text:
How would you remove sulfur hexafluoride from a gas sample?
What procedures would be available to remove SF6 from a sample (or stream) of gas?
The ideal answer would be a list of options. If it helps, please assume the rest of the sample consists of noble gases, although something that would work in presence of other impurities would be even better.
Specific example sample if it helps:

Impurities: 0.1% water vapour. Otherwise Ar 30%, Kr 39.8%, Xe 30%.
Volume: 300 mL.
Effectiveness: at least 80% after one or more passes.
Concentration of SF6 in initial sample: 0.1%.

Multiple stages of ultra-centrifugation?

How much SF6 in the stream? Really hard in general…

SF6 is very non-polar, and it must soluble in a lot of solvents (basically scrubbing process)...check in these directions as well, or what about condensation?

If you want to focus on SF6, you could have edited your previous question and include only SF6. In this way, your question could have been reopened.

You're asking much too general for any sensible answer. What other impurities?! You can obviously freeze it at -80dC. If that's not what you want, you should say so. And what concentration does it have? How effective do you want your removal?

Is this related to your other question about removing fluorinated stuff from the atmosphere? If so, on what order of magnitude do you want to operate?

Please consider this an independent question (the consensus seems to be it would have little bearing on the other). Magnitude: say 330 mL. Effectiveness: anything over 80%. Concentration: I wanted to leave that open, but say 0.1%. I couldn't resist some complicating factors, but the simple case is just with argon, krypton and xenon, as already obtained from distillation, The ideal answer would be a list of options.  Will add this.

Apparently, sulfur hexafluoride reacts exothermically with solid lithium. So dry the gas mixture you posit and run it through the lithium. Of course, if nitrogen were present, then this would not be feasible.

Per a source: "Efficient Removal of Sulfur Hexafluoride (SF6) Through Reacting with Recycled Electroplating Sludge" by Jia Zhang. A link to abstract, and quoting:

This paper reports that recycled electroplating sludge is able to efficiently remove greenhouse gas sulfur hexafluoride (SF6). The removal process involves various reactions of SF6 with the recycled sludge. Remarkably, the sludge completely removed SF6 at a capacity of 1.10 mmol/g (SF6/sludge) at 600 °C. More importantly, the evolved gases were SO2, SiF4, and a limited amount of HF, with no toxic SOF4, SO2F2, or SF4 being detected. These generated gases can be readily captured and removed by NaOH solution. The reacted solids were further found to be various metal fluorides, thus revealing that SF6 removal takes place by reacting with various metal oxides and silicate in the sludge. Moreover, the kinetic investigation revealed that the SF6 reaction with the sludge is a first-order chemically controlled process. This research thus demonstrates that the waste electroplating sludge can be potentially used as an effective removal agent for one of the notorious greenhouse gases, SF6.

Hope this helps.


text:
Mechanism of the blue bottle experiment
I have researched multiple mechanisms for this experiment yet all of them suggest different versions or are too vague. None of them are consistent since the final overall equation seems to be missing a few elements. Can someone help me out?
Step 1: $\ce{O2(g) -> O2(aq)}$ (Fast)
Step 2: $\ce{O2(aq) + MBred (aq) -> MBox (aq)}$   (Fast)
Step 3:  $\ce{MBox (aq) +C-(aq) -> MBred (aq) + X-(aq)}$ (Slow)
The general mechanism that showed up frequently is very vague. The $\ce{X-}$ is always labeled as the oxidation products of the carbohydrate. I can't seem to specifically find the product for any of the carbohydrate (glucose, fructose etc.). On top of that, from my understanding of the reaction mechanisms is that each step must be balanced and in the end, intermediate species are canceled out. However, the same issue always comes out with the leftover elementary species $\ce{O2}$ where the overall equation is then not balanced. For example:
Prior to the addition of MB: $$\ce{C6H12O6(aq) + OH-(aq) <=> C6H11O6-(aq) + H2O(l)}$$
Step 1: $\ce{O2(g) -> O2(aq)}$ (Fast)
Step 2: $\ce{O2(aq) + MBred (aq) ->  MBox (aq)}$   (Fast)
Step 3:  $\ce{MBox(aq)  + C6H11O6-(aq) -> MBred (aq) + C6H10O6(aq)}$ (Slow)
is the best possible mechanism I could put together and step 3 makes sense to me since the 2 $\ce{H+}$ ions will be used to reduce the methylene blue. but then the overall equation will look like:
$$\ce{O2 + C6H11O6-(aq) -> C6H10O6(aq)}$$
where it is now missing the balance in hydrogen and oxygen.

Amazing that this popular demonstration reaction is not better documented in readily available sources. Maybe someone should add some detail to the Wikipedia article.

According to Shakhashiri and coworker's 2012 J. Chem. Ed. paper, the linear form of glucose forms an endiolate which gets oxidized to a dicarbonyl (on carbon 1 and 2) and then further reacts with the dioxygen product, $\ce{HO2-}$ (deprotonated hydrogen peroxide), in an oxidative decarboxylation reaction to form arabinoic acid:

The blue color comes from the oxidized form of methylene blue. The "bleaching" of the colored dye through reaction with glucose is slow, while (in the presence of dissolved oxygen), the reaction of dioxygen to regenerate the dye is fast. This leads to the observed behavior of the blue bottle reaction mixture. The last step of the mechanistic hypothesis does not directly influence color changes, so visual observations don't tell us about how fast this step happens.
On the other hand, according to Wikipedia, substituting ascorbic acid or benzoin for glucose, while showing similar kinetics, results in the diketones without further reaction.
Each dioxygen needs to gain 4 electrons to yield water. Reducing the dye is a two-electron transfer yielding peroxide, which can then react with the dicarbonyl. (I guess only when it is at the end of the chain, ready to decarboxylate, so ascorbic acid does not undergo decarboxylation).
Some sources say that the product of glucose oxidation is the sugar acid gluconate, but that hypothesis seems not to be supported by experimental evidence.

See also: https://pubs.acs.org/doi/10.1021/acsomega.9b00802 (open article).

Bassam Shakhashiri's book entitled "Chemical Demonstrations, A Handbook for Teachers of Chemistry" gives a detailed study of the mechanism of the blue bottle, in his Volume $2$, page $145$ ($5$ volumes are available).
First glucose $\ce{C6H12O6}$ reacts with $\ce{NaOH}$ in a basic solution according to $$\ce{C6H12O6 + OH^- -> C6H11O6^- + H2O}$$ Now methylene blue $\ce{C14H12N3S^+}$ is added into the glucose solution. The glucose molecular ion is oxidized to gluconic acid $\ce{C6H12O7}$ (or its basic form the gluconate ion $\ce{C6H11O7^-}$)  according to :  $$\ce{C6H11O6^-  +  C14H12N3S^+ + H2O -> C6H11O7^- + C14H14N3S^+}$$ Simultaneously methylene blue is reduced into methylene white $\ce{C14H14N3S^+}$. Later on, the dissolved oxygen molecules reoxide the methylene white into methylene blue, according to $$\ce{C14H14N3S^+ + 1/2 O2 ->  C14H12N3S^+ + H2O}$$ The oxidation does not stop on the gluconic acid. This acid (and its deprotonated ion) is further oxidized into glucaric acid and other more oxidized structures that will not be studied here.
The detailed structures of the interacting molecules are more interesting than their formula.
Let's first consider glucose and its oxidized forme. In Fischer's projection, glucose is a linear molecule with an upper aldehyde group $\ce{-CHO}$ on top. The gluconic acid has the same structure except the upper atomic group which is transformed into an acidic group $\ce{- COOH}$.
Let's now describe the methylene blue structure. The methylene blue is made of two $N$-dimethylaniline structures connected by a =$N-$ bridge in their $para$ position and a $\ce{-S-}$ bridge in $meta$. One of the benzene ring has a quinonic structure, with a $+$ charge on the outer $\ce{N}$ atom. Let's now consider the reduced form called methylene white. Its rings are both benzenic. The first added $\ce{H}$ atom is attached on the internal $\ce{N}$ bridge; and the second is fixed on the outer and positively charged $\ce{N}$ atom, which destroys the quinonic structure.
Ref.: Bassam Z. Shakhashiri, Chemical Demonstrations, A Handbook for Chemistry Teachers, The University of Wisconsin Press, 1985, p.142, Vol. 2, ISBN 0-299-10130-4

This is a different description from what the same author wrote 27 years later, after studying the reaction in detail using HPLC and NMR methods to identify the product, and iodometric titration to detect peroxides as intermediates.

Thank you Karsten for the comment. I am happy to hear about this more recent report, because Bassam said in 1985 that he was not absolutely sure about the  reaction between glucose and methylene blue, as it should be a trimolecular reaction. And trimolecular reactions are often the sum of two bimolecular reactions.

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text:
Is the 'i' prefix for "iso-" italicised?
I want to abbreviate pseudoisocytosine. Should it be "ΨiC" or "ΨiC"?
I would usually write "sec-butyl", "tert-butyl", etc, and using "t-BuOH" for "t-butyl alcohol.  I usually see, in respectable places, "i-PrOH" for isopropanol, and that what I have done up till now.
The Wikipedia for chemical naming conventions notes that
"cyclo, iso, neo, and spiro are considered part of a chemical name (such as isopropanol) and not considered prefixes. No hyphens or italics are used in these cases."
IUPAC writes "isopropyl" prefix.
What is correct?

I don't have the time for a very in-depth answer at the  moment. However, if you look at P-29.6 of the 2013 Blue Book, you can see what they italicise and what they don't. (Just do a Ctrl-F for the relevant words.)
https://iupac.qmul.ac.uk/BlueBook/P2.html#2906

sec-butyl is italicised
tert-butyl is italicised
tert-pentyl is italicised
o-tolyl is italicised
isopropyl is not italicised
isobutyl is not italicised
neopentyl is not italicised

It should be pointed out that of all these, only tert-butyl is a preferred prefix. The section linked above makes this clear. But in most cases, that probably needn't concern you.
As for abbreviations, these are covered in the 2008 IUPAC recommendations on Graphical Representation Standards for Chemical Structure Diagrams. See in particular Table II on page 406, which gives the following forms: iPr, iBu, s-Bu, and t-Bu (which is in line with the Blue Book). It says that if you choose these abbreviations, then explicitly defining them is unnecessary.
(Of course, you can choose whatever abbreviation you like, as long as you define it. The text accompanying Table II makes that clear. It also says that the non-italicised forms are fine too.)

Authors are welcome to create their own abbreviations as well, but any abbreviations not included in the list below should be defined clearly when they are used. [...]
Several of the abbreviations listed below include portions of the text in italics. The italicization shown is the preferred formatting of those abbreviations. However, the corresponding forms without italicization are also acceptable.

Is this the most up-to-date version?  Sorry, I don't know this.    2. Just making sure:  So where I see "i-PrOH", when not explicitly defined, is wrong?

Yes, the 2013 recommendations are the most recent. You can see this on e.g. Wikipedia 2) It depends on how you define 'wrong'. Sure, it's not in line with IUPAC recommendations. If you're thinking about a journal, they may have their own set of rules, which may take precedence over IUPAC recommendations. And sometimes, it simply doesn't matter. Plenty of things out there are not in line with IUPAC recommendations, and people still understand them just fine.

