text:
IrO2 (iridium oxide) synthesis
How do I synthesize $\ce{IrO2}$ from $\ce{IrCl3}$ ? Please suggest a least expensive method.

On chemistry.SE, we use MathJax to format mathematical as well as chemical equations and similar expressions in questions, answers, and comments. MathJax allows us to typeset expressions using LATEX notation. https://chemistry.meta.stackexchange.com/questions/86/how-can-i-format-math-chemistry-expressions-here

The least expensive way might be to sell your IrCl${3}$ and buy IrO${2}$ - no need for any equipment.

One elder reference is Brauer's Handbook of Preparative Inorganic Chemistry, which states for this compound

"In the method of Wöhler and Streicher, $\ce{IrO2}$ is prepared from green $\ce{IrCl3}$ which can readily oxidized in a stream of $\ce{O2}$ at 600$^\circ$C giving blue-black  $\ce{IrO2}$.
The oxidation of fine iridium powder in a stream of air or oxygen does not give $\ce{IrO2}$ quantitatively. [...]"

which is accessible in archive.org's public scanned version here, p. 1590 (book's counting) / p. 1636 (the website's counting) of the 1963-edition by Academic Press.  The age of the edition should neither defer you from using it as a reference, nor inhibit you to look up additional references.

Iridium(IV) oxide can be prepared by slow hydrolysis of iridium(III) chloride at ambient conditions. A non-stoichiometric compound is obtained at first which is purified further.

Pure iridium oxide has been synthesized through a chimie douce
process. This synthesis proceeds by a slow basic hydrolysis of
hydrated iridium(III) chloride, at room temperature. A highly
disordered oxihydroxide $\ce{IrO_{1.45}(OH)_{1.10}·\frac{3}2H2O}$ is first
obtained.[...]

Another method is by adding freshly aqueous sodium borohydride to a solution of 0.1 M iridium chloride(molar ratio between them 1:4) at 80°C. The stirring was carried for 10 min till a black colored solid material floated on the surface. The colloidal solution was cooled to room temperature, the solid taken out and washed with ethanol.
References

N. Bestaoui and E. Prouzet, "A Chimie Douce Route to Pure Iridium Oxide" Chem. Mater.1997, 9, 4, 1036-1041, April 16, 1997 DOI:https://doi.org/10.1021/cm9606282
Chakrapani, K., & Sampath, S. (2015). The dual role of borohydride depending on reaction temperature: synthesis of iridium and iridium oxide. Chemical communications, 51 47, 9690-3 (link)


text:
How to create a buffer using CH3COOH and CH3COONa that has a pH of exactly 3.75?

I need to create a buffer using $\ce{CH3COOH}$ and $\ce{CH3COONa}$ that has a pH of exactly $3.75$.
A $\pu{50 mL}$ sample of your buffered solution will have to be able to withstand the addition of $\pu{25.0 mL}$ of $\pu{0.100 M}$ $\ce{NaOH}$ solution.
The buffered solution will break after the addition of no more than $\pu{35.0 mL}$ of the $\pu{0.10 M}$ $\ce{NaOH}$.

Here is what I have so far:
$$\mathrm{pH}= 3.75 \implies [\ce{H+}] = \pu{10^{-3.75} M} = \pu{1.78 \times 10^-4 M}$$
\begin{align}
\ce{NaCH3COO &-> Na+ + CH3COO-}  &&\text{(common ion effect)}\\
\ce{CH3COOH &<=> H+ + CH3COO-}
\end{align}
\begin{align}
K_\mathrm{a} &=\frac{[\ce{H+}][\ce{CH3COO-}]} {[\ce{CH3COOH}]}\\
1.8\times{10^{-5}} &=\frac{(1.78\times{10^{-4}})[\ce{CH3COO-}]} { [\ce{CH3COOH}]}\\
\frac{[\ce{CH3COO-}]}{[\ce{CH3COOH}]} &=\frac{(1.8\times{10^{-5}})}{(1.78\times{10^{-4}})}\\
\frac{[\ce{CH3COO-}]}{[\ce{CH3COOH}]} &= 0.1011 : 1~\text{(ratio)}
\end{align}
I found the ratio of $[\ce{NaCH3COO-}]$ to $[\ce{CH3COOH}]$, but I don't know what to do from here.
EDIT: "Break": The point where there's no acid left and the solution stops acting as a buffer. "Withstand": meaning that there can be a $\mathrm{pH}$ change of at most $1.0$. Thanks, guys!

The "withstand" requirement is not quantified. The "break" I understand as all acid is spent.

We have a policy which states that you should show your thoughts and/or efforts into solving the problem. It'll make us certain that we aren't doing your homework for you.
Basically any question with the wording your question has is considered homework; it needn't be literally one. Self-study questions, puzzles etc. also count as homework.
Don't worry, they're not banned. But, we require a minimal effort. Otherwise, this question may get closed. Please [edit] in your full reasoning or thoughts on this.
See Homework

Are you familiar with the Henderson-Hasselbalch equation? If you are not, this may help  https://www.khanacademy.org/science/chemistry/acid-base-equilibrium/buffer-solutions/v/buffer-solution-calculations

Can you give us the whole problem exactly as it was stated to you? Then give us your thoughts on how to solve it.

Equally, look up buffer capacity (e.g., https://en.wikipedia.org/wiki/Buffer_solution#Buffer_capacity).

"withstand" and "break" are not precise terms. How much tolerance for pH are we talking about here? A change of pH of no more than 0.1 unit? Perhaps your instructor has defined these terms quantitatively.

Added NaOH reacts with acetic acid, shifting acid/acetate ratio and therefore pH. At some point there is no acid left and solution stops acting as a buffer. This I consider a breaking point.  The meaning of "withstand" is unclear, unless it is meant some maximal oH change.

@iad22agp has raised the salient points. I know of no quantitative definition for the terms "withstand" and "break". Any solution will have to somehow define those terms exactly.

@MaxW I have merged the two posts now, you can put a custom flog on such posts, too.

@Poutnik I clarified the words "break" and "withstand" above. Thanks for your help!

@Cathy The requirements is not possible to meet, as they are mutually contradictory. The buffer cannot have as big and and the same time as low capacity as required. The difference of the spent volumes of $\ce{NaOH}$ is too small. If usage of acetate buffer was not mandatory, it would be achievable with buffers with pKa near 3.75 and not 4.75 as for acetic acid. In such a case, the buffer capacity would be decreasing and not increasing between these pH values and the final break would be faster.

In my understanding, there is considered $$\mathrm{p}K_\mathrm{a}=4.75$$ and therefore $\mathrm{pH}=3.75$ was chosen for easy calculation, as:
$$\mathrm{pH}=\mathrm{p}K_\mathrm{a}+ \log \frac {c_{\ce{A-}}}{c_{\ce{HA}}}$$
and therefore
$\frac {c_{\ce{A-}}}{c_{\ce{HA}}}=0.1$
For the buffer change to $\mathrm{pH}=4.75$, at least $\pu{25 ml 0.1 M NaOH}$ has to be spent. ("withstand")
$$\frac {c_{ws,\ce{A-}}}{c_{ws,\ce{HA}}}=1$$
Calculating molar amounts to be equal for both acetic acid and acetate:
$$0.1\cdot c_{\ce{HA}}\cdot V_\mathrm{buf}+c_\mathrm{NaOH}\cdot V_\mathrm{NaOH}\\=c_{\ce{HA}}\cdot V_\mathrm{buf}-c_\mathrm{NaOH}\cdot V_\mathrm{NaOH}$$
From that, we calculate minimal concentration of acetic acid in buffer:
$$2\cdot c_\mathrm{NaOH}\cdot V_\mathrm{NaOH}=0.9\cdot c_{\ce{HA}}\cdot V_\mathrm{buf}$$
$$c_{\ce{HA}}=\frac{2\cdot c_\mathrm{NaOH}\cdot V_\mathrm{NaOH}}{0.9\cdot  V_\mathrm{buf}}=\frac{2\cdot 0.1\cdot 0.025}{0.9\cdot  0.05}=0.005/0.045=1/9 \mathrm{~ mol/l}$$
$$c_{\ce{HA},min}=\frac19 \mathrm{~ mol/l}$$
$$c_{\ce{A-},min}=0.1 \cdot \frac19=\frac1{90} \mathrm{~ mol/l}$$
All acetic acid in $\pu{50 mL}$ of the buffer must react with not more than  $35~\text{ml}$ of $0.1~\mathrm{M~}\ce{NaOH}$.
$$\begin{align}
n_{\ce{NaOH}}&\lt 3.5\mathrm{~mmol}\\
n_{\ce{CH3COOH}}&\lt 3.5\mathrm{~mmol}\\
n_{\ce{CH3COONa}}&\lt 0.35\mathrm{~mmol}\\
\end{align}$$
As the latter 2 amounts have to be present in $\pu{50 mL}$ of the buffer:
$$\begin{align}
c_{\ce{CH3COOH}}&\lt \pu{70 mmol/L}\\
c_{\ce{CH3COONa}}&\lt \pu{7 mmol/L}\\
\end{align}$$
Summary:
Concentration of acetic acid in the buffer must be at least $\frac19=0.1111\pu{ mol/L}$ and not more then $\pu{0.07 mol/L}$.
Such requirements is impossible to meet and such a buffer is not possible to prepare.
By other words, the buffer cannot have at the same time so big buffer ("withstand") capacity and so low "break" capacity.
If usage of acetate buffer was not mandatory, the goal would be achievable with buffers with pKa closer to 3.75 than the acetic acid has.  In such a case, the buffer capacity would be decreasing and not increasing between these pH values and the final break would be faster.

See also https://chemistry.stackexchange.com/q/116186/35806, where the suitable approach is elaborated in larger details.

source: Stack Exchange

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text:
Are there processes of electric discharge to oxide As(III) into As(V)?
I want to know if there were any known experiment of electric discharge to oxyde the ion arsenic(III) into the ion arsenic(V).

The feasibility of the electro-removal of arsenate (As(V)) and
  arsenite (As(III)) from aqueous solutions via capacitive
  deionization was investigated. The effects of applied voltage
  (0.0-1.2V) and initial concentration (0.1-200 mgL(-1)) on arsenic
  removal were examined. As evidenced, an enhancement of arsenic removal
  can be achieved by capacitive deionization.[...]

Water‐soluble poly[glycidyl methacrylate N‐methyl d‐glucamine],
  P(GMA‐NMG), was synthesized and purified by ultrafiltration membranes.
  It was subsequently used for arsenic removal by coupling
  electro‐oxidation (EO) and polymer‐enhanced ultrafiltration (PEU)
  processes. In the EO‐PEU combined process, P(GMA‐NMG) was first used
  as supporting electrolyte during the exhaustive electro‐oxidation of
  As(III) to As(V) and then it was used to remove As(V) from aqueous
  solution through PEU.

Electrochemical oxidation of arsenic(III). A consecutive
  electron-transfer reaction
The reaction is based on two consecutive one-electron-transfer-step
  reaction of form:
$$\ce{ As(III) <-> As(IV) <-> As(V)}$$
$\ce{As(IV)}$ being the intermediate which has been postulated from the
  kinetic effect of the homogeneous reaction.

