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What are some easily-undertaken reactions that will precipitate or restore the iron from rust, in a non-lab context?
Unusually, this question arises in a DIY context rather than a chemistry one.
I'm doing some work in my garden with reinforced concrete (a retaining wall against a  high bank of soil). As some will know, it usually doesn't matter if the steel used in reinforced concrete  gets rusty, because it will still have its inner bulk of steel for strength, still be gripped when concrete pours around it, and won't corrode further provided it is deep enough inside the body of concrete.
There's one exception. If the steel is used to join two sections of concrete below ground level that are poured separately, without a "water stopper", water will inevitably enter the point where the two sections of concrete meet, and in favourable conditions can slowly corrode the steel there. It's a low risk, and a long term issue, but can happen.
That's the situation I'm in, and I didn't realise the issue early enough to take precautions. Having already poured one section with its steel last year, I'm trying to retrospectively prevent possible issues before pouring the second part.
To counter any corrosion issues, I plan to use a zinc based galvanising coat to retrospectively protect the steel in the region of the join between the two lots of concrete. But there's a problem. Galvanisation relies on electrical conductivity to create the sacrificial aspect, and rust doesn't conduct. Physically stripping rust back to steel is also hard on these bars, as they are designed with an irregular surface, to enhance grip.
So I'm wondering if there is a solution via chemistry - some easily accessible way to revert a rust coat back to metallic form, or remove rust to expose metal. Most rust removal products as far as I suspect, attempt to passivate or stabilise rust, not actually return a surface to metallic conductive form.
Is there anything I can easily do to return the rust layer to an electrically conductive/metallic layer? If not, any ideas how else I might chemically reduce the scope for future corrosion over the years?
Update: noting for clarity, the join is below groundwater level, with some degree of hydrostatic pressure behind it (it's a 1m retaining wall where a soil slope in the garden was cut away, and I can't add drainage below the joint because the bottom slab is already set and the joint will be 30cm below ground level even on the lower side, hence below groundwater level). Also the concrete mix is self-consolidating/self-compacting (SCC) so air bubbles as mentioned in one answer are not an issue - both mixes are luckily, specifically designed to preclude them, and the formulation sets pre-compacted without vibrating.

If you have a plastic bowl or container large enough to accommodate your item, fill it with water with with some salt (e.g., kitchen salt).  Then you need Aluminum foil lightly wrapping your rusty piece of iron which altogether are immersed in this said solution (luke warm is enough).  One important detail is that the foil is in mechanical contact to the iron, while the solution is able to be in contact with the foil and the cast iron simultaneously.
Over the course of the next hours, Aluminum will be "eaten away" -- or, more chemically speaking, Aluminum will be oxidized, and the rusty iron will be reduced.  To facilitate the process, it does not harm to scratch the Aluminum surface with a brusch on the surface exposed toward your iron pieces.
To direct the work of this oxy-reduction reaction toward your rusty iron pieces, it is important that the container for this setup itself is not made of metal.  Depending on the size of the item "to recover" things like a glass / china bowl, or a plastic bucket are suitable.  After half a day or so, remove your item, rinse it with water and a brush; dry it and protect it against new corrosion.  As already mentioned by you, this may be done by galvanizing, or by galvanic anodes.
An alternative to using a salt solution is the addition of vinegar to the water.  The key point of both (salt or vinegar) is that they increase the electric conductivity of the solution, hence accelerating the process.  If using vinegar, however, do not forget to protect your hands with gloves and eyes (goggle) against the acid.

"it usually doesn't matter if the steel used in reinforced concrete gets rusty" isn't exactly accurate. It is good if the rebar used in concrete is slightly rusty, for a better bond. But loose rust results in a poor bond. And if the steel "gets" rusty (rustier), the corrosion produces ferric oxide/hydroxides which have a larger volume than the original iron, and the expansive stresses, especially in combination with freeze-thaw conditions, result in spalling of the concrete over the corrosion.
Cold galvanizing with zinc-rich paint sounds like it might give you some extra protection of the yet-to-be-coated rebar. However, it seems like overkill, considering that you admit it's a "low risk". If you want to improve the corrosion resistance of the yet-to-be-coated rebar, how about coating it with cement paste just before casting the new concrete? Knock off any loose rust with a wire brush. Then take a heavy duty paint brush and coat the exposed rebar with a coating of cement paste (watery; w/c ~ 0.55) to soak into all the tiny crevices on the rebar surface. Also on the surface of the old concrete, to help provide a better bond and leakproofness. (There are latex additives for concrete patches for improving bond to old concrete, but this may also be overkill. Just clean off the old concrete and paste it with some of the cement paste before casting.) Then cast the fresh concrete. This will prevent (or minimize) air bubbles being trapped at the rebar surface, which help initiate corrosion.
And since you are concerned about possible corrosion at the junction of the two pours of concrete because of water ingress, you could provide a weep hole at the bottom of the concrete (above ground, but not distractingly visible).
Commercially and industrially, returning rebar rust to reinforcing status is never done. The distressed concrete is removed and replaced.
Another thing: you will have two large concrete electrodes (dry, not very active), connected by rebars. If the two concrete blocks are very similar, the voltage between them will be very small. But do not make a big change, e.g., by adding a corrosion-inhibiting admixture to the new concrete, otherwise you may set up a large galvanic cell which will favor corrosion in one or the other blocks.

Thanks James. As it happens, much of your 2nd paragraph is probably taken care of, because by lucky chance, the pours are both self-consolidating/self compacting (SCC) mixes - a formulation that simply doesn't need vibration because there aren't air bubbles when it sets. So vibration + paste may well not add anything, nor needed. However weep holes aren't practical as the join is below groundwater level. Your last para about galvanitic cell creation is a biggie, it sounds like. The concretes are identical formulation. Do you want to update your answer in light of the Infos? It would help :)

I visualized a vertical joint; a horizontal one is more troublesome. Essentially you have a dam. It may be all right. At least you won't have freeze-thaw problems. SCC concrete is usually rich, so will be alkaline as long as the water flow does not remove Ca(OH)2. Instead of making a straight butt joint, you could overlap concrete in back of the joint so water would have to travel farther to get to the rebar. This also reduces dependence on the rebar connection. It doesn't sound like galvanic interaction will be serious - just a theoretical speculation.

Insanely rich. 400-480.kg per cubic m, from memory. I'm more worried about spelling and expansion issues in years to come, than the basic loss of a strong connection.


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Is black phosphorous aromatic?
I know that graphite and boron nitride are aromatic in nature.
In a similar way, can black phosphorous too exhibit aromaticity? Since it is essentially the same having 6 membered rings with lone pair of electrons and vacant d orbitals.
If not, then it would show some double bond character also, right?

According to: https://pubs.rsc.org/en/content/articlelanding/2018/NR/C7NR05718K#!divAbstract
[...] Despite the fact that BP does not show aromaticity or single atom thickness as graphene, it has something else in common with graphene as we report below [...]
So, as to the aromatic character, it looks like it doesn't have it.
Aside from that, it would be intuitive to guess it by looking at the shape of BP: it is not planar, which is a characteristic of "common" aromatic molecules: delocalization is not favoured

There.  Are.  No.  Outer.  d.  Valence.  Orbitals.  Period!

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Nitration of aniline

(b) Nitration: Direct nitration of aniline yields tarry oxidation products in addition to the nitro derivatives. Moreover, in the strongly acidic medium, aniline is protonated to form the anilinium ion which is meta directing. That is why besides the ortho and para derivatives, significant amount of meta derivative is also formed.

Why there is more para-substituted nitroaniline formed than ortho-substituted one even though it can form hydrogen bonding?

In case of nitration in aniline, the Nitric Acid protonates the aniline to form the Anilinium ion.
Now since the nitrogen atom has no lone pair to conjugate, it has no mesomeric effect on the ring, the but as nitrogen is now protonated, it has a high negative inductive effect.
Which simply leads to the fact that the position farthest from the nitrogen substituted carbon will have the highest electron density (as inductive effect decreases as we move away from the atom causing it).

So in this case the best site for attack for the electrophile was para>meta>ortho.
The Hydrogen bonds would stabilise then ortho product once its formed, but the Rate Determining step of an electrophilic substitution on benzene is the sigma complex intermediate wherein the most stable intermediate is formed as major product. The positive charge in sigma complex stays the farthest in para position from the anilinium ion hence(most stable) is the major product.

The Diagram is wrong. $\ce{NO2^+}$ is an electrophile, not a nuleophile. It's the $\pi$-electron cloud of Benzene ring which attacks, not  $\ce{NO2^+}$. So, the arrow should be on the opposite direction. Please change the direction of the arrow or upload any different picture.

In the case of nitration of aniline intermediate stability determines products.The following figure depicts possible electrophillic attack on aniline. Structures 4 and 12 are additional stable resonance structures obtained by attack of nitronium electrophile at ortho and para positions.
Structure 8 is relatively unstable due to negative effect of Nitrogen.

In  nitration of aniline in strong acid (HNO3, H2SO4) aniline changes to anillium ion .Anillium withdraws electron density.Its effect is felt maximum at ortho followed by meta and then para position. Consequently very little of ortho nitrated product is formed.
Since electron withdrawing inductive effects are minimum at para position ,maximun nitration occurs here.

Summarizing ,
Even though there exists a possibility of hydrogen bonding in ortho nitro aniline ,strong deactivating influence of $\ce{NH3^+}$ on benzene ensures least electrondensity at ortho position in aniline leading to  only 2% of ortho nitro aniline.
Such electronwithdrawing tendencies of $\ce{NH3^+}$ are minimum at para position in aniline therefore 51% of para nitro aniline is formed.

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Why does SN1 not occur as a minor product when a strong nucleophile/strong base attacks a tertiary substrate?
The following is part of a table that appears in my Organic Chem Text (Klein)
Nucleophile Only (e.g. Strong Nucleophile/ Weak base)
$1^\circ - \text{ SN2}\\
2^\circ - \text{ SN1 + SN2}\\ 3^\circ - \text{ SN1}\\ $
Strong Nucleophile/ Strong base
$1^\circ - \text{ SN2 (maj) + E2 (min)}\\
2^\circ - \text{ SN2(min) + E2 (maj)}\\ 3^\circ - \text{E2}\\ $
Why will SN1 not also occur on tertiary substrates if the attacking nucleophile is also a strong base? That is, why doesn't it read $$3^\circ - \text{E2(maj) + SN1(min)}$$

When deciding between E2 and SN1, you could rationalize two competing key steps: deprotonation for E2, and cleavage of the leaving group for SN1 (with formation of the carbocation).
When a strong base is employed, expecially when the substrate (eg: tertiary substrate) is hindred, elimination is the only reaction which you might expect, because 1. Deprotonation is what a strong base can do very well 2. The nucleophile/base has no easy access to the electrophilic carbon atom.

@The_Vinz Thanks for your response! I understand that E2 would occur. I'm struggling with how adding the quality of "basicity" to the nucleophile stops SN1 from occurring altogether (as opposed to having both occur with SN1 the minor product). I don't understand your reasoning employing steric hindrance as my book repeatedly employs the reasoning that this is not relevant for Loss of a LG (and according SN1 as the carbocat is planar).

You should consider that in case of a strong nucleophile/strong base, the nucleophile is so reactive that it is unlikely to allow the formation of a separate carbocation: any reaction you expect is concerted (e2, sn2).

