text:
How to relate zero K to statistical analogue of entropy?
Here we have a common equation of entropy
$$ S = klnW $$
At 0 K we have only one microstate and W is 1, therefore S = 0
Now we can consider another representation of this equation
$$ S = klnQ + \frac{U}{T}\ $$
How can we use this equation to explain that at 0 K the entropy will also be equal to 0?

That would require that $Q\rightarrow1$ and $U\rightarrow0$ faster than $T\rightarrow0$ as $T\rightarrow0$.

The internal energy is defined as $\displaystyle U=Nk_BT^2\left(\frac{\partial \ln(Z)}{\partial T}\right)_V$
and the partition function $Z$ is the sum over the states
$$\displaystyle Z=\sum_i e^{-\epsilon_i/(k_BT)}\quad \text{and we shall need} \quad \frac{\partial \ln(Z)}{\partial T}=\frac{1}{Z}\frac{\partial Z}{\partial T}$$
The entropy is therefore $\displaystyle S=Nk_B\ln(Z)+Nk_BT\left(\frac{\partial \ln(Z)}{\partial T}\right)_V$
The derivative works out as $\displaystyle \frac{\partial \ln(Z)}{\partial T}=\frac{1}{Z}\frac{\partial Z}{\partial T}=\frac{\sum_i \epsilon_ie^{-\epsilon_i/(k_BT)}}{k_BT^2\sum_i e^{-\epsilon_i/(k_BT)}}$
and $\displaystyle T\frac{\partial \ln(Z)}{\partial T}=\frac{\sum_i \epsilon_ie^{-\epsilon_i/(k_BT)}}{k_BT\sum_i e^{-\epsilon_i/(k_BT)}}\tag{1}$
As $T$ tends to $0$ the exponentials tend to zero which means that $Z=\sum_i e^{-\epsilon_i/(k_BT)}\to 0$ and thus in the entropy $\ln(Z)\to -\infty$. However in the derivative eqn. 1, we are left with $0/0$ which is undefined. There may be an elegant mathematical way of evaluating eqn 1 in the limit, however, if we look at term by term we get for the first
$$\displaystyle \frac{\epsilon_0e^{-\epsilon_0/(k_BT)}   }{k_BT\sum_i e^{-\epsilon_i/(k_BT)}}=\frac{\epsilon_0}{k_BT(1+e^{-(\epsilon_1-\epsilon_0))/(k_BT)}+\cdots )}\overset{T\to 0}\to\frac{1}{0}=\infty$$
and as the exponentials tend to zero this is true of each of the terms with energy $\epsilon_1,\epsilon_2\cdots$ in the numerator. The entropy at $t\to 0$ is therefore given by $S=-\infty+\infty =0$ on the assumption that we can subtract similar infinities.


text:
Total Differential of Chemical Potential
I have seen many sources (for example this libretext) cite that the total differential of chemical potential is
$$d\mu = V_m dP - S_m dT$$
where $V_m$ is the molar volume and $S_m$ is the molar entropy. The usual justification for this formula is that chemical potential is the same as partial molar Gibbs free energy, but I was hoping that someone could make this connection more clear. My understanding is that partial molar quantities are quite literally the rate of change of a certain quantity with respect to the number of moles of a substance (please correct me if this understanding is incorrect), so saying that molar Gibbs free energy and chemical potential are the same seems purely definitional and I don't see what conclusions can be drawn from that claim or how that can be used to derive this relationship.

It's Gibbs' not Gibb's.  The chemical potential of a species tells you by how much the free energy changes if you alter the amount of that species while holding everything else constant. This is useful for predicting in which direction a reaction or process that alters the amount of the species will proceed.

Take a look at https://chem.libretexts.org/Courses/Millersville_University/CHEM_341-_Physical_Chemistry_I/07%3A_Mixtures_and_Solutions/7.03%3A_Chemical_Potential

@IanBush It is the identical link as the OP refers to.  :-)

"Purely definitional" is the best we could hope for generally, because it is not possible to derive nor proof definitions.

Whoops. But the OP should still take a look, the very first equation contradicts their assertion that the Gibb's Free Energy  and the chemical potential are the same thing. They even have different dimensions.

@JohnnySmith The partial molar Gibbs energy of species $i$ is the same as the chemical potential of species $i$, i.e. $\mu_i = \overline{g}i = \left[\dfrac{\partial (ng)}{\partial n_i}\right]{p,T,n_{k \neq i}}$. You can use either symbol. For your equation, where the fluid is pure and $g$ doesn't depend on composition, the chemical potential is just the Gibbs free energy of that single species $\mu_i = g_i$

@IanBush You have omitted "molar", referring to the article.

$$dG=VdP-SdT-\sum{\mu_in_i}$$So, $$\frac{\partial G}{\partial P}=V$$and $$\frac{\partial G}{\partial n_i}=\mu_i$$So, $$\frac{\partial^2 G}{\partial n_i \partial P}=\frac{\partial^2 G}{\partial P\partial n_i}=\frac{\partial V}{\partial n_i}=V_{m,i}=\frac{\partial \mu_i}{\partial P}$$

And the same can be done for T and  -S.

@Poutnik Yes.  Thanks.  I was hoping that the OP would realize that.

Thank you, that makes a lot of sense about how you found the coefficients on $dP$ and $dT$. However, why don't we include the terms $dn_1,dn_2,\dots,dn_k$ (excluding $dn_i$) in the total differential of $\mu_i$?

The equation assumes constant mixture composition.

Did you mean $\sum_i\mu_idn_i$ in your first equation ?

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text:
I need help for my chloride titration
I'm a 14-year-old student embarking on my first chemistry assignment. My teacher has tasked me with conducting a chloride titration of drinking water. There's just one condition: I must avoid using any hazardous substances. I've scoured the internet, but I've found only a limited number of reports. Could someone please assist me with this challenge?

Is silver nitrate regarded as hazardous by your teacher?

No, he just said the chemicals must not be carcinogenic or deadly.

For example, I am not allowed to use potassium chromate as an indicator.

Consider the Volhard or Fajans method of equivalence point detection in Argentometry. Other option could be conductometric detection.

Go to Internet Archive, search the book named Quantitative Inorganic Analysis by Vogel. It will have all the details.

The common titration methods are explained here https://en.wikipedia.org/wiki/Argentometry  Check to see what is available in the lab [That might also give a hint to the method you are supposed to use.] If you are also required to procure materials perhaps the money would be better spent buying a chloride specific electrode for use with your pH meter or a Hach Test kit. That is after you learn how each method works and can explain it to your teacher.
Chemists and people in general must learn how to work safely with hazardous materials and chemicals in general. Any school is an appropriate time to start and avoidance is not necessarily the best method. For all you know the water sample may be the most hazardous material encountered; talk to someone in Flint MI. I have worked for the NYDEC as an analytical chemist and have encountered samples with deadly cyanide content, pathological bacteria and many chemicals at levels higher than one drop of chromate indicator.

You could add an aliquot to your conical flask and add a few drops of acetic acid and dichlorofluorescein indicator along with some dextrin solution.
You can then titrate against silver nitrate solution of a known concentration. The stoichiometry of the chloride to silver ion is 1:1.
This method is callled the Fajans method.

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text:
What is something that is a strong acid and a strong reducing agent, but not a reducing agent because of the conjugate base of that acid?
So 2 answers I am not looking for is HI and formic acid. Although both of them are acids, HI and formic acid are not doing the reducing, but their conjugate forms, I- and HCOO- are doing the reducing, after protonation.
However, I would expand this question to also looking for other examples besides HI and formic acid analogies. Already taken that HCl and HBr being too weak of reducers. So therefore who else besides HI and formic acid in that category.
If the answer is, there isn't any, I would find that interesting knowing that something that is a strong base and a strong oxidizing agents exist, but without a conjugate acid form of that base. By that I mean, strong base medium-strong oxidizer or medium-strong base strong oxidizers exist, without the conjugate acid form.

How could be HI or HCOOH (or their anions) strong oxidizing agents? // You should finally learn some thorough elaboration of your questions, including compact results of prior research. Your questions often seem to be made from the same, rather weird question template.

Review the guide How to ask and Asking FAQs to prevent clarification requests, objections,  down-voting or closure.

@Poutnik oh whoops, I realized I said oxidizer in the topic, but in the body I said reducer, it's fixed.

Try to reformulate your question to be clear, it looks rather like a puzzle. Topics may be puzzles, but questions should not be. And elaborate it. If you do not put enough effort to formulate the question text, why to expect responders would put enough effort to their answers?

Poutnik, I explained they're not both at the same time. HI and formic acid are only reducers once they lose their protons 1st, become I- and HCOO-, But then, they're no longer acids.

@Mithoron that's the acidity part, once they lose their protons, then they're now reducing agents.

He means for the conjugate pair HA/A-, HA being strong acid and a strong reducing agent on its own, not indirectly because of  A- being the reducing agent. But to have any HA, it must be very concebtrated up to waterless. / /  A- will be generally always  stronger reducing agens than HA and vice versa for oxidation.

Usually acidic solutions behave as oxidant for metals M. Zinc, magnesium or iron are oxidized in acidic solutions, by the famous reaction $$\ce{M + 2 H+ -> M^{2+} + H2}$$ where $\ce{M}$ can be $\ce{Zn, Mg}$ or $\ce{ Fe}$

@Maurice But OP search of an acid as a direct reductant.

Strong acid means there only its conjugate base present, so the strong acid cannot do any reduction. And it's base is always better reductant, anyway.

Something like HCo(CO)4 would be my guess, but I have no idea how good a reducing agent it actually is, and with a pKa of 1 (https://en.wikipedia.org/wiki/Metal_carbonyl_hydride) some might debate if it is a strong acid.

IF Neal only spent all the time wasted by debating/rewriting/reopening rather on proper elaboration of his questions before posting them.....

Though I perceive the formulation of the question would benefit a revision for clarity, an answer constrained to the first half to the title of your question can hint to ascorbic acid (often named vitamin C):

For one, it is not an organic acid with a carboxylic group.  However, by its first $\text{p}K_a = 4.10$ (ref), this compound is a little bit more acidic, than acetic acid ($\text{p}K_a = 4.756$, ref).
Second, it is a known antioxidant of high relevance in (e.g. food) technology and biochemistry both in one- as well as two-electron redox reactions.  Hence it is not surprising ascorbic acid (and its derivatives) was of research interest for long time, and still is (e.g. [1, 2]).  Here, the motif of enediones adjacent to carbonyl group can yield a diketone,  a characteristic feature of the reductones.

(credit to Boorsook and Keighley, 1933)
[1] Borsook, H.; Keighley, G. Oxidation-Reduction Potential of Ascorbic Acid (Vitamin C). PNAS 1933, 19, 875–878. doi 10.1073/pnas.19.9.875 (by now available as open access).
[2] Matsui, T.; Kitagawa, Y.; Okumura, M.; Shigeta, Y. Accurate Standard Hydrogen Electrode Potential and Applications to the Redox Potentials of Vitamin C and NAD/NADH. J. Phys. Chem. A 2015, 119, 369–376. doi 10.1021/jp508308y.

