text:
Why there is a difference between atoms and chemicals
I recently asked this about why we call things "isotopes" and "elements" (atoms) instead of just having it be a bunch of particles bundled together in different numbers.
This makes me wonder about the definition of Chemical, which Wikipedia has sort of a circular definition. A chemical cannot be broken into elements (atoms) unless breaking chemical bonds (basically the definition)....
So atoms have protons and neutrons (and electrons) that are bound by "atomic bonds". I'm not sure that's correct. So strong force holds a single together. But not sure what holds protons together, maybe that's the weak force. Then "chemical bonds" are between electrons between atoms. My question is, why not just get rid of the distinction of calling things "protons vs. atoms vs. chemicals", and just say there are "particles" which bundle together. They bundle into group 1 (atoms), and group 1's bundle into group 2's (chemicals), but really it's all just particles that are all interacting.
So this makes me wonder, what exactly a chemical is. It is a stable particle system essentially, where atoms combine into larger chunks.

See, as soon as you got rid of "atoms" and "chemicals", you came up with your own "group 1" and "group 2". There is no going around it. This level of organization just is there, and you need a word for it. Other terms are too broad. Come to think of it, you are a stable group of atoms combined into larger chunks.

Chemicals is a kind of practical and commercial definition. Products that one can buy. Are Mg flakes or Fe shavings to be called chemicals? Probably, as they are listed in catalogues. Surely they aren't atoms nor fine chemicals. The definition of Wikipedia is wrong, and refers instead to molecular compounds, or even elements as they normally come as chemicals, if bonds such as the metallic one are considered. Wikipedia and this SE page seem to confuse chemicals and compounds. Thiugh the two terms aren't mutually exclusive, as it should be clear from what I said above, they refer to....

. to different contests and uses.  You pass an order to buy chemicals, not atoms or elements or molecules.... Do you get it?

I see your question - and not Wikipedia -  MIXES chemicals with chemical substances. The latter being elements or compounds.. You should change titles and text else you are asking a different thing for which my comments hold.

The distinctions matter because of what we can observe in nature and the lab
If you want to make sense of the world around you it helps to have some coherent scheme to describe the world and classify the things in it. That scheme should map onto things we can observe. We could just describe things by their nuclear properties (the count of protons and neutrons in the atoms) but this is a poor classification for two reasons:

things with the same number of protons are very, very similar and it mostly doesn't help to distinguish them unless you are doing nuclear physics
most of the substances we observe in the world have properties that their nuclear composition does not explain

We have a periodic table not a (much bigger) nucleonic table because the easily observable properties of elements (which are mostly chemical or physical unless you are a nuclear scientist) are very similar for isotopes with the same number of protons. Most of the ways those "particles" interact are highly dependent on which element they are.
We have a separate class (chemicals, molecules or chemical substances) because of the way atoms interact. Different atoms join together in different ways the mostly demand on how their electrons are configures (which, in turn, depends on the number of protons in their nucleus). There are bazillions of ways to combine atoms into molecules and each resulting chemical substance has distinct features (colour, melting point, taste, toxicity....) The observable features are dependent on how the atoms join together: different ways of joining them together result in substances with different properties. That's why we have a subject called chemistry. If we ignored that combinatorial complexity and just talked about combinations of particles we would be ignoring the obvious ways of classifying the distinct chemical substances we observe.
It is worth looking at a few examples to clarify this and to explain the Wiki definition of chemical. First note that chemical substance includes elements (everything is a chemical substance even if it is only made from one type of atom). Take carbon, for example. Diamond is one type of carbon, graphite is another. Both are forms of the element (technically allotropes). Both could be thought of as bundles of carbon atoms but that is a useless definition as it doesn't explain why their properties are so different. Chemistry does: they are different because the way the atoms are joined together and organised is different (one has flat hexagons in near-infinite planes, the other has near-infinite tetrahedral arrays of carbons joined together). Or consider ethanol and dimethyl ether. They both contain exactly the same bundle of atoms. But that doesn't explain why they are so different (one boils at -24 Celsius, the other is a pleasant intoxicating drink). again the way the atoms are connected together explains the observable difference in properties; the bundle of atoms they re made from does not.
Moreover, those observable properties of chemicals will be the same even if the atoms they are made from are different isotopes (12C and 13C are very similar and alcohol made form either will still make you drunk and taste like booze). So chemists mostly ignore nuclear properties and explain things using the periodic table which explains almost everything about how atoms connect together into molecules.
So we group different nuclei together into elements because that helps us classify the things that make up chemicals in ways that help explain what matters about how they connect together. We have a separate classification for chemical substances as they depend on the details of how atoms connect together. These are meaningful classifications that help us explains the properties we can observe in the world and the lab.

In theory you could go one step further, and describe things only by a Waveform a. la. quantum mechanics.  If you did this, you could do away with all distinctions completely, and merely have one waveform which describes the entire system!
The reasons for the different categories are practical ones.  If I'm talking at what we now call the atomic level, working with small numbers of atoms, I hardly care about the precise configuration of where the protons and neutrons are in the nucleus.  What I'm doing is almost always completely immune to changes at that level.  Thus, why waste my time looking into such small details.
I also typically don't have much control over the lower levels.  Unless I am working at a nuclear reactor, or looking into how the sun works, the number of protons and neutrons in an atom is pretty much fixed.  There's radioactivity, which sort of dabbles between them, but other than those cases, I treat the number of subatomic particles as pretty darn fixed.
Likewise, when it comes to chemicals, they have behaviors which are very reliable.  In particular, we can predict how one chemical turns into another.  If I have acetic acid and sodium bicarbonate (aka vinegar and baking soda), I can reliably predict what will happen when I combine them, without worrying about precisely how the atoms are vibrating within their general shape, nor care about the isotopic mixtures.
If you really think about it, the whole thing is rather arbitrary because it's focused around bonding.  Nothing in the laws of physics causes bonds to "form" and "break" like we claim they do.  Rather these are simplifications of much more complicated interactions.  But for all intents and purposes, it is incredibly effective to think in these terms.
So why do we have chemicals, atoms, and protons?  Because generations of scientists have found that if you choose to think in those terms, you find things behave very predictably, and if you choose other lines to draw, they are much harder to predict.  It's the human mind that put those divisions there, not nature.

I think you went a bit too much. Chemicals is just a word to refer to chemical products available for further chemical use what ever they are. Like reactives might be chemicals or made from chemicals, and like reagents can be chemicals to be used in further reactions. All practical "definitions" with little physics in it.

After I found out that the OP assumed chemicals as chemical substances. So all my comments were dictated by this. Still my comment is valid . Not to start sterile polemics but the answer is vague and mentions non necessary hard stuff. Which is not a surprise as for the question itself doesn't help a potential answerer.

@Alchimista I read the question as more of a philosophical one of why put divisions in place at all.  If the question just mentioned chemicals and atoms, I think you would be right, but when protons get into the mix, I think that shows more of a fundamental "how does science approach these problems" sort of thing.

Yes you're right. Indeed the question is a bit displacing. Also, though chemicals has its own meaning,  it is still true that all chemicals are chemical substances, after all.


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Is there a way to export detailed data about atom connectivity, dihedral angle, etc from GaussView?
As far as I know, opening the .com files as .txt files (with a text editor like simple notepad) yields a nicely formatted table of (atom) (x) (y) (z) values that can then be worked with. The problem is that this data has no information about what atom is connected to what. But within the actual Gaussview software, there seems to be a data set with more detailed information. Is there any way to easily extract that data via code?

To my knowledge, neither .xyz nor .com format preserve connectivity information by default. Gaussview (and many other programs) makes an educated guess based on atomic covalent radii and positions of atoms. Gauss .com may also store a set of internal coordinates used for optimization that gaussview derive from connectivity information. Gaussian .com may contain connectivity information if geom=connectivity option was used, but you cannot expect it. The format of connectivity section is documented in geom=connectivity suboption part of Gaussian input docs.
That said, some other formats like .pdb allow to store that data explicitly and are well documented. Also, Gaussian can be used to perform various forms of population analysis printing a lot of nice tables, some of them might be interpreted as connectivity information (then again, some cannot).

The Xmol format cannot store connectivity information at all (at least in its initial definition). It's funny that the simplest, an probably most recognised format in computational chemistry, cannot be exported by Gaussian anyway.

Just to extend on that a bit, Gaussview can export to .mol and .mol2 and .pdb, all of which store connection data. None of these formats are meant to be read by humans though, and depending on the molecule, they will not be helpful either.

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What mixing ratio of ethanol and acetone has the lowest freezing point?
I am in the lab trying to make a sample for 1H NMR. It has to remain liquid at the lowest possible temperature.
I was recommended to use a mixture of ethanol and acetone because when they're mixed in the right proportions, the mixture has a very low freezing point.
I can not find this proportion anywhere. Does anyone know where to look for these things?
Here is a picture of what I'm looking for, but with ethanol and acetone instead of water - solvent:

PD:it's very likely that its methanol instead of ethanol, please let me know... I have access to both, I just want the one that will go to the lowest temp.

Wikipedia has this table https://en.wikipedia.org/wiki/List_of_cooling_baths

if you need a glass try ethanol, diethyl ether, isopentane in 2:2:5 or 1:4;4 ratio , (EPA glass).

