text:
“Good” properties of transparent conducting oxides
I'm a computer scientist and I'm trying to figure out which are the key properties that define a “good” acting transparent conducting oxide (TCO) in order to consider several possible features to create machine-learning models for eventually discovering new TCOs.
I've already read that a “good” TCO should have a high electrical conductivity with a low absorption of visible light (which, in fact, makes it transparent), so a good measure of the TCO behavior could be defined as
$$\frac{\sigma}{\alpha} = -\frac{1}{R_\mathrm{s}\ln(T + R)},$$
where $R_\mathrm{s}$ is the sheet resistance in ohms per square, $T$ is the total visible transmission, and $R$ the total visible reflectance. By my understanding, also films thickness plays some role and it influences the values of $\sigma$ and $\alpha$. In particular, the more thickness I have, the more figure of merit $\sigma/\alpha$ increases and it results in a better behaving TCO.
Since I'm diving into chemistry for the first time, I wanted to ask you some suggestions about the topic, as well as eventually where to find interesting data about it.

You have not covered hardness, refractive index, residual stress from deposition, impurity/defect impact on color, reactivity, moisture resistance, standard solvent resistance, or likely another half dozen or more characteristics that make something 'good'.

Workability in thin films surely add to the list above


text:
Why use Stork enamine alkylation?
I recently learned about the Stork enamine alkylation method of adding electrophiles to the beta carboxyl position on a molecule. But I can't seem to find out why a direct synthesis is not possible (or perhaps favoured).
I've used an online resource to draw a mechanism that I thought would work, but clearly because we use the Stork enamine method I am missing something here and this can't be true.
Is it perhaps that the double bond is too weak of a nucleophile to attack?

Your final, hypothetical product should be a keto alcohol, not a diol. Your question needs to be more focused.

A simple Google search provides: https://en.wikipedia.org/wiki/Stork_enamine_alkylation

Aldol condensations like what you've drawn usually need some source of base or acid. The latter would probably be disastrous here, but for example, you could generate the enolate with LDA / NaHMDS etc. before adding the electrophile, and I think sometimes hydroxymethylations with formaldehyde are carried out that way (although you probably can't add formaldehyde itself, see https://en.wikipedia.org/wiki/Formaldehyde#Forms). I am pretty sure Clayden has a chapter on this.

source: Stack Exchange

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

sort: votes


text:
Install MOLDEN in MACOS Big Surf
I'm trying to install Molden in a MACOS Bigsurf.
The steps that I followed:

Install the compilers, gfortran,libraries, etc.
extract the source from molden page.
open the molden5.8 directory and from terminal run "make all".

But, in step 2 I get the following message:
Error: Type mismatch between actual argument at (1) and actual argument at (2) (REAL(8)/REAL(4)).
molden.f:6495:28:
6495 |              call parptr(16,fdum,fdum,idum)
|                            1
......
8594 |       call parptr(1,freq,fdum,idum)
|                    2
Error: Type mismatch between actual argument at (1) and actual argument at (2) (REAL(8)/REAL(4)).
make: *** [molden.o] Error 1
what is wrong?
what can I do?

This question is not on topic here, since it is not about chemistry.

There are Molden precompiled binaries for OS X.  Here is a page that shows you how to acquire molden binaries.
https://www.macinchem.org/blog/files/7a5f03265cc78741fb3cc2a86c54f41e-377.php
You ftp the server as a guest and download the most recent binary.  After you ftp in, it will open a virtual drive and you can copy it right to your Desktop.  I just tried it on Catalina and it worked right out of the box.

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


text:
How many electric vehicles can our current supply of lithium power?
This Reddit comment contends

There was 870 tons of lithium mined in 2019 in the US. That enough for under 20,000 cars, assuming you're using ~50 kg per car. So that would be 40,000 in 2023, assuming all of it is used in cars. In reality only a fraction is; most is used in consumer electronics etc.

The global car production is about 100,000,000 per year, meaning you'd need like 5,000,000 tons of lithium just for cars. Current WORLD production is 77,000 tons, although there's a large reserve.

Proliferation of electric cars is, as far as I can see, completely dependent on a large step forward in battery tech.

What exactly does there's a large reserve entail? How much Li does the "large reserve" contain?

Are these calculations correct?

See https://en.wikipedia.org/wiki/Lithium#Reserves

You assume there is X lithium and all of it cost Y. And then there would be no lithium to mine. Thats not quite right. As you are ready to pay more, harder to get sources become profitable and become part of available reserves. Sea has almost unlimited amount of lithium, but it is about x5 as expensive to mine.
The total lithium content of seawater is very large and is estimated as 230 billion tonnes
(wiki)

+1 and it's rather simple to leach practically all Li out of old batteries, so on the long run, recycling serves an increasingly big part of the demand, with battery efficiency also increasing.

Lithium isn't currently the limiting factor in battery tech. Cobalt is really the biggest Li-ion battery issue from a resource standpoint. That's why a lot of current research is focused on developing nickel-rich formulations (which, if successful, would make nickel an even bigger concern than lithium). Of course, this answer is absolutely correct (+1) and the same rationale applies to the other materials.

@PGnome I think you can go ahead and post that as an additional answer, just add a supporting link which shouldn't be hard.


text:
Why No Change in Polarizability in Asymmetric Stretch?
In order for a molecule to be Raman active, there must be a change in the polarizability, meaning that there must be change in the size, shape or orientation of the electron cloud that surrounds the molecule. And this change said to occur in symmetric stretching, but not asymmetric stretching.
Please explain using plain english (without any rules or tables) why is that in asymmetric stretching like the following, there is no change in the size, shape or orientation of the electron cloud that surrounds the molecule? Specially considering each side independent has its own electron cloud.

Is possible the polarizability change requirement applies not to particular bonds, but to the whole molecule, with the opposite bond effects  nullifying each other ?

The effect on both sides levels out, mostly. Remember that some selection rules are stronger than others: Spin must be retained, and if you just have no dipole moment, you get no IR interaction. Oh, remember you have to look at this from the view of an electromagnetic wavelenght that is a thousand times as large.

In the asymmetric stretch the molecule retains the same shape because  as one extends the other contracts and vice versa.

the shape oscillates in left and right, how can they be same shape? And why would same shape not cause change in polarizability, hope someone can Answer.

@Karl You don't need a dipole moment, you just need a change in the dipole moment.

These earlier answers may be helpful: How can I deduce the linearity of XeF2 from the IR spectrum? and Is carbon dioxide IR inactive?.

Related: https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Book%3A_Quantum_States_of_Atoms_and_Molecules_(Zielinksi_et_al)/06%3A_Vibrational_States/6.06%3A_Harmonic_Oscillator_Selection_Rules

Please explain using plain english (without any rules or tables) why is that in asymmetric stretching like the following, there is no change in the size, shape or orientation of the electron cloud that surrounds the molecule?

There is a change. The size and shape of the electron cloud does change for the individual bonds affected by the asymmetric stretch. Polarizability relates to the ease with which electrons can be moved from their equilibrium position.  Electrons in weaker (longer) bonds are more easily displaced than electrons in stronger (shorter) bonds.
From your diagram, I believe that you are discussing the case of a linear $\ce{AB2}$ molecule. Picture the asymmetric stretch in our $\ce{AB2}$ molecule with the two $\ce{B}$ atoms remaining stationary and the $\ce{A}$ atom oscillating back and forth. Simultaneously one $\ce{A-B}$ bond becomes longer and the other shorter.  One becomes more polarizable, the other less polarizable.  In the linear $\ce{AB2}$case these changes in polarizability are equal and opposite, hence for the molecule overall they cancel out and there is no change in overall polarizability (for example, see here).

In order for a molecule to be Raman active, there must be a change in the polarizability

Yes, but it must be a change in polarizability for the entire molecule, not just an individual bond. Further, keep in mind the  Principle of Mutual Exclusion. It states that, for centrosymmetric molecules (molecules with a center of symmetry, your linear $\ce{AB2}$ is an example), vibrations that are IR active are Raman inactive, and vice versa.
The asymmetric stretch in such a molecule changes the dipole moment in the molecule from zero to non-zero; hence the asymmetric stretch is IR active. Therefor by the Principle of Mutual Exclusion, the asymmetric stretch must be Raman inactive.  Even though the asymmetric stretch changes the polarizabilty of some bonds in the molecule, the polarizability of the whole molecule does not change (as explained above) and the asymmetric stretch is Raman inactive as predicted.
The links I posted in the comments show how these concepts can be applied in the linear $\ce{AB2}$ examples of $\ce{CO2}$ and $\ce{XeF2}$.

You just stated the rule about Principle of Mutual Exclusion which you quoted " It states that, for centrosymmetric molecules (molecules with a center of symmetry, your linear AB2 is an example), vibrations that are IR active are Raman inactive, and vice versa." But you didnt state the technical reasons why. Specifically, considering each side independent has its own electron cloud. So why is the polarizability in each side cancelled just because the molecules is bouncing back and forth in for example asymmetric stretching.

@Jtl Edited the answer to better address your points.

@Jtl, I think one should not imagine the polarizability as simply as you are imagining. Unfortunately, polarizability is a tensor. I don't know the answer to your query but I am trying to show polarizability is very hard picturize. See this physics query https://physics.stackexchange.com/questions/274794/difference-between-polarizability-and-dipole-moment#:~:text=Polarizability%20is%20an%20intrinsic%20property,different%20extent%20in%20different%20directions.


text:
Finding work done for a chemical reaction at non-constant pressure and temperature
I have studied the expression for work done for a chemical reaction which is basically $W=-\Delta n_{g}RT$. Along with this, the assumption that has been made is that chemical reactions take place at constant pressure and temperature.
Is this always the case? What would the general expression for work done look like if we were to consider temperature and pressure changes during the chemical reaction? Thanks.

Didn't your teacher derive that expression by starting with a general expression for work, and then applying the constraints that the only type of work is pV-work, and that the temperature are pressure are constant?

@theorist Yes it starts off with using $W=-P\mathrm dV$. So about the conditions I am asking in the OP, what would the starting point be for such a derivation?

Not quite. The starting expression for pV-work should be $\text{đ}w =  -p_{ext} dV$. Yes, you missed the differential, but the key point is that it's the external pressure that determines pV-work.  That expression does not require constant temperature or pressure.  Now what assumptions did your teacher add to go from that expression, to the one in your OP?

You have your answer.  What do you think?

