text:
pH of strong acid weak acid mixture
Question: $\ce{H2S}$ ($\pu{0.1M}$; $K_\mathrm{a} = 1.2×10^{-20} $) and $\ce{HCl}$ ($\pu{0.3M}$) with same volume are mixed together. What is the resultant $\mathrm{pH}$?
My approach:
Let the volume of each be $x \; \pu{L}$
$$
\begin{array}{lcc}
&\ce{H2S & -> & 2H+ & + & S^2-} \\
\text{initially} & 0.1 && 0 && 0 \\
\text{after} & 0.1(1-\alpha) && 2 \times 0.1\alpha && 0.1\alpha
\end{array}
$$
$$
\begin{array}{lcc}
&\ce{HCl & -> & H+ & + & Cl-} \\
\text{initially} & 0.3 && 0 && 0 \\
\text{after} & 0 && 0.3 && 0.3
\end{array}
$$
$$K_\mathrm{a} =\frac{\ce{[H]}^2\ce{[S^2-]}}{\ce{[H2S]}}$$
$\ce{[H]} = $ total moles of $\ce{H}$/Total volume $=\frac{0.2\times x\times\alpha+0.3x}{2x}=\frac{0.2\times\alpha+0.3}{2}$
$0.2×\alpha$ can be neglected and $(1-\alpha)$ can be taken as $1$.
$$K_\mathrm{a} = \left[\frac{0.3}{2}\right]^2[\alpha]$$
Calculating for $\alpha$, we get it as $5.33\times 10^{-19} $
And we get $\ce{[H]}$ as $0.15$.
$$\therefore \mathrm{pH} =0.8$$
Is my approach wrong???
Another approach:
I have been told by my teacher  that resultant $\mathrm{pH}$ will nearly be the $\mathrm{pH}$ of strong Acid but if we add Weak acid will it not dilute the strong acid. Here in this case make the volume twice.

There is nothing "another" about it. You've just described the same approach twice.

You're getting HCl with two times lower conc. - H2S is irrelevant. BTW first Ka of H2S is much higher then what you wrote but still not enough to make a difference here.

But adding  H2S will not change the concentration of the solution??

You aren't adding just H2S. You're adding H2S and a good deal of water. Of course water will dilute the solution.

So it should change the concentration of the solution. So will the pH will not be affected as the solution becomes dilute on adding h2s solution. So do we have to consider this dilution while calculationg the pH??? If yes then if we are calculating pH of hcl only and neglecting of h2s as it will be very small then also the concentration of solution will change altering the pH

Yes, that's exactly what I told you!

Question: $\ce{H2S}$ ($\pu{0.1M}$; $K_\mathrm{a} = 1.2×10^{-20} $) and $\ce{HCl}$ ($\pu{0.3M}$) with same volume are mixed together. What is the resultant $\mathrm{pH}$?

Since, $\ce{HCl}$ is strong acid, we can assume all $\ce{HCl}$ molecules are dissociated in aqueous solution to begin with. Let the volume of each solution be $\pu{1.0 L}$ and $\ce{H2S}$ is dissociated $\alpha$ amount in aqueous solution at the equilibrium after combined.
$$
\begin{array}{lcc}
&\ce{HCl + H2O & -> & H3O+ & + & Cl-} \\
\text{initial amounts} & - && \pu{0.3mol} && \pu{0.3mol}  \\
&\ce{H2S + H2O & -> & 2H3O+ & + & S^2-} \\
\text{initial amount} & \pu{0.1mol} \;(\ce{H2S}) && \pu{0.3mol}  && 0 \\
\text{amount at equilibrium} & \pu{(0.1-\alpha) mol} && \pu{(0.3+ 2\alpha) mol} && \pu{\alpha \; mol}\\
\text{conc. at equilibrium} & \frac{\pu{(0.1-\alpha) mol}}{\pu{2.0 L}} && \frac{\pu{(0.3+ 2\alpha) mol}}{\pu{2.0 L}} && \frac{\pu{\alpha \; mol}}{\pu{2.0 L}}
\end{array}
$$
Now, we can define $K_\mathrm{a}$ in terms of $\alpha$:
$$K_\mathrm{a} =\frac{\ce{[H]}^2\ce{[S^2-]}}{\ce{[H2S]}} = \frac{\left(\frac{\pu{(0.3+ 2\alpha) mol}}{\pu{2.0 L}}\right)^2\left(\frac{\pu{\alpha \; mol}}{\pu{2.0 L}} \right)}{\left(\frac{\pu{(0.1-\alpha) mol}}{\pu{2.0 L}}\right)} = \pu{\frac{\alpha ( 0.3+ 2\alpha)^2}{4.0(0.1-\alpha)} mol^2L^{-2}}$$
Since $K_\mathrm{a} = 1.2×10^{-20} $, we can assume $\alpha \lt\lt\lt 0.1 $, and hence,
$$K_\mathrm{a} \approx \pu{\frac{\alpha}{0.1}\left(\frac{0.3}{2.0}\right)^2 mol^2L^{-2}}$$
When resolve for $\alpha$, we got: $\alpha = \pu{5.33\times 10^{-20} mol\:L^{-1}}$, and hence, at equilibrium, $\ce{[H]} \approx \pu{\frac{0.3}{2} mol\:L^{-1}} \approx \pu{0.15 mol\:L^{-1}}$.
$$\therefore \mathrm{pH} =0.82$$
Although you got the anticipated answer, your setup is not correct, e.g., see after concentrations (or amounts) of $\ce{H2S}$ dissociation amounts at equilibrium.

:$K_{a1}=10^{-7},K_{a2}=10^{-19}$,How$ K_a=1.2\times{10^{-20}}$ ?

Paraphrasing from the excellent comments (even if the commenters might have been exasperated):
The pH of a mixture of $\ce{H2S}$ solution with $\ce{HCl}$ solution is about the same as the corresponding mixture of water with $\ce{HCl}$ solution. The reason is that $\ce{H2S}$ does not significantly dissociate in water, and even less so in an acidic solution (difference between pKa and pH on the order of 19). So adding $\ce{H2S}$ solution lowers the concentration of $\ce{H+}$, just like adding pure water.
In your answer, you are leaving out the units and don't say what quantities you are showing (amounts or concentrations). If it is concentrations, the concentrations of the solutions before and after mixing are different, and you are not accounting for it. For example, the total concentration (dissociated or not) of HCl is 0.3 mol/L before mixing, and 0.15 mol/L after.


text:
How do cells form a solid structure?
How do cells unite to form a solid structure. For example a plant body is made up of a number of cells. The cells are not solid structures. Their consistency is in between solid and plasma. So how come they form solids. In solids i know that the constituent particles which may be ions or atoms or molecules are present very close to each other forming an amorphous or crystalline solid structure. But in plant body there are cells. So how are the cells are arranged to form a solid structure a solid body. why not a liquid or a gaseous. I have looked up in various sources but i have not found any mention of this topic. I am reallt desperate to know more on this. So any help will be appreciated. This is not a homework question and if anybody requires me to edit the question then please say so. Thanks

Cells have containers around them, like a plastic bag that can hold liquid. Look up cell wall and cell membrane.

Arguably, cells don't form solid structures. Amoeba is pretty soft, as is human without bones.

If both OP and users answering stick to their definition of solid and or crystalline a clear answer is difficult.  A balloon filled of air can be seen as solid as well, and is even not rigid. Basically OP is asking what keep in place and shape cells as in plants skin organs etc. It might even more appropriate for Biology SE. Here we will likely refer to generic dispersive force, fiber like structure, hydrogen bonding. OP should consider the term composite materials.  It might help.  As well as the fact that a solid can be compressible, elongable, etc

So how are the cells are arranged to form a solid structure a solid body?

Plant cells have cell walls, are filled with liquid, and have strong cell-cell adhesion. This gives a fairly rigid structure that may seem like a solid. Take a carrot for example, it seems solid. But if you put the carrot in a blender, you get carrot juice - not a solid. Also, when the carrot dries out, it loses its rigidity.
Trees
Woody plants contain a lot of cellulose, a water-insoluble polymer, outside of cells. This gives them even more of a solid (and dry) appearance, but the insides of their cells are aqueous as well.
Cells other than plant cells
There are cells that lack a cell wall, they are just surrounded by a lipid bilayer. Tissues made from these cells (muscle, brain, liver, etc.) don't seem solid at all. Animals do have connective tissue (mostly protein-based) and a skeleton that holds everything together unless there is trauma (predators, accidents, etc).

How do cells form a solid structure?

Let me un-ask the question a bit. Sometimes they do form a solid-like structure, and sometimes they don't. Fibrous biological macromolecules play a large role in making tissues rigid, either directly or in combination with having cells filled with aqueous solution.

Thanks. I understand what is being said. What about skin? skin is a very solid structure as we can see. but it is made up of epithelial cells without any cell walls. Then how come the skin is a solid?

Let us say I keep a million plantcells together in a definite fashion very closely. Then will i get a solid? moreover i know in solids either crystalline or amorphous the particles are arranged in a crystal lattice or they are arranged closely in an indefinite pattern. in a plant body how does this cellular structure correspond to the general idea of solids where particles are arranged ?

@Alma It's not a solid, it just has some solid-like features. In a solid, none of the particles change neighbors by moving around. That would not support life as we know it.

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text:
Why do we use coefficients for RFM (relative formula mass) in atom economy?
Usually when calculating using moles = mass / RFM, the coefficient is not multiplied by a compound's RFM to calculate RFM.
But in calculations for atom economy, the (balanced) coefficient of a compound is used in calculating RFM. Why is this the case?

Before we come to your actual question, we should clarify a few other problems related to quantities and units in your expression:

moles = mass / RFM

It is essential to distinguish between quantities and units. For example, density is defined as “mass per volume” and not “mass per litre”. In your expression “moles” is a unit name; but it should actually be the quantity “amount of substance”:

amount of substance =  mass / RFM

Names of quantities (such as “amount of substance” and “mass”) shall not be arranged in the form of an equation. You should use the corresponding quantity symbols:

n = m / RFM

It is not permissible to use abbreviations (such as “RFM”) for unit symbols (“Mr”) or unit names (“relative molecular mass”).

n = m / Mr

There shall be a space on both sides of most operators but not for the solidus “/”.

n = m/Mr

And finally, your equation is wrong, since amount of substance is not given as mass per relative molecular mass. Remember that relative molecular mass is a dimensionless quantity. Actually, amount of substance is given as mass per molar mass:

n = m/M

Now coming to your actual question.
Atom economy is defined as the quotient of relative molecular mass of the desired product by relative molecular mass of all reactants.
Thus, for a simple reaction
$$\ce{A + B -> Y + W}$$
where $\ce{Y}$ is the desired product and $\ce{W}$ is a by-product that becomes waste, the atom economy would be
$$x=\frac{M_\mathrm r(\ce{Y})}{M_\mathrm r(\ce{A})+M_\mathrm r(\ce{B})}$$
However, for a slightly different reaction
$$\ce{C + D -> 2 Y + V}$$
you would get two molecules of the desired product $\ce{Y}$ for each molecule of the reactant $\ce{C}$ and for each molecule of the reactant $\ce{D}$. The yield of product per used amount of reactants is twice as high; therefore, the atom economy is twice as high:
$$x=\frac{2M_\mathrm r(\ce{Y})}{M_\mathrm r(\ce{C})+M_\mathrm r(\ce{D})}$$
In the same way, any other coefficients in the chemical equation would have to be taken into account.