I assume you want to use this as a solvent, right? There's a good chance that whatever it is precipitates at 140K. Otherwise still be prepared to see a spectrum that is not quite as well resolved than at room temperature.

its me again, I  want to post the answer I found because I think it might be useful for somebody in the future; it appears a 2:3 molar acetone:methanol proportion gives the lowest freezing temperature ~158 K (see attached picture below).
It appears the difficulty finding these curves was because they are usually called "solid-liquid equilibrium" curves instead of "freezing point curves", as its the same physical situation (Explanation Here)
There is a large database of these in the following link.

EDIT: As commented by "Ivan Neretin", ethanol-acetone mixtures are also good for low freezing temperatures:

Note that ethanol alone  would get you about that low, and ethanol+acetone lower yet, though not by much.

In your question, you didn't mention how low freezing temperature solvent system you are looking for instead of telling that the solvent has to remain liquid at the lowest possible temperature. And,  Ivan Neretin mentioned that ethanol alone would get you about freezing temperature close enough to get you any ethanol+acetone solvent would get. For example, freezing point of ethanol is about $\pu{159 K}$, when compared to ~$\pu{155 K}$, the reported freezing point of ethanol:acetone (83:17) system (lowest of the system; see the chart given in your answer). Yet, freezing point of pure diethyl ether is $\pu{150.6 K}$, lower than the lowest achieved in acetone/ethanol system. Thus, I decided to list some low temperature systems for your convenience (Ref.1 & 2):
$$
\begin{array}{ccc}\hline
\text{Solvent} &\text{Solute} & \text{Solvent mole fraction} & \text{Freezing point in}\; \pu{K} \\\hline
\text{diethyl ether} & \text{none} & 1.0000 & 150.61 \\
\text{diethyl ether} & \text{acetone} & 0.9737 & 149.53 \\
\text{diethyl ether} & \text{carbon disulfide} & 0.9615 & 149.53 \\
\text{diethyl ether} & \text{methanol} & 0.9586 & 149.34 \\
\text{diethyl ether} & \text{acetone} & 0.9512 & 148.85 \\
\text{carbon disulfide} & \text{toluene} & 0.9393 & 157.30 \\
\text{ethanol} & \text{none} & 1.0000 & 157.93 \\
\text{ethanol} & \text{methanol} & 0.9783 & 156.37 \\
\text{ethanol} & \text{methanol} & 0.9704 & 155.59 \\
\text{ethanol} & \text{water} & 0.88 & 151 \text{ (Ref.2)} \\
\text{methanol} & \text{water} & 0.82 & 155 \text{ (Ref.2)} \\
\hline \end{array}
$$
Note that last two entries on the table are from the charts in Ref.2, which are purely approximate values.

References:

M. Shinroku, “The Heats of Fusion of some Organic Solvents Whose Melting Points are Relatively Low,” Bulletin of the Chemical Society of Japan 1926, 1(2), 30–34 (https://doi.org/10.1246/bcsj.1.30).
K. Takaizumi, T. Wakabayashi, “The Freezing Process in Methanol-, Ethanol-, and Propanol-Water Systems as Revealed by Differential Scanning Calorimetry,” Journal of Solution Chemistry 1997, 26(10), 927–939 (https://doi.org/10.1007/BF02768051).

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Are all fatty acids monocarboxylic acids?
The answer to this question might solve a multiple choice in my previous chemistry exam, which caused an argument with my teacher.
We know that fatty acids form fats and oil due to their carboxylic groups by forming triesters with glycerol. But is it possible to have a fatty acid with two or more $\ce{-COOH}$ groups to exist naturally? If so, is the property "fats and oils are neutral" just a general property, or does it have an exception?

IUPAC's defition of fatty acids covers monocarboxylic acids only. However, some researchers treat acids like azelaic acid, sebacic acid and dodecanedioic acid as fatty acids too.

You are asking two separate questions.
Not all fatty acids are typically found in fats and oils (common name for storage lipids, which is also somewhat a common name...)  Identifying fatty acids that have an additional charged tail does not inherently mean you'll find them in your fats and oils - fatty acids do have other roles. There may be exceptions, but they're not going to be common.
From what information you have given us, the larger question your teacher is driving you towards is to think about what the physical nature of fats and oils must be given their biophysical properties and how that correlates with their biological role, and secondly what the physical nature of fatty acids would be that drives those properties.
Considered macroscopically, your storage lipids will remain neutral, and pack together. If you added charge to your storage lipids in any significant amounts, you would turn them into something that behaves like a membrane lipid.

Thanks, but that answers only part of my question...your answer would be more helpful if you add  your knowledge or at least a rational guess about the existence of such fatty acids...but thanks part of your answer was also a great help.


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Explaining hypervalency in the framework of molecular orbital theory
In preparing for teaching duties in an introductory course in inorganic chemistry, I seem to have made a digression into how we understand the hypervalent molecules.
Just to re-iterate what this concept is (most likely a bad definition, correct me if this is bad): We say that a molecule is hypervalent if an atom has more than its maximum number of valence electrons participating bonding, as drawn in Lewis structures. Three examples are $\ce{PCl5}$, $\ce{SF6}$, and $\ce{XeF2}$.
When reading up on this, it seems to me that mostly the valence bond framework is used to explain hypervalent molecules. There is talk of ionic configurations accounting for the majority of the energy, that of single electrons localized on the ligands preventing the "octet rule" from breaking.
But how do we explain this with concepts from molecular orbital theory? If I run a high level calculation of $\ce{SF6}$, my computer does not explode (presumably, I have not tried it).
Assuming that d-electrons do not participate in bonding to any significant extent, then we only have the valence electrons. If I add up the partial occupancies of the orbitals surrounding the central atom, will I then get a number very close to 8? So only 8 electrons will participate in bonding? The "remaining" electrons could then occupy nonbonding orbtils?
Is "hypervalent molecules" a non-issue in molecular orbital theory?

https://chemistry.stackexchange.com/a/49844/16683 there’s an MO scheme here. Not exactly backed up by computational work, but I don’t think it will be too different

A reasonably good overview can be found here Wikipedia - Hypervalent Molecule

That MO diagram is definitely interesting. It is still unclear to me how exactly the two remaining F atoms are bonded to the S, since the two electron pairs are localized on the F atoms and do not participate in bonding. Are they supposed to be "ionic bonds"?

In MO theory there are no hypervalent molecules. Hypervalency is a figment of Valence Bond Theory, and the apparent misrepresentation only arises from localising electrons, and the inherent inability to describe delocalisation with one structure in the VBT framework (ref. also resonance). Since MOT uses delocalised descriptions, there cannot be such limitations. It might sound strange, but when you break it down in SF6, you basically have a 2-electron-7-centre bond (completely symmetric overlap) and three degenerate 2-electron-7-centre bonds (from the p orbitals).


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Does boric acid catalyze esterifications with isopropanol?
I am going to synthetise isopropyl salicylate. From what I have read on phillysim.org (pdf) the formation of ethyl salicylate can be catalyzed by boric acid. I could use sulfuric acid but it would be much more convenient with boric acid.
Ethanol and isopropanol have some differences, so I am not sure if it works for both of them.
Has anyone tried this? Does it work? Would it be better to just use sulfuric acid?

According to
https://link.springer.com/article/10.1007%2FBF01178939
It looks like it works, but probably the "milder" conditions required by sulfuric acid (plus its dehydrating power) make it preferable.
A secondary alcohol (isopropanol) should not be a problem anyway


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What is proximity effect?
From Klein's Organic chemistry [1, p. 780]:

The 1,2-adduct is believed to form more rapidly as a result of a proximity effect. Specifically, the carbocation and the bromide ion are initially very close to each other immediately after their formation in the first step of the mechanism. The bromide ion is simply closer in proximity to C2 than C4, so attack at C2 occurs more rapidly.

Similarly, from Vollhardt's Organic Chemistry: Structure and Function [2, pp. 345–346]:

Ring size controls the speed of cyclic ether formation
A comparison of the relative rates of cyclic ether formation reveals a surprising fact: Three-membered rings form quickly, about as fast as five-membered rings. Six-membered ring systems, four-membered rings, and the larger oxacycloalkanes are generated more slowly.
Relative Rates of Cyclic Ether Formation
$$k_3 ≥ k_5 > k_6 > k_4 ≥ k_7 > k_8$$
$k_n =$ reaction rate, $n =$ ring size
[…]
A second enthalpic phenomenon, which has been called the proximity effect, operates, especially in 2-haloalkoxides.

The proximity effect was used to explain these two reactions. Is it applicable to every reaction?
What is the proximity effect?
References

Klein, D. R. Organic Chemistry, 1st ed.; John Wiley: Hoboken, NJ, 2012. ISBN 978-0-471-75614-9.
Vollhardt, K. P. C.; Schore, N. E. Organic Chemistry: Structure and Function, 7th ed.; W. H. Freeman: New York, 2014. ISBN 978-1-4641-2027-5.

