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Why water cools from outside to inside?
It is a well-known fact that

Water kept in a container, on cooling, results in freezing with initially at the circumference of surface
(1D), then on the surface (2D) and then throughout the volume (3D).

Why does it happen that way?
My attempt:
I thought that this may be due to their exposure levels.The outer surface is more exposed to cooling than the inner ones and hence cools first.However, I am not satisfied with this explanation.
I think there is some relation between this and degrees of freedom.I feel as there are more degrees of freedom in 3-D, there's more movement you should suppress to lower its energy, so maybe they freeze slower.
Is the above reasoning correct? I would also like to know different ways to reason out this.

Comments are not for extended discussion; this conversation has been moved to chat.

I was thinking that more degrees of freedom in 3D would mean that the energy that needs to be suppressed in order to cool down the body is more.Idk if this is a valid argument.Can anyone confirm?

This has everything to do with heat conduction, Setting aside freezing or melting for the time being, the two things that are happening in combination are (1) heat conduction and  (2) the capacity of a parcel of solid to store thermal energy by its temperature changing.  Heat conduction is driven by a spatial gradient in temperature, with heat flowing from a hotter region to a colder region.
So if the outside surface temperature of an object is suddenly increased, the interior temperatures will not change at all initially, and the temperature gradient will be zero everywhere, except at the surface, at which it will be very high. This will cause heat to begin flowing into the very outer layer of the object.  The heat that flows into this outer layer will cause its temperature to begin rising (to store at least part of the heat).  This temperature rise will then create a temperature gradient for heat to begin flowing into the next layer inward, where its temperature will begin rising.  This kind of sequence continues until, later, the temperature at the very center of the object begins to change.  So, as a result of conductive heat transfer, the temperature rise propagates like a wave traveling inward into the object.
This can all be described mathematically and modeled by the so-called transient heat conduction equation.

Thanks for giving this informative answer! What do you think about relating why this happens with degrees of freedom as I had mentioned in the question? Is that wrong? (Because I see none of the answers have any argument along those lines)

I have no idea what you are getting at when you refer to degrees of freedom.  If you mean that, usually with a 3D object, only part of the surface is exposed to heat transfer from the outside, that would reduce it to more of a 1d or 2D situation.  If you mean that it is easier to heat and cool a 3D object if all ifs surfaces are exposed the heating and cooling. then that, of course, would speed up the heat transfer.

I was thinking that more degrees of freedom in 3D would mean that the energy that needs to be suppressed in order to cool down the body is more.Idk if this is a valid argument.

Sorry. I still don't follow.  Maybe someone else can relate to this.

The cooling takes place at the surface provided that it is at a lower temperature than the interior, experimentally this is confirmed. The mechanism is by collisions between molecules transferring energy to one another, those with more energy transfer to those with less, this takes place in time and space and is macroscopically measured by the thermal diffusivity coefficient. We can describe this using Ficks' laws and an example is shown in the figure which I had recently calculated for other purposes. It illustrates what happens after a small circular region of a thin sample of absorbing material is excited by a laser. You can observe how the heat spreads both in time and space. The total energy remains the same so as the spatial extent increases the temperature at the centre falls but increases elsewhere. The circle shows the region initially heated.

(Ficks law was solved numerically $\displaystyle \frac{\partial U}{\partial t}=D\left(\frac{\partial^2 U}{\partial x^2}+\frac{\partial^2 U}{\partial y^2}\right)$ with $D = 0.08 \;\mathrm{mm^2/s}$ which is typical of  many liquids.)

Definitely in agreement with Chet, but does this not also have something to do with increased nucleation sites at bounds of the container?
If a container of water was super cooled and a foreign object was introduced at the center of the container, we would observe freezing from the center outward.


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Hydrolysis of phosphoryl-oximes
When acetylcholinesterase that has been inhibited by an organophosphorus compound is reactivated by an oxime, a phosphoryl-oxime is formed, which is then somehow hydrolyzed in the blood.
One study suggested that enzymes of the paraoxonase family are responsible [1], while another suggested that this may not be so [2].
What is/are the pathway(s) by which phosphoryl oximes are hydrolyzed?
References:
[1] Luo, C., Saxena, A., Smith, M., Garcia, G., Radić, Z., Taylor, P., & Doctor, B. P. (1999). Phosphoryl Oxime Inhibition of Acetylcholinesterase during Oxime Reactivation Is Prevented by Edrophonium. Biochemistry, 38(31), 9937–9947. doi:10.1021/bi9905720
[2] Kiderlen, D., Worek, F., Klimmek, R., & Eyer, P. (2000). The phosphoryl oxime-destroying activity of human plasma. Archives of Toxicology, 74(1), 27–32. doi:10.1007/s002040050648

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Why is methyl parathion less toxic than ethyl parathion?
According to the PubChem pages on ethyl parathion and methyl parathion (pages 11 and 12 of the toxicity section, respectively), methyl parathion is significantly less toxic than ethyl parathion. The difference in LD50 values depends on the route of administration and the animal to which the pesticides were administered, but it is significant.
For instance, parathion has an intraperitoneal LD50 in mice of 3 mg/kg, the LD50 for methyl parathion administered in the same way is 5.4 mg/kg, making it almost two times less toxic than parathion. Methyl parathion differs from regular parathion in that, in methyl parathion, the ethyl groups are replaced with methyl ones.
How does this structural difference contribute to a difference in toxicity? Why is methyl parathion significantly less toxic than its ethyl analog?

Both methyl parathion (O,O-Dimethyl O-(4-nitrophenyl) phosphorothioate) and ethyl parathion (O,O-Diethyl O-(4-nitrophenyl) phosphorothioate) are organothiophosphates with identical structure except for dimethoxy group in methyl analog has been replaced by diethoxy group in ethyl analog. Both analogs are acetylcholinesterase (AChE) inhibitors, although phosphorothioates ($\ce{P=S}$) are less potent inhibitors compared to corresponding phosphate ($\ce{P=O}$) analogs. The active site of AChE is depicted in following Figure (Ref.1) where acetylcholine (ACh) get hydrolyzed:

The hydrolysis mechanism invoves the $\ce{OH}$ of serine residue and imidazole moiety of histidine residue in the active site as shown. The major inhibition mechanism of all organophosphate esters is involved phosphorylation of the $\ce{OH}$ group of serine residue in the active side. The aging (breakage of the $\ce{O-R}$ ester linkages leaving $\ce{P-O-}$ groups) is followed this phosphorylation making the inhibition irreversible.
Although aging can be visualized by the process of hydrolyzing ester group by water molecule attacking phosphorus atom first, replacing $\ce{R-O-}$ group followed by proton exchange, the literature has shown preferably $\ce{O-R}$ breakage. If that's the case, ethyl breakage is facilitated over methyl breakage just by stability of the resultant products ($\ce{CH3CH2^+}$ vs $\ce{CH3^+}$). Hence, we can conclude that ethyl parathion is more potent than methyl parathion.
An early study of inhibitory activity of various batches of ethyl parathion has revealed that a dithio analog,  O,S-Diethyl O-(4-nitrophenyl) phosphorothioate is much potent inhibitor compared to ethyl parathion (Ref.2). It is possible that this dithio analog is a common contaminate during ethyl parathion sysnthesis. Thus, this factor may contributed to the higher activity of ethyl parathion compared to that of methyl parathion.
It is also worth noting that recent studies on versatility of imidazoles or in reactions with organophosphates (OPs, Ref.3). According to this study, OPs from the $\ce{P=O}$ family, imidazole attacks the phosphorus atom exclusively, making $\ce{P-N}$ bond. With the less reactive $\ce{P=S}$ family, an unusual N-alkylation predominates. Surprisingly, imidazole reacts with methyl parathion exclusively at the aliphatic carbon, whilst for ethyl parathion, imidazole reacts by two pathways: at both the phosphorus and aliphatic carbon, with predominance for the latter:

Since immidazole moiety is present in active site of the AChE, one can suggest the different reactivity of ethyl parathion over methyl parathion with imidazole may play a role in their different activity in inhibition.
References:

Hay Dvir, Israel Silman, Michal Harel, Terrone L. Rosenberry, Joel L. Sussman, “Acetylcholinesterase: from 3D structure to function,” Chemico-biological Interactions 2010, 187(1-3), 10-22 (DOI: 10.1016/j.cbi.2010.01.042).
W. M. Diggle, J. C. Gage, “Cholinesterase inhibition in vitro by OO-diethyl O-p-nitrophenyl thiophosphate (Parathion, E 605),” Biochemical Journal 1951, 49(4), 491-494 (https://doi.org/10.1042/bj0490491).
Valmir B. Silva, Elisa S. Orth, “Are Imidazoles Versatile or Promiscuous in Reactions with Organophosphates? Insights from the Case of Parathion,” J. Braz. Chem. Soc. 2019, 30(10), 2114-2124 (https://doi.org/10.21577/0103-5053.20190084 ).

Since I've posted this question, I did some searching of my own. According to doi:10.1016/j.bcp.2004.07.038  , AChE inhibited by dimethyl-paraoxon ages much faster than AChE inhibited by diethyl-paraoxon, surprisingly enough, though dimethylphosphorylated AChE is also much more prone to spontaneous reactivation. Anyways, your answer was very interesting.

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Does the depth of a lake affect how quickly its water is evaporated?
Suppose there are two freshwater lakes, each with identical surface areas and circumferences. One lake is 1 ft deep and the other is 20 ft deep. Although they exist in nature, lets assume they are both exposed to the same identical climate and sunlight. Will the difference in depth affect in any way affect the rate of evaporation of the water? Will one of the lakes lost 1 inch of water to evaporation faster?

The problem posed has no simple answer. Rates of evaporation would depend on a number of factors such as thermal layers in the water, heat exchange between between water and ground, and so forth.

What are your thoughts on this?  Generally answerers prefer is some effort is first made by the questioners.

You should consider what is different between the bodies of water in terms of thermal transport by conduction and convection. A shallower body will probably reach higher temperatures, accelerating evaporation.

This is not a chemistry question. Also for a thought experiment, it is far too unspecific.

I’m voting to close this question because it is not about chemistry.

@Karl  Agreed, the question lacks needed specifics.  But questions about evaporation rates and what affect them fall well within the purview of physical chemistry, and thus of this site.

The evaporation of water does not depend on the depth of the lake. It depends on the temperature and of the wind. Here are some values, valid for a lake of infinitive surface, with an "usual humidity"
At $10°$C, the level of the water decreases by $0.9$ mm/day if the wind is zero, by $1.2$ mm/day with a $2$ m/s wind, and by $1.7$ mm/day with a $9$ m/s wind
At $20°$C, those values are respectively $3.2$ mm/day, $4.5$ mm/day and $6.1$ mm/day, when the wind blows at $0$ m/s, $2$ m/s and $9$ m/s.
At $30°$C, the same values are respectively $6.8$ mm/day, $9.6$ mm/day, $12.9$ mm/day, when the wind blows at $0$ m/s, $2$ m/s and $9$ m/s.
If the lake is not on sea level, these values should be multiplied by ($\ce{1 + 1.08 10^{-4} h}$), if h is the height above sea level, in meter.
The present values are average values, calculated for an "average" humidity. They change with the relative humidity or the water pressure in a complicated way, with empirical parameters which would take too long to be developed here. It also changes with the dimensions of the lake. I have only given the values for an infinite surface.
Other values are given for sea water in oceans.
These values are taken from a book by M. Dorsey, Properties of the Water Substance, Reinhold Co., New York 1940.

