text:
How to remove the water from soap
I find when making soap that a difficult part is removing the water.
Currently what I do is melt the soap and then put the melted soap into a vacuum chamber. This draws the water out by vacuum. You cannot boil the water out of the soap at atmospheric pressure because that would damage the soap by the high heat.
The problem with using the vacuum chamber is that as the pressure falls, the soap rises up like an angel food cake and can easily assume 7 or 8 times its original volume. This means I have to have a gigantic container for a relatively small amount of soap. Also, the soap cools as it rises, so it solidifies before all the water is removed. Then, the process has to be repeated, over and over again, removing more of the water each time.
Is there any easier method I can use to remove the water from the soap.

Are you familiar with th salting out concept? If you add salt to a solution containing an organic substance, it separates out from water. Soaps easily separate out too. I have not tried this, but I recall reading a long time ago from a school textbook. Read more about salting out soaps.

Or use gentle heating while gradually increasing vacuum. It's more efficient and should bubble less.


text:
Separating sodium sulfate from chromium trioxide solution
I was looking at the industrial process for the production of chromium trioxide by treating sodium chromate or the corresponding sodium dichromate with sulfuric acid. This results in the formation of chromium trioxide as well as sodium sulfate. I think both of them would be mixed as a solution; so what way is employed to separate them? I read about the process in Wikipedia.
I thought of using methods which I know like distillation, but I don't know how can I check if they work practically and are employed industrially. I searched the web as well and couldn't find anything satisfactory.
I am in 12th grade and I haven't really studied this topic, so I didn't know how to approach this question.

Comments are not for extended discussion; this conversation has been moved to chat.

The overall reaction is:
$$\ce{H2SO4 + Na2Cr2O7 → 2 CrO3 + Na2SO4 + H2O}$$

How to separate chromium trioxide and sodium sulfate from solution?

Since chromium trioxide is soluble, it exist as chromic acid in the solution. So, in the reaction pot, there are compounds like  $\ce{H2CrO4, H2Cr2O7}$ along with our target compound $\ce{CrO3}$, all in equilibrium. But with the addition of a large excess of concentrated sulfuric acid, water is withdrawn from the hydrated forms and the chromium trioxide separates from the solution in the shape of red needles. Successively, the reaction pot is made to stand overnight for the chromium trioxide to precipitate slowly in form of needles and then filtered off from the solution using Buchner filter. The synthesis procedure of chromium trioxide can be found in prepchem.com:

100 grams of sodium dichromate are dissolved in 250 ml of water and
the obtained solution is filtered. 200 ml of concentrated sulfuric
acid are dropwise added with constant stirring until a slight
permanent precipitate of chromium trioxide is formed. The mixture is
cooled for half an hour or longer, then slowly, while stirring, 200 ml
of concentrated sulfuric acid are added dropwise. The reaction mixture
is left to stand over night in order that the crystal meal may become
somewhat coarser. The larger crystals (and purer) of chromium trioxide
could be obtained by heating the solution with crystal meal with
stirring to $\pu{100 ^{\circ}C}$ and allowing to cool slowly. If this process is
repeated once or twice, a more satisfactory product will be obtained.
The obtained crystals of chromium trioxide are filtered with a Buchner
funnel with a sintered glass disc instead of the usual paper filter
(chromium trioxide is a strong oxidizing agent and will oxidize
organic material).  After draining the crystals completely and
pressing the surface and treating with 15 ml of concentrated (65-70%)
nitric acid. The nitric acid is removed by suction and the operation
is repeated twice with 10 ml of nitric acid each time. Finally, the
red crystals of chromium trioxide are dried by suction as much as
possible, transferred in a dry evaporating dish and by gentle heating
the remaining nitric acid is removed. Dried product is stored in a
stoppered bottle. Yield is 60%.

Note: chromium trioxide is toxic, corrosive, a powerful oxidizer and a potent carcinogenic. Sulfuric acid is corrosive and can burn your skin if you are careless. Handle those chemicals with care. Perform this synthesis only after you become a qualified chemist. This answer is only for educational purpose only.

Thanks for the answer btw Is this the industrial method or a laboratory method?

It depends on the scale of production. The reaction is same in both methods.

@NilayGhosh, Google "Sheharbano Sangji" accident in a university lab. The OP is a grade-12 student and we do not know who the other random readers are. This safety note is critical but not complete. You cannot imagine how strict the rules are esp. in the Western universities.

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text:
Why is phosphorous pentoxide a dehydrating agent?
$$\ce{HNO3 + P2O5 -> 2HPO3 + N2O5}$$
In this reaction, both nitrogen and phosphorus have the same oxidation number before and after the product is formed. Since the $+5$ oxidation of nitrogen is unstable, it has to undergo reduction, but here it remains the same after the product is formed. So what makes the reaction go forward?
Is it because of the dehydrating property of $\ce{P2O5}$?
Why does this compound have this property and why doesn't $\ce{N2O5}$ have this dehydrating property?

Somewhat related: https://chemistry.stackexchange.com/questions/33742/what-makes-a-compound-a-dehydrating-agent/125598#125598

and: https://chemistry.stackexchange.com/questions/851/differences-between-koh-and-phosphorus-pentoxide-as-drying-agents

But they all state this just as a fact, What about N2O5? It is unstable than P2O5 because of the oxidation no it has, but in the reaction I mentioned, the product has the same oxidation number as before. What gives P2O5 this property?

N2O5  does react with water but not as exothermically compared to P2O5 who is just looking for water. So, it does not take role as dehydrating agent.

The true molecule is P4O10,not P2O5.

First of all, your reaction is wrong. $\ce{HPO3}$ does not exist as a single entity. It is actually a cylic molecule of formula $\ce{(HPO3)_n}$. The simplest such compound is $\ce{(HPO3)3}$ which is called trimetaphosphoric acid. See this answer for more details.

So, what happens is that phosphorus pentoxide is very powerful dehydrating agent and can rip off water molecules from compounds. Phosphorus pentoxide can dehydrate nitric acid to form its anhydride, dinitrogen pentoxide.
$$\ce{2HNO3 ->[P2O5][- H2O] N2O5}$$

Why dinitrogen pentoxide is not used as a dehydrating agent?

My take on this is that hydrolysis of dinitrogen pentoxide is not that exothermic as compared to hydroysis of phosphorus pentoxide. Hydrolysis of phosphorus pentoxide is very exothermic, releasing so much energy to dehydrate compounds to form corresponding anhydrides and phosphoric acid.
$$\ce{P4O10 + 6 H2O → 4 H3PO4   (–177 kJ)}$$
Phosphoric acid is hygroscopic which further sucks remaining water from compound. If we compare all the dehydrating agents, phosphorus pentoxide will be on a different league. Dinitrogen pentoxide is nothing.
Nitric acid also hydrolyses to form nitric acid but the reaction is not that exothermic. Also, nitric acid is not that hygroscopic as compared to phosphoric acid for which it is not that great of a dehydrating agent. But hydrolysis of dinitrogen pentoxide has its importance. A solution of dinitrogen pentoxide and nitric acid is a good nitrating agent and hence $\ce{N2O5}$ is known as a nitrating agent than a dehydrating agent. Also, this hydrolysis reaction is a basis of forming atmospheric aerosols ($\ce{HNO3}$ is a good $\ce{NO_x}$ reservoir).

"phosphorus pentoxide is a water-loving compound" is descriptive, but doesn't answer the core question of why the reaction with water is so favorable, especially as compared to that of N2O5.

@Andrew removed that statement and added an explanation of how P2O5 removes water and how it can be a good dehydrating agent based on that and also why N2O5 cannot achieve that.

An important experimental consideration is that $\ce{N2O5}$ is a volatile solid that melts at 41 degrees C and boils at 47 degrees C. If this reaction is run at a moderately elevated temperature (but below the boiling points or decomposition temperature of the other components), $\ce{N2O5}$ is readily removed, and Le Chatelier's principle drives any equilibrium that might be established.

The reaction given in NCERT, Shriver Atkins, and Greenwood is $\ce{P4O10 + 4HNO3 → 2N2O5 + 4HPO3  }$ Why do you think this is wrong? Can you cite a source?

@Wolgwang HPO3 is not a real entity. It is $\ce{(HPO3)_n}$. The full reaction is $\ce{P4O10 + 12 HNO3 → 4 H3PO4 + 6 N2O5}$. $\ce{P2O5/P4O10}$ is hygroscopic so it sucks water and form $\ce{H3PO4}$ Source: Holleman et. al. (2001), Inorganic Chemistry. Don't trust NCERT.

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Effect on mass when decreasing pressure in equilibrium reaction
For example, if we have the reaction:
$$\ce{A + B <=> C}$$
And the volume is decreased, the pressure would increase. Hence, according to Le chatelier's principle, the system will partially oppose the change by favoring the forward reaction to produce more $\ce{C}$. I have a few questions:

Would the mass of $\ce{C}$ increase?

I think yes, since the forward reaction is favored

Would the concentration of $\ce{[A]}$ be greater when compared to the original concentration of $\ce{[A]}$?

I think yes, since the volume is decreased.

Would the amount of $\ce{A}$ be reduced?

I think yes, since it's being consumed for the forward reaction.

Is this in gaseous or aqueous medium? If aqueous, then using pressure wouldn't make much difference since water is practically incompressible. If gaseous, there is no concept of concentration only partial pressures.

You don't mean 'mass' but the amount of C otherwise you are correct. Point 2 will be true initially but may not be as the reaction reaches equilibrium.

Assuming $\ce{A, B,}$ and $\ce{C}$ are gases and behave like ideal gases we can conclude following by applying $P_xV=n_xRT$ where $x$ is either $\ce{A, B,}$ or $\ce{C}$, and $V$ and $T$ are constants:
$$[\ce{A}] = \frac{n_A}{V}=\frac{P_A}{RT}; \quad [\ce{B}] = \frac{n_B}{V}=\frac{P_B}{RT}; \quad \text{and } \quad [\ce{C}] = \frac{n_C}{V}=\frac{P_C}{RT} $$
Since $\frac{1}{RT}$ is a constant, above relationships mean concentration of each species id directly proportional to its partial pressure.

Question 1: Would the mass of $\ce{C}$ increase?

Assuming you means amount of $\ce{C}$, your reasoning and conclusion are correct. When the volume is decreased in gaseous system, the pressure would increase. Hence, according to Le Chatelier's principle, the system will partially oppose the change by acting to reduced the pressure. To do so, it favors the forward reaction producing more C (the major reason is $\pu{2 mol} \rightarrow \pu{1 mol}$).

Question 2: Would the concentration of $[\ce{A}]$ be greater when compared to the original concentration of $[\ce{A}]$?

It would be greater just after you increase the pressure by reducing the volume. However, when the Le Chatelier's principle applies, amounts of $\ce{A}$ and $\ce{AB}$ would decrease (see answer to Q1). However, as I prove above, concentration is directly proportional to partial pressure. When $n_A$ decreases, $P_A$ decreases so does $[\ce{A}]$. However, whether $[\ce{A}]$ reduces past its original value or not can only be determine by knowing the reaction conditions, specially its $K_c$ or $K_p$ values.

Question 3: Would the amount of $\ce{A}$ be reduced?

Answer is yes as you figured out. See answer to Q2 for more details ($n_A$ means amounts of $\ce{A}$).