Acid-base reactions should be faster with a strong base present. Think about what the molecules are trying to do in the rate-determining step. The base only needs to access a tiny proton sticking out. A strong base that's actively searching for something to deprotonate will do this rapidly. Furthermore, in a tertiary substrate there can also be a statistical preponderance of sites to deprotonate. Meanwhile, in SN1, the LG has to ponder whether it's a nice day to leave the house, then do his stretching, and finally walk far enough away so that another squatter can slip in and occupy the house.

I think if you took a real example i.e., choose a model substrate and try doing the reaction in the lab, you may actually observe some substitution product (presumably via a SN1 route). My guess would be that if you tried to quantify the relative amount of the substitution product you've produced (eg.using 1H NMR), it will be less than 10%... maybe <1%, or 'trace'. It may be only meaningful to talk about 'minor' products when they are in at least some, substantial (though minor) quantity! It will also depend on the temperature at which you do your reaction.


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Quantum numbers required to specify the position of an electron
In a test today, there was this question:

How many quantum numbers are required to specify the position of an electron?
  $$(A)\ 1 \quad (B)\ 2 \quad (C)\ 3 \quad (D)\ 4$$

The answer key says that the answer is $(D)$ but I am feel that the answer should be $(C)$ as the spin quantum number describes the angular momentum of the electron.
Please let me know what should be the correct answer keeping in mind that I am a sixteen year old.
Thanks!
P.S. let me know if I should include more or change tags.
Edit: Since it was not clear previously (or so I think), my point is that spin quantum number is not necessary for specifying the position of an electron.

First of all, the question is badly formulated. It asks about "position" but clearly means "orbital", which are two different things. (An orbital is a wave function, which depends on the electrons position: $\psi(\vec r)$.) Although the difference may be difficult to explain at high school level, this choice of words is imprecise and often leads to wrong ideas about the basic concepts of quantum mechanics.
The question is obviously about the 4 quantum number ($n$, $l$, $m_l$ and $m_s$), thus including spin. Whether spin is really required to describe the spatial part of an orbital is arguable. But if we stick with high school level chemistry and consider what is typically called restricted orbitals, then both electrons ($\alpha$ and $\beta$) share the same spatial part of the orbital, and the spin quantum number is unrelated to the "position" of the electron. But the spin quantum number is still required to "distinguish" the two electrons, what seams to be what the question has in mind. (I used quotation marks, because the electrons are actually indistinguishable.)
If we go beyond high school level and into actual quantum chemistry, then it depends on what type of orbitals are considered. I already mentioned restricted orbitals. But one may also use unrestricted orbitals, where $\alpha$ and $\beta$ electrons do not share the same spatial part. Here it is obviously necessary to include the spin quantum number.
One final remark: There are actually 5 quantum numbers that specify an orbital. Analogous to $l$ and $m_l$ of spatial angular momentum, in spin angular momentum there is $s$ and $m_s$. However for electrons we always have $s=\frac{1}{2}$, therefore explicit specification of $s$ is not necessary.

Well, we don't have what you call the "actual quantum chemistry" in our curriculum, nor do we have the mention of restricted or unrestricted orbitals. So should I consider myself correct and talk to the teacher or what?

"we don't have what you call the 'actual quantum chemistry' in our curriculum" that is why I considered high school level first, the other part was just a brief outlook what is beyond. But my main point is, that the question is imprecise and therefore leads to confusion. It should be reformulated, but there are different issue that need to be considered here. Honestly I don't know the best formulation (since I am no teacher). "How many quantum numbers are required to specify an atomic spin orbital?" would be my suggestion, but might be beyond the scope of high school chemistry.

Yup. Way beyond (probably). Thanks, anyways.

"...to specify the 'state' or 'properties' of an electron" maybe


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Interpretation of Raoult law

In attachment I have tried to derive Raoult's law by using the concept of equilibrium constant. There is nothing mentioned in my text book about how this equation comes about, so I decided to do it myself. I may like to know whether there is some mistake in what I have done.

No, it looks fine.

If you want the derivation to be more rigorous, it can be done by using the concept of Chemical Potential. You can look into any standard text book of Undergrad Chemistry to find the derivation or you can search in the web too.

Hey, we were having a discussion about this in class, and I went hunting on this website for questions and found this. Could you explain why concentration cannot be ignored in the second case?

The partial pressure of a component in a mixture is always less than that of the pure component at the same temperature. (Calculating the chemical potential shows that if it were not the case the mixture would be unstable compared to the pure liquid).  The question then becomes what is the partial pressure of the component A above a mixture $p_A$ related to that above the pure liquid $p_A^*$.  Raoult's law states that the ratio is equal to the mole fraction $x_A$ of A, $p_A/p_A^*=x_A$.
Physically, this is justified by considering that the chance of finding a molecule of A in the vapour, relative to the chance for the pure liquid, is equal to its mole fraction. This is reasonable if dealing with ideal mixtures and assuming that the vapour behaves as an ideal gas. Thus there is no derivation as such, we proceed directly from the physical argument to the law.

As in the case of Henry's law, it is possible to use Raoult's law to derive a more formal relation between an equilibrium constant and the standard Gibbs free energy for evaporation of solvent in a solution containing solvent at a mole fraction $\chi$.
At equilibrium between the vapor and solution, the chemical potential of the solvent in the two phases is identical:
$$\begin{align} \mu(l)&=\mu(g)\end{align}$$
For the gas phase (assuming ideal behavior) we can write that
$$\begin{align} \mu(g)&=\mu^\circ(g)+RT\log\left(\frac{p_i}{p^\circ}\right)\end{align}$$
Raoult's law states that the partial pressure and mole fraction $\chi$ of solvent in the solution are related as
$$p_i=p_i^\bullet \chi$$
where $p_i^\bullet$ is the vapour pressure of the pure solvent at the same temperature (compare this to Henry's law). As before - and as you show - this equation suggests the definition of an equilibrium constant expression. Introducing this last expression into that for the chemical potential results in
$$\begin{align} \mu(l)&=\mu^\circ(g)+RT\log\left(\frac{p_i^\bullet \chi}{p^\circ }\right)\end{align}$$
Defining
$$\begin{align} \mu(l)^\circ&=\mu^\circ(g)+RT\log\left(\frac{p_i^\bullet}{p^\circ }\right)\end{align}$$
we can see that
$$\Delta G^\circ_m = \mu^\circ(g)-\mu^\circ(l)=-RT\log\left(\frac{p_i^\bullet}{p^\circ }\right)=-RT\log\left(K_{eq}\right)$$
As in the derivation involving Henry's law, this expression relates a standard Gibbs free energy change and an associated equilibrium constant (for the equilibrium $\ce{solvent <=> vapor}$) provided we define the equilibrium constant as
$$K_{eq} = \frac{p_i^\bullet}{p^\circ} = \frac{(p_i/p^\circ)}{\chi}$$


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How can hydrogen and oxygen coexist in high temperature? Where did the oxygen come from?
A gasification worked with the combustion of bio pallets with production of syn gases. Pure oxygen and $\ce{H2O}$ were fed into the combustion reactor to combust the pallets with no opening of any syn gas outlets. Once the pallets got ignited enough, then there was no more oxygen input. A kind of gas reformer (catalyst) was equipped in the reformer. The temperature in the reactor reached more than $\pu{1000 ^{\circ}C}$. Then all the syn gases were collected and analyzed. What I found was that the composition of the gases were methane (43%), hydrogen (23%), oxygen (20%), and other gases (carbon dioxide and monoxide). I don’t understand how such a high volume of oxygen coexisted with hydrogen together in that high temperature reactor. Is it possible that those gases stay together without an oxidation reaction occurring? Where did the oxygen come from?

The fact that there is methane and carbon monoxide suggest that there is incomplete combustion. You do not mention if pure oxygen or air was used and was heat alone used or a flame ?

Was there any H2O? Did the composition change after being at 1000C for different times?

@porphyrin I am sorry I didn't describe it well. It was pure oxygen to burn the pallets. Once the pallets was ignited enough, then there was no input of oxygen anymore. I thought that all oxygen added were used for the oxidation of carbon. The reactor was a completely closed system. I am sorry I can't let you know the whole system of the reactor in detail. Also a gas reformer (catalyst) was equipped. The pallets turned to char coals and the syn gas was collected.

@James Gaidis  I actually tried to measure them at the second round running. I asked for an institute to analyze the syn gases after it cooled down (~300 degree Celsius). They failed to analyze them. Their GC couldn't read the gases. They said that the GC coould read the standard samples but not my  gas samples. They made errors to read. However, I found that the reactor produced alcohols (I guess it's methanol) condensed on the bottom. I think that the methanol might be a mixture with H2O.

I'll answer the question in general of how Hydrogen and Oxygen can coexist in metastable mixtures at higher temperatures, though without hazarding a guess for your specific case. Like you and others, I'd be very surprised if your bulk gas is actually well mixed at $\pu{900 ^{\circ}C}$ without exploding! It's far enough beyond the autoignition temperature, that an explosive reaction seems inevitable. However, metastable combinations of methane, hydrogen and oxygen in the $\pu{550 ^{\circ}C}$ range have been reported in the literature.
Hydrogen can be a surprisingly picky burner. I was told by a colleague who specialized in rocket propulsion that in between Hydrogen's upper and lower explosive limits are a host of regions where it strongly resists ignition, even above its reported autoignition temperature of ~$\pu{570 ^{\circ}C}$. According to her, the reaction between hydrogen and oxygen is quite complex with several important intermediates. Having concentrations in these non-explosive regions apparently allows the hydrogen-oxygen mix to be metastable for long periods. This would of course not extend to long time periods at $\pu{900 ^{\circ}C}$!
More in my territory, there's also many chemicals which inhibit the slow combustion (non-explosive) of hydrogen and oxygen up through the $\pu{500 ^{\circ}C}$s. The most common is $\ce{KCl}$, which strongly inhibits the reaction of oxygen and hydrogen in non-explosive conditions allowing a metastable mixture to exist. Walls of vessels can also similarly inhibit reaction. The reaction mechanism apparently stems from $\ce{KCl}$ and other surfaces' abilities to consume reaction intermediates of the combustion such as monatomic Hydrogen:
$$\ce{KCl + H^. -> K^. + HCl}$$
$$\ce{SiO2 + H^. -> SiO(OH)}$$
or direct adsorption on metal surfaces like nickel, palladium, etc.
After doing a bit of digging, methane is apparently a superb inhibitor of hydrogen and oxygen combustion. The effect is explained in a 1959 paper by R. Baldwin, et al. (see following reference). The paper explains that when methane is above a sharp critical concentration it drastically raises the autoignition temperature of hydrogen's explosive reaction and almost completely eliminates slow combustion in non-explosive conditions. In other words, when enough methane is present, the hydrogen and oxygen mixture either does nothing or explodes completely. From data taken in systems between $\pu{450\! -\! 550 ^{\circ}C}$, there was a very clear relationship of methane concentration vs. explosive limit taking the form of (values approximate):
Explosive Limit (in $\pu{K}$) = $\frac{\pu{3350K}}{2.3-\log_{10}(X_\ce{CH4})}$
where $X_\ce{CH4}$ is the mole fraction of methane, studied in the 0.01 - 0.1 range (significantly lower than yours). For this particular heuristic, the mole fraction of hydrogen was 0.28, oxygen 0.14, and the balance was nitrogen. The paper suggests the mechanism is non-equilibrium formaldehyde formation.
As a final note, please be very cautious when handling or recreating these mixtures. Just because the hydrogen isn't burning, doesn't mean it can't explode. A lab down the hall from mine was fortunate to loose only a wall to a hydrogen evolving system which ran fine for months and then, quite suddenly, decided to explode.
Best of luck learning more about your system! It sounds fascinating.