There are a series of five assumptions you need to make to get from the general expression for work, to the expression you wrote in your OP.  With each assumption, you are applying successively more constraints to your system.  In the post below I give the assumptions, and the equation that results from each, so that you can see how the form of the equation evolves as successively more constraints are applied.
We start with the general expression for work:
$$\text{đ}w = \vec{F} \cdot d \vec{x}$$
[We use the form $\text{đ}w$ rather than $dw$ because work is path-dependent, and its differential is thus inexact; the dyet glyph, $\text{đ}$, is used to indicate this.]
Now we integrate and express the dot product in terms of scalar quantities:
$$w = \int_{path} \vec{F} \cdot d \vec{x} = \int_{path} |F| \,d |x| \cos(\theta)$$
Because work is path-dependent, it is necessary to integrate over the actual path, as indicated in the above integrals, rather than merely from the initial state to the final state.
Dropping the explict scalar notation, we can write it as:
$$w =  \int_{path} F d x \cos(\theta)$$
Now let's add some assumptions:
Assumption 1: The force is applied in the direction of motion.
As the angle between the direction of motion and the direction of force is $0$, and $\cos(0) = 1$, we have:
$$w =  \int_{path}F dx \cos(\theta) = \int_{path}F dx$$
Assumption 2:  The work is $\bf{pV\text{– work.}}$
For $pV\text{– work}$, we can take work done on the system to be either positive or negative.  I'll use the former convention. And we can only reliably use the external pressure to determine the work; the internal pressure is not always defined, particularly for rapid expansions or compressions.  To indicate this explicitly, we write "$p$" as $p_{ext}$. Then (using "$A$" for area):
$$p_{ext} = F/A;\,\,dV = -A dx\,\, \text{(negative sign is because as $x$ increases, $V$ decreases)}$$
$$\Rightarrow  \int_{path} F dx =  -\int_{path}(p_{ext}A) (dV/A) = -\int_{path}p_{ext} dV$$
Thus:
$$w = -\int_{path}p_{ext}dV$$
This is the general expression for $pV\text{– work}$.  It applies regardless of the conditions.
Assumption 3: The system is in mechanical equilibrium with its surroundings.
In this case, $p_{ext} = p_{sys}$.  By convention, we designate $p_{sys}$ simply as "$p$".
Note that this assumption restricts us to paths in which the system is continously in mechanical equilibrium with its surroundings.  This limits us to paths whose work depends only on the difference between the initial and the final state.  Thus we do not need to concern ourselves with doing an explicit path integral. I.e., because the path restriction is now implicit, we can simply integrate between intitial and final states (which I've designated as $i$ and $f$):
$$w =  -\int_{i}^{f}pdV$$
Assumption 3a: The system is at constant pressure.
Since $p$ is a constant, we can take it out of the integral:
$$w =  -\int_{i}^{f}pdV=  -p\int_{i}^{f}dV = -p\Delta V$$
I labeled this assumption as "3a" rather than 4 because we didn't need to both specify the system is at constant pressure and in mechanical equilibrium with its surroundings, since if it is at constant pressure it is also in mechanical equilibrium with its surroundings.
But I broke the steps up, since I wanted to give the expression following assumption 3 for completeness.
Assumption 4: The system is an ideal gas.
Substituting $V=nRT/p$, and remembering that the pressure is constant, gives:
$$w =  -p\int_{i}^{f}dV =  -p\int_{i}^{f}d(n R T/p)=-R\int_{i}^{f}d(n T)=-R\int_{i}^{f}ndT-R\int_{i}^{f}Tdn$$
Assumption 5: The temperature is constant.
Since $T$ is a constant, we can take it out of the integral:
$$w =-R\int_{i}^{f}d(n T)=-RT\int_{i}^{f}dn = -RT \Delta n$$
Or, equivalently:
$$w =-R\int_{i}^{f}ndT-R\int_{i}^{f}Tdn = -R\int_{i}^{f}Tdn = -RT\int_{i}^{f}dn = -RT \Delta n$$

About the dyet character, it is handy to define for it a user hot key combination in the MathJax editing script.

@Poutnik  How would I do that?  Right now MathJax doesn't recognize either of the commands that I know can make a dyet (\dj, which requires the fontec pkg, or \textcrd, which requires the tipa pkg)),so I have to just copy and paste it.

It is not  part of MathJax. The dyet
unicode literal character would be inserted as a text string by invoking a hot key defined within the script. I use it on my desktop, writing this on phone. It is the same script that e.g. wraps a chemical formula by $\ce{}$ using CTRL+ALT+E. You can customize it.

See https://chemistry.meta.stackexchange.com/questions/4264/introducing-markdown-shortcuts-for-stack-exchange   You can define extra item using the dyet character e.g. instead of \pi string


text:
How to create a Potential Energy Surface in Chemcraft?
The Chemcraft website says that it is possible to create a potential energy surface. Nevertheless, it doesn't work for my Gaussian scan job.
%chk=propylcyclohexan_scan_30.chk
#p GFINPUT POP(FULL) PM6 Opt=modredundant

 propylcyclohexan_scan_30

0 1
C         -2.32729        0.36923       -0.26826
C         -1.86086        1.79592        0.04793
C         -1.42371       -0.68619        0.35867
H         -2.35296        0.22494       -1.35567
H         -3.35311        0.22391        0.09146
C         -0.38557        2.05033       -0.34666
H         -1.99132        1.97106        1.12477
H         -2.51564        2.50378       -0.47169
C          0.03534       -0.45199       -0.00958
H         -1.73512       -1.68573        0.02974
H         -1.53451       -0.66689        1.44998
C          0.49676        0.97326        0.33008
H          0.17109       -0.63513       -1.08350
H          0.66387       -1.18248        0.51357
H          0.48096        1.10355        1.41947
H          1.53881        1.06369        0.01138
C          0.03241        3.54672       -0.05839
H         -0.30194        1.90196       -1.43267
C          1.54273        3.82216       -0.15556
H         -0.26571        3.76488        0.97694
H         -0.55219        4.21744       -0.69950
C          2.23970        5.10589       -0.65389
H          2.00099        3.01727       -0.72714
H          1.90136        3.75452        0.88334
H          2.45381        5.07336       -1.75369
H          3.23327        4.96387       -0.25958
H          2.02974        6.11114       -0.31960

D 6 17 19 22 S 11 30.0
D 12 6 17 19 S 11 30.0

It is possible to open the output file from this job and see the scan steps, but the "view scan graph" button does not work.
Does Chemcraft need information that I forgot to calculate?

FYI there is Matter Modeling [Beta].SE that is specifically suited for questions about algorithms and software used in computational chemistry. Your question is perfectly on-topic for Chemistry.SE either, just keep in mind there is a dedicated SE site for problems like this.


text:
Is there a safe way to neutralize bleach at home
I don't want to simply replace one hazardous chemical with another. The bleach is stubbornly clinging to my shower walls and I feel its effects every time I take a shower.
Ascorbic acid seems like the most natural solution to me, since it's a regular part of our diet. Online sources recommend ascorbic acid "but you should never use any other type of acid". I'd like to ask you chem experts: what's the difference? From what I've found, ascorbic acid plus bleach will create chlorine gas just like any other acid. It also forms sodium hydride, which appears to be a pretty nasty substance as well.
Are there any options? I looked into sodium theosulfate, which is recommended for fish tanks, but it seems it too is an irritant when inhaled. Why replace one irritant with another?
PS It is likely that NaOH is also present.

Stop it. Just stop. Chemophobia is a thing, and it can drive you nuts all right.

Just rinse the surfaces down thoroughly with water. All the components of bleach are fully soluble

This question it is so avulsed that should be closed. Saying so I am actually helping OP, not making fun of him/her.

I also find the smell of chlorine bleach smell annoying and have developed a rapid way to safely remove the smell, likely with more disinfecting and odor removing benefits.
Start by placing dilute hydrogen peroxide in a spray bottle. Proceed to apply to the areas that were the subject of a bleach application. Spray also into the air. Interestingly, in a few minutes, not only is the chlorine smell removed, but the air has a fresh smell to it!
Now, the chemistry is also interesting indicating the creation of the transient but powerful hydroxyl radical ($\ce{.OH}$). Per a source from the field of atmospheric science, the neutralization reaction of hypochlorite based chlorine bleach (active ingredient: $\ce{NaOCl}$), proceeds per this 2008 work: Hydroxyl radical from the reaction between hypochlorite and hydrogen peroxide in three steps. In particular, cited Reactions (1) to (3) are discussed below:
$\ce{OCl- + H2O2 -> .OH + .ClO + OH-}$ (1)
There is also the creation of an intermediary compound described as, $\ce{OH-ClO}$:
$\ce{.OH + .ClO -> OH-ClO }$  (2)
that apparently breaks down in alkaline conditions:
$\ce{OH-ClO + OH- -> ^1O2 + H2O + Cl-}$  (3)
where the net effect is the previously well known decomposition reaction given by:
$\ce{OCl- + H2O2 -> H2O + O2 + Cl-}$
Further, I suspect an interaction between the formed Singlet oxygen and oxygen gas leading to yet other transients, but active oxygen species, which are likely responsible for the fresh smell.
As a safety precaution, I would advise waiting at least an hour before returning to the treatment area (as not all reactive oxygen species are people-friendly).

For example, at 80 °C, with NaOCl and NaCl concentrations of 80 mM, and pH 6–6.5, the chlorate is produced with ∼95% efficiency. The oxygen pathway predominates at pH 10.[17] This decomposition is affected by light[18] and metal ion catalysts such as copper, nickel, cobalt,[17] and iridium.[19]
(wiki, https://en.m.wikipedia.org/wiki/Sodium_hypochlorite)
Its NaClO, ideally you want it to decompose to NaCl, table salt, without producing free Cl. Acid will likely decompose it into something that does release Cl.
NaOH isnt likely to help, NaClO becomes only more stable in base pH (reverse of an acid).
H2O2 also makes bleach decompose quicker into NaCl, without chlorine formation, but it is also deadly to microscopic life, probably even more so than bleach. And it is also decomposed by heating. But H2O2 decomposes much quicker than NaClO.
If you want to make water+bleach safe for microscopic life, without releasing chlorine gas in a process, I would suggest adding just enough H2O2 to react with the bleach, and then heating the water with copper being present. Heating and copper both make bleach and H2O2 decompose safely, releasing oxygen.
P.S. if someone knows a better answer, let me know, i need to solve similar problem too. That is, making bleach+water safe for microscopic life.


text:
Work done by system of ideal gas in isobaric expansion
Consider a system of ideal gas in a container with piston and  the isobaric expansion of gas takes place.
As the process is isobaric, so initially the pressure of gas is equal to atmospheric pressure as the piston remains in equilibrium over the container. Then, first we increase the temperature and as a result pressure of gas increases and the piston rises till the pressure reduces to atmospheric pressure. So at initial and final state pressure of gas is constant thus the process is isobaric.
So, work done by gas should be
$dW=F_{applied\;by\;gas}(x).dx$
$dW=P_{applied\;by\;gas}(x)Adx$
$s.t.\;P_{applied\;by\;gas}(x_{initial\;state})=P_{applied\;by\;gas}(x_{final\;state})=P_{atm}$
So, total work done, $W=\int_{x_{initial}}^{x_{final}}P(x)Adx$
So, we should know pressure as a function of $x$ to calculate the work done by gas.
My question is that in books, they directly use the argument that work done by gas is dependent on $P_{external}$ and as it is constant, they give the expression of work as $dW=-P_{external}Adx$.
I don't understand how they do that and use external pressure in the expression of work done.
I think we should take into account pressure of gas as a function of $x$ to find the work done by gas.
Please help me in clarifying the doubt.