For question 2, with a small decrease in volume, I can prove that, at the new equilibrium, the increase in concentration of A due to the volume decrease wins out over the decrease in concentration of A from the reduction in the number of moles of A. This conclusion is independent of the equilibrium constant.  So the concentration of A will increase with a decrease in volume.

@Chet Miller: I clearly said it depends on the condition of the reaction. So volume change isa condition. The equilibrium constant cannot be disregarded, It plays a significant role. Why don't you put your thoughts and proofs as an answer? That'd be helpful.

@MathewMahindaratne Thank you for the proof and the direct answers to my questions. Really appreciate the clarification!!! Thank you again!

Done.  See my Answer.

This is an approach to Question 2.
The reaction equilibrium condition for the system is:  $$\frac{V}{RT}\frac{n_C}{n_A n_B}=K_P\tag{1}$$Let $n_{A0}$, $n_{B0}$, and $n_{C0}$ be the number of moles at equilibrium for each species when the volume is equal to $V_0$.  Then we have
$$\frac{V_0}{RT}\frac{n_{C0}}{n_{A0} n_{B0}}=K_P\tag{2}$$.  For a small change in volume dV, we can write the new number of moles of A, B, and C at equilibrium as
$$n_A=n_{A0}-dn\tag{3a}$$
$$n_B=n_{B0}-dn\tag{3b}$$
$$n_C=n_{C0}+dn\tag{3c}$$Combining Eqns. 1-3 yields:$$\left(1+\frac{dV}{V_0}\right)=\frac{(1-dn/n_{A0})(1-dn/n_{B0})}{(1+dn/n_{C0})}$$And, linearizing this equation with respect to dn yields:$$\frac{dV}{V_0}=-\left(\frac{1}{n_{A0}}+\frac{1}{n_{B0}}+\frac{1}{n_{C0}}\right)dn$$This equation indicates that if dV is negative, then dn is positive, and the reactants decrease while the product increases.
If we apply this same methodology to the concentration of A, we obtain
$$dC_A=-\frac{n_{A)}}{V_0}\left(\frac{dn}{n_{A0}}+\frac{dV}{V_0}\right)$$Combining this with the previous equation then yields:
$$\frac{dC_A}{dV}=-\frac{n_{A0}}{V_0^2}\left[\frac{1/n_{B0}+1/n_{C0}}{1/n_{A0}+1/n_{B0}+1/n_{C0}}\right]$$The term in brackets is always positive, so, when the volume decreases, the concentration of A increases at all points along the equilibrium contour.  This conclusion is independent of the specific value of the equilibrium constant.

Thank you for the kind response. I really appreciate the proof. This has clarified a lot! Thank you again! @Chet Miller

You got my vote.


text:
Lanthanum Electronic Configuration
Lanthanum has electronic configuration [$\ce{Xe}$] $4\mathrm{f}^0$ $5\mathrm{d}^1$ $6\mathrm{s}^2$. Why is the $4\mathrm f$ sub-shell not filled even though we know that $4\mathrm f$ is filled before $5\mathrm d$ sub-shell?
I know that for some elements energy level of $4\mathrm f$>$5\mathrm d$. Why is this so?

If a question is asked on Chemistry SE site, then, in contrary to sites like Quora, it is expected the author elaborates the topic in the question by at least basic own topic review and writes what he/she understands and what is the eventual stumble stone. The "quick questions" without explicitly expressed particular effort are not very welcome, and may be closed.

Question titles should be like book titles. They distinguish the book from other books, give a hint about the content, but do not tell the full story. The content then should elaborate the topic to full depth the author is able to do. By other words, the title too long, the body too short.

Remember that energy of electrons in orbitals do not have fixed values, but depends on occupation state of other orbitals.
So if we in a thought experiment fill orbitals to get a neutral atom, 1 electron in time, some orbitals switch the order of their energies.
It has the effect the 4s orbital is filled  first, having lower energy, but after some 3d orbitals are filled, energy of filled 4s Is higher then energy of 3d.
Similar case happens between 5f and 6d orbitals for first actinoids.

Sir can you elaborate it, I didn't understand.


text:
Calculation of rupture force in a covalent bond
Is there any way to theoretically or mathematically calculate the rupture force of a covalent bond?
One of the articles I saw said that it is the maximum derivative of the potential energy curve. Is this true?

A force is the derivative of a potential energy:
$$\vec{F} = -\vec{\nabla} U$$
If you choose to apply a continuous force to stretch a bond until rupture, the force will need to exceed the restorative force (as described say by Hooke's law for small displacements from equilibrium) due to mutual attraction of the atoms. The maximum value of that restorative force is the maximum value of the gradient of the potential, therefore:
$$\vec{F}_{\text{rupture, continuous}} = (\vec{\nabla} U )_{\text{max}}$$
Note a quadratic potential (such as leading to Hooke's law) does not predict rupture. You would require a more accurate description of the internuclear potential to compute the rupture force.


text:
Effect of adding an acid or base to water with respect to concentration of ions
So I know that there's auto-ionization of water according to:
$$\ce{2H2O <=> H3O+ +OH-} \tag{1}$$
If I have pure water where the concentrations of both $\ce{H3O+}$ and $\ce{OH-}$  is $10^{-7}$
Say I add $\pu{10^{-5} M}$ $\ce{HCl}$, I'd have [$\ce{H3O+}$] = $ 10^{-5} + 10^{-7}$ (approx $10^{-5}$) and the [$\ce{OH-}$] = $10^{-7}$ (still)
but since these are in equilibrium, according to reaction ($1$), the reaction will shift to the left so this would consume $\ce{H3O+}$ and $\ce{OH-}$ until their concentrations product is $10^{-14}$, how will this happen?
Will the $\ce{OH-}$ decrease much more less than what the $\ce{H3O+}$ will decrease, in a way that what the $\ce{H3O+}$ loses is negligible compared to what $\ce{OH-}$ lost? in a way that would keep [$\ce{H3O+}$] = $10^{-5}$ and [$\ce{OH-}$] = $10^{-9}$?
In brief, how is equilibrium brought back to the concentration of ions are concerned?

There is no "how" to it. The reaction just happens according to the equation above, as blunt as that. One H3O+ consumes one OH-. Can it consume two? No. There is nothing left to choose.

But when their product is brought back to $1* 10^-14 $, will the [H3O+] remain still $ 10^-5 $ ?

Pure stoichiometry. The decrease happens in the proportion specified in the reaction, so 1:1, until the equilibrium is re-established.

If you add HCl in pure water, this will immediately reduce the amount of autoionization of water.

I just don't get it how the equlibrium restoration affect the concentration of these 2 ions. Doesn't the equilibrium derived equation (product of concentration of these ions) = 10^-14 always apply? If so, when we add HCl of 10^-5 M equilbrium will be disrupted, and when concentrations are adjusted in a way to restore equilbrium and apply the product of concentrations of ions law again, what would the new concentrations be?

To find the new concentrations, you'll have to do some math.


text:
How can t2g orbital account for extra stability in absence of ligands?
I have read in a book that t2g orbitals sometimes account for stability in 3d block transition elements. For example,

Chromium in 2+ oxidation state shows reducing properties because Chromium in 3+ oxidation state will have half-filled t2g orbitals which will account for extra stability.
Vanadium has a low value of M+3/M+2 standard electrode potential because of half-filled t2g orbital.

What I have read is that the d orbitals in the absence of ligands have equal energy, and the orbitals split up into eg and t2g due to the presence of ligands. Then how can the half-filled t2g orbitals account for stability, even in absence of ligands?

Which book is this from?

Related(Possible duplicate):https://chemistry.stackexchange.com/questions/26837/why-is-cr%C2%B2%E2%81%BA-is-a-good-reducing-agent-but-mn%C2%B3%E2%81%BA-is-a-good-oxidising-agent/26841


text:
Reaction of acyl chlorides with diazomethane and silver(I) oxide

$$\ce{R-COOH ->[SOCl2,CH2N2][Ag2O/CH3OH] X}$$

While solving Carboxylic derivatives recently, I came across this.
so following the reagents one by one-
$\ce{SOCl2}$ will form acid chloride.
$\ce{CH2N2}$ (diazomethane) on the other hand reacts with carboxylic acid to give a methyl ester. but we are currently on acid chloride, this part is a little confusing to me.
$\ce{Ag2O}$ in aqueous/alcoholic medium should give ($\ce{CH3O-}$) ions.
How to proceed? What is wrong with my reaction mechanism?

Your first reaction is correct. You do get an acid chloride. The reaction would proceed with the first step being what you have proposed:
$$\ce{R-COOH ->[SOCl2] R-COCl}$$
Now, the next step is a reaction known as Arndt-Eistert Reaction:
$$\ce{R-COCl ->[CH2N2][Ag2O/CH3OH] R-CH2-COOCH3}$$
The mechanism for this reaction is as follows:

Note: Mechanism is from Organic-Chemistry.org

What this mechanism, which is correct, does not tell you is that a minimum of two equivalents of CH2N2 is required and the acyl chloride should be added -- slowly --so that the second equivalent of CH2N2 consumes HCl as CH3Cl and N2 at the second step. Inverse addition, allows for chloride formed in structure 2 to displace N2 and form the alpha-chloroketone.

Is the slow addition to prevent the formation of the alpha chloroketone and allow the reaction of Cl- with CH2N2. just want to make sure if I've got it right..

The slow addition is to avoid a rush of N2 gas and to insure an excess of CH2N2 is present. More than 2 equivalents is advisable to insure an excess of CH2N2 at the end of the addition. The excess CH2N2 is what deprotonates your second structure. If it weren't present, then the chloride can displace N2 in the second structure by an SN2 mechanism. From a practical standpoint, I would guess that you have never prepared diazomethane. Its dangerous stuff! So the less it is handled the better. Invariably, one would add the acyl halide to the CH2N2.

(continued) The excess CH2N2 essentially scavenges the HCl produced in the reaction. Diazomethane is a stronger base than the diazoketone so it, CH2N2, gets the proton. I know that your structures are from your link but numbered structures are a big help in discussions. – user55119 5 mins ago


text:
Product of the reaction of ethyl acetoacetate with alkaline benzoyl chloride followed by addition of ammonium chloride

What does this β-keto ester give when reacted with $\ce{PhCOCl/NaOH}$ followed by $\ce{NH4Cl(aq)}?$

According to me, the $\ce{PhCOCl}$ should react with $\ce{NaOH}$ forming benzoic acid which should attack the ester on the right, forming an anhydride and a ketone.
Also, whats the function of $\ce{NH4Cl(aq)}$? Is it a proton donor?

Beta-ketoesters are carbanion sources and would undergo nucleophilic acyl substitution with $\ce{PhCOCl}$.

The intent of the question is to make the anion of AAE and then benzoylate the anion. But is it C or O benzoylation? The NH4Cl is added to neutralize any bases.

@shreya $\ce{NH4Cl}$ is used to quench the reaction (neutralize any leftover base)

If I'm not wrong, $\ce{NH4Cl}$ is used as a proton donor where you don't want anything else in your reaction flask to be affected by the $\ce{H^+}$ conc. in the medium

Reagents should be added in an order here: $$\ce{SM ->[1. NaOH][2. PhCOCl] P} \ $$ You cannot add $\ce{PhCOCl/NaOH}$ together because they react with each other as noted.

