text:
How to selectively adsorb caffeine on activated carbon filter?
As part of a project on decaffeination methods, I'm looking to see how activated carbon filters can selectively adsorb caffeine in tea and coffee solutions, and avoid adsorbing other taste-giving components.
The technique here is water decaffeination, and involves soaking tea leaves/ coffee beans in hot water, and using a carbon filter to adsorb the caffeine, thereby decaffeinating the solution.
However, I can't seem to find any documents relating to the selectivity of this process : I don't see why the activated carbon wouldn't adsorb any flavourful compounds too.
The only mention to this problem I've found is on The science behind decaf, which states :

The problem is that the extract also contains hundreds of other compounds that are critical to coffee flavor, many of which would also be removed by the carbon filter. But there is a way around this. Preload the filter with chemicals that are unlike caffeine in molecular structure, but similar to other flavour and colour compounds found in coffee. Sugar and formic acid are the chemicals used. So obviously the process isn’t “chemical-free.”. Since the activated carbon’s adsorption sites for such chemicals are now occupied, the filter will not remove coffee components other than caffeine.

Sadly, the webpage doesn't cite any sources, leaving me wondering why formic acid and sugar can increase adorption selectivity.
I'm wondering if any sympathetic coffee enthusiast can explain this to me, or cite some documents or studies that delve into this problem, how to make activated carbon only adorb caffeine ?
Any link or website would be appreciated.

The problem is, coffee and tea contains many compounds less polar than caffeine, that would adsorb themselves better. Activated carbon IMHO is not the way to go.

My best guess is that activated charcoal has holes of different sizes and the various solutions used to block everything but caffeine actually occupy the holes that don't interact with caffeine.

While the idea of using activated carbon might be possible the issue is whether it is better and cheaper than alternative methods which are widely used and seem to work well (both methylene chloride and supercritical carbon dioxide have been used in processes that selectively extract caffeine). Solutions involving activated carbon seem more elaborate and expensive to me.

The method is currently described in (nearly) complete detail in the current Wikipedia blog for "Decaffeination" (January 2024). A "flat" "filter" is saturated with other substances, allegedly preventing absorption of non-caffiene molecules.

The Swiss Water process uses hot aqueous solution (see below) to extract caffeine from green beans, and activated charcoal to remove the caffeine from the extract. This decaffeinated solution is used again for the next extraction (minimizing the leaching out of flavor components). See e.g. the "SWISS WATER® Process animation video" https://youtu.be/tAEQ4G-1jTQ
This only works if the flavor compounds are not absorbed by the activated charcoal (otherwise decaffeinated coffee would taste really bad, not just somewhat bad). There is a patent[1] that suggests that pretreating the activated charcoal can suppress flavor absorption, https://patents.google.com/patent/US5208056A/en. Patents are often written in a way that includes the method actually used, but also includes other possible methods (that might not work). This is to protect intellectual property without giving a specific recipe others could use after the patent runs out. This patent talks of using sucrose for pretreating the active charcoal. I did not find any information on using formic acid in a similar manner.
[1] U.S. Patent #5,208,056 issued May 4, 1993 to Fischer and Kummer (Chocolat Suchard Societe)


text:
Labelling of molecular motor isomeric forms as trans and cis
I am reading Koumura et al. (2002), a paper about molecular motors. It contains this figure:

Being an absolute layman when it comes to naming my molecules I'm a little confused as to why the left molecules are trans, while the right molecules are cis. Which groups are being compared to each other? The benzene ring and the MeO group?
I always learned that one should only use trans/cis if there is absolutely no ambiguity possible, and E/Z otherwise. Isn't this situation ambiguous? One could also compare MeO and Me!

Well, I guess you're right in saying it's unclear.

Here things are more clear: https://en.wikipedia.org/wiki/Synthetic_molecular_motor#Light-driven_rotary_molecular_motors

In my opinion it just show that sometimes is hard to be pedantic. But it is also pity that IUPAC rules/recommendations are neglected in works leading to nobel prizes in chemistry.

source: Stack Exchange

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/

sort: votes


text:
Why is NaCl so abundant in the ocean
Why is sodium chloride far and away the most abundant salt dissolved in ocean water. Its two constituent ions do have a very high frequency in the crust of the earth but they are far from the most common. Chlorine is (according to Wikipedia) the 21st most abundant element, and sodium 6th.
I certainly understand that a combination of their solubility and reasonably high frequency would lead one to expect them to but abundant in sea water, but they are hyper abundant, completely dominating all other salt ions. Iron, for example, is twice as abundant, and potassium only a little less abundant, fluorine more abundant than chlorine.
Moreover, if the salts are deposited in the ocean through weathering of rocks and deposition via rivers, why does the salinity not simply grow and grow? I understand that some is lost due to tectonic activity, but it seems extraordinarily unlikely that these two forces should be equally balanced, and so we would see a significant change in average salinity over time.

Umm, this is Chemistry.SE, not Earth Science and even there this would be probably too broad.

BTW this is pretty obvious that there is an equilibrium in processes that lower and increase these concentrations. Even if they were changing, it would just mean the equilibrium is shifting.

https://en.wikipedia.org/wiki/Seawater#Origin_and_history

Reactivity is also an factor. Abundance of iron might be high but it doesn't have enough reactivity to form sufficient salts.

Rather than at the abundance of the elements look for the most abundant salts at surface level. It should than fit. At least we can expect that the most abundant and soluble salt accessible to "runningc water is NaCl.

This is a very interesting question but better suited for earth science SE. An online search might help too: https://www.whoi.edu/science/GG/geodynamics/2005/images2005/holland04_treatise.pdf

I’m voting to close this question because it belongs on earth science SE

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


text:
Can the paper in thin layer chromatography be any type of paper?
I'm doing a chemistry chromatography lab that wants me to recreate anything I want. I decided to measure how far a pigment goes in each color for the same brand, but I'm stuck at the paper part. I heard there's a specific paper I can use, but I'm broke, so would regular paper or coffee filters do the trick as well? Also, how wide should the paper be? (I'm creating a line with the marker not a dot, however, feel free to give me pointers)

Coffee filters work well for paper chromatography.   See:  https://www.scientificamerican.com/article/chromatography-be-a-color-detective/

Yes in a class lab you can play with many kinds of paper. It even add some values. Note that the paper characteristics influence the experiment at different levels, at the mechanism one (interaction) as well as a practical one (robustness, ecc).

You don't need to limit yourself with ascending paper chromatography.  The lanes may easily pass beyond the 30 cm / 11 in mark if you let the elution descend instead, be circular or elute in 2D.

@Buttonwood Thank you for the help! I was just wondering, since my friend was doing a dish chromatography instead of a hanging column chromatography like me, what would be the difference and will our results differ?

Consistency may be a problem for paper chromatography if e.g., the paper varies in quality (esp. if the manufacturer did not anticipate /this/ use, e.g. here or more seriously here. While the circular PC are good to screen (out) eluents rapidly, the Rf values of circular / ascending / descending PC still differ even if all other parameters are kept fixe. E.g., gravity limits the ascending method, but not much the other two.

PC might look old fashioned, but it may run without electrical power, and without the pumps seen in HPLC.  The Czech once were well known for PC, so I recommend you look into the heavy books by Hais and Macek.  Some adjustment is necessary because the paper mills back then (1960s) are no longer the same, but these «bibles» include many hints (incl. charts, a few color photographs, literature references) how to run and eventually stain the spots if not already visible with broad coverage (incl. dyes, amino acids, etc.) An English version.

@Buttonwood Wrong question Buttonwood! :)

If possible, you may use Whatman filter paper. It has specially been designed for this purpose.
Well, that kind of a paper is preferable which has a good pore size, kind of a paper of unglossed newspaper.

Because many companies / distributors of chemicals are active in this field, this answer does not yet shed a light on plausible criteria of selection.  Agilent for example advertises chromatography papers which are "made of glass microfibers impregnated with silica gel. Agilent offers two kinds of paper: SA (contains sodium salt) and SG (contains potassium salt)". Old educational papers do not explicitly mention either silica, nor the salts added.


text:
The dependency on electron density in the reactivity order of electrophilic and nucleophilic additions
It is well-known that for electrophilic addition across an alkene, the reactivity increases when intermediate carbocation is more stable (eg. addition of $\ce{HBr}$ or acid-catalyzed addition of $\ce{H2O}$) i.e. when the electron density on carbon atom is more.
For nucleophilic addition on a carbonyl group the reactivity is opposite, i.e. more the positive charge (less electron density) on carbonyl carbon, more is the reactivity. The relation between reactivity order and electron  density is opposite in the two cases. Why is this so?
I am just finishing high school, and I have been taught that the reason for this is as follows:
For any reaction the overall rate is determined by the slowest step, called the rate determining step (r.d.s).
For electrophilic addition, the r.d.s. is the formation of carbocation. Formation of carbocation is faster if the carbocation is stable. Hence rate increases and reactivity increases for stable carbocation.
For nucleophilic addition the r.d.s. is attack of the nucleophile, which forms a tetrahedral intermediate. Attack of nucleophile is faster if the carbonyl carbon is electron deficient. Hence for carbon with greater electron deficiency rate and reactivity increases.
However, when I posted this as an answer in this question, a comment on my answer mentioned that the concept of rate determining step is not applicable and basically useless. I am open to the fact that what I have learnt might be an oversimplification of a more detailed concept (similar to how Newton's laws are taught in school first, as a simplified version of relativity at low speeds).
Since what I know is obviously wrong, I would like to know the real reason behind the discrepancy of the order of reactivity.
I searched on the site to see if a similar question has been asked before, but couldn't find any.
TL;DR - For nucleophilic addition on a carbonyl group the reactivity is opposite, i.e. more the positive charge (less electron density) on carbonyl carbon, more is the reactivity. The relation between reactivity order and electron  density is opposite in the two cases. Why is this so?

A more stable intermediate means that it has a greater chance to exist. The probability of any collision involving a more stable intermediate is greater than a collision with a rarer intermediate. Whereas in case of carbonyls the electron deficiency pulls the nucleophile towards itself.

I don't know for sure but shouldn't this be considered a duplicate of the mentioned question?

@NisargBhavsar The mentioned question does not have a correct answer so far and it's closed, so even if it is a duplicate, there is still no answer.