Isobaric means with constant external pressure. Then it depends if the process is reversible ( always considered in equilibrium ) or not. For the former, internal and external pressures are equal and constant. For more, see Reversible process on wikipedia.

Consider that for reversible isobaric expansion $p_\mathrm{int} = p_\mathrm{ext} + \mathrm{d}p$

Thanks for the reply. But if the process is irreversible then I think we have to take into account the pressure of gas as a function of x.

Because then $p_{int}=p_{ext}+\Delta p$ and $\Delta p$ can't be neglected

For example at initial and final state, pressure of gas is same as external pressure andbetween the states it varies as explained in my question. But, in high school chemistry book, it is written that work done for doth reversible and irreversible process is -$P_{ex}\Delta V$. I don't understand that.

Works is path dependent quantitiy.

Let us continue this discussion in chat.

At the interface between the gas and its surroundings, by Newton's 3rd law, the pressure exerted by the gas on its surroundings is equal to the pressure exerted by the surroundings on the gas.  So we can use either.  For a reversible expansion, the gas pressure is given by the ideal gas law.  But the ideal gas law only applies at thermodynamic equilibrium and during a reversible process (which consists of a continuous sequence of thermodynamics equilibrium states).  For an irreversible expansion, the ideal gas law doesn't work (i.e., gives the wrong answer), and we can't use it to get the gas pressure at its interface with the surroundings (where the work is being done).  But, if we know the surroundings pressure during the irreversible expansion, we can still calculate the work (because, at the interface between the gas and surroundings), the two will match.  So using the external pressure is always a valid approach to getting the work.

I think that Newton's third law is not being applied in the way you suggest. If gas particles collide with the surface of piston with force F or pressure P, then the equal and opposite force will be exerted by the piston not by the surroundings.

The piston is part of the surroundings.  The inside face of the piston is the interface between the gas and its surroundings; this is where the action-reaction pair is situated.  And, if the piston is massless, it doesn't matter whether the piston is included within the system, or within the surroundings.


text:
How to synthesize 3‐cyclopentylpropanal from (chloromethyl)cyclopentane?

I am not sure on how to start, but I think of these two possible ways:

Reacting the starting material with Grignard reagent and then oxidise the product with PCC to produce the aldehyde;
Hydrolysis the starting material to produce alcohol and then react the alcohol with PCC to form aldehyde.

But these two ways didn't give the desired product.

3 Steps:
Step 1 - react the starting material with the preformed anion of ethyl cyanoacetate (NaOEt/EtOH should do this comfortably)
Step 2 - mild base hydrolysis of the product (NaOH/EtOH) followed by acidification to remove the ethyl ester group.
Step 3 - Dibal-H reduction of the nitrile to aldehyde 1.

You can proceed with the following idea:
React the given starting material with Mg turnings in ether to form Grignard reagent. Now react this with sodium 2-chloro ethanoate. Now add lithium aluminium hydride (LAH) to this and then oxidise it to aldehyde using PCC.
Hope it helps!

Do you have a reference for the reaction of the Grignard with 2-Chloroethanoic acid?

@Waylander my organic chem teacher once gave the reaction...will look for a citable reference and share once I get it :)

Please correct me if there's any mistake in it

Grignard reagents will surely react with carboxylic acids, just not in the way you think. It's just an acid-base reaction:$$\ce{RMgBr + R'COOH -> RH + R'COOMgBr}$$

@orthocresol I agree, that's why I would like to see a reference. I don't think this will work with the acid or the sodium salt

@Waylander, my comment certainly wasn't directed at you :-) I agree that enolate + alkyl chloride is the way rather than alkyl Grignard + alpha-carbonyl electrophile.

@Waylander I couldn't find any reference online so I am certainly wrong on this one. Thank you for pointing out the mistake in my answer.

@Orthocresol Thank you for those suggestions. Now should I delete my answer as someone who may have not read the comments may be misled by my answer

Grignard with epoxide may work better.


text:
CO2 + High humidity = transperent substance?
First I want to point out that I am not a chemist, sorry for bad description.
I have CO2 incubator. CO2 concentration is 10-20% and the humidity level is around 90%. I was using a humidity sensor to measure humidity and some kind of transparent substance have formed around the sensor. Sensor becomes unprecise and had a slow response time. Substance almost looks like plexiglass.
Is it possible that carbonic acid which is forming from CO2+H20 -> H2CO3 is somehow reacting with integrated circuit case or sensor and forming some kind of glass-like substance?
New sensor:

Used sensor:

What else is in the incubator?

Nothing. Only 20% CO2 and 90% humidity. I was testing how the sensor works in a CO2 environment.

Does it stay like thay when you remove the pressure and take it to the normal air? if yes, then it is likely co2 is being absorbed into the plastic part, and that plastic part is swelling. You can avoid it by using a glass window for optical sensor. Or use H2O permeable but not CO2 permeable membrane if this sensor needs contact with the gas. Or use another type of sensor that does not contain this plastic. Or reduce CO2 pressure. I dont know any CO2+H2O compounds that would stay stable and solid at normal pressure. Methanol could be forming from those materials if energy is present, but it needs catalyst and is very unlikely to happen on its own.

Yes, It stays like this after taking it back to normal air. CO2 is at atmospheric pressure inside the incubator.  The membrane is a good idea, I will consider this option.
P.S. That bad picture is taken after one year of use in a CO2 incubator.


text:
Why is adiabatic process isentropic?
Adiabatic process is isentropic, but I can't get it why is it so. If we go by statistical method
$$S = K\ln W,$$
where $W$ is the thermodynamic probability, and if we look into adiabatic expansion the volume increases, we can say that the thermodynamic probability increases, and hence the entropy of the system should increase.
Moreover, if we look into isothermal expansion of an ideal gas in vacuum, there would be no work done, and so no heat would enter or move out of the system. But since there is a change in volume, we can write
$$∆S = nR\ln\frac{V_2}{V_1}.$$

see an answer here https://chemistry.stackexchange.com/questions/142792/will-entropy-increase-in-adiabatic-expansion/142796#142796

You left out the adjective “reversible.”  The entropy change for an adiabatic reversible process is zero.  In an adiabatic reversible process, the temperature also changes.

I get it that with the decrease in temperature due to expansion we cannot achieve the higher energy states the entropy increase due to increase in volume would be cancelled by the decrease in entropy due to decrease in temperature but temperature would also change in irreversible adiabatic expansion.

No it wouldn’t.  For an ideal gas, viscous heat generation in the irreversible expansion would cancel out the temperature decrease.

Reversible is important as you've mentioned.


text:
Air bubbles in water
Problem :
Gas.  Density. (Composition)
N2-    0.879    (0.78)
O2-.   1.32     (0.21)
Why air bubbles don't disintegrate in water into different components due to difference in apparent weight (= weight - bouncy) ?
What I think-(not sure)

it's a homogeneous mixture of gases.
pressure is more inside bubble than water(surface tension)
very slow process

Please help.

Even if you created a system consisting of the pure water, the set of pure nitrogen bubbles and the set of pure oxygen bubbles, the system would end in the equilibrium with bubbles of the same composition. Nitrogen would be dissolving in water and escaping to oxygen bubbles and vice versa.
Equilibrium of a gas dissolution follows the Henry's law
$$p =K_\mathrm{H} \cdot x$$
Where $p$ is the gas partial pressure, $K_\mathrm{H}$ is the Henry's constant and $x$ is the molar fraction of the dissolved gas in the liquid.
If $p \gt K_\mathrm{H} \cdot x$, than liquid is undersaturated and the gas is dissolving.
If $p \lt  K_\mathrm{H} \cdot x$, than liquid is oversaturated and the gas is escaping.
The consequence is, that in equlibrium, all bubbles have the same composiotion.

But I am more curious to know and  understand it from physics point of view where bodies with density greater than liquid would sink and those with less density would float (partially immerged).  Ofcource N2 and O2 molecules will be moving randomly and an equilibrium governed by Henry law will get established.   If a bubble of air is released from under water . I think according to Archemedies law it would mean that a bubble in water will separate into oxygen bubble (sink) and nitrogen bubble (float) . The nitrogen bubble will pop above water to release N2 gas

Do you observe any sinking bubble stream during water aeration?


text:
Why is HBr colorless while Br2 isn't?
Bromine ($\ce{Br2}$) has a dark reddish color. If it reacts with something like benzene, this results in the formation of Bromobenzene and $\ce{HBr}$:
$$\ce{Br2 + C6H6 ->[{Cat.}]C6H5Br + HBr}$$
It can be observed that during this reaction, the color of bromine disappears.
One can thus conclude that $\ce{HBr}$ is colorless. Why is this the case? If we look at the molecules, it seems to me that they are basically the same (speaking in how many electrons there are in each orbital, for example), except for that in $\ce{HBr}$, there is a hydrogen atom instead of another $\ce{Br}$.
So what makes the $\ce{Br2}$ bond "special", so that this molecule can absorb light in the visible spectrum but $\ce{HBr}$ can't?

I think that I have a basic understanding of the quantum mechanic model of an atom (e.g. atomic orbitals and how their energy levels correspond to different wavelengths of light being absorbed). Here are some of my ideas trying to explain the above described question:

In $\ce{Br2}$, both atoms have the same electronegativity, while in $\ce{HBr}$, there is a difference of $\ce{\Delta EN=0.76}$. The only consequence I could imagine from that is that in $\ce{HBr}$, the probability of an electron in the bond to be located closer to the $\ce{Br}$ atom is higher than being close to $\ce{H}$, while in $\ce{Br2}$, the electrons (or rather their probability) is evenly distributed

$\ce{H}$ has an electron configuration $\ce{[1s^1]}$, $\ce{Br}$ has an electron configuration $\ce{[1s^2]\,[2s^2]\,[2p^6]\,[3s^2]\,[3p^6]\,[3d^{10}]\,[4s^2]\,[4p^5]}$. So I would assume that the bond in $\ce{HBr}$ is formed by the overlapping of a $\ce{1s}$ and a $\ce{4p}$ orbital, while in $\ce{Br2}$, two $\ce{4p}$ orbitals overlap. The $\ce{4p}$ orbital has a higher energy level than $\ce{1s}$, so this might be the cause for the different absorption spectrum. However, I am not sure if this is true and how the energy levels behave when two orbitals overlap to form a new one.

The neutral Br atom has no electronic transitions in the visible range. The Br${2}$ molecule has them, because of vibrational states.

Br is not a thing at all. Color is a function of molecules and other aggregates, and not of atoms. That's just the way it is. Nearly all elements are found in variously colored compounds, and many are found in colorless compounds.

@Ivan, So you suggest there is no single molecule absorption/emission spectroscopy? It is probably a limitation of our eyes rather not a problem of atoms or molecules.