Thanks for the values, Maurice, the data can be widely useful. But temperature of the surface water depends on the lake depth. E.g. the lake bed of the lake 1ft deep would be warmed up by the direct sunlight, while the one of 20 ft deep not. There will be convection and overall warming up of the shallow lake. The Sun light will effectively warm up thinner water layer. So IMHO the evaporation rate depends on the depth indirectly.

Seems either M. Dorsey was optimistic about simplicity of the problem, or you didn't mention all his assumptions of what else has to be constant to get these values - and that's a lot of things.

@Mithoron. I should have mentioned that the values from my previous message are average values, calculated for an "average" humidity. They change with the relative humidity or the water pressure in a complicated way. There are a lot of empirical parameters which would take too long to be developed here. Why don't you open M. Dorsey's book ? It also changes with the dimensions of the lake. I have only given the values for an infinite surface.

Well, then edit this in your post. BTW I can hardly open it as it was first time I even heard about it...

M. Dorsey's book is in the library archives of my University ... I have discovered it by a bit of luck !

You haven't considered other factors.  For instance, evaporation cools the water, and a foot-deep lake will be more cooled during the evaporation of an inch of water than a deeper lake, thus causing this evaporation to be slower.  In addition, fresh water has dissolved solutes, reducing its vapor pressure, and this vapor pressure will be more reduced during evaporation from a shallower lake because the concentration of those solutes will increase more rapidly in the latter case, thus also causing reduced evaporation rate in the shallow lake.  etc. etc.

@theorist. The temperature is taken in the water, not in the air. In a shallow lake its temperature will decrease, and the evaporation will decrease, according to the numerical values from Dorsey. And fresh water does contain solutes of course, but in negligible amounts. You probably have noticed that the reported numerical values have only two significant figures. They are averages, which are not precise. You know. A rough result is better than no result at all.

@Maurice  As evaporation proceeds, the water temp in both the shallow and deep lake will decrease, but it will decrease more in the shallow lake, causing the evaporation rate in the shallow lake to decrease faster than that in the deeper lake, causing it to take longer.  That's considering just that one factor.

Freshly formed shallow pools of water evaporate rapidly also because it is  largely freshwater (seawater has lower levels of evaporation).


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Maximum distance for intramolecular proton transfer

In this reaction (formation of ester from ketone / aldehyde), intramolecular proton transfer happens, since it can stabilize the formal positive charge by resonance after that. This transfer happens across 3 atoms. ($\ce{O-O-C}$)

However, in this step (formation of acetal from $\ce{C=O}$) using alcohol, the book says it does not occur, since the $\ce{O}$ and $\ce{H}$ are 'too far' apart. This 'transfer' happens only across 2 atoms ($\ce{C-O}$). Which is less than the above reaction ($\ce{O-O-C}$). If that can occur, why can't the proton transfer here not occur?
What is the real factor that determines whether an intramolecular proton transfer can happen?

(1) What book is this? (2) Consider looking into Baldwin's rules. FWIW I would be inclined to draw the first one as two separate steps.

Electrons transfer occours through (usually double) bonds, protons transfer through space.

@orthocresol 'organic chemistry as a second language', I am struggling to figure out Baldwin's rules. It seems to apply in a ring formation (I don't know whether I am correct), is this applicable to the explanation for the second one? Thanks

@Karl sorry if I am not mistaken, if any protons can transfer through space, this means actually the transfer in the second one is workable?

What do you mean by "any"? I only said that if there is a viable (especially short enough) trajectory, a proton does not care from what (part of a) molecule it comes.  If the reaction happens of course depends on a lot of additional factors.

@Karl sorry for mistaking, can I say the first one is able to do the transfer since the atoms are somehow 'coiled up' (from the figure), thus reducing the distance between O and H. And the second one is unable due to electric repulsion of 2 O atoms in the vicinity, thus increase the separation (not short enough)?


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Writing the overall cell reaction for the Calomel electrode
I an studying about electrochemical cells, and came across the Calomel electrode. It turns out that the half cell reactions and the overall reaction are as follows.
Anode half-cell: $\text{Hg}_{2}\text{Cl}_{2}\rightarrow \text{Hg}_{2}^{2+}+2\text{Cl}^{-}$
Cathode half-cell: $\text{Hg}_{2}^{2+}+2e^{-}\rightarrow 2\text{Hg}$
Overall cell reaction: $\text{Hg}_{2}\text{Cl}_{2}+2e^{-}\rightarrow 2\text{Hg}+2\text{Cl}^{-}$
I am not sure why while writing the overall reaction, the electron was not gotten rid of. Besides why is there no net exchange of an electron in the anode half-cell reaction. Clarification would be appreciated. Thanks.

The calomel electrode is not a cell. It can be part of a cell. It is a half-cell. A cell is made of two electrodes, one calomel electrode and another one. The reaction occurring in the calomel electrode is the reaction you present as "overall cell reaction". What you described as "anode half-cell" has nothing to do with electrochemistry. It is the equation describing how $\ce{Hg_2Cl_2}$ gets dissolved in water.


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Why is a drop of chloroform stretched/pulled apart on a water surface?
I distributed drops of chloroform (with a lipid in them) on water (in a Langmuir-Blodgett trough). The chloroform drops form a layer on the water (so that the lipids are distributed on the water surface) and I need to explain why. (Afterwards, the chloroform evaporates, leaving back a lipid monolayer.)
I was told that the water pulls each chloroform drop apart and that this has to do with the fact that the surface tension of water to air is greater than the sum of the surface tension of chloroform to air and surface tension of chloroform to water.
Can someone explain how this works in more detail?

You might want to read about the contact angle.

As you pointed out, the surface tension of water-air interface is greater than the surface tension of chloroform-air interface. At $25~^\circ\mathrm{C}$, the surface tension of water-air interface is $71.99\pm0.05~\mathrm{mNm^{-1}}$ and that of chloroform-air interface is $26.67~\mathrm{mNm^{-1}}$ (source).
Now, let's understand why a drop of chloroform placed over water gets puller apart forming a layer. Surface tension is the force per unit length. More the surface tension, more is the force per unit length and vice versa. Surface tension is similar to tension in a stretched rubber sheet. As you place a drop of chloroform on water, the system looks as shown in the following diagram:

As we seen earlier, surface tension of water-air interface is greater than that of chloroform-air interface. So in the boundary separating chloroform and water there exists an imbalance of force. The force vectors are the red arrows in the diagram above and their strength is denoted by their length and weight (thickness).
Due to this, the drop of chloroform expands to form a layer on top of water. This can also be looked in the nature's tendency to lower energy. The surface energy of chloroform-air interface is lesser than that of water-air interface. So, a lower energy configuration is preferred.

Maybe you should also try to explain chemically why the surface tension of the water-air interface would be expected to be higher than the chloroform-air interface, using the idea that surface tension is dependant on intermolecular forces. Rest of the answer pretty much covers everything

Could you please explain why the surface energy of chloroform-air interface is lesser than that of water-air interface?

I understand that water has a tendency to reduce its surface, but your answer implies that the chloroform is bonded to the water, right? If yes, how?

@Filippo: Surface energy $U$, area $A$ and surface tension $S$ are related by $S=U/A$. The chloroform-air interface has a lesser surface energy than that of water-air for the same reason it has a lesser surface tension. Now an obvious question is why does the former have a lesser surface tension than the latter? This depends on the magnitude of the intermolecular forces. Generally, the stronger these bonds are more is the surface tension. This is also supported by the lower boiling point of chloroform compared to that of water.

@GuruVishnu Thank you very much :) Good to know that you can also regard this process as a minimization of (surface) energy.


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Ring expansion in cyclo compounds
In the following reaction, why does ring expansion not take place?

Mechanism that was the correct one:

In the second step when there is a protonated carbon the ring should expand to become 6 membered as that is more stable. This should be followed by a methyl shift from Quaternary to tertiary and then the $\ce{Cl}$ should get attached. But this doesn't happen. Why is that? According to my textbook's theory, what I described should occur but the mock test says otherwise. Are there some more criteria-besides.

protonated carbon next to ring,

forming of a more stable 6 number ring

for ring expansion that my text book is missing?
Any help would be appreciated.

5- and 6-member rings are fairly close in terms of ring strain. That's probably the main issue here.

Simple answer is consider the stability of carbocations. You already got the tertiary carbocation in hand that is much stable than secondary carbocation, which would be resulted by ring expantion. Sure, a methide migration would give you another tertiary carbocation, but before you get that, the reaction progress has to climb another actvation barrier, which would be steep ($3^\circ \rightarrow 2^\circ$). Bottom line is rate of clorination of carbocation is faster than rate of this rearrangement followed by methide migration ($E_\mathrm{a2} \lt E_\mathrm{a3}$):

Using which software did you draw these colorful reaction profile diagrams?

@Chem-Learner: No software. Just being creative with what available in internet.

this clears up a lot for me. Are there anymore such reactions where ring expansion could happen and it doesnt? Or does this apply only to halogenation ?

To my knowledge, it is situational. Mostly carbocation forming reactions such as $\mathrm{S_N1}$ and $\mathrm{E1}$ reactions.

@MathewMahindaratne i came across another such reaction except the cyclo hexane ring wasn't there. Instead of HCl, HI was used. This time ring expansion mechanism was followed. Why is that so ?

Can you ask a separate question refereing to this question as a difference? Make sure to include structures.


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Protonated forms of EDTA and their constants of formation with metal ions
It is widely known that the fully deprotonated form of EDTA (usually written as Y4-) is the only form that is significantly involved in complexation reactions. From what I have read, this accounts for the fact that the fully deprotonated form has the highest number of complexation sites (the other protonated forms obstruct oxygen atoms from interacting with the metal since those atoms bear protons). Hence, this form has the largest constant of formation, Kf, given the thermodynamical implications of the chelation effect. Also, correct me if I am wrong, but this is exactly the reason why complexometric titrations that use EDTA as a titrant are carried out for a specific metal ion at a fixed pH - at that pH value, the fully deprotonated form has the maximum concentration it could have while also avoiding the precipitation of the metal ion as a hydroxide.

Out of sheer curiosity I looked for any experimental or theoretical evidence to find out whether or not the other protonated forms were involved as well in this complexation equilibria. Even though their influence is neglected (because their formation constants are orders of magnitude smaller than our main Kf), I am still interested to hear to what degree do they participate in complexation with the metal.
To my disappointment, I have found no explicit answer to my question. The two analytical chemistry textbooks I have been studying from (Gary D. Christian, Purnendu K. Dasgupta, Kevin A. Schug - Analytical Chemistry-John Wiley & Sons (2013) and Douglas A. Skoog, Donald M. West, F. James Holler, Stanley R. Crouch - Fundamentals of Analytical Chemistry-Cengage Learning (2013)) covered this topic briefly, but failed at offering any insightful information.
 To sum up, are the other partially protonated forms of EDTA involved in complexation equlibria with metallic ions? If yes, to what extent, what are their formation constants? Also, feel free to correct every scientific detail that I have mentioned you might disagree with.