References

Fan CS, et al. and J Hazard Mater. "Electro-removal of arsenic(III) and arsenic(V) from aqueous solutions by capacitive deionization"
2016.(link)
Ozgur Arar, Nalan Kabay, Julio Sánchez , Bernabé L. Rivas, Marek Bryjak, Carlos Peña
"Removal of arsenic from water by combination of electro‐oxidation and polymer enhanced ultrafiltration" 15 October 2013 DOI: https://doi.org/10.1002/ep.11876
Henry A. Catherino "Electrochemical oxidation of arsenic(III). A consecutive electron-transfer reaction" J. Phys. Chem.1967,7,12,268-274, January 1, 1967 DOI: https://doi.org/10.1021/j100861a010

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text:
Magnetic properties of potassium dichromate
Is $\ce{K_2Cr_2O_7}$ diamagnetic or paramagnetic?
While näive theory predicts it's diamagnetic, but we can't be sure.
I cannot find any literature on this topic.
Nor does Wiki has that information.
Even Chem spider doesn't.
I don't know about any other reliable sources.

Related: https://doi.org/10.1002/crat.19740090109

I don't know, but ... KMnO4 is slightly paramagnetic, usually attributed to the Van Vleck mechanism. This requires a low lying excited electronic state, and KMnO4 has one, reflected in its purple colour. The colour of K2Cr2O7 is due to the same mechanism as KMnO4, but it being orange rather than purple suggests the difference in energies is larger, and so the Van Vleck paramagnetism is smaller. Now the gap needs to be quite small otherwise the diagmagnetic term is larger than the Van Vleck one. So it may be that this increase in energy gap results in K2Cr2O7 being diamagnetic - don't know.


text:
Relation between equilibrium constant and entropy change
Like the Van't Hoff equation, which relates change in enthalpy to equilibrium constant, is there a similar equation for the relation between change in entropy and equilibrium constant?
Consider the following specific case,

For a reaction taking place in a container in equilibrium with its surroundings, the effect of
  temperature on its equilibrium constant K in terms of change in entropy is described by
[A] With increase in temperature, the value of K for exothermic reaction decreases
  because the entropy change of the system is positive
[B] With increase in temperature, the value of K for endothermic reaction increases
  because unfavourable change in entropy of the surroundings decreases
[C] With increase in temperature, the value of K for endothermic reaction increases
  because the entropy change of the system is negative
[D] With increase in temperature, the value of K for exothermic reaction decreases
  because favourable change in entropy of the surroundings decreases
Answer: (B) and (D)

My attempt:
Since equilibrium with surrounding is given, the reaction must be reversible.
This implies, $∆S=0, ∆S_s=∆S_{surr}$.
Also, at equilibrium $∆G=0$.
This implies, ∆$H=T∆S_s=-T∆S_{surr}$
(where the s subscript denotes system, and surr denotes surrounding)
Writing the van't Hoff equation in differential form, I got
$= d(lnK)=∆H(dT/RT^2)$
$=d(lnK)=-∆S_{surr}(dT/RT)$
Now, aren't options A and C true too?
As on integrating, we can see directly the variation of $K$ with $∆S$.
What is the correct way to solve this?
Edit: Where ever there is a big space, it implies ∆.
Example,     '       ' H=T '       '  ,
Means, ∆H=T∆S

"the Van't Hoff equation relates change in enthalpy to rate constant" That wording sounds incorrect, depending on what you mean by "rate constant". See eg https://en.wikipedia.org/wiki/Van_%27t_Hoff_equation

@NightWriter I just realised. I messed up equilibrium with rate. But that doesn't essentially changes the problem

The ∆ is missing in all statements. Whoever edited my question thanks to you, but can you please bring back the ∆? Without it the problem is meaningless

Please try to use standard terminology and notation, otherwise you will confuse a lot of people. As it stands I don't really understand what you mean with "missing deltas" and with the meaning of extra spaces. The deltas are quite visible, just include additional ones where necessary if they are missing.

In your question's title  you do not mean 'rate constant' (or rate coefficient) but equilibrium constant which is entirely different.

You can use $\Delta G^O=-RT\ln(K_p)$  and then $\Delta G^O=\Delta H^O-T\Delta S^O$ to find the entropy change. In an exothermic reaction $\Delta H \lt 0$ and $K_p$ decreases with increasing temperature.

There are both mathematical and conceptual errors in your derivation. For instance you fail to keep track of a sign with a substitution early on. Also, you attempt to apply the equilibrium condition $\Delta G=0$ where it isn't generally valid, ie $\Delta G^0 \neq 0$ in general.

@NightWriter for an equilibrium reaction. ∆G is always 0 right? if  ∆G° isn't 0 this means the reaction will proceed in some definite direction to make ∆G=0, i.e the reaction quotient will equal the equilibrium constant at some point.

Yes but don't confuse $\Delta G$ and $\Delta G^\circ$, they are different things, see eg https://chemistry.stackexchange.com/questions/95280/why-isnt-standard-gibbs-free-energy-always-zero

@NightWriter ah, I got it. In van't hoff, I was writing ∆H instead of ∆H°

Like the Van't Hoff equation, which relates change in enthalpy to equilibrium constant, is there a similar equation for the relation between change in entropy and equilibrium constant?

A relation between $K$ and $\Delta S^\circ$ can be obtained as follows:
$$\begin{align}T\log K &= -\frac{\Delta G^\circ}{R} \\
\left(\frac{\partial(T\log K)}{\partial T} \right)_p &= -\frac{1}{R}\left(\frac{\partial \Delta G^\circ}{\partial T}\right)_p= \frac{\Delta S^\circ}{R}\end{align}$$
As regards the multiple choice question, the best way to answer it is to understand that $\Delta G$ represents a sum of the change in the entropy of the system and of the surroundings, scaled by $T$:
$$\begin{align}\Delta G &= \Delta H - T\Delta S \\&= -T\Delta S_{surr} - T\Delta S_{sys} \\&= -T(\Delta S_{surr} + \Delta S_{sys}) \end{align}$$
Exchange of heat at constant pressure (equal to the enthalpy) alters the entropy of the surroundings, so when discussion turns to the heat exchanged at constant pressure, you don't need to consider the entropy of the system, only that of the surroundings (at least for this type of problem - this alone eliminates a and c as possible choices). Next you need to remember that endothermic means the surroundings lose heat to the system, which lowers the entropy of the surroundings. The opposite holds for an exothermic process (surroundings gain entropy).
To summarize: an endothermic reaction lowers the entropy of the surroundings. The decrease in the entropy of the surroundings decreases if you increase T. The result is therefore a net increase in the total entropy. Finally, since $$\Delta S ^\circ_{\text{total}} = -\frac{\Delta G^\circ}{T}=R \log K_{\text{eq}}$$
increasing the total entropy results in a higher equilibrium constant.


text:
Henry law and equilibrium
Is Henry's law a consequence of the equilibrium between a substance S in aqueous phase and the same substance S in gaseous phase? What I mean is, is it correct to treat Henry law as a manipulation of the equation for the equilibrium constant, where
$$ \ce{S(g) <=> S(aq)}, \ \ \ K=\frac{[S_{aq}]}{p_S}$$
$$\rightarrow K \times p_S = [S_{aq}]$$
which is Henry law ($p_S$: partial pressure of S in the gas phase; $[S_{aq}]$: concentration of S in aqueous solution).

You can get lot of info in the Henry's Law Wikipedia article.

That's an appropriate interpretation of Henry's law, although as you've written it $K$ would be referred to strictly as the solubility constant $H$, see this link to the wikipedia page provided in the comments.
The following shows more formally how to derive the connection between Henry's law, an equilibrium constant and the standard Gibbs free energy for evaporation of the solute.
At equilibrium between the solute in the gas and solution phases, the chemical potential of the solute is equal in both phases, so that
$$\begin{align} \mu(sol)&=\mu(g)\end{align}$$
In the gas phase (assuming ideal behavior)
$$\begin{align} \mu(g)&=\mu^\circ(g)+RT\log\left(\frac{p_i}{p^\circ}\right)\end{align}$$
Now according to Henry's law, the partial pressure and concentration in the solution are related as
$$p_i=K\frac{m_i}{m^\circ}$$
(of course, as you point out, this can already be recast as an equilibrium constant expression, but read on)
so that
$$\begin{align} \mu(sol)&=\mu^\circ(g)+RT\log\left(\frac{K(m_i/m^\circ)}{p^\circ }\right)\end{align}$$
Defining
$$\begin{align} \mu^\circ(sol)&=\mu^\circ(g)+RT\log\left(\frac{K}{p^\circ }\right)\end{align}$$
we can see that
$$\Delta G^\circ_m = \mu^\circ(g)-\mu^\circ(sol)=-RT\log\left(\frac{K}{p^\circ }\right)=-RT\log\left(K_{eq}\right)$$
which is the familiar relation between the standard Gibbs free energy change and an associated equilibrium constant for the equilibrium $\ce{solute <=> gas}$ provided we define the equilibrium constant as
$$K_{eq} = \frac{K}{p^\circ } = \frac{p_i/p^\circ}{m_i/m^\circ }$$


text:
Why does Mg(HCO₃)₂ on heating gives Mg(OH)₂?
Normally, bicarbonates on heating forms carbonates.
Why does Mg not give this too?
In a reference book I am referring to, it is given that it is due to high solubility  product of Mg(OH)₂ compared to Mg(CO₃) .
Then shouldn't the latter form? As even small amounts of it will lead to precipitation, driving the equilibrium  forward?
Edit: This is related to removing temporary hardness of water.

Wikipedia states that both magnesium carbonate and magnesium hydroxide decompose at 350 °C to yield magnesium oxide. So what temperature do you consider "heating" to be?

I suppose he means boiling water with bicarbonates. The provided argument does not make much sense to me, as calcium hydroxide has even higher solubility product than magnesium hydroxide.

The molecules get enough energy to efficiently collide with each other.All old bonds are broken atoms and electrons are reorganized and new bonds are created to replace the old ones.Try read Arrhenius theory of chemical reactions.


text:
Leaching of copper using zinc
Why can we not use zinc for extracting copper from a solution containing $\ce{Cu^2+}$ ions even though it is a better reducing agent than $\ce{Fe}$?
I am studying metallurgy and I cannot understand why zinc cannot be used.

Just saying it, but many metallurgical process and reagents have an economic aspect to the choices made.

We can.
But I see few reasons why it is not used:

Iron is much cheaper than zinc.
There can be remaining residue of iron/zinc, coated by copper, or just being excessive. While copper can be melted away and iron stays, zinc would melt together with copper, causing unwanted impurity (unless wanted for making brass alloys)
If we remove copper for environmental concerns, then by using zinc, we would just replace one evil by another. While both metals are essential for life in small amounts, both are toxic in larger amounts. ( So does iron, but at much higher level)
Removing dissolved iron from waste is easier than removing zinc.