But I think AA is antioxidant due its anion. Its solution is more or less stable for hours, I guess. But if alkalized by NaOH, it gets intense yellow within dozens of seconds or few minutes.

I was already aware of ascorbic acid, but speaking of that, does anyone know what's the name of the tetrahydrofuran ring of ascorbic acid? Tetrahydrofuran is if it has no double bonds, whereas furan has 2 double bonds but in different locations from ascorbic acid.

I would name the ring as a lactone ring. And the idea of the antioxidant action coming from the anion is right; blood pH is three units above the pKa.

@NealConroy Based on a similarity search in the Wikipedia structure explorer, the of 2-furanone is an other one (includes the C=O, more than merely 2,5-dihydrofuran).

The best definition of acids and bases are electron pair acceptors-acids and electron pair donors-bases. These reactions can be considered as 2-electron redox reactions or possibly a quick two step one electron transfers. Redox reactions take this one step lower to one electron transfers, indeed most redox reactions are probably a sequential set of one electron transfers. My point is that the two types of reactions are hopelessly intertwined.
Oxidative or reductive strengths are relative. Your requirements are confusing. Possibilities that might match your criteria are the intermediate oxidation states of elements such as chlorine, nitrogen and sulfur and the metals in the first transition series, especially manganese and chromium. Their reactions can be a complicated mix of acid-base reactions and electron transfers.


text:
What solvents can I use to dissolve Epichlorohydrin rubber and Chloroprene Rubber
I am working with Epichlorohydrin and Chloroprene Rubbers and I would like to cast films of 2 mm at most. The solvents I have in my lab are DMSO, THF, DMF, Acetone, Ethanol and Propylene Carbonate. Has anyone made films with these rubbers/elastomers?
Update: I have dissolved epichlorohydrin rubber in Acetone, but the casted films are easily deformed when removing from my casting dish. Now the only challenge is how I can cure them in an acetone medium

Regarding chloroprene rubber, a friend dug up the following information: CR is synthesized by polymerization in aqueous solution and can be purchased as a latex. That solution is alkaline because it tends to give off $\ce{HCl}$ in storage. The solid is described as 'soluble in aromatic solvents', but the gel component may be insoluble.
Source is: Handbuch für die Gummi-Industrie, edited by Bayer AG, Geschäftsbereich Kautschuk. 1991.

For chloroprene (neoprene), you might try chlorinated solvents, e.g., methyl chloride (chloromethane).
Epichlorohydrin has "Inferior resistance to oxygenated solvents," according to River Rubber. That's a bit vague: diethyl ether? acetone? toluene? Tahco Rubber also states it has poor resistance to chlorinated solvents.
BTW, epichlorohydrin rubber is slow to vulcanize.


text:
Criterion for Spontaneity for Closed Systems at Constant Volume and Pressure
For a closed system at constant temperature and volume, the criterion for spontaneity is $dA < 0$. However, for a system at constant composition, the total differential of $A$ is given by
$$dA = -PdV -SdT.$$
This seems to imply that $dA = 0$ for a system at constant volume, temperature, and composition. The Wikipedia page for Helmholtz Free Energy claims that this relationship holds even for non-spontaneous processes. The derivation of $A$ as criterion for spontaneity at constant volume and temperature is
$$dA = dU - TdS - SdT = \delta q - TdS + \delta w - SdT.$$
If we consider only $PdV$ work, we obtain
$$dA = \delta q - TdS - PdV - SdT$$
and at constant $T$ and $V$, we get
$$dA = \delta q -TdS.$$
The criterion $dA <0$ for spontaneous processes then follows from the Clausius inequality. These two equations seem to be contradictory, so I was hoping someone could explain how this can be reconciled. Thank you so much!
Clarification:
The way I derived the $dA = -PdV - SdT$ for an arbitrary process is as follows:
For a function $f(x_1,\dots,x_n)$, the complete differential of $f$ is
$$df = f_1 dx_1 + f_2 dx_2 + \dots + f_n dx_n$$
where
$$f_i = \left(\frac{\partial f}{\partial x_i}\right)_{x_j \: j\neq i}.$$
Consider the function
$$g = f - f_1 x_1.$$
The differential of $g$ is
$$dg = df - f_1 dx_1 -x_1df_1$$
so that
$$dg = -x_1 df_1 + f_2 dx_2 + \dots + f_n dx_n.$$
Now, applying this to the special case of internal energy $U(S,V)$, if we define
$$A = U - \left(\frac{\partial U}{\partial S}\right)_{V} S = U - TS$$
then the above implies that
$$dA = -SdT - PdV$$
since
$$dU = TdS - PdV.$$

Consider review of search results for site:stackexchange.com OR site:libretexts.org thermodynamics closed system spontaneity criteria

Provide you reasoning what you think they are contradictory. as dS >= dq/T,  dq - TdS <=0.  0 for reversible processes, <0 for spontaneous ones.

For a spontaneous process at constant temperature and volume, the fact that $dQ < TdS$ implies that $dA < 0$ with a strict inequality. However, the fact that $dA = - PdV -SdT$ seems to imply that for any constant volume and temperature (and constant composition), we have $dA = 0$ since $dV = 0$ and $dT = 0$. This is the main source of my confusion.

dA=-p.dV -SdT is derived with assumption dq=T.dS, i.e. for reversible, non spontaneous changes, so TdS and -TdS mutually cancel.

The link I provided to the Wikipedia article states that this relationship holds for non-reversible processes as well. That is where I am confused

dU=TdS - p.dV only if dQ=TdS what is true only for reversible processes where dS=dQ/T

I see. Thank you so much!

$$A = U -TS$$
$$\mathrm{d}A = \delta Q - p\mathrm{d}V + \delta W_\text{nonexp} - T\mathrm{d}S - S\mathrm{d}T$$
$$dS \ge \delta Q/T \implies T\mathrm{d}S \ge \delta Q $$
Assuming $\delta W_\text{nonexp}=0$:
$$\mathrm{d}A = \delta Q - p\mathrm{d}V - T\mathrm{d}S - S\mathrm{d}T$$
As $T\mathrm{d}S \ge \delta Q$:
$$\mathrm{d}A \le T\mathrm{d}S - p\mathrm{d}V - T\mathrm{d}S - S\mathrm{d}T$$
$$\mathrm{d}A \le - p\mathrm{d}V - S\mathrm{d}T$$

This makes sense. Does the derivation of $dA = -PdV - SdT$ using Legendre transformations assume reversibility? I thought it holds in general

Because the statement that "the natural variables of A are V and T" seems to imply that A(V,T) is constant for any process at constant V and T so that dA = 0

It uses the A definition, the Clausius inequality TdS>=dQ and ordinary mathematical operations. // AFAIK, the Leg. tr. was used and the source for decision of the way to define A and G as state functions.

Aren't all of those true in general? I don't see why that derivation is only valid for reversible processes

But I as a chemist  am not familiar with Legendre transformations by the name, rather just Legendre orthogonal polynoms.

Thank you for your help! I added a clarification that details the derivation I used and this process is known as Legendre transformation: https://en.wikipedia.org/wiki/Legendre_transformation

Thank you for your help! One quick clarification with your answer: if $dA = -pdV - SdT$ is only true for reversible processes, why is it true that the natural variables of $A$ are $V$ and $T$ so that $A$ is a function of $V$ and $T$ only?

The point is, A is not function only of V and T. If the system underwent chem. reaction, phase transition or other spontaneous process, A decreased even at constant V and T.


text:
"Is there any library in Python or software for analyzing raw chromatography data?"
I have taken the data directly from the FID sensor using a DAQ, so I have the chromatography data in this format:
(2 data colummns in a txt file) (18000 rows)

Time (s) Intensity (V)
0.00367___0.0008
0.00530___0.00049
0.00672___0.00025
0.00808___0.00009
0.01014___0.00057
0.01290___0.00128
0.01545___0.00136
0.01809___0.00112

I have this data in 2 minutes cycle.
Is there any way I can process this data and obtain the gas concentrations? I understand that a calibration run is required, but what I'm most interested in right now is finding the areas, peaks, and retention times."
Thanks

For plain text formatting by StackExchange markdown, see as inspiration SE meta - formatting // SE meta -  tables  // Advanced formatting

Did you consider a search by topic e.g. on PyPI.org?  Though not yet tested myself on chromatography data, there are reports fityk equally works well on chromatograms, too

Googling ‘python chromatography’ yields multiple results.

The biggest problem here is determining peak boundaries and the baseline.. What vendor are you using? Do you have raw data files instead of the table?

I have a table of intensity and time data for each line that can be seen on the graph. I do not have a supplier since I am taking the data directly from the sensor.

@StanislavBashkyrtsev I have found that for reasonable-quality data, where you don't need to torture it too much (e.g. you are well above the limit of detection and in a linear region of a calibration curve, and where retention times don't drift too much) it is usually workable to avoid any fancy peak identification and integration algorithms: define peak boundaries by inspection and manually once per data set, and "integrate" just by summing raw intensity values in the boundary.


text:
Mechanism of deprotection of enol thioether
I was reading some enolate chemistry from Carruthers textbook and came across the selective alkylation of unsymmetrical ketones. The reaction involves blocking one alpha-position of the ketone using an enol thioether, as shown below.

I cannot understand the mechanism of the final step and while searching, I ended up with this from Sciencedirect (Ref.1)

Removal of the thioenol ether protecting group gives the aldehyde function, and under the acidic conditions the β-hydroxy group eliminates to give the mainly (E)-unsaturated enal. This reaction also works if the ketone is reduced to a secondary alcohol with LiAlH4 (69% yield).

My questions are:

Are enol thioethers the same as thioenol ethers? If not, please draw an example.
If they are the same, does the deprotection take place through an aldehyde intermediate as stated above?
Also, please explain the mechanism of this deprotection.

Ref. 1: Warren J. Ebenezer, Paul Wight, in Comprehensive Organic Functional Group Transformations, 1995
Ref. 2: Modern Methods of Organic Synthesis 4th edition, page no. 10

Mechanism:  1 -hydroxide anion adds to the =CHSBu carbon in a Michael addition giving the enolate anion of the carbonyl and -CHOH(SBu).
2- The enolate anion deprotonates the newly arrived -OH group to reform the carbonyl group
3- The alkoxide anion displaces BuS- to form the aldehyde
4- retroaldol gives the unsubstituted carbonyl

What you have here would conventionally be a thioenol ether. I've never heard the term enol thioether or enol sulfide before, but I'd assume it is the same because a thioether which is not in a enol structure would normally be classified as a sulfide.

Yes, it does. As stated it is the beta hydroxy tautomer of the beta aldehyde which is the key intermediate.

I believe the mechanism involes first hydrolysis of the thioenol ether to a thiocarbonyl. Then the thioaldehyde is hydrolysed further to an aldehyde. From here, the key step is a retro claisen condensation to remove the aldehyde group. The presence of base and glycol and water would support this mechanism. Formic acid and some formate esters would be found as by-products.