Reference:
R. R. Baldwin, N. S. Corney, R. W. Walker, “The inhibition of the hydrogen + oxygen reaction by methane,” Trans. Faraday Soc. 1960, 56, 802–814 (DOI:10.1039/TF9605600802).

Great answer. I leaned something today. I corrected it (mostly formatting), and included the details of the reference. In future, please include the references (as shown in your text) even though they were not available to public. :-)

Thank you very much for the help with the formatting! It's been taking me a while to learn.

@Michael Irving Thank you very much for the great answer. I am so excited to learn it from you. I can step forward to understand my system better. I expected from this pilot test to get a mixture of methane-hydrogen-carbon dioxide and monoxide, but never oxygen. It was quite surprising to me. I thought I got a false result. It's still a mystery of production of the oxygen, though. I carefully guess that steam reacted with the metal oxide/metal catalyst (gas reformer) and split to hydrogen and oxygen. I got to figure it out. Thank you again. This excitement won't put me asleep tonight.


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How to remove wash anthracene oil/creostae oil from stainless steel tanker?
I once loaded anthracene oil/creosate Oil in my stainless steel tanker and I had to transport it to some factory. The material was inside the tanker for throughout the journey that was of around 8 days.
The material got unloaded at the said factory. However I had another order to load Formaldehyde from the same city. I took my truck to the usual tanker service station to wash the tank with Detergents/water/diesel/scrubbing and everything which is quiet a normal procedure after and before any other liquid chemical.
However I have never had trouble with washing any other material as much as I am having with anthracene oil/creosote oil. It just doesn't seem to come off , the servicing person had severe respiratory problem when he went inside the tank.
Even after washing with several detergents, fast water streams and other things... the chemical is just not leaving the surface of my tank. One more thing, I think the service would have been much easier if the fumes/gas inside the tanker weren't so strong and scorching to the skin of the service person. He straight away told us to never come back with the same material.
Kindly let me know how to make sure to remove the essence of the gas/fumes left in the tank due to 8 days of material in the tank, how to remove the material from the surface of the tank and how to make this process well by using any other chemicals.

Do you have any information on what the components of the creosote oil are? What is its origin?

No I don't . I am not the manufacturer. Just a transporter. Do you think mixing water with caustic soda powder and let it remain inside the tank for few hours can help?

It depends on the constituents; wood-derived creosote oil contains a lot of phenols which will dissolve in sodium hydroxide solution. Coal-derived creosote oil is mainly polyaromatics which will not, it would be more likely to dissolve in toluene or gasoline. I would take sample of it and see what it does with sodium hydroxide or toluene/gasoline.

It's coal derived. My main concern is to remove the gas from the tank that gets entrapped in the tank because of several days of material inside the tanker during transit time..what could be done for that?

Safest to flush with inert gas, nitrogen or carbon dioxide.

That is a too costly procedure. Not feasible.

@Kushgra Gupta two statements   "the fumes/gas inside the tanker weren't so strong and scorching to the skin of the service person"   and  " he servicing person had severe respiratory problem when he went inside the tank " make me wondrous ,are you using chemical protection suits with face masks while cleaning.

Anthracene oil (CAS registry number: [90640-80-5]): Both European Chemical agency (ECHA) and US Toxicology Date Network (TOXNET) classify anthracene oil as:

A complex combination of polycyclic aromatic hydrocarbons obtained from coal tar having an approximate distillation range of $\pu{300 ^{\circ}C}$ to $\pu{400 ^{\circ}C}$ ($\pu{572 ^{\circ}F}$ to $\pu{752 ^{\circ}F}$). Composed primarily of phenanthrene, anthracene, and carbazole.

Anthracene oil  is also classified as a carcinogen the risks to human health and to the environment as well (though the relevant symbol is not issued).

Creosote oil (CAS registry number: [8001-58-9]): Creosote oil is a synonym for coal tar creosotes. It is a distillation product of coal tar. Creosote oil is an oily liquid consistency and range of color from yellowish-dark green to brown.  It is a mixed compound, composed primarily of aromatic hydrocarbons, anthracene, naphthalene, and phenanthrene derivatives. At least 75% of the coal tar creosote mixture is polyaromatic hydrocarbons.
Creosote oil is heavier than water (specific gravity: 1.07-1.08) and has a continuous boiling range beginning at about $\pu{200 ^{\circ}C}$  (range: $\pu{194\!-\! 400 ^{\circ}C}$).  They are slightly soluble in water, but miscible with alcohol and ether (Ref.1).
Koppers' Safety Data Sheet has following warnings for creosote oil:

Hazard Statements:

Causes eye irritation.
May cause cancer.
May cause respiratory irritation (No wonder your servicing person had severe respiratory problem when he went inside the tank).
Very toxic to aquatic life.
Toxic to aquatic life with long lasting effects.

It also listed following precautionary statements for prevention:

Avoid breathing dust/fume/gas/mist/vapors/spray.
Wash thoroughly after handling.
Wear protective gloves/protective clothing/eye protection/face protection.
Obtain special instructions before use.
Do not handle until all safety precautions have been read and understood.
Use only outdoors or in a well-ventilated area.
Avoid release to the environment.

I put all this effort to show that how dangerous your situation is and how ignorant you have been. Your first mistake is not receiving all special instructions for handling these kind of chemicals from responsible people (see above bold-face precaution). In your case, I assumed the manufacturer of the chemicals should provide these details. What you should have done now is go back to the manufacturer, and request the instruction for clean-up procedure safely. And be aware that whoever the person following that instructions, should wear with all personal protective equipment (PPE) required, specially suitable breathing equipment.
Alternatively, since the both oils are viscous, you may slant your tanks backward to make them drain out to safety container (for long time) before starting to wash them (assuming drainage is in the back). Once drained, a person wearing PPE can go inside and wipe out inner walls with cloths soaked with alcohol several times, since both oils are miscible in alcohol. Also, kerosene or gasoline or even coconut oil may work as well (please dispose those cloths accordingly). Finally, you should use appropriate detergent (or soap) with water to remove trace of remaining hazard chemicals (if there any) and washing organic material used.
References:

B. F Hibbs, J. Georges (Agency for Toxic Substances & Disease Registry), Toxicological Profile for Creosote (updates):  Wood Creosote, Coal Tar Cresote, Coal Tar, Coal Tar Pitch, and Coal Tar Pitch Volatiles; Forwarded by Dr. D. Satcher; Prepared for US Department of Health & Human Services, 1996.


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Y-axis of the reaction co-ordinate graph
In the graph of the reaction co-ordinate, is the $y$-axis that is used  $∆H,$ $∆U$ or $∆G$?

Energy, the context depends on what type.

Generally it is energy

Why though and what exactly does ∆U represent here. Like what's the difference between ∆U and ∆H , on terms of a reaction

See https://en.wikipedia.org/wiki/Reaction_coordinate and https://en.wikipedia.org/wiki/Energy_profile_(chemistry)

Single molecules have no entropy, so Gibbs makes no sense. If you want to use a molar quantity, it's ∆H.

Single molecule reaction
Some diagrams refer to a single molecule reacting. If that is the case, several quantities are not defined or ill-defined:

pressure
entropy of mixing
temperature

In those diagrams, the best label would be potential energy, and the energy difference between reactant and transition state should be labeled activation energy.
Ensemble reaction
If the diagram refers to an ensemble reacting (reaction mixture with sufficient number of molecules to have macroscopic properties), the y-axis could correspond to Gibbs energy, enthalpy or internal energy.
If the y-axis represents Gibbs energy, the energy difference between reactant and transition state should be labeled $\Delta G^\ddagger$. Those diagrams make sense when discussing how equilibrium constants remain unchanged but reaction rates change when a catalyst is added.
If the y-axis represents enthalpy or internal energy, the energy difference between reactant and transition state should be labeled $E_a$ for activation energy. Those diagrams make sense when discussing reaction rates in the context of collision theory.
What is most popular?
Looking at about 50 images of reaction energy diagrams found in a google image search, I found a wide variety, and lots of mixing and matching (i.e. incompatible quantities in a single diagram). This tells you that the diagram is conceptual, and the concepts are not always clearly articulated in diagrams.

Following is an example of Reaction Coordinate Diagram. I labelled all necessary details for your convenient. Please note that the Blue path is for the same reaction when catalyst is involved. The difference is the activation energy, which is lower in the presence of a catalyst.


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Physical meaning of tm
The definition of dead time ($t_\mathrm m$) is mentioned as the time taken for elution of an unretained species. It is also mentioned that the retention time ($t_\mathrm r$) is sum of the time analyte spends in mobile ($t_\mathrm m$) and stationary phases ($t_\mathrm s$). But how can time spent by analyte in mobile phase be equal to the time required for an unretained species to pass through (dead time, $t_\mathrm m$)?
Both seems independent of each other. An analyte's time in the different phases is dependent on its affinity while the unretained species takes the same amount of time no matter what.
Please explain how they are same.

The said 'duplicate' differs in the fact that it is with regard to the associated formula while the gravity of this question lies in what tm means.

The analyte spends the same time in the mobile phase as it tooks for the mobile phase to pass through the column.

An analyte's time in the different phases is dependent on it's affinity.

The time $t_\mathrm{m}$ is given.
The ratio of an analyte's time in the different phases is dependent on it's affinity and is equal:
$$\frac{t_\mathrm{s}}{t_\mathrm{m}}=K_\mathrm{D} \cdot \frac{V_\mathrm{s}}{ V_\mathrm{m}} $$.
where

$t_\mathrm{s}, t_\mathrm{m}$ are time is stationary and mobile phase
$K_\mathrm{D}$ is the phase distribution constant
$V_\mathrm{s}, V_\mathrm{m}$ are volumes of stationary and mobile phase.

I couldn't understand your first sentence.

The analyte spends the same time $t_\mathrm{m}$ in the mobile phase as the carrier(mobile phase) in the column.

Poutnik, I am afraid your first statement is not correct. The partition ratio determines how much time an analyte spends in the mobile phase and the stationary phase.

In fact, both, as the first ( affinity, or distribution constant) being an intensive parameter, the second ( partition ratio ) an extensive parameter. Note also I have said determines, not equals to. And also, I i have followed the OP reasoning.

@M._Farooq Whenever the analyte is in the mobile phase, it moves at the same speed as the carrier. The time to go from start to finish (not counting the time stuck in the stationary phase) is the same. Different analytes spend more or less time in the stationary phase, so $t_m$ is constant and $t_s$ is variable, down to zero for unretained species.


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What determines 3D printing material compatibility?
3D printing, specifically the type common for consumers FDM or FFF, involves laying down thin (typically 0.1 to 0.2mm) layers of molten plastic. Modern consumer printers can use various filaments: PLA, ABS, Nylon, polycarbonate, PETG, ASA, etc.
PLA printed on top of PLA will make a decently strong bond, PETG on top of PETG will make a noticeably stronger bond. PETG on top of PLA will effectively not bond at all and can be easily pulled apart.
What's going on here? What determines layer adhesion? Can polymers be "forced" to bond to one another (overheating? pressure?)? Ideally I'd like to put together a compatibility matrix for mixing 3D printing filaments, but I don't even know enough to know what to search for.


