hunds rule.
Hence for a $\mathrm d^7$ system under weak field, CFSE value is $0.8\Delta$ only. Answer for the above question is (C)

Even for weak ligands, the electrons will eventually have to pair up, once the number becomes more than $5$. So, $\mathrm{d^7}$ is indeed possible with a weak field ligand. Consider $\ce{[CoCl6]^{4-}}$, for example.

yes I understand that. But for d7, the value should be CFSE = –0.8Δ + 2P. In the question as the 2P is not given, I feel the configuration can only be restricted to d2.


text:
What does the slash in polymer nomenclature, e.g. Acrylates/ Beheneth-25 Methacrylate Copolymer, mean?
I was given a task to find SDS-Sheets for polymers/microplastic particles in cosmetics and I am confused, since there are often two chemicals divided by "/", e.g. Acrylates/ Beheneth-25 Methacrylate Copolymer. So would it be correct to look for the SDS-sheets for the both of them separately? Is it just the mixture of the two? Maybe someone can provide me with some information of how I am supposed to read these kinds of names.

Acrylates/Beheneth-25 Methacrylate Copolymer is an INCI name for a copolymer

… formed from the ester of 2-methylpropenoic acid and docosanol, ethoxylated, 25 mol EO (average molar ratio) and one or more monomers of acrylic acid, methacrylic acid or one of their simple esters

(Source: cosmetics.specialchem.com)
Also known by the trademark Aculyn™ 28.
Dow Chemical provides cumulative MSDS datasheet (pdf) for the entire Aculyn family, so I assume you can use it for your copolymer as well.


text:
Mechanism for the addition of chlorine to benzene to form hexachlorocyclohexane
I am familiar with electrophilic substitutions of benzene, with chlorine using a halogen carrier catalyst such as $\ce{AlCl3}$, to produce chlorobenzene, and I am aware of the mechanism.
But, apparently, one can get the addition of chlorine to benzene, where the delocalised structure is disrupted and hexachlorocyclohexane is formed.
I would like to know the mechanism for this addition reaction.

This is a radical addition, more here https://www.chemguide.co.uk/organicprops/arenes/halogenation.html

Thank you for referring me to the chemguide link but this only discusses the overall equation of the addition reaction, it does discuss the mechanism of the monosubstitution of chlorine into the benzene ring but I am trying to find information out about the complete chlorination of the benzene ring, i.e., the addition of 3 chlorine molecules to get C6H6Cl6, Thanks

@Waylander's link is useful if realize that the reaction is a radical chain reaction and not electrophilic aromatic substitution.


text:
How can I get xyz coordinates of atoms of a unit cell structure from CIF format files?
I'm trying to run DFT or other 1st principle simulation of Pyrope ($\ce{Mg3Al2Si3O12)}$ and Grossular $\ce{(Ca3Al2[SiO4]3)}$ unit cell structure. So, I'm looking for unit cell xyz periodic box structure of Pyrope and Grossular.
I found some data from several places (here, here and here)
CIF format files from those databases do not give me proper atomic coordinates of unit cell. I downloaded CIF files for Pyrope from those databases, and converted with Open Babel, and even visualized with VEGA ZZ.
However, it displays only 4 atoms in the periodic box, while Pyrope or Grossular should have 20 atoms in the periodic box based on their formula (Pyrope ($\ce{Mg3Al2Si3O12)}$ and Grossular $\ce{(Ca3Al2[SiO4]3)}$).
This means either CIF of database is wrong, or I am. Or the software is unable to read CIF properly. I think I'm doing something wrong, but I'm not sure what is wrong with my CIF conversion.
If anyone knows how to retrieve the unitcell structure of crystals from CIF from those databases, please help me.

CIFs typically contain just the independent part.

thank you for your comment. Could you explain a bit more if possible? Is this means I cant get unit cell structure from CIF? Then how can we get xyz coordinates of atoms of unitcell for crystals?

Have you tried Mercury from CCDC? I use it to convert crystal structures of molecules to the Xmol format.

@Martin-マーチン thank you, I installed Mercury and opened CIFs from Mercury. The results look much more promising, but I also see numerous number atoms are placed outside of the periodic box, and converted structure (xyz and pdb) is not the unit cell... Do the CCDC and Mercury provide their own database of unitcell for Mercury?