The term enol thioether is written in the book itself, and the given structure is in the enol form. And, what is enol sulfide here? Also, someone please explain the mechanism clearly.


text:
A general doubt regarding extensive/intensive properties and units in thermodynamics
I know that for reversible processes: entropy change = dq/t
Here its unit should J/Kelvin but it is mentioned in my books that its unit is J/Mol-Kelvin. I do not understand why there is mole in denominator of unit.
Also, if J/mole-Kelvin is the unit, then why it is an extensive property? Extensive properties vary based on quantity, but this is strictly per mole.

dS=dq/T, not dS=dq/dT.  When you have moles in the denominator, you are talking about molar entropy which is an intensive property (as you noted).  Since, in practice, when you use the tables, the number of moles is going to be exclusive to your system, it is necessary to tabulate the molar entropy ( or specific entropy) rather than the extensive entropy.

thanks @ChetMiller for correction

In physical chemistry/thermodynamics they remind you that:

Intensive properties are independent of the quantity of material in the system, such as pressure $p$, temperature $T$, electric potential $\phi$, composition $x_j$, etc.
Extensive properties are dependent of the quantity of material in the system, such as volume $V$, entropy $S$, kinetic energy $K$, etc.

However, for many thermodynamic functions, we can turn them into intensive properties dividing by the total amount of the system. I'll show you the notation that I like. Let $X$ denote the thermodynamic property $Q$, $S$, $G$ or $V$, then $x$ is defined as
\begin{equation}
  \boxed{x \equiv \frac{X}{n}} \tag{1}
\end{equation}
where $n$ is the total amount of the system. Here $x$ has the name of molar thermodynamic property, and may be $q$, $s$, $g$ or $v$
For example, $g$ is the molar Gibbs energy (intensive) in $\pu{J mol^-1}$ and $G$ is the Gibbs energy (extensive) in $\pu{J}$. $v$ is the molar volume (intensive) in $\pu{m^3 mol^-1}$ and $V$ is the volume (extensive) in $\pu{m^3}$.
Thus, for the following equations
\begin{equation}
  \mathrm{d}S = \frac{\unicode{x0111}Q_\mathrm{rev}}{T} \quad
  \mathrm{d}s = \frac{\unicode{x0111}q_\mathrm{rev}}{T} \tag{2,3}
\end{equation}

Eq. (2) denotes entropy $S$ in units of $\pu{J K^-1}$, and heat $Q_\mathrm{rev}$ in units of $\pu{J}$.
Eq. (3) denotes molar entropy $s$ in $\pu{J mol^-1 K^-1}$, and $q_\mathrm{rev}$ denotes molar heat (although this sounds weird) in $\pu{J mol^-1}$.

Note that not all quantities depending on material quantity are extensive. There must be proportionality. X(an)=a.X(n). By other words, there are quantities that are neither intensive nor extensive.

For differentials of W and Q, it is better to use  small "dyet"("d with stroke") Unicode symbols as for inexact differentials $đQ$ (U+0111 or by (Win)compose Compose / d ).  It is sometimes substituted in CH SE by MJ \delta  ($\delta Q$), but it is not exactly correct. OTOH, \delta looks better in context of MJ typesetting.

@Poutnik Hello Poutnik, sorry for the late reply. Thanks for the comments, you are right, the differentials are now correctly written.

Also thanks @Maurice for the modification!

So the 'mol' or mole is just a number, and so you would be perfectly valid to give entropy in units of either J/mol/K or just J/K. You can interchange between the two by multiplying or dividing by the Avigadro constant.
Also you are right about the second part, if entropy of a substance was reported in units of J/mol/K then that is a standard entropy change which is an intensive property. Entropy just generally however does depend on the number of moles of substance and so is an extensive property unless the molar quantity is specified.
Hope that helps.

1 mol is 1 of 7 fundamental SI units, so not dimensionless. Involving N_A would be a disaster.

Molar entropy $S_\text{m}[\pu{J K-1 mol-1}] = \frac{S [\pu{J K-1}]}{n [\pu{mol}]}$. Or differentially, $S_\text{m} = \left(\frac{\partial S}{\partial n_i}\right)_{p,T,n_j,i \ne j}$


text:
How can I avoid magnesium smoke during melting?
I am trying to make a magnesium-nickel alloy. I carry out the melting process in an inert gas, but during the process, the magnesium smokes. What can I do?

Ni has melting point 1455 °C, Mg has boiling point 1091 °C, so hard to advice. If Mg is immersed in liquid Ni, it would float and happily boiling. The question is, if Ni(s) sitting on the bottom of Mg below its boiling point would dissolve in Mg or not. Not sure if pressurized melting/alloying is available technology/security-wise. It could easily act as the high speed uranium anti-tank projectile when getting inside the task.

Short, laconic questions usually need a big effort and long clarification. They do not  attract useful answers, but rather votes for closure until they are more elaborated. Review the guide How to ask and Asking FAQs

This pdf might be helpful.

"Inert gas"? You don't mean nitrogen, do you? Because that isn't going to work.

Yeah, I know, I'm using argon

Is Mg and Ni miscible?  I have read somewhere Mg+Ni alloys can be made only by powder metallurgy, as the melted metals tend to separate.

Which particular phase?

I want to make Mg2Ni

First off, let us assume you are using argon as the inert gas. Nitrogen is not inert to magnesium.
You may want to look at the Mg-Ni phase diagram from Ref. 1. The horizontal axis is atomic percent nickel, with 10% spacing between grid points. At low nickel concentrations, you have an eutectic with melting point 508 °C, so you can form a melt with that composition without overheating the magnesium. If you need a higher level of nickel, you may want to start with the eutectic and incrementally add nickel and heat to "follow the shoreline".

Reference

Czerwinski, Frank. (2011). Welding and Joining of Magnesium Alloys. 10.5772/13947.

In order to avoid working at the extremes with one metal evaporating before the other melts, it's useful to start with an alloy very rich in the low-melting material and add it to the melt of the other metal. For example, ferromolybdenum  (https://en.wikipedia.org/wiki/Ferromolybdenum) is created by  aluminothermic reaction, and that concentrated Mo source is then added to molten iron.


text:
Pressure degasser of water
Currently, I am working on a project where we are using pressure degassing. The idea is to make a test setup where we lower the pressure in a flask to 'degas' the solution. The solution is just distilled water, but we plan on doing other water samples later on. The background of this project is associated with carbon capture from seawater, so we want to lower the pressure to a point where a lot of the carbon dioxide concentration in the water is moved to the gas part of the flask.
Given that we start at room temperature, and we don't want to remove ALL the carbon from the solution (due to marine life), what minimal pressure (or pressure drop in comparison with ambient pressure) would be recommended for this experiment?
Relevant papers are also very much welcome,
Thanks in advance!

Note also there are high levels of bicarbonate is sea water and intensive degasing would lead to alkalisation. There should be used pH threshold at which degassing should stop.

Also consider that at constant room temperature, water will boil at ~3kPa pressure, or if allowed not warmed, the temperature drops as pressure is reduced. See https://www.engineeringtoolbox.com/boiling-point-water-d_926.html Also consider the time for this to equilibriate. Perhaps this is best determined by experimentation with known samples.


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Electrodeposition using Ionic liquids
I haven't found to many research papers yet, but would it be possible to electroplate nickle-aluminide onto a niobium alloy using an ionic liquid? I know certain industries are already using this combination for rockets or thruster according to my research so far. I'm looking in to using something like 1-ethyl-3 methylimidazolum bis (trifluoromethyl-sofonyl)mide or [EMIM] [TG2N] as the ionic liquid in the electroplating to act as the solvent. With what I know so far, should this be a plausible process for electroplating a niobium alloy?

Please do not duplicate posts! Remove this question or https://chemistry.stackexchange.com/questions/176834/electrodeposition-using-an-ionic-liquid . If you wish to make a change, edit the question you've submitted.

Nickel-aluminum alloys can indeed be deposited from solutions of nickel chloride in an organic chloroaluminate salt. As one example, Ali and Nishikata [1] describe an electrodeposition process on platinum and mild steel. From this reference:

Electrodeposition of aluminum–nickel intermetallic compounds (particularly Ni3Al) has been carried out onto platinum and mild steel cathodes from a 2:1 (mole ratio) aluminum(III) chloride-N-(n-butyl)pyridinium chloride (BPC) molten bath saturated with nickel(II) chloride at room temperature. A single phase of Al–Ni alloy is difficult to obtain by controlled-potential and controlled-current methods; however, it can be obtained by pulse current plating. ...

For a broader eeview, consult Oriňáková et al.[2].
References

Ali, Muhammad & Nishikata, Atshushi & Tsuru, Tooru. (2001). "Electrodeposition of Al–Ni intermetallic compounds from aluminum chloride-N-(n-butyl)pyridinium chloride room temperature molten salt". Journal of Electroanalytical Chemistry. 513. 111-118. 10.1016/S0022-0728(01)00607-6.

Oriňáková, Renáta & Morovska Turonova, Andrea & Kladeková, Daniela & Gálová, Miriam & Smith, Roger. (2006). "Recent developments in the electrodeposition of nickel and some nickel-based alloys_. Journal of Applied Electrochemistry. 36. 957-972. 10.1007/s10800-006-9162-7.


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Any compound which can react with some unknown calcium compound
I have a body made of steel which needs to be cooled, and it is used tap water to do it and over the time that water has deposited some kind of calcium compound inside the cavity where it's running through and is damaging the cooling process. So I need some compound which could react to the calcium which has been deposited but not with the steel.

To your title question: yes there is such acid, it is called water. The body of your question contradicts the title, though. Why would you need something that reacts with calcium when you don't have calcium?

You seem to confuse calcium as element with calcium containing compounds.

@IvanNeretin sorry, I'll edit the question to clarify this. I have a body made of steel which needs to be cooled, we use tap water to do it and over the time that water has deposited some kind of calcium compound inside the cavity where it's running through and is damaging the cooling process, so I need some compound which could react to the calcium which has been deposited but not with the steel.

Then try diluted acetic acid.

@Poutnik yes, I messed up writing the question because the calcium compound is unknown. I was talking with my professor and we're not sure what could be, our bet it's calcium fluoride or hydroxide. Because we're running tap water through a cooling system of a body of steel and that's unknown compound is damaging the cooling process.

I have wanted to suggest calcium carbonate myself, as Maurice did in the answer.

There are many lime (CaCO3) deposit removers, some sold commercially for use at home. However, if you have no idea what the compound is, then there can be no definitive answer.

Is your water supply know to be "hard"? That is - does it contains a high level of Calcium and Magnesium ions?  A commercial kettle descaler will most likely sort this.