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Will ferrocene undergo electrophilic aromatic substitution?
Ferrocene has two structures (1 and 2). It is bonded to the aromatic cyclopentadienyl anion. Aromatic species such as benzene, cyclopentadienyl anion undergo electrophilic aromatic substitutions.
Then, will ferrocene undergo  electrophilic aromtic substitution?
Which conformer between 1 and 2, would preferentially undergo Friedel Crafts acylation reaction?

The question about which conformation is more reactive is irrelevant (and drawing the ferrocene molecule as an ionic compound is incorrect: the bonding is covalent, though complicated). The barrier to rotation is very small so there will be no significant preference between them at room temperature.

Due to the aromatic character of the cyclopentadienyl ligands, ferrocene can undergo electrophilic aromatic substitution reactions typical of aromatic compounds such as benzene. Ferrocene shows very high reactivity towards electrophiles, making it more closely comparable to phenol than to benzene, references and examples here and here.

This does not answer weather conformer 2 or conformer 1 is more reactive towards acylation..In my question i have already mentioned that cyclopentadienyl anion is aromatic.So cyclopentadienyl  does undergo electrophillic aromatic substitution.

Ferrocene is formed from two planar cyclopentadienyl rings, not anions. The two rings are staggered with respect to each other, but are otherwise equivalent so there is no answer to your question.

@Waylander true, but what matters is not that the most stable conformation is staggered but that the barrier to rotation is very small and all conformations will readily convert under normal reaction conditions.

@matt_black Thank you for the additional information

Refer Inorganic chemistry by Huheey (4th edition, Page No. 646)
They have proceeded with conformer 1 and successive acylation on both rings will result in acyl group attached on the same side of both CP rings

However, in my notes, i have proceeded with conformer 2 and the 2 equivalents of acyl are attached on opposite sides of the two CP ring.

The question about stability vs reactivity here.
Stability - The doubly acylated product of conformer 2 will be more stable (Less strained transition state as well as trans looking product and thus better packing in crystal).
Reactivity - I don't have the answer, neither did i find its explanation anywhere. So sorry i can't help you there.

Just because that is how the molecule is drawn in a textbook means nothing. The barrier to rotation is very small and, essentially, irrelevant to the ring reactivity. So, the "other side" is meaningless after double acylation.

In the Solid state, Staggered conformation will be present predominantly (since symmetry will lead to more efficient packing) while in the Gaseous state, an equilibrium between staggered and eclipsed form exists because the energy gap between them is very less (4.2 kJ/mol). So, essentially, the question comes down to which conformer will be more reactive in the Gaseous state. I haven't found the answer to that, but whatever i did find, i updated here.

Given the low barrier to rotation, it is unlikely that–even if the reactivity were different–you would notice it in a real reaction. And most reactions of this type are done in solution not solid or gas states, so those are irrelevant.

The dominance of the eclipsed ferrocene conformer in solutions at room temperature has been reported by "Differentiation of ferrocene D5d and D5h conformers using IR spectroscopy" by N. Mohammadi et al.


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How are glycosidic linkages formed?
Consider glycosidic linkages in disaccharides.
I know a condensation reaction occurs, with the loss of water.
But which OH is lost? The O in the linkage belongs to which monosaccharide?
Example: take sucrose, the O in the linkage belongs to fructose or glucose?
I couldn't find any literature regarding this. Any help is appreciated .

In sucrose molecule the  glycosidic linkage is formed by the OH group on the anomeric carbon of α-D-glucose and the OH group on the anomeric carbon of β-D-fructose.

source :
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Map%3A_Organic_Chemistry_(McMurry)/Chapter_25%3A_Biomolecules%3A_Carbohydrates/25.08_Disaccharides
As seen, in either alpha d-glucose or beta d-fructo furanose may react to give sucrose.
Let us examin.
Alpha D-gluco Pyranose gives an intermediate 1 and 2 by protonation at anomeric carbon.

Similarly  Beta D-fructo furanose gives an intermediate 3 and 4 by protonation at anomeric carbon.

If you examine , structure 3 is more stable then structure 4.

The negative inductive effect of CH2OH in fructofuranose destablizes this intermediate 4.
Hence structure 2 , an intermediate of aplha D-gluco pyranose (more stable) attacks beta D-fructo furanose togive sucrose.

The above diagrams are author drawn in CHEM DRAW

This entire answer appears to be copy-pasted from Bruice's Organic chemistry. If you are quoting third-party sources, or using any graphics/data/plots that are not made by you, put an appropriate reference and denote the part that is quoted with proper markup (>). Keep in mind that if you don't that you may eventually be accused of plagiarism.

@user226375 your question is answered now,

@andselisk you were correect .This was taken from  Bruice's Organic chemistry.(missed reference ).How ever the answer is rewritten ti the question asked .The diagrams are mine , drawn on CHEMDRAW.

@andselisk I MISSED QUOTING REFERENECE IN HURRY TO ANSWER .MY APOLOGIES .THIS WAS NOT DONE ON PURPOUSE.


text:
Oxidation of diols to dialdehydes
How could 1,4-butanediol be converted to 1,4-butandial (succinic aldehyde)? I thought of simple oxidation with $\ce{CuO}$ but apparently it yields gamma-butyrolactone according to erowid
and this paper$\ce{^{[1]}}$. Could increasing the pH result in the formation of 1,4-butandial?
$\ce{[1]}$: Romaric Gérardy, Romain Morodo, Julien Estager, Patricia Luis, Damien P. Debecker, Jean-Christophe M. Monbaliu, Sustaining the Transition from a Petrobased to a Biobased Chemical Industry with Flow Chemistry, Topics in Current Chemistry, 10.1007/s41061-018-0222-3, 377, 1, (2018). (DOI)

Alternatively, I am quite sure that PCC would stop at the aldehyde stage.
Here, a complex diol with a 4-carbon chain is oxidized to dialdehyde with the "similar" PDC
https://www.sciencedirect.com/science/article/abs/pii/S0957416610008530?via%3Dihub

DMSO-based oxidation is another valid and widely cited in literature approach (eg. Swern Oxidation)

Difficult stuff to handle as it is prone to polymerisation. Best stored as 2,5-dimethoxytetrahydrofuran which is functionally equivalent for synthetic purposes.

True; speaking of that, wouldn't it be easier to convert easily accesssible tetrahydrofuran into 2,5-dimethoxytetrahydrofuran directly? How could I substitute a methoxy group into 2nd and 5th position?

THF can be chlorinated at low temps with UV light below -10C to 2,5-dichloro-THF (ref here https://www.chempap.org/file_access.php?file=311a109.pdf) and then reacted with MeONa/MeOH to give 2,5-dimethoxy-THF. I would not call this an easy option, given that 2,5-dimethoxy-THF is widely commercially available

Chromium-based oxidations will give butyrolactone. Swern-type oxidations are a better choice because both hydroxyls are derivatized before base (Et3N) effects the formation of the aldehyde groups.

Oxone and TEMPO,  maybe with PTC if phases are immiscible?


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Why does a 4-bonded nitrogen atom appears negatively charged after an optimization?
I have a molecule bearing a quaternary ammonium and I optimized it using M062X (also Hartree-Fock) using basis sets 6-31G(d,p) and 6-31G+(d,p) respectively. After optimization, I expect that the charge on the quaternary nitrogen would be positive. However, it's negative.
I ran the Gaussian to compute NBO charges and analysed the checkpoint files as well as log files. Nowhere could I see a positive charge on any nitrogen atom.
Why is this happening? And how can I obtain a positive charge on the quaternary nitrogen?
Many thanks in advance.

Where is the positive charge, then?

What charge to the hydrogen atoms carry? How does the charge distribution differ from a neutral ammonia molecule?

Run a calculation with neutral ammonia, and you will see it, indeed. I just verified by running the same calculation.

Also, to understand the reason behind this behaviour, it could be a wise suggestion that of considering the concept of electronegativity

@IvanNeretin looks like only the total charge is positive.

Now that's quite outside the realm of things that can be. Isn't the total charge the sum of atomic charges?

@IvanNeretin yes, that's what I also assume.

@The_Vinz well, I think this is surprising. Nitrogen in neutral ammonia should be negative, if not neutral, as it's quite electronegative. For a quaternary nitrogen, however, it should be positively charged as its lone pairs are also shared by a an additional hydrogen.

Indeed, the electron pair is shared with a proton (charge +1). But nitrogen is still more electronegative than hydrogen, so two effects occur: 1.the proton sucks electrons from N's pair, and its charge tends to decrease 2.Nitrogen, electronegative, "wants" back its electrons: so its charge increases, but to a limited extent.
As you can see, in neutral ammonia, N is around -1. When ammonia is protonated, charge delocalizes over all its atoms because of the two effects described.

Another system that you could study, with different electronegativities but similar processes occurring is methylene (:CH2) vs its protonated form, the methyl cation (CH3+)

@The_Vinz thank you very much. What I later realized is that my molecule bears a cationic imine, not a quaternary nitrogen. However, the same thing happens here. I expected a positive charge, but it's negative. I think your explanation still applies.


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Can R-3-methyl-4-heptanone be enantioselectively synthesised from 4-heptanone?
I wonder if it is possible to do the following reaction with its enantioselectivity:

I am trying to do it via enolate, but I don't know how to make only the R product or if it's even possible to distinguish between R or S simply using enolates.
There's another way?
Thanks!

Enders SAMP/RAMP hydrazone chemistry will do this(1).
The Wikipedia article on this technique is a good introduction:

(1): Corey, E. J.; Enders, D. (1976). "Applications of N,N-dimethylhydrazones to synthesis. Use in efficient, positionally and stereochemically selective C-C bond formation; oxidative hydrolysis to carbonyl compounds". Tetrahedron Letters. 17 (1): 3–6.


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Does this reaction require catalyst/solvent?
Planning on synthesising 2-formylphenyl benzoate by reacting benzoyl chloride with salicylaldehyde. Since both Benzoyl chloride and salicylaldehyde is liquid, will it require a catalyst/ solvent? Or do I just add them both in a conical flask and let a magnetic stirrer do the job?

Phenols react with aromatic acid chloride in the presence of a base such as $\ce{NaOH}$ at room temperature. Such reactions are called Schotten-Baumann type reactions (Ref.1&2; both are in German).

The Schotten-Baumann reaction is an organic reaction used to convert an acyl halide or anhydride to an amide, if reacted with an amine and base, or an ester, if reacted with an alcohol and base. The reaction with the amine begins with the nitrogen attacking the carbonyl carbon of the acyl halide which rearranges to kick out the halide. Deprotonation with the base then provides the final amide product. The reaction with the alcohol would happen in a similar fashion to give an ester.

Thus, as an example, you can follow the following procedure: Dissolve phenol in 5-10% aqueous sodium hydroxide solution (choose the percentage to have alkaline solution at the end) and add benzoyl chloride. Stir the mixture until the smell of benzoyl chloride has disappeared (warm if necessary). The resultant phenyl benzoate can be separated solvent extraction (ethyl acetate/water), washed with saturated bicarbonate solution followed by cold water. Then, dry organic layer and concentrate by rotary evaporation under reduced pressure to yield crude product.
Alternatively, if you preferred to do the reaction under solvent-free conditions you may follow the procedure given in Ref.3 using same starting materials. The abstract of the reference states that:

An efficient solid-phase benzoylation of phenols and alcohols was developed under microwave irradiation. A stoichiometric amount of benzoyl chloride was sufficient to carry out the reaction. This benzoylation features short reaction time, good yields, and easy workup procedures. Furthermore, the scope of the reaction was extended to prepare 3,5-dinitrobenzoyl derivatives of alcohols.