I will not attempt to define "proximity effects" in my answer. If you are looking for a definition, this is probably not the best answer. I think the term itself is quite vague and rather qualitative if you will. I've personally seen the concept of "proximity effects" being used more in chemical biology / enzymes chemistry context, than in organic chemistry.
In the first example, I am not even sure if there is indeed any proximity effect in operation. There could be; but it really depends on the solvent effects. How does one really separate the different factors in play? Doing gas phase simulations could give you an answer. The first example you have cited, can indeed be more easily be explained in terms of relative stabilities of the "carbocation" resonance forms: it is more favourable to have the positive charge at position 2 (allylic, secondary), than to have the positive charge at position 4 (allylic, primary). In terms of resonance or MO theory, this picture translates into position 2 being more electron deficient than position 4. A nucleophile would attack at the most electron deficient position (ceteris paribus), and this would be the fastest reaction path. Attack at position 2 leads to the kinetic product, as shown in the book you've cited. Whether there is any additional "proximity effects" - such as the nucleophile being "closer" to position 2 than position 4, is up for debate and cannot be concluded from this result. On the grounds of carbocation stabilisation alone, the observed product can be explained. Note however, that under the conditions, this reaction could be reversible and eventually generate the more stable thermodynamic product.
The second example, is perhaps a more appropriate application of 'proximity effects'. However, I'm not sure if it is appropriate to call this an 'enthalpic phenomenon'. I would have guessed a more probabilistic/entropic origin of this phenomenon. In this example, I would interpret proximity effect as simply the fact that if you are doing a 1-3 ring closure (to form a 3 membered ring), your two reaction centres are closer in space, and hence the reaction is probabilistically more likely than a 1-6 ring closure. I don't know how using the term 'proximity effect' helps give a better explanation, though. If you're interested, you should read up on rates of ring closures in a book of your choice, that discusses both enthalpic and entropic contributions to analyse the relative rates of ring closure.

I think this "proximity effect" that you are referring to cannot be generalised. In fact, the "proximity effect" referred to in each example that you have provided is different from the other. My answer will just be supplementing the answer provided by Zachr as I believe he has explained the two examples with sufficient depth of analysis and clarity.
In the first case, I believe that the idea is that the product of the first step of the reaction, the protonation of the $\ce {C=C}$ bond is actually not just the carbocation intermediate in isolation, but a contact ion-pair, comprising the carbocation and the bromide counterion. This "ion pair" mechanism is discussed in greater detail with the appropriate reference to literature in this post by Orthocresol. As the positively-charged carbon atom and the bromide anion are in close proximity to each other, they can combine quickly to form the 1,2-adduct. This is certainly a kinetic phenomenon as this "ion pair" electrostatic combination just occurs more quickly.
In the second case, the "proximity effect" the author is referring to is that the nucleophilic oxygen atom is able to most easily access the electron-deficient that is closest to it. However, as Zachr has aptly pointed out, this can only be explained by analysing both the entropy of activation and the enthalpy of activation. To say that it is just an enthalpic effect, as written in the textbook source you have cited, would be myopic and more importantly, incorrect. In fact, it is the entropic factor that greatly favours the formation of the 3-membered ring. A detailed discussion of the these two factors can be found here.
While a generalisation that is useful cannot be provided, the "proximity effect" is certainly more of a kinetic phenomenon than a thermodynamic phenomenon, at least based on these two case studies.


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Why does Finkelstein reaction only work with R-Cl and NaI?
Why does Finkelstein reaction necessarily involve: R-Cl and NaI (in Acetone)? Why can it not be done with KI? How does potassium instead of Sodium affect the reaction? Why can't R-I react with KCl in the same way?
(I've not included R-I (Iodine) or Fluorine (in the ionic compound) since they are extremes and are very slow or violent in many reactions respectively.)

Also, note that the Finkelstein reaction can indeed be done with R-Br!

The classic Finkelstein reaction entails the conversion of an alkyl chloride or an alkyl bromide to an alkyl iodide by treatment with a solution of sodium iodide in anhydrouss acetone (Wikipedia). You can find here why this reaction works in this manner.
The reaction is essentially driven toward products by mass action due to the precipitation of the poorly soluble $\ce{NaCl}$ or $\ce{NaBr}$. Solubility of one specific salt (only $\ce{NaI}$) is important to the reaction. I emphasize above that your acetone has to be very dry, otherwise $\ce{NaCl}$ and $\ce{NaBr}$ are also going to be soluble large scale in acetone and reaction won't go to completion, and thus, won't be useful. For example, the solubility of $\ce{NaI}$ in acetone is found to be 29.2% by weight at $\pu{25 ^\circ C}$ but it is increased to 40.5% when 6.1% (by weight) of water added to the solution (Ref.1). Similar effect should be expected for the solubility of
$\ce{NaCl}$ and $\ce{NaBr}$ (Ref.2).
To answer your particular question about $\ce{KI}$, you need to look at the solubility of it in acetone: Unlike $\ce{NaI}$, which has maximum solubility in acetone at $\pu{25.7 ^\circ C}$, maximum solubility of $\ce{KI}$ in acetone is at $\pu{-57.5 ^\circ C}$ (Ref.3):

[....]. In this paper are listed the solubilities of potassium iodide in acetone from $+54.5$ to $\pu{-78.5 ^\circ C}$. While, in general, the solubility curves of sodium and potassium iodides are similar in form, the maximum solubility of potassium iodide is about threefold less than that of sodium iodide, and it occurs at $\pu{-57.5 ^\circ C}$, instead of $\pu{+25.7 ^\circ C}$ as was reported for the sodium salt. The solid phase which is in equilibrium with the solution below $\pu{-57.5 ^\circ C}$ is a solvated form containing five (or six?) molecules of acetone per molecule of potassium iodide.

As stated in above abstract, $\ce{KI}$ molecule is surrounded by 5 to 6 acetone molecule, instead of 3 in $\ce{NaI}$ case. That makes it crowded (steric effect) as well. Its solubility is reported as 1.330% by weight at $\pu{+25 ^\circ C}$ (Ref.3), compared 29.2% of $\ce{NaI}$ under the same conditions. These facts make $\ce{KI}$ unsuitable for used in Finkelstein reaction. Yet, there are several instants of using $\ce{KI}$ in this type of reactions, e.g., see Ref.4.
There are also instances of using substrates other than $\ce{R-Cl}$ or $\ce{R-Br}$, e.g., see Ref. 5.

References:

R. Macy, E. W. Thomas, “The System: Sodium Iodide-Acetone-Water,” J. Am. Chem. Soc. 1926, 48(6), 1547–1550 (https://doi.org/10.1021/ja01417a015).
M. Li, D. Constantinescu, L. Wang, A. Mohs, J. Gmehling, “Solubilities of $\ce{NaCl, KCl, LiCl}$, and $\ce{LiBr}$ in Methanol, Ethanol, Acetone, and Mixed Solvents and Correlation Using the LIQUAC Model,” Ind. Eng. Chem. Res. 2010, 49(10), 4981–4988 (https://doi.org/10.1021/ie100027c).
R. Livingston, R. R. Halverson, “Solubility of Potassium Iodide in Acetone,” J. Phys. Chem. 1946, 50(1), 1–6 (https://doi.org/10.1021/j150445a001).
F. G. Bordwell, W. T. Brannen, “The Effect of the Carbonyl and Related Groups on the Reactivity of Halides in $\mathrm{S_N2}$ Reactions,” J. Am. Chem. Soc. 1964, 86(21), 4645–4650 (https://doi.org/10.1021/ja01075a025).
D. J. Maloney, S. M. Hecht, “A Stereocontrolled Synthesis of $\delta$-trans-Tocotrienoloic Acid,” Org. Lett. 2005, 7(19), 4297–4300 (https://doi.org/10.1021/ol051849t).


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Method for removing CuO in an inert environment
Common oxides (i.e., CuO or Al2O3) are capable of removal via NaOH solution. However, I am interested in removing a copper oxide layer inside an inert (Ar glovebox) environment. Thus, the traditional NaOH in a water solution (or a reaction with a water by product) does not necessarily work.
Can anyone suggest an alternative chemical method which can be used to remove CuO in glovebox?

What about diluted sulfuric acid? Upon reaction it gives copper sulfate, water-soluble

Your ways are strange to us. CuO is not really eager to react with NaOH, and Al2O5 is not a thing at all.

Sorry, Al2O3. I'm not a chemist obviously. However, I did find multiple people who use NaOH to remove CuO. https://chemiday.com/en/reaction/3-1-0-1581

Polishing with Cerium oxide, electrolytic cleaning, or diluted acid solution (nitric, hydrochloric, orthophosphoric, or hot acetic acid) are recommended in the Q&A on ResearchGate.net: "How to remove native oxide growth on copper or nickel foil?".

Acids (diluted or otherwise) is not something you normally bring into an Ar glovebox.

Regardless, I don't want to use the some solution with water, because I'm in an Ar glovebox. The main purpose of the box is to remove water. Thus, also if there is some electrolytic cleaning solution that's suitable it also cant be water based..

@User2341 that's fortunate, since cerium oxide isn't soluble in water you might want to mix it with something else. There are other oxides which are basic in nature, but I'll leave you to it.


text:
Statistical thermodynamics
Entropy of indistinguishable particles $S$ is
$$S = k\ln W,$$
but in some places notation is different:
$$S = k\ln Q + \frac{U}{T}$$
Can you explain, please?