See: Org. Synth. 1938, 18, 33; DOI: 10.15227/orgsyn.018.0033 for the synthesis of ethyl benzoyl acetate. Notice that ammonia and NH4Cl is used to cleave the acyl group.


text:
What is the physical explanation driving current flow in a concentration cell?
In a galvanic cell, current is driven because the species at the cathode is reduced by grabbing some electrons from the cathode. The cathode now lacks electrons so it pulls them from the anode. Now the anode needs electrons, so it pulls them from the species that will be oxidized in the other half cell. So, now we have a current!
Now here is a concentration cell:

As you can see. the anode and cathode are made of the same metal. Only the concentrations of the solutions are different. Using my logic from above, since the metals are the same, they should be pulling on the electrons with equal magnitude. So how come current still flows? In other words, why are the nickel ions at the cathode more "hungry" for electrons than the nickel ions at the anode?
My attempt: It has to do with the rates of the metal dissolving. If I were to cut the wire connecting the two half-cells, I would notice that the beaker with the lower concentration solution would be increasing in concentration (from solvating the metal) faster than the beaker with the higher concentration. This is because the lower concentration one is "further" from the solubility equilibrium.
When I connect them back together the metal at the anode dissolves faster, thus demanding electron recompense from the cathode metal.
Attempt 2: The solution with higher Ni ion concentration has overall more ions dissolved in it. Maybe all these ions create an electric field greater than the one on the other side, thus prompting electron movement.
Attempt 3: It has to do with kinetics. Lets assume that everytime a nickel cation collides with an electrode, it steals some electrons. The half-cell with the higher concentration will have a higher amount of collisions per second. Thus, more electrons are being "consumed" at the cathode than anode per second, creating a charge differential, which in turn generates a current.

@EdV Thanks for the answer. Do you know if my physical explanation (difference in solvation rates) was correct? Don't want to be building on faulty premises.

Actually, when the Nernst equation is applied to more concentrated solutions, the reaction quotient Q should be expressed as 'effective concentrations' or in terms of the 'activities of the electroactive ionic species'. So, adding MgSO4 to one of the two Nickel sulfate ion concentration cells may alter the solution's activity and be reflected in a voltage change (an interesting experiment).

Note, per my comment, I would further comment that adding an excess of MgSO4 could salt out some of the Nickel sulfate, resulting in a significant change in dissolved Nickel.

Start with a simple thought experiment: pour 100 mL of 1 M nickel (II) sulfate solution into a beaker and very carefully layer 100 mL of 0.01 M nickel (II) sulfate solution on top of the more concentrated layer. Then, even without convection or deliberate mixing, diffusion will, sooner or later, result in the solution having concentration of 0.55 M. In what follows, it is assumed that evaporation is negligible, even on long time scales.
Now consider Fig. 1 below:

In this concentration cell, it is assumed, for simplicity, that the solution volumes are equal in the two electrode solution reservoirs. The salt bridge is 0.55 M nickel (II) sulfate solution, so its concentration is exactly half-way between that of the anode solution and cathode solution.
With the concentrations shown in the figure, the cell potential is +0.02958 V, and the anode is the nickel electrode in the more dilute solution at the left. On the cathode side, the log term is zero, in the Nernst ewuation at upper right in Fig. 1, so the potential is -0.236 V. But on the anode side, the potential is lower by 29.58 mV, due to the log term. So the anode is more negative, the cathode is more positive and electron flow is always from more negative to more positive. Therefore, Ni is spontaneously oxidized at the anode, yielding the necessary electrons, and nickel ions are spontaneously reduced at the cathode.
As a direct consequence of having an external connection, i.e., load or voltmeter, between the electrodes, net oxidation will spontaneously take place at the anode, resulting in an increase in the nickel ion concentration in that solution reservoir. Likewise, nickel ions in cathode solution reservoir will be spontaneously reduced at the nickel cathode and the nickel ion concentration will decrease. The external connection affords the opportunity for something useful to happen, i.e., using the cell as a power source. Otherwise, only diffusion happens (see below).
The cell will be "dead", i.e., have no more free energy to tap, when all three solutions are 0.55 M. There will be no concentration gradient left.
Now consider Fig. 2:

This is the same as Fig. 1 except that the cell is open circuited, i.e., there is no way for electrons to transfer from one electrode to the other. So all that is going to happen is spontaneous diffusion through the salt bridge. Eventually, there will be just one final concentration at 0.55 M.
During this process, no net oxidation occurs at the left electrode and no net reduction occurs at the right electrode. If a voltmeter is subsequently attached, the reading will be less than +0.02958 V because the concentrations are no longer the original ones: both concentrations are closer, thanks to diffusion, to the 0.55 M center value.
The measured voltage, even using a high impedance voltmeter, is not actually the open circuit voltage. It may be very close, but there is a conceptual issue. In the open circuit situation, no electrons can flow between the electrodes and no net redox processes occur at the electrodes. So the open circuit potential is an ideal potential and the measurement goal is to estimate accurately the open circuit potential while allowing negligible current to flow.
With a high input impedance voltmeter attached to the electrodes, a very small current, i.e., nA or pA, can flow. This only slightly loads down ("perturbs") the cell potential, so it provides an accurate estimate of the true open circuit potential. And it means that oxidation occurs at the anode, to a very slight extent, and reduction at the cathode, likewise to a very slight extent. Solution concentrations are negligibly perturbed.
So there are two parallel processes taking place. First, diffusion spontaneously acts to equalize the solution concentrations. Second, the concentration cell provides a way of hurrying the equalization process along while extracting useful electrical energy. So the former wastes the available free energy while the latter extracts much of it.
Finally, consider Fig. 3 below:

Now the salt bridge is gone and nothing happens: there is no DC cell potential and the solution concentrations are constant. In figures 2 and 3, there is no net redox happening: for every Ni atom that might get oxidized, there is a nickel ion that gets reduced. Everything is balanced at both electrodes. (So if the electrodes start with highly polished "mirror" finishes, they may get surface roughened by this zero net redox process.)
But once a load or voltmeter is attached to the two electrodes in Fig. 2, this opportunity causes the equilibria to become unbalanced: the solution concentrations must change by increasing (in the anode reservoir) and decreasing (in the cathode reservoir), and that immediately implies that electrons flow from the anode (at the left), through the load or voltmeter, to the cathode (at the right).
Perhaps the following will also help clarify matters. Consider Fig. 4 below:

This shows two beakers, each made of solid nickel, and containing the indicated nickel (II) sulfate solutions. Obviously, this does nothing: it is just two fancy beakers of solutions. Next consider Fig. 5:

Now the two nickel beakers are touching, i.e., in electrical contact. Does this change anything? No. Nothing happens. Next, consider Fig. 6 below:

Now there is simply a wide glass beaker with a solid nickel partition that entirely prevents the two solutions from making contact. Does this change anything? No. Nothing happens. Finally, consider Fig. 7 below:

This differs from the previous figure in having a salt bridge. It is exactly equivalent to using a nickel wire to short the two nickel electrodes in Fig. 2. So the cell is shorted out, there is no cell potential, and the cell will try to supply its short circuit (i.e., maximum) current, limited by the kinetics at the electrodes. So the nickel ion concentration will increase in the anode reservoir and decrease in the cathode reservoir, as expected.

Hi Ed, this is a great answer, thank you. One thing I still don't understand is why the Nickel at the cathode is more likely to be reduced than the Nickel at the anode. The nickel cation floating around the cathode is "dumb" and can't sense the concentration of nickel in the other cell, yet it still is more likely to be reduced. I updated my answer with another guess that has to do with kinetics.

Up vote(+1), understandable. But, per a source, https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_Chem1_(Lower)/16%3A_Electrochemistry/16.04%3A_The_Nernst_Equation , "Ions of opposite charge tend to associate into loosely-bound ion pairs in more concentrated solutions, thus reducing the number of ions that are free to donate or accept electrons at an electrode. For this reason, the Nernst equation cannot accurately predict half-cell potentials for solutions in which the total ionic concentration exceeds about 10–3 M.". Suggestion: reduce conc or introduce activity coefficent.

Thanks for the upvote and information! Funny thing, though: when I used to do the Daniell cell demo in lecture, I used 1 M copper sulfate and zinc sulfate solutions (they were already available in the lecture prep room) and the Nernst equation worked like a charm. I would routinely get +1.100 V or within 5 mV of that. This was one of the lecture demos I never had to worry would screw up at showtime! Still, in my overly long answer, I regret not discussing the (obvious) ion flows in the open circuit situation (Fig. 2) and in the externally loaded situation (Fig. 1).

Moving to equilibrium, nickel metal must be oxidized to nickel ions in the solution reservoir having the lower nickel ion concentration. This is how that solution can increase its nickel ion concentration and get to equilibrium. So that nickel electrode, shown on the left in my figures, is the anode. Likewise, nickel ions must be reduced to nickel metal in the other solution reservoir as that solution moves to its final equilibrium concentration. That reduces the nickel ion concentration as the cell moves to equilibrium. So the cathode is the nickel metal on the right.

One more thing: no need to worry at all about activity coefficients. All that matters is that the concentrations are different and that fact can be exploited to extract some useful electrical energy.

The potential of each electrode is given by Nernst law:
$$E = E^\circ + \pu{0.0296 V}\cdot\log[\ce{Ni^2+}]$$
After my tables, $E^\circ(\ce{Ni^2+}/\ce{Ni}) = \pu{-0.23 V}.$
So, in the $\pu{1 M}$ solution, the potential of the nickel electrode $E = \pu{-0.23 V}.$
In the $\pu{0.001 M}$ solution
$$E = \pu{-0.23 V} + \pu{0.0296 V}·(-3) = \pu{-0.23 V} - \pu{0.09 V} = \pu{-0.32 V}.$$
This value is more negative than in the $\pu{1 M}$ solution. As a consequence, this electrode is the anode, and $\ce{Ni}$ gets oxidized, producing electrons going to the $\pu{1 M}$ electrode.

Please don't use \ce{…} for anything other than chemical expressions and definitely not for an entire math formula. Also, keep an eye on units: they are not supposed to vanish and then appear out of nowhere.

@ andselisk. OK. And thank you for correcting the units. But why not use \ce{...} for math expression ?

No prob. \ce{…} is a macro designed and tested for typesetting chemistry only. Used improperly, things might break over time and hinder readability of your posts, not to mention it breaks code semantics (a big deal for parsers and converters like pandoc). Also, \ce{} converts every Roman symbol in math mode to the upright form (analogous to \mathrm{…}), but the variables such as $E$ must be italicized. On top of that, it uses different spacing rules, which are not standardized for math equations and make them look off.