Very good question! These two books are not meant for research so we cannot complain. For research topics, Google Scholar is your best friend. There are complete books dedicated to EDTA complexometry. Unfortunately, I do not have them now. Try to get hold of EDTA Titrations: An Introduction to Theory and Practice, Author: H. A. Flaschka

So you are right, there is indeed an optimum pH for each metal at which the stability constant has the highest value. For example, for calcium and magnesium highly basic pH is required (10-11).  So the right key word you have to search is speciation of desired metal-EDTA speciation diagram rather than EDTA speciation diagram.
The good news for analytical chemist (I believe you are studying analytical chemistry textbooks), is that EDTA always forms a 1:1 complex with the metal. We can forget if one of its coordinating site has a H attached to it.

As far I can remember, the metal EDTA formation constants are very large in each case, it is EDTA(4-) which is almost 100%.

See for example, Determination of NTA and EDTA and Speciation of Their Metal
Complexes in Aqueous Solution by Capillary Electrophoresis, Environ. Sci. Technol. 2000, 34, 885-891.


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Is the density of a two-component mixture always between the densities of its individual components?
There exist eutectic mixtures, in which the freezing point of the mixture is below those of either component of the mixture, i.e., the freezing point of the mixture is not intermediate between that of each component, but is instead lower than both.
Can the same occur for density? Are there mixtures of substances $\ce{A}$ and $\ce{B}$ such that either $\rho_\mathrm{mix}$ is greater than both $\rho_\ce{A}$ and $\rho_\ce{B},$ or $\rho_\mathrm{mix}$ is less than than both $\rho_\ce{A}$ and $\rho_\ce{B}?$
Or is the density of a two-component mixture always bracketed by (i.e., some intermediate value between) the densities of its individual components?

May I ask why this question is being downvoted? It seems a relevant question to those exploring chemistry, though perhaps technically more in the realm of physics?

No, the density of a mixture of A and B is not necessarily the weighted average of the densities of A and B. This is equivalent to saying that, when mixing two liquids, the total volume is not necessarily the sum of the two starting volumes; look Partial molar volume up. A common example is water-ethanol mixtures.

The problem is how it's written.  It wasn't sufficiently precise, plus you're not actually asking if density is additive. Densities are never simply additive, since density is an intensive property (like temperature -- you can see that temperature is not additive). You're asking whether the density of a 2-component mixture is always bracketed by (i.e., some intermediate value between) the densities of its individual components.  I've edited your question to make it clearer.

Well, optical density can often be treated as additive for the non-reacting mix, but I see no reason why mass density must ever be additive. Maybe have a look at the Wikipedia page on intensive and extensive properties

Even using the logic of density compared to the nature of compounds tells you that variations will occur. Mass will obviously stay the same but volume can be drastically different compared to the pure elements.

https://chemistry.stackexchange.com/questions/23534/how-does-volume-contraction-in-solvent-mixing-work https://chemistry.stackexchange.com/questions/23534/is-it-possible-to-have-volumesolution-volumesolvent-volumesolute https://chemistry.stackexchange.com/questions/83765/is-it-possible-to-have-volumesolution-volumesolvent-before-adding-solute

@Mithoron  Those answers all give examples of volume contraction, which is different from what's being asked. Volume contraction is a necessary but not sufficient criterion for the mixture to have a density that exceeds that of its individual components. Correspondingly, none of the volume contraction examples in those answers meet the OP's criterion:  Ethanol-water solutions never have a density greater than that of pure water (https://www.engineeringtoolbox.com/ethanol-water-mixture-density-d_2162.html) and NaCl(aq) solutions never have a density greater than that of NaCl(s) (2.16 g/cm^3).

....continued:  http://butane.chem.uiuc.edu/pshapley/genchem1/l21/1.html  [Assuming, of course, the pure components and the mixture are measured under the same conditions, particularly T and p.]

No, there is no law that requires the density of a mixture to fall between the densities of pure components. It does so most of the time, but the exceptions are not unheard of. Here's one.
Water: density $1.00\ \rm{g/cm^3}$
Hydrazine: $1.02\ \rm{g/cm^3}$
Hydrazine hydrate: $1.03\ \rm{g/cm^3}$
So it goes.

Any examples of the opposite, where the density of the mixture is lower than that of both the pure components?

Can't think of any, but there is no fundamental law forbidding that, either.

Ethanol/benzene and ethanol/toluene: https://www.researchgate.net/figure/Excess-volume-of-ethanol-benzene-top-and-ethanol-toluene-bottom-present_fig2_282982793. Benzene/m-xylene: http://www.ddbst.com/en/EED/VE/VE0%20Benzene%3Bm-Xylene.php.

Positive excess volume per se is not enough to ensure the condition in question.


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Kinetics of Sabatier reaction
I am modelling a methanation process in MATLAB, where I have a rate equation for methanation
$$\ce{CO2 + 4 H2 → CH4 + 2 H2O}$$
in which I need partial pressure of $\ce{CO2},$ $\ce{H2},$ $\ce{CH4},$ $\ce{H2O}.$
I am struggling to understand the link between the mass flow rate of $\ce{H2}$, $\ce{CO2}$ and partial pressure of all other species. The required amount of catalyst and the volume for the fixed bed reactors in accordance with the chosen stimulation parameters.
If suppose my operational pressure is $\pu{7 atm}$, $\pu{16 mol s^-1}$ of $\ce{H2},$ $\pu{4 mol s^-1}$ of $\ce{CO2},$ what will be the partial pressures of all those species and required amount of catalyst and reactor volume.
The paper in which I referred for rate reaction is this.

As a mechanical engineer you should keep in mind that "sec" refers to secant, whereas "s" refers to second.

If the gas is a stoichiometric mixture of  $\ce{CO_2}$ and $\ce{H_2}$, and if the total pressure is 7 atm, the partial pressure of $\ce{CO_2}$ before reaction is 1.4 atm, and it is 5.6 atm for $\ce{H_2}$. Now to get the corresponding values out of the reactor, some more information is needed, like the equilibrium constant, and the kinetics of the reaction.

The amount of catalyst required, and the volume of the reactor, will be dependent on the conversion of CO2 and H2 that is desirable. For example, for a stiochiometric amount of CO2 and H2, if you'd like to convert at least 95% of these reactants to CH4 and H2O, this will require more catalyst than if you were trying to convert 85%. The reaction system will be dynamic throughout the length of the Reactor. Take a look at the rate equation: as you convert more reactant to product, the partial pressures of the species important for driving the reaction forward (CO2 and H2) will  be lowered.

Not if water is removed.  Remember, the reaction is under pressure.

So you need three things:

Partial pressures of species.
Amount of catalyst.
Reactor volume.

We will design a PBR in order to calculate the mass of catalyst $ W $ needed to achieve a certain conversion $ X $.
Partial Pressures You have a heterogeneous reaction, since it takes place with a catalyst, but all the chemical species of the rxn are in the gas phase. It seems that you have no limiting reactant, as the stoichiometric ratio $ \ce{H2}/\ce{CO2} = 4/1 $ is the same as the ratio of the entering molar flow rates. I will choose $\ce{CO2}$ as the basis component, so the conversion refers to $ \ce{CO2} $. Say the reaction is $ A + 4B \rightarrow C + 2D $. The molar flow rate of any species is
$$ F_j = F_{A0} \bigg(\dfrac{F_{j0}}{F_{A0}} + \nu_j X\bigg) $$
The molar flow rates of all components, and the total molar flow rate, are
\begin{align}
  F_A &= F_{A0} (1 - X) \\
  F_B &= 4F_{A0} (1 - X) \\
  F_C &= F_{A0}X \\
  F_C &= 2F_{A0}X \\
  F   &= \sum_j F_j = F_{A0} (5 - 2X)
\end{align}
Thus, with $ P_j = y_j P = (F_j/F) P$, we have the partial pressures as a function of the conversion
\begin{align}
  P_A &= \left(\dfrac{1 - X}{5 - 2X}\right)P \tag{1} \\
  P_B &= \left(\dfrac{4 - 4X}{5 - 2X}\right)P \tag{2} \\
  P_C &= \left(\dfrac{X}{5 - 2X}\right)P \tag{3} \\
  P_C &= \left(\dfrac{2X}{5 - 2X}\right)P \tag{4}
\end{align}
Now, these equations are strictly true at the entrance of the reactor, i.e., where the conversion is $ X =0 $. Once we are in the reactor, the gas phase will suffer pressure drop.
Pressure drop I will assume that the gas flow is, at every coordinate, turbulent. The Ergun equation simplifies to
$$ \dfrac{dP}{dz} = -\dfrac{7}{4}\dfrac{(1 - \phi)G^2}{\phi^3 \rho D_P} \tag{5} $$
where $ G = \dot{m}/S $ is the mass flux (constant along the reactor), $ D_P $ is the catalyst particle diameter, and $ \phi $ the void fraction. We need to relate the density as a function of things we know, by using the ideal gas law
$$ \dfrac{dP}{dz} = -\dfrac{7}{4}\dfrac{(1 - \phi)G^2}{\phi^3D_P}
\dfrac{MRT}{P} \tag{6} $$
Now we need to relate the $ z $ coordinate to the $ W $ coordinate
\begin{align}
  W &= (1 - \phi) S z \rho_c \\
  dW &= (1 - \phi) S dz \rho_c \\
  dz &= \dfrac{dW}{(1 - \phi) S \rho_c} \tag{7}
\end{align}
where $ \rho_c $ is the mass density of the catalyst. Combining Eqs. (6) and (7)
\begin{align}
  (1 - \phi) S \rho_c \dfrac{dP}{dW} = -\dfrac{7}{4}
  \bigg(\dfrac{(1 - \phi)G^2}{\phi^3D_P}\bigg)\dfrac{MRT}{P} \\
  \dfrac{dP}{dW} = -\dfrac{7}{4} \left(\dfrac{G^2}{S \rho_c \phi^3D_P}\right)
  \dfrac{MRT}{P} \tag{8} \\
\end{align}
If you are obsessive like me, the molar mass of the gas will change along the reactor, so you should also take into account
$$ M = \sum_j y_j M_j \tag{9} $$
Amount of catalyst This is obtained by a mole balance along the $ W $ coordinate for species $ A $
\begin{align}
  F_A\vert_W - F_A\vert_{W+\Delta W} + {r_A}' \Delta W = 0  \\
  \dfrac{F_A\vert_W - F_A\vert_{W+\Delta W}}{\Delta W} + {r_A}' = 0 \\
  -\dfrac{dF_A}{dW} + {r_A}' = 0 \tag{10}
\end{align}
with $ F_A = F_{A0}(1 - X) $ or $ dF_A = -F_{A0}dX $, Eq. (10) yields
\begin{align}
  F_{A0}\dfrac{dX}{dW} + {r_A}' = 0 \\
  \dfrac{dX}{dW} = \dfrac{-{r_A}'}{F_{A0}} \tag{11}
\end{align}
where $ -{r_A}' $ is the rate of consumption of species $A$ in $ \text{mol A}/\text{kg cat} \cdot \text{s} $.
Energy Balance As a starting point, before studying an adiabatic reactor, assume that the reactor is isothermic. This translates the energy balance in the amount of heat $ Q(W) $ that you must remove at every mass coordinate in order for the temperature to remain constant. Hence, we disregard the energy balance for the first analysis, and $ T = T_0 $.
Thus, to know the mass of catalyst in the reactor, you need to numerically solve the following ODE's
\begin{align}
  \dfrac{dX}{dW} &= \dfrac{-{r_A}'}{F_{A0}} \hspace{1 cm} X(W = 0) = 0 \tag{12} \\
  \dfrac{dP}{dW} &= -\dfrac{7}{4} \left(\dfrac{G^2}{S \rho_c \phi^3D_P}\right)\dfrac{MRT}{P} \hspace{1 cm} P(W = 0) = P_0 \tag{13} \\
\end{align}
Other data we need to know

To evaluate the rate of reaction you need a kinetic law. To my surprise, there is one in the paper, and has the type of a homogenous kinetic rate law of the type
$$ -{r_A}' = f(P_A, P_B, P_C, P_D) = k'\bigg(P_A P_B^4 - \dfrac{P_C P_D^2}{K}\bigg)  $$
at constant temperature, where $ k' = k(T=T_0) $ and $ K = K(T=T_0) $ are fixed. In the partial pressures you plug Eqs. (1-4), this is why we had to do all that math.
You stop the integration depending on your objective. If equilibrium allows you, you should choose the highest conversion possible, $ X = 0.95X_{eq}(T) $ is a starting point.
Other things like the catalyst mass density $ \rho_c $, the void fraction of the PBR $ \phi $, the area $ S $, and the particle diameter of the catalyst $ D_P $.