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text:
Conceptual doubt regarding electrolytic cells involving acids
Consider the following cell
Ag(s) | AgIO3(s) | Ag+
(x M), HIO3 (1 M) || $Zn^{2+}$
(1 M) | Zn(s)
My doubt is not regarding the full calculation of the EMF , but the meaning of ' 1 M HIO3 '.
What I mean to say is , does '1M' mean the concentration of HIO3 at time 't'? (the time at which nerst equation has been applied) or at t = 0?
This is because everywhere on the internet, for the calculation of [IO3-] , the INITIAL concentration of HIO3 has been taken as 1M , which I fail to understand why. As per my knowledge , we have applied the nerst equation at any time 't' and thus involving the reaction quotient at that instant, and hence the initial concentration of HIO3 should not be 1M, but more than that (and the concentration after considering its dissociation , i.e. , c-c$\alpha$ , is 1M).
So is my understanding correct?

I think you are correct. According to me, the voltage of the cell keeps decreasing in magnitude until the reaction hits equilibrium and the cell potential is 0. So if you were to find the potential at time t (considering c=1M at t=0), the value of $t$ needs to be given. Otherwise you’ll have the value of $E_{\text{cell}}$ at the time when c=1M. It might be helpful if you link the solutions on the internet which give rise to the confusion.

Both concentrations  [$\ce{HIO3}$] and [$\ce{Zn^{2+}}$] are  to be considered as $\pu{1 M}$ at $\pu{t = 0}$.

@Maurice so is it a convention which i might have missed out that all concentrations mentioned in the IUPAC shorthand notation of the cell are to be considered at t=0?

@insipidintegrator thanks for the reply, i dont think giving the link wuld be of much use as I already mentioned what i didnt feel right about them. Anyways , there IS one video which is not in english(It doesnt matter though, the solution he has written is in english) i'll link it: https://www.youtube.com/watch?v=Rjk_U2zz6bo . See the right side of the greenboard. Clearly he has taken the written concentration as the initial one.

@Hsrah. You seem to forget that the cell tension, as determined by Nernst's law, does not depend on the time. It is measured with no current flowing out of the cell.

@Maurice thank for replying, I must admit was not at all aware of that fact. Could you please provide me with some text for further reading because I did not find any text available to me which has that concept.

@Hsrah. Sorry for having forgotten you for one month. You find plenty of texts by typing "Internal resistance of a cell" on Google. The first reference goes like this. ...The nominal potential difference (E) of a cell is always greater than the measured value V measured across a cell, if this cell is delivering a current.
Ex.: The potential difference across a cell when no current flows is 3 V. When the current I = 0.37 Amp is flowing, the measured initial potential difference falls to 2.8 Volts, if the internal resistance (r) of the cell is 0.54 ohm, according to the formula E = V + Ir

I view this as relating to a spontaneous galvanic cell involving the anodic corrosion of Zinc metal with noble Silver metal in the presence of Iodic acid.
In that event, the smallest amount of $\ce{HIO3}$ with a starting amount of 1 mole of Zinc is just 1 mole corresponding to a possible targeted formation of $\ce{IO3Zn+}$ (as, for example, noted in this source).
Note: The notation employed references an electrolysis cell involving the reverse reaction.
My comment on Iodic acid is the same 1 mole (or more, see, for example, Question (40) here or Q (16) here).
In short, the mole ratio between Zinc and Iodic acid is more variable (hence, the seeming discrepancy) depending on the possible targeted speciation of the Zn.


text:
On the definition of Molar conductivity
The textbook I follow defines the conductivity, $\kappa$, as

$$\text{Conductance} \ G=\frac{1}{R}=\kappa \frac Al$$
The inverse of resistivity, called conductivity, is represented by the symbol $\kappa$ (…) Conductivity of a material in $\text{S m}^{-1}$ is its conductance when it is $1$ m long and its area of cross-section is $1$ $\text{m}^2$.

It then introduces the molar conductivity, $\Lambda_m$, as

(…) define a physically more meaningful quantity called the molar conductivity denoted by the symbol $\Lambda_m$. It is related to the conductivity of the solution by the equation$$\text{Molar Conductivity}\ \Lambda_m =\frac{\kappa}{c}$$

and goes on to derive different equations with respect to different units of use.
But two pages later, we see this:

Molar conductivity of a solution at a given concentration is the conductance of the volume V of solution containing one mole of electrolyte kept between two electrodes with area of cross-section A and distance of unit length. Thus, $\ell=1$ and $A=V$ so$$\Lambda_m=\kappa V$$

Here, why do we need the unit length constraint? Why can’t it be unit area?
 It is rewritten a few lines later as

“at unit distance and area of cross section large enough to accommodate sufficient volume of solution that contains $1$ mole of the electrolyte.“

I have seen another casual definition of $\kappa$ as:

$\kappa$ is the conductance of unit volume of solution.

I need to confirm that this is not accurate. I think that $\ell=2 m$, $A=0.5 m^2$ also gives unit volume but from $G=\kappa\frac Al$ we get $\kappa=4G$ when $\kappa$ should equal G. The book has a similar definition, but with an added “ kept between two platinum electrodes with unit area of cross-section and at a distance of unit length.“ and I agree with this.

I have searched and not found this specific query. Most of them have discussed the variation with dilution.

I think the unit for V is $L/mol$ as it is the volume containing 1 mole of solution. @Poutnik

V is defined here (as stated in the post) to be the volume containing 1 mol of solution.

Ah, I think I can now see what you’re getting at.

You should include your info in comments in the question for clarity.

Regarding definitions(*), stick to the first 2 sections:
$\kappa = G \frac{l}{A}$ and $\Lambda = \frac{\kappa }{ c}$

The only reason for such an arrangement in section 3, with

$A[\pu{m2}] = \frac {n_1}{ c \cdot l_1 }$
$l_1$ is the unit distance
$n_1$ is the unit molar amount of solute
$c$ is the molar concentration.
$V = A \cdot  l_1$

is, that conductance is then numerically equal to molar conductivity. As multiplicators to get conductivity from conductance and then molar conductivity from conductivity have then mutually reciprocal numerical values and numerically cancel each other.
$$\kappa = G \frac{l_1}{A}$$
$$\Lambda = \frac{\kappa }{ c}=\frac{G \frac{l_1}{A} }{ \frac{n_1}{A \cdot l_1}}=G \frac{l_1^2 }{n_1}$$
As numerical values are chosen $l_1 = n_1 = 1$, then numerically, $\Lambda = G$ AND $\Lambda = \frac{\kappa }{ c} = \kappa V = \kappa  A$.
But practical measurement is not done this way, nor is molar conductivity defined that way.

"Conductivity is conductance of the volume unit" is wrong for 2 reasons:

It is numerically valid  only for the cubic shape (within rectangular shapes)
It may be OK numerically, but it is wrong dimensionally. Conductivity is never conductance, even when sharing the same numerical value  if the distance and the area have unit numerical values. The ratio of a unit distance and a unit area is not dimensionless, as the unit distance has dimension $\pu{[m]}$ and the unit area has dimension $\pu{[m2]}$.

(*) Note that conductivity can be equivalently defined also in differential form as proportionality factor between local vector quantities the current density $\pu{[A m-2]}$ and the  potential gradient $\pu{[V m-1]}$, leading to $\pu{[S m-1]}$. For anisotropic solids, these 2 vector quantities may not be colinear and conductivity is therefore a tensor (matrix) quantity.


text:
Iodine clock reaction: lower temperature but faster result?
I was told by several high school students that they experienced odd results in their iodine clock reaction lab. They did several different trials, and the lower the temperature, the faster the blue color showed.
Out of curiosity, I asked about the detail, and they told me the following:

Most students in most classes had the same problem, so I think it wasn't bad technique of some students that caused the problem.
They did experiments of different temperature with the same solutions, so I think it wasn't that they did the lower temperature experiment with higher concentration.
The temperatures were 30 °C, 20 °C and 10 °C.
Their teacher only said that maybe there's something wrong with the solution. But it made me more curious: what kind of mistake could lead to lower temperature having faster results?

The only possibility I could come up with, is that iodine clock reaction isn't a simple reaction, it includes:
\begin{align}
\ce{IO3^- + HSO3^- &-> I^- + H^+ + SO4^2-} \tag{R1}\\
\ce{IO3^- + I^- + H^+ &-> I2 + H2O} \tag{R2}\\
\ce{I2 + HSO3^- + H2O &-> I^- + H^+ + SO4^2-} \tag{R3}
\end{align}
To my knowledge, lower temperature results in lower reaction rate. Is it possible that somehow the third equation's rate is lowered to a point that it couldn't get rid of $\ce{I2}$ fast enough, so somehow the blue color showed before bisulfite is used up?
Is this possible? If there are different explanations, I would love to hear.

Wang, Could you also add some realiable literature survey using Google Scholar? This reaction must be well studied.

Is there any tips for searching for certain questions on google scholar? (I tried keywords like "iodine clock low temperature", but couldn't find paper on the point.)

I agree that the iodine clock reaction is complex, but if you need an answer for your teacher, the explanation may center around "sulfurous acid" implied in your reaction chain, that apparently does not actually exist in water (but is found just in air). For more details, per Wikipedia, to quote:

Raman spectra of solutions of sulfur dioxide in water show only signals due to the SO2 molecule and the bisulfite ion..."

At lower temperatures, there is a notable increase in dissolved sulfur dioxide (see, for example, graph provided in this source).
Next, there is the so-called Bunsen reaction, to quote Wikipedia on the reaction:

The Bunsen reaction is a chemical reaction that describes water, sulfur dioxide, and iodine reacting to form sulfuric acid and hydrogen iodide:

2H2O + SO2 + I2 → H2SO4 + 2HI

This reaction is the first step in the sulfur-iodine cycle to produce hydrogen.

Note: The products of the above Bunsen reaction matches the cited reaction products of Equation (R3).
As a consequence, this is one possible explanation as to how an increase in sulfur dioxide gas concentration itself (from cooling) may also be beneficially impacting the iodine clock reaction as it is clearly negatively temperature dependent.

Thank you for your insight! But I can't understand how more SO2 ( from lower temperature) results in faster showing of blue color. Doesn't it remove I2 faster, thus delaying the blue result?

Per a source, quoting: "Molecular iodine (I2) is not easily soluble in water, which is why potassium iodide is added. Together, they form polyiodide ions... The negatively charged iodide...acts as charge donor, the neutral iodine as a charge acceptor. Electrons... excite to a higher energy level by light. The light is absorbed in the process and its complementary color is observed by the human eye"  link provided below. So, conversion of the element Iodine to iodide results in a starch color change.  The presence of an added Bunsen reaction facilitates this event via creating iodide.

Link reference, as referenced above: https://www.chemistryviews.org/details/education/10128441/Why_Does_Iodine_Turn_Starch_Blue/#:~:text=Iodine%20Test,changes%20to%20a%20deep%20blue.


text:
Is the chemistry behind the "Anoxic Filter" actually real? [Pond/Aquarium Denitrification]
In the Aquarium hobby, there are two methods to remove NO3 from your water:

Plants
Water changes

There are a handful of people and companies that sell porous biological media that claims to create low-oxygen zones inside of it for anaerobic bacteria to grow. That seems to be snake oil.
There has emerged another proposed method called "Anoxic filtration". It involves mixing baked clay kitty litter with an iron rich substrate called Laterite to create a "biocenosis basket" and placing it inside the filter. It is said that this will convert nitrate to nitrogen gas. Below is a synopsis of how it is claimed to work:

Zone A
Negative charges in the cat litter in the baskets begin to attract molecules of ammonia (NH4) towards its centre. As these molecules pass through zone A nitrifying bacteria (nitrogen cycle bugs) will grab one ammonia molecule and four oxygen molecules; they will then excrete one nitrate molecule.
There is now far less oxygen but there will still be enough for the nitrogen cycle to continue so we will continue to journey with the ammonia molecules into zone B.