I'm going to suggest a possible interpretation in the spirit of the question, which seems to ask for a more physics interpretation than a chemical one for the creation of the electromotive force.  I believe the question comes from the idea that the ionizing of each metal atom should take about the same energy, regardless of concentration, and therefore represent the same voltage.  My thought is that it is possibly a charge density phenomenon on each electrode, where the low concentration of M+ ions in the anode allows elecrons to be pulled of in large numbers with very few nearby positive ions to "pull back" on the electrons, reducing their electromotive force.  At the cathode, the high concentration of M+ ions in the vicinity of the electrode  have the opposite effect, pulling back on the electrons and reducing the net electromotive force.  If a surface were described around each electrode including a small layer of positive ions (in higher concentration than their counter ions hypothetically because of the electrons on the electrode), the net contained charge and therefore electric potential could be different for each electrode, because of the differing densities of M+ ions.  Just another hand-waving explanation, but maybe helps with reconciling "physics intuition" with the chemical equations.


text:
Can you steam clean your range hood, by adding dishwashing liquid to boiling distilled water?
My grandparents physically can't remove their Sakura R-727II range hood that lacks any filter. Because of COVID-19, no handyperson can visit their house. But their range hood is now clogged and failing to suck out fumes, so they must de-grease their range hood's interior that hasn't been cleaned in 4 years.
Please see the pictures below. They wonder if they can de-oil their range hood by adding dishwashing liquid to boiling distilled water, then boil this mixture to evaporate the dishwashing liquid that will rise into and de-grease their range hood.

Does the dishwashing liquid vaporize as they think? Or does it remain in the saucier?

Can the resultant evaporated dishwashing liquid + steam combination de-grease their range hood? I wonder if this dishwashing liquid + steam will have enough pressure.

The two pictures above are like their range hood. Theirs isn't a Samsung range hood pictured below, but I used it to devise my collage for it shows steam.

This is not going to work.

Dishwashing liquid cannot be evaporated. It cannot work. It should not be suggested. Wait for a professional service.

There must be a way to remove and exchange the filter from the hood, if not, the thing has been worthless from the very start.

What is that nonsense that no handyman can visit their home? Just stay outside while the man is at work.

@Martin-マーチン Please see the second picture that I just added.

You see the screws in front of the fans? And, next to the rectangular lamp? And, isn't there an instruction manual for this hood available, at least on the site of the manufacturer, with a point regarding the maintenance of the hood, too? E.g., for https://www.bosch-home.com/us/productslist/cooking-baking/ventilation/island-hoods/HIB82651UC#/Tabs=tab-3324792/ there is page 9 in this .pdf: https://media3.bosch-home.com/Documents/9001126905_A.pdf

This does not look like a 4 year old range hood that has never been cleaned. Where does it exhaust to? Maybe there is a problem on that end. According to Samsung, you can remove the stainless steel baffle "filter' and put it in the dishwasher (seems to be removed in OP's second picture).

https://chemistry.stackexchange.com/questions/137073/can-you-steam-clean-your-range-hood-by-adding-dishwashing-liquid-to-boiling-distilled-water

@KarstenTheis Sorry for the confusion. Theirs is the Sakura R-727II, not the Samsung. I edited my post.

@Buttonwood Sorry for the confusion. Theirs is the Sakura R-727II, not the Samsung. I edited my post.

There is a related discussion on https://diy.stackexchange.com/questions/185676/how-remove-sakura-r-727ii-to-clean-inside

@Accounting The indicator to one of the hoods by Bosch was only a non-specific example; in part because of their push into the branch of «smart home» and «internet of things» (e.g., https://www.youtube.com/watch?v=v2kV6pgJxuo, or https://www.youtube.com/watch?v=pCoy77P-JmY (heater in the US with Celsius scale!),  a milder version https://www.youtube.com/watch?v=aSZM2iYgrfg), and in part of later surprise such hoods (now) include a WiFi interface (e.g., https://www.youtube.com/watch?v=BdPaG2W5xSU), too.  Well, not sure if their slogan «live like a Bosch» equally is used e.g., in France ...

Clogged? Failing to suck out fumes? Quantitatively, how bad is it? And in what manner did this situation arise? Slowly, over time? Or suddenly? Can you hear a fan running?
Grease has a high boiling point, and while it is theoretically possible to clog some fine filter, it is doubtful that 4 years of home cooking could stop a hood fan from operating. In commercial practice, hoods over grills are cleaned frequently from spattered grease, not so much for better air flow, but more for sanitation and prevention of possible fires from leaping flames.
If no filter is accessible from the stove end, perhaps the exit end is the cause for blockage. Exit openings frequently have flaps to prevent rain entry. Two possibilities: 1) the flap is stuck down and hinders airflow, or 2) the flap is stuck up and the tube is blocked by a bird nest or a pile of acorns. (Of course, this exit end will be so high that it is not accessible for cleaning, but that's another problem.)
There is a third situation. In my former house, the builders omitted an outdoor exit pipe for the microwave and fume hood. There was simply a fan which sucked the steam/vapor up off the stove, thru the filter and back at the cook. We generally turned the fan off. The idea of a fume hood was nice, but not really necessary.

Any time a piece of equipment goes down, you cannot get a repair person, and you are not sure what to do, try to call or email the manufacturer. This is the US contact information for Sakura.  They know their product better than anyone on this site and will probably know about common issues.
Steam will loosen, but will not get hot enough on a stove to remove grease. The only safe way to remove grease is to get long gloves and scrub it.
Have you ever seen the mess dish liquid makes when put in a dishwasher?  The results of what is suggested more likely transport the soap all over the counter-tops instead of to the grease.

I like this answer. It's like: Read the directions first. If you don't have a manual, contact the company for directions. Maybe there is a fault that they will compensate you for. Only then should you do it yourself (if there is no official help available).

First off, let's clarify something:
When you boil water in a pot, hoping to carry whatever is dissolved within the pot in the steam, that thing should have a lower or the same boiling point as water (100 ℃).
A possible slution:
As grease and other fats gets carried up into the hood it usually cools down and the polymerizes and gets denser and harder to remove. You could, however, heat up the grease and fats again by using either a steam-cleaner or boiling some water very close to the hood for a couple of hours. The heat will "soften" up the grease thus leading it to eventually loosen the hood. (ONLY TURN ON THE HOOD ON THE LOWEST SETTING TO GUIDE THE VAPOR INTO THE GREASE - A higher setting will suck the grease further into the hood).
When the heating is done, you should be able to remove the hood fairly easily without the grease holding anything into place as it is more or less "melted".

There is a way to achieve a result strictly within the bounds of your question, but it's just as insane as these bounds.
Boil a solvent, not dishwashing liquid.
Say, gasoline. Or much better dichloromethane (sold as paint remover), if there is not much
plastic at the kitchen.
Don't turn fans on. Hood must act like reflux condenser. Let dichloromethane fumes condense inside, flow through it and drip.
Needless to say, it's a way to turn the kitchen into hell and is completely insane for health and fire danger :)

By the way, you couldn't boil the dishwashing liquid because of the huge amount of foam :)

Are you out of your mind? You suggest someone to boil dichloromethane in his kitchen. While you briefly point out the health hazards, I still don't think that justifies that answer. You don't know how well the questioner, or other people who read this thread, can assess the dangers of such an undertaking. This answer is really reckless!

I strongly believe that any filtering of knowledge is evil.
My answer is pretty anecdotal, but demonstrates probably the only way to remove polymerized clogging "remotely".

Instead of boiling CH2Cl2, which probably wouldn't condense in the filter because the vapor would be diluted with too much air, or wouldn't even get into the filter if there is NO air flow, consider a safer method: squirt it in with a garden sprayer. This is safer because it  would use less CH2Cl2 and direct it more toward the problem. But then, go even safer: squirt dishwashing solution instead of CH2Cl2. Put a big pan on the stove to collect drippings. Color the liquid with food color to see where it leaks into the house - or where it exits outside.

Solvent will work (if you manage to spray it inside), but not dishwashing solution


text:
How to simulate a Lorentzian broadened spectrum in MATLAB?
I'm trying to simulate the 2nu2 band of HCN in MATLAB. I generate a stick spectrum from the data I have (taken from the HITRAN database) and then use conv() to apply my line shape function. However, when I plot the data, the broadened spectrum does not line up with the stick spectrum at all. I cannot figure out what it is that I'm doing wrong or what I need to change to align the spectra. Does anyone know what is I need to fix or a resource I could use to solve this issue?
This is the spectrum I've simulated:

and the function I use to generate the spectrum is
function f = func()
HCN2nu2 = importdata('I:/HCN 2nu2.txt');
HCN = HCN2nu2(1:111,1:2);
AX = HCN2nu2(1:111,2);
v = [];
s = [];
L = [];
q = 1;
p = HCN(q);
increase = false;
for i = 6200:0.01:6700
    for x = 1:111
        if i == round(HCN(x,1), 2)
            z = AX(x);
            increase = true;
            break
        else
            z = 0;
        end
    end
    v = [v; i];
    s = [s; z];
    if increase == true
        q = q + 1;
        increase = false;
        if q > 111
            q = 111;
        end
    end
    t = (1/(1 + ((i - p)/0.5)^2));
    L = [L; t];
end
hold on
plot(v,s)
k = conv(L,s, 'same');
plot(v,k)
ylim([0 8e-21]);
hold off
end

Any help would be greatly appreciated
Link to the data file used to generate this spectrum:
https://github.com/tkh97/HCN-2nu2.git

This is arguably off-topic for Chemistry.SE. Although the code is chemistry-related, it isn't actually asking about chemistry, it's a programming exercise. My suggestion would be to ask this on MATLAB Answers. There are lots of qualified people there. And you should probably make that txt file available, or else it's rather difficult to figure out what's going on. I know enough MATLAB to do work with it, but I am having a difficult time following your code since I have no clue what vectors HCN and AX are.

Signal Processing stack exchange is a good resource for these questions but his/her error is rather simple to fix "conceptually". It is not a Matlab issue, which is doing the convolution correctly, but rather it is a minor erroneous math.

Update : In the given data by the OP, the x scale is not evenly spaced so the discrete convolution approach will not work to generate a spectrum until and unless your data is evenly spaced! The trick mentioned by EdV is a very nice shortcut. Here, if you really want to play around with convolution via MATLAB, then read below.
Solution: You can satisfy yourself as follows:

Make sure you have a even spacing of the x-axis. Your uneven axis threw me off, and it took a while to discover why it is not working. I wish you mentioned it earlier.
Your sticks are basically scaled delta functions as Prof. Ed said.
Generate a vector for a single Lorentzian peak, whose center is zero, which is also evenly spaced. The x-axis of the Lorentzian would be symmetric [-length (wavenumber/2):Sampling rate:length(wavenumber/2)-Sampling rate].
Perform convolution of your spectrum and this Lorentzian.

Let us start with a Dirac delta centered at point $x.$ I don't know how to draw a displace Dirac delta in MATLAB. I manually made it in an Excel file. If you convolute it with a zero centered Lorentzian, then you will get a Lorentzian centered at $x.$ This is called the sifting property.
Basically, all you have to ensure is that your Lorentzian is centered at zero. First, play with convolution using single stick. Once you correct the code, so that the line position does not change, apply it on the entire spectrum.

A great book is Bracewell's Fourier Transform And Its Applications. It is all there. From MIT 2.14 / 2.140 Analysis and Design of Feedback Control Systems, Spring 2007 handout:

1.1 The “Sifting” Property of the Impulse
When an impulse appears in a product within an integrand, it has the property of ”sifting” out the value of the integrand at the point of its occurrence:
$$\int_{-\infty}^\infty f(t)δ(t - a)\,\mathrm dt = f(a)\tag{5}$$
This is easily seen by noting that $δ(t − a)$ is zero except at $t = a,$ and for its infinitesimal duration $f(t)$ may be considered a constant and taken outside the integral, so that
$$\int_{-\infty}^\infty f(t)δ(t - a)\,\mathrm dt = f(a)\int_{-\infty}^\infty δ(t - a)\,\mathrm dt = f(a)\tag{6}$$
from the unit area property.