Reactor Volume Once you solve the ODE this is just one single calculation, you just use the Eq. (7) but not differentiated, since the volume is $ V = Sz $
$$ V = \dfrac{W}{(1 - \phi)\rho_c} $$
This is an interesting problem, once I have more time, I will do some code to solve these balances. I think there is some other guy that asked for something similar.


text:
Conversion of molecular structure into Fischer projection for compounds having more than two chiral carbons
Question

What is the Fischer projection of the given compound's enantiomer?

Answer

My attempt
I tried to solve this by first converting it into the Fischer projection of the given compound and then taking its mirror image since that would provide the Fischer projection of the enantiomer
This is what I got as the Fischer projection of the given molecule.

Why is this wrong?
How do you find the Fischer projection for a compound having more than two chiral centers?

You can observe the molecule from left, then you will observe that Cl and CH3 is towards you and is toward right and left side of horizontal line respectively. OH and chain part is away from you and OH is on the top side of vertical line and chain part is toward the bottom side. Proceeding in the similar way you will get the correct Fischer projection.

As a trick you can first assign the R-S configuration to each carbon in the wedge form then make your Fischer projection accordingly. Then you can rotate the projection in the plane of paper(as it preserve configuration) to match your answer accordingly.

@Manu what do you mean by looking at the compound from the left? can you explain w.r.t the C2 carbon?

How does the F and CH3 come on the same side? since both are going into the plane and so when looking from left or right, won't that be same?

You can visualize it in the following way (for C1)- When you are looking it from above the plane of paper, then carbon chain and OH in the plane of paper and CL is toward you and CH3 is away from you. Now rotate the paper along the vertical axis counterclockwise 90° then CH3 and Cl is toward you and C-chain and OH is away from you. I hope now it is clear to you. Basically you see the molecule from that side where both wedge and dash groups are towards you.

I did what you said. However in such a scenario, how do you check the 2nd Carbon? Do we need to rotate the C1-C2 bond or is there any other way to consider?

Have you tried making physical models and rotating the bonds to align them like this? It's one thing to talk about converting projections/diagrams by a process but they are intended to show the physical structure of the molecule, so being able to visualise that molecule directly is the first step.

When we visualize the similar way you find F to be on right side and Br to be on left side.and the third carbon on below the vertical line.

My way to solve such problems is by calculating R,S configuration of all chiral carbon in wedge - dash form and then drawing the fisher projection with the same R,S configuration.

The groups to the left and right of the backbone are both essentially wedge up, if that helps.

Please note that, according to IUPAC recommendations, the use of Fischer projections in non-carbohydrates is problematic. Any drawings of this type are necessarily ambiguous since it is unclear whether the author of the drawing intended to make a Fischer projection or simply happened to draw a diagram in that orientation.

@FadedGiant I think i specified it was a fischer projection here.

To do this exercise, you must need to know how to assign stereo configuration according to Cahn–Ingold–Prelog priority rules, and what Fischer projection represent. First, get correct assignment of three stereo centers in the given molecule. That should be $(1R,2R,3S)$ as depicted in the image:

Now, sketch a Fischer projection with three stereo centers without assigning any groups but number the stereo centers as $1,2,3$. Put the least priory group of stereo center 1 (that would be $\ce{CH3}$ group) on top of the vertical line. By definition, top and bottom vertical lines are wedges-down (dashes) while all horizontal lines are wedges-up (solid wedges). If you put highest priority group ($\ce{Cl}$ group) on left horizontal line and second priority group ($\ce{OH}$ group) on right horizontal line, you get the correct clock-wise rotation for $(1R)$ configuration (See the image). You can follow same procedure or the one you are familiar to assign other stereo centers correctly. For example, for stereo center 3, the least priority group is $\ce{H}$ while the highest one is $\ce{F}$ group. The second priority group is substituted carbon chain, which is already in the projection (the last priority to consider is $\ce{CH3}$ group). Thus, I put $\ce{H}$ group on bottom vertical line (should be a dash) as did for stereo center 1. Then, to get correct counter-clock-wise rotation to give $(3S)$ configuration, you must put $\ce{F}$ group on right horizontal line so last priory group $\ce{CH3}$ would occupy the remaining left horizontal point.
Note: You can exchange any group at a stereo center with another group at same center to reverse the configuration, e.g., $(R) \rightarrow (S)$. If you repeat same with two different group, you get back the original configuration. Once I was done with assigning the groups, I have performed these operation twice to change the positions of three groups at stereo center 3 to my likeness. Still, the configuration remains same.
Finally, you may able to assign remaining two groups ($\ce{Br}$ and $\ce{F}$ groups) for stereo center 2 to get $(2R)$ configuration.
To get the enantiomer, you can sketch the mirror image of the compound, which is $(1S,2S,3R)$ as depicted in the image and also tally with the given answer for the corresponding enanthiomer.

You can arrive at the answer without assigning R and S if you build a model or do some mental contortions. The conformation should be all-syn with respect to the atoms on the vertical axis (like it is for cyclic structures, e.g. sugars in the furanose or pyranose form). Then, you can simply read off which groups are on the left and on the right.
In the starting model, the molecule is in an all-anti conformation. After changing the C2-C3 and the C3-C4 torsion angles by 180 degrees and turning the molecule, you can verify the given answer.

EDIT: I missed the part about drawing the enantiomer of the given structure, and the stereochemistry of my model does not match that of the question, but the process would be the same for the correct stereoisomer.
In general terms, if you start with a wedge-and-dash representation with the molecule in horizontal orientation, the substituents pointing towards you will end up on alternate sides of the Fischer projection, as indicated in the following illustration:

The reason is that for each chiral center, the atoms bound on the top and bottom of the atom should be behind the plane of the paper. This is only the case for atoms that are on top of the zig-zag horizontal structure after the molecule is rotated such the the "top" label is on top, and the "bottom" label is on the bottom. For the other chiral centers, left and right get switched when looking at them in the correct orientation.

See https://chemistry.stackexchange.com/questions/137145/in-the-absence-of-a-physical-model-how-can-i-use-free-softwarel-to-visualize-st


text:
Calculate molarity given density and mass percentage
Problem

Calculate molarity of a solution of $\ce{H2SO4}$ with density $\pu{1.198 g/cm3}$ and containing $27~\%$ mass of $\ce{H2SO4}.$

Answer

 $\pu{3.374 M}$

My approach
I converted the density to g/l. Then i took 27% of the density and converted it into no of moles. Thus, came moles per litre, which is actually molarity. Yet, the answer key that my teacher had provided had a different answer compared to mine.
Is my method wrong or is there any calculation mistake?

Write down the general algebraic calculation of molarity you have used. If it is right, result must be right as well. If you did trivial numeric error, it is easier to notice it this way The same if your algebra was wrong.

Sometimes a good strategy is to assume you have 1 g of solution (and therefore 0.27 g of acid), and work from there. Also, what is the molecular mass of H2SO4?

@BuckThorn I disagree with the usefulness of an approach where a physical quantity is fixed or its value taken arbitrary. When solving properly using algebraic equation, there is virtually no need to take any assumptions like that, which might be quite confusing and sometimes even wrong. P.S. also, on the nitpicking side, it's molar mass, not molecular mass you are probably referring to.

@andselisk Yes, I was thinking molar mass and wrote molecular mass. Regarding the use of 1 g, it is a very useful strategy imho because it takes an arbitrary unknown (the size of the solution as either total mass or volume) out of the problem. It also does not affect the propagation of uncertainty. So I disagree with you :-)

@BuckThorn well, to me it's a useless concept I have never seen used anywhere else beyond introductory chemistry books of questionable quality that fellow physicists would relentlessly laugh at. A new variable is introduced out of nowhere that doesn't bring any new information, but causes confusion. "What if I only have 0.1 g of NaCl, how would I solve with 1 g?"; "how many sigfigs do I use with 1 g? Is 1.0000 g enough?"; "can I take 1 oz instead?"; "do I always plug in 1 ?"; "can I plug in 1000  because there is "kilo"?" and so on. Just don't do it.

@andselisk It works for me, I use it occasionally. You can get just as confused solving equations. It's usually simplest to work backward: what do I need to compute a concentration? Hmmm, moles (sorry, amount of substance) and volume. How do I get volume from density? (divide mass by density) How do I get moles? (divide mass by molar mass) What am I missing? (molar mass) etc

Of course, for the method to work you have to recognize other things which might require a little experience: that the scale of the solution is not relevant, that you should use consistent units, that you can do some simple algebraic acrobatics in your head to save time, that power exponents often cancel, etc

@BuckThorn If it works for you, it's great. I would also use a mixing cross method occasionally in the lab and other not-exactly-correct things. But none of them should be  forced as the primary method to the students.