They have prepared 2-formylphenyl benzoate by this method and presented following spectroscopic data:

Liq. White oil, 85% yield. UV $\lambda_\mathrm{max}: \pu{265 nm}$. IR ($\ce{KBr}, \nu_\mathrm{max}, \pu{cm^{-1}}$): 3339, 3010, 2843, 2734, 1733, 1698, 1688, 1589, 1547. $\mathrm{^1H~NMR}$ ($\ce{CDCl3}, \pu{500 MHz}$) $\delta$ 6.97-7.02 (q, 1H), 7.25 (d, J 8 Hz, 1H), 7.34 (t, 1H), 7.41-7.55 (m, 1H), 7.62-7.69 (m, 2H), 7.94–7.96 (dd, J 1.5 Hz, J 1.5 Hz, 1H), 8.23 (d, J 8 Hz, 2H), 10.22 (s, 1H). $\mathrm{^{13}C~NMR}$ ($\ce{CDCl3}, \pu{125 MHz}$) $\delta$ 119.8, 120.7, 123.6, 126.5, 128.7, 130.2, 133.7, 135.3, 137.0, 152.3, 161.6, 165.0, 188.4, 196.6. HRMS: m/z calcd. for $\ce{C14H10O3}$: 227.0705 (M + H); found 227.0708.0.

References:

C. Schotten, “Ueber die Oxydation des Piperidins,”  Ber. Dtsch. Chem. Ges. 1884, 17(2), 2544–2547 (https://doi.org/10.1002/cber.188401702178).
E. Baumann, “Ueber eine einfache Methode der Darstellung von Benzoësäureäthern,”  Ber. Dtsch. Chem. Ges. 1886, 19(2), 3218–3222 (https://doi.org/10.1002/cber.188601902348).
S. Chakraborty, A. Saha, K. Basu, C. Saha, “Solid-Phase Benzoylation of Phenols and Alcohols in Microwave Reactor: An Ecofriendly Protocol,” Synthetic Communications 2015, 45(20), 2331–2343 (https://www.tandfonline.com/doi/abs/10.1080/00397911.2015.1078899).

Just three points why omitting a solvent might not be a good idea without experimental evidence that support the contrary:

There is a reason many organic syntheses are run with reagents in a concentration in range of 0.1 to 1 mol/L; to moderate both rate (kinetics) and heat (thermodynamics) of a reaction.  The omission of a solvent for an already exothermic reaction may lead to a "runaway", where the heat suddenly generated transforms your products into a tar.  The issue regarding the heat balance often becomes important when scaling the reaction to larger batches in process chemistry.
Secondly, your product target equally is an aldehyde.  At least potentially there already is a problem (chemoselectivity) discerning one of the starting materials (salicylic aldehyde) from your product (2-formylphenyl benzoate) in respect to benzoyl chloride.  In case of a highly concentrated (neat) reaction lacking any solvent, if the reaction of the acyl chloride with the aldehyde is exothermic, you may reach a temperature of your reaction mixture so elevated that both reaction pathways became accessible instead of only one (and other not envisioned here, too).
According to the information about who sells this pricy commercial product at PubChem and how -- per gramm, instead of per millilitre -- it appears safe to assume that the product of your reaction is a solid.  Having a solvent dissolving both your starting materials as well as your product equally may offer you a reaction more complete, because the progress of the reaction is not hampered by an interface of a liquid phase (your starting materials) and a solid (your product).  It equally may lead you to a product more clean, since these grains won't contain (this much) of starting materials (occlusion).


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Are there any surface treatments that would make calcium usable for metallurgy?
Calcium is the best conductor for its weight, but the soluble powdery oxide makes it impractical for use in power lines.  Are there any surface treatments or alloying processes that could help?
(Calcium phosphates are reasonably insoluble, but I don't know how well they bind to the metal underneath; lithium-aluminium alloys are used in a lot of aerospace applications despite the strong reactivity of lithium with air and water)

This has been flagged as "too broad", and I lack the expertise to decide whether or not it is such. Still, I'd appreciate it if someone could comment on their close vote.

Calcium is so reactive that (1) cladding seems to be the only chance, (2) getting it pure is going to be horrendously expensive.  Also if you were to try alloying you'd lose some of the conductivity you want.

That sounds like an interesting current surge situation, since Ca melts so low, burns oh so bright and produces toxic fumes. Using Ca for electric conductance sounds impractical, unnecessary and possibly hazardous.

Utilities have investigated the use of sodium for electrical transmission (which sounds similar). But they used thick wires coated with something like polyethylene which would ,presumably, also work with calcium. But the problem of what happens if the wire breaks still exists.


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What is the pH at both equivalence points of titration between diprotic tartaric acid and NaOH?
Here's all the info I have and my attempt in solving the question:
The chemical equation for the titration:
$$\ce{C4H4O6 + 2NaOH -> Na2C4H4O6 + 2H2O}$$
Here are the 1st and 2nd $\mathrm{Ka}$ value for tartaric acid:
$$K_\mathrm{a1} = 9.2 \times 10^{-4}, K_\mathrm{a2} = 4.3 \times 10^{-5}$$
Molarity of tartaric acid = $5 \times 10^-4$
Standard solution of $\ce{NaOH }$ was made using $\pu{0.98 g}$ of solid $\ce{NaOH}$ dissolved in $\pu{250 ml}$ of water. The molar mass of $\ce{NaOH}$ is $\pu{39.997 g/mol}$.
This is what I have done so far.
n (C4H4O6) = cv = (5 × 10-4 M) (25ml ÷ 1000L) = 0.02 mol
n (NaOH) = m ÷ M = (0.98g) ÷ (39.997 g/mol) = 0.0245 mol (4dp).
C (NaOH) = n ÷ v = (0.0245 mol) ÷ (250ml ÷ 1000L) = 0.098M
N (NaOH) at equivalence point = cv = (0.098M) (0.204L) = 0.02 mol
C4H4O6(aq)+ 2NaOH(aq) → Na2C4H4O6(aq) + 2H2O(l)
I 0.02         0.02           0            -
C - 0.02      - 0.02       + 0.02          -
E   0           0            0.02          -
C (Na2C4H4O6) = n ÷ v = 0.02 ÷ (0.025L + 0.25L) = 0.0727M (4dp)
Na2C4H4O6(aq) + H2O(l) → C4H4O4Na2^+(aq) + OH^-(aq) below is ice table, stackexchange won't allow me to seperate the values.
I  0.0727          -           0              0
C   - x            -         + x             + x
E 0.0727 – x       -          x               x
Kb = Kw ÷ Ka = (1 × 10-14) ÷ (9.2 × 10-4) = 1.087 × 10-11
Kb = [C4H4O4Na2+] [OH-] ÷ [Na2C4H4O6] 1.087 × 10-11 = x2 ÷ 0.0727 x = √ ((1.087 × 10-11) (0.0727)) x = 8.89 × 10-7
pOH = - log [OH-] = - log (8.89 × 10-7) = 1.9
pH = 14 – pOH = 14 – 1.9 = 12.1

The question is based on the wrong terminology. Rather than the endpoint, the question would seemingly be asking about the equivalence point.

Thank you, you are right. I've just changed it. Would you by any chance know how to approach this question, as I still don't quite understand it

For the 1st equivalence point, the tartaric acid $\ce{H2A}$ is titrated to $\ce{NaHA}$
For the solutions of acid salts, we can with advantage use the formula
$$\mathrm{pH}=\frac{\mathrm{p}K_\mathrm{a1}+\mathrm{p}K_\mathrm{a2}}{2}$$
that can be derived from equilibrium
$$\ce{2 HA- <=> H2A + A^2-}$$
For the 2nd equivalence point there is  $\ce{Na2A}$ solution, which is hydrolyzed with equilibrium
$$\ce{A^2- + H2O <=> HA- + OH-}$$
With neglecting OH- from water hydrolysis, $[\ce{HA^-}] = [\ce{OH^-}]$
$$\begin{align}
K_\mathrm{a2}&=\frac{[\ce{H+}][\ce{A^2-}]}{[\ce{HA-}]}\\
\mathrm{p}K_\mathrm{a2}&=\mathrm{pH} - \log{[\ce{A^2-}]} +\log {[\ce{HA-}]}\\
&=\mathrm{pH} - \log{[\ce{A^2-}]} +\log {[\ce{OH-}]}\\
&=\mathrm{pH} - \log{[\ce{A^2-}]} -\mathrm{pOH}\\
&=\mathrm{pH} - \log{[\ce{A^2-}]} -14 + \mathrm{pH}\\
\mathrm{pH}&=7+\frac12\left(\mathrm{p}K_\mathrm{a2} +\log{ c_\mathrm{\ce{Na2A}}}\right)\\
\end{align}$$
The molarity of $\ce{NaOH}$ solution $c_{\ce{NaOH}}=\frac{m}{M\times V}=\pu{0.09801  mol/L}$
The volume of $\ce{NaOH}$ solution for given volume of tartaric acid solution
$$V_{\ce{NaOH}}=V_{\ce{H2A}}\times \frac{2\times c_{\ce{H2A}}}{c_{\ce{NaOH}}}\\=0.0102\times V_{\ce{H2A}}$$
The final concentration
$$\begin{align}
[\ce{Na2A}]&=\frac{[\ce{H2A_\mathrm{init}}]\cdot V_{\ce{H2A}}}{V_{\ce{H2A}} +V_{\ce{NaOH}}}\\
&=\frac{[\ce{H2A_\mathrm{init}}]\cdot V_{\ce{H2A}}}{V_{\ce{H2A}} +
0.0102  \cdot  V_{\ce{H2A}}}\\
&=\frac{[\ce{H2A_\mathrm{init}}]}{1+ 0.0102}\\
&=5\times 10^{-4}/1.0102\\
&=\pu{4.9495\times10^{-4} mol/L}\\
&\overset{\mathrm{rounding}}=\pu{4.95\times 10^{-4} mol/L}\\
\end{align}$$

Thanks so much. I understand now. Just two questions though. For the volume, wouldn't it be 0.0102 * VH2A. Also, why is it 1.0103 in the formula with final concentration?

It was, perhaps unluckily autowrapped. The acid solution is diluted by NaOH addition. Plus, perhaps rounding issue. Anyway 5 valid digits is excessive precision.

Hello, sorry just a little confused. So volume of NaOH is meant to be 0.0102 instead of 0.0103?

and why is 1 added to 0.0103? Lastly, where is the last equation for final concentration derived from?

@Lisa As I have said, rounding error. See the fixed and updated answer.

Okay, that makes sense. I just don't get what the V (I understand it may be volume, but volume for what?) means in the equation for final concentration. I understand that V equals 1 but do not understand where the 1 value came from. I understand everything else now thank you just that V I do not know.

@Lisa Updated - better ?


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Tabulated absorption spectra of greenhouse gases?
Are there any free sources of tabulated absorption spectra of greenhouse gases from UV/Vis to far infrared (say 70 micrometers)?
This link shows the net absorption over the whole atmosphere (at least as far as I understand it), but I would be interested in the absorption by an infinitesimally small volume/distance. It would suffice to have the spectra for just one temperature, like $\pu{273K}$ or so.
On the NIST site I have found some information, for example IR spectrum of $\ce{H2O}$ vapor, but apart from the limited spectral window the spectrum is pretty useless for making a radiation simulation because concentration info was missing (as said in the disclaimer).
Is there still hope of getting accurate spectra suitable for understanding molecule specific greenhouse effects?