The only time I've ever seen that second notation is when referring to the substitution of the Helmholtz energy, as A = U-TS and therefore A = -kTlnQ. If you want an in depth explanation of how the thing is derived I suggest looking at Chapter 10 of Biomolecular Thermodynamics: From Theory to Application by Douglas E. Barrick.
Here are some photos of the key parts:

The second formula you quote follows directly from the first. Starting with $W=N!/(n_1!n_2!\cdots)$ for $N$ distinguishable particles spread over set of $i$ energy levels with energy $\epsilon_i$, then $\ln(W)=\ln(N!)-\sum\ln(n_1!)$. Using the Sterling  formula for factorials the entropy  becomes, after some fiddling around and simplifying,  $\displaystyle S=-Nk_B\sum \frac{n_i}{N}\ln\left(\frac{n_i}{N} \right)=-Nk_B\sum p_i\ln(p_i)$ where $p_i=n_i/N$ is the chance that the $i^{th}$ state is occupied. This fraction is also $p_i=e^{-\epsilon_i/k_BT}/Z$ where $Z$ is the partition function and $Z=\sum_ie^{-\epsilon_i/k_BT}$ then $\displaystyle \ln(p_i)=-\frac{\epsilon_i}{k_BT}-\ln(Z)$ and substituting gives
$$S=-Nk_B \left(-\frac{1}{k_BT}\sum(p_i\epsilon_i)-\ln(Z\sum p_i)\right)$$
which becomes $\displaystyle S=\frac{U}{T}+Nk_B\ln(Z)$
after evaluating the terms $\displaystyle \sum p_i=1$, the probability over all states is one and the total energy $\sum p_i\epsilon_i $ is  $\displaystyle \sum p_i\epsilon_i=\frac{U}{N}$ where $U$ is the internal energy.
This approach is only valid for solids because in some way the atoms in a solid can be identified. In a gas this is not possible and the number of microstates must be divided by $N!$ because after exchanging atoms/molecules  the state of the  gas  still  it indistinguishable. For a perfect gas it is found that  $\displaystyle S=\frac{U}{T}+Nk_B\ln(Z) -Nk_B(\ln(Z)-1)$


text:
Definition of early vs. late transition metal
Is there an accepted definition of an early or late transition metal? I think the late transition metals start at Group 8 (including Fe), but others have said Fe is not a late transition metal. I'm just curious if this is a matter of preference based on one's field of study or if there is a generally accepted definition.

Iron could count as an early example of the late transition metals (e.g. by https://harnedgroup.files.wordpress.com/2014/12/introductiontometals.pdf), but maybe IUPAC has deposit its definition outside the color books.

Yes, that figure is my conventional understanding of what an inorganic chemist might call early, middle, and late transition metals. Still, I'm not sure if it's IUPAC says otherwise (or if there are other fields where this classification scheme would not be accepted).

Why would you want to say that iron is a late transition metal? Groups 11&12 are very different from 6-10. I don't think that'd be a very useful definition.

The reason why is largely irrelevant to the question. But yes, Fe and Cu are extremely differently electronically (and Zn is often not even classified as a transition metal, at least according to IUPAC).


text:
Steady State Approximation final integral proof $\frac{d[P]}{d t}=k_{a}[A]$ yielding $[P]=\left(1-e^{-k_{a} t}\right)[A]_{0}$
I want to ask a question about the Steady State Approximation integral.
I understand the steady state integral works in Leyman's terms as the intermediate is formed as fast as it is consumed so the intermediate's concentration is really low and nearly always the same.
If I had a reaction:
$$\ce{A->I->P}$$
Then essentially,
$$\frac{d[I]}{d t} \approx 0$$
and therefore using this information and previous knowledge of rate
$$\begin{array}{l}{\frac{d[I]}{d t}=k_{a}[A]-k_{b}[I]} \\ {\frac{d[I]}{d t}=k_{a}[A]-k_{b}[I] \approx 0}\end{array}$$
which results after rearrangement:
$$[I] \approx \frac{k_{a}[A]}{k_{b}}$$
and hence $$\frac{d[P]}{d t}=k_{b}[I] \approx k_{b} \frac{k_{a}[A]}{k_{b}} \approx k_{a}[A]$$
The trouble is, I couldn't understand how:
$$\frac{d[P]}{d t}=k_{a}[A]$$
can be solved to give
$$[P]=\left(1-e^{-k_{a} t}\right)[A]_{0}$$
I recognise from previous work that:
$$\frac{d[A]}{d t}=-k_{a}[A]$$ gives
$$A =[A]_{0}e^{-k_{a} t}$$
from my final year at school, but I couldn't seem to apply this understanding to the final step of the integral above.
How can this be shown?

One molecule of A makes one molecule of P, so $[\ce{A}] + [\ce{P}] = $ some constant $c$, if you ignore the intermediate $\ce{I}$. At time $t = 0$, you know that $[\ce{A}] = [\ce{A}]_0$ and $[\ce{P}] = 0$...

When you make the steady state approximation species I is removed from any time dependence effectively making the problem $A \to P$ as $A=A_0e^{-kt}$ then $P=A_0(1-e^{-kt} )$ because A+P is constant.

There are a couple of ways to approach this problem. Bear in mind this problem requires some calculus.
$\boldsymbol{1^{st}}$ Way (repeated integration): To start, you were able to recognize the first order integrated rate law.
$$\frac{d[A]}{dt} = -k_a[A] \implies [A] = [A]_0e^{-k_at}$$
This result may be plugged into the rate equation for $P$ to obtain
$$\frac{d[P]}{dt} = k_a[A] = k_a[A]_0e^{-k_at}$$
This equation is directly integrable.
$$\int_0^t\frac{d[P]}{dt}\,dt = \int_0^tk_a[A]_0e^{-k_at}\,dt$$
$$[P] - [P]_0 = -\frac{k_a[A]_0}{k_a}\left(e^{-k_at} - 1\right)$$
$$\boxed{[P] = [A]_0\left(1-e^{-k_at}\right)}$$
The last step applies the knowledge from the problem that $[P]_0=0$.
$\boldsymbol{2^{nd}}$ Way (conservation law): orthocrestol in the comments above pointed out that the sum of the concentrations of all chemical species should equal to some constant in this problem. Call this constant $C$ and ignore $I$ due to the steady state approximation. Thus,
$$[A] + [P] = C$$
Now substitute this into your rate law for $P$.
$$\frac{d[P]}{dt} = k_a[A] = k_a(C - [P])$$
This differential equation is separable so we can use the neat tricks to integrate this equation.
$$\int_0^{[P]}\frac{d[P]}{C-[P]} = \int_0^tk_a\,dt$$
$$-\ln\left(1-\frac{[P]}{C}\right) = k_at$$
$$[P] = C\left(1-e^{-k_at}\right)$$
Almost done. Now recognize that at time zero, we can write the conservation law as
$$C = [A]_0 + [P]_0 = [A]_0 + 0 = [A]_0$$
This makes physical sense because since $C$ is a constant, this value cannot change at any time. Moreover, the total number of molecules at any time should be the same as the amount of molecules that you start with due to stoichiometry (everything is 1:1 in this particular problem). So make the substitution to get your answer.
$$\boxed{[P] = [A]_0\left(1-e^{-k_at}\right)}$$

Hi there! I fully understand every element of your working and I'm super grateful, but I just have a question about the 2nd to last step of your first way. You wrote $\int_{0}^{t} \frac{d[P]}{d t} d t=\int_{0}^{t} k_{a}[A]{0} e^{-k{a} t} d t$ and then gave $[P]-[P]{0}=-\frac{k{a}[A]{0}}{k{a}}\left(e^{-k_{a} t}-1\right)$ which all makes sense apart from the $-1$ inside the brackets ... how did you get that term? I might be missing something but apart from that it's a perfect answer that I can tick to confirm I understood the solution!

@vik1245 You might have forgotten to apply the limits of integration

@vik1245 The left hand side is the fundamental theorem of calculus which more or less states the definite integral of the derivative is the original function evaluated at the limits. As for the right hand side, William realized you may have missed it by forgetting to apply the limits of integration, here is some step-by-step for the integral:$$\int_0^tk_a[A]_0e^{-k_at}=k_a[A]_0\int_0^te^{-k_at}=k_a[A]_0\left(-\left.\frac{1}{k_a}e^{-k_at}\right|_0^t\right) = k_a[A]_0\left(-\frac{1}{k_a}\right)\left(e^{-k_at} - 1\right)$$

oh of course, $e^{0} = 1$ how can I be so silly?! Many thanks tho anyway for your help! @MasterYoda


text:
Why can sulfur have an expanded octet but not oxygen?
In my search to understand the bonding in structures like $\ce{SF6}$ I found many sources that said it was because sulfur has d orbitals to accommodate an expanded octet, which made sense to me. But I also found sources like the paper by Reed and Weinhold (1986) that say d orbitals contribute very little to the bonding in $\ce{SF6}$. An alternative explanation is that $\ce{SF6}$ has three-center/two-electron bonds. What I don't understand is why compounds like $\ce{SF6}$, $\ce{ClF5}$, $\ce{PCl5}$ exist for third row element centers, but not $\ce{OF6}$ if it has nothing to do with d orbitals? References would be great.