"you would get two molecules of the desired product Y for each molecule of the reactant C" i thought the coefficient represented the number of moles?

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text:
What happens if at equilibrium pressure is increased by decreasing the volume?
I think, according to Le Châtelier's principle both the conditions i.e. increasing the pressure and decreasing the volume must shift reaction towards the side on which there are less amounts of gaseous molecules. And if there are the same stoichiometric coefficients for the gaseous substances (i.e. $∆n_g = 0$) then it must not have any effect on the reaction.
But increasing the pressure and decreasing the volume increases the number of collisions between the molecules of reacting substances and hence the concentration of products must increase.
Both these thoughts confuse me.
Questions I encountered:
$$ \ce{H2O(g) + CO(g) <=> H2(g) + CO2(g) + Heat }$$
In this question $∆n_g = 0$ so there must be no effect of increasing pressure and decreasing the volume. But the answer mentions that the concentration of both $\ce{CO}$ and $\ce{CO2}$ will increase.
There was also another question of same type:
$$ \ce{SO2Cl2(g) <=> SO2(g) + Cl2(g) }$$
Here the reaction must shift backward as pressure is increased and volume is decreased.
But again the answer mentions that concentration of all the gasses increases.

When a reaction is at equilibrium, the forward and reverse rates are equal. Decreasing the volume increases the concentration of all species (both reactants and products). This will result in a higher forward rate (because the concentration of reactants increased) and in a higher reverse rate (because the concentration of products increased). If forward and reverse rate increase by the same factor, the reaction remains at equilibrium.


text:
Gibbs energy at a certain temperature

For the reaction
$$\ce{1/2 N2(g) + 3/2 H2(g) -> NH3(g)}$$
$ΔG$ at $\pu{298.15 K}$ is $\pu{-3980 cal},$ and the change in enthalpy is given by the expression:
$$ΔH = -9.19 - 7.12T + 3.182\cdot 10^{-3}T^2 - 2.64\cdot 10^{-7}T^3$$
Calculate $ΔG$ at $\pu{1000 K}$ from these data. Note: use the expression
$$\left[\frac{∂}{∂T} \frac{ΔG}{T}\right] = \frac{-ΔH}{T^2}$$
integrate the resulting expression of the $ΔH$ given.

This is what I understood so far. Using the given expression of the partial derivative I integrated on both sides:
$$∫\left[\frac{∂}{∂T} \frac{ΔG}{T}\right] = - ∫\frac{ΔH}{T^2}$$
I put the given expression of $ΔH$ on the right side of the integral.
$$∫\left[\frac{∂}{∂T} \frac{ΔG}{T}\right] = - ∫\frac{-9.19 - 7.12T + 3.182\cdot 10^{-3}T^2 - 2.64\cdot 10^{-7}T^3}{T^2}$$
Eliminate the $T$ where I could:
$$∫\left[\frac{∂}{∂T} \frac{ΔG}{T}\right] = 9.19 ∫\frac{1}{T^2}\mathrm dT + 7.12 ∫\frac{1}{T}\mathrm dT - 3.182\cdot 10^{-3}∫\mathrm dT + 2.64\cdot 10^{-7} T^3∫T\mathrm dT$$
$$\frac{ΔG}{T} + C = -\frac{9.19}{T} + 7.12\ln T - 3.182\cdot 10^{-3}T + 1.32\cdot 10^{-7}T^2$$
To get $ΔG$ I evaluated the final expression from $298.15$ to $\pu{1000 K}$ and got that
$$\frac{ΔG}{T} + C = 126.6707439$$

Since the expression from the integral is all temperatures, why I don't get units of energy instead? Am I missing something during the integration?

If I am not missing anything, the $126.6707439$, whatever unit it has, is that the $ΔG$ at $\pu{1000 K}$ I am supposed to get?

You integrated indefinitely on L.H.S. but took limits on L.H.S.. Either integrate indefinitely on both sides and put values at 298.15K or take limits on both sides. Everything is correct till the second last equation

It seems to me the initial equation implicitly supposes dS/dT=0. But, it does not seem right to me, as integral dQ/T up the temperature for reagents and the same downwards for products do not seem to me as zero, if sumed up.

@Poutnik, this is the Gibbs-Helmholtz equation and assumes $p$ is constant. Entropy is given by $\displaystyle \left( \frac{\partial G}{\partial T}\right)_p=-S$ and starting with $dG=VdP-SdT$ and $G=H-TS$ and after a few steps $\displaystyle \left (\frac{\partial G}{\partial T}\right)_p=\frac{G-H}{T}$ results which is the same as the equation in the question

Integrating the left had side of the equation wrt $T$ gives $\displaystyle \frac{\Delta G_{T_2}}{ T_2}- \frac{\Delta G_{T_1}}{ T_1}$


text:
What Value of Purity Do We Substitute in the Formula C=10PD/M
In the the formula
$$C = 10P\frac{D}{M}$$
$C$ is the concentration of a substance in a solution, $P$ is the percent concentration (purity) of the substance, $D$ is the relative density of the solution and $M$ is the molar mass of the substance.
Say $P = 60\%$, do we substitute $60$ or $60/100$ in the formula? The proofs I've read suggest $60$.

It all depends on the units other variables are expressed in.
By dimensional analysis it's obvious that you don't expect percent to appear in the final result, so $P$ is expressed as a fraction.
Using percentage number can be justified if $100\%$ is conveniently cancelled out.
The $10$-multiplier suggests that this might be case here, e.g. if $D$ is expressed in $\pu{g cm-3}$, $M$ — in $\pu{g mol-1}$, then in order to obtain $C$ in $\pu{mol dm-3}$ (or $\pu{mol L-1}$, which is the same), you must plug in the value for $P$ in percents:
$$[C] = \frac{\pu{g cm-3}}{\pu{g mol-1}} = 10^3\cdot\pu{g dm-3} = 10\cdot [P~\text{in}~\%]\cdot\pu{g dm-3}$$


text:
Trend of thermal stability of Group 2 chlorides down the group
Thermal stability of group 2 chlorides follows the order: $\ce{BeCl2>MgCl2>CaCl2}$.
Talking about carbonates of group 2 metals, the thermal stability increases as we move down the group due to decreasing polarizing power of the cations. Why is the similar trend not observed in halides?
I found an answer here which had an elaborate explanation of the fazan's rule and effect of polarizability of anions. However, that comparison was made for a single cation bonded with different halogens bonded one on one as one moves down the group. Surely, the stability in that case increases down the group as the charge density on anion decreases.
The comparison made here is different. I want to know the trend when the cation is being changed and anion is same. According to Fazan's rule, the trend should be same as that for carbonates: their stability increases on moving down because charge density of cation decreases. But this is not observed in the case of chlorides. Why is that so?

@Willim R. Ebenezer, I went through the explanation you have shared the link of, but I still can not find an answer to my question. In that answer, the trend in stability has been examined keeping the cations same and anions changing as one moves down the group. However, I want to know the trend of keeping the anions same and cations changing. I will edit my question to get more exact.

I believe that has been discussed too. I quote thence: When the ions electron cloud, is less polarized, the bond is less strong, leading to a less stable molecule.. Does that not suffice?

Ohkay. I can clearly apply this concept to my problem and I now understand why does that happen. Your quote just clicked it. Thankyou:)


text:
A broad singlet at 2.27 ppm?
I took the $\mathrm{^{1}H}$-NMR (CDCl3 solvent) of my product, 2,2-dimethoxy acetic acid. Below is what I see:

$\delta \: \pu{4.862 ppm}$ (1H), $\delta \: \pu{3.521 ppm}$ (6.78H), $\delta \: \pu{2.271 ppm}$ (5.39H)

Peaks at $\delta \: \pu{4.862 ppm}$ and $\delta \: \pu{3.521 ppm}$ indicate my product I believe, but I cannot rationalise a peak at $\delta \: \pu{2.271 ppm}$.
I did dry my organic phase over $\ce{Na2SO4}$ so this cannot be water. I have never really seen a broad singlet in 2-3 ppm region. Does anyone have any idea what this would be? Would appreciate any thoughts&ideas.

Which solvents/reactants did you use in your synthesis?
You can find the shift of some impurities in various deuterated solvents here.
https://pubs.acs.org/doi/pdf/10.1021/acs.oprd.5b00417

Definitely can be water, and almost certainly is. How broad? Do you see 13C satellites? Add a drop of H2O and see if this peak increases and moves. Or change the temperature or field.

@The_Vinz I used diethyl ether

@long very broad. like I need to increase the resolution a LOT to see it. I will try the method

Water can be present in the nmr solvent, not necessarily your sample itself


text:
Solvent for tire rubber
I want to apply tire rubber on surfaces, for various reasons - including protection against water. I think of tire rubber, as it is so widely available - helping in reducing pollution (infinitesimally) at the same time.
I am interested in a solvent which will temporarily transform the tire rubber into a (viscous) liquid - in such way that after some time, when the solvent evaporates, the tire rubber becomes solid again.
I do not need any special properties of the new rubber, there will be nothing really safety critical. If it is similar enough to the original, it is good.
Ideally, the solvent should be easy to find and I should be able to buy it without any special permits.

Think of this:

there are some granules of salt;
I add some water - and get salty water;
after some time the water evaporates and I get 1 block of salt.

Now, the salt is actually pieces of tire rubber, and the water is... what? :)

I searched on the net, and the only solvent I recognized was acetone. Does anyone have any experience with this?

Carbon disulphide is used to dissolve rubber but it will not be easy to buy nor will it be easy to work with unless you have a laboratory.  I appreciate that this doesn't answer your question but it will give you an idea of the difficulties you will encounter.

:) Thank you for the heads up. I had a good feeling about it - being a difficult task. Of course, if you will find a way to make the difficulties less impacting, I will be happy.

I think chloroalkanes would do the job though I don't have supporting info. Dichloromethane (aka methylene chloride) is the lightest and evaporates quickly. Chloroform and carbon tetrachloride are more toxic.

The environmentally friendly way to do this is shred or grind the tires to suitable sized particles and then mix them with a bonding material. No hazardous chemicals required, but you can't do it easily yourself. It is already done on a commercial scale (the product is called "crumb rubber") so disposing of your tire at a recycling site is just as good for the environment as doing something yourself.

Tire rubber is a crosslinked polymer. Doesn't really melt or dissolve as far as I know (depends on the degree of crosslinking). Crosslinked polymers can swell, but you might have trouble finding a solvent that fits your water/salt analogy. https://pslc.ws/macrog/xlink.htm

As it was mentioned in the comments, finished rubber is a cross-linked polymer, and this is the main reason why getting vulcanized rubber liquefied without destroying it seems to be a nearly impossible task.
Also, there might be other additives, e.g. tires are most likely going to contain powdered charcoal, so that even though one can dissolve the rubber by destroying the cross-linking $\ce{(-S-S-)_n}$ bonds, charcoal is likely going to form another phase, and who knows what might happen to other additives during this process.
Finally, at the end you have to somehow recreate cross-linking for the rubber to retain it's elasticity, and this doesn't happen on its own by just evaporating the solvent (see vulcanisation).
In general, the majority or rubber types can be affected by:

strong oxidizing agents (to the extreme of being irreversibly destroyed);
aromatic and chlorinated hydrocarbons (both have a tendency to soften rubber).
Visually it appears as if rubber is swelling; this can happen if one keeps rubber in chloroform or xylene for an extended period of time – for instance, this property is sometimes used to fit rubber tubing in place.