Do you know that the American Chemical Society has listed zinc as an "endangered" element? According to them, within hundred years the supply of zinc will be scarce. The best reducing agent in the world is the cathode of an electrolytic cell. Copper(II) is quite easy to reduce by electricity. Why waste an endangered element?

In fact, the similar is done during electrolytic copper purification on the industrial level.

Time to properly process primary batteries and iron/steel with zinc anti-corrosive coating.


text:
Extraction between cyanex 923 and phenol
I want to remove phenol from wastewater so I used the cyanex923 as an extractant or carrier to extract phenol from the wastewater (the name of the process is non-dispersive solvent extraction) but the problem is that I don't know what to do at the last step when the complex cyanex923-phenol is formed. I know that I need a stripping step to separate these two components and when I googled, I see some people said that NaOH can be used and some people said NaCl but I don't know exactly how it works, like what happens when NaOH or NaCl is added, please can anyone help me to explain the process because I'm totally lost

Wouild help if you told us what cyanex923 is, and what it dissolves in

I want to remove phenol from wastewater so I used the cyanex923 as an extractant or carrier to extract phenol from the wastewater but the problem is that I don't know what to do at the last step when the complex cyanex923-phenol is formed. I know that I need a stripping step to separate these two components and when I googled, I see many people said that NaOH can be used but I don't know exactly how it works, like what happens when NaOH is added, please help me

Phenol is acidic enough that it will form the water soluble sodium salt when treated with NaOH

So you mean that when NaOH is added to the complex cyanex923-phenol, water soluble sodium salt will form? Can you please explain a bit more? I'm not good at chemistry

Phenol is really pretty acidic with a pKa 10, so any basic solution over pH11 or stronger will completely form the sodium salt. Dissolve your complex up in a water-immiscible solvent (see my answer below) and wash that solution with aqueous sodium hydroxide

Thank you so much for your explanation and suggestion, I really appreciate it

Please accept the answer if you are happy with it

So far as I can see from Google, Cyanex 923 is a liquid mixture of trialkyl phosphine oxides. This should be soluble in the standard range of organic solvents (ethyl acetate, diethyl ether, methyl t-butyl ether) so washing a solution of the cyanex923-phenol complex with aqueous sodium hydroxide will remove phenol without the cyanex going into aqueous. Usual drying, filtration and solvent evaporation under reduced pressure will give the cleaned cyanex

Based on the original question, I can only speculate if further detail about the composition of cyanex923 may be helpful for the OP, or not.
On the other hand, there is at least one scientific publication about this complexant's compositon by a Polish group deploying GC-MS.  (Sounds like determining the secret of WD40.)  To paraphrase the publication, 18 peaks were identified, four of them about these major components: trihexylphosphine oxide (8.5%), octyldihexylphosphine oxide (30.4%), dioctylhexylphosphine oxide (37.4%), and trioctylphosphine oxide (16.1%) summing up to 92.4%, and mostly n-alkyl groups.
Beside cyanex923, sibling formulations cyanex925, 921 and Topo -- at time of the study all tradenames of products by then American Cyanamide Company (defunct since 1994) are equally surveyed.
Reference:  Dziwinski and Szymanowski in Solvent Extraction and Ion Exchange,
16, 1998, 1515-1525, doi 10.1080/07366299808934592 at Taylor & Francis (or here at ResearchGate).
Addition

There is an open-access publication by the same group about properties of  cyanex923, including the determination of extraction isotherms for various phenols between water and octane or toluene like depicted in the following figure (unsubstituted phenol is described by the squares):

(reference: Polish Journal of Chemical Technology, 12, 2010, 19-23,
doi 10.2478/v10026-010-0027-9)

Still by same group, the extraction performance of cyanex923 deployed in membrane extractors was compared with amberlite and trioctylamine in hollow fiber modules.  Cyanex923 performed much better than the other two, as for example depicted in the following figure:

(reference: Environmental Science & Technology, 36,  2002, 2088-2093, doi 10.1021/es010910w)

After I've done some researches I think it works like the following:
We need an extraction module and a stripping module. The organic stream in the extraction module containing the phenol–cyanex complex is sent to the stripping module. The complex diffuses across the microporous wall filled with organic solution to the stripping–organic interface and I think in the stripping module we have NaOH which reacting to the phenol and form PhONa (sodium phenolate) (according to the article, this step occurs in a hollow fibre membrane contactor and I wonder why can't we just use a tank?).
This means that at the interface cyanex is regenerated and recycled back to the extraction module. The solute (sodium phenolate) is then released into the aqueous stripping solution. I'm not sure what they mean by "the aqueous solution" but anyway, we have the PhONa and need to be converted back to the phenol so that we can use phenol again and I think we can do it in acidification condition.
Some study researches mentioned that we can add HCl to lower pH and when the the pH drops the phenol associates and forms an organic phase separated from a saline aqueous phase.
This is what I've understood from some articles that I've read, please correct me if I'm wrong

Among the papers around the original one co-authored Szymanowski, there are at least two potentially relevant to your question.  Have a look at them.


text:
Understanding "Gay-Lussac's law of Gaseous volumes/combining volumes"
Understanding "Gay-Lussac's law of Gaseous volumes/combining volumes". So I was reading Wikipedia and stumble across this. It defines it as follows "The ratio between the volumes of the reactant gases and the gaseous products can be expressed in simple whole numbers" and then give an example of
$$\ce{2H2 + O2 -> 2H2O}$$
The problem is that example does not fit the definition. The volume of reactants is $3x$ where $x$ is a variable. The volume of product is $2x$. The ratio is $\frac 3 2$, which is not a whole number. Clearly, something is wrong here or something is wrong with me. Either way, can anyone help?

Instead of "ratio as a whole number" should be "ratio of whole numbers"

Or, mathematically, it is a rational number.

@Poutnik  Should I edit the Wikipedia?

The ratio between the volumes of the reactant gases and the gaseous products can be expressed in simple whole numbers

Wikipedia is correct.
Expressed in, not expressed as.
$3/2$ can be expressed in simple whole numbers $3$ and $2$.


text:
What is the pale green solution from dissolving sodium hydroxide in a stainless steel flask?
I have previously used drain/sink cleaner gel based on sodium hydroxide but noticed that caustic soda crystals were a lot cheaper.  I decided to experiment with making my own cleaner.
I bought a 1 litre stainless steel flask and placed 500 ml of tap water in it.  Then I added 500 g of caustic soda (from the hardware store) over maybe a 3 hour period.  I stirred it with a new, unpainted wooden spoon.  The solution got very hot!  Then I poured it into a HDPE milk bottle, there was no free solid left and it had all dissolved.  However the solution was pale green!  When we used sodium hydroxide solution years ago in the lab, it was colourless.  Also the end of the spoon which was in contact with the alkali was discoloured to a dark brown.

What is the source of this green?  Is it a chromium compound, having stripped this out of the stainless steel?
The stainless steel flash looks normal, and has been thoroughly rinsed. Is it safe to use for holding tea/coffee?

I believe the sodium hydroxide attacked some organic molecule present in the wood of the spoon, and it ended up dissolved in the solution (the spoon discoloration is a sign this could be the case). About the second question, as a safety rule I'd always keep separate flasks for food and for experimentation.

To second IanC:  once a container is dedicated to chemical experimentation, this container either stays for chemical experimention, does not come close to food containers and must be clearly different from a food container -- or, after appropriate cleaning, goes into recycling.  You won't drink from it anymore either, both because of even potential residues left as well as your concentrated lye might irreversibly dammaged the inner coating of your container, too.


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Naming this iron complex: Fe2Cl6[H2O]5
This is the iron complex in question:
$\ce{Fe2Cl6[H2O]5}$
I'm pretty sure according to the nomenclature it should be: pentaaquahexachloroiron(III) but I wanted to check somewhere to be sure.

In coordination chemistry square brackets denote inner coordination sphere; $\ce{[H2O]5}$ makes little sense in this context.
It looks like the correct molecular formula for this compound appears to be $\ce{[FeCl2(H2O)4]+[FeCl4]- * H2O}$ according to the crystal structure determined by Szymański [1].
The IUPAC-compliant name proposed by the author is tetraaqua-cis-dichloroiron(III) tetrachloroferrate(III) monohydrate.

Figure 1. Asymmetric unit $\ce{[FeCl2(H2O)4]+[FeCl4]- * H2O}$ (ICSD-8198) with added hydrogen atoms.
Color code: $\color{#EEEEEE}{\Large\bullet}~\ce{H}$; $\color{#FF0D0D}{\Large\bullet}~\ce{O}$; $\color{#1FF01F}{\Large\bullet}~\ce{Cl}$; $\color{#E06633}{\Large\bullet}~\ce{Fe}$.
References

Szymański, J. T. The Crystal Structure of Tetraaqua-cis-Dichloroiron(III) Tetrachloroferrate(III) Monohydrate (Ferric Chloride $2\frac 1 2$-Hydrate). Acta Cryst. B 1979, B35 (9), 1958–1963. https://doi.org/10.1107/S0567740879008244.


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Prolonging the exothermic reaction between calcium oxide and water
Is it possible to generate 10-minutes of heat using very small amounts of CaO and water? This is an exothermic reaction.
I've done this experiment at home, but it is very short-lived. I'm trying to understand the ratio of reactants I need to generate heat that lasts long (like heat packs.)

You can make heat release semi-continuous by adding stoichiometric amount of water drop-wise for 10 min.

Perhaps a include a slower reaction whose product is CaO?

Use larger particles of CaO. Or, get cement clinker (which is rich in CaO), and grind the 1 inch chunks to whatever size works for you.

There are funnels with an Archimedic spiral at the bottom -- in principle quite similar to a truck with a concrete mixer.  So, provided you are able to grind CaO into a fine powder, slowly turning this screw will add CaO into a receiver flask below.  Like this one:

(reference)


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Measuring pH in ferric oxalate solutions
I have a problem with measuring $\mathrm{pH}$ in ferric oxalate solutions ($\sim 5~\text{to}~\pu{10 g L-1}$ $\ce{Fe},$ $\pu{0.8 M}$ oxalic acid, $\pu{25 °C}$). While I am sure about the $\mathrm{pH}$-meter device and buffer solutions ($4$ and $7$) and also the good function of my electrode (Sentek) in other solutions, the $\mathrm{pH}$ value does not stabilize at all in such a way that it continuously decreases as far as the pH becomes negative and eventually, under-ranged.
Is there any interference or any requirements for $\mathrm{pH}$ measurement in oxalate solution?

What is your $\ce{Fe^3+}$ source?

Can be interfering eventual draging out calcium ions from the electrode glass ?


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Distribution of polymer lengths
Suppose we have a chemical reaction where a linear chain can form (reversibly) from building blocks (e.g. atoms or molecules):
$$\ce{X_$i$ + X_$k$ <=>[$k_\mathrm{fwd}$][$k_\mathrm{rev}$] X_$i$X_$k$}$$
where $k_\mathrm{fwd}$ and $k_\mathrm{rev}$ are rate constants.
What is the distribution of polymer lengths (at equilibrium)? What are some canonical references that derive these distributions under variations of the above polymerization process (e.g. step versus chain growth)?