I do suspect you are quite right; however, it's really hard for me to verify this (and thus upvote) since those vectors are coming from a black box .txt file. Very nice suggestion on Sig Proc SE though.

@orthocresol, I am recently writing a paper in on a new approach towards Fourier self-deconvolution for resolving spectral lines. These shifting issues during convolution came up again and again and it took me at least a year to sort them out. OP is seeing exactly the same issue. Many thanks to Prof. Ed as well. I cannot parse his Matlab code. I am rather an elementary user of Matlab.

(+1) Funny thing about this one: the stick figure spectrum is just a scaled set of “delta functions”, and convolution with a “delta function” is the identity operation, so it looks like all that is necessary is to place a “stick height”-scaled Lorentzian (with 1 wavenumber FWHM) at each of the sticks in the raw spectrum.

This is an elaborated explanation of what commented upon. As stated by others, the question likely should have been posted to another stack exchange, e.g., signal processing. Furthermore, the OP really just wants to know why their Matlab code results in a shifted HCN spectrum, but is otherwise correct. The answer by M. Farooq, which I upvoted, provides the OP with what they need to consider as they revise and test their program. Here, however, I simply show how to do solve the problem without doing explicit convolution.
To begin, consider a unit height Lorentzian line profile function:
$$Lor(x;L,W) = \frac{1}{1+4[(x-L)/W]^2} \tag{1}$$
where x is the independent variable, L is the peak's centroid location and W is the FWHM (full width at half maximum height) of the peak. Units: x, L and W are in wavenumbers, i.e., $cm^{-1}$. From the legend of the OP's figure, W = 1 $cm^{-1}$, so equation (1) reduces to
$$Lor(x;L,1) = \frac{1}{1+4(x-L)^2} \tag{2}$$
If L = 0, then the Lorentzian peak is zero-centered:
$$Lor(x;0,1) = \frac{1}{1+4x^2} \tag{3}$$
The data file linked by the OP gives 111 L,h pairs. These are the respective locations and heights of the 111 "sticks" in the OP's stick figure HCN spectrum. Plotting these yields the following result:

Now for the short cut. Each of the "sticks" in the HCN spectrum may be considered as a scaled and shifted "delta function". Thus the $i^{th}$ "stick", having $L_i$ location and $h_i$ height, is
$$stick_i = h_i \times \delta(x-L_i) \tag{4}$$
But convolution with a "delta function" is the identity operation, so convolution of a zero-centered Lorentzian peak profile, with a scaled and shifted "delta function", simply results in an $h_i$-scaled Lorentzian peak at location $L_i$. Hence
$$\frac{h_i}{1+4(x-L_i)^2} = Lor(x;0,1) * h_i \times \delta(x-L_i)       \tag{5}$$
where convolution is denoted by "*".
Now all that is necessary is to evaluate 111 Lorentzians, one per $L_i,h_i$ pair, as per the left-hand side of equation (5), and then add them all up. I did this with a simple computer program (not Matlab), with x ranging from 6200.00 to 6700.00, in increments of 0.01 $cm^{-1}$. The resulting spectrum is

Even though the $h_i$ values are unevenly spaced and given to 5 decimal places in the OP's linked file, the Lorentzian function evaluation spacing of 0.01 $cm^{-1}$ introduces negligible error.

As usual the OP has vanished despite giving him useful information.

@M.Farooq Yes. And yet other OPs quickly accept so-so or half-baked answers, before an actual thoughtful answer gets posted.

The problem (noted in another answer) is that the line broadening function is not centered, therefore the effect of the convolution, in addition to modifying the peak widths, is to shift the spectrum. An off-center Lorentzian (such as used by the OP) is itself a convolution of a centered Lorentzian and a shifted delta function. If you  ignore the Lorentzian for a moment, the effect of the shifted delta function is to shift the spectrum.
If a centered LB function is used, as shown in the following figure, the problem is largely resolved:

I did this by using the following convolution function:
L = 1./(1 + 20*(vsim - mean(vsim)).^2);
where vsim are the frequencies in the interpolated spectrum.

Aside
The code in OP leaves much to be desired: not commented, unused blocks of code and variables, non-optimal use of Matlab's intrinsic matrix handling properties and inbuilt functions. However this is not a subject for this SE - perhaps better to ask over at code review or another SE.
The following performs the desired convolution. It is not ideal (see previous comment) but will get the job done:
HCN2nu2 = importdata('./HCN 2nu2.txt');
HCN = HCN2nu2(1:111,1);
AX = HCN2nu2(1:111,2);
v = []; s = []; L = [];
range = 6200:0.01:6700;
p = mean(range);
for i = range
    for j = 1:111
        if i == round(HCN(j),2)
            z = AX(j);
            break
        else
            z = 0;
        end
    end
    v = [v; i];
    s = [s; z];
    t = 1/(1 + 4*(i - p)^2);
    L = [L; t];
end

figure
plot(v,s)
k = conv(L,s, 'same');
hold on
plot(v,k,'k')

BuckThorn, Mathematics never fails to surprise us. I am wondering the why your mean centered Lorentzian also works. Suppose we had one stick at w= 6500, now the convolution property is that the two centroids add. For example, your Lorentzian is 6450 and the stick is at 6500, the resulting convolution is 6450+6500. Not sure, why you do not see a shift.

BuckThorn, What is the source of your 20? Please see the Lorentzian equation (1) which Ed and I am using.

@M.Farooq In essence, by subtracting the mean in the equation shown the Lorentzian is centered in the window, which is exactly what you suggest the OP do to fix the problem. I happened to implement it without seeing your answer first. After I posted my answer I read yours more carefully and realized that other than providing  guidance with the code my answer was not going to enlighten anyone on the finer points of FT and convolution, which you already covered. I am tempted to post more code but think the OP has to conduct a substantial code review.

The 20 is proportional to the inverse linewidth.

PS The detailed values of the frequency scale shown in the plot of the LB function are not meaningful. What's important is registry with the interpolated stick spectrum, as you also suggest.

BuckThorn, I am still puzzled. Why your mean centered Lorentzian is working? What is the mathematical basis? I have posted the question. https://dsp.stackexchange.com/questions/69498/sifting-property-of-shifted-impulse

@M.Farooq I am sure you'll get a more thorough answer at DSP. :-)

Sadly, so far their is an irrelevant answer.

@Farooq focus on the second comment by Matt L in the DSP answer, that is all the explanation you need. It's not that mysterious if you (1) understand where the center of the discrete FT spectrum lies and (2) how convolution works. For (2) start with the delta function example by Matt (the shifting property). Unfortunately the routines in Matlab are sometimes black boxes although sometimes you can inspect their inner workings.

@M.Farooq After reading your DSP post: the origin is at the center of the spectrum, in other words $\mu = 0$. Note that this is the origin of the "zero-filled" stick spectrum, not of the original data set.

I qualitatively get your point...but mu is not at zero, it is at the mean of the x axis. This is the only problem I have. As you have read the reply, the poster is saying that MATLAB is fooling us there. You can try the same code. Just keep in mind we are doing convolution in the time domain. No involvement of FT so let us keep that out.

Matlab doesn't know what units you are using for the frequency (or time or conjugate dimension), neither does the DFT. All you need to realize is that the zero frequency (no modulation) point is halfway in the spectrum. If I have time I will take a fresh look at this later but that's all I have to suggest for now.

can you explain why p which is equivalent to L in Eq.(2) (Ed V's answer)?


text:
Mechanism of R-OH to R-X using SOCl2
https://youtu.be/ANlXQdpv6cc?t=1006
In this lecture, I am learning on the reaction of alcohol to alkyl halide using SOCl2 and pyridine. However, there are 2 different approaches on the first step.

The 1-step initial step, which is printed on many reference books.

The 2-step 'inital step' (forming the same intermediate), which the lecturer says is correct. This is because during a SN2, the sulfur (atom bonded to the leaving group - Cl) must be sp3 hybridized, otherwise, SN2 cannot happen (so figure 1 is wrong).
I have no idea why figure 1 is wrong though. Sulfur in SOCl2 is sp3 hybridized. (1 lone pair, '3' bonds), while the lecturer says the sulfur is actually dsp3 hybridized, not sp3. So my questions are:

is sulfur dsp3 or sp3 hybridized?
Why can't SN2 reaction happen if it is not sp3 hybridized?

(3.) Which step above is the most plausible?

Hybridisation in its simplest form ("sp3/sp2/sp") starts to fall apart for non-second row elements. (Don't even think about using it for transition metals.) In any case, there should be very negligible d-orbital participation, so sp3 is probably a better description. This dsp3 thing really shouldn't be taught anymore. For mechanism of substitution see Substitution at thionyl chloride sulfur.

So it is actually dsp3 hybridized instead of sp3? And when I encounter similiar situations later, I just need to think of s & p orbitals and ignoring others e.g. during SN2?

Quite the contrary; you shouldn't use d-orbitals to describe bonding in "hypervalent" molecules, as usually it isn't needed. When you call something "dsp3" you are basically saying that d-orbitals play a critical role in the bonding. Why did you get the impression I was saying it was "dsp3"?

Also, this gives good overview.


text:
Why do subshells and orbitals exist?
I'm just a curious high school student. Sorry if this sounds dumb.
How exactly did the concept of atomic subshells and orbitals come about? And why exactly are there n-1 subshells and 2l+1 orbitals?
Based on what I've heard, these things were a step forward after the Sommerfeld atomic model. But orbitals???
Again, sorry if this sounds dumb.
I just want to know how these things came about.

It is a simplification - the raw quantum mechanical description would be extremely complex. But sometimes it is needed (for example, to calculate the physical/chemical properties of unknown elements/compounds).

There's no such thing as a sub-orbital. Maybe you meant subshell. Basically, all of this comes about from quantum mechanics (to be precise, the orbitals are solutions of the Schrodinger equation for the hydrogen atom). The extension to other atoms (helium, lithium, ...) is an approximation that generally works well enough for chemistry.

Oh yeah thanks for pointing that out I fixed it

If this is a serious question and you wish to get a conceptual grasp of it, there is a lovely thin book called "Hydrogen" by Rigden (from Harvard). It walks you through the conceptual development.
Briefly, the need of better and better model of hydrogen atom arose because of the need to completely rationalize the experimental hydrogen spectrum. By the 1920s, experimental atomic spectroscopy was very advanced and people already started to very high resolution spectrum of atomic hydrogen. If Bohr's theory could explain every experimental aspect of hydrogen spectrum, there would be no need of another approach. For example, in hydrogen atom spectrum you could see closely two spaced lines which under low resolution appear as a single line (search doublet of hydrogen).
The oldest was the planetary model and Bohr's model is similar. Then Sommerfeld improved the model but still it was not adequate.
Contrary to the common opinion, orbital was not coined by Schrodinger. Robert Mulliken (another Nobel laureate) proposed this term.
An orbital is not a physical object or structure inside an atom, no matter how many useless general chemistry books or websites show them as balloons and other drawings. Even leading journals fell prey to this non-sense saying that orbitals have been observed or imaged, only to be embarased later. Orbital is a short form name of "one-electron orbital wave function". It is a mathematical construct which exists on paper and in the mind of a physicist.
Roughly you can imagine orbital as a pattern of electron density around a nucleus. But again, there is no actual "white cloud" of an electron surrounding the nucleus (if we were able to see it). Do not take the small world of particles in a literal sense. Much is known and much more is not known.