What I suggest is to use a so-called rubber mix or rubber compound: a semi-finished mix of [natural] rubbers, vulcanizing agents, plasticizers, fillers and additives (carbon, chalk, kaolinite etc.).
Rubber mix is used ubiquitously in the industry to produce all sorts of rubber products, including composite materials such as rubberized fabric.
Rubber mix is a versatile material which is highly viscous at the room temperature, but softens upon heating to approx. 100 °C and can be dissolved in volatile organic solvents (even benzene and gasoline).
The drawback is that at the end you still have to use vulcanization to crosslink the rubber.

I know this is old but I stumbled across it because I was intrigued by a question on a game show. I found a patent from the 30s that has some interesting information related to your question. In fact, it almost seems like the guy was doing exactly what you are trying to do. As others have pointed out, it’s definitely not a process I would want to try at home, but it is definitely an interesting read!
https://patents.google.com/patent/US2019207A/en


text:
Rearrangment during alkyne synthesis

What I have got as (G) is this.
$\hskip1in$
My questions

How the rearrangement from (G) to $non-1-yne$ can happen i.e. its mechanism?
Why Sodium Amide in particular yield terminal alkyne, unlike the fused Potassium hydroxides why almost always yield internal alkyne?
Is there any limitation by how much this alkyne formation site can shift?

Please note that this is not my homework, I have done this question and I am interested in the mechanism.

An internal alkyne is formed followed by migration of the triple bond to the terminus of the chain. See Alkyne Zipper at ursula.chem.yale.edu/~chem220/chem220js/STUDYAIDS/Varrentrapp/varrentrapp.html

Your questions:
Q1) How the rearrangement from (G) to non-1-yne can happen (i.e., its mechanism)?
For example, see the following mechanism suggested in Ref.1 as the Acetylene Zipper reaction:

Q2) Why Sodium Amide in particular yields terminal alkyne, unlike the fused Potassium hydroxides why almost always yield internal alkyne?
To answer your question, I have to dig down literature to the 19th Century. To my knowledge, sodium amide ($\ce{NaNH2}$) is not the best base to use in this purposes. The best base, in particular, is the monopotassium salt of 1,3-diamminopropane or simply, KAPA (see mechanism), a difunctional “superbase.” According to the Ref.1:

KAPA produces exceptionally rapid migrations of triple bonds from the interior to the terminus of the carbon chain in seconds at $0^{\circ}$.

On the other hands, from a synthetic standpoint, $\ce{KOH}$ as a dehydrohalogenating agent, tends to promote the shift of the triple bond away from the end of a chain, which is sometimes considered as a disadvantage (Ref.2).
Q3) Is there any limitation by how much this alkyne formation site can shift?
There are no limitations. Yet, if one substitution of disubstituted acetylene is branched, the triple bond migrate through the straight chain to the end (Ref.2). Mean time, the reaction has some interesting facts:

An equilibrium between alkynes and allenes can be established using basic catalysts (see the mechanism). For example, when very strong bases such as sodium metal ($\ce{Na}$) or sodium amide ($\ce{NaNH2}$) are used, the rearrangement of a triple bond of a disubstituted acetylene tends to migrate to a terminal position. This isomerization equilibrium shifts toward forward direction in favor of 1-alkyne formation due to the fact that formation of the terminal metal acetylide salt, which is stable, precipitated out from the solution, thus automatically removing itself from the reaction mixture.
Favorskiî was the first to discover that 2-alkyne could be irreversibly removed from solution by isomerization to the 1-alkyne and resultant precipitation of the sodium salt in 1887 (Ref.3). This procedure used metal $\ce{Na}$ as the strong base, and thus, has the disadvantage that some olefin would produce  as a result from hydrogenation of the alkyne by the hydrogen liberated upon acetylide formation. On the other hand, $\ce{NaNH2}$ when used in inert solvents, was shown to accomplished the same rearrangement but avoids the reduction problem (Ref.3,4). Temperatures greater than $\pu{100 ^{\circ}C}$ are employed with $\ce{Na}$ metal while the best results with a $\ce{NaNH2}$ catalyst are obtained at $\pu{150-160 ^{\circ}C}$. The data are summarized in the Table (vide infra). The rates of the rearrangement differ with the structures of the compounds isomerized. In general, 2-alkynes rearrange faster than 3-alkynes, and 1,2-alkadienes (allenes) rearrange faster than 2-alkynes.  The amount of 1-alkyne formed is proportional to the reaction time, and equilibrium is obtained earlier at higher reaction temperatures (Ref.3).
Later studies have found that $\ce{NaNH2}$ brings about the propargylic rearrangement in liquid ammonia at room temperature with essentially quantitative hydrocarbon recovery. (Ref.3). The ratio of 1-alkyne, 1,2-alkadiene, and 2-alkyne varied with the amount of $\ce{NaNH2}$ used. High $\ce{NaNH2}$ concentrations favored the formation of 1-alkynes. Equivalent or excess amount of $\ce{KNH2}$ in liquid ammonia, sometimes in admixture with HMPA, have also been shown to convert 2-alkynes to 1-alkynes (Ref.3,5).
The advent of newer strong bases has made the isomerization to terminal acetylenes a very facile reaction. The addition of dialkylacetylenes to a slightly excess of KAPA (the monopotassium salt of 1,3-diamminopropane) results in extremely rapid, quantitative isomerization of the triple bond to the terminal position where it is trapped as the acetylide ion (Ref.1).

Table:
$$
\begin{array}{ccc}
\hline
\text{Disubstituted alkyne} & \text{Base} & \text{Temperature} & \text{1-Alkyne} & \text{Disubstituted alkyne} & \text{Ref} \\ \hline
\ce{C2H4C#CCH3} & \ce{Na} & \pu{100 ^{\circ}C} &  \pu{4\! -5 h} &  n\text{-}\ce{C3H7C#CH} & 3\\
n\text{-}\ce{C3H7C#CCH3} & \ce{Na} & \pu{100 ^{\circ}C} &- &  n\text{-}\ce{C4H9C#CH} & 3\\
n\text{-}\ce{C13H27C#CCH3} & \ce{Na} & \pu{200\! -220 ^{\circ}C} &  \pu{15\! -20 h} &  n\text{-}\ce{C14H29C#CH} & 3\\
n\text{-}\ce{C4H9C#CCH3} & \ce{NaNH2} & \pu{160 ^{\circ}C} &  \pu{12 h} &  n\text{-}\ce{C5H11C#CH} & 3,4\\
n\text{-}\ce{C5H11C#CC5H11} & \ce{NaNH2} & \pu{210 ^{\circ}C} &  \pu{7 h} &  n\text{-}\ce{C10H21C#CH} & 3,5\\
n\text{-}\ce{C4H9C#CC2H5} & \ce{NaNH2} & \pu{170 ^{\circ}C} &  \pu{9 h} &  n\text{-}\ce{C6H13C#CH} & 2\\
n\text{-}\ce{C6H13C#CC6H13} & \ce{K(NH2C3H6NH)} & \pu{20 ^{\circ}C} &  \pu{1 min} &  n\text{-}\ce{C12H25C#CH} & 1\\ \hline
\end{array}
$$
Entry 4: The catalyst used was finely divided $\ce{NaNH2}$ suspended in mineral oil.
References:

C. A. Brown, A. Yamashita, “Saline hydrides and superbases in organic reactions. IX. Acetylene zipper. Exceptionally facile contrathermodynamic multipositional isomeriazation of alkynes with potassium 3-aminopropylamide,” J. Am. Chem. Soc. 1975, 97(4), 891-892 (DOI: 10.1021/ja00837a034).
T. L. Jacobs, “The Synthesis of Acetylenes,” Organic Reactions 1949, 5, Chapter 1 (78 pages) (https://doi.org/10.1002/0471264180.or005.01).
Hermon Pines, Wayne M. Stalick, In Base-Catalyzed Reactions of Hydrocarbons and Related Compounds; Academic Press, Inc.: New York, NY, 1977, “Chapter 3: Isomerization of acetylenes and allenes,” pp. 124–204.
H. H. Guest, “Rearrangements of the Triple Bond,” J. Am. Chem. Soc. 1928, 50(6), 1744–1746 (DOI: 10.1021/ja01393a036).
L. Brandsma, In Preparative Acetylenic Chemistry (Studies in Organic Chemistry Series: Vol. 34); 2nd Ed.; Elsevier Science: New York, NY, 1988, “Chapter XI. Base-Promoted Interconversions of Acetylenes,” pp. 231–246.
T. H. Vaughn, “A New Reaction of 1-Iodoacetylenes and Some New Mercury Acetylides,” J. Am. Chem. Soc. 1933, 55(8), 3453–3458 (DOI: 10.1021/ja01335a073).

The first "zipper" reaction was observed by Varrentrapp in 1840. Fused KOH converted oleic acid to palmitic and actec acid. http://ursula.chem.yale.edu/~chem220/chem220js/STUDYAIDS/Varrentrapp/varrentrapp.html

@user55119 : You are correct, but that was for alkene "Zipper." Favorskiî 's finding were for alkyne "Zipper" (both ways: terminal towards substituted alkynes with $\ce{KOH}$ and substituted towards terminal alkynes with $\ce{Na}$ metal as catalysts).

All I said was that Varrentrapp observed the first zipper reaction. I did not imply that he was involved in the later alkyne zipper. Susan Abrams work on perdeuterating cycloalkynes is a classic. Of interest is that enantiomerically pure, due to branching, alkynes do not racemize during the zipper. This implies that the allene does not form at the chiral center. Nice  review.

@user55119: No harm done! Thanks for educating me. I learned myself during time spend on this. :-)

We can envisage the first steps of the reaction from the dibromide form the internal alkyne, then this undergoes the zipper reaction to isomerize to the terminal alkyne due to $\mathrm{p}K_\mathrm{a}$ differences. See e.g. Alkyne zipper reaction. The fused KOH is not sufficently strong a base to effectively catalyze this isomerization.


text:
Why are triple covalent carbon to carbon bonds drawn linearly in skeletal structure?
In the link see that cyclic C-C structure with a triple covalent bond is drawn with an angle, probably the only way to draw it but why is it drawn linearly in non-cyclic and not like single or double bond? I am wondering about the reason for this not that its a convention.

Alkyne Wikipedia

A triple bond is drawn this way because it is this way. There is nothing conventional about it.

I'm afraid I don't quite follow. Are you asking why $\ce{-C#C-}$ group is linear, or why no stereospecific descriptors such as Cram bonds were used to depict structure f (cycloheptyne)?

I am asking why the 180degree angle that is drawn in carbon to carbon skeletal structure and not the 120 or what is it in single and double bonds. I saw in many yt videos and other websites that this is the way to draw triple bonds but why is it? Is it the actual geometry of the compunds? If so why is that? Do you know of any source that could explain to me the why? In the cyclic compund the structure has an angle. I dont understand why in non-cyclic it is a 180 degree

@ŽanSpehonja The carbon atoms in $\ce{-C#C-}$ are linked by one σ- and two π-bonds, the maximum electron density of which is located in two mutually perpendicular planes. Two intersecting planes form a straight line.

The structure (f) is therefore not very accurately drawn  because the molecule will be highly strained, the $\ce{-C\equiv C-}$ bond wanting to be straight and the others at 109 degrees.