Possibly related:
https://chemistry.stackexchange.com/questions/42750/how-can-the-size-of-atom-be-the-cause-of-octet-expansion
https://chemistry.stackexchange.com/questions/13949/what-allows-sulfur-and-phosphorus-to-expand-their-octet

Also related Hypervalency of Elements.  See also Hypervalent Molecule.

Very related: https://chemistry.stackexchange.com/q/100911/44877

@Cell https://chemistry.stackexchange.com/q/100911/44877 is quite relavent to answer your question.

Neither sulfur nor oxygen have an expanded octet. Sulfur is capable of forming hexacoordinate compounds like SF6 (whereas oxygen cannot), but this does not involve d-orbitals, and there is little need to go beyond the standard 8 valence electrons. I think this is conceivably a duplicate of the question linked by @TanYongBoon.

@orthocresol So the answer is that period 2 elements are too small to bond to that many molecules? I'm surprised why none of of my undergrad textbooks mention such a simple reason for why SF6 forms but not OF6 etc. instead they discuss expanded octets hypervalency etc.

Another relevant answer that perhaps qualifies this as duplicate: https://chemistry.stackexchange.com/questions/4760/why-do-compounds-like-sf6-and-sf4-exist-but-sh6-and-sh4-dont

Neither oxygen nor sulfur "expands its octet". Instead, sulfur in compounds such as $\ce{SF6}$ delocalizes the bonds formed with the usual octet over multiple linkages. As described in this answer, this delocalization in a neutral molecule comes with occupying ligand-based orbitals that are roughly nonbonding with a large central atom, but antibonding with a small central atom because of ligand-ligand interactions. Thus forming a compound with delocalized bonding and increased coordination number is more favored for third period elements than for second period elements.
We can get such coordination with second-period elements if the ligand-based orbitals noted above are unoccupied, for then the net ligand-ligand overlaps with the remaining orbitals are bonding. This usually requires a cationic species such as the six-coordinate carbon complex reported by Yamaguchi et al.[1], pictured below from the reference.

References
1.
Torahiko Yamaguchi, Yohsuke Yamamoto, Daisuke Kinoshita, Kin-ya Akiba, Yun Zhang, Christopher A. Reed, Daisuke Hashizume, and Fujiko Iwasaki. "Synthesis and Structure of a Hexacoordinate Carbon Compound", J. Am. Chem. Soc. 2008, 130, 22, 6894–6895. https://doi.org/10.1021/ja710423d

Hopefully the pictures answers the question as to why oxygen cannot form an octet, when covalently bonding with other atoms, such as if it was possible for it to form OF6, but sulfur can, to form SF6. But a brief added explanation below:
It's to with the working out oxygen and sulfur's electon configuration, and then also representing this configuration using arrows in boxes. Realising that oxygen only goes up to the 2nd quantum principal shell number, so 2s and 2p, and ends here.
Sulfur has electrons in 3s and 3p, but has an additional 3d sub shell as well, that's currently empty. It can excite some its paired electrons into this empty orbital, spreading them out, so each can now form bonds and expand an octet to 12 electrons.
Oxygen has no empty orbital in quantum 2, so cannot excite and 'spread' its electrons, so cannot bond these to form an expanded octet.
Hope this helps.
Hannah S

Please make sure the image is properly oriented and is easily readable (yours lack contrast and handwritten text is hard to read). Besides, about 3/4 information is text, not graphics, and must be input as such, not as photo.

Regardless of some adjustment on your illustration, it still is at the level of a doodle to prepare an answer anticipated for chemistry.se. Even if the doodle were prepared by with typewriter or a computer to ease reading, it is an image and as such the text is not searchable, the content is locked away.  But this is just the opposite of chemistry.se's approach to retrieve answers easily by entering a few keywords.  Please consider this perspective, and edit your answer accordingly.


text:
Why is formyl chloride unstable while higher acyl chlorides are stable?
From Bruice's Organic Chemistry, section 18.6 Friedel–Crafts acylation of benzene [1, pp. 876–877]:

Friedel–Crafts acylation places an acyl group on a benzene ring, and Friedel–Crafts alkylation places an alkyl group on a benzene ring.
An acyl chloride or an acid anhydride is used as the source of the acyl group needed for a Friedel–Crafts acylation.

[…]
The synthesis of benzaldehyde from benzene poses a problem because formyl chloride, the acyl halide required for the reaction, is unstable and must be generated in situ.
The Gatterman–Koch reaction uses a high-pressure mixture of carbon monoxide and HCl to generate formyl chloride and uses an aluminum chloride–cuprous chloride catalyst for the acylation.

From this excerpt we infer that formyl chloride is unstable.
Is it just a remark, or does it affect the reaction?
If formyl chloride is unstable, why is acyl chloride stable?
The answer by TheSimpliFire to Stability of formyl chloride suggests vigorous decomposition of formyl chloride by water, but the Gatterman–Koch reaction proceeds under anhydrous conditions. So, why is formyl chloride unstable even in water-free environment?
Reference

Bruice, P. Y. Organic Chemistry, 8th ed.; Pearson: Upper Saddle River, NJ, 2016. ISBN 978-0-13-404228-2.

Formyl chloride may be much less stable then higher acylchlorides. Similarly acetic acid is stable, but formic acid is unstable ( HCOOH-> H2O + CO) and should be stored with open vent

The textbook has a typo: it's Gattermann, not Gatterman (https://doi.org/10.1002/cber.18980310281).

The simplest stable acyl chloride is ethanoyl chloride or acetyl chloride; methanoyl chloride (formyl chloride) is not stable at room temperature, although it can be prepared at –60 °C or below. (Wikipedia)

The instability of $\ce{HCOCl}$ is caused by ease of elimination of HCl from its molecules. Cl is decent leaving group and after it's gone, remaining acylium cation has a very acidic hydrogen instead of alkyl present in other acyl halides. Thus easier breaking of C-H bond vs C-C bond causes drastic difference in thermal stability.
Note also the formyl chloride can be considered not only the derivate of formic acid, but also a derivate of formaldehyde.

The question got me curious about the decomposition mechanism, every one I thought of involved some weird intermediates. Maybe it's a concerted reaction? The deprotonation, dehalogenation and oxygen dative bonding all happening at once? I think it's worth adding to the answer that very likely the reason formyl chloride is more unstable than other acyl chlorides is the fact the carbon is very positively charged, has a great leaving group and is connected to a very acid hydrogen (due to the charge)

Anybody, feel free to write a better answer, I am not an organic chemist, I am not sure now why I did not left it to them....

More stable formyl compounds
While formyl chloride is unstable, formyl fluoride, with a stronger bond from the carbonyl carbon to the halogen, is marginally more stable. It can be made by a double displacement reaction:

$\ce{HCOONa + C6H5C(O)F → FC(O)H + C6H5COONa}$[1]

The compound is still unstable enough at ambient temperature to require special procedures to prepare and store, as it decomposes autocatalytically. Low temperature preparation and storage over alkali metal fluorides (which absorb the hydrogen fluoride catalyst) are employed to stabilize the product.
Formyl cyanide, which does not have a good proton-acceptor atom, is stable enough to be made at high temperature. It is made by flash pyrolytic methods [2][3][4].
Cited Reference

Olah, G. A.; Ohannesian, L.; Arvanaghi, M. (1987). "Formylating Agents". Chemical Reviews. 87 (4): 671–686. https://doi.org/10.1021%2Fcr00080a001.

Lewis-Bevan, Wyn; Gaston, Rick D.; Tyrrell, James; Stork, Wilmer D.; Salmon, Gary L. (March 1992). "Formyl cyanide: a stable species. Experimental and theoretical studies". Journal of the American Chemical Society. 114 (6): 1933–1938. https://doi.org/10.1021/ja00032a001.

Bogey, M.; Destombes, J.L.; Vallee, Y.; Ripoll, J.L. (May 1988). "Formyl cyanide: Efficient production from allyloxyacetonitrile and its millimeter-wave spectrum". Chemical Physics Letters. 146 (3–4): 227–229. Bibcode:1988CPL...146..227B. https://doi.org/10.1016/0009-2614(88)87435-9.

Bogey, M.; Demuynck, C.; Destombes, J.L.; Vallee, Y. (August 1995). "Millimeter-Wave Spectrum of Formyl Cyanide, HCOCN: Centrifugal Distortion and Hyperfine Structure Analysis". Journal of Molecular Spectroscopy. 172 (2): 344–351. Bibcode:1995JMoSp.172..344B. https://doi.org/10.1006/jmsp.1995.1183


text:
Will strong concentrated acid destroy litmus paper and pH paper if they are dipped in acid?
I am curious to know what will happen if I have litmus paper and pH paper, and I dip them into strong concentrated acid like conc. $\ce{H2SO4}$ to test its acidity or basicity. Will acid destroy papers? if YES, then how will I determine the acidity? do I need something else to carry my experiment?

There are $\mathrm{pH}$ meters too, unless you're also concerned about dissolving glass electrodes.

Side note: never dip paper indicators into the solution; this is a waste of material (more paper is soaked) and introduces impurities (indicator, cellulose fibers, dust etc.) into the batch. Instead, put a glass stick in a thoroughly mixed solution, then touch the paper strip with the tip of the glass stick.

What is the sense in testing acidity of concentrated sulphuric acid in the first place ?