Why was this question closed? It is clearly not a homework question and I did share my thoughts

We see something colorful if there is a permitted electronic transition between a ground state and an excited state corresponding to a wavelength of about 200 to 800 nm. 2) Isolated atoms' electronic transitions do not match this range, so they appear colorless to our eyes. 3) If atoms join to yield molecules, conceptually speaking, atom orbitals may mix to yield molecular orbitals. 4) Among these molecular orbitals, there may be permitted electronic transitions corresponding to a wavelength in the visible range; which depends on the atoms contributing. So $\ce{Br}$ is not $\ce{Br2}$.

(continued)  I speculate, those voting to close the question, did not recognize if these concepts already were considered by you, e.g., when comparing the molecule $\ce{HBr}$ with the molecule of $\ce{Br2}$. | Editorial hint:  Don't use MathJax formatting in the title of a question or answer; some programs don't understand this.  But, you may use it in the body of questions, answers, and comments; use it when suitable. Note, for chemistry (e.g., formulae, reaction equations) in the body of text, see how mhchem may help you in addition (https://mhchem.github.io/MathJax-mhchem/) to MathJax.

The colour of molecules come from electronic transitions that are very sensitive to the nature of the molecule
To simplify a little, molecules absorb light when the energy difference between electrons in an occupied orbital and an unoccupied orbital match the energy of a photon of light (and that light will have a specific colour). This much you already understand.
But the energy levels of the electrons that matter for colour are usually the bonds in the molecule that join one atom to another, not the other electrons in the atom that don't get involved in the bonding. So atoms are rarely thought of as having an inherent colour that is independent of what they are attached to (some isolated atoms do have colour: sodium vapour has narrow absorption bands in the yellow, for example, but this isn't important in sodium compounds).
The assumption you make that is false is that "If we look at the molecules, it seems to me that they are basically the same". If this were true there would be very little interesting in the whole subject of chemistry. When bromine reacts with benzene a Br-Br bond is replaced by a Br-C bond and a Br-H bond. Those are very, very different bonds. A Br-Br molecule has a readily accessible electron transition that results in a dark red colour. That this particular transition is very sensitive to the exact nature of the molecule is obvious as other dihalogens (with very similar electronic structures) have very different colours (iodine is dark purple, chlorine is light yellow). But the orbitals available in other Br-X bonds will be very different as will their energy. Often the corresponding colour of light will be in the UV and there will be no visible colour.
The essential point is that colour comes from electronic transitions that vary a great deal with the atoms involved in the bond. And in many colourful compounds those transitions involve molecular orbitals including multiple atoms across the whole molecule. There is no simple relationship between atoms and the colours of molecules. Every bond is (very) different.


text:
Make Acetic Acid from wood
I would like to know if it would be possible to make Acetic Acid from Wood. I know that you can charcoal wood to get the gases, cool it, and get the liquid. This would yield a bunch of chemicals in a mixture. I also know that you can steam wood to get wood vinegar. Would it be possible to extract Acetic Acid from either the Pyrolysis Gas or Wood Vinegar? Or would I need a different method of extraction or just not be possible at all?

Any "home chemistry" attempt to do this and get a reasonably pure solution of acetic acid would likely be dramatically inferior to buying the same solution, at a higher purity and lower cost.

Plus the effort, to assemble raw materials and tools $\ce{->}$ running $\ce{->}$ disassemble the setup for you compared to just go to the shop, picking the jar from the shelf.

Acetic acid is present in 'wood vinegar' that you get by thermal destruction with a lack of oxygen.
Steam can be used as a source of heat, but to be of any help in your desired production it would have to be overheated or under very high pressure, about 100 atmospheres. It is still useful as a way to increase the yield a bit.
Likely the best way is to heat the wood to 300*C without oxygen, collect and cool the gaseous products to condense them. You will get wood vinegar.
https://en.wikipedia.org/wiki/Pyroligneous_acid
After this you can separate the products using distillation.
https://en.wikipedia.org/wiki/Distillation
About 70 % of wood mass will be converted during the thermal destruction into wood vinegar.
About 15 % of wood mass will become non-water wood vinegar.
About 5 % of wood mass will become acetic acid.
Small yield, toxic waste, energy intensity of the process, are reasons why this process is not used anymore.


text:
How can you extract palladium from palladium chloride
Hey guys I am working on a school project, focused on how precious metals can be recycled from industrial products. I am very much not an expert on the subject of chemistry. But I am just trying to get basic understanding on how to extract palladium from a catalyst like palladium chloride.
Thanks in advance,
Ilias

From what I have read, one good way to reduce Pd(II) to its metallic Pd(0) form is to "paint" the solution of Pd(II) onto a paper and heat in a reducing atmosphere of hydrogen. You should be able to recover the metallic palladium as a fine black powder. If you prefer, this can then be melted into a metal ingot.

Yes, there are several methods of reducing palladium chloride to elemental chloride.

By converting Pd(II) salt to palladium acetate ($\ce{[Pd(CH3CO2)2]3}$) and decomposing it at 200-300 $\pu{^\circ C}$. (link)

By using reducing agents

carbon monoxide
$$\ce{CO + PdCl2 + H2O → Pd + CO2 + 2HCl}$$
This is a qualitative reaction to determine the presence of carbon monoxide in the
air. A paper moistened with a solution of palladium chloride turns black in the presence of carbon monoxide due to the formation of finely dispersed metallic palladium. The reaction proceeds at r.t. The yield is high (84%).

Hydrogen
$$\ce{PdCl2 + H2 -> Pd + 2HCl}$$
Pd(II) salts can be reduced to finely divided elemental palladium by using
hydrogen. The reaction proceeds at r.t. but the reaction rate can be increased by
increasing the temperature. See this paper for more details.

Using $\ce{SnCl2}$

The reaction proceeds by formation of a complex. Yield is also high (90%). See here for more details.

Other reducing agents ($\ce{TiCl3, Fe, Cu, Zn}$)

Decomposition of palladium chloride. But the reaction is unreliable. The reaction proceeds at high temperature (590-740 $\pu{^\circ C}$). May release fumes of chlorine and hydrogen chloride which might be dangerous. Palladium obtained may be impure because of formation of various types of oxides. (link)

Decomposition of potassium tetra/hexachloropallade

Calcination in air. Dissolve the salt in a suitable solvent (usually water), "paint" the solution on a substrate, allow the solvent to evaporate and then start calcination. Pd(II) would become Pd(0), leaving a thin metal film on the substrate.

Binding of Pd-salt into an organic matrix (e.g., a zeolitic inorganic-organic polymer electrolyte), and then calcinate the product. However, using this approach, calcination should be done under vacuum or under an inert atmosphere (otherwise, oxygen might yield Pd oxide systems). In this case, Pd ends up as metal nanoparticles into a carbonaceous matrix.

Solution phase reduction. Impregnate silica with your metal salt solution, add a reducing agent like formic acid, ethylene glycol, $\ce{NaBH4}$ or hydrazine, heat at $\ce{150 ^\circ C}$ for ethylene glycol and at $\ce{80 ^\circ C}$  for the rest of the reducing agents, you will get Pd decorated $\ce{SiO2}$ particles. Nano particles are prone to oxidation at higher temperatures, so performing the reaction in inert atmosphere like $\ce{N2}$ or Argon will ensure the Pd(0) particles.

by Emeraldine base to form palladium nanoparticles. See here

Details about the points can be found in this researchgate article. But since this is a school project, points 3-7 are out of scope. The best method would be the "reduction through carbon monoxide".

Thank you very much for all the information!!

If this answered your question, please click the check mark so that the answer gets accepted. Thank you.


text:
Enthalpy of mixing of a non-ideal solution
I have this equation of Flory–Huggins model:
$$\Delta_\mathrm{mix}G = RT(n_1\ln\phi_1 + n_2\ln\phi_2) + \chi n_1\phi_2,$$
where $\phi$ is the volume fraction, $n$ is the amount of substance and $\chi$ is the parameter of solvent-polymer processing that depends only on temperature.
I need an expression for $\Delta_\mathrm{mix}H.$

In that model, think about which term represents $\Delta S$ (hint: think about the functional form of the entropy of mixing for a two-component system), and which represents $\Delta H$.

There are two ways (at least) to solve this:
(1) Evaluate $\Delta S = -\left( \frac{\partial \Delta G}{\partial T} \right)_{p,n} $
and add $T\Delta S$ to $\Delta G$.
or
(1) Evaluate directly $\Delta H = -T^2 \left( \frac{\partial (\Delta G/T)}{\partial T} \right)_{p,n} $
Either way I obtained the result
$\Delta H = \chi n_1 \phi_2 - n_1\phi_2 T \left( \frac{\partial \chi }{\partial T} \right)_{p,n} $


text:
Why is dilution exothermic?
The breaking of bonds usually absorbs energy and forming bonds usually releases energy.
Then, why is the dilution of acids and bases exothermic?

Except when it is endothermic;  I have water soluble fertilizer that is strongly endothermic.

Dissolution can be both exothermic like for $\ce{NaOH}$ or $\ce{H2SO4}$, or endothermic like for $\ce{CaCl2 . 6 H2O}$, $\ce{KNO3}$ or $\ce{KClO3}$.
The net thermal effect sign is determined by combination of endothermic breaking of a ionic lattice or chemical bonds and of exothermic ion hydration.
$$\Delta H_\mathrm{dissol} = \Delta H_\mathrm{ioniz} + \Delta H_\mathrm{hydr}$$
Compounds with the greater absolute value of lattice or bond energy ( the former term ) than hydration energy ( the latter term )   have endothermic dissolution. Compounds with the opposite relation have exothermic dissolution.
To address dilution after original misreading:
Dilution of concentrated solutions is exothermic, as the exothermic ion hydration is gradual, depending on H2O : ion  molar ratio.
For liquid concentrated acids like $\ce{H2SO4}$ or $\ce{HNO3}$, the process of dissolution gradually switch to dilution, with acid molecules being gradually ionized and hydrated.
I suppose a complicated process with a complicated PE graph, as it is not a single reaction like e.g. SN2 nucleophilic substitution. E.g. hydration of protons comes with multiple steps and several layers.
$$\ce{H+ -> H3O+ -> H3O+ . H2O -> H3O+.(H2O)2 -> H3O+.(H2O)3 -> \\ ..   -> H3O+.(H2O)6 -> .. -> H3O+.(H2O)20 }$$
See Wikipedia - hydronium - solvation. Additionally, dissociation and hydration are not subsequent processes, but are ongoing simultaneously.
Similar would happen with anions.

The question was about dilution, pretty sure that's no typo.

@Karl  Ah, I see. I know difference, of course, I have just read something else than what was written. :-)  It was before my morning coffee.

Probably all the steps described by Poutnik 2 days ago are exothermic. I repeat them  H+⟶H3O+⟶H3O+⋅H2O⟶H3O+⋅(H2O)2⟶H3O+⋅(H2O)3⟶⋅⋅⟶H3O+⋅(H2O)6⟶⋅⋅⟶H3O+⋅(H2O)20

@Poutnik In the case of bases, I think, there is no hydration going on during dilution. So, how is that exothermic?