Consider the answer introducing strain

Why mark my question as duplicate to "How should a triple bond be depicted in skeletal structures?" When I clearly wanted to know WHY so the reason for it and clearly alredy knew HOW it is drawn. Before I posted this question I alredy saw the supposed original question. This is a completely different question. The user on the other question was asking how a C to C triple bond is drawn since he probably drew it with a 120 degree angle and it got market as incorrect by his professor, I could have answered his question before I posted this one

The −C≡C− is normally drawn as linear about each C atom as that is what is found in nature for most alkyne x-ray crystal structures. We can use hybridization theory to explain this as the C atoms would be expected to adopt an sp hybridization, which gives two hybrid orbitals along the axis of the p orbital hybridized with the s to make the sp hybrids:

https://courses.lumenlearning.com/introchem/chapter/sp-hybridization/
These orbitals will then overlap with orbitals on adjacent atoms to make two sigma bonds in a linear arrangement about each C atom, giving alkynes their overall linear shape:

https://chem-is-you.blogspot.com/2012/09/organic-chemistry-for-level-structure.html
Cycloalkynes tend to be strained molecules as the ideal bond angles of $180^o$ about the alkyne C atoms are not consistent with a small cyclic molecule. This causes cycloalkynes with less than 10 C atoms to have measurable strain and the strain increases as the ring gets smaller:

https://en.wikipedia.org/wiki/Cycloalkyne
Chemical drawing programs will tend to keep the general shape of a cyclic structure (a heptagon in the case of molecule f) and just add the triple bond, but it does do a good job of indicating to us that the molecule will suffer from this strain as it naturally tends to deviate the angles away from $180^o$!

Upvoted because refers to strain.

Triple bond in carbon means that the carbon atom is sp hybridized. sp hybridized carbon has the two bonded groups at 180 to each other. Therefore it is displayed linearly which is closest to the actual structure. Single and double bonds are at an angle of nearly 109.5 and 120 degree therefore are not drawn linearly.


text:
Why F- is more stable than CH3-
When I was reading Organic textbook, I saw a statement of 'F- is more stable than CH3- because the F has a greater electronegativity'. I agree with the explanation 'because of its high electronegativity, F can hold the electrons very well'. But I am just confused about: as the F has great electronegativity, it can react with other cations/electrophile very easily, making it reactive thus unstable. So I am just very confused about this contradiction.

True, F has great electronegativity and hence is very reactive, but what happens when it reacts  with something? Does it still remain the same?

High electronegativity does not imply high nucleophilicity or high basicity, but it is a bit difficult to grasp these concepts.

CH3-  as typified by methyl lithium is an exteremely strong base and will tear protons off most things except other alkanes.


text:
Is it possible to raise pH in RO water without Na, Mg, K, Ca, S addition?
Maybe the question in the title is simple (=stupid) and the answer is 'simple No' but I have the urge to ask.
Some background, I have an aquarium which I use RO water and I use some portions of $\ce{KHCO3, MgSO4, CaSO4}$ to raise my GH and KH and in result, I get also an amount of $\ce{K, Mg, S, Ca}$ + (bi)carbonates in the water in a specific $\pu{ppm}$ quantity and ratio between them which I want to keep it stable.
Still, after the additions of the compounds above, I get something like 1 KH, 3 GH and pH of 6.4 but I need to raise it around 7.
Is there any other way to do it without another compound? (preferable no baking soda or soda ash)
Thanks!

Try replace part of KHCO3 by K2CO3, keeping the same total amount of K. The needed ratio is best to determine experimentally.

Strontium carbonate, maybe? Not sure about the effects of $\ce{Sr^2+}$ on aquarium inhabitants...

NH4OH, which is given off by aquarium wastes, anyway.

Adding ammonium ions to an aquarium water is not good idea.


text:
Aldol Reaction in Water Not Working
I am conducting an Aldol formation reaction in water, and below is the experimental protocol I am following (from J. Org. Chem. 1994, 59, 3590—3596):

Formation of Aldol in Water using Yb(OTf)3 Catalyst
To commercial formaldehyde solution ($\pu{1 mL}$) and THF ($\pu{3 mL}$) were successively added $\ce{Yb(OTf)3}$ ($\pu{0.04 mmol}$, $\pu{10 mol\%}$) and silyl enol ether 1-(Trimethylsiloxy)cyclohexene ($\pu{0.4 mmol}$) in THF ($\pu{1 mL}$) at rt. The mixture was stirred for $\pu{24 h}$ at this temperature and then THF was removed under reduced pressure. Water was added and the product was extracted with dichloromethane. After the usual workup, the crude product was chromatographed on silica gel to give the pure adduct.

Despite multiple attempts, I was unable to get the aldol product. The procedure itself is not even that convoluted to follow. NMR shows bare starting materials. Can anyone please help me to troubleshoot my issue?
Post edit: JOC reference as follows:
S. Kobayashi, I. Hachiya, “Lanthanide Triflates as Water-Tolerant Lewis Acids. Activation of Commercial Formaldehyde Solution and Use in the Aldol Reaction of Silyl Enol Ethers with Aldehydes in Aqueous Media,” J. Org. Chem. 1994, 59(13), 3590–3596 (DOI: 10.1021/jo00092a017).

Your JOC reference is a link to a paper on a different topic in Tetrahedron.

Also, we have no idea what the reference enolate is or what your enolate is. Depending on what the identities of these substances are, there's no reason for it to work a priori.

@Zhe: I put the info about JOC paper for your convenience.

@chemrese: You have described the procedure for hydroxymethylation of O-silyl enolates (ethers).  If you are using other aldehydes other than formaldehyde, there is an other procedure. Thus, it's better to described one of your failed attempts so that we have a clear picture.

@Zhe fixed it. The silyl enol ether I've used is: 1-(Trimethylsiloxy)cyclohexene

@chemrese I don't have journal access anymore. What substrates are given in the substrate table in the paper?

@Mathew Mahindaratne Apparently it worked under Nitrogen (done by my colleague) what's the point of running a reaction under N2 if water is used as a solvent?

@chemrese: Apparently, moisture is not the problem here. But, probably atmospheric $\ce{O2}$ is. Usually, we do reactions under $\ce{N2}$, mostly to prevent unnecessary oxidations by $\ce{O2}$, not moisture, unless your reactants are moisture sensitive such as organolithium.


text:
Consumer choice; harmful environmental aspects of polyethylene vs polypropylene
What is the difference between polyethylene and polypropylene in terms of their aspects which are harmful to the environment during manufacture and disposal.

Define "better" and "environment", e.g. criteria to compare plastics and where they are going to be used.

Recycled glass containers like soft drink and milk bottles of old.

Obviously the one which is technically better suited for the application, because you need less of it.

@andselisk I made an edit to the question, then noticed your comment. I didn't mean to circumvent your question. The word "better" can be a "downvote attractor", so I sometimes try to rescue a new user from too many of them.

Every material is terrible if not properly collected and processed after use.
Wether you re-use it (like glass bottles), or recycle the material, or burn it to make electricity and district heating, is small potatoes.
What material you use is also not very important, only the actual energy and environmental costs of "bio"materials like paper or cotton are very high. You have to re-use them very often before they make sense!


text:
Fehling's solution - What is the shelf life of the unmixed A & B components?
If you prepare two separate solutions of Fehling's A and Fehling's B, what is the shelf life likely to be? Will the effectiveness of the solution diminish over time? The literature recommends preparing these solutions separately and storing them in rubber stoppered bottles. I assume this means that atmospheric air will affect the solution. My thought is that carbon dioxide will dissolve in the solution and form an acid, which would lead to a false positive result later in the Fehling's reagent test for carbohydrates.

Copper (II) sulfate solution (Fehling's A) is completely stable.


text:
Neutralisation Reaction Confusion
What will happen when $\pu{1 mol}$ of $\ce{H2SO4}$ reacts with $\pu{1 mol}$ of $\ce{NaOH}$?
Will $\pu{1 mol}$ each of $\ce{NaHSO4}$ and water be formed?
Or will $\pu{0.5 mol}$ of $\ce{H2SO4}$ will remain yield $\pu{0.5 mol}$ of $\ce{Na2SO4}$ and $\pu{1 mol}$ of water?
I believe it will be the second way since H2SO4 is a strong acid and would again quickly react with the $\ce{NaHSO4}$ formed. But I am not quite sure how the mechanism will be and whether the Rate constant of the second reaction will be high enough to facilitate this.
What will happen and why?

@andselisk Those two ways are my thinking only. It wasn't asked in any homework question. As I am confused which will happen, I asked the same

well, compared to $\ce{H2SO4}$. $\ce{HSO4-}$ is only a weak acid so I'd assume that $\ce{NaHSO4}$ will be formed first, but i guess and say and equilibrium is probably stablished between this and $\ce{Na2SO4}$

First, let's assume these reaction happen in aqueous solution and everything is soluble. The two reaction you are interested in are:
$$\ce{H2SO4(aq) + NaOH(aq) -> HSO4-(aq) + H2O(l) + Na+(aq)}\label{rxn:1}\tag{1}$$
$$\ce{H2SO4(aq) + 2NaOH(aq) -> SO4^2-(aq) + 2H2O(l) + 2Na+(aq)}\label{rxn:2}\tag{2}$$
Your saying that reaction \eqref{rxn:2} might happen even when $\ce{NaOH}$ is limiting, and you would end up with some sulfuric acid and some sulfate. In a thought experiment, let's say that is what happened. Now, we can discuss whether sulfuric acid and sulfate, which are present in equimolar amounts, will react with each other according to:
$$\ce{H2SO4(aq) + SO4^2-(aq) <=> 2HSO4-(aq)}\label{rxn:3}\tag{3}$$
Sulfuric acid is a strong acid, and sulfate is a weak (very weak) base. This is like mixing hydronium ($\ce{H3O+}$) and hydroxide ($\ce{OH-}$) ions, which (mostly) form water. So just like water, which occurs mostly undissociated, the equilibrium of reaction \eqref{rxn:3} lies on the side of the products when the reactants are present in equimolar amounts.
Of course, if you add more $\ce{NaOH}$, you would start making a lot of sulfate.
If you wanted to know all of the products and their concentrations, you would have to consider water auto-dissociation as well. The argument I made is not meant to be quantitative, but to show why we don't expect sulfuric acid to be left over. In aqueous solution, there is never much sulfuric acid (even in the absence of $\ce{NaOH}$) because, as a strong acid, it dissociates.

Tip on formatting references to the equations/reactions: you can label them with \label{<label>} (I use eqn:<equation #> and rxn:<reaction #> as <label>), and then refer to the corresponding numbered item via \eqref{<label>} (see my edit). This way the reference is clickable and consistent (as long as labels are correct).

@andselisk Yes, I just saw that - very cool. And congrats on becoming a moderator!

Thank you very much! That was one of the "pro tips" Martin taught me a year or so ago:)

@andselisk: I agree with Karsten, it is very cool tip! So, I checked it myself, but accidentally save as an edit. I did nothing but it save you edit as as my edit anyway. Sorry. But, since you are now a moderator (Congratulations!), you may able to save it! Thanks for the great tip.

@MathewMahindaratne I'm not sure I understand the editing part (yours and mine are still two separate edits), so I'm probably going to leave it as is – the answer looks perfectly fine to me:)

@MatthewMahindaratne You and I were editing at the same time, and I could not save my edits (I saved them locally and then tried again when the lock on the file was gone, so I did not lose anything). Anyway, I have seen this once or twice where the software is unhappy because too many things are going on at the same time.