Very nice question. I also had the same question. These orbitals are nothing more than functions (wavefunctions). They are solutions to Schrodinger's equation, a second order differential equation. So you solve the equation and the solutions you get are determined by $4$ numbers also called quantum numbers $n, l, m_l, m_s$. A shell consists of all orbitals with same value of $n$ whereas a subshell consists of all orbitals with same values of $n$ and $l$. As others have pointed out the solutions are exactly only for the hydrogen and hydrogen like atoms.


text:
Reagent to reduce aliphatic aldehyde to alcohol while protecting double bond
I have a straight chain, aliphatic aldehyde (liquid) with a double bond in it. I want to reduce the terminal carbonyl group to an alcohol but without touching the C=C double bond. The double bond is at a beta position with respect to the end carbonyl group.
Any suggestions whether there are any reactions that may do this?
Below I am showing the configuration of the double bond relative to the end group to avoid confusion.

Does NaBH4 not work? Is this on paper, or in practice?

@orthocresol: Will try. This is at the brainstorming stage. Basically, there's an enzymatic process and I am wondering why they are bothering with expensive finicky enzymes and low yields for this step when we could do it chemically.

Dibal-H will do this  https://sites.google.com/site/anthonycrastoreagents/diisobutylaluminium-hydride?tmpl=%2Fsystem%2Fapp%2Ftemplates%2Fprint%2F&showPrintDialog=1

When you say -position, do you mean conjugated or deconjugated?

@user55119 I was not sure of the question so I posted an image to clarify the position of the double bond.

You have the (Z)-deconjugated aldehyde. NaBH4 is a good choice but the alkaline conditions in alcohol or water may cause conjugation to the (E)-aldehyde prior to reduction to an (E)-allylic alcohol. I would opt for adding the aldehyde to LiAlH4 in ether.


text:
Factors affecting rate of β-elimination reaction with alcoholic KOH
The question asks for the rate of reaction in $\beta$-elimination using alcoholic $\ce{KOH}$.

Arrange the following alkyl halides in decreasing order of rate of $\beta$-elimination reaction with alcoholic $\ce{KOH}$:
(A) $\ce{CH3-CH(CH3)-CH2Br}$
(B) $\ce{CH3-CH2Br}$
(C) $\ce{CH3-CH2-CH2Br}$

The answer I got was:

B > C > A

The answer given is:

A > C > B

In many sources it is given that the order depends on the stability of the double bond formed, that the more substituted alkene is more stable.
But for alc. $\ce{KOH}$ shouldn't we consider it as an $\mathrm{E2}$ mechanism?
By this I mean, shouldn't the reaction rate depend on the steric hindrances caused by the beta carbon group to the attacking base?
Is this wrong? What would be the correct explanation?

Are you aware of the $\beta-$elimination mechanism? The base abstracts the H-atom here, followed by the leaving of the Br atom. It's easiest to remove the H-atom in A, which is why it's the fastest. Refer here.

The RDS in an E$2$ mechanism is the transition state:  The more stable the transition state, the faster the rate. So  the more substituted the more number of $\alpha$-hydrogens, which makes it more stable.

bases have nothing to do with size, they can abstract a proton regardless of size as they do not have to attack. In $\ce{E2}$ a stronger base is used and stability is decided by transition state ($\ce{alpha}$-hydrogen)

@Aniruddha Deb: Your terminology that "the base abstracts the H-atom" and "followed by the leaving of the Br atom". The description makes the process sound as though it is stepwise rather than concerted.


text:
Most favorable position of attack of an electrophile on 4-trifluoromethyl-1,2-diphenylbenzene
Problem
This problem is from a 2016 FIITJEE AITS paper:

Which position is most favorable for the attack of an electrophile?

Answer

 Position 2

My attempt
The more favorable position for attack should be the on the para ring (considering ring with $\ce{-CF3}$ group as the reference ring), because of the lower inductive effect as compared to meta position.
If the electrophile attacks at position $1$ or $6$ (which means the carbocation will be formed at the carbon connected via a double bond), the carbocation would be equally stabilized by resonance, also the charge can move to the other 2 rings forming 9 Resonating structures.
If carbocation is formed at position $6$ by the electrophilic attack on 2nd position, the carbocation won't be able to move out to other rings for resonance implying that there would be only 3 resonating structures.
So my answer would be:

 Position 1 or 6

Why is my answer is wrong? What is the correct explanation?

This is an another way to answer the question: On the way, I also want to show OP that why central ring bearing $\ce{CF3}$ group has the highest electron deficiency at ortho- and para-positions (meaning, the highest electron density is at meta-position in this ring). Let's look at the nitration of trifluoromethylbenzene:

According to this University of Columbia website, the highest percentage of nitration is at meta-position. If the rate of nitration of benzene is considered to be $1$, that for trifluoromethylbenzene is $6.7 \times 10^{-5}$ for meta-substitution while that for ortho- and para-positions are about $4.5 \times 10^{-6}$. That means, $\ce{CF3}$ group is a strong deactivator (c.f., rate of nitration of toluene is 58, 42, and 2.5 for para-, ortho-, and meta-substitution, respectively compared to benzene being 1).
The reason for preference for meta-substitution of trifluoromethylbenzene is depicted in the scheme. At least one of resonance contributors of the relevant Wheland intermediate during each of ortho- and para-substitution is destabilized by the neighboring $\ce{CF3}$ group ($\ce{F3C-C^+}$ destabilization). Meantime, none of resonance contributors of Wheland intermediate corresponding to meta-substitution do not deactivated and enjoys lower activation energy. All of these directed to the conclusion that the highest electron density is at meta-position when the aromatic nucleus is deactivated by +I effect (Ref.1-4). However, keep in mind that this highest electron density at meta-position is still way lower than the electron density of benzene.
Based on these facts, the substitution happens in the aromatic ring with 3, 4, and 5 positions (Keep in mind that there are no resonance contribution what so ever due to aromatic nuclei being non co-planner). When you consider the aromatic ring with 3, 4, and 5 positions, it has a substitution next to 3-position, which is an electron deficient biphenyl with $\ce{CF3}$ group on one of the phenyl ring. This substituted biphenyl is, therefore, meta-directing (a similar way as described for the substitution of trifluoromethylbenzene in the scheme). Thus, up coming substitution on the sought ring would be at meta-position, which is marked as position 4.
References:

C. Sandorfy, "Note on the Inductive and Mesomeric effects in Substituted Benzene Derivatives," Can. J. Chem. 1958, 36(12), 1739-1741 (https://doi.org/10.1139/v58-251).
F. A. Matsen, "Molecular Orbital Theory and Spectra of Monosubstituted Benzenes. I. The Resonance Effect," J. Am. Chem. Soc. 1950, 72(11), 5243–5248 (https://doi.org/10.1021/ja01167a127).
W. W. Robertson, F. A. Matsen, "Molecular Orbital Theory and the Near Ultraviolet Absorption Spectrum of Monosubstituted Benzenes. IV. The Phenyl Halides and the Inductive Effect," J. Am. Chem. Soc. 1950, 72(11), 5252–5256 (https://doi.org/10.1021/ja01167a130).
Christopher K. Ingold, "Principles of an Electronic Theory of Organic Reactions," Chem. Rev. 1934, 15(2), 225–274 (https://doi.org/10.1021/cr60051a003).

So is the method used by the OP correct?

@ Safdar: No, OP chose para-position of central nucleus as the position with the highest electron density (relatively speaking). That is not correct at the first place.

I mean the part where he attacked the ring and then checked for stability. I was asking if that was correct?

@ Safdar: OP's method is not correct because of thinking of mesomeric effect. The rings are not co-planner so no resonance contribution what so ever. The resonance contributor with a positive charge on ipso-carbon in both cases (substitution on 1 and 6 positions) will be destabilized by +I effect of central ring.

last 2nd para.. it should be -I effect instead of +I effect..looks like a typo.

Great answer Mathew! Also, good find on the UoC lecture slides: they explain what is happening perfectly. +1

TL;DR - The ring with positions 1, 2, and 6 on it is more deactivated than the ring with positions 3, 4, and 5, so the correct answer cannot be position 2 (as marked) or positions 1 and 6 (as you determined). The correct answer must be on the other ring. Since the central ring imposes a -I effect, the correct answer is position 4.

There are two factors to consider in this question. The first is that the trifluoromethyl group ($\ce{CF3}$) is electron-withdrawing by induction. Fluorine is the most electronegative element, and the $\ce{C-F}$ bond is polarized toward $\ce{F}$. The $\ce{CF3}$ group has a -I effect on the ring, and that effect is strongest at ortho and para.
The second is that the three rings in this structure are not coplanar. The steric interactions between hydrogen atoms on the two rings with labeled positions in your image causes the rings to twist out of coplanarity. This feature of the structure means that the central ring cannot exert a resonance effect on either of the other rings. The central ring will is electron poor thanks to the $\ce{CF3}$ group, and so it will provide a -I effect to the other two rings. Since the -I effect on the central ring is stronger at the position para to the $\ce{CF3}$ group than at the position meta, the ring with positions 1, 2, and 6 will be more deactivated than the ring with positions 3, 4, and 5.

Since the lower ring is experiencing a -I effect, electrophilic attack at meta is most likely, so position 4 is the true correct answer.

One query, how did you get the fact that the -I effect is stronger at ortho and para when it is distance dependant?

-H effect would be more true here.

@Safdar Try to form the arenium ion by attack of an electrophile on tert-Butylbenzene. You will see that when the attack is on ortho and para position, the +ve charge will appear on the carbon where the tert-butyl is present(through resonance).

In this question, we need to find the most electron-rich carbon out of positions $1,2,3,4,5,6$ in the compound given below:

We proceed using the process of elimination.
Note: Not sure how scientific this is, however an attempt I made in finding the effect of the inductive effect was to run a DFT B3LYP/3-21 G on trifluoromethylbenzene. The result showed the m-carbon to have a charge of $-0.173$ and the p-carbon had a charge of $-0.172$. The original value in the benzene molecule is $-0.180$ and so inductive effect is present due to $\ce{-CF3}$. This meant that the m-carbon was more electron rich compared to the p-carbon. I do not have a reason for this however.
Elimination $1$
We reject positions $1,2,6$ since as stated above the meta-carbon is more electron rich than the para-carbon.
Elimination $2$
Now, we are left with the ring containing positions $3,4,5$. In order to proceed further, we assume a $\delta +$ charge on the carbon para to the $\ce{-CF3}$ group. This would mean that the phenyl ring resonates with the $\delta +$ to give five different possible resonance structures. In order to simplify the resonance structures, I am considering only this ring.

Answer
Due to the two eliminations made, we can say that position $4$ would be the most electron rich carbon and so the most likely to be attacked by an electrophile.

How did you assume a +ve charge para to CF3? What's the effect making up the charge?