CIF contains coordinates for the atoms in asymmetric unit only (what Ivan called "the independent part" in the comments), plus the set of symmetry operation to fill the remaining space with those atoms.
What you want to do is to open CIF in any viewer supporting that format, make sure the unit cell is packed, e.g. the symmetry operations are applied along all three crystallographic axes $a, b, c$ from $0$ to $1$ and export the data to *.xyz format.
There are several programs that are capable of doing this.
For consistency I suggest using crystallographic data by Meagher [1] for both minerals at 25 °C from AMCSD:

Pyrope CIF
Grossular CIF

CSD Mercury
As Martin suggested, Mercury (Free, available on Windows, Linux, MacOS) is one of the most user-friendly and reliable crystallographic viewers.

Open the program and load CIF (I loaded pyrope.cif).
An asymmetric unit pops up:

Go to Calculate → Packing/Slicing… and under Packing area tick Pack option, and under Include Atoms area select … that Fit.

Now you have the content of the entire unit cell:

Save the data as XYZ by clicking File → Save As… → Save as type: XMol files (*.xyz).
This produces a file with the coordinates of all atoms of the unit cell.

Unfortunately, I'm not aware of a good method to eliminate atoms located on the periodic boundary with Mercury as packing coordinates can only be changed by a 0.1 Å step, which is rather crude.
However, there is a workaround I'm aware of with another program, VESTA.
VESTA
VESTA (free, available on Windows, Linux, MacOS) is yet another functional crystallographic viewer.

Open the program and load CIF (again, I loaded pyrope.cif).
By default VESTA not only recognizes atomic connectivity, but also populates the unit cell and the first coordination spheres of all atoms on top of that.

Since the goal is to cut off the content of neighboring unit cells and the boundary atoms, the interatomic bonds should be removed.
Go to Edit → Bonds… and delete all [three] bonds from the list.

Now we have the same atomic set as in Mercury after packing with all the boundary atoms present.

To eliminate the boundary atoms, you can decrease the boundary box by a tiny fraction.
On Style tab click Boundary… button and adjust upper and lower limits for  all three axes ($x_\mathrm{min}$, $x_\mathrm{max}$, $y_\mathrm{min}$, $y_\mathrm{max}$, $z_\mathrm{min}$, $z_\mathrm{max}$) making the box smaller by 0.001 Å on each margin.

After applying new boundaries you get the unit cell content without boundary atoms.

Save the data as XYZ by clicking File → Save As… → Save as type: XYZ file (*.xyz).

References

Meagher, E. P. The Crystal Structures of Pyrope and Grossularite at Elevated Temperatures. American Mineralogist 1975, 60 (3–4), 218–228.

Thank you so much!!! Looks like 1.0 packing also includes the atoms in periodic boundary, which I think is not a proper unit cell, but I will read more manual to check the periodicity of unit cell. Thank you!

@exsonic01 No prob, the thing with Mercury is that I don't know how to change the packing box more precisely. I added a workaround for VESTA as an alternative though, which, if I understood you correctly, does precisely what you asked for.

Thank you so much. For my purpose, range of fraction should be 0 to 0.999 for 3 directions, because I need to include atoms of periodic boundary at only 3 surfaces, not 6 surfaces neither 0 surfaces. This gave me 160 atom unit cell, I may able to decrease the unit cell size even more by tweaking fractional coord range. But I need to calculate the unit cell size carefully in this case. Thank you to let me know about VESTA.

@exsonic01 Ah, I see now. Yep, then you need to use 0, not 0.001 for the lower limit.

You can either leave the boundary atoms and set their occupancy to a fractional value (2 if on the face, 4 if on the line, 8 if on the corner), or you can take the interval [[0,1 as suggested in the comment.

Possibly Avogadro may deserve a test complementary to the suggestions by @andselisk, too.  Similar as @andselisk, I accessed the 25 degC *.cif data set about Pyrope by Meagher1975 at AMCSD.  Already the very first automatic display of the data was the one about an unit cell depicted below:

The program may be of interest for you since it allows to save this packed cell both as *.xyz, as well *.gamin / GAMESS input, *.gau / Gaussian cartesian input, or *.gzmat / Gaussian Z-matrix input.  Perhaps equally of interest for you is the entry "Slab builder" in the section of crystallography, where you may define (and eventually export) little "islets" of the structure of predefined dimension oriented along the Miller indices.

For this test, I used Avogadro 1.2.0 in an instance of Linux Xubuntu 18.04.2 LTS.

That's a great one, I totally forgot about Avogadro! For some reason I always think of it as of molecular editor and I forget to mention it when it comes to crystalline matter.