@KarstenTheis I wondered a similar thing a while ago (question on Meta), and it doesn't look like much has changed since: simultaneous edits still cause conflicts.

@andselisk That's fine, once you have encountered it once, you know what to do next time. I have to say, otherwise the user interface is pleasant and powerful.

@KarstenTheis So will reaction two happen first or reaction 1?


text:
What would superionic water ice look like?
Phys.org's Giant lasers crystallize water with shockwaves, revealing the atomic structure of superionic ice links to

Nature Physics: Experimental evidence for superionic water ice using shock compression
Nature: Nanosecond X-ray diffraction of shock-compressed superionic water ice

Question: What would superionic water ice look like if it could be maintained at ultra-high pressure long enough? Would it be fairly transparent and have a low index of refraction like glass or "normal" water ice? Or would it be opaque or shiny, or perhaps transparent but high index of refraction like diamond?
With a description like "solid lattice of oxygen and liquid-like hydrogen superionic ice" it certainly sounds "shiny/metallic" to me!

The Phys.org article says:

In 1988, scientists first predicted that water would transition to an exotic state of matter characterized by the coexistence of a solid lattice of oxygen and liquid-like hydrogen—superionic ice—when subjected to the extreme pressures and temperatures that exist in the interior of water-rich giant planets like Uranus and Neptune. These predictions remained in place until 2018, when a team led by scientists from LLNL presented the first experimental evidence for this strange state of water. (emphasis added)
[...]
The researchers performed a series of experiments at the Omega Laser Facility at the University of Rochester's Laboratory for Laser Energetics (LLE). They used six giant laser beams to generate a sequence of shockwaves of progressively increasing intensity to compress a thin layer of initially liquid water to extreme pressures (100-400 gigapascals (GPa), or 1-4 million times Earth's atmospheric pressure) and temperatures (3,000-5,000 degrees Fahrenheit).
"We designed the experiments to compress the water so that it would freeze into solid ice, but it was not certain that the ice crystals would actually form and grow in the few billionths of a second that we can hold the pressure-temperature conditions," said LLNL physicist and co-lead author Marius Millot.
[...]
"Water is known to have many different crystalline structures known as ice Ih, II, III, up to XVII," Coppari said. "So, we propose to call the new f.c.c. solid form 'ice XVIII.' Computer simulations have proposed a number of different possible crystalline structures for superionic ice. Our study provides a critical test to numerical methods."

related https://chemistry.stackexchange.com/questions/26863/what-is-the-meaning-of-superionic

It would shine like a lightbulb, so color could be difficult to notice, but it's supposedly black.

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

@Mithoron I think your linked question could use a better answer. I hadn't thought about temperature, by "shine like a lightbulb" you mean it will be hot enough to glow? If it were transparent you could probably still see through it, but what makes you think it is " supposedly black"?

"With a description like "solid lattice of oxygen and liquid-like hydrogen superionic ice" it certainly sounds "shiny/metallic" to me!" - Remember AgI amongst others is a superionic aka fast-ion conductor above 420K. That doesn't look very metallic.

Also accurate colour prediction via calculation of macroscopic samples of condensed matter phases is extremely difficult, especially when you have mobile charges. "Shine like a lightbulb" is probably the best you will get, along with "black" as in black-body radiation.

@IanBush somehow I may not have been at the top of my game when I posted this; I think I imagined that it had cooled off enough to see it in reflected light, but ya, if it only exists in the 2000 to 3000 K temperature range, all we're going to see is blackbody radiation and if we're lucky and careful be able to argue about it's emissivity.

@IanBush: Source of your article is https://en.m.wikipedia.org/wiki/Silver_iodide,

As suggested in the comments, "what does superionic ice look like" in terms of its "natural" (reflected/transmitted) color is difficult to access because the phase seems to exist only at high temperatures where its radiant "heat" would glow yellow/white. While the "natural" color can still be measured with the proper experimental setup (eg. using a tunable chopped laser with a photodiode and lock-in amplifier (samples amenable to such a setup are not readily available. If we were to compress water at a "non-glowing" temperature, even several hundred degrees Celsius, then at about 60 GPa or more we would see not superionic ice but Ice X. Ice X is a macromolecular phase in which the hydrogen and oxygen atoms are held by delocalized, but fixed bonds.
Ice X might thus be considered as (fully) frozen superionic ice. It has cubic crystals, but information related to color/absorption spectra seems limited to infrared wavelengths. Chemistry LibreTexts reports:

Ice-ten looks identical as seen from the x-, y- or z-direction. Using a first-principles DFT code, ideal crystals of ice X have been shown to have a Raman peak at 998 cm$^{−1}$ and two infrared absorption peaks at 450 cm$^{−1}$ and 1507 cm$^{−1}$ [1].

(Visible light would be 12500-25000 cm$^{−1}$.)
Cited Reference

L. Jiang, S.-K. Yao, K. Zhang, Z.-R. Wang, H.-W. Luo, X.-L. Zhu , Y. Gu and P. Zhang, "Exotic spectra and lattice vibrations of ice X using the DFT method", Molecules, 23 (2018) 2780.

Also my experience is steel slabs look orange at only about 1500 K (when they exit the furnace going into a hot rolling mill). So I choose a lighter color for the "glow" at Ice XVIII temperatures.

Looks great, thanks!  Well if we're in a room lit by fluorescent or LED white lights to set our personal "white point" then it would be at least a little orangeish https://en.wikipedia.org/wiki/Color_temperature or https://brandon-lighting.com/color-temperature/ for example


text:
In a galvanic cell, why do electrons move through the wire from zinc to copper?
I understand that there is aqueous $\ce{Cu^2+}$ in this scenario which pulls on the electrons from the zinc causing the electrons to move through the wire, but how does the electronegativity of $\ce{Cu^2+}$ pull on the electrons from the $\ce{Zn}$ when they aren't "connected".
What I mean by that is the $\ce{Zn}$ metal and the wire are only connected to the $\ce{Cu}$ metal and not the aqueous $\ce{Cu^2+}$ solution. This would make more sense to me if the $\ce{Zn}$ metal was connected to $\ce{Cu^2+}$ solution via a wire rather than being connected to the $\ce{Cu}$ metal.

The wire is made of a conducting material, usually copper! So the wire and the Cu electrode are all part of the same conducting system.

$\ce{Cu^2+}$ ions in the bulk solution don't react. Only if they are in contact with the electrode will they be able to gain electrons. So in that sense the Zn metal is connected with the $\ce{Cu^2+}$ ions via conducting metal.

For an extremely thorough discussion of the findamentals, see this paper: K. Schmidt-Rohr, "How Batteries Store and Release Energy: Explaining Basic Electrochemistry", J. Chem. Ed., 95 (10) (2018) 1801-1810. The Zn and Cu Daniell cell is addressed at great length (way too long for an answer here) and the metal cohesive energy is the source of more than 75% of the energy in the Daniell cell (p. 1805.).

If we consider the classical Daniel cell
$$\ce{Zn|Zn^2+||Cu^2+|Cu|}\text{(wiring  to Zn)}$$
then due higher tendency of zinc ions to    leave the metal, getting hydrated in solution, the zinc electrode obtains more negative electrostatic potential due extra electrons, compared to cupper electrode.
As electrodes are galvanically connected, electrons flow from more negative potential of the $\ce{Zn}$ electrode toward more positive $\ce{Cu}$ electrode, increasing potential of the former and decreasing the potential of the latter.
The $\ce{Cu}$ electrode, with the potential lower then its equilibrium potential, reduces $\ce{Cu^2+}$ to $\ce{Cu}$.
The $\ce{Zn}$ electrode gets the higher than equilibrium potential, as electrodes leave it by the wire, so more $\ce{Zn^2+}$ ions are transferred to $\ce{Zn^2+}$ salt solution, to approach the equilibrium potential.
Ions in both solutions, connected by diaphragm or salt bridge, migrate along the potential and concentration gradients, caused by unequal potentials and electrode reactions.
The whole cycle is closed.

I've also searched multiple places for the answer and came across several similar questions.
What causes electrons to move from zinc to copper?
Why do Electrons leave the Zinc in a Galvanic Cell
However, none of the answers inside seems to be on point or understand the intention of the people who asked the question.
I've thought about the problem and have an explanation myself. Why does the electron leave zinc and flow into copper? It doesn't make much sense since both sides are neutral and where does the attraction force come from then?
The answer I believe lies in the way that a piece of solid zinc reacts with copper sulfate solution. Copper sulfate is also neutral then why the reaction happens when you throw in a piece of zinc? I think it's because although copper sulfate solution is overall neutral, they are not locally. The copper ion close to zinc will show as a positive charge locally and attract the electron.
When it comes to the battery, the copper ion in the solution will also attract the electrons in the copper sheet which makes the sheet have a polarity even though neutral overall. When a wire is connected from zinc to copper, this polarity will then polarize the wire and propagate all the way to the zinc end. So zinc now feels the attraction from the wire on their end the electron will flow.


text:
Why is the $\mathrm d_{z^2}$ orbital so different from the rest?
$\mathrm d_{z^2}$ orbital appears to be special:

Although degenerate with other d orbitals, it has no nodal planes, instead it has two nodal "cones".
Instead of having four lobes, it has two lobes and one ring.
Its electron density is prominently distributed in all $x, y$ and $z$ directions unlike others.

I know the wave function is what determines the shape, but what makes $\mathrm d_{z^2}$ orbital different?
Is there any fundamental reason?

Well, $d_{x^2-y^2}$ is sort of special too...

It is no more 'special' than any of the other solutions to the Schroedinger equation.

Note the degeneracy is true in absence of magnetic fields.

@NightWriter and electric fields too, right?

My understanding is that E-field interactions occur only with the right symmetry (to first order),  see eg https://en.wikipedia.org/wiki/Stark_effect

The Wikipedia is helpful in explaining why radial variations should arise in the density of non-s orbitals:

The non radial-symmetry properties of non-s orbitals are necessary to localize a particle with angular momentum and a wave nature in an orbital where it must tend to stay away from the central attraction force, since any particle localized at the point of central attraction could have no angular momentum.

What's unique about the $\mathrm d_{z^2}$ orbital (see the table above, from the Wikipedia) compared to the other $l=2$ angular momentum wavefunctions is that the $z$-component is zero $(m=0).$ This further constrains the geometry of the wavefunction.
The functions describing the angular dependence of the hydrogenic wavefunctions are Spherical Harmonics $Y_{lm}(\theta,\phi)$. In the case of d-orbitals, they satisfy
$$ \hat{L}^2Y_{lm}(\theta,\phi)=\hbar^2l(l+1)Y_{lm}(\theta,\phi)$$
with $l=2$, where $ \hat{L}$ is the angular momentum operator. Since the $z$-component of the angular momentum is also quantized, the following eigenequation also holds:
$$\hat{L}_zY_{lm}(\theta,\phi)=\hbar mY_{lm}(\theta,\phi)$$
with $m=0$ in the case of the $\mathrm d_{z^2}$ orbital, and this last equation leads to the following condition:
$$\frac{\partial\psi}{\partial y^2} = \frac{\partial\psi}{\partial x^2}$$
which implies that the solutions must be cylindrically symmetric about $z.$ However, the condition $l\neq 0$ implies that the solution is not spherically symmetric. The result is the unexpected shape of the $\mathrm d_{z^2}$ orbital.