Related: https://www.ncbi.nlm.nih.gov/pmc/articles/PMC6174548/

Remote sensing of the simple (few atoms) greenhouse gases works by comparing the measurement spectra with calculated spectra. These calculations are very precise and probably the reason why you do not find so many measured reference spectra.
The reason for calculation is that the atmosphere has pressure, concentration and temperature gradients which even depend on path angle(s). So sensing by comparison with lab-measured spectra requires many spectra under different conditions and sophisticated interpolation schemes (Which is nevertheless done for the more complex greenhouse gases or the HFCs)
One important database for greenhouse gases is HITRAN. Since the simulation code is not so trivial to implement I would suggest you to use a website which does the calculation for you. e.g.:
Spectral Calc or HITRAN on the Web
There you enter your gas parameters (pressure, temperature, concentration, path-length) and so on, and it will calculate the spectrum in (almost) any wavelength range for you. Of course the accuracy may vary with wavelength range, there you would need to check the accompanying literature. Nevertheless, the strong absorption bands of the greenhouse gases are well researched and the data is of high accuracy. E.g. for the 4.3 µm absorption band of $CO_2$ we probably have the best researched spectral data of all gases, with accuracy of probably better than 1% if not 0.1 %.
For the more complex molecules, where a line-by-line simulation is not available, you find high-quality measured spectra on the HITRAN website (-> "Absorption cross sections")

So when you say "simulation code is not so trivial" you probably don't mean the part of calculating vibration spectra in GAMESS for example, do you? I don't do spectroscopy on a professional basis, so bear with me. But I remember from my physics studies back then, that there are several line broadening effects (pressure, doppler, impact), some more important than others. Is this what you mean is difficult, calculating line width from single molecule spectra?

@oliver Yes there are multiple broadening effects. The coefficients for this are also tabulated in the database, but putting everything correctly together with the correct temperature corrections requires some physical understanding. It is not impossible of course as there is documentation on the HITRAN website, but it is incomplete with gaps needed to be filled in on your own. If you want to understand how simulation is really done, you need to download some source code (e.g. HAPI found at the HITRAN website) that performs this simulation and look at it.

So I would say the difficulty is really putting everything correctly together.

Looks/sounds promising and about as close to my goal as possible. So if I understood you correctly, I could just fill in the query masks on the HITRAN website, get the ready absorption spectra (for example for H2O and CO2) and then solving the radiation transport equation (either by writing my own code for that, or finding an open source code that does it) to understand total absorption by the atmosphere (given the actual/current concentrations of H2O and CO2), right?

@oliver Yes. That should work. Note that there plenty of codes for radiative transfer already (https://en.wikipedia.org/wiki/Atmospheric_radiative_transfer_codes) so option 2) would be probably it.

Cool! Thanks for the link. Though I often prefer to write my own code because it helps with understanding the physics, it is always a question of spare time of course. ;-)

Contributing to the first half of the answer: The MPI-Mainz UV/VIS Spectral Atlas of Gaseous Molecules of Atmospheric Interest (public access here) allows you to search by chemical name, sum formula, CAS registry number, InChi key or author.
The output is a plot, annotated with the literature reference, e.g.

(reference)
It does not cover IR data to cover the second half of your question.  However, why not asking them if they in turn would know about a complementary database and letting the users of ChemSE know about their answer?
(Otherwise, if affilated with a research institution with a department of chemistry, knowing a chemical identifier like the CAS RN could help you to retrieve spectral data in Elsevier's Reaxys [rooted in Beilstein and Gmelin database], too.)


text:
Adding pure solid/liquid in heterogeneous equilibrium
If I have the heterogeneous equilibrium
$$\ce{aA(s) + bB(g) <=> cC(g) + dD(g)}$$
I understand that the equilibrium expression is
$$K_c = \frac{[\ce{C}]^c[\ce{D}]^d}{[\ce{B}]^b}$$
and that $\ce{A}$ is left out of the equilibrium expression since it is a pure solid and pure solids/liquids are left out of the equilibrium expression in heterogeneous equilibria.
However, does this mean that adding or removing substance $\ce{A}$ has no effect on the reaction at all? For instance, is it true that adding $\ce{A}$ does not produce more products and removing $\ce{A}$ does not produce more reactants (as Le Chatelier's principle would predict for a reactant included in the equilibrium expression, such as $\ce{B}$)?
On a side note, adding/removing the other substances could potentially affect $[\ce{A}],$ right? For example, adding $\ce{C}$ or $\ce{D}$ would cause $[\ce{A}]$ to increase (because more $\ce{B}$ would be formed and with that more $\ce{A}$ would be formed as well) and adding more of substance $\ce{B}$ would cause $[\ce{A}]$ to decrease (because $\ce{B}$ would be consumed and with it $\ce{A}$ would be consumed as well), correct?

In order for an equilibrium to exist A must be in excess.

@MaxW Why is this?

"Excess" is not quite the correct word. The idea is more that there needs to be sufficient A so that when the equilibrium is established then there is still some A remaining. // If there is no A remaining then there can't be an equilibrium. In that case A is the limiting reagent.

[A] does not change when the amount of solid increases or decreases (the distance of particles in the pure solid is a constant). It will become zero when the solid is used up, otherwise it is a constant.


text:
What Ion gets reduced in this aqueous electrolysis?
In an electrolysis of tin(II) iodide ($\ce{SnI2}$) with led cathode and nickel anode, why is it that the tin ion gets reduced and not the nickel ion?

What nickel ion?

the nickel that gets oxidized

ok, but if the nickel gets oxidized, and then reduced, there is no net reaction, right?

yes, this is from a school assignment. I did a similar one where there was no net reaction also but I understood why. In this one I dont know why one takes precedence over the other

are you posting the full description of the problem?

The problem asks to write the ionization equations, the reduction and oxidation equations and the global equation


text:
Reaction between copper(II) iodate and iodide
I was attempting the following question from IChO 1989, which involved balancing of redox equations. One may just expect this to be a typical iodate-iodide redox reaction, usually employed in iodometry. However, the presence of $\ce{Cu(II)}$ makes things a bit more complicated since it can also be reduced to give $\ce{CuI}$.

To determine the solubility product of copper(II) iodate, $\ce{Cu(IO3)2}$, by iodometric titration in an acidic solution ($\pu{25 ^\circ C}$), $\pu{30.00 cm3}$ of a $0.100$ molar sodium thiosulfate solution are needed to titrate $\pu{20.00 cm3}$ of a saturated aqueous solution of $\ce{Cu(IO3)2}$.
1.1 Write the sequence of balanced equations for the above described reactions. [...]

After realising that $\ce {Cu^2+}$ is reduced as well, I wrote the following overall equation :
$$\ce{2Cu^2+ + 2IO3^- + 12H^+ + 14I^- -> 7I2 + 2 CuI + 6H2O} \tag{1}$$
which is based on
$$\begin{align}
\ce{2Cu^2+ + 4I^- &-> 2CuI + I2} \tag{2} \label{eq:cuii} \\
\ce{2IO3^- + 12H^+ + 10I^- &-> 6I2 + 6H2O} \tag{3} \label{eq:iodate}
\end{align}$$
However, it turns out that the equation given in the answer scheme seems to have multiplied eq. $(\ref{eq:iodate})$ by $2$ and the overall equation given is:
$$\ce{2Cu^2+ + 4IO3^- + 24H^+ + 24I^- -> 13I2 + 2CuI + 12H2O} \tag{4}$$
Usually, this is a trivial thing and does not matter. But in this case, it does since the reaction stiochiometry is completely different based on the equation given. Have I gone wrong somewhere?

Well, I get what you got. Iodate to iodide is +6e balanced with protons and water. Cupric to cuprous is + 1e. A total of +7e for reduction. Two iodide to iodine is -2e oxidation. Multiply reduction by 2 and the oxidation by 7 and it works out. Can't speak for the "answer scheme".

This question is clearly not closeable as homework because there is effort provided. If you want to close it, please describe your reasoning in a comment.

In your eq. (4), the copper/iodate ratio was adjusted to match the stoichiometry present in copper(II) iodate. 1 copper(II) ion to 2 iodate ions equals 2 copper(II) ions to 4 iodate ions.


text:
Identifying a chloride of the unknown element from the masses of chlorine and compound as well as vapor density

1 g of a chloride of an element contains 0.835 g of chlorine. If vapour density of the chloride is 85, find weight of the element and its valency.

I tried to solve it but couldn't get to the answer then I saw its solution and I also could not understand what they want to explain. Can anyone please help me out.
By the way, it's not my homework.

First find empirical formula of the chloride. Then, use vapor density to find the molecular formula. According to Wikipedia vapor density is the density of a vapor in relation to that of hydrogen. So, you can tell what is the valency of the element.

I agree with the starting point of the Matthew's solution, namely the definition of vapor density, but I fail to understand the follow-up math, so I just post my approach.
$$M(\ce{ECl_x}) = D\cdot M(\ce{H2}) = 85\cdot\pu{2 g mol-1} = \pu{170 g mol-1}\tag{1}$$
By definition, molar mass $M(\ce{ECl_x})$ is also
$$M(\ce{ECl_x}) = M(\ce{E}) + x\cdot M(\ce{Cl})$$
$$\implies M(\ce{E}) = M(\ce{ECl_x}) - x\cdot M(\ce{Cl}) = (170 - x\cdot 35.5)~\pu{g mol-1}\label{eqn:2}\tag{2}$$
The unknown variable $x$ is defined by the molar ratio between the elements:
$$x = \frac{n(\ce{Cl})}{n(\ce{E})} = \frac{m(\ce{Cl})\cdot M(\ce{E})}{m(\ce{E})\cdot M(\ce{Cl})} = \frac{\pu{0.835 g}\cdot (170 - x\cdot 35.5)~\pu{g mol-1}}{(1 - 0.835)~\pu{g} \cdot \pu{35.5 g mol-1}} = 24.14 - 5.04\cdot x\label{eqn:3}\tag{3}$$
Solving \eqref{eqn:3}:
$$6.04\cdot x= 24.14 \quad\implies\quad x = 4.0 \tag{4}$$
Using $x$, we can now determine the element from \eqref{eqn:2}:
$$M(\ce{E}) = (170 - 4.0\cdot 35.5)~\pu{g mol-1} = \pu{28.0 g mol-1}\tag{5}$$
which appears to be silicon, valency 4.
The unknown compound is silicon tetrachloride $\ce{SiCl4}$, a fuming volatile liquid at NTP.