Liquid pH testing solutions or drops are also available. Other indicators like phenolphthalein, methyl orange , methyl red can be used if you expect your test solution to fall in the pH range of the respective indicators. Also it will be wise to take a small sample of the solution to test it rather than testing the whole thing.

Doesn't this answer circumvent the main question?


text:
Can Rydberg constant be in joules?
In my textbook (Chemistry Part - I for Class XI published by NCERT), there is an equation for the energy of an electron in an energy state: $$E_n = -R_\mathrm H\left(\frac{1}{n^2}\right)$$ and there is a paragraph below it with the following text:

where $R_\mathrm H$ is called Rydberg constant and its value is $2.18\times10^{-18}\ \text{J}$.

There is another section with the expression for the wavenumber ($\overline{\nu}$):
$$\overline{\nu}=109\,677 \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)\ \text{cm}^{-1}$$
with a paragraph with the following text:

The value $109\,677 \space\text{cm}^{-1}$ is called the Rydberg constant for hydrogen.

I checked online and found that in most (all) websites (incl. Wikipedia), the value of Rydberg constant is $109\,677 \space\text{cm}^{-1}$. But when I searched for its value in joules, I found this website with the value of Rydberg constant $= 2.18\times10^{-18}\ \text{J}$.
How can Rydberg constant be written in joules?

How can a price of a cheeseburger be written in ¥? Well, just like that. Rydberg constant is energy, and Joule is energy, so what's the problem?

Look at the two different equations. Consider the dimensions in each case, and why that means in one case the Rydberg constant must have dimensions of Energy, and in the other it must have dimensions of wave number. Now what relations do you know relating Energy and wave number?

Although wavenumbers are not strictly energy you will see from the value in joules that it is much more convenient to use wavenumbers. The conversion is $ 1 ,\mathrm{cm^{-1}}\equiv 1.986\cdot 10^{-23}$ J.

Authors may be sloppy about notation in this matter. I recommend considering $R_\ce{H} \approx \pu{10973 cm-1}$ and $Ry \approx \pu{2.18e-18 J}$, noting $Ry = hc \cdot R_\ce{H}$. Units of wavenumbers $(\pu{cm-1})$ and energy are often considered interchangeable in practice because they are proportional to each other by the constant value $hc$.
In my notes, I would always be sure to write $R_\ce{H}$ or $Ry$ to explicitly remind myself "which" Rydberg constant I was using (in fact I merged the R and y into a single symbol because I didn't like the suggestion of multiplication.)
Note also that there is a unit of energy known as a Rydberg, with $\pu{1 Ry} = Ry = hc \cdot R_\ce{H}$.

Check those numbers, maybe a digit is missing in $R_H$?

Rydberg constant $R_∞$ is usually given in reciprocal length units historically and because it's determined from hydrogen and deuterium transition frequencies [1].
Current value (in $\pu{m-1}$) is listed at NIST [2] website (accessed 2019-06-05):
$$R_∞ = \pu{10973731.568160(21) m-1}$$
Since it's an energy unit, one can convert it to SI rather trivially via multiplying the value in reciprocal length units by $hc$ ($h$ is the Planck constant; $c$ is the speed of light in vacuum):
$$E = hν = \frac{hc}{λ} \quad\text{or}\quad R_∞[\pu{J}] = hc\cdot R_∞[\pu{m-1}]$$
resulting in the following value:
$$R_∞ = \pu{2.1798723611035(42)e-18 J}$$
References

Mohr, P. J.; Newell, D. B.; Taylor, B. N. CODATA Recommended Values of the Fundamental Physical Constants: 2014. Reviews of Modern Physics 2016, 88 (3). https://doi.org/10.1103/RevModPhys.88.035009.
Tiesinga E.; Mohr P. J.; Newell D. B.; Taylor, B. N. "The 2018 CODATA Recommended Values of the Fundamental Physical Constants" (Web Version 8.0). Database developed by J. Baker, M. Douma, and S. Kotochigova. Available at http://physics.nist.gov/constants, National Institute of Standards and Technology, Gaithersburg, MD 20899. 2019.

In spectroscopy and related fields it is common to measure energy levels in units of reciprocal centimeters (e.g., IR and Raman spectroscopy). Strictly speaking, these units ($\pu{cm^{−1}}$) are not energy units, but units proportional to energies, with $hc$ being the proportionality constant (Wikipedia). In general, $hc$ can be attributed to the value $\pu{1.986E-23 J cm}$. Hence:
$$R_∞ = \pu{109677 cm^{−1}}$$
$$\pu{1 Ry} = \pu{109677 cm^{−1}} \times hc = \pu{109677 cm^{−1}} \times \pu{1.986E-23 J cm} = \pu{2.178E-18 J}$$


text:
Reaction of potassium with water
In general, when we put potassium into water, it will produce potassium hydroxide and hydrogen:
$\ce{2K + 2H2O ⟶2KOH + H2}$
However isn't the following reaction also possible?
$\ce{2K + H2O ⟶K2O + H2}$
Why isn't the oxide produced?

That's a chemical equation, not a formula (just nitpicking). Welcome to the site! People may say "$\ce{K2O}$ might be transiently formed", but it's too unstable to be isolated, especially in water. We might as well say that $\ce{KOH}$ is directly formed. Side note: A chemical equation can always be balanced, but that is no indication of whether the reaction is a fairy tale or not.

Thanks for your answering. I am really appreciate your answer and manner. You even spot some mistakes in the question and let me have a good chance to learn. Thanks :)

Thermodynamics comes into effect.

The reactions are ongoing this way:
Relatively free electrons of  potassium reduce water:
$$\ce{2 e- + 2 H2O -> H2 + 2 OH-}\tag{1}$$
That leaves metal positively  charged.
Liquid ammonia, if exposed to alkali metal, reacts with electrons much slower than water, forming a dark blue solution of solvated electrons. As electrons progressively kick out protons from ammonia, forming hydrogen, the solution finally turns to a colourless solution of NaNH2.
But back to water.
The potassium ions get hydrated, decreasing the charge.....
$$\ce{K(s)^{n+} -> K(s)^{(n-m)+} + m K+} \tag{2}$$
forming $\ce{KOH}$ solution in form of mixture of hydrated ions $\ce{K+ + OH-}$
But heavy potassium ions cannot keep pace with light and fast electrons and the drop of melted metal progressively gains positive charge and finally ends up by - as authors call it -  Coulombic explosion.
The hydrogen gets eventually ignited by microsparcs due the charge instability  even before the explosion. As "Terminator T1000-like" spikes of liquid metal eventually pierce isolating vapour+hydrogen layer, getting into contact with ignitable hydrogen-air mixture.
It was recently theoretically predicted by quantum chemistry simulation for several dozens of alkali atoms by Czech chemist Pavel Jungwirth And col. Chemistryworld-Alkali metal explosion explained
They have experimentally verified it by high speed 10000 f/s camera, using sodium/potassium alloy forming an eutectic with low melting point.
I knew that from the popular science radio broadcast interview, finding backwards some reference for it.
See also their article in Nature ( which I forgot about and found later):
Coulomb explosion during the early stages of the reaction of alkali metals with water

Abstract
  Alkali metals can react explosively with water and it is textbook knowledge that this vigorous behaviour results from heat release, steam formation and ignition of the hydrogen gas that is produced. Here we suggest that the initial process enabling the alkali metal explosion in water is, however, of a completely different nature. High-speed camera imaging of liquid drops of a sodium/potassium alloy in water reveals submillisecond formation of metal spikes that protrude from the surface of the drop. Molecular dynamics simulations demonstrate that on immersion in water there is an almost immediate release of electrons from the metal surface. The system thus quickly reaches the Rayleigh instability limit, which leads to a ‘coulomb explosion’ of the alkali metal drop. Consequently, a new metal surface in contact with water is formed, which explains why the reaction does not become self-quenched by its products, but can rather lead to explosive behaviour.

This is very insightful. Never thought about such details of reaction mechanism. We take so many things for granted in science. But how to prove the equation number 1 i.e. "Relatively free electrons of potassium reduce water:"?

@M. Farooq Note that eq 1 was supposed even earlier. The article was about mechanism of droplet explosion and hydrogen ignition. Hehe, what about testing of flying curvature of metal droplets in strong electrostatic field ? :-)

You know about Taylor cone, even water jet becomes a spray in a strong electric field.

That reminds me Kelvin water dropper. When the apparatus gets charged enough, streams of drops start to diverge, repulsed by the same charge.

@M. Farooq See also Nature article link. About the electrons, consider blue solution of solvated electrons in liquid ammonia with alkali metal, as ammonia reacts with electrons much slower than water, finally forming a colourless solution of NaNH2.

Very interesting, it is very similar to electrospray phenomenon which we teach in mass spectrometry.

The solvated electron is another mystery!


text:
How do I remove tetrabutylammonium perchlorate from an organic synthesis?
I am trying to remove tetrabutylammonium perchlorate (TBAP) from an organic synthesis where TBAP was used as an electrolyte. I am working with compounds containing the amide group. I have used flash column chromatography (FCC) (gradient) but this is laborious.
To give you a better understanding of the experiment this is the path I follow

Once the electrosynthesis of the compounds has completed, the solvents are removed (MeOH & MeCN) in vacuo.
Now for step two, I usually carry out FCC. But I have tried separation using Sodium hydroxide but this did not give me the desired result, the electrolyte was still present after NMR.