It isn't a full positive charge. Due to the $\ce{-CF3}$ group there would be a delta positive charge formed since it is highly EW. So, to show the delocalisation of the charge i took it to be a positive charge.

CF3 will cause a positive charge. But then since CF3 doesn't show -M / hyperconjucation, the positive charge should only be due to inductive effect? But as far as I know, it becomes almost insignificant at the 3rd carbon.

I suggested eliminating 3,4,5 because of inductive effect because when no other effects (M/ hyperconjucation) are differtiating, it's the last choice. But the positive charge you suggested, is so minute, I think won't be able to participate in resonance as such. We usually see cations doing such stuff. And given that it is question at +2 level (not research stuff) , I somewhere think that it must some concept which hasn't been pointed out yet. :)

The funny thing is if you try to calculate the values of the actual charges on the trifluoro methyl benzene C-4 is more positive than C-3.. minute difference so may be due to approximations but still.. Also, the nucleophilic centre would be the most electron rich carbon..

I wonder how is it so? which electonic effect? Alothough this won't still answer the given question. :)

Not sure how scientific this is, however an attempt I made in finding the effect of the inductive effect was to run a DFT B3LYP/3-21 G on trifluoromethylbenzene. A phone book might have given you more accurate results. If you are tight on resources, try semi-empirical methods (xtb, pm6/7). If you have slightly more, try bp86, pbe, m06l, (so called pure functionals) in combination with def2-svp. For a qualitative approach that should suffice. Hybrid methods with a small basis set take longer and are less reliable. Have a <3 for semi-empirical methods!

Also, I don't know how you derived your charges, so that is worth looking into. Expanding resource: https://mattermodeling.stackexchange.com/q/1439/49

@Martin-マーチン Thanks for this.. I'll look into it..

You're welcome. This is definitely quite a nice molecule to play around with, so you can use it to play around a little bit with quantum chemistry.

@Martin-マーチン btw how small is a small basis set? just curious.

Anything with fewer functions than 6-31G is small, with STO-3G being minimal (tiny). Pople basis sets are quite old and not as well balanced as newer ones, which might contain more functions but still perform better (=faster). (They have a lot of educational value though.) I'm currently a bit afk, but you can try an ping me in chat if you want to discuss further.

$\ce{CF3}$ loves to pull electrons, hence it is easier to polarize the molecule in this way (which is inaccurate because you cannot polarize it completely, but considering partial effects). This polarized structure is made considering the requirement of negative charge at the carbon positioned adjacent to CF3.
position 3 and 5 maybe a dispute but, as a known fact, $-\ce{I}$ is distance dependent, 3,5 will be lesser favourable that 2.
Your reason is apparently wrong because you considered an attack first and stabilised the product of your attack.

Optimised geometry{tinker:force field mm3}
Then exported as xyz form and performed Hartree Fock using TeraChem, the following conclusions were made after the chat in comments.

The inductive effect is dependent on distance and has nothing to do with
indivisual bond polarities as in the first diagram.

The distance between CF3 , (2,3,5) is maximum for 2.

How do you decide the dispute is between 3,4 and 2???

I remarked the polarisation, please read the first line

This is more of speculation than a proper answer. First, you have compared 3,5 and 2 which are on different rings. Second, the most your answer has done is to say that what OP did is wrong . And haven't properly explained your reasoning..

if so you wanted not to compare any positions you might proceed by leaving the question. OP has not asked for reasoning yet. If it's your question, ask it as your own

I think explanation and reasoning mean the same thing. Last line in Q asks for the explanation

My statement meant reasoning for "Your reason is apparently wrong because you considered an attack first and stabilised the product of your attack."

Let us continue this discussion in chat.


text:
Mechansim of reaction between 1‐bromo‐2‐fluorobenzene and furan in the presence of Li/Hg
Problem
From FITJEE AITS test series:

What is the mechanism of the following reaction?

Solution

My approach
It looks like Birch reduction with $\ce{Li}$ converting to $\ce{Li+}.$ But I can't make anything much out it.
Maybe there's a typo given in solution because $\ce{Br}$ and $\ce{F}$ should be present initially, or do they get removed by any means?

Nothing against Safdar, but this is technically not a two-step reaction. When lithium amalgam is added to 2-bromofluorobenzene in presence of furan, the reaction goes directly to the product, a Deals-Alder adduct:

However, in the absence of furan, the reaction proceeds to give biphenylene and triphenylene (Ref.1&2). The tentative reaction mechanism can be as follows:

History: During their investigation on the reaction of sodium with boiling chlorobenzene (a Wurtz–Fittig reaction), Bachmann and Clarke at the Eastman Kodak Co. were the first to propose benzyne ($\ce{C6H4}$) as a reactive intermediate (Ref.3). Further advances were made when it was discovered that benzyne could be generated from ortho-dihalobenzenes (e.g., 2-bromofluorobenzene) with lithium amalgam (Ref.1) or by forming the Grignard reagent with magnesium (Ref.4&5). The first Diels–Alder reaction of benzyne, with furan, was performed in this manner, giving a 76% yield of the DA-adduct, which was converted to 1-naphthol with acid (Ref.1). In the absence of the furan trapping agent, biphenylene and triphenylene were obtained in yields of 24 and 3%, respectively (Ref.2).
The halogen-lithium exchange followed by Diels–Alder reaction in one-pot has been studied recently (Ref.6):

References:

G. Wittig, Liselotte Pohmer, “Intermediäre Bildung von Dehydrobenzol (Cyclohexa‐dienin),” Angew. Chem. 1955, 67(13), 348-348 (https://doi.org/10.1002/ange.19550671306).
Curt Wentrup, “The Benzyne Story,” J. Aust. Chem. Soc. 2010, 63(7), 979–986 (https://doi.org/10.1071/CH10179).
W. E. Bachmann, H. T. Clarke, “The Mechanism of the Wurtz-Fittig Reaction,” J. Am. Chem. Soc. 1927, 49(8), 2089-2098 (https://doi.org/10.1021/ja01407a038).
Harry Heaney, Frederick G. Mann, Ian T. Millar, “781. The reaction of o-bromoiodobenzene with magnesium and lithium,” J. Chem. Soc. 1957, 3930-3938 (https://doi.org/10.1039/JR9570003930).
Georg Wittig, Erhard Knauss, “Dehydrobenzol und Cyclopentadien,” Chem. Ber. 1958, 91(4), 895-907 (https://doi.org/10.1002/cber.19580910502).
Shohei Eda, Toshiyuki Hamura, “Selective Halogen-Lithium Exchange of 1,2-Dihaloarenes for Successive [2+4] Cycloadditions of Arynes and Isobenzofurans,” Molecules 2015, 20(10), 19449-19462 (https://doi.org/10.3390/molecules201019449).

Why would Li attack Br first? In the rate determining step F- is a poor leaving group so reaction should be slower if Br is attacked first. I know that C F bond is much stronger than C Br bond so which halogen will be attacked first? Also would rate be faster if both substituents were Br?

This is a question based on two reactions.
The first reaction is the formation of benzyne using $\ce{Li/Hg}$. After this a Diels–Alder reaction takes place between furan and benzyne. The reaction mechanism would be as follows [1]:

A more generalised form of this reaction, dealing with the conditions needed and the yield of reaction would be [2]:

Reference

Shashidher, B.; Bajjuri, R.; Guguloath, V. Formation and Trapping of Benzyne. Pharm Anal Acta 2011, 02 (07). DOI: 10.4172/2153-2435.1000137. (PDF)

Comprehensive Organic Synthesis; Wolfgang Oppolzer; Université de Genève, Switzerland 1991. DOI: https://doi.org/10.1016/B978-0-08-052349-1.00128-1


text:
Types of chromatogram shapes
I'm trying to generate a testing set of different chromatograms to check peak detection (whether peak boundaries are detected correctly) and smoothing algorithms. Currently I'm figuring out which features of chromatograms I should vary for my tests. Hopefully someone could help fill my blanks. So far I'm thinking of all combinations of all bullet points:

Peak shapes: Gaussian, ...?
Sharp & wide peaks
Different levels of skewness
Baseline drifts: any types? Should it also resemble Guassian peak?
2 peaks next to each other:

Peak resolution e.g. from .2 to 1. Allows to test valleys & round peaks.
Different peak heights. Together with different levels of resolution allows to test valleys, shoulders, rider peaks.

Tailing and fronting: are there some known distributions that resemble them?

As for smoothing:

High frequency noise: uniform distribution?

So far this is what I've got for each resolution that I choose: chromatogram-shapes.pdf. Want to add baseline drift next, but haven't decided which distribution(s) it should resemble.

What exactly is your background? What chromatograms are we talking about, an artificial test set for testing what exactly?

@Karl, I'm a software engineer, though I'm learning chemistry/analytical chemistry/DSP to create chromatogram processing software. This is going to be an artificially generated test set to check algorithms like ApexTrack or maybe we'll come up with a new one some day. Currently I'm interested mostly in MS & DAD detectors and the types of signal they generate.

I strongly recommend you find yourself a cooperation partner with a chemistry degree, who helps you to firstly define a scope for your project.  DAD and MS are detectors, one for HPLC usually, the other for gas chromatography. They work fundamentally differently.

@Karl I understand the principles of how they work and I understand when/why/how they are used (not as well as a professional chemist of course). Though I'd love to have a chemist to talk to from time to time :) But note that this question is not about spectra - it's about chromatograms - they don't differ as much between detectors - it's still chromatography (I'm not talking about TOF either) hence it's still a time vs. intensity plot. Though sure - they may have some features of their own (absolute intensities, noise profiles) and that's one of the things that I'd like to know better.

Well, you brought up the detectors. ;-) And the detector type is certainly a major issue for your question. They have different noise levels, kind of disturbances, and dynamic range, linearity, ... But you also need to choose a chomatography. HPLC, GC, SEC, .... And what kind of samples?

There are plenty of databases with chromatograms in them, especially for MS.  You could mine those datasets for chromatograms...one example of such a database is GNPS/MASSive.

@CurtF. wooooow, I've been looking for such database for a long time, many thanks! I probably won't be using real samples for peak-detection testing, it's more convenient to control exactly all the variables. But real samples will help me a lot too. If you know other places with MS (or other detectors) - I'd be very grateful if you pointed me there :)

Good to know that someone is working in DSP and chromatography but there are many software out there like PeakFit, why a new one?

Your first test could be peak detection as function of resolution as well their height ratios. Find out under what thresholds your peak detection fails. This is the most important thing.

Peak shapes: Gaussian, Exponentially modified Gaussians are very very common. Tailing is far more common than fronting. There is a very nice equation that covers all three shapes. Search BI-EMG in Google Scholar and you would find it. "BI" here is bi-directional."

Sharp & wide peaks: It is a relative term. You have to check what maximum efficiencies are achievable on instruments on GC, SFC and LC.

Different levels of skewness: Asymmetry of 2 is usually the maximum allowed number. Asymmetry of 1 is highly desirable. In practice it is possible to obtain 1.0 to 1.1.

Baseline drifts: Look more like linear ramps. You have to familiarize yourself with gradient elution.

Peak resolution: Start from peak resolution of > 0.5

Tailing and fronting: There is a very nice single equation that covers all three. See point no. 1.