Suppose the element is $\ce{Q}$, its chloride is $\ce{QCl_x}$ (since the oxidation number of $\ce{Cl}$ is $-1$, the oxidation number of $\ce{Q}$ would be $+x$), and the atomic weight of $\ce{Q}$ is $W$.
Since $\pu{1.0 g}$ of $\ce{QCl_x}$ contains $\pu{0.835 g}$ of $\ce{Cl}$, the mass of $\ce{Q}$ is $\pu{(1.0-0.835)  g} = \pu{0.165 g}$. Also:
$$\frac{x}{1} = \frac{n_\ce{Cl}}{n_\ce{Q}} $$
But, $n_\ce{Cl} =  \frac{\pu{0.835 g}}{\pu{35.5 gmol^{-1}}}= \pu{0.0235 mol}$ and $n_\ce{Q} =  \frac{\pu{0.165 g}}{\pu{W gmol^{-1}}}= \pu{\frac{0.165}{W} mol}$. Therefore,
$$\frac{x}{1} = \frac{n_\ce{Cl}}{n_\ce{Q}} =  \frac{\pu{0.0235 mol}}{\pu{\frac{0.165}{W} mol}}=0.1424W$$
$$\therefore x=0.1424W$$
Now, it is known that $\text{vapor density} \approx \frac12 \times \text{molar mass}$ (Wikipedia).
$$\therefore W + x \times 35.5  = 2 \times 85= 170 $$
Apply, $x=0.1424W$ here, and hence,
$$W + 0.1424W \times 35.5 = (1+5.055)W = 6.055W = 170 $$
$$\therefore W=\frac{170}{6.055}=28.08$$
$$\therefore x=0.1424 \times 28.08=4.0 \approx 5$$
Thus, the valency of the element is $+4$. Since its calculated atomic weight is $\pu{28.08 gmol^{-1}}$, it should be silicon ($\ce{Si}$). Thus, I conclude that the compound is $\ce{SiCl4}$.

The answer looks logical, but I think you might want to check the math part. For some reason I arrive at $\ce{SiCl4}$; also the molar mass of $\ce{PCl5}$ (208 g/mol) is way off from 170 g/mol in my opinion.

@andselisk: I checked and corrected the answer. Thank you for the careful reading. My problem was not counting the forth decimal point. I have been sleepy that night and been sloppy not to check back the calculation. Thanks again.


text:
Which functional group gets reduced first?
The following is a reduction reaction. In the reduction of 2 to 3, keto group undergoes reduction by nucleophillic addition.
Ester could undergo nucleophillic acyl substitution to give an aldo group, further reduced via nucleophillic addition again.
My question is, during such reductions, is keto group reduced first or ester group gets reduced first?

I believe you understand $\ce{LiAlH4}$ is stronger base than $\ce{NaBH4}$. In the reduction of 2 to 1, $\ce{NaBH4}$ failed to reduce ester function at all, but reduce 3-keto function to 3-hydroxy function. That result tells us that isolated carbonyl group is easy to reduce than that in ester (not even to relevant aldehyde). Thus, you can conclude that isolated carbonyl function reduced first before hard to reduce ester function in the presence of stronger reducing agent such as $\ce{LiAlH4}$, which can able to reduce both.

I agree with your instincts but the only sure way to know is to reduce the ketoester with LAH as the limiting reagent (5-10 mol%) and then analyze the products. In which case there will be recovered ketoester, hydroxyester and perhaps some diol in that an intermediate aldehyde would reduce like a ketone.

Great point and I agreed. But, I answered according to the data given. :-)


text:
Could you form quartz crystals by melting and cooling sand very slowly?
Since glass is amorphous quartz and when you slow the cooling of something enough it will crystallize, couldn’t you melt $\ce{SiO2}$ (sand) and then very slowly cool it to cause spontaneous nucleation and form quartz crystals?
If so, then why hasn’t it been used before? It would seem to be a better method than flux because in flux you need many chemicals and a dissolving chemical or mix and it is a complicated and very long process to grow very small crystals throughout the whole mix. If you simply cooled a molten mixture slowly enough to cause complete crystallization, then the entire mix of molten crystal would need to form crystal because it all is cooling down.

About the second part of your question, both the α- and the β-phase of quartz have what we call an enantiomorphic space group. It forms spirals that can be 'enantiomers' to each other, which I guess also gives it its properties. For many applications you often need the pure enantiomer which will grow in a seed of the correct orientation under for example hydrothermal conditions.

Melting and cooling will give you a bunch of crystals lumped together. Why would you want that? Then again, I've seen single crystals of quartz as big as a tabletop, and they were not made by melting.

see eg https://www.ndk.com/catalog/AN-SQC_GG_e.pdf

In principle, maybe. But that isn't how big crystals are formed in nature. Most deposit from solutions not from melted pure silica.

@NightWriter Your example post is about hydrothermal growth of a crystal.

@Buttonwood Sure, but it includes a partial explanation of why that process is employed, which is at the root of the question


text:
In HPLC, how does column diameter affect the retention time?
I am using a smaller diameter column than my previous one keeping all other factors such as stationary phase, solvent, solvent gradient constant. (Obviously, flow rate will be less in smaller column). I would like to know will changing the diameter affect the retention time of my target?
I searched and got results like this and this But they do not mention about the effect of column diameter on retention time. Moreover, I am weak in understanding mathematical equations.

I do not see the way how it could directly affect the retention times. But it may affect them indirectly, e.g by affecting the column packing, or the provider using different production process. It may slightly affect HETP via column/particle diameter ratio.

The key idea is that when you change column diameter, you keep the superficial linear velocity the same. For example if I have an analyte which elutes on a 10 cm x 4.6 mm i.d column at 7 minutes at 1 mL/min. Now we change the diameter to 3 mm, we have to use a flow rate of 0.425 mL/min to keep the retention time the same. Now we wish reduce the diameter further, say 2.1 mm i.d., now you should use 0.21 mL/min so that your analyte elutes at 7 minutes.
Search the keyword: superficial linear velocity, HPLC


text:
Role of zinc chloride in reaction of ether
I am trying to do a reaction between acetyl chloride $\ce{CH3COCl}$ with diethyl ether. I know that upon using anhydrous $\ce{ZnCl2}$ and heat, I'll get an ester and an alkyl chloride, but I'm not able to guess the mechanism.
Does $\ce{ZnCl2}$ break the $\ce{C-O}$ bond in ether as it would've done to alcohol? Please help me with the mechanism.

Note that on Chemistry SE site, it is adviced not to use MathJax in titles.When it is used in the body, use \ce{} structure for chemical formulas and equations.

Yes I believe the ZnCl2 is used to cleave the ether first to give the alkyl chloride and an alkoxide. This alkoxide then goes on to attack the acyl chloride, giving the ester.

I agree, the Zn2+ binds to the oxygen polarising the C-O bond enough for nucleophilic attack by Cl- at C.

I think that it reminds Fischer esterification at the beginning. Zinc dichloride acts as a Lewis acid by coordination at the carbonyl oxygen in AcCl. It activates C=O bond because of larger contribution of carbocation form. And this cation undergoes coordination of Et2O with formation of zwitterionic tetrahedral intermediate. This particle eliminates our catalyst and cloride-anion. The product of this step is acetyl diethyl oxonium. Cloride-anion is quite nucleophilic, but we have an excellent leaving group without charge (AcOEt), so Sn2-reaction occurs and lead to the final products


text:
Pressure dependent rate coefficient
I am trying to study how the pressure will affect the formation of ozone at around 5 bar. So, I am looking for a simple relation between the reaction rate coefficient and the pressure to use it in my simulation. Is there any suggestion?

Why would there be a relation in the first place?

If you already have the model of reaction mechanism and kinetics, you just plug in the pressure into the initial conditions.

@IvanNeretin In some papers, it seems to depend on the pressure. see for example  https://en.wikipedia.org/wiki/Lindemann_mechanism or https://www.sciencedirect.com/science/article/pii/0010218084901093

@Poutnik I do not know how the pressure affects the reaction in the first place. So, I need an equation that relates the pressure with the rate coefficient.

Well, if you have a set of reactions and kinetics, you already know how pressure affects them. If you do not have them, then you do not know. How about to add to your question the reaction and kinetic model you are analyzing ? Then other may put their hand on it and help you analyze the pressure effects.


text:
Oxidising nature of hydrogen peroxide
It is said that $\ce {H2O2}$ behaves like a reducing agent in an external oxidizing environment but along with $\ce {H2SO4}$ a strong oxidizing agent it oxidizes $\ce {FeSO4}$ to $\ce {Fe2(SO4)3}$ in acidic medium. How and why?

Why what? Yes, H2O2 (like a good many other compounds) can act as a reducing agent or as an oxidizing agent, or even as both at once, depending on circumstances. There is nothing special about it.

Hi maybe you can clarify a little bit your question! You are asking How hydrogen peroxide can be an oxidizing agent and a reducing agent depending on the pH? I think it's a good question... We can help you to edit it so it's more clear...

Also note that $\ce{H2SO4}$ is usually not an oxidizing agent in water chemistry.

One way you can look at it, at least based on what my teacher said, is that the oxidation state of O in H2O2 is intermediate between 0 and -2. Thus, it shares the reducing ability of oxides as well as the oxidising ability of dioxygen.


text:
Where to find compounds IC50 on an enzyme?
I am looking for half maximal inhibitory concentration (IC50) values on a enzyme.
I already searched through ChemBL, PubChem and BindingDB. The three seem to give consistent results.
However, looking in the literature, there are lots of compounds which are not in these databases. These later may provide higher level of curation. I am still interested in the data available in literature. Any other available repository or text mining approach will be great.

To my knowledge, there are no data bases accumulating  half maximal inhibitory concentration values on a enzyme ($\mathrm{IC_{50}}$). According to Wikipedia:

Half maximal inhibitory concentration ($\mathrm{IC_{50}}$) is a measure of the potency of a substance in inhibiting a specific biological or biochemical function (i.e. an enzyme, cell, cell receptor, microorganism, etc.).

$\mathrm{IC_{50}}$ is commonly used as a measure of antagonist drug potency in pharmacological research. According to the FDA, $\mathrm{IC_{50}}$ represents the concentration of a drug that is required for 50% inhibition in vitro. It is comparable to an $\mathrm{EC_{50}}$ for agonist drugs.
The major reason for not to accumulate $\mathrm{IC_{50}}$ values is their abundance. For example, for given specific enzyme, there may be vast number of inhibitors present. All of these inhibitors have their specific $\mathrm{IC_{50}}$ value. Thus, if you need to find a $\mathrm{IC_{50}}$ value of specific inhibitor on a particular protein (or enzyme) or a receptor or other biological system, you should search using both names of the inhibitor and biological system as key words. Thus, you would find any literature, which may have $\mathrm{IC_{50}}$ values for sought combination.
For example, Galectin-3 is  a protein that is a member of the lectin family (carbohydrate-binding proteins). Galectin-3 is approximately $\pu{30 kDa}$, and contains a carbohydrate-recognition-binding domain with about 130 amino acids that enable the specific binding of $\beta$-galactosides, such as N-acetyllactosamine (14; the parent substrate; $\mathrm{IC_{50}} = \pu{158 \mu M}$ under the conditions; see insert red box). Several derivatives of N-acetyllactosamine (15 - 18) have displayed inhibitory activities with different $\mathrm{IC_{50}}$ values ranging from $13$ to $\pu{31 \mu M}$ (See the plot below):

On the other hand one inhibitor can be active against more than one biological systems with different $\mathrm{IC_{50}}$ values.