There is actually a very simple reason the $\mathrm d_{z^2}$ orbital looks so different compared to the other d orbitals.
You can imagine it as being two d orbitals in one: the $x^2-z^2$ and the $y^2-z^2$, so to speak. These two "orbitals" would look just like all the other d orbitals and would lie on the x and z axis, and the y and z axis, respectively. The same way that the $x^2-y^2$ orbital lies on the x and y axis.
But since there can only be 5 d orbitals, these two orbitals are combined into one. This is why the lobes on the shared z axis are preserved, but the lobes from the x and y axes combine into a ring.

It's a matter of counting, really, if you are aware of what you are counting.
We may regard the angular dependence of the four-lobe functions at any radial distance $r$ as follows:
$(a_1x+b_1y+c_1z)(a_2x+b_2y+c_2z)$
where the nodes between the lobes are described by either factor being zero and $x,y,z$ vary according to $x^2+y^2+z^2=r^2$ for our chosen $r$. Different radial distances would have different valyes for the coefficients $a_1$, etc.
It looks like there are six independent parameters for the coefficients at any radial distance. But:

Changing them all by the same constant factor gives back the same orbital; this cannot change the nodal planes.

The nodal planes must he perpendicular to each other to match the solutions to the Schrödinger equation.

So with these limirltations there are only four independent combinations that satisfy tgehe "normal" four-lobed appearance of $d$ orbitals. To have the actual five $d$ orbitals one must be structured differently no matter what basis set you try.


text:
Do TDS meters measure HCO₃⁻/CO₃²⁻?
I am trying to add $\pu{2 g}$ of $\ce{KHCO3}$ to $\pu{20 L}$ of RO water $(\mathrm{pH} = 5.2)$ which theoretically it will add
$$
\begin{array}{lr}
\hline
\ce{K+} &\pu{10.32 ppm} \\
\ce{HCO3-} &\pu{16.1 ppm} \\
\hline
\end{array}
$$
Will my TDS measure only the $\ce{K+}~(\pu{10-11 ppm})$ or it counts also $\ce{HCO3-}~(\pu{26-27 ppm})$ (or even $\ce{CO3^2-})?$

In my understanding, TDS meter estimation is based on conductivity measurement, so it takes into account all ions.


text:
Additive to an aluminum sulfate and water solution to lower pumping pressures
I am currently looking for a surfactant (powder form) that I can add to a solution of aluminum sulfate and water.  My current solution is comprised of 34 gallons of water and 4 (four) 50 lbs bag of aluminum sulfate.  It has a specific gravity of 1.275
It is used as an additive to a gypsum based fireproofing product to lower the density as well as help with a quick set of the fireproofing material.  It is injected into the fireproofing material at the base or a 1" hose that is 20 feet long.  I was hoping to find an additive (surfactant?) that would reduce the pumping pressure of the fireproofing material thru the 1" hose without affecting the chemical nature of the aluminum sulfate and water combination.
The aluminum sulfate and water are mixed in a 50-gallon drum with a high propeller blade that is fairly aggressive.  I worry that a normal surfactant will foam up too much and make the product unusable.
If a non-foaming agent can be added to the water that will offer to lower the pumping pressures of the material thru the 1" hose that would be what I am looking for.

The gypsum-based fireproofing material probably also contains fireproof/insulating additives like vermiculite or perlite. A commercial material like this is sold in bags, dry, and dumped into a mixer where water is added. Mixing might generate some air, and it is advisable to keep the air developed under control, but a lot of the air is eliminated during the spraying operation. In any case, it is necessary to keep the wet fireproofing material fluid enough to pump easily, but "dry" enough after spraying to stick to the surface to be fireproofed.
The aluminum sulfate solution viscosity is not the issue here. When it is added to the wet fireproofing mass, it begins to accelerate the set of the plaster of Paris ("gypsum"), and the mass begins to thicken. One way to reduce pumping pressure is to increase the hose diameter, say to 1.5 inches; or to shorten the hose a bit. Another way is to add a bit more water to the dry fireproofing in the mixer - make it just a bit too wet. As it begins to set, it may reach an optimum spraying consistency with a somewhat reduced hose pressure.
Controlling the air content is extremely important, because it makes the wet material less viscous, but when put under pressure, the air compresses and fluidity is lost. This is a problem that occurs before spraying and is not cured by air loss during spraying.
These items are a matter of experience and "feel". It can get better if you keep looking for ways to adjust, although there is some amount of variation from bag-to-bag, and in the exact amount of water added, and the exact amount of mixing that occurs.
One possibility for development of a fireproofing material, which might require fire-testing on a large scale (at UL labs), would be addition of long-chain polyethylene oxides (https://www.industrialcellulosics.com/products/polyox)  - the materials that produce pituosity in water and are used in concrete pumping applications to reduce pump pressures. Since they do act at the surface interface, they might affect the adhesion of the fireproofing material to the intended substrate.

The needed pressure depends on liquid viscosity. The significant decrease of water solution viscosity cannot be achieved by adding of minor additives.
It could be achieved by methods that may not be applicable:

Decreasing concentration of the solute
Decreasing viscosity of solvent by different solvent composition
Increasing temperature of solution. A rule of thumb says viscosity decreases by 3% by increasing temperature by $1 ^{\circ}\mathrm{C}$.

Brother Aluminium Sulfate is a coagulant. Its used in order to bind to molecules suspended in water. You add any insoluble ions to the solution, it will bind to them and get thicker. Only thing you can do is increase temperature


text:
How can you deduce from the structure that probenecid is an acid and ephedrine is a base?
Probenecid contains a carboxyl group, indicating an acid. But it also contains a nitrogen atom, which indicates a base? How can I decide that it is an acid and not a base?

I have the same problem with the ephedrine structure.

The nitrogen atom in Probenecid is not part of a basic functional group (eg: amine), but of another kind of functional group: sulfonamide.
Sulfonamides are, just like amides, quite acidic: the presence of an electron-withdrawing group next to the nitrogen atom makes the electron pair less available, and makes the functional group quite acidic (see Lewis' definition of acids and bases).
For comparison, ammonium has a pKa of around 9, amides have a pKa above 15.
Sharing with amides the presence of an electron withdrawing group, sulfonamides can be expected to share the same acidic behaviour of amides.
You should check pKa list for the functional groups you have studied.
An example of such chart:
http://evans.rc.fas.harvard.edu/pdf/evans_pKa_table.pdf
(Be careful: values between parentheses are referred to the acidity in DMSO).
To better understand the delocalization of the electron pair in amides, see:
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Amides/Properties_of_Amides/Structure_of_Amides

Probenecid, also sold under the brandname Probalan ,has two functional groups Sulfonamide and carboxylic acid.In Sulfonamide , lone pair on N are involved in resonance with sulphur.Such resonance gives 1 and 2 .In these N possess positive charge rendering it less basic.Coupling with stearic hindrance factors N is considerablly less basic.

However,
Carboxylic acid is stabilized by resonance (structure 3) and also an electron withdrawing group at para position that enhances stability of carboxylate ion.

Taking the above factors into account ,with stearic hindrance and conjugation decreasing basic nature of amine and resonance stabilization of carboxylate ion ,Probenecid has acidic nature.
Referances
1 https://en.wikipedia.org/wiki/Probenecid
2 https://www.drugbank.ca/drugs/DB01032


text:
Could an ester react with magnesium oxide?
Esters undergo saponification in the presence of hydroxides. Assuming anhydrous conditions, would they react with magnesium oxide?

In what solvent? Or are you referring to MgO in neat ester?

Magnesium oxide can be an effective catalyst in transesterification reaction of vegetable oil samples mostly containing high carbon esters:

$\ce{MgO}$ is used as catalyst to increase the rate of transesterification reaction of sunflower oil in supercritical methanol.
The reaction in case of soybean oil has also been reported. This method was developed using metal-chitosan complex leading to higher yield of fatty acids and ethyl ester (biodiesel).
Sometimes alkaline supported catalyst like $\ce{MgO-K2CO3}$ ($\ce{K2CO3}$ fused $\ce{MgO}$) has also been used in the transesterification reaction of canola oil.

Another reaction using magnesium oxide as catalyst is the carbon-acylation of malonates with either acyl chlorides or chloroformates. Various malonic esters (methyl, ethyl, isopropyl and isobutyl) were easily alkoxycarbonylated to give the corresponding methanetricarboxylic esters.

References (in the order of mentioned points)

A. Demirbas, "Biodiesel from Vegetable Oils with MgO Catalytic Transesterification in Supercritical Methanol" Pages 1645-1651, 22 Jul 2008 DOI: https://doi.org/10.1080/15567030701268401
Gizelle I. Almerindoa, Luiz F., D.Probsta Carlos E. ,M. Campos Rusiene , M.de Almeidac , Simoni M. P.Meneghettic, Mario R.Meneghettic , Jean-Marc Clacens, Humberto V.Fajardoe "Magnesium oxide prepared via metal–chitosan complexation method: Application as catalyst for transesterification of soybean oil and catalyst deactivation studies" Journal of Power Sources Volume 196, Issue 19, 1 October 2011, Pages 8057-8063 DOI: https://doi.org/10.1016/j.jpowsour.2011.05.030
http://www.asianjournalofchemistry.co.in/User/ViewFreeArticle.aspx?ArticleID=28_2_6
Jacek Skarżewski "Carbon-acylations in the presence of magnesium oxide. A simple synthesis of methanetricarboxylic esters"Tetrahedron Volume 45, Issue 14, 1989, Pages 4593-4598 DOI: https://doi.org/10.1016/S0040-4020(01)89094-3


text:
Are there materials permeable to oxygen and carbon dioxide but not nitrogen?
Are there materials permeable to $\ce{O2}$ and $\ce{CO2}$ but not $\ce{N2}$?
Would this even be possible in theory? Can it be done at even at pressures of 300 atm?

Permeable is not a 0/1 quality. For three unpolar small gas molecules, the difference in permeability for each will never be very much different, but certainly it will be different.


text:
Retention of configuration in Hofmann rearrangement
Why is there a retention of configuration in Hofmann bromamide degradation/Hofmann rearrangement?
I would think that there should be an inversion of configuration due to the migration of alkyl group to the other side. Why is this not so?

The only group which can have a stereocenter in a substrate for the Hofmann rearrangement is the "R-" group to which the primary amide (-CONH2) is attached.
In all the steps of the reaction, the R group is never involved (no electron transfer is made from or to the R group), so the configuration is not touched.
Which step is not clear to you?

I mean the absolute configuration not the stereocentre. Basically, why the Fisher projection remains same.

I have added two examples from the book Solmon and Fryle.