Smoothing: I have written a couple of papers on smoothing. You can play with wavelets (most complex), Hamming filter, Savitsky Golay, simple central moving average, and Gaussian weighted moving average is the most common. I cannot disclose which instruments use which method. You can assume Gaussian noise rather than uniform. To the best of my knowledge wavelets or FFT filtering is not used in commercial GC, HPLC etc.

Unfortunately, PeakFit development appears to be stagnant for over a decade and it is a proprietary software. Out of curiosity, are you aware of any comparable open-source tools for peak fitting in chromatography? I've heard Fityk is a pretty good general-purpose program, but I never used it extensively in the field. And thank you for the insightful answer!

You are right, PeakFit is stagnant but it is still very powerful. I personally like it because it has a plenty of peak functions in spectroscopy+chromatography, which are openly visible in the help section. So anyone who is developing a new software, should have something equivalent or better. I haven't used open software.

Many thanks! I'm using SG for now, but it seems to perform poorly on my artificial samples, I'll see if it's better w/ Gaussian noise (it performed better w/ real samples though). I thought moving average widens signals too much and thus wasn't a popular choice, but I'll try it and others too (couldn't defeat WT last year though - was too complicated for me). As for baselines - I've seen cases when it's not linear at all, maybe it was a problem with the column or leftovers in it, don't know.. Wasn't sure how common this is.

As for why I'm building the software: I see how chemists who I work with suffer and I can't bear it. I've been working in Pharma for more than 4 years and all software that I've seen is horrible. Since I wanted to be more involved w/ science and I liked Analytical Chemistry the most - that decided it. I mean scientifically the software could be good, but the actual software part (how convenient, fast, intuitive it is) - it's all very bad. I on the other hand am very good at software part, but I'm week at Math/DSP/chemistry (frankly when I started I didn't expect it to be so complicated).

@StanislavBashkyrtsev, Do you work for a company or is it an academic project?

@M.Farooq I have a day job for a company, but this chromatographic software is a side project that I develop with yet another guy, you can check it out here: https://elsci.io/peaksel/. BTW, is this the paper that you reference for BI-EMG: https://onlinelibrary.wiley.com/doi/abs/10.1002/jssc.201801131?

@StanislavBashkyrtsev, Looks like a good effort. Hope you succeed. Regarding the paper, BI-EMG is written there as equation 14, BI-EMG can also be seen from PeakFit help function.


text:
Explanation of terminology in equation used for energy calculation using Hartree–Fock method
I found in the literature that the energy of the Hartree–Fock method is given by the following equation:
$$E_\mathrm{HF} = \int (\Psi^* \hat{H}\Psi)\,\mathrm d\tau$$
The term $\Psi$ is the wave function, but what would the term $\Psi^*$ be?
Does anyone have a simple of explanation of what this may be?

Something with an asterisk in QC usually refers to a complex conjugate. See Wikipedia and a related question on Physics.SE.

I had seen it on wikipedia but thought there would be another explanation for that (physical meaning or something like that). Thanks for the references @andselisk

I highly recommend that you review the basic math of quantum mechanical operators before tackling the HF method. It will make a lot more sense that way.

@Andrew would you have review articles in the area to kindly recommend reading?

Levine's Quantum Chemistry textbook has a great section just on the math that only assumes familiarity with standard calculus.

For every $x$ and $t,$ $\Psi(x,t)$ is a complex number. $\Psi^*$ is the conjugate of that number, no more, no less. The reason it seems like sometimes it's only the $t$ part that gets conjugated is simply that often it is the only part of the wavefunction that is complex. Let's use an example:
$$\Psi = \sqrt{\frac{2}{a}}\sin\left(\frac{n\pi x}{a}\right)\mathrm e^{-\mathrm iE_n t}.$$
We want to calculate $\Psi^*$. Well, since the conjugate of the product of two numbers is the product of their conjugates (that is, $(zw)^* = z^* w^*),$ let's do it step by step.
First we need to conjugate $\displaystyle\sqrt{\frac{2}{a}}$, but since it's a real number, it is equal to its conjugate. So we leave it alone and move on. Now we need to conjugate $\displaystyle\sin\left(\frac{n\pi x}{a}\right),$ but again, this is a real number, because $\sin x$ is real whenever $x$ is real. The last part is $\mathrm e^{-\mathrm iE_n t}$. This is actually complex, so we need to conjugate it, and its conjugate is $\mathrm e^{\mathrm iE_n t}.$ So putting it all together, we have
$$\Psi^* = \sqrt{\frac{2}{a}}\sin\left(\frac{n\pi x}{a}\right)\mathrm e^{\mathrm iE_n t}.$$


text:
Factors dominating forces between acetone and chloroform
What type of interaction is present between acetone and chloroform?
According to me, acetone has weak hydrogen-bonds with chloroform, the hydrogen on an electronegative atom has electron density that is getting pulled on by an electronegative atom nearby.
While there is weak hydrogen bonding present, are dipole interactions not more prevalent?

even weak Hydrogen bonds are way stronger than dipole interactions

@AniruddhaDeb Not necessarily - there is no threshold of how weak h-bond can be. While in past weak hydrogen bonds were ignored this case here is relatively weak, but quite real.


text:
Quantum mechanics electron probability
Schrödinger's equation yields us the solutions of 90% probability of presence of electrons in orbitals. Hence, the electrons in theory should be able to exist elsewhere in space as well.
In that case, they could exist in another atoms' vicinity, possibly in its orbital as well. So, what would be the quantum numbers in this case? Shouldn't they be affected by possible interference, hence changin quantum properties like spin, etc.
P.S. I have very limited knowledge on quantum mechanics, so please excuse me if any statement of mine is incorrect or sounds rudimentary.

This is incorrect from the very start: "Schrödinger's equation yields us the solutions of 90% probability of presence of electrons in orbitals." Schrodinger's Eq gives us wavefunctions, not just probabilities, and especially not "90% probabilities". Thought this the statement is unrelated to your question, which is about the quantum numbers if I understand well.

Orbitals aren't "90% probability regions". Orbitals are wavefunctions, which are functions that have a given value at every point in space, let's say $f = f(x, y, z)$, (although the conventional symbol is $\psi$ not $f$). The shapes that you see are our mere human attempts to visualise, or graphically represent these functions, and because the functions extend to infinity, we have to choose some cutoff for our visualisation. The 90% thing is related to this cutoff and is solely for visualisation purposes; it does not actually have much to do with the orbitals themselves.

Heisenbergs hydrogen atom is a hypothetical entity, living in a universe that is otherwise completely empty. Its orbitals properties differ a lot from those of  a single hydrogen molecule, which still differ a bit from those of a molecule in a drop of liquid hydrogen.

Ok, misconception cleared. I knew the fact that Ψ^2 represents the probability density and that orbital is a hypothetical region of space that we assume to be present for better understanding, however didn't seem to correlate the two facts together.

The issue here is that Schrödinger's equation doesn't yield us a "solution of 90% probability". It doesn't yield a probability at all. It yields a wave function, $\psi(x,y,z,t)$, which has units $\frac{1}{\sqrt{m^3s}}$. To get a probability density, we square its norm: $|\psi|^2=\psi^*\psi$. Note: even this is not a probability. A probability is dimensionless, whereas this has units $\frac{1}{{m^3s}}$. We can represent the probability of the particle described by $\psi$ being in the range (x, y, z, t) and (x+dx, y+dy, z+dz, t+dt) by multiplying the norm squared by dxdydz*dt: $P(x,y,z,t) = |\psi(x,y,z,t)|^2dxdydzdt$. Of course, these increments are infinitesimal, so this alone would evaluate to zero. However, this approach allows you to integrate over a range of x,y,z,and t:
$P(x+\Delta{x},y+\Delta{y},z+\Delta{z},t+\Delta{t}) = \int{dx\int{dy\int{dz\int{dt|\psi(x,y,z,t)|^2}}}}$.
A better understanding of this will help you understand the issues with your second question. However, we can also rely on the Pauli exclusion principle to assert that there cannot be two Fermions that share the same set of quantum numbers (note that some of the quantum numbers may be shared, but not all simultaneously.

Why do I think you're wasting time on this OP?

Ideally, the answer will help more than just the OP

Yeah, it sure could, but under better question (or at least more popular) chances would be better. There was lots of q. "what are these orbitals" already.


text:
reactions for alcohols into alkyl halides with 100% retention
I learnt that alcohols can be converted into alkyl halides in the following ways with 1/2 steps:

adding H-X (X=Cl/Br/I): primary alcohol: inversion (SN2), secondary / tertiary alcohol: racemization (forming Carbocation)

PBr3 + pyridine / SOCl2 + pyridine: primary / secondary alcohol: inversion (SN2), tertiary alcohol: no reaction as unavaliable for SN2

TsCl + pyridine: primary alcohol: inversion (SN2), secondary / tertiary alcohol: racemization (forming Carbocation, if I am correct on this mechanism)

Is there any way for me to make a 100% retention alkyl halide from any degrees of alcohol with 1/2 steps?

Use just $\ce{SOCl2}$, it gives an SNi mechanism. Also 100% retention is bit hard to achieve practically since there may be different sub reactions. and you also need an enantiomerically pure substrate inorder to verify whether it worked.

For 100% retention you need to do a double inversion


text:
Feasibility of a Reaction
I understand for a reaction to be feasible, $\Delta$H needs to be negative and $\Delta$S positive. But in terms of Gibbs free Energy Change, without considering kinectic factors, If $\Delta$H is negative and $\Delta$S is negative the reaction is not feasible at a certain temperature but for a higher temperature the reaction can occur.
Secondly with $\Delta$H as a positve value and $\Delta$S as a postive value, this reaction is also not feasible but with a certain high enough temperature the reaction occurs. I think this because $\Delta$G is always dependent on T$\Delta$S and if $\Delta$S is positive then $\Delta$G will be negative which means it is feasible.
Is this correct?

The answer would have bigger value for you, if you elaborate in the question, why you think it may be right and why you think it may be wrong.

What do you mean by a bigger value? A bigger value for Delta G?

Bigger value figurally. Like yes/no and why > yes/no only.

$\Delta H$ can be positive if $\Delta S$ is negative enough.

In short, the real dictator of whether a reaction is spontaneous or not is the sign of the change in total entropy, as per the second law of thermodynamics. All the specific conditions we memorize ($\Delta{A}<0$ for constant T,V; $\Delta{G}<0$ for constant T,P) are just convenient simplifications of this requirement, as it's very difficult to control for/measure the change in entropy of the surroundings. For example, for a reaction to be spontaneous at fixed temperature and pressure, the change in the system's Gibbs free energy must be negative. This, in turn, can be derived from the requirement that a spontaneous process result in a negative total change in entropy.
Now, onto your question! For a reaction to be feasible, we only need to consider the sign of $\Delta{S}$, where S is the total entropy. What you are referring to as S is the entropy of just the system (and not of the surroundings). This alone is not enough information to decide whether a reaction is spontaneous for any circumstance. If we do not have information about the total entropy change, then we must rely on these simplifying cases. At constant T and P,
$\Delta{G}=\Delta{H}-T\Delta{S}$ must be negative for a spontaneous reaction.
∆H could be positive, as long as T∆S were larger in magnitude (and positive). Conversely, ∆S could be negative, as long as T∆S were smaller in magnitude than ∆H.
Your overall sentiment about temperature dependence is correct, but the statements about feasibility is not (as described above).