Zone B
As more and more ammonia is converted to nitrate, the ammonia level drops even more and the nitrate level rises. So much oxygen has been used in the process that this area can no longer be called truly aerobic, (oxygen rich), but there is still a little oxygen left to sustain some nitrifying bacteria so we will follow the remaining ammonia as it journeys into zone C

Zone C
The ammonia is still being pulled remorselessly toward the centre of the basket but almost all oxygen in the water has already been used. The nitrogen cycle, as we know it, ceases. Nitrification cannot occur if dissolved oxygen levels fall too far below about 2 mg/L and it will be lower than that in zone C where facultative anaerobic heterotrophic bacteria live.
A facultative anaerobic heterotroph can easily make use of the three oxygen atoms and leave just the nitrogen atom which is initially dissolved in the water but will migrate out of the basket and gas off to the atmosphere.

Full article on Anoxic Filtration

Questions:

Is this method of nitrification actually possible?

There is an article by a hobbyist with a BA in Chemistry that is dedicated to refuting the entire idea:

The science behind anoxic denitrification of nitrate to nitrogen gas dictates that biocenosis clarification baskets and plenums simply cannot work.

I don't know if I trust the article though because it reads more like a hit piece toward the person who came up with the idea.
I'm looking for clarity on this topic from someone outside of the aquarium/pond hobby.

I would feel more confident about it if they had not described ammonia as NH4

@Waylander I believe the NH4 depiction is because at that stage the NH3 has ostensibly become positively charged.

The idea that positive ammonium ions are attracted to the negatively charged center is wrong. While there may be negative charges, it is overall neutral, balanced by positive charges.

Playing with anaerobic environment may lead to forming dangerous nitrites and sulfane.

@MikeCiffone  if that is the case then they should be refering to ammonium ions

Denitrification happens in natural anoxic environments due to the right bacteria exploiting available nitrate in the absence of significant amounts of molecular oxygen. This question however is about whether such a process can be exploited in the suggested design of aquarium filter, which may not be a question for this site.

This may be more appropriate at Biology SE.


text:
Is it possible to store a Tween 80 stock solution?
I am making a stock solution of Tween 80 (0.05 % w/v) and each experiment requires just 15 ml of the stock solution. Is it possible to make a large batch (about 100 ml) and store it somewhere? If so, what should I look out for when storage?

Please note that nonconforming units such as "% w/v" should not be used in science and technology.

The Sigma Aldrich/Merck product information for "TWEEN® 80 solution" states the following:

Storage/Stability
Aqueous solutions of polysorbates as well as the neat liquid will
undergo autoxidation over time, with changes being catalyzed by light,
increased temperature, and copper sulfate.[9]
Solutions are reasonably stable at 2 - 8 °C for short periods. For
special applications, storage under argon or nitrogen may be
preferred.
[9.] Donbrow, M., et al., Autoxidation of polysorbates. J. Pharm.
Sci., 67, 1676-1681 (1978).

While this is not a very specific time span, it clearly indicates that TWEEN solutions are not long-term stable. If you need to store it, then in a fridge, protected from light and preferentially under an inert atmosphere for a few days max.

If you do this be sure to leave the date the solution was prepared on the bottle.


text:
Photoelectric effect: What happens when you irradiate the metal strip with a wave of frequency exactly equal to threshold frequency of the metal?
My teacher said that if we irradiate the metal strip (used in generating the photoelectric effect) with  light having frequency equal to the metal's threshold frequency, then the electrons will have enough energy to ionize and get separated from the metal atoms. However, they won't have any kinetic energy left, and so will remain stationary.
But why does it matter that the electrons won't have any kinetic energy? Shouldn't the electrons get repelled from the electrons behind them and go forward?

BTW, Einstein got for the photoelectric effect (officially mainly for that, to avoid SR/GR critics) his Nobel price in physics, not chemistry.

@Poutnik do we assume all the electrons that have been removed from the atom to be near stationary?

@Poutnik And why do those near-stationary electrons attach themselves to the matrix instead of getting repelled ahead?

Also, should I delete this question and post it again at the physics stack exchange?

@Poutnik so as far as I have understood, the reason behind the electrons not being repelled ahead is that they have a higher tendency to  form anions. Am i correct?

Electrons have always some kinetic energy, it is requirement of QM principles. Plus there is distribution of the threshold photon energy due Doppler effect, because of moving atoms. Near stationary electrons will probably attach themselves back to the solid matrix, possibly leading to local charge disbalance. We are already used to that in static electricity context.
In the above scenario, electron will have like thermal speed when released. what for low electron mass and 298 K is typically $v_\mathrm{RMS}=\sqrt(\frac{3kT}{m_\mathrm{e}}) \approx \pu{116 km/s}$.
For mettallic conductors with metallic bonds, where electrons can freely move, one electron less or more as the structure deviation play no significant role, similarly as electrons and "holes" in semiconductors.
For covalent bond structures, electrons as point charges will tend to bind to many neutral molecules, forming anions. It is similar to "naked" protons.
Near all elements (but noble gases and few exceptions) have positive electron affinity is spite of dense clouds of electrons around their nucleus. It should not be therefore surprising electrons can attach itself to molecules.
An electron can also return to its original place while releasing its energy, or to break a bond and attach to one of fragments.
E.g.:
$$\ce{e- + R3C-H -> R3C- + H^.}$$
or
$$\ce{e- + R3C-OH -> R3C^. + OH-}$$


text:
Mechanism for reaction of beta-ketoacid with AcONa, AcOH with heat
I am unsure of how to go about designing a mechanism for the following molecule's reaction with AcONa and AcOH with heat.
I thought perhaps it may work as a cyclisation reaction, forming an ether; or perhaps a decarboxylation. I've tried many of these mechanisms and none of them seem to give an answer that seems correct.
Any hints/tips would be much appreciated.

Sodium acetate in AcOH is acting as a base, so my first thought is that these are decarboxylation conditions


text:
Why is the pH of this ammonia solution more acidic than another ammonia solution?
I made two similar ammonia solutions in different ways, and they resulted in different pH, but i'm not sure why.
Solution 1
Added 5.84 g NaCl to a 500 mL flask. Then added 10 mL 25% ammonia solution. Filled to 500 mL with distilled water to make a 500 mL 0.5% ammonia solution (0.267 M) with 0.2 M NaCl.
pH: 11.75
Assumed chemistry:
NH3 + H2O + NaCl --> NH4 (but mostly NH3) + OH- (but mostly H2O) + Na + Cl-
Solution 2
Added 14.8504 g NH4Cl to 722 mL distilled water. Added 278 mL 1 M NaOH. This leaves me with a 1 L solution of 0.5% ammonia (0.278 M) and 0.278 M NaCl
pH: 10.01
Assumed chemistry:
~ NH4Cl solution~
NH4Cl + H2O --> NH4 + Cl (pH.5?)
NH4 + Cl + H2O --> NH3 + H3O+ + Cl (pH 5?)
~NaOH solution~
NaOH + H2O --> Na + OH- (pH 10)
~Adding NaOH to NH4Cl solution~
Na + OH- + NH3 + H3O+ + Cl --> NaCl + NH3 + H2O
Both solutions contain roughly similar the molar amount of ammonia (~0.2 M) and NaCl. Yet, the pH of the solutions greatly differ.
Is it that in solution 2, ammonia is existing mainly as NH4, or perhaps there are remaining hydronium ions?
Any advice would be appreciated.

The solution 2 is essentially a buffer solution.

@MathewMahindaratne Thanks Mathew! What part of it makes a buffer? I thought adding a strong acid and a strong base will produce salt and water (as above). But I suppose NH3 is a weak acid?

NH3 is not a weak acid. It is a weak base. And the final solution is not really a buffer solution, as the amount of ammonium ion $\ce{NH4^+}$ is equal to the amount of $\ce{OH-}$ ions.

The theoretical pH value should be $11.34$, if the concentrations are equal to the activities. So there is something odd in your second solution, where the pH was $10.01$. This pH is too low to be credible.

@Maurice Thank you. So by your comment, have I got the chemistry correct that theoretically both solutions should be a similar pH, and at the end have the same chemical composition? I thought this would be the case so was surprised when it came back so different.

Your solutions 1 and 2 have the same final composition, if I read correctly your descriptions. This is why I don't understand the pH difference. There must be some unknown impurity, or some mistake somewhere...

Both the concentration of your 25% ammonia and your 1N NaOH are suspect especially the NaOH.  How did you measure the 10 mL? Attempting to do quantitative analysis without using good quantitative technique is a recipe for disaster. Also please write correctly balanced equations. Trying to maintain ionic strength with constant NaCL is very good.


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Dissolving Pyrite with HCl: Will dissolving KNO3 in HCl give it a similar oxidizing effect to HNO3?
I'm trying to dissolve small pyrite (FeS2) grains. It's supposed to be done in nitric acid, but my lab doesn't have any. It does have a lot of 37% HCl and various salt oxidizers, such as KNO3, KMnO4, or NH4NO3.
I've tried dissolving the pyrite in full-strength HCl with no success. I have read that the dissolution comes from the oxidizing effect of the nitric acid's conjugate base, and not necessarily the H+. So if I dissolve some of the aforementioned salts into HCl, would it be able to then dissolve the pyrite?

Properties of the solution containing particular concentrations of hydrated ions H+, K+, Cl- and NO3- do not depend on origin of these ions. What is essential here is low activity of water and high activity of H+ and NO3^-. When based on HCl, there may be too much of water.

Dissolving $\ce{KNO3}$ in concentrated $\ce{HCl}$ will produce some $\ce{HNO3}$. But as soon as it is produced, $\ce{HNO3}$ will react with $\ce{HCl}$ in excess to produce "aqua regia", according to the equation : $$\ce{HNO3 + 3 HCl -> Cl2 + NOCl + 2 H2O}$$ And $\ce{Cl2}$ will attack pyrite

Per this 1991 study "Effect of Additives on Pyrite Oxidation" by By G. W. Reimers and D. F. Franke, apparently the presence of chloride reduces pyrite oxidation with air/oxygen and water.
The cited products of this air oxidation includes soluble ferrous sulfate and also sulfur dioxide. The latter forms bisulfite with water, and with transition metal tap water (with increased ferrous content) upon vigorous aeration, the sulfite radical is introduced (from Fenton hydroxyl radicals acting on bisulfite). This sulfite radical can apparently rapidly react with oxygen creating a long chain of reactive species (see, for example, this ScienceDirect account and even more recent work by Rivas discussed here).
So, my recommendation, do not use nitrate or chloride source, just vigorous aeration (like with an air pump) of tap water containing the pyrite should suffice especially with optional strong sunlight.
However, if you want to add some acid along with the aeration actually consider a combination of ascorbic acid and citric acid per the 2009 work of Vidrio et al.,"Generation of Hydroxyl Radicals from Dissolved Transition Metals in Surrogate Lung Fluid Solutions" (where an available but  impure source would be just common lemon juice) that would act mainly as an initial driver of hydroxyl radicals to further induce sulfite radicals, etc.