Any help would be appreciated.

Welcome to Chemistry SE.  Do you have references for your procedure?  What is the compound you are isolating?

Hi Michael,
Thank you.
Reference for the procedure in the paper below
“An experimentalist’s guide to electrosynthesis: the Shono oxidation”
But to give you an overview of the procedure.
In a glass cell the following are added An analyte selected to undergo oxidation, MeOH and MeCN used as solvents and TBAP as electrolyte (does not take part in the reaction).
Galvanostatic method (2 electrode) is used to achieve the oxidation
A charge is applied until desired coulombs is reached.
Right now I am using this method on diethyltoluamide (DEET)

What are your products soluble in? T-butylammonium perchlorate is insoluble in diethyl ether so you could precipitate it out.

How about treating the whole with an aqueous solution of sodium metabisulfite to reduce the perchlorate all the way down to chloride.  You'll have tetrabutyl ammonium chloride and sodium sulfate.  Extract this mixture with an organic solvent to get your 'methoxy DEET' compound.  Take care when adding the reducing agent to the oxidising mixture, ensure water is present prior to the addition and cautiously add your metabisulfite solution so that the water acts as a heat sink.  Ensure that the reaction is addition rate controlled before doing this at scale.  Report back if it works.

It's tempting to use phase separation with an aqueous phase, but you're clearly having issues.  One option is to evaporate the volatile solvents off, add toluene, cyclohexane or heptane and try water or another additive to boost the ionic strength. The approach using NaOH suggests this may not be successful.
A less conventional approach may be to partition between two organic solvents e.g. acetonitrile or methanol with heptane or methylcyclohexane. The solubility of TBAP appears good enough in alcohols to dissolve well in that phase, while your amide seems more likely to have less polar solvent solubility. Your methanol-MeCN mix may convolute things,  so maybe try and strip off the methanol (and likely some MeCN will be lost, but you can top up) or if all else fails,  evaporate both solvents and redissolve from scratch.


text:
Can I use litmus paper to test unknown liquid that contains a concentrated strong acid?
Can I use litmus paper to test unknown liquid that contains a concentrated strong acid (example $\ce{H2SO4}$)? Can concentrated strong acid destroy litmus paper?

To test for concentrated sulphuric ac in such a way you can use also ham or bananas.... an unknown liquid cannot contain a concentrated acid, at most can contain a strong acid in high concentration. Regarding your last line: what destroys litmus paper and bananas is not an acid-base reaction but water withdrawal (of water already in the sample) and oxidation. Here sulphuric ac. acts as a strong oxidizer in acidic medium as well as dessicant.

Yes, concentrated strong acid can destroy litmus paper.
Paper is mostly cellulose, which is a carbohydrate polymer. Concentrated sulfuric acid will pull the hydrogen and oxygen out of it as water, leaving carbon behind. The indicator dye that's supposed to change color probably won't fare too well, either.
You can dilute the unknown by adding a drop or two of it to a small amount of distilled water, then use the litmus paper to check that.
A search for "concentrated sulfuric acid on paper" will yield many images and videos showing what the acid does to paper.

thanks jeffB . will the result be same if I replace litmus paper by ph paper?

Paper.  Yes the result will be the same.  And even if the paper and the dye were not irrevocably damaged, you could get unexpected results with the dye in either strong acid of strong base.  See what can happen to phenolphthalein at extreme pH.  Again use a diluted solution to keep things where your textbook tells them they are.

The "dilute" solutions to which Oscar refers are still strong enough to be a safety concern. A 0.1 molar solution of either is still nasty stuff requiring appropriate safety precautions.

@maxw as do all chemicals, really.  Even supposedly innocuous orange oil is not a fun experience if it gets in your eye.


text:
Why is Potassium permanganate a better catalyst than Potassium Iodide?
I did an experiment measuring the activation energy of Hydrogen Peroxide with different catalysts. Potassium Permanganate seems to be a better catalyst as the activation energy was lower than the one with Potassium Iodide. However, I would like know why that was the case. Was it because of the different structures?

Hi, welcome to SE chem. Try to add some more detail to you question, are you talking about the decomposition of hydrogen peroxide to water and oxygen gas?

It's because of the mechanism of catalysis, they both react with hydrogen peroxide as follows.
$$\ce{2KMnO4 + 3H2O2 => 3O2 + 2MnO2 + 2KOH + 2H2O}$$
$$\ce{KI + H2O2 => KIO + H2O}$$
$$\ce{KIO + H2O2 => KI + H2O + O2}$$
The most important bit is that in the first equation the peroxide is actually a reducing agent and the permanganate doesn't actually catalyze the decomposition of peroxide it just happens to react with it to produce oxygen and is not found unchanged after the reaction. Potassium iodide is a true catalyst, it reacts with the peroxide but is regenerated everytime oxygen is produced. Based on that, you can't actually compare the two as catalysts because only the iodide is one.


text:
Calculating pH when weak base is added to an strong acid

$\ce{NH3}$ solution of $\pu{0.1 mol dm-3}$ is being added to a $\pu{25.0 cm3}$ of $\pu{0.1 mol dm-3}$ $\ce{HCl}$ solution. Calculate the pH of the solution when volume of added $\ce{NH3}$ solution is $\pu{25.0 cm3}$ and $\pu{26.0 cm3}$. The $\mathrm{p}K_\mathrm{a}$ of $\ce{NH3}$ is ${9.25}$.

So I started with the equation:
$$\ce{HCl + NH3 <=> NH4Cl}\tag{1}$$
EDIT #1:
As the initial concentration of both base and acid are same, I calculated how much of base is left in the solution:

$V_\text{miscon} = \pu{26.0 cm3 - 25.0 cm3 = 1.0 cm3}$

That's how I tried to calculate pH of the solution when $\pu{26.0 cm3}$ of $\ce{NH3}$ is added:

$\ce{H+}$ concentration:
  $$[\ce{H+}]_\text{miscon} = [\ce{H+}]_\text{initial} \frac{V_\text{initial}}{V_\text{miscon}}\tag{2} $$ $$=  \pu{(0.1 mol dm-3} \times \pu{0.025 dm3)/0.001 dm3} = \pu{2.5 mol dm-3}$$

But, in this way, the results I am getting are completely different from marking, and even negative pH values.
But, when I tried to find $\ce{H+}$ concentration (to calculate pH) of the solution when $\pu{25.0 cm3}$ of $\ce{NH3}$ is added, I found out that initial $\ce{HCl}$ volume is completely reacted. Only ammonium chloride is left in the solution.
The marking only gives the answers, $\ce{pH = 5.28}$ and $\ce{pH = 7.85}$.
How can I calculate pH of this solution in above two conditions separately?
I can't figure it out, I am stuck in the above equation. Please give me some hints to start solving. Thanks in advance.
EDIT #2:
I finally figured it out (Adding $\pu{25.0 cm3}$ of $\ce{NH3}$).
$$\ce{[NH4+(aq)]}=n(\ce{NH4+})/V_{\mathrm{\ce{total}}}$$
$$=(\pu{0.1 mol dm-3}\cdot\pu{0.025 dm3})/\pu{0.05 dm3}$$
$$=\pu{0.0025 mol}/\pu{0.05 dm3}$$
$$\ \ \ =5.0\times 10^{-2}\  \pu{mol dm-3}$$
$\ce{NH4+}$ reacts with water to produce $\ce{H3O+}$ ions.
$$\ce{NH4+(aq) + H2O(l)  <=> H3O+(aq) + NH3(aq)}$$
Using acid dissociation constant for $\ce{NH4+}$:
$$\ce{K_\mathrm{a} = [H3O+(aq)] [NH3(aq)] / [NH4+(aq)]}$$
$$\ce{[H3O+(aq)] = \pu{( K_\mathrm{a} \cdot [\ce{NH4+(aq)}] )0.5}}$$
$$\ce{= \pu{( 5.62\times 10^{-10}\pu{mol dm-3} \cdot 5.0\times 10^{-2}\pu{mol dm-3} )0.5}}$$
$$\ce{= \pu{( 2.81\times 10^{-11}\pu{mol2 dm-6} )0.5}}$$
$$\ \ \ =5.301\times 10^{-6}\  \pu{mol dm-3}$$

$$ \text{pH} = -(\log 5.301 - 6\log 10) = -(0.7244 - 6) = 5.28 $$

You should do some basic stoichiometry.  Figure out how many moles of each reactant there is, and which species is in excess in either case.

The task description has an error. pKa=9.25 is for NH4+, not for NH3. For NH3, there is relevant pKb=pKw-pKa=14-9.25=4.75.

Why do you think one pH value is acidic, and the other one basic? "Only ammonium chloride is left" - what would be the pH of ammonium chloride, acidic or basic?

There are given 3 directly related equilibrium constants for reactions $$\begin{align}\\ce{NH4+ &<=> NH3 + H+}\ \ce{NH3 + H2O &<=> NH4+ + OH-}\ \ce{H2O &<=>H+ + OH-}\ \end{align}$$

Just as a hint: Adding a small amount of weak base to a larger (both in volume AND molarity) amount of strong acid changes basically nothing.

@MichaelLautman I already tried, but NH3 partially dissociates unlike strong base NaOH, so how can I calculate how much is left?