I allowed myself to edit your question a bit by merging both forward and reverse reactions into one (since you are speaking of equilibrium) and I also changed notations for the rate constants to the standardized ones $k_\mathrm{fwd}$ and $k_\mathrm{rev}$ to avoid ambiguity.

When this exercise come from?

As the expression is written at present you have numerous  pairs of species in equilibrium. There is no way for them to grow. You should look for references to kinetics of radical and of condensation polymerisation. Many chemical kinetics texts cover these topics.

Concentrations are a geometric series
First, let's focus on the reactions where one of the reactants is a monomer $\ce{X_1}$:
$$\ce{X_1 + X_n <=> X_{n+1}}$$
and call the equilibrium constant for this reaction $K_{\mathrm{extend}}$.
From this, we can figure out the equilibrium constant for making a polymer from monomers:
$$\ce{n X_1 <=> X_n}$$
$$K_n = (K_{\mathrm{extend}})^{n-1} = \frac{[X_n]}{[X_1]^n}$$
So we can express the concentration of a polymer $X_n$ dependent on the monomer concentration:
$$[X_n] =  (K_{\mathrm{extend}})^{n-1} [X_1]^n = \frac{1}{K_{\mathrm{extend}}}(K_{\mathrm{extend}} \cdot [X_1])^n$$
This is a geometric series (if you take the concentration of polymers as a series, the ratio of neighbors in the series is constant). Because the amount of matter is finite, $K_{\mathrm{extend}}\cdot [X_1] < 1$ for an equilibrium scenario (as opposed to ending up with a single chain containing all the monomers). So the concentration of $[X_1]$ is the highest, and concentrations drop with increase of $n$.
Expression for $[X_1]$
Let $[X]_\mathrm{tot}$ be the total concentration of monomer, no matter whether in a polymer or not.
$$ [X]_\mathrm{tot} = \sum_{j=1}^{\infty} j[X_j]$$ $$ = \frac{1}{K_{\mathrm{extend}}} \sum_{j=1}^{\infty} j (K_{\mathrm{extend}}\cdot [X_1])^j $$
Using $$\sum_{j=1}^{\infty} j \beta^j = \frac{\beta}{(\beta - 1)^2}$$
we get
$$ [X]_\mathrm{tot} = \frac{[X_1]}{(K_{\mathrm{extend}}\cdot [X_1] - 1)^2}$$
which you can solve for the monomer concentration.
Relationship to real polymerization reactions
The polymerization reactions in chemistry and biochemistry that I am aware of are kinetically controlled. There are fibers (such as actin) in biochemistry that are dynamic (i.e. there is polymerization and depolymerization), but the processes are affected by nucleotide binding, complicating matters. I would be interested to hear whether the OP has an example that behaves like this.

here there is polymerization/depolymerization too because the end of the polymer can come off. can you say what properties of actin fibers change the dynamics you outlined?

Can you point to a book or article which contains this derivation?

I did the derivation from scratch, but here is a more comprehensive treatment that has the result as the simplest 1D case: https://hal.archives-ouvertes.fr/hal-01565995/document. @TonyRuth

@TonyRuth Here is another reference that considers real systems: https://pubs.acs.org/doi/10.1021/jp981592z


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Hill equation with inhibition and activation
The Hill equation for an activating enzymatic interaction with cooperative multiplicity $n$ is
$$\frac{\mathrm d[\ce{P}]}{\mathrm dt} = V_\mathrm{max}\frac{[\ce{S}]^n}{K_\ce{S} + [\ce{S}]^n}\tag{1}$$
and for an inhibitory interaction
$$\frac{\mathrm d[\ce{P}]}{\mathrm dt} = V_\mathrm{max}\frac{K_\ce{I}}{K_\ce{I} + [\ce{I}]^m}\tag{2}$$
What would the equation be for competing activating and inhibiting compounds $\ce{S}$ and $\ce{I},$ respectively?
My sort of ansatz is that it would be of the form
$$\frac{\mathrm d[\ce{P}]}{\mathrm dt} = V_\mathrm{max} \frac{c_\ce{S}[\ce{S}]^n}{K_\ce{SI} + c_\ce{S}[\ce{S}]^n + c_\ce{I}[\ce{I}]^m}\tag{3}$$
I spent about an hour trying to derive the actual answer, but I'm no expert in physical chemistry/biochemistry and increasingly have the sense I should turn to someone who is.

What type of inhibition (competitive or otherwise) are you interested in? What do you mean by "roughly" in the context of the Hill equations? Hill-ish? Are the $K$s the same?

@KarstenTheis Let's go with non-competitive inhibition for simplicity's sake, though if there exist modifications for competitive and uncompetitive inhibition I'd be interested in those as well. The Ks are not the same, and I definitely should have been clearer about that. I say "roughly" because I've absorbed the $K_i$ from the canonical form of the inhibition equation into $V_{max}$ in the numerator of the inhibition equation.

I also believe it would be more accurate to have some constants $c_i$ and $c_s$ to determine the proportional binding strengths of each compound. In the first two equations those could simply be absorbed into K, but in the third equation they might be important.

Could you edit the question to incorporate the information in the comments? And maybe use two different subscripts for the two different Ks?

Should all be clearer now. Apologies on the original imprecision. I'm a physicist by training, and I suppose that preference for rough approximations slipped in.

Equation 2 has a weird limit for [I] = 0. It should be Vmax, but it is Vmax/Ki.

As mentioned in my earlier comment, I had absorbed a $K_I$ into the $V_{max}$ for that equation to simplify it. You just get a new effective $V_{max}$ when you divide. I understand the units have to be modified slightly to account for that, but it's quite simple. Regardless, I'll add a $K_I$ to the numerator if that's a sticking point. I've also removed the comments about the equations "roughly" being correct, since I now give the exact Hill equations. This was essentially what I meant about "rough" correctness originally. The equations had the right form all along, up to constant factors.

Mechanism of cooperative enzyme activity
The easiest way to explain cooperativity is the example of hemoglobin, where four nearly identical subunits switch cooperatively between two states T and R of different binding affinity. When no ligand (L) is bound, hemoglobin is in the T state; once at least one ligand is bound, it is predominately in the R state.
$$\ce{T4 + 4 L <=> R_4L + 3 L <=> R4L + 3 L <=> R4L2 + 2 L <=> R_4L3 + L <=> R4L4}$$
The states with 1-3 ligands are less populated (because the R state has higher affinity for the ligand L thant the T state), and we will ignore them:
$$\ce{T4 + 4 L <=> R4L4}$$
From this, you can write the law of mass action and the binding isotherms.
If we apply this to an enzyme with a Hill coefficient of $n$, and call the active state of the enzyme E and the inactive state F, we would have the equilibrium:
$$\ce{F_n + nS<=> E_nS_n}$$
For sufficiently high concentration of substrate S, all enzyme would be in complex with substrate, leading to a rate of $v_\text{max}$.
Mechanism of cooperative inhibition
In the OP's question, the Hill coefficient for inhibition $m$ is different from that for enzyme activity ($n$). So we could have a regulatory protein with $m$ subunits attached to the enzyme. It binds cooperatively to inhibitor, which allosterically affects enzyme activity. The regulatory protein would also have two states, T and R, either free or bound to $m$ inhibitors:
$$\ce{ T_m + m I <=> R_mI_m}$$
When the allosteric inhibitor is bound to ligand (i.e. in the R state), the enzyme lacks activity.
Combination of the two
Only a certain fraction of enzyme is bound to substrate, and only a certain fraction of enzyme-bound substrate is not inhibited by the regulatory protein. For this scenario, the enzyme activity is $v_\text{max}$ times those two fractions.
$$\frac{\mathrm d[\ce{P}]}{\mathrm dt} = v_\mathrm{max}\frac{[\ce{S}]^n}{K_S + [\ce{S}]^n} \cdot \frac{K_I}{K_I + [\ce{I}]^m}\tag{2}$$
What happens when there is competitive inhibition?
It gets more complicated. Will the inhibitor also cause a switch in the enzymes conformation? In which direction? For the simpler (non-cooperative) Michaelis Menten model, a competitive inhibitor changes the apparent affinity of enzyme to ligand in a concentration-dependent manner (the enzyme partitions between free, ligand-bound and inhibitor-bound state). For the cooperative case, you would have to decide whether mixed states (such as $\ce{ES2I2}$) are populated; if they are, they would contribute to product formation, but less than $\ce{ES4}$ because not all active sites are turning over substrate.


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Rate law of the reaction of propane with oxygen
In class, I have a demonstration where I light a bubble filled with propane. As I add oxygen, the bubbles combust faster and more quickly. I have already balanced out the chemical reaction like so:
$$\ce{C3H8 + 5 O2 → 3 CO2 + 4 H2O + heat + light}$$
I have reasoned that the reaction must go faster because the limiting reagent is oxygen, and since the air is only about 20% oxygen adding more oxygen to the bubble will speed the reaction process. I have figured out the kinetics, the oxidation, the enthalpy diagram, etc.
However, the only thing I don't know how to do in this demonstration is how to find the rate law. How would I find the rate law and equation of a combustion reaction? Wouldn't it just be second order since it's a combustion reaction? I have no idea where to even begin or why it is so.

The reaction has a multistep mechanism and the kinetics will not be trivial.

The oxidation of hydrocarbons ($\ce{RH}$) such as this one proceed via a chain mechanism. The initiation step is
$$\ce{RH + O2 -> HO2^. + R^.}$$
and the propagation
$$\ce{ R^.+O2 -> RO2^. \\ RO2^. +RH -> ROOH +R^.      }$$
But it does not stop there as the hydroperoxide decomposes $\ce{ ROOH -> RO^. + OH^.}$ and there are other reactions such as
$$\begin{align}
\ce{ R^. + O2 -> RO^. + O^. \\ RH + OH^. -> H2O + R^. \\ RCH2OO^. + O2 -> RCH2O + O3  \\ RH + O3 -> RO^. +HO2^.}\end{align}$$
and chain branching $\ce{ \quad O^. + RH -> OH^. + R^.}$.
The exact mechanism depends on temperature and pressure, and in a plot of $T$ vs $p$ there are regions of slow reactions, explosion and hot and cold flames. (An original paper is by Newitt & Thorne, J. Chem. Soc 1937, p1656). Thus you can appreciate that sorting out the mechanism is no simple task and will need much specialised equipment.

In order to determine the rate law, you need to be able to measure either the rate at which the reactants are consumed or the products are formed.
A very common error is to base the rate law on the balanced chemical equation.  In fact, many high school level teachers will teach kinetics this way.  Unfortunately this is a serious error.
It is possible for the exponents in the rate law to match the coefficients of the balanced chemical equation, but only when the reaction occurs in a single step.  We can't know how many steps are in a reaction until we investigate the kinetics.
In terms of determining the rate law, you would need to do the following:

Vary the amount of propane systematically
Vary the amount of oxygen systematically
Measure the amount of time it takes for the reaction to finish with each variation

Since there are only two reactants, you should be able to do this by running three separate trials; one which we will consider as a standard (or reference), one where the amount of propane is increased, and one where the amount of oxygen is increased.
On a practical level, your instructor may be working off the assumption that the reaction is first order in propane ($\ce{rate = k[propane]^1[oxygen]^x}$).  If this is the case, then you only need to vary the amount of oxygen. This may be simply be for practical reasons.