@NisargBhavsar Regarding the answer you mentioned in your comment (about stable intermediate and pulling the nucleophile), isn't it the same as what I already know (and what I've mentioned in my question?)

@TRC You might want to put your final question in bold or put a TL;DR in bold so the readers don’t get confused.

@TRC I think what Martin wanted to point out was the inaccurate concept of r.d.s and not that the reactivity order is wrong.

@Rishi I included my main question in bold, thanks for the suggestion :) (What is TLDR, though?) Also, I did realize that Martin wanted to say that the reason I gave is wrong, not the order. I accepted his word - he is obviously much more knowledgeable than me. But I still don't know the real reason for the orders being as they are. If you have any idea, please consider posting an answer. Thank you!

A tl;dr is a summary of the main points in something such as an online post or an article. tl;dr is an abbreviation for 'too long; didn't read'.

After not having any answer for over a week, I wrote about this question in the comments of an answer I had posted to another question. There another user confirmed that the reasoning I have given in my question is not wrong, and is valid -
For any reaction the overall rate is determined by the slowest step, called the rate determining step (r.d.s). Reactivity is determined by rate of reaction, since more reactive implies it reacts faster.
For electrophilic addition, the r.d.s. is the formation of carbocation. Formation of carbocation is faster if the carbocation is stable. Hence rate increases and reactivity increases for stable carbocation. Thus the reactivity of electrophilic addition increases with increasing electron density.
For nucleophilic addition the r.d.s. is attack of the nucleophile, which forms a tetrahedral intermediate. Attack of nucleophile is faster if the carbonyl carbon is electron deficient. Hence for nucleophilic addition, reactivity increases with decreasing electron density.


text:
d orbital irreducible representations in metal in ML4 model
In ML4 (Metal—4 ligands)model which has a square planar shape and D4h point group, what is the irreducible representation for the 5 kinds of d orbitals in the central metal?

You will need to place x,y,z vectors on each atom and work out what happens to them as you perform each of the operations in the point group. Count 0 if the atom moves and $\pm 1$ if it unchanged or reversed.  Most book on symmetry go through examples.

I almost know the procedure but probably wrong with first finding out the reducible representations because I cannot get an integer for the number of A1g or B1g in the later step.

@user111996 Please take a site tour and review our policies for asking questions and including your efforts within.

I had this example worked out. You should be able to copy the method, it is a bit long but easier than using the equations as errors are easily made there.
To find the symmetry of the vibrations of a C2v molecule such as H2O or SO2. (Note that  the method only works this way for 180 deg rotation groups if not you need to use cos/ sine to find amounts.)
The method is:
(1) Place three (unit) vectors on each atom and along x, y, z axes.
(2) Move each vector in turn and according to each of the symmetry operations; in C2v these are:
identity E, rotation about the principal axis C2, and two reflections $\sigma_{xz}$ and $\sigma_{zy}$.
(3) Add up the number of vectors unmoved, moved or inverted and this
generates our reducible representation which we label $\Gamma_R$. We then have to use this to produce an irreducible representation
(4) if a vector is unmoved we add 1, if it is inverted we add -1, if it is moved we add 0.
(5) Place the total count under each column
of the point group table operators and this is the $\Gamma_R$.

example
Rotation C2: The C2 axis lies along z1 therefore z1 remains unchanged so we count $z_1 \equiv 1$, $y_1$ and $x_1$ are both inverted by the $180^\text{o}$ rotation so $y_1 \equiv -1,\;  x_1 \equiv -1$. Each vector on the H atoms is moved to the opposite side, so exchange places, and as they have completely moved they each count 0.
So the total for $C_2$ rotation  is $+1-1 -1=-1$.
The reduced representation looks like
$$\displaystyle \begin{array}\\
 C_{2v} & E & C_2 & \sigma_{xz} & \sigma_{zy}\\
 \hline
\Gamma_R & 9 & -1 & 3 & 1\\
\hline
\end{array}$$

Now we need to reduce this representation into its irreducible representations present in the character table.
Build a new table;
(1) add the classes on the top line E, C2 etc.
(2) The next row is the reducible representation $\Gamma_R$ plus
S as the sum of the characters in that row and also
S/h as the result we want which is the number of times that representation is
present in $\Gamma_R$.
( $h$ is the order of the group which for C2v = 4; its different for other groups, you will need to calculate it from the point group table. The order is the number of columns times the number in each class . This is 1 for each of the operations in C2v )
(3) subsequent rows start with each Mulliken symbol in the Character table
To calculate the irreducible representation start each row with the Mulliken labels (A1, A2 etc) from the point group (character) table then for each entry in the table:
Multiply the number in the class (1 for C2 and each of the $\sigma$'s) by the number under it in GR and by the entry in the point group table: e.g.
$$ \text{( Number in class)} \times \text{( value in } \Gamma_R) \times \text{( entry in character table)} $$
for the $A_2-\sigma_{xz}$ entry the product is $(1)x(3)x(-1)=-3$. The $-1$ comes from the point group table entry at position $A_2 : \sigma_{xz}$

The reduced representation is therefore formed from $3A_1, 1A_2, 3B_1$ and $2B_2$ irreducible representations;
$$\Gamma_R=3A_1 + A_2+3B_1+2B_2$$
Because we have used (unit) vectors on each atom we have 3N coordinates, 3 are from whole molecule translations and 3 from rotations leaving 3N-6 vibrations.
The translations are in each of the x, y, and z directions and also rotations about x, y and x axes and these have to be removed to reveal the 3N-6 vibrations.
The translations and rotations are easily found as they are described by the irreducible representations in the point group table under x, y, z and Rx, Ry and Rz.
The values for C2v are:
Translations  $A_1 + B_1 + B_2$ Rotations    $A_2 + B_1 + B_2$
Total trans+rot $A_1 +A_2 +2B_1 +2B_2$
If we subtract these from the total $\Gamma_R = 3A_1 + A_2 + 3B_1 + 2B_2$ we have calculated we obtain the symmetry of the remaining vibrations:
Vibrations $= 2A_1 + B_1$.


text:
How do I label stereogenic centers R or S when there are 3 bonds in the plane and 1 bond coming out?
I need help with question (d). I labeled the O the first priority, then the C that's bonded to an O the 2nd, and the C above that is attached to 3 other C's the 3rd priority. So, I ended up getting S, but because the 4th priority was not in the back of the plane, I switched it to R. However, this was not the correct answer, as the answer key told me it is actually S.
I'm confused on how to find R or S in this specific type of situation where there are 3 bonds in the plane and 1 bond coming out. Would appreciate help and pointers on where I went wrong.

Imagine that the bond coming out is actually in the back. That would change the configuration. Now switch the places between OH and the 4th priority. That would change the configuration again.

Build a 3D model if you can't see the structure around the centre in your head. BTW it is not by any stretch in the plane.

Recognize that carbon "e" lies below the cyclohexene ring.  This post may be of help. https://chemistry.stackexchange.com/questions/151831/r-or-s-configuration/151842#151842

You can also assign based on the highest priority. After assigning all priorities, if the highest priority group is forward and the direction from 1->2->3 is clockwise, the configuration is R. If counterclockwise, the configuration is S.
In this case, the absolute configuration of d is straightforward. You do not switch from S to R because the low priority is not backward. You keep it as S because the highest priority is facing forward.


text:
What is the solubility trend of alkaline earth metal fluorides?
I was going through JD Lee and I found certain contradictory statements.
First:

With most compounds, on descending the group, the hydration energy decreases more rapidly than the lattice energy: hence the compounds become less soluble as the metal gets larger. However, with fluorides and hydroxides the lattice energy decreases more rapidly than the hydration energy, and so their solubility increases on descending the group.

Whereas,

Beryllium fluoride is very soluble in water, owing to the high solvation energy of Be in forming $\ce{[Be(H2O)4]^2+}$. The other fluorides MF2 are are all almost insoluble.

and

The sulphates of calcium, strontium and barium are insoluble, and the carbonates, oxalates, chromates and fluorides of the whole group are insoluble. This is a useful factor in qualitative analysis.

I'm confused about what the order is. Wikipedia says $\ce{BeF2}$ is highly soluble, whereas $\ce{CaF2}$ is insoluble. Are these statements contradictory because of the toxicity of beryllium compounds, which is why they have been studied less and are not spoken about frequently? What is the actual order of solubility of alkaline earth metal fluorides?

Please take a site tour and familiarize yourself with how to typeset here.

As Todd mentioned, please do not post images with text, rather type text into your post. I transcribed for you this time but might be less generous in the future.

Sure, I'll keep that in mind next time and thanks @BuckThorn

https://chemistry.stackexchange.com/questions/127687/how-can-you-explain-the-solubility-of-bef2

The order of solubility of the alkaline earth fluorides is, after $\ce{BeF2}$ :
$\ce{MgF2 : 0.026 g/L}$
$\ce{CaF2 : ~0.016 g/L}$
$\ce{SrF2 : ~~~0.11 g/L}$
$\ce{BaF2 : ~~~1.7 g/L}$
$\ce{BeF2}$ itself is mentioned as infinitely soluble in water.
So, the minimum solubility is in the middle of the series. It is $\ce{CaF2}$
Ref : C. D. Hodgman, R. C. Weast, Handbook of Chemistry and Physics, Chemical Rubber Publishing Co. Cleveland, 1962.


text:
Is it possible to filter Stille catalysts using syringe filter?
I'm conducting an organic synthesis involving Stille coupling.
Stille coupling uses $\ce{Pd2(dba)3}$ and $\ce{P(o-tol)3}$ as catalysts.
Both catalysts dissolves in toluene which is the solvent of my experiment.
I have considered several ways to filter these catalysts (e.g. Celite filtration), and I just noticed there are syringe filters in my lab. Is it possible to filter Stille catalysts using syringe filter? I have not used the syringe filter, so I searched for its usage. It says that syringe filter can be used for removing particulates from solvent (also for HPLC preparation) according to its material and pore size. Syringe filter which I have is Whatman™ puradisc 25 TF 0.45 μm PTFE membrane with polypropylene housing. Diameter is 25 mm.