@Shub Because you think wrong here.


text:
Is 6-ethenyl-4-(prop-2'-enyl) octa-6-en-3-ol a valid name?
I recently study organic chemistry and I read a book by Sana Jamshaid. I came across this compound and I cannot understand why it named 6-ethenyl-4-(prop-2'-enyl) octa-6-en-3-ol:

I understand that the longest chain is 8, hence octa (but if we go up on the 4th longest can be 9) but I don't understand the 4-(prop-2’-enyl). Are there any explanations to why the compound named in such way? The double bond also sits on the 7th and not the 6th.

The name looks wrong. The longest chain is 10 the second longest nine.

I cannot rationalize that name either, and the drawing looks off (angles at tertiary carbon are nonsensical). Probably it would help to draw the same molecule differently and proceed from there. It also wouldn't hurt if there were a complete reference to the book. "The country should know its heroes", as they say in Russia.

The suggested name
“6-ethenyl-4-(prop-2'-enyl) octa-6-en-3-ol”
is not correct; the closest valid name is
6-ethenyl-4-(prop-2-en-1-yl)oct-6-en-3-ol.
However, this name does not describe the compound that is shown in the picture. Note that the parent chain actually is non-7-en-3-ol instead of oct-6-en-3-ol, and the longest side chain actually is but-3-en-1-yl instead of prop-2-en-1-yl.
Therefore, the correct name for the compound that is shown in the picture is
(7E)-4-(but-3-en-1-yl)-6-ethenylnon-7-en-3-ol.

I also agree with @Loong, the suggested name "6-ethenyl-4-(prop-2'-enyl) octa-6-en-3-ol" seems incorrect. The correct name for this compound should be (7E)-4-(but-3-en-1-yl)-6-ethenylnon-7-en-3-ol, as @Loong mentioned.
In this compound, the longest carbon chain has 10 carbon atoms, but we select the carbon chain with 9 carbon atoms as it has a functional group (-OH) attached. So, parent chain will be non- as there are 9 carbons.

Now, we number the chain from left to right, as the main functional group is always assigned the least possible locant.

So, we see that we get -OH at locant 3, -CH2CH2CH=CH2 at locant 4, -CH=CH2 at locant 6 and double bond at locant 7. So these substituent/functional groups can be named accordingly:

3-ol
(but-3-en-1-yl) (as carbon atom attached to parent chain (alkyl) is always numbered 1)
ethenyl (vinyl group)
7-ene (NOTE: the ending -e will be dropped while writing the complete IUPAC name)

Also, since it is a trans- (w.r.t double bond at locant 7), we put (7E)- in the beginning of the name. (Z is put for cis- )
So, now, the IUPAC name should be (7E)-4-(but-3-en-1-yl)-6-ethenylnon-7-en-3-ol
Here are a few resources:
NCERT Class XI Organic Chemistry - some basic principles and techniques
Systematic IUPAC nomenclature - Aditya Vardhan V
cis- and trans- nomenclature in organic compounds


text:
Exception to haloform reaction
While explaining haloform reaction my teacher told me that Me3C-(C=O)-CH3 does not give haloform reaction.
What is the reason behind it since it does have a CH3 group?

The carbon in "MeC" looks like it has only two bonds. Can you clarify?

Do you mean Me3C-(C=O)-CH3 ?

@Adiboy sorry for the error, yes what you said is the one i am asking for

I suggest that you read this well-written answer, especially the second paragraph https://chemistry.stackexchange.com/a/111543/94220

Pinacolone is the common name of Me3C-(C=O)-CH3


text:
Simulation of electronic potential from nanoparticles interacting with molecules
I work with SERS, where the Raman signal gets enhanced by the electric field of the nanoparticles. Now I like to simulate this kind of spectrum with Gaussian. I found a paper[1], where it looks very nice how they simulate it. They don't get a bond between the molecule and the silver nanoparticle. If I try it on my own, then the nanoparticles are thrown away by Gaussian.
Has anyone of you an idea of how they simulate it? My idea is the following workflow: Frist, they optimize the attached molecule alone; after this, they made a scan with one variable. The variable is the distance between the part of the molecule where the nanoparticle should attach and the nanoparticle itself. And after this, they put the nanoparticle in the distance with the lowest energy and make a single point calculation with options to get the orbitals and the ESP mapping. It makes sense, or is it not an ordinary workflow for gaussian? I learned Gaussian myself self so I'm not sure at all.
[1] Gao, Y.; Xu, M.-L.; Xiong, J. J. Environ. Sci. Heal. Part B 2019, 54, 665-675
Additional Information:
This is my route section: #p opt=vtight int=ultrafine oniom(PW6B95D3/aug-cc-pvtz:b3lyp/lanl2dz) nosymm geom=connectivity
I used the OMNIC layer model and used the high layer for the molecule with PW6B95D3/aug-cc-pvtz and the low layer for the gold atom with b3lyp/lanl2dz. I have no idea how to fix the gold atom in a specific region.

You may want to try posting the question at Materials Modeling SE, which is designed to address this kind of question. Either way, you might want to elaborate on some unclear points. For instance, "If I try it on my own, then the nanoparticles are thrown away by Gaussian." What do you mean by "throw away"? How? Why? Otherwise the workflow seems fine.

I place the gold atom close to my nitro group about 2.5 Angström distance between them. When Gaussian starts the optimization, Gaussian toggles the gold atom wild around the nitro group about 10-20 Angströms and does not converge.

Ok, so the position of the gold atom wrt nitro is not properly constrained? How are you applying constraints? Maybe post the Gaussian route or other info.

@BuckThorn: No I do not know how honestly. My route section is: #p opt=vtight int=ultrafine oniom(PW6B95D3/aug-cc-pvtz:b3lyp/lanl2dz) nosymm geom=connectivity I directed the pyridine molecule to the high layer with the aug-cc-pvtz basisset and the gold atom to the low layer with lanl2dz basisset. Do you got a tip for me how I can perform it in a better way?

No but add this info to your question and somebody here probably can. Try setting geometry constraint between the gold and nitro. Check maybe https://wanglab.hosted.uark.edu/g03guide/G03Guide/www.gaussian.com/g_ur/k_oniom.htm or more recent manual


text:
Why flame testing different ions give different colors?
In my textbook, there was an experiment that described how to flame test different ions and I was just curious why they give off different colors. My knowledge is very basic being just a high school-er, but at what academic level do we learn the reason different ions give off different colors? Would appreciate if someone could explain in a very basic way, if it is possible.

You can surely search here for more. Each species, atomic, molecular or ionised, has its own electronic structures. Within this structure, only certain jumps are allowed, and they happened at once. Like you go from the first floor to one allowed superior floor without using all the steps of the stair, but in a sort of jumping. The same is valid for the way back to the ground floor. The distance between floors corresponds to an amount energy. Again, now see this energy as carried by one discrete speck of light, surely you know already the word photon... This can be seen as a dose or packet of

of energy. It turns out that each energy correspond to a colour. That is why each species absorbs or emits characteristic colours. Not sure if this helps or it is too elementary. In all cases it can be useful for a future reader at elementary levels.

I was sort of able to understand, but the jumping thing still seemed a little advanced, but thanks I now have at least a basic understanding hoping to learn more in the future

When you heat an atom the electrons present in it gets excited by absorbing energy. So it jumps to a higher energy level or shell. And when they return to their original positions they emit a quantum of energy(i.e., a discrete quantity of energy). The wavelength of the emitted light depends on the difference in the two energy levels.
This energy is given by, $$E=(hc)/λ$$ Where $λ$ is the wave length of electromagnetic wave. The released energy is fixed for the orbitals of each element and therefore the wavelength too are unique for every element(and h&c are constants).
This is why different elements give off different colors.
Hope you understand:)


text:
How long do rechargeable Batteries last vs Standard Batteries?
Ordinary Batteries will deplete most of their Power in a 10 year period. What happens to rechargeable Batteries? How long would they still functin properly?
Hope this is the right community for this question. If you need a specific example:
Powertoll Battery:
https://www.metabo.com/ch/de/zubehoer/zubehoer-akkumaschinen/akkupacks/lihd/akkupack-lihd-18-v-10-0-ah-625549000.html
1.5v Battery
https://www.energizer.com/batteries/energizer-rechargeable-batteries
Stored in an airsealed container, decent temperature & humidity. And lets say we charge & deplete them 3x per year.
Just to be sure, I'm not asking how many charges they will survive, but after how many decades they are still usable.

There is no answer to this question. Because the answer depends on you. It depends on the care you used. If you short-circuited it, only once, it will be unusable anymore.

How about storing the components of a battery separately for 10 years, and then assembling it? It could be a simple zinc/copper Daniel cell, for example.

@Maurica I literally gave 2 examples of batteries and 1 example of its use & care.

@KarstenTheis I would have to get into that, thanks for the idea

It mostly depend on battery type, and after that current. Batteries usually dislike very high currents, that would make them charge or discharge faster than in 1 hour, because of heating and secondary reactions, and very low currents, that would make them discharge in a year, because of dendrite formation.
Worst type of battery for decade long work are lithium-ion batteries. Their decay is mostly due to time and temperature, rather than cycles used. Exact opposite of your goal. NiCd, NiMH are less energy dense, they also will need additional 3 recharges per year due to self-discharge. All of them lose about half of their maximum capacity in 3 years. Li-ion because of degradation due to temperature. NiCd and NiMH bevause of large self discharge.
LiFePO4 is significantly longer living battery than li-ion. Lead-acid bartery is somewhat similar for your usage. They will lose about half of their capacity in 10 years. They achieve it in a different way. Lead-acid has significant self-discharge, needing an additional recharge every year, but its bulky and degradation takes a long time to affect it all. LiFePO4 is the opposite, very delicate, low self discharge, and degradation is just slower in general. At this timeframe your device will become obsolete.
Lithium titanate, LTO, is probably the best of what we have today for your usage. It is very durable and degrades very slowly. Its hard to put an estimate on this, but i guess you can expect about 25 years till half of it maximum capacity is gone. At this time frame wires will rust away first.
If you need even longer working time, you could check supercapacitors. They have 10 times less energy density, but they will work till their shell is rusted away. About 50 years till half of of them are rusted. They also need 100 recharges per year due to self discharge.
If you want even longer life, check ordinary capacitors and compound filled PCB. Capacity is 100 times less than batteries. Those will work till UV, ozone and moisture will destroy the compound or wires going to the board. At this time frame your smart PCB components are likely to fail first. About 100 years. Will need a few recharges per year due to selfdischarge.
For longer still, will need something exotic, like flywheels and magnetic levitation in a vacuum chamber.

Thanks for the answer however it is not accurate according to my observations. My 5 year old Li-Ion 4AH Metabo power tool battery still has practically 100% of its maximum capacity. According to you it should be down to 30-40%. I tested by using them in my battery powered vacuum cleaner and timed with a stopwatch to find out how long it will run (still 15 1/2 minutes). They are rated 72wH and fully charge in less then 5 minutes. I am doing this now for every battery pack and will keep track on how long the vacuum cleaner runs every consecutive year.