Your unknown calcium compound is probably calcium carbonate $\ce{CaCO3}$, which is produced spontaneously when tap water is heated. Tap water contains the ions $\ce{Ca^{2+}}$ and $\ce{HCO3^-}$ which are both slightly soluble in water. When heated the ions $\ce{HCO3^-}$ are decomposed according to $$\ce{2 HCO3^- -> CO3^{2-} + H2O + CO2}$$ And then the ions $\ce{CO3^{2-}}$ react immediately with the calcium ions $\ce{Ca^{2+}}$ present in tap water to produce calcium carbonate $\ce{CaCO3}$, which is insoluble in water $$\ce{Ca^{2+} + CO3^{2-} -> CaCO3}$$
This deposit of $\ce{CaCO3}$ can be removed and dissolved by any dilute acidic solution, like acetic acid, as proposed by Ivan Neretin. The reaction is due to the ion $\ce{H+}$ present in the acidic solution : $$\ce{CaCO3 + 2 H+ -> Ca^{2+} + H2O + CO2}$$
If the acidic solution is diluted, the iron atoms of the steel are not corroded significantly by the $\ce{H+}$ ions of the solution.


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Why do grignard reagents add to carbonyls at the carbonyl carbon, and not deprotonate at the alpha carbon to form an enol?
I thought that since grignards are such strong bases that they would be immediately protonated by the alpha hydrogen since it is an acidic hydrogen. However, I keep seeing answers that are saying that they just add to the carbonyl. I understand that they are strong bases, and therefore make bad leaving groups (forming a stable tetrahedral compound) once they add to the carbonyl, but I don't understand why they would add rather than deprotonate.

https://chemistry.stackexchange.com/questions/134270/grignard-reagent-will-act-as-a-base-or-a-nucleophile https://chemistry.stackexchange.com/questions/113023/grignard-reagent-and-acidic-hydrogen


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Which enantiomer of 3-Chloro-2-pentanol is this?

I think that there is a chiral centers at $C_2$ and at $C_3$ (Numbering starting such that the carbon with hydroxy attached is number 2.
Then, for $C_2$:

Priority would be: OH > $C_3$ > $C_1$ > H
This makes it clockwise, so R

Then, for for $C_2$:

Priority would be: Cl > $C_2$ > $C_4$ > H
This makes it anticlock wise, so S

Therefore, the molecule becomes 2R, 3S 3-Chloro-2-pentanol.
Is this correct?

Clockwise/counterclockwise change their meaning depending on whether the H is facing towards you or away.

@IvanNeretin I know, however here I do not have any information about the H? Do I?

@IvanNeretin I just realised, that in $C_2$, H is actually, at the front if CH3 is at the back. Then, I would have to switch them up, and it would actually become 2S, 3S. Is this correct?

Now that looks correct to me.

@IvanNeretin Thanks

@NaitikMundra A program like ChemDraw (sample page) can provide a check: draw the structure, mark them by the lasso, use the structure2name function (bottom row of the icons, the fourth counting from the right hand side). But: don't rely blindly on a computer program (which may include an [accidental] error of implementing the CIP rules, and nomenclature in general) -- continue your training to be able to apply the rules by yourself.

@NaitikMundra The site's resources actually compiled reference material about chemical nomenclature, too.

@Buttonwood Thanks! The answer was in front of me and I never noticed.


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Acid hydrolysis and alkaline hydrolysis
Alkaline hydrolysis is not called "alkali hydrolysis", while acid hydrolysis is not called "acidic hydrolysis". Why? I am asking this question as I believe it may not be a simple naming problem. Instead, I think it may be related to some characteristics of the reactions.

What is alkaline (adjective, basic) need not to be alkali (noun, hydroxide of alkali metals). Alkali does not hydrolyze to write "alkali hydrolysis". Alkaline hydrolysis is hydrolysis in alkaline environment. We could write "hydrolysis by alkali" (or possibly less precise "alkali hydrolysis" in sense of "hydrolysis by alkali") as specific alkaline hydrolysis using a hydroxide of an alkali metal.
Acid hydrolysis is hydrolysis of an acid in sense of acid decomposition by water, e.g. less known autocatalytic hydrolysis of otherwise quite stable (gaseous at room temperature) carbonic acid $\ce{H2CO3->[H2O]CO2 + H2O}$. An acidic hydrolysis is hydrolysis in acidic environment.


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pKa and pKb in solvents other than water
I know that when you have aqueous solutions, then the pKa and the pKb are linked to each other (pKa + pKb = 14). Is there an equivalent relationship between pKa and pKb in non aqueous environments? I'm currently looking at a paper where they have measured pKb values in glacial acetic acid (https://pubs.acs.org/doi/epdf/10.1021/ic50084a055), but I'm interested in finding the pKa values for a different set of compounds in glacial acetic acid. Thank you.

Closely related, possibly duplicate:  https://chemistry.stackexchange.com/q/28022/35806

Bruckenstein and Kolthoff give some values in the abstract, which is available in the free preview. Notable among acids: perchloric acid, $pK_a=4.87$; hydrogen chloride, $pK_a=8.55$.
Also given is an autodissociation constant, whose base-10 logarithm is $-14.45$. Thus whereas we render $pK_a+pK_b=14$ in water under ambient conditions, the corresponding constant in glacial acetic acid would be $14.45$. As an example, citing the hydrogen chloride acid dissociation constant given above, we can compute the $pK_b$ of the conjugate base chloride ion as $14.45-8.55=5.90$ (significant zero).
Reference

S. Bruckenstein and I. M. Kolthoff (1956). "Acid-Base Equilibria in Glacial Acetic Acid. III. Acidity Scale. Potentiometric Determination of Dissociation Constants of Acids, Bases and Salts". J. Am. Chem. Soc. 78, 13, 2974–2979.
https://doi.org/10.1021/ja01594a014

Wow, that is unexpected. I have been assuming perchloric acid is still strong in acetic acid, considering it is used for titration of weak bases there. But I admit I have zero experience in the area.

This is brilliant! Thank you for the reference. I'm wanting to use perchloric acid for the non-aqueous titration, so this is also a really useful resource for that value.

From the autodissociation constant aso given in the abstract, we calculate that the neutral "pH" in acetic acid is about 7.2. Perchloric acid with $pK_a$ between two and three units less can reach that condition with most bases; hydrogen chloride can't (unless you go far beyond the equivalence point) and sulfuric acid is marginal at best.


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Why does burnt milk on bottom of pan have 'vaguely' cork-like pattern?
I burnt milk while boiling it. I noticed that the burnt milk on the bottom of the pan, formed a pattern that vaguely resembled the pattern of cork-cells . Why is this pattern being formed?

Can you say how such patterns are cork-like?

@RobbieGoodwin I am just using it as a description...It does not need to exactly match the pattern found in cork cells- https://www.researchgate.net/figure/Cork-cellular-structure_fig1_26635525

Milk comprises sugars and fats, with the sugars being prone to caramelization upon exposure to heat from stovetop. You can see the developing brown color and if you scrape further, you will see burnt organic material (black). The observed markings result from bubbles that were unable to ascend to the surface and consequently became entrapped and collapsed at the base of the pan during the boiling process.

So you mean that that when the bubble was being formed at the base, the milk under it had kind of thickened and so when the bubble collapses it causes an indent?

The pattern has everything to do with the interacting of opposing forces such as in convection, combined with mass transport, precipitation, and agglomeration of particles in the fluid. The same pattern can be formed regardless of the chemistry involved and is certainly not a crystal.
See Turing patterns, which exemplify "how patterns in nature, such as stripes and spots, can arise naturally and autonomously from a homogeneous, uniform state."

https://en.wikipedia.org/wiki/Turing_pattern
Also check out a convection specific instance of pattern formation, Bénard cells, but similar: https://en.wikipedia.org/wiki/Rayleigh%E2%80%93B%C3%A9nard_convection
In this case, the bubbles form hundred/thousands of times, from the same locations called nucleation sites. Each time pushing a little more flocculated milk protein to the edge of the nucleation site. The patterns of the nucleation sites themselves are random at first and then become evenly spread by a process of symmetry breaking, where the nearby sites make it harder to form a site so they become an average distance apart, making the pattern. The physical remnants of the pattern are what you see.

Does this convection pattern apply in this case as there is no convection happening?

What do you mean? Heating the fluid from below 100% caused convection.

I mean, those benard cells are formed during convection right? In this case, convection is not happening at the moment right?

@Pumpkin_Star See my update, you are just seeing them remnants of the process that formed the pattern. The bubbles physically push coagulated milk solids away from where the bubble forms.

If you watch carefully boiling of milk or water at lower level of pan , bubble form due to the high heat at bottom rapidly ,the shape you see there is just the result of that the cavity is result of complete evaporation of water and fats get stick to the corner of bubble shape.

The other answer says that the indentations were formed because the bubbles collapsed. So is it because the bubble collapsed or is it because when the bubble escaped to the surface, it left behind the solid in a cavity like shape? Or can it be both?

What u mean by bubble escape when their is overheating at last min there's no water to form bubble all left is that circular ring that form as remain of milk fats being pushed away at circumference.imagine like when water boil it spread hit a circular boundary of your pan . There also happen something like that.

Please read the answer too. That answer suggests that the bubbles collapse and form those cavities, while your answer states that the cavities are formed 'while' the bubble is being formed... There is a slight difference between these two statements, I think.


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Is mercury(I) chloride ionic or covalent?
Is mercury(I) chloride ionic or covalent? If I look here, I see arguments either way. The article doesn't mention anything about a cation or an anion.
Some might think there's two mercury ions each 1+ but it seems to me that's not the case, as  the article mentions that the chemical is also known as "mercurous chloride". And I see there is such a thing as a mercurious ion, which I understand to be a polyatomic ion with charge of 2+, so that would suggest it's an ionic compound.
This Wikipedia article also mentions that it's "Also known as the mineral calomel" and it says calomel is a salt "mercury salts (such as calomel)". So that indicates to me that it's an ionic compound.
But the article on mercury(I) chloride also says  "Hg2Cl2 is a linear molecule". Well, that suggests it's a covalent compound.
I've also heard that it has "a relatively low sublimation point which also implies molecular".
The Wikipedia page also mentions "the Hg–Cl bond length in the linear Hg2Cl2 unit is 243 pm"
I suppose that might suggest covalent, because as I understand it, "bond length" is normally mentioned for covalent bonds rather than ionic bonds.. Though with ionic bonds the concept of bond length could apply though i've heard it's usually called "interionic distance".
Looking at the pictures, it has this one

And I understand that that style of diagram is usually used for ionic compounds..
It shows a unit cell  (which seems to show maybe linear molecules)

And as I understand it, a unit cell could apply not just to an ionic crystal, but to a molecular crystal or a network covalent crystal
So I'm really confused as to what the structure is. I can't tell whether it's ionic or whether it's covalent molecules?

related https://chemistry.stackexchange.com/questions/37141/why-does-mercury-form-polycations

https://chemistry.stackexchange.com/questions/86918/the-impossibility-of-100-ionic-bond https://chemistry.stackexchange.com/questions/17064/is-kf-the-most-ionic-compound?noredirect=1&lq=1

There is no clear-cut division between the two. Let that sink in.

Hg2Cl2 crystals are definitely molecular, and HgCl2 too. They still dissociate in water to some degree.