The short answer is no, as such this task is not possible.
You mention $\ce{Pd2(dba)3}$ and $\ce{P(o-tol)3}$ dissolve in the solvent of reaction; this describes an example of homogeneous catalysis, where the reagents share a phase in common.  A simple filtration to remove the Pd catalyst will not work.  It is like passing water of the ocean through a paper filter; the filtrate still contains the dissolved salt.  The separation by Celite (or similar, by silica) added works because the catalyst adsorbs to the surface of grains of Celite in first place.  And if these grains are larger in diameter than the pores of the filter in second place, the filtrate will contain less of the catalyst.
Thus passing the raw reaction mixture across a short plug of Celite, preferentially with pressure from above of the short column, rather than by gentle suction from below the glass filter / Buchner filter is your likely next unit operation ahead.  Equally, it is better to start a filtration with a coarse filter in first place (which the pad of Celite may be suitable, too); large particles may rapidly clog the fine pores of a syringe filter.  Because said adsorption of the catalyst to Celite competes with any other constituent of the reagent solution, you may loose some of your starting material / reaction product by this operation.

If you are interested to reuse the catalytic system (not a priori excluded from the present question), you may consider to change for heterogeneous catalysis instead.  Then, catalyst and reagent do not share the same phase and the reaction takes place at the interface of the two.  Either
a) perform the reaction under conditions of phase-transfer catalysis with water soluble catalysts.  This may be achieved e.g., by addition of $\ce{-SO3H}$ groups to the ligand (e.g., TPPTS), or using otherwise charged catalysts (an example).  Ideally, after completion of the reaction, similar to a liquid-liquid extraction, you simply let the two layers separate from each other.
.or.
b) You chemically bind your catalyst to a solid support, e.g., a polymer resin.  Depending on the reaction conditions, some of the Pd will bleach out into the reaction mixture, yet the larger fraction of the catalyst will be removed from the reaction mixture by filtering-off the beads (an example) across a coarse filter.
The transition between homogeneous and heterogeneous conditions may require additional work to identify optimal reaction conditions.  It depends on the scale of reagents to process and other parameters (e.g., acceptable concentration of remaining Pd in the product) if such an investment is justified.

You said that I had better use column rather than glass or Buchner filter. Is it similar to column chromatography, but without glass wool or sea sand? Is it okay to pack only Celite in the column and just drop reaction mixture? (my reaction mixture is very small scale)

Glass and Buchner filter retain material because of its granularity, column chromtography by physio-/chemisorption. Celite is more coarse than the silica you purchase to pack your (low pressure) column / flash chromatography; the consists of fine grains (around 40 to 60 µm in diameter) which frequently are fine enough to pass a filter, then trapped in the product; it remains undetected by 1H and 13 C NMR and may cause (especially on small scale [50 mg reactions]) yields greater 100%.  Glass wool / sand pad below the silica + caution with the pressure applied aim to prevent this breakthrough.

If you work on real small scale, you may pack a column in a Pasteur pipette (an example).  The Baran lab showcases a step-by-step procedure if you choose test tubes as column (entry), just in case you already have more material at disposition.  The later takes some training for the later, but once you are used to it, it is quick and good enough.

Thank you for your kind and detailed answer, and sorry for additional question. English is not my mother language, so there was some difficulty on translating your answer. So you said that grain of silica is finer than Celite, so if I do filtration using silica, it can pass through filter and mix with my reaction mixture, but Celite is okay. (am I right?) but the links (Pasteur pipette and test tube column example) all use silica. Does it mean that I should replace silica with Celite in the procedure of the links? or the procedure of the links are different process from using Celite?

If both Celite and silica for chromatography are in the shelf and for volumes of reaction greater than say 25 mL, I usually choose Celite (don't forget to wet the pad with the solvent of reaction prior, and to rinse a little after the filtration; both increase the volume of the filtrate compared to the initial volume) and then carefully concentrate the filtrate (reduced pressure) close to dryness. The examples with the Pasteur pipette column originally are about chromatography / separation of multiple fractions (at small scale); if you elute well, they may be used as a small scale filter, too.

Aha, so you mean that if the volume of reaction is greater than 25 mL, then I can filter mixture using Celite (like using Buchner filter?), and if the volume is much less than 25 mL, then I can use small scale column chromatography using silica like the links you provided? Did I understand well? Thank you sincerely for your kind answer and sorry to keep bothering you.

If you lack experience for certain manipulations (perhaps especially at smaller scales),  train yourself multiple times with a cheap mix in first place. Once you acquired the dexterity (when you may proceed without need to go back to the lab notes to read the next line, perform the work, and return to the notes to read the next line, etc.) you apply this on your «real stuff», expensive because you spent x hours of reaction + x amount of money. Here, why not mix toluene, vanilin (e.g., from the TLC stain) and active carbon / fine charcoal and test filtering and determine yield of recovery.

Thank you for your advice! You are a big help. I will try filtration next week right away.

And see if your hosting group / scientific library has a primer / an introduction about typical lab techniques.  As an example, I recall Zubrick's book as a gentle introduction in fairly accessible English, but your librarian / colleagues may indicate to similar references.  On occasion, you find good tutorials on the web, too example.


text:
Do superoxide salts of heavier alkali metals contradict the principle that lattice energy depends inversely on cation-anion distance?
I have a very basic understanding of quantitative lattice energy calculations: I only know that the lattice energy is directly proportional to the product of charges on the ions and inversely proportional to the distance between cation and anion.
Based upon the direct application of the above facts, lattice energy follows the order:

LiCl > NaCl > KCl > RbCl > CsCl

But I have also read in JD Lee:

Large cations can be stabilized by large anions, since if both the ions are similar in size the coordination number will be high, and this gives high lattice energy

This was given to explain that superoxide salts exist only with K, Rb, Cs. Does this not contradict the basic principle that lattice energy is inversely proportional to the distance b/w cation and anion?

TL, DR: They do not contradict. Although larger ions decrease the lattice energy by increasing ion distance, it is compensated by another factor that increases the lattice energy. And like Nisarg Bhavsar said, product formation does not solely depend on lattice energy. Other factors such as the electron affinity and ionization energy of the element should be accounted too.
From Ref. 1:

The answer has to do with the crowding of alkali ions around oxygen,
as illustrated for $ \ce{K2O}$ at the right. Because eight large $ \ce{K^+}$
ions surround each $ \ce{O^{2-}}$ ion in the structure, the cations
are essentially in contact. Indeed, the metal-oxygen bonds are
"stretched" in $ \ce{Na2O}$, $ \ce{K2O}$, and $ \ce{Rb2O}$ relative to
$ \ce{M-O}$ bonds with the same bond order in other structures.The
situation is so extreme for $ \ce{Cs2O}$ that it finds an
(electrostatically unfavorable) structure in which the coordination is
lowered to 6:3. This packing problem is relieved somewhat in the
peroxides, where the coordination is still 8:4 but the anion is
larger, and especially in the superoxides where the cation:anion ratio
is 1:1 and the coordination is 6:6. Thus the larger alkali ions ($
 \ce{K^+}$, $ \ce{Rb^+}$ and $ \ce{Cs^+}$) tend to form superoxides.

Above: $ \ce{K2O}$ structure, below: $ \ce{Cs2O}$ structure. Oxides are red, potassiums are blue and caesiums are magenta.
Thus, the reason why heavier alkali metals tend to form superoxides is because the stoichiometry of the compound and ionic radius of the anion allows the large cation to arrange so that they won't touch each other as much. Since like-charges repel, having cations in contact with each other is a crime in forming crystal lattices because it would greatly destabilize the crystal.
In case you don't know, the Born-Lande equation to calculate energy is actually derived from Coulomb's formula of electrostatic potential energy, i.e
$$E_\mathrm{pair} = \frac{z_1\cdot z_2\cdot e^2}{4πε_0r} $$
Where $z_i$ is the charges of one ion and $r$ is the distance between the point charges (ions in this case).
Specifically, assume a theoretical compound $\ce{M^{2+}X^{2-}}$ that crystallizes in the rocksalt ($\ce{NaCl}$) structure. The Born-Lande equation is derived from the sum of all electrostatic forces that one ion in the lattice experiences. Let's use the $\ce{M^{2+}}$ cation.
The rocksalt lattice has 6:6 coordination. It means that one $\ce{M^{2+}}$ cations is equally surrounded by six $\ce{X^{2-}}$ ions. Thus the expression for the electrostatic energy between the $\ce{M^{2+}}$ in question with the six $\ce{X^{2-}}$ surrounding it is
$$E_{1} = 6\frac{2\cdot (-2) e^2}{4πε_0 r} $$
The $\ce{M^{2+}}$ is also equally surrounded by twelve $\ce{M^{2+}}$ with a distance of $\sqrt{2}r$. Thus the expression for the electrostatic energy between the $\ce{M^{2+}}$ with the 12 $\ce{M^{2+}}$ surrounding it is:
$$E_{2} = 12 \frac{2\cdot 2 e^2}{4πε_0 \sqrt{2}r} $$
Etc. You can do another $\pu{E_{i}}$s of other sets of further ions surrounding the $\ce{M^{2+}}$ ion.
Adding all of this electrostatic energy equations together we get the sum of all the electrostatic energy on one particular ion:
$$E_\mathrm{all} = \frac{2\cdot (-2) \cdot e^2}{4πε_0 r} \cdot \pu{(6 - \frac{12}{\sqrt{2}}+...)} $$
The sum of $(6 - \frac{12}{\sqrt{2}}+...)$ in the equation is the Madelung constant. Since those numbers derive from the geometry and coordination number of the lattice, it is dependent on the crystal system rather than the nature of the ions, as you might have learned in textbooks.
Finally, combining this with the account for the electron-electron and nucleus-nucleus repulsions between the ions and multiply with the Avogadro constant since we are using $\pu{kJ/mol}$, we arrive at the Born-Lande equation:
$$E_{lattice} = \frac{N_A \cdot z_1 \cdot z_2 \cdot e^2 \cdot M}{4πε_0 r} \cdot ( 1-\frac{1}{d}) $$
As you can see, since the lattice energy involves the interaction forces of the ions, having too much repulsion would decrease it. The benefit of increased lattice energy from this factor is more than enough to compensate for the increased ion distance.
References:

The Strange Case of the Alkali Oxides. (2021, June 7). Retrieved June 11, 2021, from https://chem.libretexts.org/@go/page/189354

Energetics of Crystalline Solids- The Ionic Model. (2021, June 7). Retrieved June 11, 2021, from https://chem.libretexts.org/@go/page/183348


text:
Does folic acid contain a benzyl or a phenyl group?
I'm studying for my MCAT currently and I stumbled upon this question. It gives the structure of a folic acid molecule and asks what functional groups can be found within the molecule. I understand that there is an amine, amide, imine, and carboxylic acid group, but I don't understand why the benzene ring in the center of the molecule does not count as phenyl or benzyl group. If anyone could explain this to me I would really appreciate it.