There are about 10 li-ion types, and I ment the most popular one. And Metaboo is using some special battery type that they call LiHD, with improved lifetime and other parameters. I cant find what they use exactly. Another thing to account for is that powertool may work with reduced speed if battery voltage is lower due to aging.


text:
What is the precision of intensity data in XRD measurement?
I made some relative crystallinity measurements on a zeolite (crystalline aluminosilicate). Which is measured for a specific peak area and compared with a reference sample.
The XRD of powder patterns were recorded using Malvern PANalytical X’Pert PRO Diffractometer with CuKα radiation. The patterns are collected over a range of 3-60 °2θ with a time per step of 0.017 °s-1 using a quarter° divergence slit. In this setup the sample is rotated during measurement. I use Origin for integrated area peak calculation for 21-33 °2θ and the reference sample is made to 100%. We try to reduce the sample thickness error by putting the same mass quantity in all the holders.
I obtained some discrepancy within the data, within a 4% margin. I'm trying to find in the literature what is the margin of error when measuring intensity with XRD.

In short: what does the manual of your diffractometer claim?  Were you able to replicate its standards in first place?
The peak area depends both on the peak's height and width.  But this is difficult to answer because your question does not detail how data were recorded in first place.  Beside content (elements) and quality of your sample (remaining mechanical stress, orientation / size / shape of the crystallites, sample thickness), temperature and pressure while recording the data, the peak area will depend on the source (e.g., wavelength monochromators, slits widths on source and detector), type of detector, i.e. if a point detector was used moving around the sample, or a fixed image plate / area detector and data processing (e.g., how do you account for background intensity, (how) do you deconvolute peaks in question).

(modified from Stoe's IP-PSD)
If you opted for an area detector recording all intensities along $2\theta$ simultaneously, was it the same as for your reference sample?  You are set by the geometry and the inner design of this detector (pixel size and response to the radiation used) on $2\theta$-resolution and may improve intensity resolution only by change of the time of the exposure.
With a point detector, you are less limited for angular resolution because the detector may travel as far around the capillary or rotating dish as mechanics permit in tiny steps.  Large machines

(credit)
claim increments down to steps of 0.0001° (reference) which may equate to only one experiment per day but likely better peak resolution (think peak deconvolution), potentially with a much better signal/noise ratio than the area detector, especially if the sweep is recorded more than once.

@user65414 Then my suggest is to put your comment, including your initial point («I made some relative crystallinity measurements on a zeolite (crystalline aluminosilicate). Which is measured for a specific peak area and compared with a reference sample. I obtained some discrepancy within the data, within a 4% margin.» as question to either a) a user group for PANalytical (https://www.malvernpanalytical.com/en/about-us/contact-us) or b) the wider (still Xray specific) audience of https://www.xrayforum.co.uk/

Thank you @user65414 for the suggestion.


text:
Why doesn't octahedral d8 system exhibit Jahn-Teller effect?

Chemistry LibreTexts stated clearly that (under the section 'Large $\Delta$' and 'Small $\Delta$') octahedral d$^8$ system does not exhibit Jahn-Teller distortions. But then it stated also:

...Some common examples (that exhibit distortions) include Cr$^{3+}$, Co$^{3+}$, and Ni$^{2+}$

Isn't Ni$^{2+}$ d$^8$(Group 10, +2, so 10 - 2 = 8) and should not exhibit distortions?

I believe this question is asking the similar thing. But then when the energy level diagrams drawn out as shown below, I saw that even for d$^8$, if singlet favoured over triplet, the distortion does lower the energy (and so should happen).

So what is wrong with the two arguments above?

Your distortion doesn’t preserve spin. As drawn, you should have one electron in z2 and one in x2-y2. Something is wrong with the sentence that you quoted. Perhaps they meant Ni(III), which does exhibit JT distortions. I also don’t know where they got Co(III) from. Co(III) is nearly always low spin and hence no JT distortion. Likewise, Cr(III) is d3, which should not have JT distortions.

Having looked at the paragraph in more detail, I think it is just poorly written and these are supposed to be examples of ions which don’t exhibit JT distortions, rather than ions which do. They literally picked out d3, d6, and d8 examples.

@orthocresol I just remember the spin must be preserved! But based on that, does that mean that, all d8 axial elongated tetragonal complexes will be paramagnetic (as total spin cannot be 0)?

You can put the electron in the lower orbital, thus getting a singlet state, basically what you've drawn. The only question is whether it's more stable than the alternative, which is to have a triplet state with one electron in $z^2$ and $x^2-y^2$. That would depend on the classical considerations of pairing energy and orbital energy gap: for example, in square planar complexes, the $z^2$ and $x^2-y^2$ are so far apart that both electrons sit inside the $z^2$. Thus, for example, $\ce{[Ni(CN)4]^2-}$ is diamagnetic. But that is no longer the original topic of Jahn–Teller distortions. :-)

I know I shouldn't say thank you here, but I still want to thank you to @orthocresol for often answering my question so pertinently.

No problem at all. I should probably should convert my comments to an answer...


text:
What are some convenient examples of so-called dissolved organic matter (DOM) for lab work?
Background, I have come across a chemical reaction system, occurring in nature, that apparently involves the presence of dissolved organic matter (in a catalytic role).
My research returns, for example, the following as examples of DOM per an educational source:

Dissolved organic matter (DOM) consists of soluble organic materials derived from the partial decomposition of organic materials, including soil organic matter, plant residues, and soluble particles released by living organisms, including bacteria, algae, and plants.

So my question is basically other than grinding up a plant,  are there other examples of dissolved organic matter that I can conveniently employ?
One idea is perhaps organic tea (which is still a bit general) as a dissolved organic colored solution.
But,  possibly better (in terms of convenience and reactivity) suggestions would be appreciated.
The cited reaction, by the way, is as follows:

$\ce{ O2 + 2 .CO3- + DOM -> 2 .O2- + 2 CO2 + DOM }$

Namely, the conversion by the action of oxygen on the carbonate radical anion to the superoxide radical anion and CO2, all in the presence of DOM.

So you have no leads with regard to which class of compound (e.g. polyphenol) this might be? How about tannic acid?

Do have leads, for example, dark color and fluorescence, but these most likely alude to the photocatalytic properties of the DOMs as well. Also, mention of Fulvic acid. Interestingly, one reference mentions tree leaves (which of course, includes tea leaves, albeit, the latter are more processed). Likely, have to experimentally verify efficacy.

For example, did find an interesting reference on dissolved organic carbon (DOC )  here https://www.researchgate.net/topic/Dissolved-Organic-Matter. To quote:
"...the effect of DOC (made from tree leaves and peat leachate) on macrophytes. In our lab the concentration of DOC can be measured, but the DOC cannot be characterized"

Did find related work, to quote results: The results from this study can be chemically rationalized by a reaction scheme of (a) photodecarboxylation/ regeneration of carboxyl: CxHyOz(COOH)m + aO2 + (metals, hv) --> bCO2 + cH2O2 + Cx-bHy'Oz'(COOH)m-b(COOH)b or of (b) nondecarboxylation photooxidation: CxHyOz(COOH)m + aO2 + (metals, hv) --> bCO2 + cH2O2 + Cx-bHy'Oz'(COOH)m. at here https://pubmed.ncbi.nlm.nih.gov/15352449/ , which apparently could include vinegar plus a colorant for photo hv effect.


text:
Does CH3COOH + H2O (vinegar) lose its antibacterial and antiviral properties when exposed to air?
I have been reading about using CH3COOH + H2O (vinegar) as a mild antibacterial and antiviral agent.  For example: https://pubmed.ncbi.nlm.nih.gov/15698693/
Note, before anyone gets confused (or concerned), this has nothing to do with the SARS-CoV2 virus which causes COVID-19.  I strongly doubt vinegar is effective at killing a coronavirus, and everything I have read confirms that doubt (the pH is too high, for starters).
Now, back to the topic at hand.  Does CH3COOH + H2O lose its antibacterial and antiviral properties when exposed to room temperature air for a long duration (over 8 hours, but less than 500 hours)?
I've been trying to find scientific articles regarding this, and so far, have not found anything helpful.
Ideally, I would love to see one or more graphs that plot effectiveness along one axis and time duration of air exposure along the other.  Of course, tables with data points would be equally welcome.
Note that I'm not looking for anything exact, as there are too many variables to be precise without getting into minute details.  I'm looking for a general idea of the changes (if any) over time as CH3COOH + H2O is exposed to room temperature air.

Besides evaporating?

@JonCuster Good question.  Correct, besides evaporating.

In what kind of setting do you want to use vinegar to kill what kind of bacteria or virus? Salad sauce? Yes, except that those pathogens that give you the shits can even survive the hydrochloric acid in your stomach. Otherwise, ... well, it's obvious.

Acetic acid boils at higher temperature than water, at 118 °C. Acetic acid does not react with the air.
It is more likely that acetic acid antimicrobial properties will improve as more water evaporates, because it evaporates faster.

Thank you for your answer.  I think they key to your answer is the part where you wrote "Acetic acid does not react with the air". Assuming you are correct (and I have no data to the contrary), why does the smell of vinegar decrease the longer it sits exposed to air? As a simple test, you can pour 0.5 litre of common white vinegar into a wide bowl (to maximize surface area exposed to the air). After 1min, smell the air around it at a distance of 20cm.  Repeat smelling the air after 1hr, then 12hr, then 24hr, then 48hr.  When I did this, I noticed a significant decrease in the odor over time.

If you are nearby this bowl, you will lose ability to smell this particular scent. Human body usually reduces percieved intensity of persistent stimulus, not just scents. You could test it by asking your friend to pour another bowl with new vinegar and let you try to distinguish it blindfolded, if im right you will not guess it better than 8\10. Another option is flavorings, if it is kitchen vinegar it may have something other than vinegar itself, and you sense that instead.

Interesting thoughts... thank you.  To reply, I am not regularly within 3m of the test bowl, so I don't think that's it.  The blindfolded test you propose (is that a single-blind test? LOL!) is a great idea, and I may give that a try.  Regarding flavorings, I will take a close look at the bottle, but the CH3COOH + H2O I tested was not cooking or cleaning vinegar.  It should be nothing more than CH3COOH + H2O, but I need to confirm.


text:
SN2 reaction with alkyl bromide and pentanenitrile?
Is it possible to do a Friedel-Crafts alkylation on a plain alkane such as the one I've attached below, or is it only reserved for EArS mechanisms?
Also, I was wondering what would happen if I tried this reaction with another halogen on the starting substrate (if I had 1-bromoethane for example)...

Edit 1: I realize now that the top reaction doesn't make sense. Rather than a Friedel-Crafts Alkylation, I think I found a mechanism to make this type of reaction work (SN2) with Br as a leaving group and CN as a strong nucleophile, but I'm not sure...

Edit 2: I miscounted the carbons.

Edit 3: This is what I understand, from the suggestions given...