@permeakra yes I think it is ionic and molecular.  Ionic because it's composed of ions  $Cl^-$  --- $Hg_2^{2+}$ --- $Cl^-$     And molecular because those $Cl^-$  --- $Hg_2^{2+}$ --- $Cl^-$  entities are separate/discrete, molecules.  No covalent or ionic bond between one of those entities and another - only within. And I suppose if attempted to dissolve in water then it'd separate into $Cl^-$, $Cl^-$, and $Hg_2^{2+}$. The  $Hg_2^{2+}$ ion remaining intact as it's bonded covalently.

@barlop Stop. A crystal is molecular if it contains clearly distinguishable molecules which Hg2Cl2 (and HgCl2) clearly do. Mercury in general doesn't favor ionic bonding due to its high ionisation energy (making it hard to produce a ion) and low charge density of the cations (making it hard for ion to stabilize by ion-dipole or ion-ion interactions). This is evidences by many compounds with mercury-alkyl bond that are stable in water solutions (!!!)

See: https://chemistry.stackexchange.com/questions/168056/why-is-hg2cl2-less-soluble-in-water-than-hgcl2

@permeakra I don't understand your comment.. e.g. you write "Stop. A crystal is molecular if it contains clearly distinguishable molecules which Hg2Cl2 (and HgCl2) clearly do" <-- I am saying it is molecular!
Also, I said, "it is ionic and molecular". Ionic doesn't mean an ion. Ionic means composed of ions.  Each molecule is composed of ions, three of them!

A ionic crystal is one with structure dominated by ionic interactions. The charge separation in $\ce{Hg_2Cl_2}$ is small, with Mullicken charges about 0.2e at most [1].  By comparison, alkali halides have Mullicken charges at about 0.8e [2] It is further evidenced by direct $\ce{Cl-Cl}$ contacts that would be impossible if there was significant charge separation in the molecules.
[1] https://doi.org/10.1016/j.matchemphys.2021.125379
[2] https://doi.org/10.1039/C9RA05190B
Background. Mercury is a weird metal that has unusually high for metal first and second ionization energy at 10.4 and 18.7 eV respectively. This is comparable with heavier halogens with first ionization energies in 10-15 eV range. At the same time, mercury cations are large at about 110-120 nm, which is comparable to calcium with ~110 nm and ionization energies at 6.1 and 11.2.
This means, that mercury strongly favors covalent bonding despite its relatively low formal electronegativity. This is evidenced by various alkylmercury compounds that are stable in contact with water.
Of course, there is a catch in that first and second bonds formed by mercury atom are different. When first bond is formed, it uses mercury's s-orbital that is very low in energy since it has considerable electron density near highly charged nucleus. When mercury forms two bonds, it by necessity utilizes at least one p-orbital, which is higher in energy. Same actually applies to zinc and cadmium but to much, much lesser degree. This is evidenced by formation and relative stability of $\ce{M2}$ moieties and annoying persistence of $\ce{CH3Hg+}$ cation in environment.
So, returning to your question. $\ce{Hg2Cl2}$ and $\ce{HgCl2}$ are mostly covalent compounds with molecular crystal structures. There is a degree of charge separation in those molecules, but it is fairly low. Both only partially dissociate in water. This simple picture is complicated by fairly diffuse outer shells of halogens and mercury, meaning there are strong inter-molecular dispersion interactions between individual molecules in the solids. But they are clearly not ionic.

Well, I  commented that it's ionic and molecular. You agree that it's molecular. So the question is then if it's ionic.   Ionic means composed of ions, it doesn't mean that the thing is an ion. e.g. NaCl is ionic but not an ion. (And indeed NaCl isn't molecular).  You write "individual molecules in the solids. But they are clearly not ionic." <-- Each molecule consists of three ions  $Cl^-$---$Hg_2^{2+}$---$Cl^-$  so it is ionic by definition(or at least is in that regard!)!

@barlop No, the molecules are covalent, that is they ARE NOT composed of ions. I edited the answer to give hard charge numbers, those were realatively easy to find.

Thanks. What do you make of the fact that it has a reported ksp https://en.wikipedia.org/wiki/Mercury(I)_chloride   ksp of 1.43×10^-18  .  and the idea that a reported ksp would suggest that it breaks into ions when in water?

just calculate the ionic product using solubility data on the same page.

I know that a ksp can be calculated from solubility data, and somebody could do that whether or not there are ions. The calculation does treat things as if there r ions.  I wondered if maybe if a reliable source states something as having a ksp then theyd have checked that it's ionic, but I guess not? eg i've heard that atomic absorbtion spectroscopy can be used to measure ksp(and that'd really only apply where there are ions as it'd detect them). I suppose in this case it's (clearly) just calculated  from solubility data not via AAS. So shouldn't be taken as an indication of there being ions?

Just calculate it. It isn't hard.

I did and it came to the same number hence I said "I suppose in this case it's (clearly) just calculated from solubility data not via AAS. So shouldn't be taken as an indication of there being ions? " (I expected you might say yes, but you haven't yet!).

I doubt you did. ghci> let x = ( 0.002/472) in x * 2x2*x evaluates to
3.0431543633964523e-16 . This is about two orders of magnitude off.


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Higher or lower temperature to produce maximum yield of NH3(g) in Haber process
Question: In order to get the highest yield of NH3(g) in Haber process, should you use high or low temperature?
The answer in the textbook says lower temperature produces more yield of NH3(g) in Haber process since it is exothermic. Doesn't lower the temperature also lower the reaction rate which results lower yield? My teacher says higher temperature produces more NH3(g) even though the the reaction shifts to the reactant side.

Industrial processes have to trade off yield and speed for many reactions. Even if a higher yield could be obtained at a lower temperature, this is not helpful if that equilibrium takes a very long time to achieve.

Does this answer your question? Why is the Haber process carried out at such high temperatures?

The book is right, and so is your teacher.

The reaction is exothermic, $\ce {N2 + 3 H2 -> 2 NH3}$, ΔH° = -91.8 ~kJ/mol, so higher temperature leads to decomposition, i.e., makes $\ce{2 NH3 -> N2 + 3H2}$ the preferred reaction.
The rate of reaction would be rather slow at STP (i.e., not coming to equilibrium within the lifetime of the universe), so even at high pressure (commonly 20 MPa), using a good catalyst, it is run at ~775 K because that optimizes profitable throughput.

Think also of carbon: at STP, graphite is more stable than diamond, but have no fear your gems and jewelry will change allotrope in your lifetime.

The oversimplified and probably expected answer is the higher yield is provided at lower temperature, as the reaction is exothermic and its equilibrium constant decreases with temperature.
But the scenario is more complicated and there are multiple factors to consider, as the Haber-Bosch process is more than just the reaction $\ce{3 H2 + N2 -> 2 NH3}$. (Considering temperature, pressure, flow, production rate, gas flow cost, catalyst efficiency, production loopbacks..)

The initial reaction rate of synthesis of $\ce{NH3}$ exponentially raises with T.
The reaction rate constant of $\ce{NH3}$ decomposition raises with T faster than the reaction rate constant of its synthesis.
Above some reaction yield, the net reaction rate (initially faster)  becomes slower for higher T than for lower T.
Above some even higher reaction yield, the net reaction rate becomes slower for higher T due progressively fast backward reaction.
As the Haber process is about $\ce{NH3}$ production, lower reaction yield with higher gas flow leads to higher production rate.
For a given pressure, there exists such a "neutral gas flow" passing the catalyst providing a product "neutral yield" that more or less does not depend on T.
If there is chosen gas flow lower than "neutral gas flow", the yield is higher than the "neutral yield" and than lower temperature leads to the higher yield.
If there is chosen gas flow higher than "neutral gas flow", the yield is lower than the "neutral yield" and than higher temperature leads to the higher yield.
Overall, tuning up parameters to optimize Haber-Bosch process is not a trivial task.


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What chemical species is really acting as oxidizer in hot concentrated sulfuric acid
In some japanese website it is written that hot conc sulfuric acid has greater oxidation power than that of normal temperature since it contains sulfur trioxide formed by decomposition of the sulfuric acid.
（like this)
But I couldn't find any source about this theory.
Does anyone know anything about this?
I must say that this is not a duplicate question.
Since I am asking about something different from someone else's previous questions.
What I am concerned is what chemical species is actually involved in the oxidation reaction of hot concentrated sulfuric acid. Not the existence of sulfur trioxide.


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Amidine synthesis from nitrile and dimethylamine using n-Buli as a base

I am making LiNMe2 using n-Buli (5.5 equiv.) and dimethylamine (5.0 equiv), -15 ºC, dry THF and under nitrogen atmosphere.
I have been doing the reaction and I can observe conversion of starting material (nitrile compound) by HPLC analysis. However there are multiple peaks observed.
The article I am following reports using NaHCO3:Na2CO3 1:1 buffer for the work up. The reaction mixture was added to the buffer with stirring under 5ºC and precipitation was observed. Been attempting to do that, but no precipitation is observed. Instead it is forming an oil that floats on top. Anyone who has worked on something similar or who can advise?

Does the procedure you are using describe the preparation of this exact compound?  Do you have any evidence of formation of your required product?

so far i have no evidence. I can only see conversion of starting material by HPLC. The NMRs i have run so far, only one of it where i could see 3 all the methyls accounted for on the non aromatic side. But the aromatic side order of splitting was not as expected

Do you have access to LC/MS?


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Soil chemical Anaylsis/CAL-P
I have added two levels of Phosphorus (P) to the soil, one at 30 mg/kg and the other at 130 mg/kg, representing low and high P levels, respectively. After thoroughly mixing the fertilizer with the soil, I analyzed whether the intended amounts of P were present in the soil. The analysis results are presented in a table, and I am seeking assistance in interpreting the data to ascertain the actual P content in these two soil conditions.I appreciate your help.
CAL-Extraktion
(P2O5 mg/100g)
Low P soil sample=6.57
High P=17.1

So you want to know what unit belongs to the numbers 6.57 and 17.1

Also, what does the fertilizer packet exactly mention for phosphorus analysis?

@Karl, units are already mentioned P2O5 mg/100 g.

@AChem Ah, got it. 6.57 mq P2O5 equivalents per 100g soil. Well, that's the answer.

The first sample of soil has $6.57$mg $\ce{P2O5}$ in $100$ mg soil, or $65.7$ mg $\ce{P2O5}$ in $1$ kg soil. As $\ce{P2O5}$ contains $43.6$ % P, the soil contains $0.436 · 65.7$ mg P = $28.5$ mg P in $1$ kg soil.
The second sample of soil has $17.1$ mg $\ce{P2O5}$ in $100$ mg soil. $17.1$ mg is $2.61$ times $6.57$ mg. So this sample contains $2.61$ times as much P atom, or $2.61·28.5 = 74.55$ mg P in $1$ kg soil.

The answer from your analytics contractor means if all the phosphorous in 100 g of your low P sample was turned into $\ce{P2O5}$ and separated, it would amount to 6.57 milligrams.
To turn that number into "amount of pure P per 100g", you simply multiply by the mass fraction of P in $\ce{P2O5}$, i.e. with the relative molecular weight of 2 $\ce{P}$ and $\ce{P2O5}$, which is $2 \times 30.97 / 141.96 \approx 0.4$.