Solution
By definition, molarity $c$ is the amount of substance $n$ per volume of solution $V$:
$$c = \frac{n}{V}\label{eqn:1}\tag{1}$$
The volume $V$ can be found using total mass of solution $m_\mathrm{tot}$ and its density $\rho$:
$$V = \frac{m_\mathrm{tot}}{\rho}\label{eqn:2}\tag{2}$$
The amount of substance $n$ can be found using mass of solute $m$ and molar mass of solute $M$.
Knowing mass fraction $\omega,$ one can switch to the mass of solution $m_\mathrm{tot}$ already used in \eqref{eqn:2}:
$$n = \frac{m}{M} = \frac{\omega m_\mathrm{tot}}{M}\label{eqn:3}\tag{3}$$
By plugging both \eqref{eqn:2} and \eqref{eqn:3} into \eqref{eqn:1}, one receives the final expression:
$$
\require{cancel}
\begin{align}
c &= \frac{\omega \cancel{m_\mathrm{tot}}\rho}{M\cancel{m_\mathrm{tot}}} \\
  &= \frac{\omega\rho}{M} \\
  &= \frac{0.27\times\pu{1.198 g cm-3}}{\pu{98.08 g mol-1}} \\
  &= \pu{3.30E-3 mol cm-3}~\text{or}~\pu{3.30 mol L-1}\tag{4}
\end{align}
$$
Notes

There is little to no point of converting anything prior to calculations as if the dimensional analysis is done right, you'll get the answer anyways.
Density is an intensive property: you cannot use mass fraction to determine the density of solute alone.
Regarding "no of moles": first, don't use non-standardized abbreviations like "no", and second, it's amount of substance, not "number of moles".
"Number of moles" is an appropriate term in a dialogue with your neighbor about the population of talpids in your garden, but not in science.
Don't forget to define each physical quantity properly and always try to solve the problem algebraically first, and plug in the values at the end when you have the ready-to-use equation for the quantity you were asked to find.
Keep the track of significant figures (slang: "sigfigs"), e.g. you are given the density with four sigfigs, so it would make sense to also use molar mass with four sigfigs as well.
Yet due to two sigfigs of mass fraction, the accuracy of the final answer is reduced.

There is something to be observed that the answer is different from the given answer in my question. Thus, there can be two explanations, either my teacher made a mistake jn the answer key or u might've made a mistake. I am not sure but I too got the same answer as you


text:
Can pthalic acid or 1,2-Benzenedimethanol be possible products of Cannizzaro reaction of phthalaldehyde?
I know the main product that is ortho-hydroxymethyl benzoate ion but I think that phthalic acid and catechol must be formed also in very small amounts but I am unable to get any confirmation from the internet. Please help me with the products.

As far as intramolecular Cannizaro is concerned, I believe that one part oxidizes and the other gets reduced only. Can you shed some light on your intuition (like mechanism) ?

what I am thinking is that in the case of intramolecular Cannizzaro what you write is correct but take an example, if there are two molecules of phthalaldehyde and one molecule oxidizes and molecules reduce completely that is no intramolecular Cannizzaro then some other products will be formed.

You won't form catechol (aka, pyrocatechol). I believe you are thinking of the diol, 1,2-phenylenedimethanol.

https://www.nrcresearchpress.com/doi/abs/10.1139/v81-157#.XwI4UpOpHlw

@user55119 you are correct catechol is not formed.

The Cannizzaro reaction is the disproportionation of an aldehyde, which do not have an $\alpha$-hydrogen, to an equimolar mixture of corresponding primary alcohol and carboxylic acid salt. For example, original Cannizzaro's experiment:
$$\ce{2C6H5-CHO ->[aq NaOH] C6H5-CH2OH + C6H5-COO^-Na+}$$
Since there is no $\alpha$-hydrogen, the mixture does not undergo an aldol condensation, which is the characteristic of this reaction. The Cannizzaro reaction is an example of a typical redox reaction, which is considered to be one of the most important synthetic reactions of organic chemistry prior to the discovery of $\ce{LiAlH4}$ in 1946 (Ref.1).
The proposed reaction mechanism is as follows (Ref.1):

According to Ref.1, the hydride transfer must be from one molecule of benzaldehyde to the other molecule (when the reaction is performed in $\ce{D2O}$, no deuterated $\ce{CD2-O}$ is found in benzyl alcohol). The formation of benzyl benzoate was suggested and supported (e.g., Ref.2), evidence in Ref.1 doubts this proposal.
The Cannizzaro reaction of phthalaldehyde is studied in aqueous base ($\ce{NaOH}$) to obtain sodium o-hydroxymethylbenzoate as the product. The studies include the kinetics of the reaction, which have been measured over a range of $\ce{NaOH}$ concentration at $\pu{40 ^\circ C}$. This kinetic study proposed the hydration of phthalaldehyde to hemi-ketal monooxoanion is one of the first steps in the mechanism (hemi-ketal formation is depicted in the insert box of above image). The dioxoanion formation is depend on the concentration of $\ce{NaOH}$ solution (Ref.3). The hydride transfer from mono- or di-oxoanion follows, which is the rate determining step (determined by the isotope effect). The observed rate is based on following equation:
$$k_\mathrm{obs} = \frac{a[\ce{OH-}]}{1 + b[\ce{OH-}]}$$
where $a= \pu{(1.38 \pm 0.02)\times 10^{-3} M-1s-1}$ and $b= \pu{0.99 \pm 0.02 M-1}$ (the reaction follows this equation for $\pu{0.05-2.0 M}$ $\ce{NaOH}$). There are no other product has been observed.
I understand your concern of intra- versus inter-molecular hydride transfer. If intermolecular transfer occurs, at least 1,2-bis(hydroxymethyl)benzene and 1,2-benzenedicarboxylic acid should form as minor components. Since hydride transfer is the rate determining, intramolecular hydride transfer is much faster than that of intermolecular version. Therefore no chance of minor product formation is possible since all other steps are in equilibrium.
Note: I assume your suggestion of forming phthalic acid and catechol as byproduct is mistaken for 1,2-bis(hydroxymethyl)benzene and 1,2-benzenedicarboxylic acid (phthalic acid). Keep in mind that there is noway of breaking two aromatic-CHO bonds and making two aromatic-O bonds to form catechol molecule.
References:

C. Gardner Swain, Arnet L. Powell, William A. Sheppard, Charles R. Morgan, "Mechanism of the Cannizzaro reaction," J. Am. Chem. Soc. 1979, 101(13), 3576–3583 (https://doi.org/10.1021/ja00507a023).
Arthur Lachman, "The Benzyl Rearrangement v. Cannizzaro's reaction," J. Am. Chem. Soc. 1923, 45(10), 2356–2363 (https://doi.org/10.1021/ja01663a017).
Robert S. McDonald, Christine E. Sibley, "The intramolecular Cannizzaro reaction of phthalaldehyde," Canadian Journal of Chemistry 1981, 59(7), 1061-1067 (https://doi.org/10.1139/v81-157).


text:
How to predict the reason for color in a transition metal complex?
I know about two reasons for the color of transition metal complexes that are dd transition and charge transfer phenomenon but I am confused when a complex can show both of the above-mentioned phenomena then how to choose which can be the dominating reason for the color of the complex.

How to predict the reason for color in a transition metal complex?

Nobody can do it just by "looking" at the color of any compound. Spectroscopists have to do calculations to figure that out. There are at least 15 reasons for the "color" of chemical compounds.
Now qualitatively one can "predict" whether it is a d-d vs. it is charge transfer transition for transition metal compounds.
Keep in mind that d-d transitions are "forbidden" (Laporte selection rule). Therefore, majority of the d-d transitions correspond to low molar absorptivity. For example, Cu(II) solution is light blue, manganese (II) is almost invisible pink, chromium (III) is light green.
Charge transfer transitions are very intense! Add two three crystals of potassium permanganate to water, and it is intensely purple.


text:
Chemical elements or compounds with yellowish white color
Are there chemical elements or compounds that have a yellowish white (cream) color in their natural state?

Aluminium(III) chloride - often described as white but samples is sometimes contaminated with iron(III) chloride giving it a light yellow color
Antimony(V) oxide - pale yellow solid
Bismuth(III) oxide - light yellow solid
Rhodium(III) oxide - lemon yellow
Cerium(IV) oxide - pale yellow-white compound
Dysprosium(III) nitrate - white to pale yellow
Holmium(III) oxide- Color depends on lighting condition. In daylight, it is tannish light yellow; Under trichromatic light, it is fiery orange red (Thanks @EdV)
Samarium(III) oxide - light yellow
Silver(I) iodide - samples can have a range of color(white to pale yellow to yellow) due to impurity and photosensitization
Selenium(IV) chloride - white to faint yellow volatile solid
Gold(III) chloride - pale yellow to yellow
Potassium Telluride - pale yellow
Thallium(III) bromide - pale yellow
Some vanadate salts are light yellow

Bromium nitrate is pale yellow but not confirmed. Silver(I) bromide and lead(II) bromide are light yellow and white cream colored respectively.
Above data source: http://lanthanumkchemistry.over-blog.com/article-the-colors-of-chemicals-complexes-104821449.html

(+1) Excellent list! I have numbers 5, 8 and 11 and can confirm that the color of holmium (III) oxide depends on the illumination, as per here.

@EdV Thank you. I am editing that information onto the answer. I was hoping for someone to confirm the list as I compiled the data from a single source.

In addition to the various inorganic compounds listed above, there are countless organic compounds (and mixtures/emulsions/suspensions/etc. of organic compounds) that satisfy this criterion. Also, sulfur contamination of something that's normally white will give you something yellow.

Calcium hypochlorite is an example of such a yellowish white compound.
Another example would be Silver Bromide

I am unable to comment and so this is the only way I can answer a question

Could the downvoter please explain the reason for downvoting?

The only reason I can think is that your link to Ca(OCl)2 takes you to AgBr

Thanks for that. @IanBush I had both links marked in the answer but had a mix up in alloting them. I fixed the issue.

Article you link says calcium hypochlorite is white/gray. Also such questions are supposed to be closed as too broad, not answered.

The reaction of any acid on a thiosulfate solution makes a precipitate of sulfur $\ce{S_8}$ which is pale yellow, nearly white.

Undyed ("natural") nylon is yellowish-white:

Source of picture: aiplastics.com


text:
Why is cell potential an intensive property?
I've heard that cell potential is an intensive property. However, we know that an intensive property does not depend on number of moles, mass, or concentration of the substance involved. However, when we calculate cell potential using the Nernst equation, we definitely have to use the concentration of the substances. Therefore, how is cell potential categorized as an intensive property?

An intensive property does not depend on the total mass, or total volume. But it may change with the concentration.

Welcome to Chemistry! Please note that the proper term for "number of moles" is amount of substance. The former would be the same as referring to the mass as "number of kilograms".

The simple mental test for the property being intensive or extensive is, what happens if you make the all the system twice as big.
If the property remains the same, like concentration, or the cell electromotoric force ( voltage ), it is intensive. If you use the double volume of the solution, its  concentration is still the same. The open voltage of Li-ion cell does not change with the size. You can notice from the Nernst equation the potential depends on concentration, but not on the mass nor volume.
Intensive properties must not depend on extensive properties, unless they mutually cancel their extensiveness, forming an intermediate intensive property, like e.g. specific conductivity: $$\kappa_\mathrm{intensive} = const_\mathrm{intensive} \cdot  \left( \frac{m_\mathrm{extensive} }{ V_\mathrm{extensive}}\right)_\mathrm{intensive} $$
If the property changes additively, like mass, energy, heat capacity, it is extensive. The cell capacity does change with the mass of the active material and electrolyte. Extensive properties may depend on any other properties.
If the property changes, but not additively with the system scaling, it is neither intensive neither extensive.


text:
Why do alkylarenes react with alkaline KMnO4, but not benzene?
Benzene($\ce{C6H6}$) does not react with alkaline $\ce{KMnO4}$, however alkylarenes react when there are benzylic hydrogens. What could be the reason for this?
Thanks in advance

Benzene does not have any benzylic hydrogen it's formula is C6H6 while alkyl arenes 1 degree or 2 degree have benzylic H atoms which get oxidized in presence of strong oxidizing agents like alkaline KMnO4
The mechanism given in this link is as follows.


text:
What is the CNC bond angle in methyl isothiocyanate?
According to VSEPR theory, for $\mathrm{sp^2}$ hybridisation the bond angle is approximately $120^\circ$. In $\ce{H3CNCS}$, the nitrogen has one lone pair. As lone-pair-bond-pair repulsion is greater than bond-pair-bond-pair repulsion, the bond angle of $\ce{CNC}$ should be less than $120^\circ$,  but it is actually greater. How can this be rationalised?