@andselisk Using Avogadro more often than Mercury often is by mere "infection" / "seeding" in a mixed ecosystem of chemists and physicists with Xray *.cif and computational results on the desk.  Maybe the newer branch (Avogadro2) allows to pack crystallographic cells in a range ]0...1[ as shown in your example with Vesta I should look at.

Just in case, VESTA works just fine on Xubuntu 18.04 (screenshot) ;)

Thank you!! I heard about Avogadro before, but I didn't know it has such options. I will check this out too!

@andselisk To prevent potential misunderstanding: a) I meant that I did not see (yet) the newer branch Avogadro2 listed in the default repositories of Xubuntu 18.04 or Debian testing (but Fedora, for example, offers it for multiple years). b) Vesta's developers and potential users might find your findings valuable since currently the list of Linuxes known to work with it stops at Ubuntu 16.04, justifying a note to vesta.dev(at)gmail.com.


text:
Mechanism of ring formation of an epoxide under acidic aqueous conditions
I am trying to figure out the mechanism of this reaction:

I think the $\ce{H+}$ will first attack the oxygen of the epoxide and another $\ce{OH-}$ will be added to the most substituted carbocation, which is the secondary carbon of the epoxide. This will form a glycol, but don't know where to go from there.
What is the mechanism that leads to that product?

This is a compilation from the comments (some of them deleted by now) showing the thought process in finding the reaction product and eliminating some of the steps that might seem plausible. None of the insight is mine.
Formation of a cation

I think the H+ will first attack the oxygen of the epoxide

Yes, that is the first step.

No attack by a nucleophilic oxygen at this point

[...] and another OH− will be added to the most substituted carbocation, which is the secondary carbon of the epoxide. This will form a glycol,but don't know where to go from there.

This does not happen.

H3O+ means H+ in water so OH- is available.

The solution is acidic, so there is very little hydroxide available.

H20 will act as a nucleophile and attack the carbocation, this will result in a H+ being released from +(H2O)

This does not happen.

Why is the H2O not attacking before the formation of 5 member ring?

Compared to hydroxide, water is not a strong nucleophile. ANBENZENE comments: "[...] the molecule will always try to generate a stable carbocation [...]".
Rearrangement to make a more stable carbocation
As Waylander states in a comment: "Have the double bond electrons attack the protonated epoxide to give the most stable cation. Remember, forming [5] rings is favorable".
This is an intramolecular reaction, going from a carbocation on a secondary carbon to one on a tertiary carbon. Also, a five-membered ring is formed and a three-membered ring was opened.

Final step: water reacts with the carbocation
Finally, the oxygen on water does attack a carbocation, resulting in the second hydroxyl group found in the product.

Oxygen atom being under strain(sp3 hybrid orbitals at an angle of 60°!!)  will be more reactive and hence  3° carbocation is formed as the  last attack is performed on the sp2 (2°) carbon atom.
Hope it helps.

Why is the H2O not attacking before the formation of 5 member ring?

That is not thermodynamically favourable, because the molecule will always try to generate a stable carbocation.[From another view: If you consider HSAB theory , H2O is not a hard base and hence it won't attack comparably hard carbocation ;#NOTE: hyperconjugation decreases the hardness of the carbocation]

It's unclear whether cation formation and attack of the double bond should be concerted or not (as shown). Mechanistic studies would be required to distinguish.

Considering the angle strain, I think it would follow unimolecular pathway.

@ANBENZENE Unclear if that's actually the case. Though given the existence of the Eschenmoser-Stork hypothesis, your claim is not unreasonable.


text:
Why is the major product of the reaction of 1-naphtol with benzene diazonium chloride not the ortho product?
If 1-napthol reacts with benzene diazonium chloride, it gives the para product with respect to the hydroxyl group. However, shouldn't the major product be the ortho one, because there will be intramolecular $\ce{H}$-bonding between the $\ce{H}$ atom and $\ce{N}$?

When benzene diazonium chloride attacks the ortho position with respect to -OH it(the benzene ring) experiences a lot of steric hindrances,along with that,the triple bond also repels the lone pairs of oxygen,therefore the nitrogen atom will attack para with respect to the -OH group.If the temperature is increased i think that an ortho product will be formed as it is the thermodynamicly stable product,while the para product is the kinetically stable product.


text:
Do all reactants have appear in the first mechanism of the rate law?
I have the following reaction,
$$\ce{CO + NO2 -> CO2 +NO}$$
The "second" mechanism is given:
$$\ce{NO3 + CO -> NO2 + CO2}$$
So when searching for the first mechanism, I found this one to work:
$$\ce{NO2 + NO2 -> NO3 + NO}$$.
Would this be considered a valid mechanism? Does it matter at all that one of the reactants does not show up in the first reaction but does so in the second reaction?