Tip: Following the mechanism for the rearrangement given by your book, write with the curved-arrow formalism a mechanism of the rearrangement for those two molecules.


text:
Procedures for Syringe Pump and organometallic solution
For dispensing a pyrophoric organometallic acid into a reactor under argon flow, is it permissible to place the loaded syringe (glass 10ml, 9ml solution) on the syringe pump, with a long needle (36" septum piercing) through a septum on the reaction vessel (500ml, 50% volume solvent and reactant) set at 1.5 ml/h?
I am asking because after a couple trial runs I suspect an indeterminate amount of acid is staying in the long needle after drawing from the sure seal bottle (two penetrations: 36" needle and the argon needle) and despite holding the syringe upright and pushing the excess into the headspace and drawing an inert "cushion" back into the syringe I am somehow still overdosing.
My epiphany came as I was looking into schlenk storage flasks that in order to utilize a syringe pump for proper dose rate with pyrophorics one would run a nitrogen filled syringe in the pump with a needle going into a pressurized schlenk flask with a cannula coming out from the bottom of the liquid level to the reactor vessel and as the syringe pump pushed the empty syringe the displacement would equate to the amount of acid being pushed through the cannula.
If this all seems obvious I apologize, but I have read many of the Sigma Aldrich technical bulletins regarding pyrophoric handling and they do not mention slowly dispersing from a syringe. So far, besides the overdosing problem, I also suspect the solution is not entirely insulated within the syringe and it may be less reactive once it gets to the reactor. I am considering a Teflon pusher for the glass syringe and going to a PTFE-lined vinyl tube with luer lock barbs.
If anyone can point me in the right direction in terms of syringe pumping to a reactor with pyrophoric acids that would be awesome.

Using a second Schlenk flask and transfer needle is a standard procedure, I'd definitively try that. However I don't get how you can be overdosing. Are you sure about the concentration of the storage vessel?

So I should cannula into a schlenk flask from the sure seal bottle, and then push from the schlenk flask via cannula to the reactor? The syringe draws in argon from the schlenk flask, then it's pressurized, and as the syringe pump runs the liquid is forced through the cannula at the given rate through a septum on the reactor?

I'm not sure how well it will work to push gas from the syringe pump into the Schlenk. How vital is the slow addition, and how long should it take? The usual technique is to control the transfer by carefully creating a small pressure difference, but that might not be practical in your case. Very thin transfer needle, and perhaps dilute the reagent?

I have it pumping at 1.5ml per hour, a rate I came up with based off of the reference experiment's data (4.5ml over 3 hours)

the technique your referring to would involve diluting the stock TiBAl solution in more hexane and then using a pressure equalizing addition funnel atop the reactor?

Basically you partially close the argon line on the reactor, that should be enough. The transfer needle is in the air (argon) atop the reactor solution, so you can observe the solution drop. Make a test with pure solvents to see how it runs. I've seen many labs who tried out syringe pumps, and they practically always ended up in some corner collecting dust.


text:
Purifying cysteine containing peptides
I would like to purify my peptide having cysteine residue by using HPLC. It is known that thiol containing peptides undergo oxidation and form disulfide linkages. In order to prevent it I used proper cocktail solution during cleavage from the resin. However, I am still worried what if my peptide undergo oxidation during HPLC purification? How can I prevent it?
I searched in research gate, but I couldn't find suitable answers.

One way is to add reducing agent to the running buffer. I don't know of another way.

What reducing agent I can use? Any experience?

This might be useful as a comparison: http://agscientific.com/blog/2019/04/side-by-side-comparison-dtt-vs-tcep-preferred-reducing-reagents/ Also, I recommend searching a little more, it's not an uncommon problem.

Most HPLCs have degassers, so oxygen tension in the apparatus should be low and oxidation should be minimal.  You can inject from a TCEP (or DTT or BME)-containing sample, as recommended by @CuriousTree, and add these same agents to your fraction collector's tubes ahead of time, but you shouldn't need to add them to you chromatography buffers.

I dont know about HPLC column tolerance to various reducing agents, this would need to be checked. However, in biochemistry we routinely add reducing agents to avoid disulphide-bridge formation in the proteins.
Overall there are three main reducing agents used, to avoid this problem. The choice of which one to use depends on your needs. They include:

TCEP: Considered the golden standard, it is relatively odourless compared to the other reducing agents. Once you have dissolved it, it should also remain very stable over time. Typically used at a 0.4 mM concentration.

DTT: Cheaper than TCEP, although I have heard that it is not as stable over time

2-Mercaptoethanol: The absolute cheapest, stinks like crazy. Typically to a final of 5 mM.

Rather than worry about reducing agents in the HPLC column, why not oxidize your cysteines with Ellman's Reagent and reduce them after purification?
As long as you use an excess of Ellman's Reagent, it should completely oxidize the thiols and prevent disulfide bond formation.


text:
Consequences of ortho effect on resonance
Consider any carboxylic acid with any group at ortho position. According to ortho effect, the resonance of the $\ce{C=O}$  with the ring is inhibited.
Is this resonance entirely lost? Or some little amount still exists?
Can resonance be present when the participating orbitals are not in the same plane?

Well, the "extent of overlap" of the orbitals will decrease but it won't completely vanish, so yes, there will be some resonance stabilisation.

If they are off the plane due to steric effect,  then resonance wouldn't occur, hence the (-)R effect of - COOH group vanishes and hence "generally" acidity of the respective acid increases.

Wouldn't occur at all?

Also it took me a minute to realize OFF is not oxygen difluoride. Please don't use CAPS for emphasis (italics or bold is much better).


text:
How to determine the conductivity of different chemical compounds?
I am doing an experiment that involves testing the conductivity of three different chemical compounds dissolved in water. I will be testing Potassium Chloride, Calcium Chloride and Zinc Chloride. All three will be of equal concentration. Is there any way to determine the conductivity of different chemical compounds dissolved in water? Out of these three chemical compounds, which one will be the most conductive, and why? Is there any correlation between reactivity and conductivity of chemical compounds?

In my days the word "experiment" was used to denote the act of grabbing some actual chemical compounds  with your hands and doing something with them. If you went halfway through this process only to realize that you don't know how to perform the next measurement or whether it is possible at all, this was considered a flaw in your planning. Things might be different these days, though...

@IvanNeretin I'm just asking about if there is a way to determine the conductivity by looking at the electron configuration or reactivity or something similar.

Yes, but not in the way you think. Electron configuration is irrelevant. Reactivity is not a thing at all, and irrelevant, too. Most importantly, are you doing an experiment or not? If yes, you are supposed to obtain some real measured values, and then maybe explain why they differ the way they do.

@IvanNeretin Yes, I am doing the experiment, and this is completely unrelated to it. I'm just curious if there is a way to determine the conductivity.

Well, see which of the salts produces more ions, and that would be it.


text:
Paramagnetism of group 2 elements
As far as I know, paramagnetism in normal substance arises due to unpaired electrons. I believe that the valence electrons of group 2 elements (Be, Mg, Ca, Sr, Ba) are paired.
But, here in the list of given compounds, all of the group 2 elements (except Be) have positive susceptibility, and it is specified in the same document that positive susceptibility implies paramagnetism.
My question:
How can these elements show paramagnetic behaviour?

When you have https://en.wikipedia.org/wiki/Metallic_bonding, you don't have the electrons paired the same way as you would with isolated atoms.  The real question may be why this doesn't work for beryllium.  Lithium, the only other (ambient) metal in Period 2, is paramagnetic.

I suspected that, but I thought I was wrong. As I can't reason through the fact that why gallium is diamagnetic and aluminium is paramagnetic despite the fact that both have metallic bonding and both belongs to the same group, one hand-wavy thought might be that Gallium go extensive of it. And for Be, one thought might be Be is not-that metallic.


text:
Why do nitrile gloves turn yellow from sweaty hands?
When I take off my nitrile gloves after some hours in the lab, the inside of them has turned yellow. This happens especially in the regions around joints.
I suspect this is due to a reaction between some compounds in the sweat and the nitrile, but I couldn't find any definite answer online about what the responsible mechanism for this is.
What exactly is reacting and turning yellow?

Are the gloves powdered? If so, it might be an accumulation of the powder lubricant.


text:
How would this reaction look like in a potential energy diagram? (SN1 Reaction)
My task is to explain, why molecule A reacts in a SN1-reaction faster than molecule B. (I would do this by comparing their potential energy diagrams)
My idea is to have a closer look at the reaction mechanism at first:

TS1 is here the intermediate with the carbo-cation. That's the step, where the reaction is slow, where one needs a lot of energy to remove the Bromine.
Maybe I guess one does not need this lot of energy in molecule A compared to B because the Bromine is in the background (trans- isomer) there and not in the foreground? Furthermore, in SN1 reactions the reaction rate is determined by the substrate (the first step).
So, the activation energy of molecule A is smaller then the activation energy needed in B and that's the reason, why the reaction ist faster?

An interesting question is how (or if) one really knows if both reactions proceed via bonafide SN1 type mechanism. Under the assumption that both reactions proceed via bonafide SN1 mechanism, we must infer that the carbocation (carbenium ion) intermediate 1 in both cases must be the same. While the previous answer correctly assumes that this intermediate may be described well as planar (at the carbon centre), the attack by water is the faster step and hence should not affect the overall rate of the reaction. Since both A and B must proceed through the same intermediate, (although not rate determining!), both must react at the same rate with the incoming nucleophile.
A better explanation can be deduced by looking at the most preferred conformations in the two starting materials. These are designated 2 and 3 below, for A and B respectively. In both cases, the bulky tertiary butyl group will likely prefer to be sat at an equatorial position to minimise 1,3 diaxial interactions. Once tertiary butyl group is "locked" at the equatorial position, we have to place the bromine axially (up) in A (2), and equatorially (down) in B (3).It is possible to argue (although still very qualitative and hand-wavy) that 3 is likely more stable than 2, since both groups are equatorial in 3.
Now, the rate of a reaction depends on the activation free energy barrier, which is the difference between the free energy of the reactant and that of the transition state. At this stage, without computation it is impossible to predict which transition state will be at a higher or lower energy. However, speculatively, to a first approximation the energies of the transition states would be very similar. Under this assumption, the activation energy barrier for A (higher energy conformation) should be less than that for B, which should explain the observed reaction rates.
Having said that, I would not necessarily think that energy diagrams are the best way to explain this difference in reactivity, or indeed if there is a single, simple explanation.

Note: I have not shown the subsequent transition states and the intermediates in my energy diagrams, since these will not be rate determining, and hence irrelevant to the question (about rates).

At this level of abstraction it probably is fair to speak of a rate determining step, as it is (given that it actually is SN1) probably an elementary step. In more complex reactions this approximation will break down. A particular example for this is a Diels-Alder reaction with buta-1,3-diene, where the trans conformation is lower in energy. Only the cis conformation is viable for the TS though. The rate will therefore be determined through the energy span between trans and the TS. In any case, your analysis is good and should hold for this level of abstraction. ↑

I would rather not solve this by comparing  potential energy values.
Instead there is another theoretical way of doing so.
When SN₁ happens, either the attacking reagent( In this case water) can attack from above or below the planar carbocation.
In the first case, Br⁻ leaves from above, and now water can easily attack from below.
In the second case, Br⁻ leaves from below, and water has to attack from above(to avoid steric repulsions between the exiting Br⁻ and attacking water), but now there are steric repulsions between C(Me)₃ group and the attacking water, leading to a reduced SN₁ rate as compared to the first case.

But if you would solve it by comparing potential energy values how would it look like then? Like I did above?

You are just speculating about the energies required. You don't know it for sure unless an experiment is conducted. Therefore, a potential energy diagram cannot be plotted.

@user226375: Your explanation that considers from which side the nucleophile (water) attacks, would be more suited if OP described the reaction as SN2. See answer below for details.


text:
Stereochemical dichloride treatment

In this question there is no inversion in A to B, but there is one in C to D. I feel like this must be due to the lone pairs on the $\ce{S}$ (in A) being able to open the ring, whereas this isn't possible in C to D so this can only occur via an $\mathrm{S_N2}$ reaction, but I just can't seem to make the mechanisms work for either.