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Which Reaction is More Thermodynamically Favourable?
I have a gaseous mixture containing $\ce{CH4}$, $\ce{CO2}$ and $\ce{H2O}$ that is being reformed via the Steam-Methane Reforming Reaction (SMR), the Dry-Reforming of Methane (DR) or a combination of both. The mixtures are:
Mixture composition
$\ce{CH4}$ = 2.075 mol/s, $\ce{CO2}$ = 2.075 mol/s, $\ce{H2O}$ = 1.0375 mol/s
At 800C, the Gibbs free energy change of the reactions are:
SMR = -45418 J/mol and  DR = -44578 J/mol
Is it possible to predict which reaction will take precedence?  I.e. can I assume that the SMR reaction will consume all of the $\ce{H2O}$ after the remaining methane will be reformed by the dry reforming reaction because the delta G value of the SMR reaction at this temperature is more negative?

With the minimal relative difference of the reaction Gibbs energies, "both take preference". There would be just minor difference in their equilibrium reaction quotients. SMR will have it slightly higher.

The reaction rates are most likely to determine the outcome, the fastest reaction could consume all the reactants before the slower one has reacted by much.


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Automatically extracting SMILES of sub-group of larger SMILES structure?
Is there any software tool that can automatically extract SMILES of fatty acid legs of phospholipids if the SMILES of the parent lipid is given? Or any tool that can do it in InChI ?
Any advice will be really appreciated.


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Apparent color of nickel(II)chloride and nickel(II)chloride hexahydrate
Why does anhydrous nickel chloride (NiCl2) appear yellow, even though chloride ions are considered weak field ligands, which should result in a smaller ligand field splitting energy and the absorption of longer-wavelength light, while nickel chloride hexahydrate (NiCl2·6H2O) appears green, despite the presence of water, a strong field ligand that should lead to a larger ligand field splitting energy and the absorption of shorter-wavelength light? Can someone clarify this apparent contradiction?
In this example, NiCl2 absorbs in the violet region of Vis-light (high energy) as a solid, so appears yellow, although chloride ions are weak field splitting ligands according to the spectroscopic series. NiCl2 hexahydrate appears green in solid form, so absorbs in the red region of Vis-light (low energy), although water is a comparatively strong field splitting ligand. Why?

Maybe the chloride ions are not acting as ligands in anhydrous nickel chloride, like in $\ce{NaCl}$


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Question about computed equilibrium graph

When I was reading a paper about sulfuric acid decomposition,I couldn't understand this graph very well.(link)
The caption says concentration but the vertical axis says mole(not mole/L or something related to concentration.)
And there's no mention on pressure. I think equilibrium is affected by the pressure.
Can someone explain this graph?

And there's no mention on pressure. I think equilibrium is affected by the pressure.

You have three reactions:
$\ce{H2SO4(l) <=> H2SO4(g)}\tag{1}$
$\ce{H2SO4(g) <=> H2O(g) + SO3(g)}\tag{2}$
$\ce{SO3(g) <=> SO2(g) + 1/2O2(g)}\tag{3}$
All of these are affected by pressure. From the context of the paper quoted by the OP, ambient pressure would make sense for the theoretical calculation. That would mean that at equilibrium, the sum of the partial pressures is 1 atm. You could envision starting with pure sulfuric acid in an airtight but expandable vessel, and heating it up.
For a liquid that just boils, you would have sharp transition between liquid and gas phase at a given temperature. Because of the additional reactions in the gas phase, the situation is more complicated.

The caption says concentration but the vertical axis says mole(not mole/L or something related to concentration.)

Yes, that is strange. There is not enough information in the paper to figure out what assumptions they are making in their calculation, but they are showing amount of substance in the graph, not concentration.

Thank you so much for clarifying the graph. Your explanation helped me a lot.

Using the term concentration in the chart legend is misleading. The chart represents molar fraction of particular (phase of) compounds as a function of absolute temperature, taking as the initial composition $\pu{1 mol}$ of pure sulfuric acid. It also means the compound molar amounts are numerically also their molar fractions.
Concentrations of gaseous compounds can be approximately  calculated using ideal gas approximation:
$$c_{i,\text{g}}=\frac{n_{i,\text{g}}}{V}=\frac{n_{i,\text{g}}p}{\sum{(n_{i,\text{g}})}RT}=\frac{x_{i,\text{g}}p}{RT}$$
Where

$c_{i,\text{g}}$ is amount concentration of gaseous compound $i$.
$n_{i,\text{g}}$ is the molar amount of gaseous compound $i$.
$\sum{(n_{i,\text{g}})} \le \pu{1 mol}$ is the sum of amounts of
gaseous compounds.
$1 - \sum{(n_{i,\text{g}})}$ is the amount of liquid sulfuric acid.
$V$ is the total volume of gaseous compounds at the assume pressure $p$ and absolute temperature $T$.

The pressure is obviously atmospheric pressure, there is no reason for otherwise.
Molar fractions from the chart cannot be used directly in the formula, as the formula relates only to gaseous compounds, so the total amount of gaseous compounds is less than $\pu{1 mol}$ by the amount of remaining liquid sulfuric acid.
Therefore:
$$x_{i,\text{g}}=\frac  {x_{i,\text{chart}}} {x_{\ce{H2SO4(l)}}}$$
$$c_{i,\text{g}}=\frac{x_{i,\text{chart}}p}{x_{\ce{H2SO4(l)}}RT}$$

Thank you for your answering.But I still can't understand this graph well. There's H2SO4 in the graph. It seems to represent liquid not gas(because there's gas H2SO4).

And that is the reason of the last paragraph of my answer.

It seems my question was something different. If we think about ,say water,  and temperature is 100℃ then there would be gap in the equilibrium graph.(water(l) and water(gas) (water in box with 1 atm pressure). I wonder why there's no such a thing in this graph of sulfuric acid. This seemed very odd to me.

It is because H2SO4 never reaches true boiling point. It's partial pressure tries to raise, but this is progressively overruled by its decomposition.

Thank you so much. I understood. I didn't  distinguish the difference between true boiling point and the point when total vapor pressure is 1 atm.


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Is *.xyz a standard file format in computational chemistry?

https://youtu.be/f6K8k0sH2TY

The above YouTube tutorial talks about a file format called XYZ.
Later, the tutor uses this format to compute the bond length and bond graph of a molecule.
Is *.xyz a standard file format in computational chemistry? Or did this tutor just make it up for the convenience of his computations?
If this is a standard format, then is he following the standard in his XYZ files?

The tutorial series elected the .xyz file format because this utility format is syntactically simple plain ASCII, without a patent limiting its use, and widely used.  By custom in this community, the Cartesian coordinates are (implicitly) set in Angstrom ($1 Å = \pu{10E-10 m} = \pu{0.1 nm} = \pu{100 pm}$) because it is a unit of length close to typical bond lengths (the $\ce{C=C}$ double bond in ethene fore example $\pu{134 pm}$, or $\pu{1.34 Å}$, ref).

the organization of information eases to extract/parse the information with a script or program you might want to write for yourself / one of the many structure viewers around

in comparison to other file formats to represent a molecular structure, its content is very basic.  Some information, for instance which atoms are connected with each other, or the relevant bond order (e.g., single, double, aromatic, triple) organized in blocks (see for example chemical table formats as in .mol and .sdf (an entry on Wikipedia)) are missing. No indicator if the structure consists of one, or multiple molecules, nor if there is a charge, either.

it is not the input file format to launch one of the more advanced computations in typical quantum chemical programs (MOPAC, Gaussian, GAMESS-UK/GAMESS-US, etc.).  In addition to the atoms' coordinates and the intended computation (and level of theory, keyword basis set), they expect an input explicit about spin multiplicity (singlet, doublet, triplett) and  overall charge of the structure submitted.
Departing from for example a .sdf file, there is help to prepare these input files.  If not already provided by the program itself, openbabel (most recent documentation here, specific to quantum chemistry formats here), input generators of Avogadro, or gabedit
are among the assistants available.

To best of my knowledge, there is no (ISO like) standard regarding the representation of the (x,y,z) coordinates as 3F10.5.  If you would like a consistent representation of your data, you always can request openbabel in lines of
obabel input.xyz -O output.xyz

(additional examples of application here) to «tidy up» .xyz files to stick to the F10.5 format for each column.


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Choosing energy levels for finding quantum numbers
When an element has multiple energy levels (i.e Titanium has $(Ar)3d^24s^2$), what energy level do we use to find the quantum numbers? Is it generally assumed the LAST energy level is used? or is it the orbital that has the highest energy?

Review the guide How to ask and Asking FAQs to prevent clarification requests, objections,  down-voting or closure.

Each orbital has its own set of quantum numbers. Without external fields,  each orbital group has its own energy level, otherwise each orbital. It is your choice, for which orbital you want to know it's quantum numbers and energy.

Right. Thing is, we had a midterm on this and it wasnt specified on what orbital to compute the Q.N from. Was there an underlying assumption I missed?

You can always provide the answer for multiple choices. // A asking B "Why did C say X?" does not usually provide a satisfactory answer if C had multiple valid choices. As B does not usually know why C did/said so either,  A should ask C .


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Queries regarding Lanthanoid Contraction
Consider these two statements regarding lanthanoid contraction and their explanation as understood by me.

Lanthanoid contraction is the greater than expected decrease in the atomic radii of the lanthanides varying from lanthanum to lutetium. This causes the atomic radii of 2nd and transition series to be identical

My Reason
The reason is attributed to the introduction of the $f$ orbital which provides very less shielding to the outer electrons in the $s$ orbital.
Now the second one

Lanthanoid contraction is the gradual and not the abrupt decrease in the atomic and ionic radii of the lanthanides varying from lanthanum to luteitium

My Reason
This cause is attributed to the fact that the size of $f$ orbitals go on decreasing when we proceed towards right due to increasing nuclear charge. This would increase shielding and hence the drop in radii becomes less abrupt.
I implore you to advise on which statement and explantion is correct. Thanks
EDIT
The second explanation comes from NCERT CLASS 12 as (It is the text which I use)

The factor responsible for the lanthanoid
contraction is somewhat similar to that observed
in an ordinary transition series and is attributed
to similar cause, i.e., the imperfect shielding of
one electron by another in the same set of orbitals.
However, the shielding of one 4f electron by
another is less than that of one d electron by
another, and as the nuclear charge increases
along the series, there is fairly regular decrease
in the size of the entire 4f  orbitals.

The first statement comes from wikipedia
here

The lanthanide contraction is the greater-than-expected decrease in atomic radii and ionic radii of the elements in the lanthanide series, from left to right. It is caused by the poor shielding effect of nuclear charge by the 4f electrons along with the expected periodic trend of increasing electronegativity and nuclear charge on moving from left to right

Review the search results for site:stackexchange.com OR site:libretexts.org lanthanoid OR lanthanide contraction explanation

According to this account, both, or in fact any unexpected trends in radii that can be explained by “poor shielding of f-electrons”.

@Karsten Thank You very much, is the second explanation correct?

What makes you believe the two explanations are not both correct? What information do you see as contradictory between one and the other?