Do all teams of the Super Bowl have to appear in the first game?

@IvanNeretin While the answer to both question is 'no,' somehow, this analogy isn't as good as your usual ones...

It can be even worse than that. A catalyst participates in the mechanism, but because it's regenerated at the end of the reaction, it cancels from both sides of the equation and doesn't appear in the reactants at all.

Your net equation is
$$\ce{CO + NO2 -> CO2 +NO}.$$
To explain the mechanism, the steps have to add up to the net equation. If you add up the two reactions that you suggested,
$$\ce{NO2 + NO2 -> NO3 + NO}$$
$$\ce{NO3 + CO -> NO2 + CO2}$$
and cancel species that appear on both sides, you get the net reaction. Which step you call first or second is not really important, but in the order I wrote them, you see how the $\ce{NO3}$ shown as a reactant in the second step is produced by the first reaction step.
As Zhe said in his comments, if there is a catalyst you would see species in the equations of the steps that don't even appear in the net reaction. So, for example, if your reaction were catalyzed by platinum, you would write
$$\ce{CO + NO2 ->[Pt] CO2 +NO}$$.
Then, nobody would be surprised if platinum appears in one of the steps of the proposed mechanism. In fact, you would expect it to play some part in the reaction.


text:
Finding the n-factor of C₂O₄²⁻

In the reaction $\ce{C2O4^2- → CO2}$ the $n$-factor of $\ce{C2O4^2-}$ is:

So I think the answer to this question is $2$ because in $\ce{C2O4^2-}$ the oxidation number of carbon is $+3$ and in $\ce{CO2}$ the oxidation number of carbon is $+4$. Now according to the formula of calculating $n$-factor, i.e.:
$$|(\text{no. of atoms})\times(\text{change in oxidation no.}) + (\text{do the same for other atoms in the compound})|$$
So we get $n$-factor of $\ce{C2O4^2-} = |2\times(-1) + 4\times(0)| = 2$.
I guess I have used the right formula and got the right answer but still I think that my answer is wrong. Can somebody please clarify that is my answer wrong or right and if wrong why?

You may be using a cannon against sparrows. What happens if you cut away the net charge and cut the rest in 2 pieces ? How many electrons is  there, aside of the product. ?

@poutnik Hi thanks for responding to my question but I am not able to understand what you are trying to say.

I mean, why to count  or even use a formula, when you must see it. $\ce{C2O4^2- -> 2 CO2 + 2e-}$

@Poutnik Oh... I had to use that formula because that was the one taught to me. But overall my answer is correct, right?

Well there are basically 2 methods to know, how many apples were taken from the box. 1/ Count apples at the start and at the end, and then calculate the difference. 2/ Count just the taken apples. 1/ is the summing the differences of oxidation numbers. 2/ is counting the exchanged electrons. BTW, I have not studed chemistry in English, so I may have a gap, but I have not heart about the term "n-factor".

You should elaborate on "I think that my answer is wrong" to avoid having questions like this closed as "Am I right" question. See https://chemistry.meta.stackexchange.com/q/3396

By using formula:
$\ce{\overset{(+3)}{C}_2O4^2- ->[oxidation]~\overset{(+4)}{C}O2}$
$n$-factor $=$ change in oxidation number per $\ce{C}$ atom $\times$ total number of $\ce{C}$ atoms in $\ce{C2O4^2-}$
$=1 \times 2=2$
By using logic (as Poutnik suggested in comment):
$\ce{C2O4^2- ->[oxidation]~2CO2 + 2e-}$
$n-\text{factor} =2$


text:
Point group from mono to bulk MoS2
Many articles say
even numbers of layers or bulk crystal MoS2 have D6h point group
and odd numbers of layers (including monolayer) MoS2 have D3h point group
But whenever looking at the figure of MoS2, I can't recognize there is a symmetry representing the number '6' which means 6-fold(?). With the point group flow chart,

The final answer is C3h for the bulk MoS2.
If I included the inverse symmetry with C6 symmetry, then it would give D6h? Is the inversion symmetry necessary, then why? Especially, if you can point out the six ford axis in bulk case or two layers MoS2, please let us know.
I got the hind from the reference document that Mo has D3h and S has C3v. And these two are factor group of D6h. But I don't much know about group theory... So definitely it is not D6h?
Here the unit cell of bulk 2H-MoS2

Here the top and side view of 1H-MoS2

It would help a great deal if you would add a picture or two of MoS2.