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Why is the dissociation reaction of acetic acid in water initially endothermic and then exothermic?
Why is the dissociation reaction of acetic acid in water initially endothermic as temperature is increased and then becomes exothermic as it is increased further?
My understanding from Le Chatelier's principle is that a rise in temperature "adds" heat to the system and a drop in temperature "removes" heat from the system. Correspondingly, a temperature increase would favour the endothermic (heat absorbing) direction and a temperature decrease would favour the exothermic (heat releasing) direction. Why is this contradiction? My thoughts are that maybe the autoionisation of water is endo or exo, which somehow dominates the reaction, but I don't know. Could someone please provide a possible explanation?

Le Châtelier's principle is applied to a given value of $\Delta_r H$. It does not offer predictions for the evolution of $\Delta_r H$.
As $c_p = \left(\frac{\partial H}{\partial T}\right)_p$, the evolution of $\Delta_r H$ is given by $\Delta c_p$.
The answer to your question probably lies in that negative value of $\Delta c_p$.

@ThomasJungers so are you saying the change in delta Cp dominates the change in temperature, causing the delta H to switch sign? why does the heat capacity change anyway and why does it get lower as temp increases in this case?

Be aware that by the Hess law,  $\frac{\mathrm{d}H}{\mathrm{d}T} = \sum{nC_{p,\mathrm{reactants}}}  -  \sum{nC_{p,\mathrm{products}}}$

@Poutnik could you please explain specifically how that has an effect?

If the total heat capacity of reagents is higher than the total heat capacity of products, $\Delta H$ of the reaction is decreasing with temperature. And vice versa. It is the direct consequence of the energy conservation law. $\Delta H$ value is function of the initial and final states only, not of the path between them. Note the previous note should have flipped the sign.

@Poutnik How did you get that formula?

Have you followed the Hess law link ?

The total $\Delta H$ is the same, if you let chemical react at T1, or if you first first change T1 to T2, let it react at T2, and then change T2 back to T1.

@Poutnik Ok, I think I get the math. Could you explain conceptually, what's going on though? Like in terms of bonding or energy

My interpretation is at higher temperature water molecules need less energy to break mutual hydrogen bonds to hydrate the proton.

I have found a paper which does this same experiment and provides a discussion of this phenomenon[1]. I will begin by just quoting from the discussion section:

It has been stated in the introductory section that a
  maximum in the formal Cp° for a pair of thermally
  stable free ions can be ascribed to an endothermic
  change in the solvent shell around one of the ions.
  However, the physical description of that change can
  cover a wide range of phenomena. At one extreme,
  the change might be a genuine chemical reaction: for
  instance, an octahydrate becomes a hexahydrate. At
  the other extreme, it might be merely an artifact resulting from a peculiar energy level diagram for a single
  species.

Intuitively, one would expect the reaction to stay either endothermic or exothermic with only small changes due to the relative thermal populations of various vibrational modes in the reactants and products. This picture is more relevant to the gas phase. Thus, the authors are stating that this kind of large change in the behavior of a reaction must be related to the environment which each ion experiences, namely, the solvation environment.
The authors go on to give diagrams for each of the cases they describe above. They stress that in the case where a genuine chemical change occurs, you are essentially moving from populating one set of states to another set of states entirely.
On the other hand, and I suspect this is more commonly the case when water is the solvent, the density of states might increase quite markedly as energy increases. That is the solvent may begin exciting librational modes quite strongly or even move to a regime where the solvents can more or less freely rotate. This behavior of the solvent would be very sensitive to temperature. The authors are very careful to note that each of these situations can result in the same thermodynamic outcome, so one can determine which is happening on the basis of thermodynamic data alone. Nowadays, we could easily run a molecular dynamics simulation or carry out some kind of 2-dimensional IR experiment and extract meaningful kinetic information to supplement the thermodynamics.
So, to summarize, this large change in enthalpy which crosses from exothermic to endothermic is almost certainly due to large changes in the solvation environment and not large changes with respect to the reactants and products.

References:
[1]: Leung, C. S., & Grunwald, E. (1970). Temperature dependence of. DELTA.-Cp. deg. for the self-ionization of water and for the acid dissociation of acetic acid and benzoic acid in water. The Journal of Physical Chemistry, 74(4), 687-696.


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Acid etching of float glass
While we have done some etching of glass with hydrofluoric acid alone, to achieve a matte white finish on float glass, I am curious to know what formula and procedure is required to create a more transparent and smooth finish ?
(Lead crystal polishing is achieved after diamond grinding, dipping in a mixture of $\ce{HF}$ and $\ce{H2SO4}$, and alternating with rinsing.)
Can anyone advise on the possible advantages of mixing another acid with the $\ce{HF}$?

I find the "matte white finish" and "a more transparent and smooth finish" contradicting. If it's the latter you are after, why etch glass at all? The purpose of etching is to reduce transparency and smoothness, not to increase it. Or are you using etching for shaping/printing on the glass surface (e.g. something like a PCB)?

what about using potassium hydrogenfkuoride KHF2 instead of HF?https://en.wikipedia.org/wiki/Potassium_bifluoride "Potassium bifluoride is the inorganic compound with the formula KHF2. This colourless salt consists of the potassium cation and the bifluoride (HF2−) anion. The salt is used in etchant for glass. Sodium bifluoride is related and is also of commercial use as an etchant as well as in cleaning products."

I appreciate the responses. The desired result is for the surface to be etched either with acid or sandblasting while leaving raised dots by masking with a resist. The etched surface is required to be semi transparent and not matte white. We are considering either doing the etching by sandblasting then 'acid' etching the sandblasted surface to render it more smooth, or doing all the etching with a liquid etchant which will also achieve the required finish.

Alkaline etching of glass tends to be less rapid than HF and leaves a smoother surface. This may not be what you do for the first etch on fresh glass, but it may smooth out the etch somewhat after an HF etch. The smoothing may also be caused by silicate ions that form in the alkaline etch fluid, so perhaps a solution of potassium (or sodium) silicate (ortho or meta) would be gentler. A polysilicate might be too slow.


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8a-methyl-1,2,3,4,4a,8a-hexahydronaphthalen-4a-ylium carbocation rearrangement

In this reaction after the attack of lone pairs on $\ce{H+}$ ions, a stable $3^{°}$ carbocation is formed. But seeing the six membered ring and the double bonds already present, I can't help but think that there's some way of obtaining a benzene ring through rearrangements. Can someone suggest a mechanism for this?

Be careful with your thinking.  H+ ions do not attack.  The lone pair on the alcohol attacks (or accepts ) H+.  You should always think of mechanisms in terms of the movement of electrons.  Have you attempted a mechanism?  Can you show us what you have tried?

Yes I phrased it wrong. The lone pair on alcohol attacks on the H+ ions to convert it into a good leaving group ie. water and the water leaves to form a carbocation as shown.

https://chemistry.stackexchange.com/questions/98588/rearrangement-of-8a-ethyl-1-3-4-8a-tetrahydronaphthalen-4a2h-ol

I think Sameer Thakur was in right track when started to write the mechanism. But the path got lost at the end. I don't see reason to have a methide shift followed by a hydride shift and then proton abstraction. The 1,2-methide shift gives you a very stable $3^\circ$-carbocation, which is compatible with the initial $3^\circ$-carbocation given by elimination of water. Thus, I think the following mechanism is a very reliable one for gaining aromaticity:

You are right. The hydride shift is redundant in my solution.

While there are examples of 1,2-methide shifts, this question has an uncanny resemblance to the dienone-phenol rearrangement whose mechanism was first elucidated by Woodward and Singh in 1950. Dienone 1 under acidic conditions undergoes rearrangement to phenol 6 and not, based on earlier speculation, to phenol 4. The direct 1,2-methide shift (2b --> 3) does not occur but rather the reaction proceeds through the spiro carbocation 5. continued

Applying this approach to the carbocation 7 generated from the alcohol in this question, spiro carbocation 8 forms the predicted tetrahydronaphthalene 10. One way to distinguish between the two mechanisms is via a labeling experiment. All of the carbon label in 7 will retain its location as the red star in 10 if the 1,2-methyl shift mechanism applies. The spiro mechanism will partition the label ~50:50 between the two ring benzylic carbons. For related studies, see reference 2.
1) R. B. Woodward and T. Singh, J. Am. Chem. Soc., 1950, 72, 494.
2) A. J. Waring, J. H. Zaidi and J. W. Pilkington, J. Chem. Soc, Perkin Transactions I, 1981, 1454.

It will not work the way the OP likely envisions.  To make an aromatic ring you need to move the methyl group into the bottom ring, but all of these bottom positions are saturated.  No place for the proposed rearranging group to land.
There is, however, another possibility if the system from a nonclassical carbocation.  To make the nonclassical carbocation, the methyl group moves over the bridging bond but not all the way across so that the bond between the methyl group is delocalized between the two bridgehead carbons.  In this structure, if it forms, the delocalized electrons are also conjugated across the bridge like an additional pi bonding pair so that the top ring (6 conjugated electrons) and the bridge structure (2 electrons) are both aromatic.  Possibly other users more familiar with the system can indicate whether this structure forms, leading to a new question.

Why do we need to move the methyl group in the lower ring?

Upper ring otherwise cannot be fully conjugated.

Okay I figured it out. First I drew the resonating structure for the carbocation and then performed a methyl shift followed by a hydride shift. Then the H+ ion leaves via the dehydration mechanism making the ring aromatic in the process.

Your first two "resonance structures" are confusing. Have you migrated the methyl group (not resonance) or turned the molecule over?

You can get the second structure, just by resonance among the 4electron, 5 center pi system on the top ring. Then turn the molecule over.

Not sure about this.  I'm still curious about the possibility of a nonclassical ion.

@Withnail I moved the double bonds towards left through resonance and then turned the molecule over. It is confusing but it would proceed the same way even without turning it.

@OscarLanzi what do you mean by a nonclassical ion?

Read my reference to it, from the IUPAC Gold Book.


text:
Possible nonclassical ion from a bicyclic system
In this question a bicycling alcohol is reacted with acid to make what appears to be a tertiary carbocation, and the OP asked whether it could become aromatic.  The given answer suggests it could be, but only if a nonclassical ion is formed.  So, does that happen?
The substrate is 8a-methyl-1,3,4,8a-tetrahydronaphthalen-4a(2H)-ol.

If classical carbocation is already heavily stabilised, it would be surprising if additional "nonclassical" stabilisation would be added. In this case it's a difference between sigma and pi complex between substituted benzene ring and methyl group - it would be rather a pi complex then nonclassical carbocation, even if it was an actual intermediate here.