Relevant References:

P. Sörme, B. Kahl-Knutsson, U. Wellmar, B.-G. Magnusson, H. Leffler, U. J. Nilsson, “[12] Design and Synthesis of Galectin Inhibitors,” In Methods in Enzymology, Volume 363 – Recognition of Carbohydrates in Biological Systems, Part B: Special Applications;   Y. C. Lee and R. C. Lee, Eds.; Academic Press: Amsterdam, The Netherlands, 2003, pp. 157–169.
P. Sörme, Y. Qian, Per‐G. Nyholm,  H. Leffler, U. J. Nilsson, “Low micromolar inhibitors of galectin-3 based on 3'-derivatization of N-acetyllactosamine,” ChemBio Chem 2002, 3(2-3), 183–189 (https://doi.org/10.1002/1439-7633(20020301)3:2/3<183::AID-CBIC183>3.0.CO;2-%23).
J. Hynes, Jr., S.  K. Nair, “Chapter Nine - Advances in the Discovery of Small-Molecule IRAK4 Inhibitors,” Annual Reports in Medicinal Chemistry 2014, 49(1), 117–133 (https://doi.org/10.1016/B978-0-12-800167-7.00009-2).


text:
Can a protein complex be a machine?
Discussing an article with friends we began discussing whether a protein complex that punches holes in cells (membrane attack complex) would be considered a machine.
Now, my friends are biochemistry students while I am a computer science student so this is far from my expertise. In my imagination this would be a mechanical action, hence the definition of a machine found on Wikipedia.
My friends counter argument is more of an emotional one, saying it feels wrong to call it a machine but also couldn't come up with a better word.
So:
Is it a machine? Is it something else? Or is a complex just a complex?
Does chemistry/biology have its very own definitions for things that do stuff in coordination without being "alive"?

If some gaseous molecules collide, they can react to form a new compound, given they have enough energy and the right orientation. Would you call the reactants a machine?

I might argue that for the distinction to be meaningful the issue would be whether a protein is manipulating something chemical (eg breaking bonds selectively) or physical (eg moving some other object or punching a hole in something without breaking specific bonds).

@M.A.R. Not necessarily, I guess. But what about a protein that fold under light. Is that not technically converting energy to mechanical motion?

I would generalize that a machine can transform chemical energy (or other form) into mechanical energy, and can operate in a cycle, eventually returning to its original state.

@NightWriter so you would consider it a machine given it can revert and repeat the process?

@Harri That's how I think of one. Then again I suppose you could have a one-off (disposable) machine, probably I am imposing an unnecessary constraint with the cyclic business. https://en.wikipedia.org/wiki/Molecular_machine#Biological

Ah, perfect article. If you were to post it as answer I'd mark it the correct one

Look for articles on the crystal  structure of ATPase, this has a rotor in the centre of the protein that performs chemistry as rotates and opens and closes different sites.

Not sure about the specific case but molecular machines do exist. Then is matter of some subtle definition. Muscles and bones are machine, small molecules can be seen as machines, proteins are just in between.  Besides the merely chemical bonding aspect, changes in geometry results in motion. Normally chemistry is non dealing with that. But once things are anchored or analysed at the proper scale,  they certainly become machines. Is the emphasis on this machine aspect that is not often needed. Exception are indeed biochemistry and (nano)material science.

@Harri — I think that the key here is that a machine is generally understood as "man-made" which is clearly not the case for a natural protein. However, man-made proteins might be considered as true machines.

@SteffX that is precisely the issue at hand in the discussion with my friends. We decided to call it a mechanism but based this solely on personal preference. I had addid this as an answer but it was pojnted out to me that personal preference does not fit the criteria for a useful answer on this site. Putting this general undestanding aside and going purely by (Oxford) definition, it should be understood as a machine I think.

According to the Oxford dictionary, a machine is:

An apparatus using mechanical power and having several parts, each with a definite function and together performing a particular task.

Based on that definition, I think some proteins and enzymes could definitely be considered machines. A good example for that, in my opinion, would be ATP Synthase and all of the proteins that participate in the oxidative phosphorilation process.

According to to the Oxford dictionary, an apparatus is 1. technical equipment or machinery [....], 2.the complex structure of a particular organization or system. 3. A collection of notes [...] - I do not really see a good fit for proteins here. But if a protein isn't an apparatus, it's not a machine either...

Well, we can just keep going down the rabbit hole. According to the Oxford dictionary, technical is: "Relating to a particular subject, art, or craft, or its techniques." Seems to fit. I guess there is no clear answer because it's a question that relies heavily on semantics.

:) Fully agree that this question heavily relies on semantics.

@Arsak Apparatus has much narrower meaning.


text:
Calculation of resultant pH

A solution with $\mathrm{pH}=4$ is mixed with a solution with $\mathrm{pH}=10$ in equal volumes. Calculate resultant $\mathrm{pH}$.
  Please note both solution consists of strong acids and/or bases i.e. They will completely dissociate.

$Method 1:$
Let volume be 1L of each
pH of solution 1=4
Concentration of H is $10^{-4} $
pH of solution 2=10
Concentration of H is $10^{-10} $
Resultant concentration =$\frac{10^{-4} +10^{-10} }{2}$
Therefore calculting pH we get 4.301
$Method 2:$
pH of solution 1=4
Concentration of H is $10^{-4} $
pH of solution 2=10
pOH=14-10=4
Concentration of OH=$10^{-4} $
Hence the H and OH neutralizes each other so it will result in neutral solution ph=7
Edit:
I found another method which is kind of same as method 2
Method 3
In Solution 1:
pH=4. Concentration of H is $10^{-4} $
Since pOH=14-pH. Hence pOH=10. Therefore concentration of OH=$10^{-10} $
In solution 2
pH=10. Concentration of H is $10^{-10} $
Since pOH=14-pH. Hence pOH=4. Therefore concentration of OH=$10^{-4} $
So the total concentration of H=concentration of OH
Hence neutrilization reaction would occur and resultant pH will be 7
Which method is right and why??

The second method, so to put it, is slightly less wrong. But in fact we can't calculate the resulting pH at all. We don't have the necessary data. Knowing the pH of original solutions is simply not enough.

Can you please explain why is it so??

What else is required??

We need to know what compounds are there.

Why do we need that?? I couldn't get your point. I think if it is given that solution2 has pH 10 it will be basic and that means we have a basic compound and the other will be acidic as its pH is 4

A dilute solution of a strong acid and a concentrated solution of a weak acid might have identical pH, but very different neutralization behavior.

Thanks for reminding me Both the compound are strong acids and base

Then the second method will do.

And is the third method i mentioned in edit correct??

How is it different from method 2?

Method 1 would work for a 'normal solution' but in the acid/base case there is the equilibrium $\ce{H^+ + OH^- <=>H2O}$ and so recognising is what you do in method 2.

The specification that strong acids and bases are used is enough to calculate that the final pH is 7. No further decimals are used in the description of the problem, so the fact that the final pH may be 0.01 or 0.02 off is not significant. The concentrations of H+ and OH- are equal; the volumes are equal; the strengths are equal. Nothing more is needed.


text:
What are the half reactions for the redox reaction between SO2 and iodine?

What are the half reactions for the redox reaction between $\ce{SO2}$ and iodine? The relevant chemical equation:
  $$\ce{ SO2 + H2O + I2 -> H2SO4 + 2HI}$$

This is what I think it is, not sure if it's correct.
Oxidation: $\ce{SO2 + 2H2O -> SO4^2- + 4H+ + 2e-}$
Reduction: $\ce{I2 + 2e- -> 2I-}$

Your initial comment exchange made me realize that you have some difficulties in balancing redox equations. Thus, I decided to include some clues for your benefit.
An important part of writing half-reactions is making sure they're balanced by mass and charge. Since most redox reactions are done in aqueous medium, you can always balance $\ce{O}$ by $\ce{H2O}$. By doing so, you contribute extra $\ce{H}$ to the equation so you can balance them by $\ce{H+}$ since it is mostly an acid medium reactions (if they are in base or neutral conditions, you may cancel $\ce{H+}$ by adding the same amount of $\ce{OH-}$ to both sides of the reaction). Finally, cancel the net plus charge by adding $\ce{e-}$s to the appropriate side.
Let's see the easy one first, in your case reduction half reaction where $\ce{I2}$ reduce to $\ce{I-}$: $\ce{I2 -> I-}$. Balance its mass, which gives you: $\ce{I2 -> 2I-}$. Now, balance the negative charges by $\ce{e-}$s. So, you got balanced reduction half-reaction ($\ce{e-}$s are in LHS):
$$\ce{I2 + 2e- -> 2I-} \qquad \mathrm{E^\circ = \pu{0.536 V}} \qquad \text{(1)}$$
Now, see the more difficult second equation, the oxidation half reaction where $\ce{SO2}$ oxidizes to $\ce{SO4^2-}$: $\ce{SO2 -> SO4^2-}$. Its $\ce{S}$ is already balanced, but $\ce{O}$ is not. So, balance it with $\ce{H2O}$, which gives you: $\ce{SO2 + 2H2O -> SO4^2-}$. Now, balance additional $\ce{H}$ by $\ce{H+}$, which gives you a mass-balanced equation: $\ce{SO2 + 2H2O -> SO4^2- + 4H+}$. Now, balance the negative charges by $\ce{e-}$s. So, you got balanced oxidation half-reaction ($\ce{e-}$s are in RHS):
$$\ce{SO2 + 2H2O -> SO4^2- + 4H+ + 2e-} \qquad \mathrm{E^\circ = \pu{0.157 V}} \qquad \text{(2)}$$
If you add (1) and (2) together in order to cancel $\ce{e-}$s, you get the redox reaction you are looking for:
$$\ce{SO2 + 2H2O + I2 -> SO4^2- + 4H+ + 2I-} \qquad \mathrm{E^\circ_{cell} = \pu{0.693 V}} \qquad \text{(3)}$$
The value of $\mathrm{E^\circ_{cell}}$ is positive means the reaction is spontaneous.
(Note: The value of $\mathrm{E^\circ_{\ce{SO2/SO4^2-}}}$ is from Ref.1)
Reference:

J. A. O’Brien, J. T. Hinkley, S. W. Donne, “Electrochemical Oxidation of Aqueous Sulfur Dioxide II. Comparative Studies on Platinum and Gold Electrodes,” J. Electrochem. Soc. 2012, 159(9), F585–F593 (DOI: 10.1149/2.060209jes).

One path I accept is as proceeding via the created presence of HOI from the interaction of iodine and SO2 with water in a series of reactions:
$\ce{I2 + H2O = H+ + I- + HOI}$
$\ce{SO2 + H2O = H+ + HSO3-}$
$\ce{HOI + HSO3- -> H+ + I- + HSO4-}$
However, the reaction could also proceed to some extent via the following reactions following a disproportionation of hypoiodous acid:
$\ce{HOI -> 2/3 HI + 1/3 HIO3}$
$\ce{1/3 HIO3 + H2SO3 -> 1/3 HI + H2SO4}$
Bottom line, the system is in reality, I suspect, a little more complex as compared to the reported action of HOCl on aqueous SO2 as only concentrated hypochlorous is likely to form HClO3 (although UV light may provide a radical path with dilute hypochlorous acid and warming), which is not the case with HOI. For my proposed pathway, a supporting related reaction source:
"A further mechanism for SO2 oxidation in solution with sea water proposed via halogen compounds HOCl and HOBr (Vogt et al., 1996....
Simulations by Keene et al. (1998) found reactions with HOCl more rapid than with H2O2 at pH 3–5.5 and HOBr more rapid..."
The above reference also cites radical pathways.
Here is another HOCl/HOBr interaction with SO2 source 'Heterogeneous reactions of SO2 with HOCl and HOBr on ice surfaces'. Note, a reference to possible surface chemistry afoot as well at low temperatures.
The question "What are the half reactions for the redox reaction between SO2 and iodine?" should be more correctly stated as "Postulating a simple reaction pathway, speculate on a suggested half cell reactions for the REDOX of SO2 with aqueous I2 at room temperature."
On the electrolysis based response, I would caution students that electrolysis is based on net reactions, and not a proper source for understanding a reaction system mechanics, but can give measurement-based guidance.