I took a liberty to correct formatting, notations and added compound's name. But where did you find a C−N−C in $\ce{\ce{CH3CNS}}?$ What angle are you talking about, exactly? Have you seen https://doi.org/10.1071/CH10303, especially its SI available for free?

Actually a typing mistake. It would be CH3NCS. Sorry for the inconvenience.

No prob, I corrected the name and bond order accordingly. I'm pretty sure answers from Bond angle in CH3NCO would apply here as well: partial delocalization of the lone pair on N into a π orbital on C. This effect is even more pronounced in silyl isothiocyanate, which is almost linear.

@andselisk the answers aren't really good though. To all who are tempted, I would let this question open for the time being.

Related: What is the Si–N–C bond angle in silyl isothiocyanate?

I have just answered What is the CNC bond angle in methyl isocyanate? Pretty much everything written there also applies to methyl isothiocyanate, just a bit more extreme.
Here is the structure calculated on the RI-BP86/def2-SVP level of theory:

The bond angle is $\angle(\ce{CNC}) = 156^\circ$, so we expect a much larger contribution of the right side in the following resonance description:
$$\ce{H3C-N=C=S <-> H3C-N^+\bond{3}C-S^-}$$
Reasons for this are not trivial, but that would exceed the scope of this answer.
TL;DR From VSEPR we would expect a trigonal coordination and a $120^\circ$ angle. Refining with resonance, we'll expect a widening of the bond angle. Everything beyond that  needs much more involved models.

Similar to an other comment, not about methyl isothiocyanates, but $\ce{R-N=C=S}$ there are at least these 15 entries in the COD with coordinates to compare with: http://www.crystallography.net/cod/7116329.html and (just substitute the numbers), COD7116328, 7116327, 7116326, 7116325, 7103098, 4117347, 4108942, 4108356, 4075734, 4024672, 4024671, 2234166, 2233483, and 2214501.


text:
Existence of Hartree–Fock limit
I'm reading McQuarrie's Physical chemistry textbook [1]. On p. 281, the author explains:

If we use a more flexible trial function of the form in which $\psi(\mathbf{r_1},\mathbf{r_2})$ is a product of one-electron functions, or orbitals,
$$\psi(\mathbf{r_1},\mathbf{r_2})= \phi(\mathbf{r_1})\phi(\mathbf{r_2})\tag{8.15}$$
and allow $\phi(r)$ to be completely general, then we reach a limit that is both practical and theoretical. […]
This limiting value is the best value of the energy that can be obtained using trial function of the form of a product of one-electron wave equations (Equation 8.15). This limit is called the Hartree-Fock limit.

But why does Hartree–Fock limit exist? If we increase the number of parameters of the trial function we use, we may expect better calculation results. Nevertheless, why is the wave function multiplied by the two trial functions limited to the accuracy that can be achieved?
My guess is that the functional form does not properly consider the interelectronic correction term by treating the two trial functions independently. But can't this difference be overcome by using more parameters? I wonder if I can theoretically explain the existence of the Hartree–Fock limit.
In addition, if this limit exists, the function will have to be altered to produce better calculation results. I wonder what theory computational chemistry has developed about this.
Reference
McQuarrie, D. A.; Simon, J. D. Physical Chemistry: A Molecular Approach; University Science Books: Sausalito, California, 1997. ISBN 978-0-935702-99-6.

This question might be very well suited for https://mattermodeling.stackexchange.com/.

@Felipe S. S. Schneider I agree that mattermodeling is more specialized community related to the concept of my question. But first of all, because the book that I was asking was related to physical chemistry, I put it up on here. (It seems to span the boundary.) If I can't receive any informative answers, I'd post it on there. Thanks for letting me know such website :)

HF is variational; that ensures that the energy is higher of equal to the exact energy of the system. Couple this with the fact that HF cannot reproduce all of the correlation energy, electron-electron interactions are treated in a mean field approximation, you can see, that the energy of HF does converge to a higher value than the exact energy. Advances are, to name but a few, perturbation theory, configuration interaction, and other so called post-HF methods. That is a broad field though and a more detailed text on computational chemistry so be more helpful than this site.

The issue is: what parameters can we introduce? As for changing the form of the trial wavefunction to something else to achieve a lower energy, this is prohibited by the variational principle, which HF theory satisfies.
In short (and within the context of Hartree-Fock), the variational principle guarantees that any given trial function will produce an energy equal to or greater than the solution provided by Hartree-Fock.


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Are atomic mass unit (amu), unified mass unit (u) and Dalton (Da) the same unit but with different names?
I recently read up on some chemistry and found these units of mass:

atomic mass unit (amu)
unified atomic mass unit (u)
Dalton (Da)

Are they all the same unit but just with different names or are they different? I'm looking for a simple explanation as I'm new to chemistry.

Well for one thing it isn't the unified mass unit but rather the unified atomic mass unit.

It is a simple question that requires a rather complicated answer.
Origins

For technical reasons, in 1898, chemist Wilhelm Ostwald and others proposed to redefine the unit of atomic mass as 1/16 of the mass of an oxygen atom. That proposal was formally adopted by the International Committee on Atomic Weights (ICAW) in 1903. ... This suggestion was made before the discovery of the existence of elemental isotopes, which occurred in 1912. ... This definition remained unchanged until 1961.$^1$

I think the Wikipedia article is wrong about the name. I think the notion was initially conceived as "atomic weight," not "atomic mass."
atomic mass unit(amu)

The discovery of isotopes of oxygen in 1929 required a more precise definition of the unit. Unfortunately, two distinct definitions came into use. Chemists choose to define the amu as 1/16 of the average mass of an oxygen atom as found in nature; that is, the average of the masses of the known isotopes, weighted by their natural abundance. Physicists, on the other hand, defined it as 1/16 of the mass of an atom of the isotope oxygen-16 ($\ce{^{16}O}$).$^1$

unified atomic mass unit(u)

...in 1961 the International Union of Pure and Applied Chemistry (IUPAC), which had absorbed the ICAW, adopted a new definition of the atomic mass unit for use in both physics and chemistry; namely, 1/12 of the mass of a carbon-12 atom. ... The new unit was named the "unified atomic mass unit" and given a new symbol "u", to replace the old "amu" that had been used for the oxygen-based units. However, the old symbol "amu" has sometimes been used, after 1961, to refer to the new unit, particularly in lay and preparatory contexts.$^1$

I'll be less diplomatic. Since 1961 "atomic mass unit" has been conflated with the "unified atomic mass unit."
dalton

In 1993, the IUPAC proposed the shorter name "dalton" (with symbol "Da") for the unified atomic mass unit.$^1$

"And now..., for the rest of the story" as Paul Harvey used to say.
Until the 2019 redefinitions of SI units, the number of "entities" in a mole was determined experimentally. Since 1889 the kilogram was the mass of a particular platinum/iridium cylinder. With the 1961 definition of the unified atomic mass unit, exactly 12 grams of $\ce{^{12}C}$ had exactly 1 mole of carbon atoms. So the experimental measure of a mole was based on how many $\ce{^{12}C}$ atoms would be in exactly 12/1000 of a kilogram.
However in 2019 both the kilogram and the mole were redefined to give them particular values that would no longer depend on experimental measurements. This was in a continuance of an effort by scientists to define a set of primary physical constants that didn't depend on particular physical standards such as scratches on a bar to determine the meter or a blob of metal to determine the kilogram.

The mole is now defined as exactly $6.02214076\times10^{23}$ entities.
The kilogram "is defined in terms of three fundamental physical constants: The speed of light c, a specific atomic transition frequency ΔνCs, and the Planck constant h."$^2$

The point of this is that with the mole and kilogram now being given definite fixed values, exactly one mole of carbon-12 atoms no longer has a mass of exactly 12 grams by definition. Thus now $u≈\pu{1.66053906660(50)\times 10^{−27} kg}$.

To chemists any difference between the pre and post 2019 values will be infinitesimally small. This is because the average atomic masses of the elements depend on the relative abundances of the isotopes of the element which vary. Thus the isotopic variation greatly limits the precision of the average atomic mass of the elements.
To physicists the difference in measurements will make a difference since the mass of a particular isotope is known with much greater precision that the average atomic mass of the element.

tl;dr For practical purposes the three terms are synonymous as user Maurice noted. It would seem that the preferred term is now the dalton.
$^1$ Wikipedia's article on the dalton (Da).
$^2$ Wikipedia's article on the kilogram.

There is a saying in project management The nice thing about standards is that there are so many to choose from.

Obligatory xkcd.


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Do some alkyloxiranes possess surfactant properties?
As an example, i notice that decyloxirane has a long non-polar "tail" and a tiny "head" that features two polar bonds. This molecule looks not so different from a common soap to me.
Would it act as a detergent when added to an industrial lubricant or hydraulic oil?

Being insoluble is quite a problem...

It´s surely a better surfactant than decane? And a lot worse than, say, SLES. Without an application or a specific property in mind, categorising sth as a surfactant is quite pointless.

@Karl Thanks for pointing that out. Edited to narrow it somewhat to a specific application.

I have no idea why you would add surfactants to an oil, especially now you said it should make a detergent. The central property is to stabilise immicible mixtures by forming micelles. What do you want to achieve?

@Karl The purpose is only to understand if alkyl oxiranes are able to function as detergents (even a gentle detergent action) or not. What i have in mind is that if such a molecule were to be added to a polar base oil it might be able to scrape any dirt/old oil degradation product (carbonized oil) off of pipe walls (via the non-polar tail) as the oil flows about and also able to "seize" such particles (thanks to the micelles) and keep them from depositing anywhere (except upon a filter element). But i could be wrong. That's why i'm asking


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Isomerization of phosphorylated 2-imino-1,3-dioxalanes into phosphorylated cyclic carbamates
I found a patent dating back to the Soviet Union. The patent number is 256764, and the patent describes the synthesis of phosphorylated cyclic carbamates. Here is a section of the patent, followed by a translation:

....A method of obtaining n-phosphoryl cyclic urethanes of the general formula [Picture] where R=alkyl, R1= H or alkyl, n=1,2, by the interaction of dichloromethyleneamides of phosphoric acid with glycols in the presence of organic bases, for example triethylamine, in organic solvent under cooling of the reaction mixture, preferrably to 0 C, is offered. In the case of using ethylene glycol or 1,2 propyleneglycol, and also 2,3- or 1,3 butyleneglycol, first, it seems, n phosphoryl 2-imino-1,3-dioxolanes are formed, which are converted by means of elevated temperatures to the desired products.
The rest of the patent gives an example of such synthesis.
It seems to me that the authors are suggesting that this reaction happens after the glycol is reacted with a dichloromethyleneamide of esters of phosphoric acid:

This looks similar to a Chapman Rearrangement, but, to my knowledge, a Chapman Rearrangement requires the presence of aromatic groups, so this isomerization reaction is not a Chapman Rearrangement. How and why does this isomerization reaction occur? What is the reaction mechanism?
Link to patent: https://patents.su/2-256764-sposob-polucheniya-n-fosforilirovannykh-ciklicheskikh-uretanov.html


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Can E2 reactions be reversible?