I assume you've seen the FC alkylation mechanism before. How do you think it could be extended to this? Can you draw out a proposed mechanism?

For a moment, I thought it were an example of ~Wurtz-Fittig reaction~, but your reaction does not include elemental sodium.

yes, I don't think that my first attempt makes sense. I tried another mechanism in my edit to form a different product.

The edit also does not make sense. Count the number of carbons.

oops I missed that. but can that mechanism still work?

I still don't think so. There's branching in the product and not in your substrates. The general reaction works, but not the specific one you wrote.

You need strong base to generate an anion next to the -CN then the general reaction can work (but not the example you have drawn)

Why are you showing a dashed C-CN bond? Any product will be a racemate as described.

@user55119 that was the stereochemistry of the given product

@Waylander Oh ok, I thought that the CN- could act as the strong base, can I add any strong base like NaOH for example or...?

@Zhe sorry, I'm not understanding, my logic was that the pentanenitrile would rearrange to form the product that I put (like as a major product)... is it my substrate that is wrong? or am I missing something else?

@chemquestionquest09  NaOH is not strong enough, try lithium diisopropylamide

If you do the reaction the way you describe, then only 50% of the product will have the correct stereochemistry. Your alkylation should have a normal line.

@user55119 I'm not sure I understand what you mean by 'normal line'... this might be wrong, but I thought that it would simply be an SN2 reaction (not an alkylation) since I'm not using Cl, Br etc. nor AlCl3. -Referring to Edit 2

A "normal line" is like all other single bonds in your structures. Such a bond is the equivalent of 50% dashed and 50% bold, a racemate. The alkylation you are doing will not give only one of them.

You need to re-edit the placement and number of carbons in your nitrile reactants.

@user55119 ok, I'll try and I'll make another edit, thank you for your help!

Edit 3 is still wrong. The starting nitrile is wrong. You started with 6 carbons and end with 7. Would you like an Answer?

@user55119 oh ok... and yes please I'm really unsure where exactly I'm going wrong. thank you!

It appears that you are having some difficulty with this synthesis. The suggestion in the Comments that you employ lithium diisopropylamide (LDA) to generate the anion of a nitrile is a wise one. The anion is usually formed a -78oC in tetrahydrofuran (THF). The bromide is added and the reaction mixture is allowed to warm to ambient temperature. There are two possible bonds that can be made in what is termed a retrosynthetic analysis. One by method A and another by method B. The bonds formed are shown in red in structures 2. Method A is better than method B because an SN2 reaction is being conducted on a primary halide, ethyl bromide. Method B requires an SN2 reaction on a secondary halide, 2-bromopropane, a process that is hindered and in competition with an E2 elimination that would produce propene 4 and the starting nitrile 3. Nitrile 2 must be a racemate given its method of formation. It is comprised of equal amounts of the two enantiomers, (R)-2 and (S)-2. By the methods employed here, a single enantiomer cannot be formed. Thus, a "normal bond" (not dashed nor bold) is shown in nitriles 2.

yes, I am struggling with this, thank you so much. so in order to get the desired product I absolutely have to start with the nitrile on my substrate?

If you must start with nitrile 3, add the anion to acetone and dehydrate to the conjugated nitrile. Then the double bond has to be reduced without affecting the nitrile. Perhaps Pd/C and H2 in ETOAc at atmospheric pressure. But that's another question.

thank you!! I really appreciate all your help!


text:
Why does the plot of volume and temperature have a kink in a second order phase transition?

The $(b)$ part of the figure shows how thermodynamic variables change in a second order phase transformation.
We observe there is a kink in the plot of volume with temperature.
Why is that so?
The plot is from http://faculty.chem.queensu.ca/people/faculty/mombourquette/Chem221/5_PhaseChanges/PhaseThermo.asp

Because they are second order phase transitions? GO back to the definition of a second order phase transition, such as in https://en.wikipedia.org/wiki/Phase_transition.

Thank you Mr Jon Custer. I was interested in the actual mechanism as to why the kink was there.

An example for a), first order of phase transition, is the one about liquid water $\ce{->}$ gaseous water / steam; you have both a sudden jump in $\Delta{}V/\Delta{}T$ and the energy needed for this transition ends in latent heat.  Guess what, not all transitions involve latent heat (i.e., compare with your lecture notes, references at disposition, etc.).

Thank you buttonworth. I was interested in the second order transition in which there isn't a jump but just a bend in the plot of V with T

@YasirSadiq - because instead of a jump in volume, there is a jump in the derivative of volume (for example). And for a third order phase transition there would be a jump in the second derivative of volume... That is how the order of a phase transition is defined.


text:
How do you break down <10% H202 and <10% NAOH
I have a waste discharge that will generate 40 litres of <10% H202 and <10% NAOH at separate times.  I have authorization to discharge to the sanitary drain, but I am concerned about degradation of the cast iron sanitary line.  What is the best way to break down these two compounds?

I think you shouldn't be concerned unless this will be a recurrent task.

I would dilute the hydrogen peroxide in a large open container with water. Then, add a piece of iron metal that was previously heated in air. The $\ce{FeO}$ coating will cause the breakdown of the $\ce{H2O2}$ (reference, see this work and also Wikipedia comments here).
An alternate path that liberates oxygen is via the addition of a dilute solution of Sodium hypochlorite ($\ce{NaOCl}$) the active ingredient in chlorine bleach. The residual product is salt water.
Also, to address the safe disposal of the caustic Sodium hydroxide, I would similarly dilute the $\ce{NaOH}$. Proceed to add a convenient acid source, say vinegar or lemon juice to effect a safe neutralization.

Apparently it is not $\ce{FeO}$ that catalyzes decomposition of $\ce{H2O2}$. It is either $\ce{Fe3O4}$ or goethite $\ce{FeO(OH)}$

@Maurice: Any ferrous salt, as noted in the Wikipedia reference, will work. It is Fenton chemistry, see https://www.pnas.org/content/111/2/623. The particular compounds you noted have mixed oxidation states of iron including Fe(II).

I would add further that the particular mixed iron salts you mentioned are especially effective (as in energetic). No likely the best choice here, especially if the H2O2 is not well diluted, but thanks for the comment as it does highlight a potential safety issue.

If it is possible to store one waste stream until the other is generated, simply mixing the two waste streams might be effective. Peroxide solutions are usually maintained at acidic pH because the hydroperoxy anion is unstable and breaks down to $\ce{HO-}$ and $\ce{O2}$ through the reaction
$$\ce{H-O-O- + H-O-O-H -> H2O + O2 + HO-}$$
So adding base to your peroxide solution until the pH is high enough to generate hydroperoxy anions will result in their destruction (and neutralize the base you are adding). NOTE THAT THIS REACTION SHOULD BE DONE SLOWLY AND NOT WITH HIGHER CONCENTRATIONS OF PEROXIDE, as a fair amount of heat is generated. The vessels must be vented as well, as the gas production can build up pressure rapidly. But with your low concentrations, the safety hazards are fairly low.
If you still have NaOH solution leftover or the mix is too high pH, simply bubbling carbon dioxide or adding solid sodium bicarbonate will neutralize it.
You can find more about the above reaction here

My experience with dilute 3% H2O2 (which is normally acidic for stability reasons) to which is added some NaHCO3 results in a mild reaction. The H2O2 now, however, per testing with NaOCl, displays a much-reduced shelf-life. So, this is an alternate safe path as suggested by Andrew (namely, raising the pH of a diluted H2O2 solution).


text:
Synthesizing 2-bromoaniline from bromobenzene
I am trying to synthesize 2-bromoaniline from bromobenzene. I came up with a reaction path, but I want your thoughts on whether this is feasible and can produce a decent yield.

Sulfonation of bromobenzene (to block the para position)
Nitration of the product from (1). This should produce 4-bromo-3-nitrobenzenesolfanic acid.
Reduction of the nitro group via Zn(Hg) and HCl (to turn the NO2 group into NH2).
Reverse sulfonation (to get rid of the sulfonic acid group).

And the final product from this should give 2-bromoaniline. Should this theoretically produce a decent yield even if bromobenzene is slightly deactivated?

It will be much quicker to run the direct nitration of bromobenzene and isolate the ortho nitro product by chromatography. It is the minor nitration product, but your overall yield will be better by cutting out two steps which will give only moderate yields.
Nitration and chromatography procedure here 1


text:
What is the form of the Hamiltonian for solids?
For atoms, and even molecules, I can  understand how the Hamiltonian would be constructed, but what of solids (as in, for electronic structure calculations)? Thank you for any help.

What difference do you expect in the Hamiltonian of solids compared to molecules? Also, tagging this with DFT makes even less clear that what kind of Hamiltonian you are talking about.

Look into Bloch functions for crystals - the symmetry of the crystal is explicitly introduced into the Hamiltonian. Perhaps more solid state physics than chemistry...

There are multiple possibilities. First of all, you can construct them like the ones for molecules, since a solid is also made from nuclei and electrons. So the Hamiltonian looks identical, but with much more terms.
On the practical side, you would do something different. If, for example, your system is periodic (think of a crystal), then you would choose a single elementary cell from the crystal and build the Hamiltonian for that (ignoring all the other cells), so you arrive at a small Hamiltonian.
Then to get the wavefunction for the whole system you do not change the definition of the Hamiltonian but the one of the wavefunction. While for molecules the wavefunction has to vanish at infinity, you now want the wavefunction to have special properties, called periodic boundary conditions:
For example, consider a one-dimensional crystal with unit cell from 0 to L. You now want the wavefunction $\psi$ at 0 and L to be equal, as well as the derivatives of $\psi$ to be equal at these points:
$\psi\left(0\right)=\psi\left(L\right), \frac{\partial\psi}{\partial x}|_{x=0}=\frac{\partial\psi}{\partial x}|_{x=L}$
With this information you can determine the wavefunction in this cell with any method you want. Now you could put this exact wavefunction into the cell from L to 2L and and the given wavefunction would be a solution for this second cell, since the cells are identical. This way you can get a wavefunction $\Psi$ for the whole system, by periodically putting the wavefunctions for the unit cells $\psi$ together:
$\Psi\left(x\right)=\psi\left(x-n\cdot L\right), x\in\left]n\cdot L,(n+1)\cdot L\right], n\in \mathbb{Z}$
The interaction of the wavefunction with the other nuclei in the crystal is taken care of by the periodic boundary conditions, so the wavefunction will be the real solution to the Problem.


text:
What functional groups can 900-2500nm IR spectrometers detect?
3700 cm-1 corresponds to about 2,700 nm
650 cm-1 corresponds to about 15,000 nm.
Many cheap IR spectrometers ($7000 or so like the ones at Stellarnet) has range of 900nm to 2400nm (or 11,111 cm-1 to 4166 cm-1). In the following table or list of the molecules functional groups, the range is between 2700nm to 15,000 nm (3700 cm-1 and 650 cm-1). Outside of the range of cheap IR spectrometers. So what functional group or stuff can you see with IR spectrometers that has range below 2400nm (above 4000 cm-1)?
https://www.sigmaaldrich.com/technical-documents/articles/biology/ir-spectrum-table.html

see the link above for complete list

You recall vibrations IR spectrometers detect in the range of $4000\dots\pu{800 cm^{-1}}$ typically are either fundamental vibrations and combinations.  Tuning for shorter wavelengths / higher wavenumbers however renders recording absorptions about overtones instead of fundamentals more likely.  On occasion, you may encounter this part of the electromagnetic spectrum accessed by the larger UV-Vis-NIR spectrometers (e.g., PerkinElmer Lambda series) with a spectral range from $175\dots\pu{3300 nm}$); here, NIR is the label about near infrared.
The typical overtone is less probable than the fundamental, thus -- in parlance of Beer-Lambert's law, $\mbox{Abs} = \varepsilon \cdot c \cdot d$ -- the molecular absorption coefficient $\varepsilon$ is smaller.  While this may appear on first sight as a drawback (at same analyte concentration $c$ and  pathlength $d$, this lowers $\mbox{Abs}$ recorded), it very well may become an advantage if you want to record absorption bands about matter in high concentration, e.g., to check the quality of grain and crops in general  where NIR spectroscopy is used as such.
IR spectroscopy has a long tradition of chemometrics, e.g. in the on-line process control of petroleum fractions where whole spectra are used to determine concentrations of components (e.g., keyword principal component analysis) allowing to map a sample for its chemical composition as tool in histology (a review).  Again, tuning the senors to record NIR instead of the range of $4000\dots\pu{800 cm^{-1}}$ allows to see some things literally in a different light (e.g., Landsat satellites, farming).