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Gibbs Free Energy of Ideal Gas in Carnot Cycle
Let's just say we have 2 moles of monoatomic (maybe helium) ideal gas that is doing a Carnot Cycle with reservoir temperature of 300 and 750 K.

Now, here's a simple question, what is the Gibb Free energy change from point A to point B and from point B to C in this process?
Because (do correct me if I'm wrong), I believe that we need to know the absolute value of the entropy in this case.

Generally, along a reversible path, dG=-SdT+VdP. But, for the isothermal path between A and B, $$dG=VdP=nRT\frac{dP}{P}=-nRT\frac{dV}{V}$$

What about from B to C?

That would involve the S at B (orC).

Yes. But how do we calculate that? I mean, its not that we need $\Delta S$, rather we need S

S is specified relative to some well-defined reference state.  We don't need to know the absolute value of S in order to use it in practical calculations.


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Differential equation in consecutive reactions
I'm blocked in a step of getting to the equation $(7)$, more precisely in the step between $(5)$ and $(6)$. I show my way to solve it down, but I don't get the same as $(6)$, so I wanted to know if I was wrong and an explanation of the correct aproach.
$$\ce{A ->[k_\mathrm{1}]I->[k_\mathrm{2}]P}$$
$$\frac{\mathrm d[\ce{A}]}{\mathrm dt} = -k_\mathrm{1}[\ce{A}] \tag{1}$$
$$\frac{\mathrm d[\ce{I}]}{\mathrm dt} = k_\mathrm{1}[\ce{A}] -k_\mathrm{2}[\ce{I}] \tag{2}$$
$$\frac{\mathrm d[\ce{P}]}{\mathrm dt} = k_\mathrm{2}[\ce{I}] \tag{3}$$
$$[\ce{A}]_\mathrm{0}\neq0 ; [\ce{I}]_\mathrm{0}=0; [\ce{P}]_\mathrm{0}=0$$
$$[\ce{A}]=[\ce{A}]_\mathrm{0}e^{-k_\mathrm{1}t}  \tag{4}$$
$$\frac{\mathrm d[\ce{I}]}{\mathrm dt} = k_\mathrm{1}[\ce{A}]_\mathrm{0}e^{-k_\mathrm{1}t} -k_\mathrm{2}[\ce{I}] \tag{5}$$
$$[\ce{I}]=\frac{k_\mathrm{1}}{k_\mathrm{2}-k_\mathrm{1}}(e^{-k_\mathrm{1}t}-e^{-k_\mathrm{2}t})[\ce{A}]_\mathrm{0} \tag{6}$$
$$[\ce{A}]_\mathrm{0}=[\ce{A}]-[\ce{I}]-[\ce{P}]$$
$$[\ce{P}]=[\ce{A}]-[\ce{A}]_\mathrm{0}-[\ce{I}]$$
$$[\ce{P}]=(\frac{k_\mathrm{1}e^{-k_\mathrm{2}t}-k_\mathrm{2}e^{-k_\mathrm{1}t}}{k_\mathrm{2}-k_\mathrm{1}}+1)[\ce{A}]_\mathrm{0}  \tag{7}$$
And here is what I've done:
$$\int_{0}^{[\ce{I}]} \, \mathrm{d}[\ce{I}]=k_\mathrm{1}[\ce{A}]_\mathrm{0}\int_{0}^{t}e^{-k_\mathrm{1}t} \, \mathrm{d}t-k_\mathrm{2}[\ce{I}]\int_{0}^{t} \, \mathrm{d}t \tag{A}$$
$${[\ce{I}]}=-e^{-k_\mathrm{1}t}[\ce{A}]_\mathrm{0}-k_\mathrm{2}t[\ce{I}] \tag{B}$$
$${[\ce{I}]}=-[\ce{A}]_\mathrm{0}(\frac{e^{-k_\mathrm{1}t}}{k_\mathrm{2}t+1}) \tag{C}$$
Anyone could help me? Thanks :)

This symbolic integrator to resolve both infinite and finite integrals may help you: https://www.integral-calculator.com // It is very powerful.

Thanks a lot, but I actually know how to solve thos integrals and I've just checked them. The question was how to solve the differential equation to reach equation (6).

It looks to me you need to separate terms to integral equation for I and t on respective sides.

I end up with the same because I can't pass "dt" to the right without affecting "k2[I]".

Related

@Rodrigo de Azevedo, is NOT the same question, even if it's the same topic.

@JorgeBonifaz I claimed it was related. I did not claim it was a duplicate.

Sorry, a question I posted last week was closed because it was "duplicated" but it wasn't that.

This is a form of a linear first-order differential equation, and you can solve it like this (or just plug it into wolfram alpha).
We have
$$\frac{d[I]}{dt} = k_1[A]_0e^{-k_1t} - k_2[I]$$
$$\frac{d[I]}{dt} + k_2[I] = k_1[A]_0e^{-k_1t}$$
If we let $P(t) = k_2$, $Q(t) = k_1[A]_0e^{-k_1t}$, and $I(t) = e^{\int P(t)dt} = e^{k_2t}$ the solution to this sort of differential equation is
$$[I] = \frac{1}{I(t)} \left( \int I(t)Q(t)dt + C \right)$$
where C is a constant. Plugging in the values set for this problem, we have
$$[I] = \frac{1}{e^{k_2t}} \left( \int e^{k_2t}\cdot k_1[A]_0e^{-k_1t} dt + C\right)$$
$$[I] = Ce^{-k_2t} + k_1[A]_0\cdot\frac{1}{k_2-k_1}e^{(k_2-k_1)t}\cdot e^{-k_2t}$$
$$[I] = Ce^{-k_2t} + k_1[A]_0\cdot \frac{1}{k_2-k_1}e^{-k_1t}$$
Now plugging in $t = 0$, we know $[I]_0 = 0$. So:
$$C = -\frac{k_1[A]_0}{k_2-k_1}$$
Reinputting C to the previous equation gives:
$$[I] = -\frac{k_1[A]_0}{k_2-k_1} e^{-k_2t} + \frac{k_1[A]_0}{k_2-k_1} e^{-k_1t}$$
$$[I] = \frac{k_1}{k_2-k_1}(e^{-k_1t} - e^{k_2t})[A]_0$$
We have to do this because [I] depends on t as well, so you can't just take it out of the integral like you did.

Thanks a lot. I'm not in university yet, so I didn't know how to solve differential equations, but now I understand it.


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Why the electronic configuration of samarium is ending with 4f⁶6s²
What is coming in my mind is that 4f^7 6s^1 should be more stable as it allows two orbitals to be half filled rather than 4f^6 6s^2 in which only one orbital is completely filled....can someone please help me in this and tell that where I am thinking wrong....

In the f-block elements is more stable to have a d1 orbital at the f7's next element (for instance, gadolinium, which has 6s2 4f7 5d1) than having the previous one with s1 configuration.

While this is perhaps true (the electronic configurations prove it), this doesn't actually explain anything. Could you add any more reasoning, or physical data, for this?


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Why is diamond not the most dense substance?
Diamond is the hardest substance on earth which can be attributed to its tightly-bonded crystal lattice. Hence, assuming there were a substance more dense than diamond (which there are), mustn't it be harder than Diamond?

The tightly-packed crystal structure of diamond is only one of the factors involved in determining its hardness, but I do recall reading that diamond is the material with the highest atomic density (atoms per cubic centimeter) in ambient conditions, though graphene probably beats it when it comes to area density (atoms per square centimeter).

Mercury is about 4 times denser than diamond and it it a liquid. Graphite is also made of carbon, but is relatively soft. Gold is nearly six times denser but is also so soft when pure you can mark it with your fingernail. Perhaps hardness is more complex than you think.

There is no general relation between hardness and density. One thing is the packing and bonding character ( what matters mostly for hardness), other is the idealized atomic ball radius and the atomic mass, what matters mostly for density.
Osmium has the highest density among chemical elements, being more than 6 times denser than diamonds with $\pu{22.6 g cm-3}$ versus $\pu{3.52 g cm-3}$. Gold with $\pu{19.5 g cm-3}$ should be one of hardest materials, what is obviously not true as it is very soft.

I have asked in physics.SE: Is there any empirical relation between hardness and density? (cc @CannedOrgi)

@NilayGhosh There will be close relation hardness vs density within a very narrow group of similar materials, in special cases. Like binary alloys as extreme example. But not generally.

@NilayGhosh There is obviously many soft and hard materials across the density scale. Perhaps more interesting would be comparing hardness and molar amount density, with the latter addressing density of atomic packing.


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Why is AgOH brown when silver(I) ions are colorless?
As Wikipedia states, $\ce{Ag+}$ ions are colorless in solution: https://en.wikipedia.org/wiki/Color_of_chemicals
Compounds that I know of containing $\ce{OH-}$ are also colorless or white (such as $\ce{Mg(OH)2}$, which just like $\ce{AgOH}$ is a precipitate). How is it possible then that AgOH has a brown color?

Well, afaik it is not really a hydroxide - formula is misleading. Either way, you can only use such reasoning when cation and anion are reasonably "separate" and that's not the case.

Wikipedia:Don't abbreviate "Wikipedia" as "Wiki"! Otherwise, you can use some fairly simple syntax to correctly typeset chemical formulae: see https://chemistry.meta.stackexchange.com/q/86/16683 for an introduction

$\ce{AgOH}$ does not exist. It is spontaneously decomposed into $\ce{Ag2O}$ and $\ce{H2O}$. Anyway, the question is not solved : Why is silver oxide brown ? The same problem can be discussed with $\ce{MnO2}$ which is brown-black, or $\ce{Cu2O}$ which is orange-yellow, $\ce{CuO}$ which is black, $\ce{HgO}$ which is yellow (sometimes orange), and probably a lot of other examples. Why is it so ?

https://chemistry.stackexchange.com/questions/32019/why-are-hgo-and-cds-compounds-colourful https://chemistry.stackexchange.com/questions/94249/why-is-a-cuscn2-complex-black

It also depends on the physical state: finely divided materials may have different color than a bulk sample. For example, Ag metal is shiny, (obviously, 'silvery ;-) en masse, but black when finely divided in photographic emulsion. Colloidal gold may be red when coloring glass.

There is no $\ce{AgOH}$. It is rather $\ce{Ag2O.xH2O}$


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Can the relative velocity of two colliding molecules be too large to allow reaction to occur?
In a collision between two molecules, can the relative velocity of one with respect to the other be too large for a reaction to occur? For example, suppose that two molecules collide with (a) a perfect orientation to enable a reaction and (b) kinetic energy in excess of the activation energy, but their relative velocity is very large in magnitude. Will the two molecules simply bounce and part ways chemically unchanged?

If molecules are colliding with good orientation and a little more than the activation energy, they will react.

... but it may happen, if they cannot release the the reaction energy by other way, they revert the reaction back. Like 2H <=> H2* -> H2 + energy

@Poutnik - did you mean the last $H2$ to be $2H$?