Similarly as there is ethyl CH3-CH2-  and ethylene -CH2-CH2-, there is phenyl C6H5- and phenylene  -C6H4-. ( E.g. either (o,m,p-)diaminobenzene either phenylenediamin )

Did you look up, e.g., in Wikipedia about benzyl bromide, benzyl chloride for their motif of structure in common vs. phenol, phenylamine (equally known as aniline)?

It does have a phenyl group. if you search up the IUPAC name of folic acid this turns up: (2S)-2-[(4-{[(2-amino-4-hydroxypteridin-6-yl)methyl]amino} phenyl )formamido]pentanedioic acid

Does folic acid contain a benzyl or a phenyl group?

This is the question asked in the title. At the first glance to the structure, one would say folic acid consists of phenyl function but not benzyl function because the question did not define what is the phenyl group. In reality, phenyl group $(\ce{Ph})$ is $\ce{C6H5}$. As Poutnik pointed out in the comment, folic acid really contains a phenylene $(\ce{C6H4})$ function. On the other hand, the Benzyl function is strictly $\ce{C6H5-CH2}$ or $\ce{Ph-CH2}$. It is debatable to call $\ce{X-C6H4-CH2}$ a benzyl group, unless a definition is given upperhand. Nonetheless, folic acid does not contain either $\ce{C6H5-CH2}$ or $\ce{X-C6H4-CH2}$, and therefore no argument there for the presence of benzyl group. What it contain is $\ce{X-C6H4-C=O}$ group, which can tentatively called benzoyl (strictly when $\ce{X = H}$).
The more generic term for a substituted benzene ring, without restrictions on subatituents, is aryl.
Albeit, without giving definitions for the function naming, the mCAT question assigned $\bf{(A)}$ as a correct answer. However, there are clear-cut carboxylic and amine groups visible, the imine group is a part of Pteridine ring system:

Thus, it is a part of aromatic nucleus and therefore not a clear-cut isolated imine function. If the authors considered it as an imine, then phenylene system can be considered as a phenyl function. Accordingly, the mCAT question has possible two answers: Answers $\bf{(A)}$ and $\bf{(C)}$ as depicted in the image.

I tried to post an answer but found it makes no new contributions except mentioning "aryl" as a more generic term. So I deleted that answer and added the allusion to "aryl" here. Feel free to roll back.

@Oscar Lanzi : I glad you did. I think it is a great addition. I didn't think about that.

I would argue that the imine functionality you have assigned is incorrect as they are part of an aryl system. The imine I would say is in the other half of the pteridine within the 'guanidine' moiety, but honestly none of them are good representations.

@Tragamor: That was my argument in my answer. None represents a clear cut imine. If they are considering one of them are imine, they should have consider the aryl nucleus as phenyl.  On guanidine, carbin with two geminal nitrogen can't be a imine (are you saying enamine?)

It's not an enamine but I'm referring to the double bonded nitrogen within the 'guanidine' as that would have more "imine" like character than the pyridazine ring candidates.


text:
First ionization energy and electron gain enthalpy of dihydrogen / stability comparison of dihydrogen ions
Subset of: How can antibonding orbitals be more antibonding than bonding orbitals are bonding?

Stability of $\ce{H2^{-}}<\ce{H2^{+}}<\ce{H2}$

Stability of $\ce{N2^{-}}<\ce{N2^{+}}<\ce{N2}$

Source: Online lecture Etoosindia.
I am more concerned of $\ce{H2^{-}}<\ce{H2^{+}}$.
Does any satisfactory explanation exists (perhaps which can be generalized) for it? That is to add electron in anti-bonding molecular orbital more difficult than removing electron from bonding molecular orbital?
Observation indicates that First ionization energy of $\ce{H2}$ is greater than electron gain enthalpy (magnitude).

Yes, that's a general rule that antibonding electrons "more antibonding" than bonding electrons are "bonding".

@Jay  As Mithoron mentioned, this may help https://chemistry.stackexchange.com/questions/4517/how-can-antibonding-orbitals-be-more-antibonding-than-bonding-orbitals-are-bonding


text:
Why does an improper rotation axis preclude chirality?
Can anybody prove that any molecule with an axis of improper rotation is nonchiral? I still do not understand why this is so, any proper mathematical proof or visualisation of a generalised molecule would be appreciated!

Do you understand why mirror symmetry means no chirality, to begin with?

https://chemistry.stackexchange.com/questions/142244/example-of-an-achiral-molecule-without-a-plane-of-symmetry-or-inversion-center https://chemistry.stackexchange.com/questions/51854/does-an-axis-of-symmetry-determine-chiralty

https://chemistry.stackexchange.com/questions/145084/what-are-the-criteria-for-a-molecule-to-be-chiral

Still none of them have a proper proof why an axis of improper rotation means no chirality, the most these answers have is examples


text:
Relation between Ionization energy and reactivity
So I was learning about the periodic table where I came across the topic of ionization energy. As a general trend the Ionization energy decreases as we move down a group with a few exceptions such as copper having a greater value of Ionization energy than silver and a few more examples. But on going through the reactivity series Copper is placed above Silver this can also imply that a lot of energy is required to remove an electron from copper than silver so it be said that copper should be less reactive than silver but the opposite is true. Is there any other reason affecting reactivity?

All the chain of elementary - real or formal - processes has to be considered, both thermodynamically and kinetically. E.g.  Cu(s) -> Cu(g) -> Cu^2+(g) -> Cu^2+(aq), versus Ag(s) -> Ag(g) -> Ag+(g) -> Ag+(aq)

If you’re just getting started on the properties of elements, I’d say stick to whatever is being taught for now. Because chemistry is a pretty vast subject that seldom has rules that are followed throughout. There’s always exceptions and many different things that come into play. When you have a pretty good view of all other topics, you can ask yourself such questions, and have a much better chance of finding the solution. +1 for the curiosity though!

There's isn't really much of a relation between the two, whatever you mean by "reactivity" which isn't some parameter like ionization energy, that could be compared.


text:
What would happen if the alpha particles directly hit the electrons in the gold foil in Rutherford's experiment?
What would happen if the alpha particles directly hit the electrons in the gold foil in Rutherford's experiment?
Would it get ionized or is there no probability of such a thing happening ?

An energetic charged ion in a solid will come to an equilibrium charge state, based on the Z of the ion and the velocity of the ion. There are plenty of ion-electron interactions (they are Rutherford scattering as well) that result in energy loss.

The alpha particles used in this experiment have an energy of about $5.59\ \mathrm{MeV}$. This energy is too high; the alpha particle cannot simply capture two electrons and form a stable helium atom. By way of comparison, the first and second ionization energy of helium is $24.6\ \mathrm{eV}$ and $54.4\ \mathrm{eV}$, respectively.
Nevertheless, alpha particles cause frequent direct ionizations within a narrow diameter around a relatively straight track. This transfer of energy from the alpha particles to electrons causes an approximately continuous deceleration of the alpha particle. For $5.59\ \mathrm{MeV}$ alpha particles in gold, the initial electronic stopping power is about $220\ \mathrm{MeV\ cm^2\ g^{-1}}$. With a high density of $19.3\ \mathrm{g\ cm^{-3}}$, this means the initial linear energy transfer is about $4200\ \mathrm{MeV\ cm^{-1}}$. Therefore, the alpha particle loses its energy very quickly so that the range of the alpha particle in gold is very short. The continuous slowing down approximation (CSDA) range of $5.59\ \mathrm{MeV}$ alpha particles in gold is about $0.020\ \mathrm{g\ cm^{-2}}$. Considering the density of $19.3\ \mathrm{g\ cm^{-3}}$ again, the resulting range in gold is about $0.0010\ \mathrm{cm}$. (Clearly, the gold foil in the experiment had to be very thin.)
Only after the alpha particle has slowed down like this, it can finally capture two electrons and form a stable helium atom.

IS IT POSSIBLE FOR ALPHA PARTICLE TO HIT THE GOLD ELECTRONS AND GET IONIZED AS THEY MOVE IN A VERY HIGH SPEED AND THEY DON'T GET THE SUFFICIENT TIME FOR ATTRACTING EACH OTHER ?

@RitabantaGhosh No. It is obviously not possible for alpha particles to get ionized any further since they are already fully ionized $\ce{He^{2+}}$ and don't have any electrons left. And please don't use all caps.

Can the alpha particles get neutral by accepting electrons ?

An alpha particle is basically a helium nucleus, with two positively charged protons and no electrons, the particle would react with any electrons it comes into contact with. However it likely depends on which electron it interacts with, and how close the particle comes that determine the outcome.
As ScienceDirect describes here -

In some cases, this force is not sufficient to separate the electron
from the atom, but the electron is raised to a higher energy state and
the atom is said to be “excited.” In other cases, the attractive force
is sufficient to remove the electron from the atom (ionization). The
closer an alpha particle passes near an electron the stronger the
force and the higher the probability an ionizing event will occur.

In Layman's words, as the alpha particle gets closer to the electron, the more likely it is to take the electron from its orbital for itself.
Electrons further form the nucleus of the gold atoms in the sheet would have been easier to ionize, as they'd have les ionisation energy.

The alpha particle causes many ionizations, but that doesn't mean that it can take the electrons for itself. Otherwise, the interactions would already stop after two ionizations.

The sciencedirect aggregation things are a bit dubious, I must say. I really don't like it. Sure, you are getting text from papers, but you don't see any of the context that surrounds it, and that's often critical to understanding a paper. I firmly believe that you can't learn stuff by reading random paragraphs from random papers.