First, atomic radius is not as well defined as one might hope (from wikipedia):

The radius of an atom is not a uniquely defined property and depends on the definition. Data derived from other sources with different assumptions cannot be compared

If we plot the theoretical atomic radii by period, we can see that there is a general trend of a decrease of the radius within a period (source of data: https://en.wikipedia.org/wiki/Atomic_radii_of_the_elements_(data_page)):

(The value shown in the x-axis is the column number in the long form periodic table, i.e. the number of protons added since the "last" noble gas element.)
The term "lanthanide contraction" can be applied to the transition metals of periods 4 (blue), 5 (orange) and 6 (gray). 5th period elements have a larger radius than 4th period elements in the same group, which can be rationalized by having outer electrons in the 5th rather than the 4th shell (e.g. Ni vs Pt).
Perhaps surprisingly, the 6th period elements have about the same radius as the 5th period elements in the same group (e.g. Pt vs Pd). This is explained by the lanthanide contraction (increase in number of protons with inefficient shielding by the increase in f-electrons).
You can see how efficient different electrons are in shielding by looking at the slope of the decrease in atomic radii in the main group, transition metal and f-block group parts of the graph.

where did you get the graph

@TheCuriousOne I plotted the data in a Wikipedia data table. I added the source.


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Confusion regarding spontaneous and non-spontaneous reactions
According to Wikipedia: "A chemical reaction is endergonic when non spontaneous. Thus in this type of reaction the Gibbs free energy increases."
My problem with that statement is the wording of "non-spontaneous" reactions. According to my knowledge, for every chemical reaction there is a ΔG, be it positive or negative, and is conncected to its equilibrium constant:

So in essence, if we only start with reactants, every reaction should yield some products, even if its ΔG is >0. Basically, every reaction is spontaneous to some extent. So is it only about semantics?
For example, in the biological context, even "non-spontaneous" reactions (with ΔG>0) can be driven to completion if its products are coupled to an exergonic reaction that consumes them and removes them from the endergonic equilibrium of the first reaction. That would mean that under those cirumstances the "non-spontaneous" reaction still produced some products (when starting with only reactants), is that right?
(What I also read though is that when stoechiometric molar quantities of reactants and products are mixed, more reactants are formed when ΔG is positive, so the reaction goes "in the wrong direction".)

It's only a "rule of thumb" for neophytes.  But it's taken on a life of its own from repetition by irresponsible authors.  You are entirely right in your assessment;  all reactions will proceed to some extent.

You might be confusing $\Delta G$ and $\Delta G^\circ$

See https://chemistry.stackexchange.com/questions/133585/deriving-the-equation-for-reaction-spontaneity?rq=1 and many many more....

Taking a thermodynamic approach to whether a reaction is spontaneous needs to be compared to an approach that considers kinetics. In normal english usage "spontaneous" means the reaction will happen if the components are mixed. But many chemicals do not. Boranes catch fire when mixed with air; hydrogen and oxygen mixtures do not, though both reactions are exothermic. Kinetic barriers prevent many reactions from happening without a kick to push them over the activation energy barrier.

See also https://chemistry.stackexchange.com/questions/43258/spontaneous-non-spontaneous-reactions-and-reversible-reactions and https://chemistry.stackexchange.com/questions/47262/is-every-chemical-reaction-in-equilibrium and https://chemistry.stackexchange.com/questions/31047/chemical-kinetics-of-a-reversible-reaction, and, as Buck Thorn says, many others.

For evaluation of spontaneity, two possible approaches can be chosen:

A reaction is spontaneous, if $\Delta_\mathrm{r} G^\circ =  -RT \ln{K} \lt 0$.

This is to be interpreted as a system with all components in their standard state with unit thermodynamic activities undergoes spontaneously the net reaction in forward direction.
This is the evaluation of spontaneity of reaction regardless of the initial system state, as the implicit presumption.

A reaction is spontaneous, if $\Delta_\mathrm{r} G = \Delta_\mathrm{r} G^\circ + RT \ln{Q} = RT \ln{\left(\frac{Q}{K}\right)}\lt 0$.

This is to be interpreted as a system with components in general states and component activities undergoes spontaneously the net reaction in forward direction.
This is the evaluation of spontaneity of reaction considering the initial system state.
If $Q \lt K$ then the equilibrium is in the forward direction of the reaction and the reaction is spontaneous in the latter sense regardless of if it is spontaneous in the former sense.

"Spontaneous" is a confusing term, especially because different sources use it differently (either it makes a statement about the standard Gibbs energy of reaction, or the Gibbs energy of reaction with respect to the current state).
The term "spontaneous reaction" has been around since about 1910:

In John Parker's 1891 Elementary Thermodynamics, the term "spontaneous" is used in the following way:

(found via this page and accessed here)

At equilibrium the forward and reverse reactions have equal rates and the activities [approximated by concentrations] satisfy the equilibrium constant and the cell potential when the reaction is done as an electrical cell equals zero volts. If the activities are changed from the equilibrium condition in EITHER direction by a change in chemical activities, temperature, or applied voltage, the reaction will be spontaneous to reestablish equilibrium. This is approaching the situation from an active equilibrium, the equilibrium concentrations, appropriate temperature, mixing and provided there is a possible mechanism and catalysis.
A reaction: Ag+ + Cl- = AgCl or AgCl = Ag+ + Cl-.  When approached from the first equation adding Ag+ to a solution of Cl- will produce a precipitate of AgCl as soon as the equilibrium constant is satisfied. This will be equilibrium and does not imply completion of the reaction. The reaction is driven by the enthalpy change in forming the crystal; adding more chloride or silver ion will increase the amount of AgCl and give a new equilibrium position. Removing either chloride or silver ion will reverse the reaction; the reaction will become spontaneous in the reverse direction. This is why ammonia dissolves AgCl, it removes silver ion.
When AgCl is added to pure water there are originally no silver or chloride ions. The second equation is now spontaneous and is entropy driven until the equilibrium concentrations are established. The interplay between enthalpy and entropy can be complicated. At equilibrium for a simple process one direction is enthalpy driven the other entropy driven; at equilibrium they balance: Delta H = TDelta S.
Stoichiometry is WHAT HAPPENS not necessarily what is present.  Changing activities may also change the chemical reactions taking place. This does happen in the AgCl reaction because higher concentrations involve complex formation.


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Where does the ionisation energy for a reaction to take place comes from?
Can anyone tell me exactly what happens during a chemical reaction in terms of ionisation energy? I know that during a reaction there is an exchange of electrons (except for those where electrons are shared), more specifically the valence electrons. But where exactly does the ionisation energy come from? Does it come from the charge of the other atom or...?
I'm asking this because I'm being asked to describe the reaction of sodium and water in terms of valance electrons and ionisation energy - and I just can't seem to understand where the ionisation energy comes from for there to be a reaction. I tried putting the same question as above to some AI chatbots and they seemed to suggest that the ionisation energy comes from the positive charge of the water attracting the single valance electron of the sodium. But water is neutral, so that can't really be the case, can it?

Why would you think that in the actual reaction there would be some single atoms of Na? Surprisingly, quite a lot of electrons are actually lost from a chunk of sodium reacting with water, but it's whole phase getting charged and I think you're over your head here. https://chemistry.stackexchange.com/questions/2606/what-exactly-is-happening-when-sodium-potassium-or-other-alkali-metal-explodes

Does this answer your question? Definition of activation energy

Thank you Mithoron and DrMoishe Pippik :D!! The answer provided by @anon below proves to be very useful, but the threads which you have linked is also quite interesting. Maybe I will check them out in the future!

In most reactions, the activation energy -- that is, the energy required to start the reaction -- comes from the motion of the molecules. In other words, it comes from heat.
(There are other possibilities. In photochemical reactions, some or all of the activation energy is provided by light. But the reaction of sodium and water is not photochemical. It happens in total darkness.)
There is an important difference between ionization energy and activation energy. As Poutnik explains in his answer, ionization energy is like escape velocity. It's the amount of energy needed to take an electron from sodium and fling it away to an infinite distance. This is a large amount of energy: 496 kJ/mol for sodium. That is way too much to be provided by the ambient heat at room temperature.
The reaction between sodium and water proceeds at room temperature anyway. This is because the electron does not actually need to be ripped away and sent to an infinite distance. It only needs to be pulled far enough away to create a transition state: a situation where the electron is balanced between the sodium it came from and a water molecule. In the transition state, the water molecule is quite close to the sodium, not at an infinite distance. Therefore only a small fraction of the ionization energy is required to reach the transition state.

Thank you for your answer! I got it now :D!


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What are the differences between an alpha-particle and a helium 2+ ion?
What are the differences between the alpha particle and helium 2+ ion? Or are they the same? Do differences arise because of the process by which they are formed?

They are the same think: $\ce{^4_2He^{2+}}$ nuclei. Guess you would call them $\alpha$ particles when the source is the radiactive decay of a heavy atom.

It is similar as the difference between proton and "naked" H+, or a beta particle and an electron. The same thing but different context.

Wikipedia: Alpha particles have a typical kinetic energy of 5 MeV (or ≈ 0.13% of their total energy, 110 TJ/kg) and have a speed of about 15,000,000 m/s, or 5% of the speed of light.

This is a good thoughtful question. Alpha particles and $\ce{He^{2+}}$ have the same composition but their origins are different. There are no "helium nuclei" units or packets in a large radioactive nucleus. Suppose someone hands over two red apples to you, one from America and one from Japan, then just by looking at them you will not be not able tell their farm origins. However, the key point is that if you to measure the energies of alpha particles, then one might say something about the source. Once alpha particle is out of the nucleus of a heavier nucleus, it is no different chemically or composition wise than a doubly ionized helium atom.
Historically nobody knew that alpha particles have the same composition as the helium nuclei until an elegant experiment was done. Rutherford and his students passed alpha particles through a thin sheet into an empty glass tube. After a while, they passed electric current through the glass tube, lo and behold, the spectrum of the "material" was identical to helium gas. Hence the connection is made.

Thanks for differentiating them.

All alpha particles are $\ce{^4He}$ nuclei, but not necessarily vice versa. (and not all $\ce{He^2+}$ ions belong to helium-4 as alpha particles do, as permeakra noted.)
Alpha particles have been named so historically as particles of the unknown, high energy radiation named alpha, observed at some heavy elements, originally for uranium minerals like uraninite $\ce{U3O8}$, later for newly discovered radium, radon, polonium, thorium and others.
Alpha particles are product of an alpha decay of some unstable atomic nuclei. $\ce{^{$x$}_{$y$}A -> ^{$x-4$}_{$y-2$}B + ^4_2He}$
$\ce{^4He}$ nuclei may have origin of being initially an alpha particle, or they can be product of helium ionization.
Alpha particles are called so while they keep substantial part of their original high kinetic energy in order of $\ce{MeV}$(while typical thermal particle energies are in order of $\ce{0.01 eV}$) . Their mean flight length until becoming neutral helium atoms is typically fractions of a milimeter in condensed matter or $\pu{1-2 cm}$ in air.