@JonCuster No, I did not. If H2* is not able to pass the energy, the bond breaks. If it is, H2 stays. Notice the first arrow is bidirectional.

Too high kinetic energy may lead to alternative, destructive reactions.  O may react with O2 to form O3, but can with enough energy also form 3 O instead.

If say H+H collide and the energy is above the dissociation limit the bond will form, temporarily, i..e one half of a vibrational period at most and dissociate again since no energy is lost. A 3rd body needs to collide during the half period mentioned to remove some energy then an vibrationally/rotationally excited H2 will form. With polyatomics its possible to soak up some collisional energy in vibs/rots so more collisional energy will be needed to prevent reaction.

One important thing to remember is that rarely is there ever only one possible reaction for two molecules to engage in; we often make distinctions between "kinetic" reactions (reactions that occur because their activation energies are low) and "thermodynamic" reactions (reactions that occur because the products have favorable energies). However, every reaction we'd reasonably do in a lab is technically a "kinetic" reaction, because extremely high-energy velocities will eventually surpass the activation barriers to extremely unstable products like carbanions and random molecular fragments. Basically, go fast enough and your molecules will just explode on contact. We like to think of gas-phase molecules as billiard balls bouncing around, and that can be a pretty good model! Especially considering that if you hit two billiard balls together fast enough, they will also just explode.
So we're asking "are there kinetic energies below the activation energies of competing reactions that will make collisions faster than the time-scale of bond formation?" And the answer is a very confident "no"—bond formation and rearrangement occurs on the femto- and picosecond time-scales (mere billionths of a second long) and there is simply no way to make them go fast enough such that they'd spend too little time in the regions where suitable electron-wavefunction overlap for a transition-state occurs, especially if you were hoping they don't just explode.
If you're interested in high-energy reactions, I recommend exploring some of the theoretical mechanisms behind fragmentations in mass spectrometry. Much of mass spectrometry leverages kinetic energy to access high-energy fragmentary reactions for structural analysis of compounds.

What about molecules travelling at 99% speed of light passing close to each other? Close enough for chemical reactions to occur.

basically all our models for chemistry break down at relativistic speeds. the Schrodinger equation—which dictates all modeling for orbitals that engage in bonding—is not Lorentz invariant, and thus cannot be applied in cases of relativistic speeds. one could attempt solving for a simple molecule at relativistic speeds using the Dirac equation or QED, but these computations might be insoluble and would almost definitely not yield interactions relevant to chemistry of large enough molecules. interesting thought experiment though, I might play around with it.

so we don't know the answer yet for relativistic speeds?

@suyashsingh234 i don’t mean to imply that—i’m uncertain, but this may have been investigated before. treatments with the Dirac equation are not unheard of; it’s roughly used to approximate the behavior of core electrons in large molecules, but i personally am unaware of treatments for valence/bonding electrons (i’m sure it’s been looked at)

It takes time for molecules to rearrange, dispose of excess energy etc. This could be a reason why exothermic reactions become less favorable as temperature is increased and molecules decompose at high temperatures. The potential energy for bond formation must be released either by radiation or transfer to colder molecules. When this cannot happen, the bond does not form.


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Do chemical bonds impact half-life of elements?
Let’s say we have element X with half life of 100 years. Can chemical bonds like X−X or X−Cl increase/decrease half-life of X?
As a follow up question, can it increase/decrease radioactivity of X?

See https://physics.stackexchange.com/questions/713918/can-a-beryllium-7-nuclei-be-stable-if-it-has-lost-its-electrons/713923#713923 which has one answer (full disclosure - it is mine), and a link to another related post. A fully stripped $^{7}$Be ion can't decay by electron capture since there isn't an electron to capture. Further, electron density in compounds with $^{7}$Be$ do impact the measured half life.

https://chemistry.stackexchange.com/a/62042/9961

https://chemistry.stackexchange.com/a/85435/9961

It might depend on the type of decay undergone by the element, but certainly for some decays it is possible. That being said, these are the exception rather than the rule. Typically, nuclear decay is controlled by the weak nuclear force, which is mediated by the W and Z bosons in the Standard Model of Particle Physics. This is in contrast to chemical bonds arising from interactions between electrons, resulting from the electromagnetic force which is mediated by photons in the Standard Model. Any impact is primarily going to relate to the interactions between the pertinent quantum fields. Electrons can undergo processes guided by the weak nuclear force, so implicitly any impact to the electron's wavefunction will subtly—and occasionally noticeably—affect its ability to interact with the nucleus.
Different chemical bonds perturb the electronic environment of the nucleus, and thus any decay processes that are in some way dependent on the local electronic environment will have a shift in decay-rate, and thus a shift in half-life. The easiest example is electron capture. Electron capture is a decay process where an electron is captured by a proton and converted into a neutron. If the electronic wavefunction doesn't sufficiently overlap with where it needs to be for the weak force to mediate electron capture, then the decay rate will change. Here is a paper where the authors lengthen the half-life of a beryllium-7 atom by encapsulation with palladium atoms.
While I didn't see any examples in my (admittedly surface-level) search just now, I would also put forth that this likely holds for some other decay processes, such as beta decays. Beta decay occurs when a quark in a nucleon decays into a quark of the opposite type (called "flavor") and emits a W boson that itself decays either to a positron or to an electron* depending on if it was a proton $\rightarrow$ neutron decay (positron emission) or a neutron $\rightarrow$ proton decay (electron emission). Both of these processes produce a charged particle, and thus I imagine that screening of the local electron environment affects their rates (similar to how a chemical reaction is less favorable if there is a high activation energy, etc). Brief aside, all of this isn't particularly relevant for alpha decays (emission of a helium-4 nucleus) due to the fact this decay is due to interactions with the strong nuclear force and thus the impact of nuclear binding energy is significantly greater in magnitude than impact from the Coulombic interactions.
*and the relevant antineutrino
In reference to your final question, radioactivity is determined explicitly by decay process and decay rate—just like half-life—and so a shift in half-life would imply a shift in radioactivity a priori. The magnitude of change likely isn't enough that it would practically impair or accelerate radioactivity though, especially for the heavier elements where chemical bonding would be less relevant due to the larger role of valence electrons in chemical bonding than core electrons.

I'm not convinced on beta decays, but have requests in for a few review papers from some time ago. In electron capture you need an electron to capture, and the electron density near the nucleus provides them. Considering it as an overlap integral between initial state and final state, you can impact the initial state. In beta decay, extant electrons around the nucleus really don't impact the decay channel (the final state), so might not impact the rate.

And, the second comment by Mithrion pointing to https://chemistry.stackexchange.com/questions/85425/is-it-possible-to-speed-up-radioactive-decay/85435#85435 contradicts my assertion on not impacting the final state, so, yes, beta decay is another option. Good call!

While the answer outlines some of the rare exceptions, it should also be worth emphasizing that these are very rare. Most radioactive nuclei will have the same rate of decay whatever chemical environment they are found in.

@matt_black yeah I tried to mention it a bit, but you're right I should place more emphasis on it. I'll update it now!

This makes me think of a hypothetical machine which can switch on/off radioactivity by making and breaking chemical bonds.

The half-life of a radioactive nucleus is not affected by the molecule or the ion containing this nucleus. It has been demonstrated plenty of times in the beginning of the 20th century.

This is not true, see https://physics.stackexchange.com/questions/713918/can-a-beryllium-7-nuclei-be-stable-if-it-has-lost-its-electrons/713923#713923 - elements that decay by electron capture are sensitive to the electronic environment. Note the paper on experimental measurements of the half-life of $\ce{^7Be}$ in different chemical compounds, and the ability to see differences in the half-lifes.

@Jon Custer. OK. I did not know about $\ce{^7Be}$. Is Beryllium-7 the only nuclid where half-life can be modified by change of electronic environment ?

basically any electron capture lifetime can be changed. The extreme is still fully stripped ions leading to infinite lifetime (seen in cosmic rays). $^{7}Be$ has the nice properties of (1) being proposed as a target by Segre and Wiegand in 1949, and (2) having a tractable lifetime of a bit less than 2 months allowing fairly short experiments. Citations for Segre's paper indicate a variety of longer review articles over the decades, so it seems unlikely that $^{7}Be$ is the only one.

@EdV - indeed, per ENSDF, normal $\beta^{-}$ decay is 4.33E10 years, but fully ionized is only 32.9 years, called bound state $\beta^{-}$ decay as referenced in https://chemistry.stackexchange.com/questions/85425/is-it-possible-to-speed-up-radioactive-decay/85435#85435 in a comment above. Still waiting on my library to get me a review article I've requested...

@EdV - I'm learning a lot here too, so no harm done at all, in fact pushing to find more is turning out to be quite interesting...

@EdV - Even more interesting, $^{163}$Dy is stable as a neutral atom, but has a half life as $^{163}$Dy$^{+66}$ of 48 days because of bound-state $\beta$ decay.


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Does the Darzens reaction work with an alpha-chlorocarboxylic acid and an aldehyde?
Is it possible that a reaction of an alpha-chlorocarboxylic acid with an aldehyde work for the Darzens reaction to produce an epoxide (with NaOH as the base, and H3O+ workup)?

Welcome to ChemSE! What are your thoughts?

Hi, I've seen many examples of aldehyde reacting with esters in this reaction, but none so far of an aldehyde with a carboxylic acid. The mechanism I have written seems to work on paper, but I am a little unsure about it. Also, I am not sure whether NaOH would be a strong enough base. Any help would be much appreciated. Thanks

Any base you would use would form the carboxylate which would risk intra- or intermolecular displacement of chloride. Secondly, to form the dianion you need a much stronger base in an aprotic medium. If step one is feasible, 2 equivalents of lithium diisopropylamine (LDA) in tetrahydrofuran (THF) might do the trick at low temperarure. Then add the aldehyde and warm to room temperature.


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Do disulfide bonds determine the 3-dimensional structure of a protein or do they just stabilise the 3D structure?
I noticed that most disulfide bonds occur when two cysteine side chains exist in close proximity to each other. Do those cysteine side chains "look out" for each other during folding, that is do they cause the polypeptide to fold so that a disulfide bond can be formed? If not, does this mean that they contribute to the tertiary structure of the polypeptide by stabilising only?

I do not suppose SH groups have abilities to look for each other. They are brought together to bond themselves by protein forming mechanisms, probably in ribosoms. What does not exclude incidental bonding.

The introduction and conclusion of this paper contains relevant information and citations, including the classic Anfinsen work.


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Help understanding Isothermal titration calorimetry ITC
Trying to understand Isothermal titration calorimetry: ITC , from a public online forum (Off-topic question related to ITC binding studies) I got this question:
We performed an ITC binding study (using Affinity ITC, TA Instruments) to evaluate protein-DNA interaction which resulted in a perfect sigmoidal curve. We used the ‘one set of sites’ binding algorithm (“independent” model) for curve fitting and to calculate binding and thermodynamic parameters. The study suggested two copies of the protein binding to a single duplex DNA, i.e., the stoichiometry of protein:DNA is 2:1 (N=2). The ITC calculated the KD (equilibrium molar dissociation constant) in μM (micromolar). But the reviewer is asking to report the KD in (micromolar)^2 instead of micromolar mentioning that the binding reaction is 2A + B <-> (A2)B and the complex is (A2)B and not AB. Though we're trying to explain to the reviewer that we couldn't find any software that can compute the KD in (micromolar)^2 for the stoichiometry of 2 but he is not agreeing to it. We have used the NanoAnalyze software from the TA instrument. This software does not have a model to measure the KD in (micromolar)^2.
I would be grateful if you could help me to resolve this problem or at least let me know what explanation might be appropriate to answer the reviewer’s concern that it’s a general practice to report the KD in the Molar irrespective of stoichiometry
ADDENDUM:
Based on the curve fit, we also believe that the two proteins bind independently to the DNA duplex with the same affinity. The DNA indeed contains 2 sites for protein binding. As far as the dimerization of protein is concerned, we have already confirmed, through different approaches, that the protein doesn't behave as a dimer in the solution.