I see... apologies for any inconveniences, I'll use multiple more reliable sources in any future answers.


text:
To which extent is protein folding conserving subprotein structures?
I am actually a pure mathematician, who stumbled over this paper «Protein-Folding Analysis Using Features Obtained by Persistent Homology» by Ichinomiya et al. (Biophys. J. 118, 2020, 2926-2937; link), and then red some introductions on protein folding, as well as articles on Deepminds progress on the issue.
Assume we are given a protein $p$ in primary structure, that is $p = [a_1,...a_n]$ is a finite, non empty list of amino acids denoted by $a_i$. A protein $p' = [b_1,...b_m]$ is called a subprotein of $p$ if there is some number $l \in [1:n]$ such that $a_{l+i} = b_i$ for all $i \in [1,...m]$.
Let $Q(p)$ denote the quaternary protein structure of $p$. It can be viewn as a subset of vertices (and edges if one needs that) of $\mathbb{R}^3$.
Question 1: Are there known examples (or even a database?) of proteins $p$ in primary structure, such that a subprotein $p'$ even exists?
Question 2: Assume we are given $p$ and a subprotein $p'$. Are there known examples for which $Q(p')$ is a subset of $Q(p)$? Note that we do not just need some injection, but a distance and angle preserving injection!
Do not forget that i am a noob. So these questions may be anything, from trivial to impossible to answer. I can not guess the outcome. Otherwise i would not ask.
For the interested reader: By a new result of Yuri Manin, persistence homology is linked to Nori-motives from algebraic geometry. As the paper cited above did come calculations on proteins via persistence homology, it may be possible to somehow "pull back" to the Nori category and derive a motivic decomposition of the folded protein but also of the protein in primary structure embedded in a plane. If question 2 has an "yes" for at least some cases, i expect for some isomorphic motivic summands to occur in both decompositions of $p$ and $p'$. If this is the case, then there could even be some common summand in the the motive of $p$ and $Q(P)$.

Sorry, but I have a feeling question made like you did wouldn't be useful or clear to pretty much anyone :/ If you just made up this "subprotein" term than it's even more likely. That's the reality - without proper jargon and wording in general, one can hardly make a good question.

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

@ Mithoron Well, I properly defined what i mean. I now have read through the definition of protein domains, to check if the definitions coincide.
@ Karsten, of course. I corrected it.

If Question 1) asks for the primary structure for the proteins, i.e., the mere sequence of the proteins, then this is a task entries from a database like uniprot.org may be used. E.g., BioPython contains functions to identify such similarities. But the concept isn't limited to proteins; you equally may apply this similarity check on nucleic acids, too. Perhaps a detail if a forensic lab (e.g., paternity test) / Craig Venter's human genome project use(d) an interpreted or a compiled language for performance, though.

I think you mean tertiary rather than quaternary structure for Q? Most proteins do not have quaternary structure, as they are free monomers.

@Andrew I still recall one of my biology teachers at high school level stating «quaternary structure of a protein is putting multiple proteins already folded in tertiary structure into a complex in common».  Thus, like a higher organization of multiple entities into one.

@Guestlee Equally see genome sequence analysis.

@Buttonwood That's a reasonable definition. Some people even go as far as 'quinary' structure for complexes of quaternary structure ...

If you want to apply math to the folding of macromolecules, you should also investigate (single-stranded) RNA folding. In some ways it is more amenable to a theoretical approach than protein folding, because its folding is hierarchical:  You can treat the creation of the secondary structure (base pairing) and tertiary structure (how the molecule arranges itself in 3D space) as separate problems.  A lot of mathematical models have been created for RNA folding for this reason.  Protein folding, by contrast, is less tractable, because the secondary and tertiary structures form cooperatively.

To question (1), the short answer is yes, such scenarios do exist, but they are rare for m greater than 10 or so due to the large number of possible sequences and the high frequency of substitutions even in very closely related proteins. Those that do exist are often encoded by the same gene, and the deletion of parts of one polypeptide (producing a "subprotein" by your definition) result from either posttranscriptional processing (deletion of parts of the encoding mRNA) or posttranslational processing (deletion of part of the synthesized polypeptide by proteolytic cleavage or intein excision).
For question (2), I'm going to assume you meant tertiary structure rather than quaternary, and your question, simply stated, is whether a subprotein has the same 3D folded structure alone as it does when within a larger protein. There are some cases where this is true, such as when short signal sequences are cleaved off. You can also have cleavage in a linker region between two folded domains, such that the folds of the domains are not affected. This result, however, cannot be universally assumed, as there are also many cases where deletion even of short segments of the primary sequence results in changes to the folded structure.
A well known example is trypsinogen, from which 15 amino acids are cleaved to produce the subprotein trypsin. The trypsin rearranges so that its overall structure is slightly different from the structure it had when part of the trypsinogen molecule. The new fold is catalytically active, whereas the trypsinogen is not. This type of activation that results from proteolytic cleavage is a common theme for enzymes, and the larger inactive forms are called zymogens.
Hopefully this is enough to point you in the right direction to read up more on this topic.

Thats a great answer. I have no idea which databases and software to consult (there some mentioned in the paper, but fiddling out how much of the whole picture they present will be very hard for me), but Biopyhton sounds like a start. I didnt even know that it existed.

The protein structural terminology in your question needs some work. For example "protein in primary structure embedded in a plane" - the primary structure is the sequence of the protein, and it is not clear how that would be embedded in the plane. Perhaps you mean embedding the graph corresponding to the atomic structure?
Pulling back, however, the larger question is "Can abstract mathematics (such as algebraic geometry) be applied to protein structural analysis?". The answer is that only if you can translate from the mathematics back to biology.
Consider this quote from the paper you link to:

Although PH is a highly effective tool for the analysis of nonlocal structures, it has several inherent limitations. First, PH results are sometimes difficult to interpret. In the original PH analysis, we obtain two values called “birth” and “death” for each loop or cavity and make decisions based on the distribution of these values. Frequently, these two values are insufficient to understand the physical relevance provided by PH.

They go on to describe how to fix the problem of this analysis being incomprehensible, but I think you see the point. A mathematically 'neat' description of something like protein structure might be elegant and amenable to powerful tools to analyse and transform that description. However it may be difficult to 'translate' back to conclusions about the actual biology or chemistry.
So using Nori-motives to say something about protein structure might 'work' in some sense, but the important thing would be what question in protein structural analysis it would answer. Also important is whether it would answer it better than simpler techniques that might be easier to understand and extend.

Yes you are right. I consider the 3d model of the protein before folding. I quietly assume that a protein starts folding only after it is totally generated. I fear that this is wrong and the bending starts, when the generating process is still going. Interpreting the result is anorher story of course, thats true. But since i am a pure mathematician, that isnt my biggest concern :) Establishing some correspondence beetween some categories or objects, is good enough for me.


text:
Significance of the ground state energy in quantum chemistry?
In quantum mechanics, quantum chemistry, and condensed matter physics courses, it seems like finding the ground state energy of a Hamiltonian is incredibly important. I can understand why having the ground state wave function (or behaviors of the wave function) could help, but I’m struggling to understand how the ground state energy helps us.
In Quantum Computing pitches by huge companies, I always hear something along the lines of “Quantum Chemistry simulations will allow for breakthroughs in the pharmaceutical industries,” but the quantum simulation algorithm in quantum computing they refer to is the Variational Quantum Eigensolver, which returns an estimate for the ground state energy. How does knowing this energy tell us anything about the material? I’m struggling to bridge this gap.

To understand the excited states, you need to understand the ground state. If you can't get the ground state correct, why would you believe any excited states?

I can understand this argument for the states themselves, but does the ground state energy on it's own tell us anything?

DFT for narrow-gap semiconductors quite often results in a ground state with a negative band gap - that is, DFT says it is a metal. So, how do you trust any DFT results on the material? Well, you don't until you get a proper ground state. In other situations, what you 'know' the ground state to be might not be what it actually is, particularly for complex systems with long range interactions.

So is the claim that knowing the GS energy tells us nothing, other than the fact that we can use the results to check against our other calculations? As in, knowing the true GS allows us to use and trust other computational methods?

Minimising the ground state energy WRT geometry gives us the equilibrium geometry, which is pretty important. Further Calculation of the ground state in a quantum calculation doesn't just give us the Energy, it gives us an (approximation to) the wavefunction as well, from which any ground state property can be calculated. Further perturbation theory of various forms applied to the ground state gives us a way into excited state properties, but you need the ground state first.

The ground state energy is used ubiquitously throughout various computational chemistry disciplines. I suppose a single ground state energy isn't particularly useful, but that's because all we can measure experimentally are relative energies.

We can probe the thermodynamic stability of a reaction (or set of reactions) by determining the energy difference between reactants and products. If you want to know if a drug will actually bind to a target protein/membrane site, you need to determine if the bound complex is actually stable relative to the separated form.

Differences in energy between reactants/products and transition states gives us insight about the kinetics of a process. If you want to know the timeframe for a drug to take effect or break down in the body, you need have some understanding of it's reaction kinetics.

A number of molecular properties (dipoles, polarizabilities, forces, etc) can be determined as derivatives of the ground state energy. These require either having an analytical formula for the ground state energy which can be differentiated or, for more complex methods where such a formula isn't available, using numerical/automatic differentiation, which requires evaluating the ground state energy (multiple times in the case of numerical derivatives). Molecular dynamics studies of diffusion of a drug in the blood or through a cell membrane require knowing the forces on the molecule.

The most common experimental observable for an excited state is the excitation energy, which to calculate requires both the energy of that state and the ground state. We don't really know the excited state energy experimentally, we only know it's energy relative to the ground state.

Speaking as a code developer one of the most important uses of the Energy is as the very first hurdle code has to pass before you actually believe the results - "if you can't get that right how can you get anything else right?" tends to be the view.

In addition to the excellent (thermodynamic) arguments listed by Tiberius, there is also a more technical reason why you want to know the energy. The variational principle implies that a trial wave function is a better approximation of the real wave function if it has a lower energy. Most methods in quantum chemistry take an expansion of eigen functions and try to find the combination of expansion coefficients that give the lowest overall energy in accordance with the variational principle. You therefore get the energy from the process of finding the best representation of your wave function.


text:
Are ekaconjugated polymers used as dielectrics? If not, why so?
So I came across an article called "Giant polarization in high polymers" and I was wondering if there was a reason why these types of polymers haven't found use in capacitors at this point.  This being ekaconjugated polymers which can store energy using nomadic polarization.  Perhaps the issue is with dielectric loss?  The paper also reports results at high pressures.  I've found surprisingly little on this concept in the last twenty years (with basic searches).