Helium 2+ ion is any $\ce{He^{2+}}$ ion, meaning both $\ce{^3_2He^{2+}}$ and $\ce{^4_2He^{2+}}$ ions qualify as well as heavier ones.
In contrast, $\alpha$-particle means specifically $\ce{^4_2He^{2+}}$, usually in context of nuclear decay or nuclear and particle physics.

@Poutnik I meant exactly what I wrote. Isotopes up to He-10 are characterised, though only He-6 and He-8 live long enough to speak about (absence of) chemical properties.

Yeah, I remember them. I just did not have them on mind, mentioning them myself elsewhere.


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When a gas is expanded, why doesn't the number of moles of a gas increase despite it's volume increasing (doubt in JEE sample question)
This question might be elementary, but it is genuine. I realise that some users might downvote the question for being so basic, but I urge them to be a little merciful, and help out a confused high school student,
If a gas is expanded at a constant temperature then,
according to Boyles' Law, $P \propto \frac{1}{V}$ (if $n$ an $T$ are constant).
According to Avogadro's Law, $V \propto n$ (if $P$ and $T$ are constant).
We know that $T$ is constant, but what about $P$ and $n$?
If $n$ is constant, then Boyles' Law is valid. $P \propto \frac{1}{V}$ ie. the pressure decreases on expanding the gas.
If $P$ is constant, then Avogadro's Law is valid. $V \propto n$ ie. the number of moles increases on increasing the volume. However, this is found to be false.
Furthermore, intuitively, we know that for an ideal gas, $n = \frac{V_{gas}}{V_{molar}}$, and here, $V_{gas}$ continuously increases, indicating that $n$ should also continuously increase.
This doubt stemmed from a question from a question bank which prepares Indian students for a popular competitive engineering exam called JEE.
The question, which has multiple correct options, reads:
If a gas expands at constant temperature :
(A) the pressure decreases
(B) the kinetic energy of the molecules remains the same
(C) the K.E. of the molecules decrease
(D) the number of molecules of the gas increase
I know that B is certainly correct, as the average kinetic energy of each molecule $A.K.E. = \frac{3}{2}RT$, and temperature is constant. My doubt was in options (A) and (D). For this question, (B) and (A) are the only correct answers and (D) is wrong, but I didn't understand the rationale of (A) being correct and (D) being wrong.
If someone could explain why the number of moles of a gas doesn't increase despite the volume increasing i.e. it is expanded, as well as the reasoning of answering the JEE question bank question, it would be very helpful.
Edit: My doubt has been clarified by user Maurice. Here, the number of moles remains constant as no chemical reaction is taking place for them to increase. So Boyle's Law is valid, and the pressure decreases due to the increase in volume.

:/ While gas might break down to smaller molecules, this is lame gas law question, not chemical reactor problem.

Could you please elaborate a little bit? I did not understand the context of this with respect to the question.

Well, how you imagine getting more moles of gas? Another option is a reaction creating new gas, but being created doesn't exactly mean expand.

One of the sentences Bongo Man writes has a wrong conclusion. I repeat it here in bold :  If   is constant, then Avogadro's Law is valid.  is proportional to   i.e. the number of moles increases on increasing the volume. However, this is found to be false. NO ! This is not "found to be false". If, in a syringe, the number of gaseous molecules increases without changing the pressure, the volume of the gas does increase.

@Maurice I was referring to the context of the question, and how Avogadro's Law seemingly doesn't apply here.

From the ideal gas state equation pV=nRT can be derived all 4 fundamental gas laws and how the quantities are related when 2 of 4 (n,p,V,T)  are kept fixed.

There is a chatroom for JEE: https://chat.stackexchange.com/rooms/74807/jee-chemistry-club

Your last question may have two different correct answers. It depends on the parameters to be constant in the transformation. These constant parameters are forgottten in the question. I will give here the two interpretations. All depends of the first sentence of the question : If a gas expands at constant temperature Well, in order of a gas sample to expand at constant temperature, information must be given on the number of moles $n$ of this gas. Is this $n$ constant or not ?
$1$) The number of gaseous moles stays constant during the transformation. So now, if the volume increases, the pressure must decrease, and (A) is correct, and (D) is wrong. One should add that this transformation is not spontaneous. Usually it goes in the other sense. The increase of volume of a gas sample is produced by a decrease of pressure
$2$) The number of moles increases during the transformation, at constant temperature and pressure. This is not easy to imagine how this increase of moles is carried out. As Mithoron states, It may be due to a chemical reaction producing a gas like $\ce{Zn + 2 HCl ->  ZnCl2 + H2}$. But this produces practical difficulties about how to measure this changing volume. If we suppose these practical difficulties are solved, the number of moles of gas must increase, and the total gases volume increases. So (A) is wrong, and (D) is correct

Thank you! I understood the concept now. Essentially, the number of moles cannot increase, unless there is some sort of reaction (which is clearly not taking place here), so it has to be constant.

Answers A or D could be correct. If you keep $n$ constant, A is correct. If you keep $P$ constant, D is correct. You still would have to explain how $n$ changed. It could be by adding particles (pumping up a flat tire) or by a chemical reaction (e.g. $\ce{N2O4 <=> NO2}$).
From the comments:

While gas might break down to smaller molecules, this is lame gas law question, not chemical reactor problem.

From the question prompt "If a gas expands", it does not sound like you are adding molecules or that a reaction is going on. For most multiple choice question, the best answer depends on what we know about the topic, and what we know about the author of the question (i.e. the answer key).
All the laws mentioned by the OP are ideal gas laws. The question does not state that the gas is ideal. You can still make qualitative statements ("increases", "decreases"), however, like in the answer choices.

Your answers makes sense. But why isn't Avogadro's law valid here? If the volume increases, so should the number of moles. I appreciate the effort, but I still feel dissatisfied, as my original question is unanswered, and I still haven't understood the logic of (A) being right and (D) being wrong.

@BongoMan (D) is not wrong, and Avogadro's law is valid, answer key notwithstanding.

The answer is most certainly right, as it was supervised by a very experienced and senior Chemistry faculty (named Vishal Jain, who has a degree from the best college in India, IIT-B). If you could formulate your reasoning, such that it aligns with the answer given, and answers my question, I'd gladly upvote and mark it as an answer

@BongoMan That is not how science works.

@ Karsten. Please explain. How does science work, according to you ?

@Maurice: Claims and evidence, in a nutshell. Models supported by data, and updated if there is new data the model no longer fits to. It does not work like this: "Here is the answer I want, please give me the scientific reasoning that fits".

You are right about how to approach multiple choice questions for qualification exams. Use the information available. Here you are only told that you have a gas, not the type. You might say that in the absence of more information on the nature of the gas, assume it is ideal and non-reactive, and assume the vessel is closed.

The OP is relying on someone they regard as trustworthy (reputable) for the correct answer. When they are starting out from scratch and assembling basic facts in preparation for a qualifier, asking them to rely on the "scientific method" (beyond thinking logically, using a process of inference and elimination) might be more than necessary.


text:
Is it possible to make a carbon icosahedron?
Cyclopentadienyl $\ce{C5H5^-}$ is a hydrocarbon anion with a 5 fold symmetry. The carbon atoms can be replaced with nitrogen, oxygen, sulfur, etc. to yield diverse aromatic 5-membered rings. For example, because a nitrogen is isoelectronic to a carbon and a hydrogen, replacing CH with nitrogen yields pyrrole, imidazole, triazole, tetrazole, and even pentazole.

Here is an interesting question. Boron and hydrogen can make a divalent anion $\ce{B12H12^{2-}}$ with an icosahedral symmetry. Since a carbon atom is isoelectronic to a BH, can we replace some or even all BH with C like what we did on cyclopentadienyl?

Some, but not all. https://en.wikipedia.org/wiki/Carborane

https://chemistry.stackexchange.com/questions/4027/carbon-with-5-bonds https://chemistry.stackexchange.com/questions/99893/iupac-name-of-c%e2%82%86ch%e2%82%83%e2%82%86%c2%b2%e2%81%ba

You know, this looks quite like two pyramidal dications sandwich... Still, there would have to be some central atom.

https://en.wikipedia.org/wiki/Sandwich_compound

No exactly an inverse sandwich I thought similar to this, but to cool to ignore https://www.sciencedirect.com/science/article/abs/pii/S0277538716000395

This coodination is not possible with $C$, otherwise it will be very unstable.

https://en.wikipedia.org/wiki/Platonic_hydrocarbon

Probably not an icosahedron. We might have better results with a smaller cluster, such as four atoms.
One major difficulty is the fact that isoelectronic replacement of boron with carbon introduces a positive charge, and so the equivalent to $\ce{B12H12^{2-}}$ would be $\ce{C12H12^{10+}}$ with a high positive charge density. Such a structure would be expected to break up into simpler species to disperse the charge, react even with relatively benign counterions or nucleophiles, or both.
Given this charge difference between isoelectronic species, full interchange of boron and carbon in the cluster is more likely with smaller clusters than the icosahedron. Here the case of a four-atom cluster is considered, as this allows both the carbon-ring cation and the boron-ring anion to have the same charge per atom ($\ce{B4H4^{2-}}$ vs $\ce{C4H4^{2+}}$).
The $4n+2$ rule would predict aromatic coupling of these species were planar, but they apparently are not. Both are instead calculated to have puckered structures. For the boron species, Shen et al.[1] study ions of the form $\ce{B_nH_n^{2-}}$ by computation; for $n=4$ they report the aforementioned puckered structure with $D_{2d}$ symmetry. This work does not directly examine experimental detection, but reports that $\ce{B4H4^{2-}}$ has been seen.
In the case of $\ce{C4H4^{2+}}$, Firme et al.[2] report that the cation has a calculated puckered structure. The calculations find a similar puckering in the tetramethyl-substituted species $\ce{C4(CH3)4^{2+}}$, but a planar structure for the more strongly conjugated phenylated cations. Experimentally Olah and Staral[3] experimentally prepared and studied these species via NMR spectroscopy (tge unsubstituted $\ce{C4H4^{2+}}$ ion is nit yet experimentally known). The experimental detection of both $\ce{B4H4^{2-}}$ and $\ce{C4(CH3)4^{2+}}$ with apparent puckered, $D_{2d}$ ring geometries indicates that with conjugated four-atom clusters full interchange between carbon and boron is indeed possible.
References

Yan-Fang Shen, Chang Xu and  Long-Jiu Cheng (2017). "Deciphering chemical bonding in BnHn2− (n = 2–17): flexible multicenter bonding".
RSC Adv. 2017,7, 36755-36764.
https://doi.org/10.1039/C7RA06811E.

Firme CL, Antunes OA, Esteves PM (2007). "Electronic nature of planar cyclobutenyl dication derivatives". J Phys Chem A. 111(46):11904-7. doi: 10.1021/jp075869j. Epub 2007 Oct 19. PMID: 17948974.

George A. Olah and John S. Staral (1976). "Novel aromatic systems. 4. Cyclobutadiene dications". J. Am. Chem. Soc. 98, 20, 6290–6304.
https://doi.org/10.1021/ja00436a037

Cool, but OP thought about exhanging BH for C, not CH.

Maybe so, but we tend to see more B/C interchange experimentally.