I don't think it is necessary to consider a mechanism with a nonclassical carbocation. It would undergo normal rearrangement with a 1,2-methide shift within the same ring. See the mechanism I posted in previous question you were directing to:

Answer to this question would support for my mechanism.

So the answer seems to be "No".  Fair enough.

While there are examples of 1,2-methide shifts, this question has an uncanny resemblance to the dienone-phenol rearrangement whose mechanism was first elucidated by Woodward and Singh in 1950. Dienone 1 under acidic conditions undergoes rearrangement to phenol 6 and not, based on earlier speculation, to phenol 4. The direct 1,2-methide shift (2b --> 3) does not occur but rather the reaction proceeds through the spiro carbocation 5. continued

Applying this approach to the carbocation 7 generated from the alcohol in this question, spiro carbocation 8 forms the predicted tetrahydronaphthalene 10. One way to distinguish between the two mechanisms is via a labeling experiment. All of the carbon label in 7 will retain its location as the red star in 10 if the 1,2-methyl shift mechanism applies. The spiro mechanism will partition the label ~50:50 between the two ring benzylic carbons. For related studies, see reference 2.
1) R. B. Woodward and T. Singh, J. Am. Chem. Soc., 1950, 72, 494.
2) A. J. Waring, J. H. Zaidi and J. W. Pilkington, J. Chem. Soc, Perkin Transactions I, 1981, 1454.


text:
Lattice energy while bond formation
My chemistry lecturer taught us that lattice energy is the energy required to break an ionic bond between two oppositely charged ions in a crystal lattice.
He then told us that lattice energy is also the energy released when separate oppositely charged ions come together to form a bond. He told us that the ions "must" be in gaseous state to combine.
I thought that the reason was that there is no way for the ions to combine other than in the gaseous state. But the reason he gave appeared quite different than mine. I couldn't grasp the reason he told us.
Can you please help in making me understand the reason properly?

Ions may combine in pretty much any state; if you have the reason to think otherwise, you may just as well throw that reason away as nonsense. Gaseous state, though, surely gets mentioned in the definition of lattice energy, and that for a good reason.

Can you share  the reason Ivan?

Well, it's simple: you need to measure the energy released when the bonds form, hence you start from the state with no bonds and go to the state with bonds. The latter is the crystal; now what about the former?

Former is the ionic state of atoms. I'm not sure if they must be in gaseous state but I certainly think that ,it being that way will make bond formation easier.

"Easier" is not a concern at all. The process is imaginary anyway.

You should read about the Madelung constant and the Born–Haber cycle.

It looks like you have two separate questions here. I would recommend creating two separate questions so that each question can get attention.

The actual, "real" state of the atoms (and then ions) that combine to form an ionic compound is not particularly significant to the question at hand.
When we want to calculate the lattice energy of an ionic compound, one method used is a Born-Haber Cycle.  In essence, we use Hess' Law to combine a series of elementary steps that results in the overall transformation we are interested in. The enthalpies of each of these steps can be combined and used alongside the enthalpy of the overall reaction to determine the lattice energy.
Let's take the formation of sodium chloride from its elements as an example. The overall reaction is:
$\ce{Na(s) + 1/2Cl2(g) -> NaCl(s)}$ $\Delta H_f = -411 kJ/mol$ .[ 1 ]
Remember that it's OK to use fractions in chemical equations when we are doing energy calculations.
In many texts, the lattice energy is defined as the energy released when gas-phase ions are combined to form a solid compound. $^{[2]}$
One of the reasons that we have to have gas-phase ions is that the thermodynamic data for things ionization energy and electron affinity are for gas-phase particles. This also means that I have to make sure that all the particles in the systems are transformed into gas-phase atoms which can then be ionized.
The complete cycle for the formation of sodium chloride would be:
$\ce{Na(s) ->Na(g)\quad \Delta H_{sublimation}}$
$\ce{1/2Cl_2(g) -> Cl(g) \quad \Delta H_{bond \quad energy}}$
$\ce{Na(g) -> Na^+(g) + e^- \quad \Delta H_{ionization}}$
$\ce{Cl(g) + e^- -> Cl^-(g) \quad \Delta H_{electron \quad affinity}}$
$\ce{Na^+(g) + Cl^-(g) -> NaCl(s) \quad \Delta H_{lattice \quad energy}}$
To sum this up; we can think of the lattice energy as holding ions to each other.  When the ions come together to form the compound, this energy is released.  If we want to separate the ions (for example, if we want to vapourize the compound) we need to add the same (slightly more really) amount of energy.  The requirement that the ions be in the gas phase comes from the mathematical model that we use.

Footnotes
[2]:Atkins; et al. (2010). Shriver and Atkins' Inorganic Chemistry (Fifth ed.). New York: W. H. Freeman and Company. ISBN 978-1-4292-1820-7

Can you explain what you mean by the thermodynamic Data for things ionisation energy and electron affinity?


text:
Electrode potential for metal-metal ion and metal-metal insoluble salt-salt ion electrode
So I read it in a book that

The potential of metal-metal ion electrode and metal-metal insoluble salt-salt anion electrode is same while their standard potentials are not same. Standard potential are related by the following equation:
$$E^\circ_{\ce{X-/MX/M}} = E^\circ_{\ce{M+/M}} + \frac{2.303RT}{F}\log_{10}K_\mathrm{sp}$$

My question is that $E^\circ$ is just the potential at some given standard condition, i.e. 1 bar and 25 °C, but when $E$ is same for both at all other conditions, why is it that it is not valid for the standard  particular case/condition?

My question is that $E^\circ$ is just the potential at some given standard condition, i.e. 1 bar and 25 °C....

No, $E^\circ$ is the potential at the given standard condition 25 °C, 1 ATM and activities of involved compounds equal to 1.
This may be the source of your confusion.
Standard potentials are for the unit activities. In my example, once for silver ions, once for chloride ions.
The activity of $\ce{Ag+}$ in 1M $\ce{AgNO3}$ Is very different to activity of $\ce{Ag+}$ in 1 M $\ce{KCl}$ with the $\ce{AgCl}$ precipitate.
You cannot find pressure in the Nernst equation. The pressure plays only very marginal role here, like being big enough to avoid water boiling at low pressure.
Usage of each of the standard potentials implies usage of activity of different ions to compute the final potential, or concentations as approximations.
One is using the metal cation activity, the other the anion activity.
One is using the metal cation activity directly, the other indirectly via the salt solubility product.
$$\begin{align}
E_\mathrm{Ag/Ag+}&=E^{\circ}_\mathrm{Ag/Ag+}+0.059\log{\left(a_\mathrm{Ag+}\right)}\\
&=E^{\circ}_\mathrm{Ag/Ag+}+0.059\left(\log{\left(K_\mathrm{s,AgCl}\right)}
-\log{\left(a_\mathrm{Cl^-}\right)}\right)\\
&=E^{\circ}_\mathrm{Ag/AgCl/Cl^-}
-0.059\log{\left(a_\mathrm{Cl^-}\right)}\\
\end{align}$$

I'm not asking the proof , All i ask is if E of both the cells is some for all conditions, why is it different at the standard condition?

Say if I calculated E for both at 1.1 atm , then according to the statement , they are same. Same even for 1.01 ,1.00000001.... so why at 1 ATM is it different

See the answer update.

SE sites focus on high quality answers to high quality, not easily answered, questions, to be a gain for all.
Quick questions, where one can easily find answers in any relevant textbook or within few clicks on Wikipedia, do not count as high quality questions.
They will be answered only if there is explicitly provided failed elaboration effort to raise their quality a little. This applies to literal homework, self-studies, puzzles, worked examples etc - forming the Homework class of questions.


text:
What is the final molar concentration (molarity) of the nitrate ions in the resulting solution?

A chemist mixes 16.3 mL of 1.23 M sodium nitrate with 6.28 mL 0.823 M solution of calcium nitrate. What is the final molar concentration (molarity) of the nitrate ions in the resulting solution?

What I know: I know I can add the volumes to use it to find the final molarity. After I multiply both volume and molarity and get amount of substance, I multiply by the ratio to get the amount of product and use those amounts to calculate molarity. But I don't know how to isolate the amounts of nitrate from the overall amount of product. Can someone explain how I do that?

Could you edit your question to show the chemical formula of sodium nitrate and calcium nitrate? That should help with figuring out the stoichiometry.

First, when solving problems like this one, you have to make sure that both salts possess good solubility and that no double salt is precipitating:
$$
\begin{align}
\ce{NaNO3 &<=>> Na+ + NO3-}\label{rxn:R1}\tag{R1}\\
\ce{Ca(NO3)2 &<=>> Ca^2+ + 2NO3-}\label{rxn:R2}\tag{R2}
\end{align}
$$
Both nitrates are indeed well soluble in water (equilibrium is shifted to the far right), and no poorly soluble side-product is formed, e.g. we are likely to have a true solution consisting of hydrated cations $\ce{Na+}$, $\ce{Ca^2+}$ as well as $\ce{NO3-}$ anions.
As you already suggested, one can indeed use total amounts of substances $n_i$ and the preservation of volumes $V_i$ to find the final molar concentration on nitrate (here indices $1$ and $2$ refer to initial sodium and calcium nitrate solutions, respectively):
$$c(\ce{NO3-}) = \frac{n_1(\ce{NO3-}) + n_2(\ce{NO3-})}{V_1 + V_2}\label{eqn:1}\tag{1}$$
According to the stoichiometry (see \eqref{rxn:R1} and \eqref{rxn:R2}):
$$n_1(\ce{NO3-}) = n(\ce{NaNO3}) = c(\ce{NaNO3})\cdot V_1\tag{2.1}$$
$$n_2(\ce{NO3-}) = 2\cdot n(\ce{Ca(NO3)2}) = 2\cdot c(\ce{Ca(NO3)2})\cdot V_2\tag{2.2}$$
Finally, we can plug all the known quantities back to \eqref{eqn:1}:
$$
\begin{align}
c(\ce{NO3-}) &= \frac{c(\ce{NaNO3})\cdot V_1 + 2\cdot c(\ce{Ca(NO3)2})\cdot V_2}{V_1 + V_2}\tag{3}\\
  &= \frac{\pu{1.23 mol L-1}\cdot\pu{16.3 mL} + 2\cdot\pu{0.823 mol L-1}\cdot\pu{6.28 mL}}{\pu{16.3 mL} + \pu{6.28 mL}}\\
  &= \pu{1.35 mol L-1}
\end{align}
$$


text:
What could be causing my cyclic voltammograms to have a sloped scan?

I am running a CV (cyclic voltammogram) in a two compartment bulk electrolysis cell with pH 7 phosphate buffer (0.5 ion concentration) and a napheon membrane separating the two compartments. I'm using a platinum microelectrode with a Ag/AgCl psudo reference electrode and a large platinum flag counter electrode. Currently I am running the setup all in one side of the two compartment cell in order to isolate the distance issue.
I took a first order derivative to show the slope of the line is rather constant.