Can E2 reactions be reversible?

A lot of E2 reactions I have seen are irreversible. Is this true in general?

(Image source: Organic Chemistry 11e by Solomons)
I tried to reason this using the following argument; The transition state requires the base, the $\beta$-hydrogen, $\alpha,\beta$-carbons, and the leaving group to be simultaneously co-planar. In the forward direction, it is usually possible due to the stability of the staggered conformation. In the reverse, we would require both the leaving group and the conjugate acid of the base to become simultaneously co-planar with the double bond, which would have nearly $0$ probability.
Is this reasoning correct? Or is there an example of a reversible E2 reaction?

An objective way of quantifying reversibility is simply to take a look at the values of the equilibrium constants for such elimination reactions.

The principle reason why it's at least rare in the reverse direction is that, by microscopic reversibility, you would have a transition state with three components, decreasing the likelihood of forming the transition state for entropic reasons. Whether or not it's true in general is difficult to say, since enzymatic processes may overcome the previous point.

@Zhe so can we say that non-enzymatic E2 processes are irreversible in general?

The E2 does only have one transition state, however, the energy required to move in the reverse direction for it is quite large.

In general, sure.


text:
Are there any tasteless edible salts?
I saw on Youtube a video where caffeine was extracted from Red Bull using dichloromethane. However, due to dichloromethane tendency to form emulsions with liquid, an inefficient method was used to mix the dichloromethane and the Red Bull to minimize the emulsion.
The video maker stated that an emulsion can be broken by adding a salt, but a salt would also drastically affect the taste of the final product. I was wondering, are there any tasteless edible salts that could have been used to break the emulsion without affecting the flavor?

Interesting, I found that magnesium salts taste bitter, also this would give you diahrrea

I found boron phosphate and (KCl?) from the comments of this previous question as an example of an odorless an tasteless salt (is it edible? Don't think so!)

SAFETY - I wouldn't taste the extract, and I surely wouldn't drink it. There will be some DCM in the aqueous phase.

@MaxW They extracted the dichloromethane using a vacuum. An hour under the vacuum seemed to be enough to evaporate it all.

The protocol and the title question are two different things. Adding a salt, even a salty tasting one, helps the extraction by increasing the density of the aqueous fraction and/or via electrostatic effects, depending on the density of the solvent used. The salt is supposed not to be in the targeted organic phase, when the latter is properly washed.

@Alchimista Proper washing cannot be used without compromising the quality of the final product because red bull contains lots of flavorings

I am speaking of chemistry operation. And (most) the flavors would be in the caffeine containing phase as well. I do not see any reason to extract caffeine from Red Bull. Not to say it has a terrible taste itself. If you want to drink something caffeine free buy that. Or make a version of it from scratch. Also I wonder why you are concerned with a bit of salt while using dichloromethane.

@Nilay Gosh KCl is a bit salty mostly bitter.

@Alchimista that's why I put a question mark ;-)

That is why I answered :) @Nilay Gosh

Giving a cursory Google e-book search gave me results particularly from old textbooks of 19th century:

Phosphate of soda (Sodium phosphate, $\ce{Na3PO4}$) - was sold under the name of tasteless salt
Sulfate of lime (calcium sulfate, $\ce{CaSO4}$) - it was named agustine because this salt was considered to have no taste. It was considered to mix with acids to form tasteless salts
Phosphates of barytes (barium phosphate, $\ce{Ba3(PO4)2}$) - considered tasteless
Phosphate of strontium (strontium phoshate, $\ce{Sr3(PO4)2}$) - considered tasteless
Superphosphate of lime (calcium superphosphate, $\ce{Ca(H2PO4)2}$) - at high temperature, it melts to a semi-transparent glass which is tasteless.
Sulfate of strontium (strontium sulfate, $\ce{SrSO4}$) - considered tasteless

@Ivan said in a comment(deleted) that calcium sulfate, barium phosphate, strontium phosphate and strontium sulfate are insoluble in water, so they cannot serve the purpose of breaking the emulsion. Finally as OP clarifies, calcium superphosphate is corrosive and hence not safe to eat, it all comes down to sodium phosphate which is both tasteless and edible.
References(in order of mentioned points)

London Medical Gazette: Or, Journal of Practical Medicine, Volume 18, 1836
The Critical Review: Or, Annals of Literature, Volume 33
Tobias Smollett, 1801
Pantologia: A New Cabinet Cyclopaedia, Comprehending a Complete Series of Essays, Treatises, and Systems, Alphabetically Arranged, J. Walker, 1819
The Cyclopædia, Or, Universal Dictionary of Arts, Sciences, and Literature, Volume 39, Longman, Hurst, Rees, Orme & Brown, 1819

Sodium phosphate is indeed edible and is sold food grade.
Calcium superphosphate turned up this page: https://pubchem.ncbi.nlm.nih.gov/compound/Calcium-superphosphate calling it "corrosive" and an "irritant" so probably not safe to eat or drink

Taste is rather tricky: I have a vial of "P.T.C Paper" test strips "for use in medical genetics", as it says on the vial's label. The strips contain phenylthiourea-phenylthiocarbamide and it is said that people either find the taste extremely bitter (as I do) or completely tasteless. Most people I have tested this on find the test strips tasteless (no pun intended).

@EdV  I think that's the exception that proves the rule. The PTC is famous because there is such a difference in reaction to it, and such striking differences are unusual.   Of course, there is individual variation in taste.  But, for the most part, the perceptions of  bitter/sweet/salty/sour are universal.   Very few people don't perceive sugar as sweet, NaCl as salty, lemon juice as sour, etc.

No argument! Still not sure if I ever want to try eating durian, if I ever get the opportunity! ;)


text:
Which make HCO3- to show two pH values at two scenarios?
According to the below two titrations,

Image Reference
If we consider the reactions, at the first reaction (in first figure after adding 1.0), there is $\ce{HCO3-, NaCl,}$ and $\ce{H2O}$ at the first equivalence point. At the second one (after adding 1.0), there is $\ce{HCO3-,}$ and $\ce{H2O, Cl-}$ at the first equivalence point. $\ce{HCO3-}$ only affect to the $\mathrm{pH}$ of the medium. But,

According to the first figure, $\ce{HA- + H2O → H2A + OH-}$    it acts as a base
According to the second figure, $\ce{HA– + H2O → A2– + H3O+}$    it acts as a acid

I am doubtful on why do $\ce{HCO3-}$ perform as two types at these different situations? Do anything in the medium  manipulate $\ce{HCO3-}$ to work as a base or an acid? How to describe these two scenarios? Generally I can't see a theoretical method for describe it.
I searched this in so many resources, but I couldn't find a satisfied answer. So, I hope a theoretical answer, for this problem, not a mathematical description.

The attached graphs have been taken from https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_Chem1_(Lower)/13%3A_Acid-Base_Equilibria/13.05%3A_Acid%2F%2FBase_Titration

The problem is that both titration curves are wrong: at the first equivalence points, the $\mathrm{pH}$ does not equal $\mathrm{p}K_\mathrm{a1}$ and the $\mathrm{pOH}$ does not equal $\mathrm{p}K_\mathrm{b1}$. As well, the two provided dissociation constants for carbonate ion, in the OP's upper figure, are also wrong.
The figure below shows two titration curves: 1) for $\pu{10 mL}$ of $\pu{0.1 M}$ "carbonic acid", with $\pu{0.1 M}$ $\ce{NaOH}$ titrant (black rising curve) and 2) for $\pu{10 mL}$ of $\pu{0.1 M}$ sodium carbonate solution, with $\pu{0.1 M}$ $\ce{HCl}$ titrant (red falling curve):

The $\mathrm{pH}$ of the starting $\pu{0.1 M}$ carbonic acid is 3.67 and the $\mathrm{pH}$ of the starting $\pu{0.1 M}$ sodium carbonate is 11.65. At the first equivalence point, for both titration curves, the $\mathrm{pH}$ is approximately 8.34. The $\mathrm{p}K_\mathrm{a}$ values are as shown in the figure and are from (or derived from) the reference at the bottom.
The next two figures show Excel spreadsheet $\mathrm{pH}$ calculations. The first one shows that $\mathrm{pH} = 8.338$ at the first equivalence point, for the neutralization of sodium carbonate solution with $\ce{HCl}$:

At the first equivalence point, the solution is $\pu{0.05 M}$ sodium bicarbonate plus $\pu{0.05 M}$ $\ce{NaCl}$. Ignoring ionic strength effects, the $\ce{NaCl}$ does nothing to the $\mathrm{pH}$.
The next one shows that $\mathrm{pH} = 8.339$ at the first equivalence point, for the neutralization of carbonic acid with $\ce{NaOH}$:

At the first equivalence point, the solution is $\pu{0.05 M}$ sodium bicarbonate. In both cases, the $\mathrm{pH}$ is approximately 8.34 and is consistent with this answer.
The two $\mathrm{p}K_\mathrm{a}$ values for carbonic acid are from D.G. Harris, In Quantitative Chemical Analysis, 7th Edition; Appendix G; W. H. Freeman and Company, ©2007. The two $\mathrm{p}K_\mathrm{b}$ values for carbonate ion are obtained from the two $\mathrm{p}K_\mathrm{a}$ values for carbonic acid via $\left(\mathrm{p}K_\mathrm{b1} = 14.00 - \mathrm{p}K_\mathrm{a2}\right)$ and $\left(\mathrm{p}K_\mathrm{b2} = 14.00 - \mathrm{p}K_\mathrm{a1}\right)$.

Your answer has described only half of my doubt. What about the titration of Na2CO3(Same concentration of your considered H2CO3) with HCl at the first equivalence point (after added 10ml of HC)l. Will the pH same to 8.34 as this?Or If it is not then what is the reason to having different pH values?

I have shown the image reference in the question.(after edit). But while HCO3- acts as a base(according to the 8.34 pH value) how do HCO3- moreover react with NaOH?

But while HCO3- acts as a base(according to the 8.34 pH value) how do HCO3- moreover react with NaOH? While there is no excess H+ in the medium?(8.34 shows there is excess OH-) What is that exception?

Let us continue this discussion in chat.

@MathewMahindaratne Thanks, it looks much better now!

@Ed V: It is the icing on the delicious cake! :-)

The equation shown in the second spreadsheet is the same, with slightly different notation, as the equation here: https://chemistry.stackexchange.com/a/43422/79678.