I'm certain we have members here who can address your query. That said, you might get more traction on Electrical Engineering SE if you don't get something satisfactory or complete, just a heads-up.

Well, I for one didn't know that eka-conjugation means "extended long-range electron orbital delocalization" in stuff like conductive polymers, much less whatever "nomadic polarization" is...

<ruling-out @Mithoron> on this one :)

If ekaconjugated polymers means conjugated polymers, there is work considering them in almost every electronics and optics scenario's. The broad class can be further divided. However, they can have very high dielectric but are prone to charge injection. I would resume that in many cases they aren't enough insulating. Exception and specific arrangement may differ, of course. But at the chemical physical level I think the answer is "they are too conducting". Also stability toward degradation is a rather general issue.


text:
Why is it advised to use neutral ferric chloride solution while performing confirmatory test for acetate ions?
The reaction of $\ce{FeCl3}$ with $\ce{CH3COO-}$ is
$$\ce{3Fe^{+3} +6CH3COO- +H2O<=> [Fe3(OH)2(CH3COO)6]^{+} +2H^{+}}$$
Recently I read that a solution of $\ce{FeCl3}$ hydrolyses very quickly to form $\ce{Fe(OH)3 +3H^{+}}$ as described by the following equilibrium:
$$\ce{FeCl3(aq) +3H2O <=> Fe(OH)3(s) +3H^{+}} \tag{1}\label{eqn:1}$$
with hydrolysis constant $K_\mathrm{h}$. I found a $K_\mathrm{sp}$ value of $\ce{Fe(OH)3}$ of $\pu{4E-38}$, so the equilibrium
$$\ce{Fe(OH)3 <=>Fe^{+3} +3OH-}$$
has $K_\mathrm{sp} =\pu{4E-38}$ and after a bit of rearrangement I got $K_\mathrm{h}$ of $(\ref{eqn:1})$ as $\frac{K_\mathrm w^{3}}{K\mathrm{sp}}$.
Hence at $\pu{25 ^\circ C}$, since the ionic product of water is $10^{-14}$, calculation gives $K_\mathrm{h}= \pu{2.5E-5}$. If we maintain neutral pH (ie. $10^{-7}$) we then get $\alpha$ (dissociation constant) for  $(\ref{eqn:1})$ as almost 1.0.
How is this possible? If this is true then how can $\ce{Fe^3+}$ ions be in solution for complex formation even if $\ce{CH3COO-}$ ion is not hydrolysed to acetic acid because of such low concentration of $\ce{H^+}$ in solution?

Theory guides and experiment decides (I. M. Kolthoff). Sometimes thermodynamics predicts a reaction but it does not occur very fast because its kinetics is slow. If you dissolve iron (III) chloride in pure denionized water, assuming it is from a fresh bottle, it will take hours to form a precipitate of iron hydroxides. Therefore, in order to perform this qualitative test, iron (III) chloride should be freshly made in water. Ten or 15 minutes won't make a difference.
As the linked answer here is slightly incorrect. The main reason to use neutral pH iron (III) chloride is that the carboxylic acid to be tested should be present as carboxylate ion in order to chelate iron (III) rather readily (chemical kinetics).
Same goes for ethylenediaminetetraacetic acid, is a powerful chelating agent. However its chelating power is not that great at low pH. It will also not dissolve in water until you add a few drops of a strong base. Then it is converted into ethylenediaminetetraacetate ion, and this anion can rapidly chelate with many metallic ions.

thankyou so much for the answer sir, just one thing ,can you confirm if  i am understanding this correctly? so acc to you this making soln neutral say by adding few drops of K2CO3 will prevent acetic acid formation thus will enable fe^+3 ions present in soln to effectively form a chelating complex(end product of my eq 1) much faster than it would take for fe+3 ions to hydrolyse thus making the test feasable to perform

If you add carbonate ion to iron (III) soln, you would see immediate ppt. So, carbonate is out of the question.

so what do we use to keep the ph of soln neutral

ow so acc to you freshly made fecl3 is a neutral solution hence no need to add anything else

i was thinking of something like a buffer to mentain ph if test is to be performed after an hour or so

Buffer will have ions which will interfere with analysis!

I am sorry but I read this somewhere that FeCl3 is neutralized by adding na2co3 then its used to perform the test to maintain neutrality.Your argument seems correct that it would catalyse the formation of fe(oh)3 giving red brown ppt and would interfere with test so i suposse freshly made soln will do right'


text:
What is the major product in the reaction between 4-chloropentanal and lithium aluminium hydride?
What would be the major product between 4-chloropentanal and $\ce{LiAlH4}$?
After the reduction of the aldehyde, would the hydroxyl group formed at $\ce{C-1}$ attack $\ce{C-4}$ causing cyclization to occur?

There could be an intramolecular attack on the chlorine bearing carbon once the hydride ion attacks the carbonyl carbon and the oxygen gets a whole negative charge. This will lead to the formation of methyl substituted oxolane. But this route will be followed if the halide part isn't reduced before the aldehyde part. In terms of reactivity, LiAlH4 will most likely attack the aldo group before the carbon bearing chlorine as the carbonyl carbon is more electrophilic.

It will depend on strict control of the equivalents of reductant, your work-up conditions and your control of temperature during the workup.
The intial product of the reduction is the aluminium alkoxide. If you use a proton source such as $\ce{HCl (aq)}$ or $\ce{NH4Cl (aq)}$ to quench the alkoxide and keep the reaction mixture cold then you will get the 4-chloropentan-1-ol product. If you use the commonly used $\ce{NaOH (aq)}$ workup then you may get the 2-methyltetrahydrofuran.
It is worth noting that 4-chlorobutan-1-ol is commercially available in large quantities.

I took the opportunity to add 2-methyl part to THF. You can rall back if you don't like my editing.

The reduction of alkyl halides to the corresponding alkane by $\ce{LiAlH4}$ is well known reaction (For example, see Ref.1):
$$\ce{R-X + LiAlH4 -> R-H + AlH3 + LiX}$$
The mechanism of this reaction has also been studied in some details (For example, Ref.2). Accordingly, the reaction is preferably $\mathrm{S_N}$2 and faster if the halogen is primary alkyl iodide and the hydride is $\ce{LiAlH4}$. However, the reaction of $\ce{LiAlH4}$ with a primary alkyl chloride is significantly slow compared to corresponding iodide. Therefore, the expected major product from the reaction of 4-chloropentanal and $\ce{LiAlH4}$ would be 4-chloropentan-1-ol after shorter reaction time and immediate workup. Now, if the substrate used is 4-iodopentanal, the major product would be 1-pentanol after comparably fast simultaneous carbonyl and halide reductions. However, if $\ce{LiAlH4}$ is replaced by the milder $\ce{NaBH4}$, the major product you would obtained is 2-methyltetrahydrofuran (2-Me-THF) due to comparably longer reaction time. The reduction of carbonyl group would result in 4-chloropentoxide intermediate, which would readily undergo intramolecular $\mathrm{S_N}$2 reaction to give the final product.

References:

S. Krishnamurthy and Herbert C. Brown, "Selective reductions. 27. Reaction of alkyl halides with representative complex metal hydrides and metal hydrides. Comparison of various hydride reducing agents," J. Org. Chem. 1980, 45(5), 849–856 (DOI: https://doi.org/10.1021/jo01293a018).
E. C. Ashby, T. N. Pham, and A. Amrollah-Madjdabadi, "Concerning the mechanism of reaction of lithium aluminum hydride with alkyl halides," J. Org. Chem. 1991, 56(4), 1596–1603 (DOI: https://doi.org/10.1021/jo00004a047).


text:
Finding freezing temperature and osmotic pressure of an ethylene glycol and glycerol solution in water

A water solution is prepared by mixing $\pu{200 g}$ of a ethylene glycol (ethan-1,2-diol; $\ce{C2H6O2}$) solution (20% by weight) with $\pu{500 g}$ of glycerol (propan-1,2,3-triol; $\ce{C3H8O3}$) solution (15% by weight). Calculate the freezing temperature and the osmotic pressure at $\pu{25 ^\circ C}$ knowing that the density of the prepared solution is $\pu{1.25 g/ml}$ (For water,  $K_c = \pu{1.86 K kg mol-1}$).

I know all the formulas but how do I proceed to start with these types of exercises?
I know, for example, that $\Pi=MRTi$, with $M$ being molarity. But how do I find the volume of the solution?
And what about $i$, how do I calculate it here?

Hint: You know that masses are always additive. You also know the density of the total mixture. Determine the volume (L).

The volume you need here is to calculate the molarity, that is for finding osmotic pressure. Use the weight % to calculate molality, which needs to calculate freezing point.

First, the $i$ in the equation you cited is the van't Hoff factor of the solute(s) (ethylene glycol and glycerol in this case). The van't Hoff factor accounts for how many species are generated by the solute. Specifically, it is the important for electrolytes such as $ \ce{NaCl} $. Because $ \ce{NaCl} $ dissociates into $ \ce{Na^{+}} $ and $ \ce{Cl^{-}} $ in the solution, its van't Hoff factor should be 2. Similarly, you can see that $ \ce{CaCl2} $'s factor should be 3 (1 $ \ce{Ca^{2+}} $ and 2 $ \ce{Cl^{-}} $ )
However, since it is obvious that neither ethylene glycol nor glycerol dissociate in the solution, you can safely ignore it because their van't Hoff factors are all 1.
About the volume, it is actually fairly easy. They gave the density of the prepared solution to be $ \pu{1.25 g/ml} $, which is data you need to calculate the volume. Since density equals mass divided by volume, you will need mass and density to calculate the volume. The mass of the resulting solution is equal to the sum of the masses of the original solutions, i.e
$$ m_{\text{result}} = \pu{(200 + 500) g} = \pu{700 g } $$
From this we can calculate the solution's volume:
$$ \text{Volume} = \frac{\text{mass}}{\text{density}} = \frac{\pu{700 g}}{\pu{1.25 g/ml}} = \pu{560 ml = 0.56 L} $$
As you said you know all further formulas I will leave the rest for you.

Thank you! I guess i was just confused from the %wt. DO they not play any role in here at all?

The %wt is the concentration of the original solutions. From this you can calculate the masses and moles of the two solutes. For example the mass of ethylene glycol should be 200 * 20% = 40 g.