This was a background taken, after degassing with both stirring and helium sparging, using a polished 25 micron platinum microelectrode. Any ideas what could be causing my background scan to be a literal diagonal line, rather than flat like a regular baseline?

Just related.  Labelling should be A • 10E7 if you want those values. I know that is normally wrong also in professional literature but it is so. Your background is perhaps a straight flat line depending on your requirements. It is matter of scale. It should rather pass by 0,0 but somehow obeys a linear relation with resistance related to the electrolytes in the cell.

@Alchimista Thanks for the insight, I'll change that. If it was due to electrolytes the slope of the line should remain constant though, correct? The slope of the line tends to change quite a bit between runs. I suppose it could be that I am generating new electrolytes between runs however.

Normally you should get basically noise without electrolyte and a straight line with it . All the rest depends on the actual set up (el. conc. and electrode size. Again, and as in general, what is crucial is what current do you expect later. If it is milliA or more, your line is a correct background, at least for the standard application of CV. And yes, you might have some electrochemistry going on, influencing the subsequent run. But you have tenths of microampere so it should not matter at all.


text:
How do I convert salinity measured in mg/L Cl into parts per thousand as NaCl?
I have collected water data using a refractometer that records salinity in ‰ (parts per thousand) as $\ce{NaCl}$. I want to compare my data to other data but every journal article I read  has the results in $\pu{mg/L }\ce{Cl-}$
How do I convert my $\pu{ppt}$ data into $\pu{mg/L}$ or vice versa so I am comparing apples to apples?

There is problem with ppx values they are ambiguous. It may be w/w, w/v, v/v, n/n. But salt water has density significantly different to 1 g/ml, e.g 1.04. So 1o/oo may mean 1000 mg/L or 1000 mg/kg. The former is more probable, but check it. E.g prepare a salt solution of known concentration and check, in which of promile variant Is your refractometry measurement calibrated.

Related: Is 1 ppb equal to 1 μg/kg?

$$\pu{1 mg L-1} = \pu{1 ppm} = \frac{1}{1000}\cdot\pu{ppt}$$
or
$$\pu{1 ppt} = \pu{1000 ppm}$$
For example:
$$\pu{10 ppm} = \frac{10}{1000}~\pu{ppt} = \pu{0.01 ppt}\tag{1}$$
$$\pu{5 ppt} = 5\cdot\pu{1000 ppm} = \pu{5000 ppm}\tag{2}$$

Please visit this page, this page and this one on how to format your future posts better with MathJax and Markdown. Briefly, you do't want to use > for highlighting (> is reserved for quoting), also note that there is a \pu{…} macro for physical units (proper spacing between the unit and symbol; upright notations).

@andselisk Thank you, I'll follow it my future posts.

This is a good answer but neglects to adjust the values from a $\ce{Cl-}$ basis to a $\ce{NaCl}$ basis.

There is problem with ppx values they are ambiguous. It may be w/w, w/v, v/v, n/n.
Salt water has density significantly different to $\pu{1 g/ml}$, so $\pu{1 ppt(parts per thausand) }$ may mean $\pu{1000 mg/L}$ or $\pu{1000 mg/kg}$, with the recalculation factor of the solution density.
The former ($\pu{ppt w/v as 1000 mg/L}$) is more probable, but check it. E.g prepare a salt solution of known concentration and check, in which ppt variant is your refractometry measurement calibrated.
$$\pu{1 ppt(w/v)} = \pu{1000 mg/L}$$
$$\pu{1 ppt(w/w) = 1000 mg/kg =  \frac{\pu{1000 mg/L}}{ \rho(\pu{g/mL})} }$$
where $\rho$ is density of solution in $\pu{g/ml}$
$\pu{1000 mg NaCl}$ is equivalent to $1000\cdot \frac {M_{\ce{Cl}}}  {M_{\ce{Na}}+
M_{\ce{Cl}}}=\frac {35.453}{22.990+35.453}=\pu{606.6 mg Cl }$
$$\pu{1 ppt NaCl(w/v)} = \pu{606.6 mg/L Cl}$$
$$\pu{1 ppt NaCl(w/w) =  \frac{\pu{606.6 mg/L Cl}}{ \rho(\pu{g/mL})} }$$


text:
Relation between size of ion and hydration energy
Suppose I take 1 tablespoon of an ionic crystal and put it in excess water. We know that shall the hydration energy of the crystal be more than the lattice energy, the crystal will separate in ions and water molecules will surround each ion, making it soluble.
I wonder if the relation is direct or inverse in:
i) Anion's size and hydration energy
ii)Anion's size and lattice energy
iii) Cation's size and lattice energy
iv) Cation's size and hydration energy
I think the Cation's size is directly proportional to the hydration energy, since the bigger the atom, more easily shall it form the bond with the water molecules. The Anion's size should be inversely proportional to the hydration energy as the bigger the atom, more difficult would it be to form a strong bond between the water molecules. Similar reason goes for the other two.
PS: I'm new in chemistry so please try to explain a little.

For the lattice energy, both cation and anion radii release more potential energy, when they are smaller.
It is similar as if a satellite goes to the lower orbit, or an  electron in an atom goes to the quantum state with lower energy.
For multiatom anions, it is more complicated due their structure and charge distribution.
For hydration energy and cations of the similar kind, like alkali metals, smaller the radius, bigger hydration energy, as there is stronger electrostatic fields.
It has e.g. direct impact on ion mobility and conductivity.
For different kinds of captions, the electrophilic  affinity of cations plays role as well.
Hydration of anions does not have as big effect as for cations. The anions are often rather big, charge of anions is often delocalized and hydration via water H atoms is less effective.

Can you explain that why is their stronger electrostatic field if radius is smaller?

$$E = \frac{q}{4 \pi \varepsilon_0 r^2}$$

That $r$ is distance between two charged particles and not the radius of it.

$r$ is distance from a (point-like or spherically symmetric) charge. The smaller distance (because of the smaller ion radius) means stronger field.

How smaller distance? How can radius of an ion decide it's distance from another charge?

@An_Elephant Think longer. E is the intensity of the field as the function of the distance from a point charge (or a center of a spherical charge). It says nothing about other charges. The smaller the ion is, the stronger is the field at its boundary, if simplified by a solid charged ball model. Stronger field means stronger field gradient and strong attraction of water dipoles.

If I understand you correct, you meant the field is stronger at boundary of atom if radii is small (charge is kept constant ) and hence the potential at boundary will be larger because $V= Kq/r$. And oxygen of water will be more attracted toward the more potential surface of ion. Thanks very much. I was searching for this from many days but couldn't got it.

@An_Elephant It is based on very basics of classical electrostatics of charges and dipoles.

Yes, but I am a newbie and in the early stage of learning to apply concepts.

@An_Elephant Rather, water molecule as  dipole is attracted to the ion due non homogenity of the field, which is proportional to 1/r^3, being dE/dr.

We can look at solubility as a competition between lattice energy and hydration.
$\Delta H_\text{solution} = \Delta H_\text{hydration} + \Delta H_\text{lattice}$
If the enthalpy of solution is exothermic or slightly endothermic, the compound will be soluble (we are ignoring entropy here).
If we take a simplified view of both lattice energy and enthalpy of hydration, we can see that they are both generally proportional to both the size of the ions as well as the charges, that is:
$$E \propto\frac{|q_1q_2|}{r} $$
Where $q_1$ and $q_2$ are the charges on the species in question and $r$ is the inter-nuclear distance.  So, ions with higher charges are more attracted to each other AND form strong ion-dipole interactions with water.  At the same time, larger ions experience weaker attractions to each other AND form weaker ion-dipole interactions with water.
It is important to remember that this simplified analysis does not take lattice structure and crystal packing into account at all.
The differences in solubility between different ionic compounds follows two general trends.

Mismatch in Ion Size
In general, when there is a difference in the size of the ions in a compound, the compound is more soluble.  This probably arises from poorer packing in the solid state.  For example at room temperature, $\ce{MgSO4}$ has a solubility of 1.16 g/100mL while $\ce{BaSO4}$ has a solubility of $2.45*10^{-4}$ g/100mL$^{[1]}$.  The barium ($159$ pm) and sulfate ($242$ pm) ions are much closer in size than magnesium ($86$ pm) and sulfate$^{[2]}$.
Covalent Character
In general, the more covalent character present in an ionic compound, the lower it's solubility.  We can estimate covalent character by looking at the difference in electronegativities of the ions in the compound.  The larger the difference in electronegativities, the less covalent character in the compound.

For example, $\ce{AgF}$ has
$\Delta \mathrm{EN} = (4.0 - 1.93) = 2.07$
Solubility = $\dfrac{180}{100} \mathrm{~g~mL^{-1}}$
While $\ce{AgCl}$ has
$\Delta \mathrm{EN}= (3.5 - 1.93) = 1.57$
Solubility = $\dfrac{0.193}{100}\mathrm{~g~mL^{-1}}$
In summary, the solubility of ionic compounds is greater when there is a significant size difference between the ions and when there is a greater difference in their electronegativities.
Footnotes

[1] Tro et al, Chemistry, A Molecular Approach, 2nd Canadian Edition
[2] https://en.wikipedia.org/wiki/Ionic_radius

Energies are proportional to radius of ions? How? Isn't the $r$ the distance between two centre of charge instead of radius of an ion?


text:
Molecular Orbital Hybridization vs Delocalization for positive charge stabilization
While studying Molecular Orbital Theory, I ran into a comparison of basicity between pyridine and piperidine. The latter was concluded to be more basic because of the sp3 hybridization of the positively charged nitrogen opposed to the sp2 hybridization of the pyridine nitrogen. The p orbitals are higher in energy, hence, the higher the p influence on the final hybrid orbital the bigger it will be in energy. Higher energy orbitals result in higher stability for positive charges.

On another moment, when reading about carbocation stability, it was stated that tertiary carbocations are more stable due to the delocalization of the positive charge.
Both statements regarding the positive charge stabilization makes sense, but here's where I started getting confused:

If positive charge delocalization plays a big role on carbocation stability, wouldn't the π-system of the pyridine help delocalize the positive charge through resonance? Wouldn't that compensate the effect of the orbitals lower energy (due to higher s orbital character)?
Considering the answer to the above question was that the orbital hybridization actually plays a bigger factor in positive charge stabilization than delocalization, then why alkynes like methylacetylene are attacked in the secondary carbon (which are sp hybridized in the carbocation form?) and not in the sp3 hybridized carbon? (Consider R = CH3 in the image below)

Could someone give me a light on those matters? I assume I might be making a confusion regarding the carbocation hybridization of methylacetylene, would the secondary carbon have a sp geometry when it's positively charged?

Basicity is the ability to donate a pair of electrons.
In the case of pyridine, the lone pair is in a sp2 orbital. This orbital does not participate in any resonance, and is smaller than the sp3. This means the electrons will be closer to the nucleus due to stronger attraction, making the pyridine a weaker base, since it's harder for it to donate these electrons.
Tertiary carbocations are more stable, that is correct, but you should look closely on what that means. A tertiary carbon is carbon atom that is to 3 other alkyl groups , not just any other atom. Alkyl groups are weak electron-donating groups, therefore stabilizing the carbocation, because they increase the electron density in the electron-poor center.
That is why according to Markovnikov's rule, the electrophile, in this case the proton, will bond to the less substituted carbon - in order to create a more stable secondary carbocation, which the nucelophile (the chloride) will attack.