@Poutnik would you be so kind to create the Isothermal titration calorimetry or ITC tag ? Cant edit equation right know, running from tablet, old too, it ll be on my to do list for this forum

I am not sure if it is a good idea to create a tag until there are multiple cases of the ITC topic. I suppose using the general calorimetry is OK, having 31 cases total, 3 this year.

It might help if you post some real or mock data and fit parameters, and the figure for the fit, as an example so that readers that want to can check the problem.

@BuckThorn any free package that can run simulation with models thatt could suit the question above ? First time I read a lot about ITC but not sure their question is right. To me sounds like the DNA ligand could be the titrate with 2 binding site and the protein the titrant so could use the 1:2 model identical sites Association of a monovalent ligand with a bivalent receptor having two identical sites  of Affimeter software suite, but dont have the software available

@BuckThorn, its not my question I just came accross it and was trying to understand it. All I knew about ITC was about the basic 1:1N=1 model to calculate kd 1/ka

@BuckThorn I found a free package to analyze ITC experiments: https://pubs.acs.org/doi/full/10.1021/acs.biochem.7b01264 , https://pytc.readthedocs.io/en/latest/. Any free source of dataset example to be analyzed ??

@user836049 have you tried looking within the package itself? Often they bundle test data to verify that it works as it claims.

@BuckThorn I looked at it didnt find them. Problem is reads only .DH files or out of NITPIC : " The user provides files that hold the heat per injection for each experiment, determined by integrating raw power curves using the instrument software or an alternative such as NITPIC. " BUt what gets me mad is that wasnt able to find anything about .DH file format. Tried to get it from pytc code but wasnt able to find where and how it reads the .DH file

well found some .DH files into : https://github.com/harmslab/pytc-demos/tree/master/ca-edta that is a  a collection of demo jupyter notebooks that show how to use the pytc API to fit Isothermal Titration Calorimetry experiments.  .. Dont have jupiter (dont even know what is it but now I have some .DH examples


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What material(s) can offer similar performance to polyurethane in a paint coating, but with less hydrogen?
We need to coat parts of an neutron instrument with a boron shielding coating. This currently is achieved using an airbrush gun with a paint mix of demineralized water, polyurethane and boron powder.
Ideally we would like as little hydrogen as possible in the coating, as the hydrogen causes neutrons to scatter which causes undesirable noise in detectors. The polyurethane gives decent adhesion and toughness but contains a fair amount of hydrogen. Are there any alternative binders which could offer similar performance with less hydrogen content, or zero hydrogen content?
Water uptake is not too much of an issue, nor is outgassing as we are not at particular high vacuum levels.

It would help to know what the problem is with "too much" hydrogen and what, specifically you mean by too much.

Are you concerned about water uptake, outgassing in vacuum, what?


text:
Molarity 1% ammonia solution?
I have a stock solution of 25% ammonia, with a density of 910 g/L. I used this to make a 1% ammonia solution by mixing 20 mL 25% ammonia into 480 mL distilled water with 5.84 g NaCl (0.2 M), density ~1000 g/L.
To calculate the molarity of the 25% stock solution, I used the following equation:
Molarity = (Percentage concentration * Density in g/L) / (Molar mass * 100)
Molarity = 13.36 M
To calculate the molarity of the 1% solution, do I need use the density of the 25% solution, or use the new density from the mixed solution?
If I calculate molarity by C1V1=C2V2 so that C2 = 13.36 M * 20 mL / 500 mL I get a molarity of 0.5344 M, the same as if i'd used the density of the 25% ammonia in the former calculation. But i'm unsure if this is the correct approach.

Is the percentage w/w, v/v?

@Zhe i'm unsure. The bottle doesn't mention anything like that, it just says %: https://uk.vwr.com/store/product/2345606/ammonia-solution-25-analar-normapur-analytical-reagent

It is w/w (unfortunately not often stated).

w/w is the easiest to deal with. For a fixed mass of solution (easy to compute from the volume and density), you can compute the mass of ammonia and this mass is fixed regardless of the final volume of solution from dilution.


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Confusion in Van der Waals Equation
I understand that the behavior of ideal gases deviates largely from that of real gases in terms of pressure exerted by the gas molecules on the container in which it is present, space available for each gas molecule to move, and other factors.
I know the van der Waals equation includes the necessary correction terms in pressure and volume so that one could fit the behavior of a real gas using ideal gas equation.
In all the examples given in my textbook,for a real gas,the ideal gas equation would be:
$$\left(P + \frac{an^2}{V^2}\right)(V - nb) = nRT$$
where $n$ is the amount of gas, $T$ the temperature, $R$ the gas constant, and $(a, b)$ van der Waals' constants.

What exactly is $V$ in the equation? $P$ in the equation has to be the pressure exerted by the gas molecules in an nonideal state since due to attraction of gas molecules in real state, the pressure exerted by the gas would be lesser than what it would be in the case of an ideal gas. Hence we are adding $an^2/V^2$.
But, the $V$ is representative of the space available for gas molecules to move. In an ideal case, since the size of the molecules is insignificant, we take the space available for the gas molecules to be that of the volume of the container. However, in a nonideal situation, the gas molecules occupy a significant portion of the container, and hence the available space for them to move is less than the volume of the container.

Is $V$ the volume of the container? If so, then why are we even including a correction term to the volume part of the equation? Doesn't the van der Waals equation attempt to fit the behaviour of a real gas using an equation similar to the ideal gas law?

Useful links for text and formula formatting:
Notation basics  /  Formatting of math/chem expressions  /
upright vs italic // Use plain texts in CH SE titles. // For more, see Math SE MathJax tutorial.

i accidentally typed "large" for "real".apologies!

Please consider reading the links in Poutnik's comment. They will let you typeset maths professionally. Also, please, put spaces after punctuation marks. Spaces after commas, spaces after full stops. Lastly, I took the liberty of correcting your equations from $an/V^2$ to $an^2/V^2$: I assume what you really meant was $a(n/V)^2$, but if you want to write it like that, you need the parentheses.

@Poutnik thank you, I was looking everywhere for that and just gave up !

The van der Waals equation is represented as such:
$$\left(P + \frac{an^2}{V^2}\right)(V-nb) = nRT$$
Here, $P$ is the pressure of the container and $V$ is indeed the volume of the container. All the symbols that come from the original ideal gas law maintain their definitions. The best way to conceptualize the van der Waals equation is as a perturbation of the ideal gas law that brings it closer to real gas behavior. Let's start with the ideal gas law:
$$PV = nRT$$
This equation simply states that pressure and volume are inversely proportional to each other at a constant number of molecules $n$ and a constant temperature $T$; that their product is directly proportional to number of molecules $n$ and absolute temperature $T$; and that these parameters are the only considerations to model gases' behaviors. It is not simply that the equation conforms to evidence from the world (empirical) but also that it applies in the fictional but reasonably approximate case that molecules 1) do not interact and 2) do not occupy space. It is often more useful to conceptualize an ideal gas as a gas where gas particles pass through each other; this is equivalent both to excluding particle interactions and to allowing each gas particle the entire volume of the container (because nothing gets in the way). If gas particles all passed through each other, then the ideal gas law would be true, but alas this is not the case.
Since this is not the case, we need to add parameters that account for particle interactions. An entirely new equation is not necessary though: instead of starting from scratch, we can add a term to the pressure and volume variables that corrects for the existence of particle interactions. It is simplest to demonstrate with the case of the $(V-nb)$ term:
Let's say that particles do not interact except they collide with each other and bounce off elastically like billiard balls. The volume term in the ideal gas law $V$ stands for the volume of the entire container; however, if two molecules simply collide with each other, then for every molecule added to the container, there is a molecule's volume less space for every other molecule to be. So, while we still see the volume of the container as $V$, from the molecule's perspective there isn't a full $V$ volume worth of space to go, because other molecules are in the way.
For an analogy, if you walk into an empty room, you can navigate every square foot of the room easily. However, if you invite a bunch of friends over, suddenly there is much less space to navigate even though the room did not literally shrink in size—you effectively have less space to move. We'll call the amount of space that the molecule can actually move "effective volume" or $V_\text{eff}$.
We expect for every additional molecule we add that the volume $V$ stays the same, but the effective volume $V_\text{eff}$ decreases a little bit. So, our effective volume should be 1) linearly related to the number of molecules proportional to the molecule's volume and 2) smaller than the original volume. We can use this to make an equation, where $b$ depends on the volume of the molecule:
$$V_\text{eff} = V - nb$$
A similar argument is made to account for molecular attractions with the pressure parameter, but this is much more complex and includes a squared volume term for complicated reasons. Regardless, we can construct an effective pressure $P_\text{eff}$ which is just the original pressure term with an additional correction for particle interaction included. If we let:
$$P_\text{eff} = P + \frac{an^2}{V^2}$$
and
$$V_\text{eff} = V - nb$$
and simply let $a$ and $b$ be corrections to the original pressures and volumes, then we get the van der Waals equation looking like this:
$$P_\text{eff}V_\text{eff} = nRT$$
which looks exactly like the ideal gas law, but corrected for additional factors.

so the P in van der Waal's equation is actually the pressure exerted by the gas molecules without taking into account the mutual interactions between other molecules?and not the pressure that the gas molecules exert on the container in a non ideal situation?

@anotherhyooman P is pressure on container walls. Internal gas pressure is higher by an^2/V^2. (according to the van der Waals real gas model)

@anotherhyooman yes—by "effective pressure" I mean it, similar to effective volume, as pressure from the molecular perspective. you can ~sorta~ conceptualize P_eff as the average amount of forces experienced by a molecule, which are for essentially all purposes attractive forces. If you rearranged the equation to solve for P, you'd see that P is LESS than the term linearly related to temperature. More attractive forces means the molecule experiences more forces, but that also means a smaller portion of the total forces goes to hitting the container.


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what are dinitrophenyl (DNP) derivatives?
what are dinitrophenyl (DNP) derivatives? Reading "The Amino-acid Sequence in the Phenylalanyl Chain of Insulin" paper by Sanger (1951). I am not a Biochem major.. yet

It’s the product of the reaction of an amino acid with fluorodinitro benzene: https://en.m.wikipedia.org/wiki/1-Fluoro-2,4-dinitrobenzene

See: https://chemistry.stackexchange.com/questions/46359/dinitrophenyl-dnp-derivatives-of-amino-acids-specifically-epsilon-dnp-lysine


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Osmotic Pressure definition clarification
currently reading Legninger's Principles of Biochemistry (8th edition if you absolutely needed to know) and under "Osmotic Pressure" it gives the definition to be:

Osmotic pressure, $\Pi$, [is] measured as the force necessary to resist water movement...