In the answer elsewhere, ĐỨc Lê Hồng has given how to find the volume of the solution and why van't Hoff factor is not important the case of  ethylene glycol and glycerol are involved as solutes in the solution. However, you must know how to find the molality to find the freeding point depression and molarity to find the osmotic pressure.
To do so, you need to find the amounts of these two solutes in the solutions. First look at the amount of ethylene glycol (ethan-1,2-diol; $\ce{C2H6O2}$) solution. It is said that $\pu{200 g}$ of a $20\%$ $(w/w)$ ethylene glycol solution is mixed.  That means $200 \times \frac{20}{100} = \pu{40 g}$ of ethylene glycol in there. The rest, $\pu{160 g}$ is water. The molar mass of ethylene glycol is $\pu{62.07 g mol-1}$. Thus amount of ethylene glycol in the final solution is $\frac{40}{62.07} = \pu{0.644 mol}$.
You can do the same calculations to find the amount of glycerol (propan-1,2,3-triol;  $\ce{C3H8O3}$) in the final solution. It is said that $\pu{500 g}$ of a $15\%$ $(w/w)$ glycerol solution is mixed with ethylene glycol solution to obtain the final solution. That means $500 \times \frac{15}{100} = \pu{75 g}$ of glycerol in the mixture. The rest, $500 - 75 = \pu{425 g}$ is water. The molar mass of glycerol is $\pu{92.09 g mol-1}$. Thus amount of glycerol in the final solution is $\frac{75}{92.09} = \pu{0.814 mol}$.
Thus, the amount of solute in the final solution $= \pu{(0.644 + 0.814) mol} = \pu{1.458 mol} $.
And, the amount of pure water in the final solution $= \pu{(160 + 425) g} = \pu{585 g} = \pu{0.585 kg}$.
Therefore, molality $(m)$ of the final solution $= \frac{\pu{1.458 mol}}{\pu{(0.585 kg}} = \pu{2.492 mol kg-1} $.
Since you know the equations you can calculate the $\Delta T$ of melting point, and hence freezing point of the solution.
Using same data, you can calculate the molarity of the solution as well. The final volume of the solution was found to be $\pu{0.560 L}$ (see ĐỨc Lê Hồng's answer).
The amount of mixed solutes in the final solution $= \pu{(0.644 + 0.814) mol} = \pu{1.458 mol} $.
Therefore, the molarity $(M)$ of the final solution $= \frac{\pu{1.458 mol}}{\pu{(0.560 L}} = \pu{2.604 mol L-1} $.
Since you know the equations you can calculate the $\Pi$ (osmotic pressure) of the solution as well.

thanks a lot! Unfortunately i cant vote yet cuz i dont have 15 points yet sorry!


text:
How to determine the equivalents of OH- in a titration curve
How does one go about determining the equivalents of $\ce{OH-}$ needed for the titration of a weak acid. For example here is the titration curve for Aspartate:

Why is the equivalents for $\mathrm{p}K_\mathrm{a1} = 0.5$? Why not $0.75$ or $0.80$? How was the answer 0.5 derived. How were the equivalents for $\mathrm{p}K_\mathrm{a2} = 1.5$? derived? I understand the $\mathrm{pI}$ is 1 because there's a balance of charges but why the 0.5, 1.5, and 2.5

It is trivial, as by definition of Ka and the derived pKa,  pKa is pH where the ratio of respective concentrations ( exactly rather activities ) is 1:1, therefore the half of the equivalence.

The answer can also be:  Because log(1)=0

Let's say you are titrating a weak acid $\ce{H3A}$ with a strong base.
$$\ce{H3A + OH- -> H2A- + H2O}$$
$$\ce{H2A- + OH- -> HA^{2-} + H2O}$$
$$\ce{HA^{2-} + OH- -> A^{3-} + H2O}$$
When all $\ce{H3A}$  reacts with $\ce{OH-}$ to form $\ce{H2A-}$, the first equivalence point is reached.
Before the first equivalence point is reached, the $\ce{H3A}$ and $\ce{H2A-}$ buffer exists, whose pH is given by the Henderson-Hasselbalch equation.
$$\mathrm{ pH = pK_{a1} + log\frac{\ce[H_2A^-]}{\ce{[H3A]}}}$$
When $\mathrm{pH = pK_{a1}}$, $\ce{[H2A-]}$=$\ce{[H3A]}$ which means the first half-equivalence point is reached, which requires $\pu{0.5 equivalents}$ of $\ce{OH-}$.


text:
What's the purpose of significant figures in measurements?
I have read about significant figures in my chemistry and physics books, but none of them pointed out the significance of the significant figures.
Searching the Internet I found two possible answers: Some people are saying that it is to denote the accuracy and others are saying that it is to denote the precision of the measurement.
Both of these arguments are unsatisfactory, because in my opinion significant figures are not the only thing that governs the accuracy or precision of a measurement.
Layman's definition of accuracy and precision
Given a set of data points from repeated measurements of the same quantity, the set can be said to be accurate if their average is close to the true value of the quantity being measured, while the set can be said to be precise if the values are close to each other.
My arguments regarding significant figures, accuracy and precision:
Accuracy.
Suppose I measure the weight of a candy using some device and get the measurement as 1.2 grams and then I measure it again using some other sophisticated device and got the answer 1.436575383 grams.
The first measurement is having two significant figures whereas the latter one is having 10 significant figures.
How can we say which measurement is accurate, if we don't know the actual mass of the candy? Here many might go for the latter measurement and will do all the calculations with that value. However, what if the second device was faulty, which was not known at the time of measurement.
Precision.
Does a higher number of significant figures make a measurement more precise? How?
The larger the object I measure using the same device the higher is the number of significant figures displayed, but that does not make the measurements more precise.
Don't both, the precision and accuracy, depend on the operator of device?
So what is exactly the purpose of significant figures?

Significant figures and the related rules for arithmetics have been invented as the layman way how to address and process quantitative aspect of resolution/precission/accuracy errors and the error propagation. Alternative way are the rules of propagation of standard error estimation for artihmetic operations, or, more generally, with involving partial derivatives of dependent variable.

Some people are saying that it is to denote the accuracy ( method bias ) and others are saying that it is to denote the precision ( method random errors ) of the measurement.  Both are true. And some people are saying that it is to denote the resolution ( obtainable primary data )

You may want to read this Wikipedia article.

Moreover, since the term "precision" is somewhat ambiguous for historical reasons, nowadays metrology-aware people tend to use the terms "accuracy" (closeness to the "true" value) and "repeatability" (closeness to each other of repeated measures of the same quantity).

Related, possible duplicate: Significant Figures Interpretation

While estimating uncertainty can be quite involved, the purpose of significant figures is quite simple:  They are a shorthand method of communicating the uncertainty of a measurement from those who are in a position to be able to estimate the uncertainty (those who made the measurement) to those who need to know the uncertainty, but can't determine it themselves (the reader/audience).
I say shorthand because the uncertainty is more completely communicated by explicitly specifying it, e.g.:
$1.0342(86)$, which is equivalent to  $1.0342 \pm 0.0086$

You wrote:

Both of these arguments are unsatisfactory because in my opinion significant figures are not the only thing that govern the accuracy or precision of a measurement.

You seem to have it backwards. Significant figures do not govern the accuracy or precision of a measurement, in any way. Rather, it's the accuracy/precision that governs the number of significant figures.
I.e., it's not significant figures ⟹ measurement uncertainty.  Rather, as explained above, significant figures communicate what the measurement uncertainty is. Thus it's measurement uncertainty ⟹ significant figures.
First you determine the uncertainty.  Then you determine the appropriate number of significant figures to use to express that uncertainty.  The fact that your scale reads out, say, $\pu{0.285746 g}$ doesn't mean that the uncertainty is in the last digit (which, here, would be micrograms), and thus doesn't mean your measurement should be expressed to six significant figures.  It could be that you know the uncertainty is 100's of micrograms, regardless of the number of digits in its readout.  In that case, you'd express the answer as $\pu{0.2857 g}$.  Again: Uncertainty ⟹ significant figures.
To give a car analogy: Reviews of a new sports car in various enthusiast magazines don't determine the car's performance.  Rather, assuming the reviews are sufficiently competent and unbiased, it's the car's performance that determines the content of the reviews.  The reviews don't determine performance, they merely communicate it.

You are absolutely correct in that there are several kind of uncertainties (often called errors). These are referred to as systematic and statistical errors and the difference between them can be conveniently be described in terms of a target with a distribution of shots. The distribution of the shots with respect to the bullseye represents the kind of errors as also illustrated in the figure below. The spread of shots represents the statistical error and is related to the precision of the measurement. The less spread the more precise. This says something about the reproducibility of the measurement. The deviation of the average of the shots is a measure of the systematic error (e.g. a calibration error). This is related to the trueness of the measurement. If both precision and trueness are high, the measurement is said to be accurate. (Confusingly, sometimes accuracy is used for trueness instead and no separate term is used for a measurement that is both precise and true).
Whereas you can get an idea of the magnitude of the statistical error by doing many measurements and using standard statistical analysis methods, the systematic errors are very difficult to spot. One way is, as you pointed out with the example of weighting candy, is to change the measurement instrument. Other means are by repeating measurements over long time scales or manipulating your measurement in a known way and check the response of the thing you want to measure. For the determination of fundamental constants for instance, several independent measurement methods are required.
Now to come to your question about significant figures. Personally I think the best way to address this issue is to not only state a certain number of significant digits, but also the error in this digit. Often the statistical and systematic errors are combined into one by adding them in quadrature (in which case it is implicitly assumed they are independent)
$$
\Delta_\text{tot.}=\sqrt{\Delta_\text{stat.}^2+\Delta_\text{sys.}^2},
$$
But even better, in my opinion, is to give both the systematic and statistical error separately.

Image source

Once upon a time, calculations were done on paper and on slide rules. It would take considerable effort to carry 10 digits in a calculation instead of 3 (which is what most slide rules are capable of). Then, students started using calculators, and would just copy the result they got, with 10 or more digits. Science teachers thought that was ridiculous, and started teaching error propagation using significant figures.
Usually, when students start working in the lab, they have cheap or moderately expensive glassware and instruments, and there is always a measurement going into the calculation not known to more than 3 or 4 significant figures (unless you just determine a mass, where you might get 6 or 7). Some teachers, for that reason, implement a 3 significant figures rule, and move on to more sophisticated treatment when the students gain experience and access to better instrumentation.