@porphyrin I added. Hope to be helpful.

This is not a proper answer but can give you a clue. I arrived here looking for the same answer.
I didn't arrive to find the C6 rotation axis in bulk 2H-MoS2. On the other hand, if you look for the symmetry operations of the space group of 2H-MoS2 (P63/mmc, No. 194, D46h):
http://img.chem.ucl.ac.uk/sgp/large/194az1.htm
You can see that there are not any C6 rotation axis for this space group. Nevertheless, there are 6-fold screw axes (http://img.chem.ucl.ac.uk/sgp/misc/sixscrew.htm) and 6-fold inversion axes (http://img.chem.ucl.ac.uk/sgp/misc/barsix.htm).
You are right that the point group corresponding to this space group is D6h (https://en.wikipedia.org/wiki/Space_group#Table_of_space_groups_in_3_dimensions). And according to character tables a D6h point group should have C6 (http://symmetry.jacobs-university.de/cgi-bin/group.cgi?group=606&option=4).
Both things are contradictory. My background is Chemistry, and I am used to group theory applied to molecules rather than crystals. My guess is that the 6-fold screw axis, which is a symmetry element that is only possible in crystals (as it requires translational symmetry) might be related with labelling as D6h. But this is just my speculation.
In this article, they also talk about the symmetry operations of single-layer, multi-layer (odd and even layers) and bulk MoS2, and could be usefull for you too:
http://www1.spms.ntu.edu.sg/~qihuagroup/data/Xiong/Papers/nano%202013%20yanyuan.pdf

Thanks a lot. I never thought the 6-fold screw and inversion axis. It's helpful but quite strange yet. There is 6-fold subgroup as you referred the 5th site. Well... It is trivial if we assume it is 6Dh. But if we apply the operator (C6), then mismatch with original positions. And I'm trying to read the last paper, thanks again.

I read a part of the paper, and I thought if the C6 involves translation symmetry along the z-axis for the bulk (not one, two, or finite layers), the atoms lie on their original positions. Because Dh6 symbol doesn't consider translation, so we are not supposed to ignore that. And I also found another example on wiki: en.wikipedia.org/wiki/Space_group See the first figure of bulk H2O.


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Table for magnetic moments for all elements in periodic table?
I'm trying to look for a table for magnetic moments for all elements of the periodic table for my research project in computational chemistry, can anyone direct me to a suitable one?

Magnetic moment is not a property of elements.

Unpaired electrons will give rise to a magnetic moment equal to the 'spin only' value $\mu_s=g\sqrt{S(S+1)}$ in units of Bohr Magnetons, $e\hbar/2m_e$, for example H atoms  with $S=1/2$. $g$ is the magnetogyric ratio $\approx 2$.
There may also be an orbital contribution (from orbital angular momentum) depending on what levels are filled then the equation becomes $\mu_{s+L}=\sqrt{4S(S+1)+L(L+1)}$ where $L$ is the orbital angular quantum number, e.g. $V^{3+},\,S=1,\,L=3$ gives $\mu_{s+L}=4.47$.
(Many transition metal complexes have values differing from this formula  because of bonding between metal and ligand restricts the orbital angular momentum. You will need to consult texts on Ligand Field Theory and come to grips with the Van Vleck equation to understand this topic in more detail.)


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SpectroVis spectrophotometer values changing after pulling cuvette out
I am required to write a research paper for school in a specific subject and I have chosen to write mine in chemistry.
I am investigating the kinetics of the Briggs-Rauscher oscillating reaction.
I am using a spectrophotometer to take samples of the reactants and determine the absorbance at 460nm every 45 seconds to calculate the rate at which iodine is produced.
However, I am using a spectrovis spectrophotometer, who's calibration values change without me doing anything and sometimes do not register the fact that iodine concentration has increased. Is there anyway I can fix this problem and prevent the readings from changing?

One should be there to see exactly what is going on. Are you following the reaction in the cuvette or just taking sample at regular interval? By the way I try to find an answer that migh help you and can be done in parallel.  Comparing the two might really makes a great paper. .....

Here it is. Get a decent baseline and do both. https://chemistry.stackexchange.com/questions/112263/taking-kinetic-readings-for-the-briggs-rauscher-oscillating-reaction/112276#112276

"However, I am using a spectrovis spectrophotometer, who's calibration values change without me doing anything and sometimes do not register the fact that iodine concentration has increased." Please share an example of the data. What is the meaning of "calibration values change"?  Are you putting the cuvet  properly each time i.e. the transparent side is in the light beam path?