I find this way of thinking a little confusing, why should I care about resisting the water's force across a membrane? I want to know if it's equivalent for me to interpret this as the force the water EXERTS as it passes through a membrane. Learning from Mr. Isaac Newton, I would take this to be a safe assumption. Please let me know if I am wrong!

*SIR Isaac Newton,  forgive me

Rather as "...the pressure necessary...", not force.
The full speed osmosis means zero differential pressure on the membrane. The non zero water net flow is caused by differences of water molecule collision frequencies and therefore of water diffusion rates in both directions.
Applying pressure ( externally or hydrostatically ) on the membrane from the side of the (more concentrated) solution compensates lower diffusion speed by counteracting pressure gradient flow.
The particular value of the counteracting pressure, that would cause the zero net water flow, is called the osmotic pressure of the solution.
If the pressure is high enough, it may cause a rupture of the membrane. One of possible demonstrations can be putting a tomato in deionized water.
OTOH, if no such pressure is applied and there is full osmotic water(or other solvent) flow, there is no unidirectional mechanical pressure on the membrane.

In summary, osmosis is the flow of solvent across a semipermeable membrane from a region of lower to higher solute concentration. Thermodynamically the process is driven by the increase in entropy on mixing more solvent with the solute. In solution, the solvent passes both ways through the semipermeable membrane but the solute cannot. Osmosis is not restricted to solution and can occur in the gas phase.
Using the word 'force' is confusing as is 'resistance'. A pressure difference exists across the semi-permeable membrane. This is balanced by applying external pressure, such as from a vertical capillary tube into which the solution moves as its volume increases, and equilibrium thereby restored.  The external pressure produced by gravity acting on the vertical column of solution is equal to the osmotic pressure. The thermodynamic description was first given by Gibbs in 1897 but this does not explain anything at a molecular level. In fact many textbooks avoid this aspect completely.
One experimental observation is that the flow due to osmosis is greater than that possible from diffusion and is due to hydrodynamic flow through the membrane. This would be damped out very quickly so there has to be a process that keeps this going. Kramers et al. (Kramers & Myers, Am. J. Phys. v80,p694 2012) suggest the following. The solute collides with the  membrane and is repelled, its momentum (now away from the membrane) is then dispersed between the solvent molecules, these being in greatest numbers, and this produces an effective potential barrier to solvent flow which by Boltzmann distribution means that fewer solvent molecules enter the membrane from the solution side, i.e. the membrane partly repels solvent molecules. This does not happen on the pure solvent side so more solvent flows into the solution than flows out increasing the hydrostatic pressure in the solution.
Another explanation based on MD simulations (Lion et Al. J. Chem. Phys. v137, p244911, 2012) is that osmosis is due to the density imbalance of solvent particles across the membrane. They write
'the density imbalance of solvent particles across the membrane is maintained by a balance between an outward force-driven flux (due to the higher total density in the solution) and an inward diffusive flux (due to the lower solvent density in the solution). We show that a simple calculation, based on the mechanics of hopping of solvent molecules across the membrane, can be used to derive the same relation between the solvent density gradient and the osmotic pressure as is obtained by standard thermodynamic arguments.'
So it seems that although osmosis is well studied its actual mechanism is still in dispute and much remains unresolved.


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IUPAC Names for trihalomethanes
The drinking water section of the Massachusetts DEP mandates the electronic submission of analytical results. The compound Chlorodibromomethane must be entered when reporting just trihalomethanes and must be entered as Dibromochloromethane when reporting a full list of volatile organic compounds (EPA Method 525).
The IUPAC system says that the substituents should be alphabetized. I would believe the substituents to be alphabetized are bromine and chlorine and that the "Di" prefix is not considered in the alphabetization logic.
Can any cite a specific reference to the proper naming convention?

Here's the relevant rule.
P-14.5.1 Simple prefixes (i.e., those describing atoms and unsubstituted substituents) are arranged alphabetically; multiplicative prefixes, if necessary, are then inserted and do not alter the alphabetical order already established.
You are right, the di-, tri-, etc prefixes do not matter and only 'chloro' 'bromo' are considered.

You may also add https://iupac.qmul.ac.uk/BlueBook/P1.html#160501 P-16.5.1.3.1 which states that ‘chloro’ should be in parentheses.

@orthocresol I dont think that rule is relevant in this case, parentheses are used when you would want to write multiplied complex  substituents as 'tri' instead of 'tris' or 'di' instead of 'bis'.


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Additives to increase water density and solubility
What additives do I need to add to a water-based drink (coffee or tea for example), to increase its density and to make the drink insoluble (like olive oil for example) in salt sea water or inland water, and the drink with additives is still consumable?
A little bit like some bar-cocktails with layered liquids. The idea is to serve and drink tea or coffee under water, whilst scubaing with students or tourists, or to kill boredom during mandatory decos. Not sure why I can't find such an answer elsewhere on the internet, given how cool the demand might be..

Problem solved.

Ok- this is going somewhere- almost. Most of these are alcoholic (cant serve to kids) and aren't indicating if they drop below water? Has anyone tried anything of these?

You can make a drink thicker with food-safe thickening additives, such as pre-gelatinized starches, gelatine, or agar(-agar). Make your beverage just thick enough to stay in the glass and resist mixing, and add food coloring, if desired.
There's no safe way to make a water-based drink insoluble in water, though. And this doesn't answer the question of how safe it would be to try to drink while in SCUBA gear under water. Consider, for example, that at 10 m depth, theoretically, you could exert a force of 2 atmospheres sucking on a straw in the concoction -- likely to pull it up the straw, and also likely to cause damage to the body. After all, if you're in decompression, and form a partial vacuum in your mouth, aren't you inviting air embolism?

Pooring a thickened water liquid into a cup won't work in water, it will just dissolve but not fall into a cup. This does not answer density or solubility, or solve the purpose of the question. (As for safety: The liquid will be stored in a plastic flexible bottle. Everything will be at the same pressure as the depth- no safety concerns.)


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Are the np orbitals of light group 2 elements considered as valence electrons in basis sets?
An old mail archive says that, in the usually used basis sets, polarisation functions are treated by a single-Gaussian treatment even for valence double-zeta basis sets.
A priori, I've done a few NBO calculations on some simple magnesium compounds, say the cubic magnesium oxide tetramer anchored to the sum of ionic radii, on Gaussian.
The results seem to indicate that the 3p electrons on magnesium are treated as Rydberg, i.e. polarisation, functions on that package. This effect exists on Q-Chem as well, leading to my general conclusion that the np orbitals of lighter group 2 elements beryllium and magnesium are generally treated in the usual basis sets as polarisation functions that are not necessarily accurate and represented by a single Gaussian function, no matter the rest of the basis set's accuracy(i.e. the number of zetas).
However, there do exist circumstances where e.g. a 2p orbital of beryllium act as "legitimate" valence orbitals(such as these cases), in which case the treatment of the 2p orbitals of beryllium as Rydberg orbitals should not lead to accurate results. The last paper is behind a paywall that my institution is not affiliated with so I could not check if the authors of the article used "specialised" basis sets(no prizes for guessing out the meaning of the word "specialised" here); even if they did, these basis sets would likely be not directly accessible in the "vanilla" versions of the commonly used quantum chemistry packages and are thus hard to actually use.
My question now follows- do the basis sets included in (some or all) the "vanilla" versions of the commonly used quantum chemistry packages treat the np functions of the two lighter group 2 elements by the same way as they treat the rest of the valence(e.g. by a double-zeta for def2-SVP and by a triple-zeta for 6-311G(d))?; if not, are there any basis sets that do treat them as such available over-the-counter, should that be the right word here, in the literature?
P.S. My question also holds for the np functions of transition metals, which Gaussian treats as valence but Q-Chem treats as polarisation. To quote the quotes on Wikipedia, "Fe(−4), Ru(−4), and Os(−4) have been observed in metal-rich compounds containing octahedral complexes [MIn6−xSnx]; Pt(−3) (as a dimeric anion [Pt–Pt]6−), Cu(−2), Zn(−2), Ag(−2), Cd(−2), Au(−2), and Hg(−2) have been observed (as dimeric and monomeric anions; dimeric ions were initially reported to be [T–T]2− for Zn, Cd, Hg, but later shown to be [T–T]4− for all these elements) in La2Pt2In, La2Cu2In, Ca5Au3, Ca5Ag3, Ca5Hg3, Sr5Cd3, Ca5Zn3(structure (AE2+)5(T–T)4−T2−⋅4e−), Yb3Ag2, Ca5Au4, and Ca3Hg2; Au(–3) has been observed in ScAuSn and in other 18-electron half-Heusler compounds. See Changhoon Lee; Myung-Hwan Whangbo (2008). "Late transition metal anions acting as p-metal elements". Solid State Sciences. 10 (4): 444–449. Bibcode:2008SSSci..10..444K. doi:10.1016/j.solidstatesciences.2007.12.001. and Changhoon Lee; Myung-Hwan Whangbo; Jürgen Köhler (2010). "Analysis of Electronic Structures and Chemical Bonding of Metal-rich Compounds. 2. Presence of Dimer (T–T)4– and Isolated T2– Anions in the Polar Intermetallic Cr5B3-Type Compounds AE5T3 (AE = Ca, Sr; T = Au, Ag, Hg, Cd, Zn)". Zeitschrift für Anorganische und Allgemeine Chemie. 636 (1): 36–40. doi:10.1002/zaac.200900421.", surely some of these entities, such as Ag(-II) and Au(-II), indicate meaningful participation of the np orbitals of the transition metals silver and gold as "genuine" valence orbitals.

Some formatting may help the post readability.

It is certainly commonplace for orbitals outside the usual valence set to be included in molecular orbital calculations these days. One of the most famous examples actually involves calcium as the alkaline earth metal, with $3d$ as an "extra" subshell [1]. In the "inverse sandwich" complex $\ce{[(thf)3Ca\{μ-C6H3-1,3,5-Ph3\}Ca(thf)3]}$, calcium $3d$ orbitals with the correct symmetry are held to overlap with otherwise  antibonding orbitals of the organic ligand surrounding the dicalcium core, thus stabilizing these orbitals through forming calcium-carbon bonds. This enables electron transfer into these orbitals and the resultant emergence of calcium(I) in the core.
Reference

Sven Krieck, Helmar Görls, Lian Yu, Markus Reiher, and Matthias Westerhausen (2009). "Stable 'Inverse' Sandwich Complex with Unprecedented Organocalcium(I): Crystal Structures of [(thf)2Mg(Br)-C6H2-2,4,6-Ph3] and [(thf)3Ca{μ-C6H3-1,3,5-Ph3}Ca(thf)3]".
J. Am. Chem. Soc. 131, 8, 2977–2985. https://doi.org/10.1021/ja808524y

Is the 3d orbital of calcium represented by the same number of zetas as the 4s one in the basis set used? The whole point of my question was that.