The (equivalence)points, which you have marked in the graphs, are where all of the acid present in the sample has been neutralized by the base added with the titrant –and vice versa.
But, pKa equals pH at the point where only half of the acid is neutralized. In other words: [HA] = [A-] -the concentration of acid equals the concentration of the corresponding base.
At these points, according to the Henderson-Hasselebalch equation, pH = pKa.
They are called inflection points, and is where the titration curve is horizontal.
See e.g. Wikipedia – titration curves

I think this is not the answer for which I have asked, I asked about the equivalence point which have every H2CO3 has converted into HCO3- and every Na2CO3 has converted into HCO3-(First equivalence point: Which has green color pointed pH=7.6(1) and pH=6.4(2) ).


text:
Rigorous reason behind internal energy change being zero while mixing

A container is divided in two parts: one part contains oxygen gas $(n_1$ moles, at temperature $T_1)$ and the other part contains helium gas $(n_2$ moles, at temperature $T_2).$ The partition separating them is removed. Find the final temperature of the mixture $(T).$

While solving this (multiple choice exam) question, since I was constrained by time, I started to look for hints within the question. I realized the question has explicitly mentioned the gases, i.e. it gave us information about their degrees of freedom (since oxygen is diatomic and helium is monoatomic). So my instinct was that this had to do something with $C_V.$ At the time, I hypothesized $\Delta{U}_\mathrm{sys} = 0.$ Thus:
$$n_1\frac{5R}{2}(T - T_1) + n_2\frac{3R}{2}(T - T_2) = 0$$
$$T = \frac{5n_1T_1 + 3n_2T_2}{5n_1 + 3n_2}$$
Which was indeed one of the options, and in fact, the correct one.
However, is there any rigorous method/reasoning that tells us why exactly is it that $\Delta{U}_\mathrm{sys} = 0?$ I can't think of a satisfactory explanation. I tried to analyze it using the first law of thermodynamics, but that clearly seems to be the wrong approach.

The system here is defined as the container, initially partitioned and then united.
The process in question is presumably adiabatic ($q=0$), since we are told that the temperature changes and are not given any information about the temperature of the surroundings (adiabaticity can sometimes also be assumed if gases are mixed rapidly).
Also, the final volume of the system (that of the entire container) is the same as its initial volume. It follows that $w=-\int_{V_\textrm{ini}}^{V_\textrm{fin}} p_{\textrm{ext}}dV_{\textrm{system}}=0$.
Therefore overall neither heat nor work exchanges between system and surroundings during the process, which means that in accordance with the first law the change in the internal energy of the system $\Delta U = q+w=0+0=0$.
Additional comments:
The internal energy of the individual gases changes during mixing. That is described by the individual terms in the sum $$n_1\frac{5R}{2}(T - T_1) + n_2\frac{3R}{2}(T - T_2)$$ The internal energy of one gas (the colder one) increases at the expense of the hotter one. The volumes available to the individual gases also changes (both gases expand freely but this is ultimately not critical to the analysis). Accordingly the entropy and free energy content of the individual gases changes (guaranteeing that the process of mixing is spontaneous).
Regarding the reversible Carnot process: this is not an isolated process. In this case both heat and work are exchanged with the surroundings. Since it is a cyclic process it is also true that $\Delta U=0$ so that $w=-q$. This is true in the present case also, it just happens that here $q=-w=0$.

@BuckThorn. This is in reference to your statement $\delta(V)$=0 and therefore no total work.  Consider a cyclic process (for instance a carnot cycle). There too, $\delta(V)$=0, but obviously there is work done.


text:
Fries Rearrangement Products
During Fries Rearrangement, oxygen donates its lone pair to $\ce{AlCl_3}$ forming $\ce{-O^+-AlCl3}$. Since this is a strong electron withdrawing group, shouldn't the final product be meta-substituted?

Phenoxide ions are typically electron donating though?

The above is the mechanism for Fries Rearrangement for a brominated ester( doesn't matter really, for any ester the mechanism essentially remains to be the same).As per what I infer from your question, I think you mean to ask why even in the presence of $\ce{O^+-AlCl3}$ which is a strongly deactivating group, the major product is not meta-substituted.
As you can clearly see, in the next step itself a stable acylium is formed and the formal charge on $\ce{O}$ is no longer positive.So, as phenoxide ion is an activating group and o,p-director, two possible products are expected which are ortho and para substituted.However, at higher temperatures, the major product is ortho substituted which is driven by chelation.
NOTE:
As far as I know, both intramolecular and intermolecular reaction mechanisms are able to explain the major products and it is still not confirmed which one is the actual mechanism.
Image source:
https://www.synarchive.com/named-reactions/fries-rearrangement

@Niescte Does this answer your query?

Such ping doesn't work https://chemistry.meta.stackexchange.com/questions/3889/pinging-in-comment-to-suggest-answering-question

@Mithoron Niescte has asked this question and I was not sure if I understood Niescte's question correctly.So, wanted to confirm with the author of the post, whether I interpreted in the correct way.If this is against the policy of the community, then I can delete that comment

sigh You'd have to comment under his question or comment, otherwise it just doesn't ping, that's my point - check out the meta stuff I linked for more info about pinging. BTW asking about clarification, before answering would be better...


text:
Mechanism for the chlorination of Fluorene with NCS?

From literature, it was described that fluorene can react with NCS and conc. HCl to produce the 2,7-diCl derivative. Please help me understand the mechanism.

I'd always suggest checking the original paper, rather than relying on the one-liner summary from SciFinder / Reaxys. Since you're on SciFinder, it shouldn't be too many extra clicks.

The source of the reaction conditions presented by the OP is: https://www.thieme-connect.de/products/ejournals/abstract/10.1055/s-1994-25668, which predates the link provided by Waylander.

According to this paper here

it has been described recently that at very low pH ($\mathrm{pH <  3}$)  $\ce{Cl2}$  functions  as  an  active  chlorinating  agent

In the authors' view the system works by $\ce{NCS}$ oxidising chloride to chlorine which then provides $\ce{Cl+}$ as the chlorinating agent.


text:
Shape of PEG in water
According to the chemical formula $\ce{H−(O−CH2−CH2)_n−OH}$ of polyethylene glycol, or PEG, I assume that this organic compound has a long chain structure when it is dry and in powder form. (This is also what guys in the chemistry field say.)
But when PEG is dissolved in water, would it form some structure like a cluster or would it remain a long chain?
Can we somehow calculate the diameter of a dissolved PEG molecule when $n$ is given? Or the dissolved PEG molecules band randomly so there are no calculations for an average diameter statistically at all?

Polymer molecules form individual "random coils" in solution. The end-to-end distance of the molecule and thereby also its coiled "diameter"  (radius of gyration) scales linearly with molecular weight. The absolute numbers depend of course on the stiffness of the molecule, kind of solvent, solute-solvent interaction, and temperature.

PEG is the term usually reserved for shorter chains, while PEO is  used to refer to longer chain molecules. When polyethylene glycols are dissolved in water, they increase the viscosity of the solution. While it might be easy to model short-chain polymers in dilute solution as some kind of folded chain approaching a spherical volume, as the chains get longer and the solution becomes more concentrated (>1%), there is significant entanglement.

The picture above is for a long-chain polyacrylate, which can be adjusted to be very hydrophilic by adding alkali to generate anions all along the chain. The images show various shapes in water: some fairly tight, some looser (spread out), some spread out but very much entangled. At some point in increasing chain length or increasing concentration, it becomes difficult to project the image of a single molecule of PEO into useful predictions of solution properties. Measurements of dilute or shorter chain molecules would include viscosity; as the solution becomes more concentrated or the chain longer, the property includes some elasticity, so we have viscoelasticity, and at very high chain lengths, we have pituosity (more commonly known as stringiness or ropiness). A video shows this effect dramatically:
https://www.neatorama.com/2017/10/09/Fun-with-Polyethylene-Glycol/.
Since PEGs and PEOs are so soluble and compatible with water, the first molecules to be added just fill all the volume of water; you could calculate how long a PEO molecule would be if completely stretched out. Stirring can extend the molecule, showing effects due to viscoelasticity and pituosity.

Your last paragraph is blatantly wrong.

The probability for an ethylene or oxymethylene chain with N > 25 or so to exist in all-trans conformation at room temperature is, for any practical purpose, zero. If you pull at its end groups, it will rip long before you can stretch it to its contour lenght.

@Karl: Experimentally, if you allow a moderately concentrated solution of PEOs to stand, it seems fairly isotropic. If, however, you stir it, a weak structure will form in the liquid, which is evidenced by viscoelasticity, that is, when stirring is suddenly stopped, the liquid continues to flow: BACKWARD. A backwards flow of 45 degrees is easy to obtain. While only an idealist would consider that a polymer chain could be stretched out completely, I think that one molecule in a liter wouldn't necessarily bunch up - and if it did, I could stir the solution a bit and s t r e t c h it out a lot.

You will be able to detect the prolate deformation of the polymer coil in a shear flow, for example by light scattering, and by the effect of its mechanical relaxation, but you are nowhere near stretching the coil out. Not least because the viscosity drops remarkably, and you cannot stir very fast before the flow becomes turbulent. Last but not least, what you describe proves that the polymer forms a compact, spherical coil. (The latter only by the fact that the solution becomes isotropic again after stirring).

Btw. the contour length of a 1000 kg/mol PEO is in the range of ten microns.


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How to produce Acetylene or Ethylene for Artificial Fruit Ripening in Sachets? Without calcium carbide
Please advise what chemicals (powder + salts) can produce Ethylene or Acetylene Gas for Artificial Ripening of fruits in packed fruit boxes?

I have never heard about acetylene as fruit ripening agent, only about ethylene.

@Poutnik As an entry: https://en.wikipedia.org/wiki/Ethylene_as_a_plant_hormone

@Buttonwood I do know about ethylene as a plant hormone, but the article does not contain the word acetylene.

@Poutnik Well ethylene is a naturally inducing ripening agent produced by plant hormones. Calcium carbide is an artificial ripening agent because it comes in contact with moisture to produce acetylene gas which has similar effects to the natural ripening agent, ethylene. See the additional note in my answer.

Good to know, as I have originally expected rather a link talking about the substitution.

@Poutnik Gotcha, sorry I flipped the two.  Yet both seem to be used extensively enough to justify a studies like «Comparative effects of acetylene and ethylene gas on initiation of banana ripening» (https://doi.org/10.1111/j.1744-7348.1982.tb00837.x), too.

Acetylene is produced by any of three methods:

by reaction of water with calcium carbide

$$\ce{CaC2 + 2H2O -> CH#CH + Ca(OH)2}$$

by passage of a hydrocarbon through an electric arc

$$\ce{coal + H2 ->[\Delta] CH#CH + non-volatile salts}$$

by partial combustion of methane with air or oxygen.

$$\ce{6CH4 + O2 ->[1500 ^\circ C] 2CH#CH + 2CO +10H2}$$
Note that the calcium carbide procedure is easily the most controllable reaction and hence it is favored for small commercial purposes. The other two methods are very drastic as well as expensive. Maintaining an electric arc at high temperature is no joke. Hence, they are called as industrial level methods. So, if you are trying to produce acetylene to use it as a ripening agent, the best reaction would be the one with calcium carbide.
In the Wikipedia article of ripening, it is written that:

Calcium carbide is also used in some countries for artificially
ripening fruit. When calcium carbide comes in contact with moisture,
it produces acetylene gas, which is similar in its effects to the
natural ripening agent, ethylene. Acetylene accelerates the ripening
process.

Ethylene is a naturally inducing ripening agent produced by plant hormones, so you don't have to worry about it.

I agree. Calcium Carbide is by far the easiest method