So the stabilization of charges require the orbitals interacting to be the ones actually holding the charge? I thought maybe other orbitals could help stabilize the charge if, for example, they held pi-system electrons for more/less time, even if they were not the actual orbitals responsible for the charge. But thinking better it does make sense that the orbitals with the excess/deficit of electrons are the ones that really matter on stability through resonance.

By the way, just confirming, the positive carbon on the carbocation that has a double bond, would still have sp hybridization right?

Yes, as it still has only two "hands" (That's an easy way to remember hybridization.)


text:
Salt bridge confusion :(
How exactly does a salt bridge close the circuit? I understand its role of neutralizing the solustions but I can't see how it forms a complete circuit. Do charges flow back to the anode through it? But why? Isn't the purpose of a battery to carry charge from high potential (anode) to low potential (cathode)? Isn't the external connecting wire enough?

Do you know the force, by which there would be attracted the opposite charges in 1m distance, caused by the current 1A by just 1s ? About 9 GN. Do you you know about any battery able to withstand comparable force ? Without a salt bridge or equivalent feature, the current in the wire would ceased after passing just very small charge.

SE sites focus on high quality answers to high quality, not easily answered, questions, to be a gain for all.
Quick questions, where one can easily find answers in any relevant textbook or within few clicks on Wikipedia, do not count as high quality questions.
They will be answered only if there is explicitly provided failed elaboration effort to raise their quality a little. This applies to literal homework, self-studies, puzzles, worked examples etc - forming the Homework class of questions.

I dunno, this question seems conceptual enough not to count as homework. If you think it's poorly researched, you're free to downvote, but I'm voting to leave it open.

Well, it is rather a basic concept for physics SE about consequences of big charge displacement. Displace electrons by current 1 A for 1-2 minutes to the distance 10 km away and then observe the GC+ lightning 10 km long with the potential difference 1 billion volts. :-)

It is the same principle as why there is not going any current from a voltage source to a powered device, if there is currupted galvanic connection anywhere. The initial current very quickly balances potentials of power outlets with potentials on the wires and no current is flowing after charging the wire parasitic capacitance.
Let consider the classical Daniel cell $\ce{Zn|ZnSO4||CuSO4|Cu}$ with $||$ being a salt bridge ( or a diaphragm permeable for ions, but avoiding mixing of solutions).
$\ce{Zn}$ anode has more negative potential than the $\ce{Cu}$ one and provides excess of electrons by dissolving and hydrating $\ce{Zn^2+}$ ions.
The electrons flow via the wiring to the copper cathode with higher potential, where they reduce $\ce{Cu^2+}$ to the metallic copper.
In case of the absence of the bridge or the diaphragm, the accumulating negative charge at the cathode compartment very quickly creates solution potential difference that balances the electrode potential difference and the current vanishes.
In case of the presence of the salt bridge or the diaphragm, the potential difference between the solutions makes ions to migrate to the respective electrodes, according to their charge and direction of the potential gradient.
This keeps the compartment potential difference lower than electrode  potential difference  and the current is free to flow via the wire.


text:
Converting SMILES to .sdf files
I have around 1000 SMILES entries in Excel and I have to generate .sdf file for all. Is there any way to automate this whole process, preferably with an offline tool?

Datawarrior lets you do it.
Export a csv sheet with two columns: The name of the compound and its SMILES code.
Be sure to use a comma as a delimiter.
Your output file, opened with a text editor, should look like this:
Name,SMILES
Hexazine,n1nnnnn1
Furazan,c1nonc1
Benzisoxazole,c1noc2c1cccc2

From DataWarrior, File>Open and open your file.
Structures are automatically generated from SMILES codes.
Then, File>Save Special>SD-File, and you are done.

Working with this many SMILES you should consider to have a look at Open Babel, both, because it may transcribe many chemistry relevant formats as well as extract and filter information from them.  This being said, you have at least two options:

One time use, preferentially with shorter lists to convert:  Use Open Babel's GUI which equally is freely available.  On the left hand side, select the input format as .smi and that you would like to paste the SMILES, rather than to read them from a file. Select on the right hand column the export format and enter the directory where your permanent .sdf shall be written.  Helpful for a human is to add the option (center column) "Append output index to title" as a counter, and highly recommended for future work with the .sdf to activate "Generate 3D coordinates" a bit lower in this column, too.

Depending on the operational system at your disposition, Open Babel has multiple interfaces to languages like Python.  Let us assume you saved your data in the .xlsx format (e.g., Excel, LibreOffice Calc or gnumeric) in the second column of your spreadsheet:

Even without entering into the inner tools of Open Babel, you then may already benefit from a module like openpyxl to access the SMILES directly, relaying them to Open Babel, e.g.
# name:  reader.py

import openpyxl
import subprocess as sub
import sys

try:
    source_file = sys.argv[1]
    workbook = openpyxl.load_workbook(source_file)
    sheet = workbook['Sheet1']
except:
    print("Call the script on the CLI by 'python reader.py [my_data.xlsx]'.")
    sys.exit()

def reading():
    """ retrieve the SMILES from the .xlsx """
    global smiles_collector
    smiles_collector = ""
    for cell in sheet['B']:
        smiles_collector += str('-:"') + str(cell.value) + str('" ')

def writing():
    """ provide a container .sdf """
    reading()
    conversion = str("obabel {} -osdf > output.sdf --gen3d".format(smiles_collector))
    sub.call(conversion, shell=True)

reading()
writing()
sys.exit()


text:
Freezing point of solution
It's stated in my textbook that at the freezing point, the vapour pressure of the solid phase equals the vapour pressure of the liquid phase. In a mixture, since the vapour pressure of the solution is lower than the vapour pressure of pure solvent, the temperature has to be lowered in order for the vapour pressures to equalize and hence freezing to take place.
According to what I understand after reading my textbook, when a solution freezes, it's the solvent which converts into solid and leaves behind liquid phase consisting of both the solvent and solute. Thus, how can the pure solid solvent and liquid phase mixture have the same vapour pressure at the same temperature, since the vapour pressure of pure solvent at any temperature is greater than the vapour pressure of mixture?

The phrase "vapour pressure of the solution is lower than the vapour pressure of pure solvent"  has one crucial word missing.

Can u tell me what is that word..

It is "liquid". Liquid solution and liquid solvent, correspondingly. (Or solid and solid, but that's less important, as solid solutions are relatively rare.) Most importantly, it is definitely not about liquid solution vs solid solvent.

What they mean having same vapour pressure for liquid and solid phase at equilibrium

How can a solution have vapour pressure equals to solid solvent at same temperature

Well, just like that. What's strange or unexpected about it?

Could you please provide a reference to the textbook?

My textbook is class 12 ncert textbook..

@user79504 - consider a liquid and a solid, connected only through the vapor phase. If the solid and liquid are in equilibrium, the vapor pressure must be the same. Otherwise the vapor would be the path to move the higher free energy phase to the lower free energy phase (i.e. the vapor would come off the liquid and condense onto the solid if the solid is the stable phase).

The figure shows the relationship between the solid and liquid solvent and a solution.
Instead of freezing at the freezing point of the solvent $T_1$ a solution freezes at $T_2$, where the vapour pressure of the solution equals that of the pure solid solvent.
Using the Clausius-Clapeyron equation we can show that the solution curve will always intersect the solid solvent curve. The vapour pressure of the solution follows the line $\displaystyle \frac{d\ln(p)}{dT}=\frac{\Delta H_{vap}}{RT}$ and the pure solid solvent $\displaystyle \frac{d\ln(p_{ss})}{dT}=\frac{\Delta H_{sub}}{RT}$ where $\Delta H_{sub}$ is the molar heat of sublimation. This in turn is given by $\Delta H_{sub}=\Delta H_{vap}+\Delta H_{fus}$ where 'fus' means heat of fusion. Thus $\Delta H_{sub} \gt \Delta H_{vap}$ which in turn means that the slope of the solid solvent curve is always greater than that of the solution $\displaystyle \frac{d\ln(p_{ss})}{dT} \gt \frac{d\ln(p)}{dT}$.  As the freezing point of the pure solvent is higher than that of the solution the vapour pressure curve of the solid solvent and solution must always intersect.


text:
How does increase in volume change the speed of reaction in production of NO2?
$$\ce{2 NO + O2 <=> 2 NO2}$$
I understand that increasing the volume $×2$ means lower pressure so production of $\ce{NO2}$ is slowed down, but when asked how to express this mathematicaly I can't get to the correct value, which is either $4$ or $8.$ I don't see how inserting $1/2$ in $v = Δc/(2×Δt)$ can give anything close to the offered solutions.

The reaction rate isn't necessarily proportional to the concentration in reactants: it depends on the order of the reaction.

@ThomasJungers Yes, but concentration is the only value I can change with the increase of volume. I tried placing 2V in the formula for the constant of the reaction k=([NO2]^2)/([NO]^2×[O2]) but i got different results for every option I tried and non of them are 4 nor 8

Your $k$ is an equilibrium constant. That is entirely different from the kinetics of the reaction.

I'm not an expert but this is how I would do it.

Determine the reaction rate formula from the reaction equation
$$r = K[\ce{NO}]^2[\ce{O2}]$$
Changing the volume by a factor $2$ means the concentration of $\ce{NO}$ and $\ce{O2}$ will become half
$$r' = K\cdot\frac{[\ce{NO}]}{2}^2\cdot\frac{[\ce{O2}]}{2}$$
Compare the reaction rates
$$r' = K[\ce{NO}]^2\cdot\frac{1}{4}\cdot [\ce{O2}]\cdot\frac{1}{2}$$
$$\frac{r}{r'} = \frac{K[\ce{NO}]^2[\ce{O2}]}{K[\ce{NO}]^2[\ce{O2}]\cdot\frac{1}{8}} = 8$$

Conclusion: increasing the volume by $2$ will decrease the concentration (or pressure) of $\ce{NO}$ and $\ce{O2}.$ This causes the reaction rate to decrease by a factor of $8$.

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The partial orders in the rate law are equal to the stoichiometric coefficients of a reaction only when that reaction is an elementary reaction. I sincerely doubt that $\ce{2 NO + O2 <=> 2 NO2}$ is an elementary reaction.
As it happens, the rate law for this reaction is indeed $r = k [\ce{NO}]^2 [\ce{O2}]$, but it is a coincidence.

thanks for the feedback. Thomas I agree with you on that. It's impossible for the reaction to be elementary because the reactant contains 3 molecules, I didn't think about that. I searched for the reaction mechanism on google and this: https://resources.saylor.org/wwwresources/archived/site/wp-content/uploads/2013/03/CHEM102-1.4-Reaction-Mechanisms.pdf page 6 suggested that it's a two step reaction forming an intermediate N2O2. And it happens to match the observed rate law.