Suppose I measure the weight of a candy using some device and get the measurement as 1.2 gram and then I measure it again using some other sophisticated device and got the answer 1.436575383 gram.

An analytical balance that reads off to the nanogram will be calibrated back to the SI unit definition, otherwise you would not pay for all those digits (so you expect it to be accurate). Also, there will be more sophisticated measures to prevent drafts and vibrations than in a cheap balance (so you expect it to be precise). So for an instrument that is not broken and is used properly, you expect the precision and accuracy to correllate to a certain degree, and you would not average results from a cheap and a sophisticated instrument.
If you are wondering about the skill of the operator, you would measure a standard multiple times and compare results to an experienced operator or to the specs that came with the instrument.

A number itself doesn't inherently precision or accuracy. It is the representation of a measurement: the output of a measuring instrument. That instrument is what has precision and accuracy. We accept the number representing an instrument's output on the basis that we trust the calibration and quality of the instrument.
Establishing a certain number of significant figures is a way of capturing the precision of the measurement based on what we know about the properties of the instrument. If we know that some scale weighs to the milligram, and we weigh some object that weighs 230 milligrams, then we have three significant figures. To clarify that the 0 counts as significant, and isn't merely a placeholder, we use scientific notation: 2.30 × 102 mg.
This figure could be wildly inaccurate. Suppose someone weighed a 1000 mg object on the scale previously and zeroed it out, so that it reads -1000 unloaded. Our 230 mg object is actually 1230 mg. We have misinterpreted the output of the instrument based on a false belief about its calibration state.
Suppose the scale is of poor quality. If we measure the same object multiple times, we get different figures: 220, 227, 239, 231, ... in that case, our acceptance of three significant figures of the 230 measurement had been based on the false belief that all digits on the scale's display actually mean something.
The absolute value of a measurement, and the number of significant figures, are just representations, which are based on beliefs about the instrument. If those beliefs are true, the measurement's value is useful, and the number of significant figures is an additional important, meaningful attribute which gives a coarse idea about the limits of the measurement's precision.
The primary value of the concept of significant figures is that it provides an at-a-glance sanity check on the honesty of scientific and engineering measurements and calculations. We know that if some calculation has several inputs, the least accurate of which has only two significant figures, but the result is being reported with four significant figures, that something is wrong. Suppose we see some final or intermediate result $x$ being calculated via some function $f$ like this:
$$x \approx f(1.1\times {10}^2, 5.763\times 10^{-3}, 8.59\times 10^3)$$
$$x \approx 2.253\times {10}^{-2}$$
We know right away, "wait a minute, how is this researcher getting four significant figures when the least precise input has a mantissa of 1.1? Oops; the honest result is actually 2.2 something, not 2.253". Note that we can perform this sanity check at a glance just by looking at the list of input arguments and the output; we do not have to know anything about $f$.
That's the value of significant figures. Numbers do not attest to the accuracy of a measurement, and significant figures do not vouch for it accuracy. What significant figures do is provide a representation of the measurement's precision and help eliminate or reduce false inflation of precision in subsequent calculations.  They cannot ensure that the precision of the measurement itself isn't falsely inflated relative to the actual abilities of the instrument, but that's a separate problem which doesn't negate the value of significant figures.
Like all computing mechanisms, significant figures obey the GIGO principle: garbage in, garbage out. If a measurement appears to give us 10 significant figure, but 7 of them are garbage, then the same is true of our calculation; it may look like we have 10 significant figures in the result, but at least 7 of them are garbage.

So in short I can say, that significant figures are used to denote both accuracy and precision given that the person reporting it is trust worthy and experienced, he had not used a faulty machine etc.


text:
How to compare the bond angle of ozone and nitrite anion
How do I compare the bond angle between $\ce{O_3}$ and $\ce{NO_2^-}$ ?
I've studied how to compare the bond angle when normal covalent bonds are present where the electrons are shared in the bond and so the concept of electron repulsion is used to judge the bond angle. But what about the case when one of the bonds is a coordinate bond? Is the process of finding the bond angles the same or is there anything else that happens in this case?

Your "the bond angle between $\ce{O3}$ and $\ce{NO^-_2}$" reads like the two were in close contact with each other.  At ambient conditions, this is unlikely as ozone is a gas, and the nitrite either dissolved in a solution, or in a salt, then in solid state.  More plausible is "the bond angle of $\ce{O3}$ and $\ce{NO^-_2}$", accounting for each individually.  Second, do you know about resonance?  The negative charge is not fix and isolated to stay on one of the oxygen atoms of the nitrite.

Changed the title as you said. I wanted to create a scenario where there was a coordinate bond and this "bent" sort of system and so I chose $\ce{NO_2^-}$. What I mainly wanted to know was that how these rules apply in case there are coordinate bonds?

To record an angle $\angle{} (A,B,C)$ along A, B (the apex), and C, you need 3 points (here: atoms); regardless if these belong to an ion or neuter molecule. At present, your question lacks some clarity to see if it addresses VSEPR theory or / and AXE rules (illustrations incl. ions) where however lone pairs (= potentially a charge) and overall charges may matter.

@Buttonwood I wish to compare the O-O-O and the O-N-O angles. However, my primary concern is not with these specific compounds but rather the idea. In case of normal compounds with standard covalent bonds the bond angle can be compared using the quoted theory here. And so I wanted to know what would happen if one of these bonds were coordinate. Please elaborate on where clarity is needed and I'll modify the question accordingly

Your question becomes even more confusing for me because VSEPR may be applied (with the same limitations as for compounds with "standard covalent bonds") equally on transition metal complexes, e.g., the diaminesilver(I)-complex (depicted, e.g. here).  Count me out of this question.

@Ashish Consider a coordinate bond like a covalent bond and when you do draw the structure of $\ce{O_3 ,{ NO_2}^-}$ you will notice that the only apparent difference is the central atom. Considering a electronegativity arguement it can be shown that Bond angle of Ozone should be more.

This question is a complete riddle to me, and since I'm not the first, I'm going to close it. If you want to compare something, you measure it, then you compare it. I think what you are looking for are models of prediction for bond angles. The other problem is that there is no coordinate bond in either of these molecules, do your question doesn't make sense. Please include your understanding of bonding into your question. I guess there is a general misunderstanding there. Have you heard of molecular orbital theory?


text:
When does reaction of alcohol with alumina give an ether and when an alkene?

What is the main product when an excess of ethyl alcohol vapour is passed over heated alumina at 250 °C?

I read that this reaction gives an ether (here, diethyl ether) at relatively lower temperature than when it gives an alkene (here, ethene). This didn't bother me much as I had also read that higher temperatures favour elimination reactions, whereas lower temperatures favour nucleophilic substitutions.
So I was pretty sure that at 250 °C it would give ethene. But according to the book, it's diethyl ether. Why does this happen?
One thought that I had was maybe it's because “excess of ethyl alcohol” is mentioned, it would follow the formation of ether by the reaction of two alcohol molecules. Is that right?

As you say the key here is excess ethanol. Enough ethanol that one ethanol molecule that is interacting with the alumina surface is captured by a second ethanol molecule before it loses H+ to become ethene

What book is this question from?

According to Paul Sabatier in "Catalysis in Organic Chemistry",

With ethyl alcohol [on alumina], ether is formed above 240° and at 290° pure ethylene is evolved regularly, this evolution becoming rapid at 340°.

so at 250°C the product should be primarily diethyl ether. I don't think "excess" has the normal organic chemistry meaning in this context of heterogeneous catalysis. My supposition is that the question is just lazily trying to get you to recall this exact statement from the preceding chapter, rather that testing your conceptual knowledge. I would be interested to know what book this is!


text:
Why is the bond angle of sulphur difluoride greater than that of hydrogen sulphide?
Why is the bond angle of $\ce{SF_2 (98.05^\circ) > SH_2 (92.11^\circ)}$?
Isn't this contradicting Bent's rule or otherwise electron repulsion rule ? Fluorine is more electronegative and hence it should have bonds with higher $\mathrm p$ character than those which hydrogen will form. Which means $\ce{F-S-F}$ bond angle should be smaller than $\ce{H-S-H}$ bond angle. Why is it not the case?
Source:

$\ce{SH2}$ bond angle
$\ce{SF2}$ bond angle

Electrostatic repulsion?

Extent of hybridization is very low in these compounds and thus Bent's rule doesn't hold.

Moreover sulphur has a positive formal charge in $\ce{SF2}$ whereas in case of $\ce{SH2}$ it has a negative formal charge.

@NisargBhavsar Any reason for not correcting “flourine” and “diflouride”?

Please note that names of chemicals or chemical elements (like 'hydrogen') should not be capitalised, unless occurring at the beginning of a sentence. Additionally, IUPAC (and most people) use the spelling 'sulfur' nowadays.

@SafdarFaisal could it possibly be back bonding by p orbital of fluorine to vacant d orbitals of sulfur?


text:
Heat released at constant pressure and temperature and heat released at constant density and temperature

Is there a relation between heat released at constant pressure and temperature and heat released at constant density and temperature?

I was reading some books on propulsion and combustion, and the author suggests the following exercise:
the heat released of the reaction $\ce{H_2 + 1/2O_2 -> H_2O_{(g)} }$ at constant-temperature and constant-pressure is $q_{_{(P,T)}} = \pu{57.8 kcal}$.
What is the heat released $q_{_{(\rho,T)}}$ of the reaction $\ce{ O_2 + 5N_2 + 2H_2 -> 2H_2O + 5N_2 }$ at constant-temperature and constant-density?
As a matter of fact, I have been wondering: what is the general relation between the heat released $q_{_{(P,T)}}$ at constant $(P,T)$ and the heat released $q_{_{(\rho,T)}}$ at constant $(\rho,T)$ for a same reaction $\sum  \nu_i C_i \rightarrow 0$ ?

The relation would be too complex to be any useful. Furthermore it would be too reaction specific to generalize. It would depend on how the average molecular mass changes with progress in the reaction.

@NisargBhavsar And about this specific reaction, how can I calculate the heat released at constant $(\rho,T)$ ?