(credit)


text:
Stress-Strain Graph Under Uniaxial and Biaxial Strain
I was going through a recently published journal paper by ACS (Applied Nano Materials) about gas separation using nanoporous Graphene.
DOI No. - https://dx.doi.org/10.1021/acsanm.0c02340
Supporting Information - https://pubs.acs.org/doi/suppl/10.1021/acsanm.0c02340/suppl_file/an0c02340_si_001.pdf
I'm quoting the exact part where I'm facing some problems-

Figure 2 shows the stress-calculated strain curves of graphene under
uniaxial and biaxial strains.   Herein,
nominal strain and stress are used, which are defined as the length
change divided by the original length and the tensile force divided by
the original area of cross-section, respectively (nominal thickness of
3.4 Å is adopted).
Figure 2a shows that the mechanical responses of nanoporous graphene
stretched along the armchair direction become distinct from that along
the zigzag direction with the increasing strain, resulting from the
strain-induced symmetry breaking. The strain to failure of this
nanoporous graphene loaded along with the armchair and zigzag directions
are 11 and 17%, respectively, which defines the maximum strains that
can be applied to the structure. In addition to uniaxial strain, we
also investigate the elastic responses of biaxial strain and the
strain to failure can also exceed 10% (Figure 2b). Considering the
structural stability based on these investigations, we explore the gas
separation performance of nanoporous graphene under uniaxial and
biaxial strains below 10%.

However, I understand that this is a Stress vs Strain graph. But I have some doubts regarding this -

What they mean by uniaxial and diaxial strain
What exactly they mean by armchair direction and zigzag direction
Lastly, it'll be very helpful for me if one can explain the graph clearly (I'm getting a vague understanding about that)

AC and ZZ: https://commons.wikimedia.org/wiki/File:Graphene_edge_names.svg To find the zig-zag direction find a direction where you can move in almost a straight line, with only a small alternating left-right-left-right zigzag motion. Perpendicular to that is the armchair direction where you have two atoms in a straight line, then a larger jump left-right, then two atoms in a straight line... It looks like hexagonal armchairs where people are seated looking alternately to the left and to the right of the direction of motion.

I've always assumed that in the honeycomb net context "uniaxial strain" is strain in either AC or ZZ direction onlybut not both, and that biaxial strain in this context means strain in both AC and ZZ directions but not necessarily the same amount. If it's the same in both I believe it's called "uniform strain" or "equi-biaxial strain" but I don't know for sure. https://www.ncbi.nlm.nih.gov/pmc/articles/PMC4678326/figure/f1/ In that image $Exx$ is in the AC direction and $Eyy$ is in the ZZ direction.

Thanks, @uhoh...you helped me a lot...Those answers cleared all my doubts...

That's great! But let's wait for some confirmation from others. I'm no expert in this, it's just my current understanding.

Sure!!!...but I think your understanding is right as it successfully explains the graph as well as the results...


text:
Organic Nomenclature: But-1-en-3-yne or But-3-yne-1-en
Just wondering whether "but-1-en-3-yne" or "but-3-yne-1-en" follows the correct IUPAC naming convention for organic molecules. I'm going through a text book that uses the former in an example, but in the answer to one of the questions it uses the latter, so is it interchangeable or is one just wrong?

The IUCN Blue Book says that:

P-31.1.1.1 The presence of one or more double or triple bonds in an otherwise saturated parent hydride [...] is denoted by changing the ending ‘ane’ of the name of a saturated parent hydride to ‘ene’ or ‘yne’. Locants as low as possible are given to multiple bonds as a set, even though this may at times give ‘yne’ endings lower locants than ‘ene’ endings. If a choice remains, preference for low locants is given to the double bonds. In names, the ending ‘ene’ always precedes ‘yne’, with elision of the final letter ‘e’ in ‘ene’. Only the lower locant for a multiple bond is cited, except when the numerical difference between the two locants is greater than one, in which case the higher locant is enclosed in parentheses.

Therefore the double bond gets priority whatever be its locants (numbering) i.e., '-ene' comes before '-yne'. And in this case it's "but-1-en-3-yne".

I would like to know why I got downvotes. It seems like there's nothing wrong in my answer.

If someone else too have misunderstood the meaning of my answer or @An0n1m1ty's question I would like them to cancel their downvotes. If it's something else, please tell me so that I can improve my answer.

@andselisk , I've added the link to Wikipedia. The reason I didn't add it earlier was that it would be quite hard to find that single sentence in that big document (just like looking needle in a haystack). Is there any more changes required?

Your answer is very terse. Depending on the question, sometimes that is fine, but such cases are quite rare, and you will find that well-received answers on Stack Exchange are nearly always quite thoroughly elaborated. (And conversely, one-sentence answers are likely to draw downvotes from passers-by, regardless of whether they are correct.)

Regarding what you could do to improve this particular answer, there are two things: (1) quote the relevant section; that also avoids forcing the user to "find a needle in a haystack". (2) Use the IUPAC Blue Book as a reference, rather than Wikipedia or other websites which often have outdated or simplified rules (after all, the Blue Book is >1000 pages long). I acknowledge this is not possible for most people, but the bar for IUPAC nomenclature answers on SE has already been set at a very high level, see e.g. the answer to the duplicate question, which does both of the things I suggested.

@orthocresol, I did search it in the IUPAC website but they showed me no results and that's why I added the link to Wikipedia. And believe me, I never knew there was something like a Blue Book until you told me. At school, we are just taught how to name these compounds and not where these things are written. Moreover, when I thought of quoting it from the book it was a kind-of-impossible for me to find that in the book, as you told. It would be quite helpful if you could explain what you meant this 'but the bar for IUPAC nomenclature answers on SE has already been set at a very high level'.

Many of the existing IUPAC nomenclature answers are generally very high-quality. To live up to that standard, therefore, you need to produce quite a good answer. Anything less may not be received so well (hence the downvotes you got). In a way, the community is spoilt, but in my opinion that's a net benefit. BTW, it's IUPAC, not IUCN. Unless we are considering endangered molecules ;)


text:
Carbon monoxide treatment
When CO is the by product of certain reactions( burning of coal under limited air supply), how is it treated before releasing in the atmosphere or is it converted to some other form? Can that byproduct be used to synthesize some other compound ?

AFAIK, production scale processes are highly optimized, producing just traces of $\ce{CO}$. There traces are either not managed, or can by passed through catalyzators.
On the other hand, burning coke with limited air or oxygen $\ce{2 C(s) + O2(g) -> 2 CO(s)}$ or passing water vapour through burning coke $\ce{C(s) + H2O(g) -> CO(g) + H2(g)}$, are used intentionally, to produce $\ce{CO}$, eventually with hydrogen, for further synthesis.

Scrubbing by a highly corrosive copper chloride solution. The industrial application of this method has long been discontinued. (See this patent for more information)
Catalytically converting $\ce{CO}$ into $\ce{CO2}$, according to the water gas    shift reaction.
Catalytic methanation of $\ce{CO}$ with hydrogen to methane, forming a    synthetic natural gas.
Scrubbing with liquid nitrogen

Source: https://www.eolss.net/Sample-Chapters/C09/E4-14-02-01.pdf

The need to do something about CO emission and the methods used to reduce the emissions depend strongly on the circumstances.
Poutnik is right that large scale processes will be both optimised and designed to minimise emissions, there are other, different, mechanisms and solutions in other processes.
A major example is vehicles whose engines produce notable amounts of CO because combustion is never fully controlled and where the optimal power output, efficiency and emissions of other pollutants leave notable amounts of CO in the exhaust gases. Modern cars use a variety of types of catalytic converter in the exhaust path to minimise the ultimate CO released.

To answer the question: "how is it treated before releasing in the atmosphere or is it converted to some other form?", a direct answer is that a significant part (if not a majority) of the created carbon monoxide is not treated.  Hence, reported significant poor air quality issues in select locales, as per a featured story, Helping China Fight Air Pollution, as an example, to quote:

Coal is the single-largest source of air pollutants and greenhouse emissions in China. The country heavily relies on coal to meet its energy demand at 66% of the energy mix. China consumed approximately 4 billion tons of coal in 2014— more than the rest of the world combined. Half of China’s coal is used for decentralized boilers in the industrial and residential sectors.

However, courtesy of the following photochemical reaction naturally occurring in the atmosphere, some mitigation occurring, to quote from a Science Direct source:

The prevailing view is that $\ce{CO}$ is oxidized by $\ce{.OH}$ via

$\ce{ CO + .OH -> CO2 + .H}$

The $\ce{.OH}$ itself is ultimately derived from the photolysis of water vapor. Thus, even though water vapor is a minor constituent, it may play a very important role in maintaining $\ce{CO2}$ as the dominant constituent. Support for the importance of this water chemistry comes from the detection of molecular hydrogen ($\ce{H2)}$ in the atmosphere and the distribution and abundance of ozone.

I would also note more commonly occurring in the surface waters on earth as a matter of academic interest (as detailed here and not commercial paths), reactive oxygen species including the hydroxyl radical can be formed in a Fenton (or Fenton-type) reaction of transition metal salts (like with a soluble cuprous complex) with acidic hydrogen peroxide (or $\ce{H2O2}$ as sourced from oxygen). Or relatedly, per the above-cited atmospheric chemistry, one can even replace photolysis with sonolysis of the water (which jointly creates $\ce{.H}$ and $\ce{.OH}$).