Remember S is a significantly larger atom than O, and episulfonium ions are well known

Instead of opening (i.e. breaking) rings, think about forming rings, maybe

So is the sulphur lone pair able to attack via SN2 to form an episulfonium ring, which is then attacked by SN2 with the CN ions, meaning there is no inversion?

That is exactly so

Because of the absence of a lone pair, C undergoes usual $\mathrm{S_N2}$ transformation under the conditions. However, the availability of a lone pair in A undergoes the nucleophilic substitution of cyanide ion via neighboring-group participation (see above comments by Waylander and J. Deans) by the sulfur atom under the conditions (Ref.1). A very reliable mechanism is given in Ref.1 for the substitution of azide ion (work the same way for cyanide ion):

Reference:

D. D. Díaz, A. Converso, K. B. Sharpless, M. G. Finn, “2,6-Dichloro-9-thiabicyclo[3.3.1]nonane: Multigram Display of Azide and Cyanide Components on a Versatile Scaffold,” Molecules 2006, 11(4), 212–218 (doi: 10.3390/11040212).

Could you check out the duplicate of the question, and maybe post another (essentially the same) answer there?

@Martin - マーチン: I can do that but is it ethical?

I think it's fine. You can add a disclaimer that you have posted the answer on another question before. Or if you feel more comfortable, you can post your answer there and then flag this one for deletion (you can't delete yourself as it is accepted). I think it would be beneficial to have the duplicate link point to a question that is actually answered. We could also have a meta discussion about it, of you want to make absolutely sure.

@Martin - マーチン: Thank you Martin for your lengthy explanation. Let me think about it. Previous question is asking to show the stereochemistry to prove the out come of the products. I think this may do that. :-)

Thank you very much for your answer. Very sorry I was unaware the question had been posted before

@Martin - マーチン: I put a different answer for duplicate but with same figure. Thanks.

@MathewMahindaratne I've seen and cited on it. Thanks a lot for doing this.


text:
Determine direction of mass transfer

For part a) I don't really know what to look at. Diffusion occurs from low to high concentrations so by looking at $t_1$ I would say that the direction of the mass transfer is from phase II (organic phase) to phase I (aqueous phase) since the organic phase starts at a lower concentration than the aqueous phase.
But apparently it is supposed to be the opposite. And the answer is the same at $t_2$.
Could anyone please explain how to determine the direction of the mass transfer and how it can be the same at $t_1$ and $t_2$. Thank you!

You start from the wrong premise, "Diffusion occurs from low to high concentration" It's just the opposite.

What's the source of what you copy/pasted?

You start from a false premise, "Diffusion occurs from low to high concentration". Within a phase the expected behavior is exactly opposite. However between phases things can get more complicated.
The direction of diffusion at the interface (on the resolution scale of your diagram) is based on relative solubility. At higher resolution, the higher solubility in one phase encourages transfer of solute into it from the other phase. Solute accumulates in the interfacial region of the preferred phase and from there diffuses into the remainder of that phase, following the usual dependence on the concentration gradient. In the less preferred phase the opposite occurs: solute is depleted from the interfacial region and replenished by solute diffusing from the remainder of the solution.
So you should follow these rules:

When you compare phases consider the relative solubility. The solubility dictates in which direction solute will be transported.
When you look within a phase consider the concentrations at different positions in the phase. Diffusion is always from higher to lower concentration (or chemical potential, strictly speaking).

In your problem the solubility is higher in the organic phase so transport is into that phase, until equilibrium is established.

I'm sorry but I still don't quite understand how to determine the direction from Fig. 4. For example, if I look at $t_1$ should I look at the left side of the interface and see that phase I has a higher concentration, and then on the right side phase II has a lower conc., and draw the conclusion that the mass will transfer from the aqueous phase to the organic phase? Because then if I look at $t_2$, the mass transfer should be from phase II to phase I, which according to our answer key is wrong.

Could anyone please explain how to determine the direction of the mass transfer [...]

In order for concentrations to change (given constant volumes and no chemical reactions), there has to be mass transfer. If the concentration in a solvent increases over time, mass transfer is to that solvent, if it decreases over time, it is away from that solvent.
So you have to look at the slopes in the diagram at the different time points. For the organic solvent, the concentration increases at both times, so mass transfer is from aqueous to organic phase.
This question is not about what you expect to happen, but what you observe to happen. It is great to check afterwards to see whether your observation matches what you expect, but you should interpret the observation in an un-biased manner. If you don't, there might be a temptation to see what you expect to see.

[...]and how it can be the same at t1 and t2

The slope has the same sign, so the direction is the same. This is quite commont - most processes don't oscillate as they approach equilibrium. In most cases, you see large changes at the beginning and smaller and smaller changes as you approach equilibrium. This is exactly what we see here.
As for why equilibrium is not when concentrations in the two phases are equal, see the answer by Night Writer.

+1 The fact that the two concentration lines cross is spurious, the slope directions are relevant.  If you changed the scales you could lift the one line above the other and they would not cross, also we don't know what the scales are or if they are the same any way.  All we know is that there is transfer and because it looks like an asymptotic curve we determine that the transfer is slowing down.  However the curves are not flat so we have not reached the point of no further transfer hence transfer is always in the same direction as described in the text, from aqueous to organic.


text:
Which carbocation from following is stable by hyperconjugation?

These are the 4 figures, how to tell which one of them is stable by hyperconjugation? A B C all are stable by resonance if im not wrong but which one is by hyperconjugation too?

Hyperconjugative stabilization is such that it cannot be depicted using delocalization of $\pi$ electrons, but instead uses the breaking of a $\sigma$ bond in the resonance structure (if you use resonance structures to depict at all). I'll let you try to figure out which of the options fits this pattern.

Another way to look at it is that in the limit, hyperconjugative stabilization may just be a group migration.

It should probably be option C as in one of its resonating structures, there is an alpha hydrogen(as it is saturated) wrt the carbocation formed. Hence, it can show hyper conjugation

I wouldn't say C is totally wrong, but it is definitely not the best answer.


text:
Henry's Law Constant in Freon
I am a physics student trying to calculate some values for a project I am working on, and am confused about Henry's constant.
I want to dissolve certain noble gases into freon in a pressurized chamber with temperature control, and want to know some of the numbers ahead of time.
I have seen the equation for the constant: $$K=\dfrac{p}{c}$$
My question is this: Where do the solvent characteristics come into play? I don't see how using this constant equation would get me a different constant for say, argon, in water vs. $\ce{C3F8}$?

Why, they come into play in that constant K. You switch to another solvent, you'll have another K.

Also, $p$ is the partial pressure of the gas which is dissolving (which is distinct from the vapour of the liquid) and not the total pressure in the container.

To elaborate on another comment, it is better to write $K_{12}=p_2/c_2$ which makes explicit that K depends on the solute (subscript 2) and solvent (subscript 1) pair.

Thank you all for your input. As it turns out, Henry's Law seems to only be valid is the solvent is close to the density of water, so I will need to search for another method to solve this problem.


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Getting radial Kohn-Sham orbital from the output of an electronic structure code using gaussian basis sets
As we know that in solving the Schrodinger equation (Kohn-Sham equation), we expand the wavefunction in terms of some basis sets and them optimize the energy (the expectation value of Hamiltonian) by varying the coefficients of the basis in the expansion. Most of the electronic structure codes use the basis sets. What I need is that if I use gaussian basis sets in NWCHEM for example, how would I get radial Kohn-Sham orbitals or the optimized coefficients of the basis? I will be grateful if someone can tell any other option to get these orbitals from different electronic structure code but the one which uses Gaussian basis sets.

Every qc code should have some documented option to print this information. Did you check the manual?

I am aware of getting molecular orbitals in NWCHEM ("http://www.nwchem-sw.org/index.php/Release66:DPLOT#GAUSSIAN_--_Gaussian_Cube_format") in Gaussian cube format but do not know how to get the radial orbitals directly or from the corresponding cube file.  I know how to get the radial density. This I learned from a similar question "https://chemistry.stackexchange.com/questions/70021/how-to-obtain-the-radial-probability-distribution-function-from-a-quantum-chemic" but for radial density.

I have trouble understanding the question, particularly I do not know what is meant by radial Kohn-Sham orbital. With the optimised coefficients don't you just mean one of the density matrices?

Okay, I understand your point. The systems I am interested in have spherical symmetry (like closed-shell atoms, so the wavefunction can be separated in radial and angular part).  Even if it's not spherically symmetric, it should be the wavefunction which is expanded in terms of the basis set chosen (I am interested in knowing the optimized coefficients of this expansion) and the density matrices then be calculated from the wavefunction.

For example "https://physics.stackexchange.com/questions/420899/density-matrix-approach-in-density-functional-theory-interpretation".

I think you are asking about the MO coefficients, this should be included in the output file, but it might require some special keyword.

Yes, thanks. I have got some clue here "http://www.nwchem-sw.org/index.php/Special:AWCforum/st/id510/How_to_print_full_DFT_MO_Analysi....html". I will check whether I may get what I wanted.


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Does 2-chloro-1,3-dinitrobenzene show resonance?
I am preparing for my exam, and then I stuck up here. The answer to this question given by my teacher was

No, this doesn't show resonance because there are two bulky groups alongside chlorine, due to this chlorine will leave the plane and as we know resonance is a planar phenomenon, it won't show resonance

My question to you is:
Why did chlorine leave the plane on the first place? And what is the whole concept behind it? Please help me with it.
My guess to this is that since these two groups are really big and bulky, and are present near Chlorine, they take up almost all the space around which causes chlorine to leave the plane.

In one line, it's Steric Inhibition of Resonance.

2-chloro-1,3-dinitrobenzene is shown below:

I believe your question is referring to the lack of conjugation between the nitro groups and the aromatic ring. this is best shown in an X-ray crystal structure such as this one:

https://www.ccdc.cam.ac.uk/structures/Search?Ccdcid=722415
Due to the sterics between the nitro groups and the chloro substituent, the nitro groups cannot adopt a conformation planar with the ring. This prevents the $\pi$ system of the nitro group from achieving substantial overlap with the $\pi$ system of the ring and prevents full conjugation of the electrons. The sterics are better seen in a spacefilling view:

So, it is not the chloro group that is out of the plane, but rather the nitro groups.

That 3D rendition of the molecule's structure is not exactly an "X-ray crystal structure".

@William R. Ebenezer. I will take it down if I am mistaken. It came from the Cambridge crystallographic database. Check the link and I will remove if you could explain my mistake.

@Withnail thankyou so much. I understand it now !

@Withnail But hey, doesn't it go against common sense?

@Withnail I can't find it, could you be more specific?

@william R Ebenezer. The link to CCDC is in the answer. https://www.ccdc.cam.ac.uk/structures/Search?Ccdcid=722415

@Withnail yes, I have visited the link, but couldn't find that term.

@William R. Ebenezer. The CCDC is an X-ray crystallography database. That image is from a structure determined by Marten et al. and deposited in 2010

Yes, but that does not mean that image is an "x-ray crystal structure". Do you know what XRD/XRC is?

The image is based on calculations (maybe aided by XRD etc), but afaik it is wrong terminology to say that image is an "xray crystal structure".

I am not a crystallographer, but I have had structures of compounds that I have synthesized determined by crystallographers. So far as I can tell, that entry in the CCDC comes from a single crystal XRD experiment. I have not read the original paper, just the entry in the CCDC. Please let me know if you know otherwise about where it comes from.