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A new green method for purifying acetylene with the help of Zeolite molecular sieve
I'm investigating for a green method for purification of acetylene in order to reduce the environmental impacts in manufacturing plants. In the current method mercuric chloride is used in the purification material that is indeed harmful to the environment.
The main impurities consists of:
1- Phosphine           (Maximum concentration must be 200 ppm)
2- Arsine,
3- Hydrogen sulfide,  (Maximum concentration must be 100 ppm)
4- Ammonia
5- water.
Recently, I read that Zeolite molecular sieves are used to purify many gases in industry. Further information can be found in the article below:
https://www.honeywell-uop.cn/wp-content/uploads/2011/07/UOP-Adsorbents-for-purification-of-olefin-and-polymer-process-streams-brochure.pdf
If I noticed correctly, this porous material absorbed different gases with respect to its amount of pore diameter. Gases that their molecular sizes are similar or smaller than the pore diameter will be absorbed while other gases will pass from this material. I wonder that if we could use this material for purifying acetylene.
Since the size of the gas molecules in the acetylene gas mixture are different, this seems practical.
What is your suggestion?
Do you think that Zeolite molecular sieves can be a solution for purifying acetylene?
Do you know any papers or articles that is relevant to this subject?
Thank you very much.

I get some 70 hits on Web of Science for 'Acetylene AND Zeolite'...

Thanks for your comment. Yes, but I didn't found any resources about purifying acetylene using Zeolite sieve. One example of using zeolite for separating acetylene is in this article: https://epubs.stfc.ac.uk/manifestation/51157539/STFC-AAM-2021-016.pdf

With respect to "What is your suggestion?", I note that apparently hydrogen peroxide does not readily react with acetylene, see comments here, for example, to quote:

A hydrogen peroxide (H2O2) solution reacts with calcium carbide (CaC2), so as to generate the mixed gas of oxygen (O2) and acetylene (C2H2).

However, the created mix of acetylene and hydrogen peroxide appears per a recent 2018 work to present an enhanced explosion hazard (see "Water Vapor and Hydrogen Peroxide as Promoters of Acetylene Explosive Decay"), where "the acetylene explosiveness is largely determined by humidity" implying the further employment of a drying agent.
Also, on phosphine with hydrogen peroxide, to quote a source:

Reactive oxygen species for lipid peroxidation may therefore be derived from direct reaction of PH3 with H2O2 as an alternative hypothesis to their respiration-linked formation.

Also, $\ce{H2S}$ and $\ce{AsH3}$ both react with $\ce{H2O2}$ see, for example, "Catalyzed oxidation of arsenic(III) by hydrogen peroxide on the surface of ferrihydrite: an in situ ATR FTIR study".
As such, I would suggest a possible hydrogen peroxide scrubbing step, per the manner described in the literature, as a feasible green but apparently, not completely safe, route.

Thank you very much for your idea about using hydrogen peroxide for oxidizing agent as a new green method. It can be  a very valuable and green method. Thanks. I will investigate about it.
After I read your message, I was wondering that can Ozone (another oxidizing compound of oxygen) be used too? (for example ozonated water scrubber) thanks.

Just an idea similar to the method you have suggested. In the patent below you can observe the purifying method that used sulfuric acid as the absorbing material ( currently used in acetylene production). But the idea of the writer is similar to your suggestion. That is recovering sulfuric acid by using oxidizing agents such as hydrogen peroxide. Actually in this procedure, the impurities(in this article only phosphine), first absorbed by sulfuric acid and then oxidized. So, they will be prevented from returning back to gas stream. In this article solvated phosphine is converted to phosphite.

https://www.freepatentsonline.com/2313022.html

No, ozone is too reactive (see this https://www.youtube.com/watch?v=ViMocu_iPOU ). Also, ozone and water creates the hydroxyl radical which attacks acetylene (see "Ozonation of Water: Role of Hydroxyl Radicals as Oxidizing Intermediates" at
https://www.science.org/doi/10.1126/science.190.4216.782).  Also, one came not use a precursor to H2O2 like the HO2 radical (see https://digital.library.unt.edu/ark:/67531/metadc1706551/m2/1/high_res_d/manus.pdf).


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Issues with Chlorine Liquefaction
A common problem that occurs when releasing chlorine from cylinders during industrial chlorination processes is that of liquefaction of the gas and ice deposition in and around the cylinders which slow the rate of withdrawal of chlorine from the cylinders. Thermodynamically, its a legit process when chlorine will undergo expansion, with the help of Joule Thompson coefficient, it will cool down.
My issue with the reasoning is that since the expansion is occurring at the valve itself and adiabatically, there shouldn't be cooling inside the cylinder yet I do have pictures of a cylinder with ice deposited all over its surface due to expansion via cooling. How do we explain the fact that the cooling effect is travelling back through the expansion nozzle and more importantly, how do we calculate the cooling rate and total energy removed from the cylinder via expansion through nozzle?

Gases, that at room temperature condense before reaching sufficient storage pressure, are provided in the liquified form with gaseous headspace, what allows much better storage capacity.
Liquid chlorine has the vapor pressure $\pu{5830 mm Hg}$ at $\pu{25^{\circ}C}$, what is about $\pu{7.7 atm}$.
Having cylinders with just few atmospheres of gaseous chlorine would not make much sense.
Adiabatic or Joule-Thompson expansion cooling has minor impact here. Additionally, considering what the output chlorine pressure may be, there may not be much expansion at all, depending on application.
The major effect is evaporation cooling of liquid chlorine, based on its evaporation enthalpy. The similar effect is observed on the outdoor camping LPG burners. When cooking, you can feel how cold the thin wall LPG cartriges become.

The chlorine boiling point is $\pu{−34.04^{\circ}C}$.
The propane boiling point is $\pu{−42.25}$ to $\pu{−42.04^{\circ}C}$.

Therefore chlorine is even more easily liquified than propane. With an intense output flow, the boiling chlorine may reach deep frost temperature.
If the pressure regulator has equal or lower temperature than the liquid chlorine - e.g. for the minor expansion effect, then the gaseous chlorine will condense there. Similarly as water vapor condenses on glasses, if you enter a vapor chamber in a spa, or if you enter a warm room from cold outside during winter. Mild heating of regulators may help.
Dew or frost deposition on cylinders indicates another issue, as low cylinder temperature significantly decreases the gas pressure. (LPG cartridges are often kept in warm sleeping bags overnight, as butane boiling point is near $\pu{0^{\circ}C}$.) It would help if there was applicable some technical way to keep cylinders at room temperature.

So you are saying that, in the case of chlorine, under typical starting conditions, most of the chlorine will be liquid?

Poutnik is right on this one so I downvoted your answer Chet. I calculated the total cooling by plotting taking adiabatic work done for the duration and the cooling effect from adiabatic expansion is nowhere near the cooling effect I observed but I started talking to the guy who does HVAC on site and he also hinted at the liquefaction of gas and cited the LPG example. I checked the pressure of the Cl2 and it was comfortably above 10 kg. So, the gas was in liquid state inside the cylinder. I calculated the evaporative cooling and the answer was legit.

@Poutnik Now you've made me look stupid in front of the whole plant. I overdesigned my chlorine handling system badly. I was more comfortable with other compressed gases like N2 CO2 etc and they have higher pressures like 80 bar or above so I got a 35 bar Pressure relief valve. I apparently have installed a show piece in the line. I will plug it tomorrow. First the overdesign with cooling tower pumps and now this

I am sorry for that. :-)

Hypothetically, putting aside chlorine reactivity and chemical weapon quality, it could be used in fridges or AC, similarly as halocarbons or ammonia.

You have it backwards.  The major part of the cooling occurs within the cylinder itself, with very little cooling (if any) occurring in the valve due to Joule Thomson.
The gas remaining within the cylinder at any time has expanded adiabatically and reversibly to do work in forcing the gas exiting the cylinder ahead of it into the valve.   On the other hand, the gas passing through the valve is experiencing both expansion cooling and viscous frictional heating, so that the net effect is much smaller.
To calculate the cooling effect within the cylinder, one merely needs to apply the open system version of the 1st law of thermodynamics to this adiabatic expansion into the valve.  For an ideal gas, this would read $$nC_vdT=RTdn$$ where n is the number of moles of gas within the cylinder at any time.

But Cl2(l) evaporation must play role as well. It's boiling point is just about 6 deg C below b.p. of propane. The pressure cannot be very high, so gas expansion thermal effects would be much weaker than e.g. for air, N2 or O2.

Yes, that would have to be taken into account that too.  I just gave the ideal gas as a simple example.

I see... Ideal gas examples never hurts.

Correction, remembering chlorine and propane b.p. wrong. 8 deg C above propane.


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Does acrolein give positive iodoform test?

Which of the following compound(s) give(s) a positive iodoform test?
A. Acrolein
B. Acetone
C. Acetoacetic ester
D. 2-Propanol

According to me, the answer should be options B, C, and D as B is a methyl ketone, C is a β-ketoester and D can be converted to methyl ketone by tautomerization.
But the answer in the key is given as A, B, C, and D. I cannot figure out why A is there. Is the key correct or does it need correction for A?

I thought of the above mechanism for reaching the answer. Is this reasoning alright? Please approve it, whether it is right or wrong.

Your reasoning "D can be converted to methyl ketone by tautomerization" is incorrect. It is an oxidation by the testing reagent $(\ce{I2/OH-})$, which converts 2-propanol to acetone.

Thank you for correcting me @MathewMahindaratne that really made me clear about secondary alcohols giving the IODOFORM test

Thank you @adselisk for presenting my question in a better way.


text:
How does temperature affect ionic Ca in milk
I’m doing an investigation on the effects of temperature on ionic calcium content in milk through a classic EDTA titration method and have a few questions:
— what is the scientific theory behind the reason why ionic Ca decreases from increased temperature? (with respect to how it is measured through EDTA chelation)
— does this permanently affect Ca content
— how relevant is this with regard to health (e.g. does precipitated Ca have health benefits?)

Would you mind showing your results? How does the calcium concentration change and what is the exact titration procedure?

I mean what temp. are you talking about? What is the indicator etc?

Asking for answers and not providing details, purpose or background of questions are contradictory decisions.

related https://chemistry.stackexchange.com/questions/74135/does-changing-edta-concentrations-affect-the-amount-of-calcium-ions-in-milk


text:
Why does the chlor alkali process work?
Chlorine is a stronger oxidiser than oxygen, and sodium is a much stronger reducing agent than hydrogen. Therefore, in the chlor alkali process, when voltage is applied to the electrodes and the solution is one of brine, why don't we end up electrolysing the water instead? We should, in fact, start seeing chlorine gas formation only when the brine is extremely thick, almost like a semi solid.

Welcome to the chemistry site of the SE network.
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Comparison of standard redox potentials compares redox systems at standard conditions. Consider how pH value changes the oxygen redox potential and high Cl- ions activity the chlorine redox potential. For sodium, consider hydrogen overpotential on mercury, high activity of Na+ ions and low activity of Na metal solution in the amalgam.


text:
How to convert phenylacetic acid to 3-phenylpropionitrile?
I understand that phenylacetic acid reacts with $\ce{KSCN}$ can form phenylacetonitrile plus the side product ($\ce{CO2}$ and $\ce{KSH}$). This is Letts nitrile synthesis
However, the problem is to convert phenylacetic acid to 3-phenylpropionitrile.
Is that possible? What reagents do we need?

You need to add one carbon, the easiest source of this is as CN-.
Three steps.

Reduce phenylacetic acid to 2-phenylethanol (borane.THF is a good way)

Form the tosylate of the alcohol (TsCl, pyridine)

Treat the tosylate with NaCN in DMSO reference here.