@hpm0615ngvup You have not involved the equilibrium constants in your  analysis. Why ?

@KarstenTheis I think the pH of ammonium chloride changes with its concentration.

@Poutnik I already involved the equilibrium between ammonia, hydrochloric acid and ammonium chloride in the question.

But you have not included it in your expected initial effort of this homework-like question.

@Poutnik I can't think how to proceed. But I included how I tried to calculate concentration of hydronium ions.

Are not you expected to know at least the basics, like mixing an acid with a base cannot lead to an acid with 25 times higher concentration??

@Poutnik I know it, but I cannot understand the way to solve. Give me an starting point.

Adding 25 ml of 0.1 M NH3 to 25 ml of 0.1 M HCl  forms solution of NH4Cl.  No HCl is left, neither there is an excess of NH3. But NH4+ ion is a weak acid. Calculate pH of a weak acid of given concentration and pKa. Adding 1 more extra ml of 0.1 M NH3 forms solution of NH4Cl and NH3. Calculate pH of a a mixture of the conjugate acid (NH4+) and base (NH3), knowing the acidity or basicity constant ( pKa + pKb = 14 ).

You know, unless you express effort to understand the underlaying concepts, your question would remain off topic and may be closed. Start with https://en.wikipedia.org/wiki/Henderson%E2%80%93Hasselbalch_equation

The question
This is the question, corrected according to Poutnik's comment, and adding (i) and (ii) to distinguish the two scenarios:

$\ce{NH3}$ solution of $\pu{0.1 mol dm-3}$ is being added to a $\pu{25.0 cm3}$ of $\pu{0.1 mol dm-3}$ $\ce{HCl}$ solution. Calculate the pH of the solution when volume of added $\ce{NH3}$ solution is
(i) $\pu{25.0 cm3}$ and
(ii) $\pu{26.0 cm3}$.
The $\mathrm{p}K_\mathrm{a}$ of $\ce{NH4+}$ is ${9.25}$.

The chemical reactions
The chemical equation (1) in the question is incorrect and missing something. It should reflect that in aqueous solution, hydrochloric acid is completely dissociated (strong acid), and ammonia is partially dissociated (weak base). One way of writing this is:
$$\ce{HCl(aq) + NH3(aq) -> NH4+(aq) + Cl-(aq)}\tag{A1}$$
$$\ce{NH4+(aq) <=> NH3(aq) + H+(aq)}\tag{A2}$$
Equation A1 shows that the reaction of a strong acid with a weak base goes to completion (no HCl will be present, chloride does not get protonated). Equation A2 shows that ammonium is a weak acid.
If you find it confusing that ammonia first picks up a proton, and then partially loses it again, you could also write it like this:
$$\ce{HCl(aq) -> H+(aq) + Cl-(aq)}\tag{B1}$$
$$\ce{NH3 + H+ <=> NH4+(aq)}\tag{B2}$$
If you learned not to write $\ce{H+}$ but instead  $\ce{H3O+}$, you could write the B-equations slightly differently:
$$\ce{HCl(aq) + H2O(l)-> H3O+(aq) + Cl-(aq)}\tag{B1'}$$
$$\ce{NH3 + H3O+(aq)  <=> NH4+(aq) + H2O(l)}\tag{B2'}$$
In the end, each pair of equations says the same thing: Hydrochloric acid is as a strong acid, ammonia is a weak base. If ammonia is a weak base, we also know that the ammonium cation is a weak acid (with ammonia its conjugate base).
The calculation
There are two steps. You missed one (the one that involves the equilibrium). Let's work with equations A1 and A2 to work through the calculations.
You had the correct intuition about what happens with the hydrochloric acid: When adding 25 mL ammonia solution to the hydrochloric acid solution, the reactants are present at stoichiometric ratio, and the only species that is left to influence the pH is ammonium cation. When adding 26 mL ammonia solution, however, the ammonia is in excess, so some of it turns into the ammonium cation, and some remains. Scenario (i) is called a weak acid in solution, scenario (ii) is called a buffer solution. We can ignore the chloride ion, it is a spectator ion.
The amount of hydrochloric acid present initially can be determined from initial concentration and volume:
$$n_{\mathrm{\ce{HCl},initial}} = c_{\mathrm{\ce{HCl},initial}} \cdot V_{\mathrm{\ce{initial}}}$$
$$\ \ \ =0.10\  \frac{\mathrm{mol}}{\mathrm{L}} \cdot 0.025\  \mathrm{L}$$
$$\ \ \ =2.5\times 10^{-3}\  \mathrm{mol}$$

Scenario (i): weak acid

The amount of ammonia present initially can be determined from initial concentration and volume:
$$n_{\mathrm{\ce{NH3},initial}} = c_{\mathrm{\ce{NH3},initial}} \cdot V_{\mathrm{\ce{initial}}}$$
$$\ \ \ =0.10\  \frac{\mathrm{mol}}{\mathrm{L}} \cdot 0.025\  \mathrm{L}$$
$$\ \ \ =2.5\times 10^{-3}\  \mathrm{mol}$$
You calculate the concentration of species by first figuring out the amount, and then dividing by the volume of the solution, which is 50 mL. In scenario (i), all of the ammonia was turned into ammonium, so you know:
$$n_{\mathrm{\ce{NH4+}}}=2.5\times 10^{-3}\  \mathrm{mol}$$
$$c_{\mathrm{\ce{NH4+}}}= \frac{n_{\mathrm{\ce{NH4+}}}}{V_\text{solution}} = \frac{2.5\times 10^{-3}\  \mathrm{mol}}{\pu{50 mL}} = \pu{0.05 mol/L}$$
This concentration is lower by a factor two compared to the initial concentration of ammonia because the volume doubled when we added hydrochloric acid.
Now you can calculate the pH (pH of a weak acid with a pKa of 9.25 and a concentration of 0.05 mol/L:
$$ \text{pH} = \frac{1}{2} 9.25 - \frac{1}{2} \log 0.05 = 5.28 $$

Scenario (ii): buffer

$$n_{\mathrm{\ce{NH3},initial}} = c_{\mathrm{\ce{NH3},initial}} \cdot V_{\mathrm{\ce{initial}}}$$
$$\ \ \ =0.10\  \frac{\mathrm{mol}}{\mathrm{L}} \cdot 0.026\  \mathrm{L}$$
$$\ \ \ =2.6\times 10^{-3}\  \mathrm{mol}$$
HCl is limiting in reaction A1, so that determines the amount of product:
$$ n_{\mathrm{\ce{NH4+}}} = n_{\mathrm{\ce{HCl},initial}} = \pu{2.5e-3 mol}$$
Some of the ammonia is still present:
$$ n_{\mathrm{\ce{NH3}}} = n_{\mathrm{\ce{NH3},initial}} -  n_{\mathrm{\ce{NH4+}}} = \pu{2.6e-3 mol} - \pu{2.5e-3 mol} = \pu{0.1e-3 mol} $$
You can plug these concentrations into the Henderson-Hasselbalch equation (buffer equation) to get the pH:
$$ \text{pH} = 9.25 + \log \frac{0.1}{2.5} = 7.85$$

I hope that the answers to all the questions are similar to this clear, explicit answer that takes into account the individual differences and different cultures of the students. Blessed efforts

I agree, but there is also fine distinction between giving fish versus teaching fishing. Solving the whole task may be like just giving fish.

@Poutnik I already calculated the pH when 25.0 cm3 of NH3 added. Then I only left to solve when 26.0 cm3 of NH3 added......in this case, some NH3 is left, this is difficult as NH4Cl is a weak acid and NH3 is a weak base. Both of them don't dissociate completely in water. See - pH of WA+WB

Sorry if I missed it, I have not noticed it yet, reading Q multiple times. The key point in answer is helping to understand principles, not solving tasks. I am not sure if that succeeded. the NH3/NH4+ buffer case Is textbook example for the H.H. equation I have previously provided .

The point is, they are not just "a" weak acid and "a" weak base. They are the conjugated pair acid-base by NH3 + H+ <=> NH4+, with equilibrium defined by the given constant. Such pairs are bases of almost all pH buffers - solutions keeping desired pH. If you know the ratio and the constant you know pH.

@Poutnik re Fish... I would have liked to just post the first part of the answer, but we have a no hint policy. I thought the way the OP got stuck was interesting, and the question was on the verge of getting closed.

I see. I am just thinking the participation of the OP could avoid closure.

@KarstenTheisNeglcting water autoionization and taking into cosideration $\ce{NH3}$ and $\ce{HCl}$ ionization . Which charge balance equation is correct for the above scenario(i) and why?:
$$\ce{[NH4+] + [H+] = [Cl-] + [OH-] }$$ or$$\ce{[NH4+] + [H+] = [Cl-]  }$$

@AdnanAL-Amleh NH4+ cannot accept protons. As it is an acid, it will only donate protons. (Brønsted-Lowry definition of an acid) Its conjugate base NH3 will accept a proton because of its lone pair. Also keep in mind that water is amphoteric.

@hpm0615ngvup: Can we write the charge balance equation of the above scenario(i)?

@KarstenTheis Hi, how does the pH become $1/29.25−1/2log0.05=5.28$? Where does the $1/2$ come from and why do you do $1/2*log0.05$? Thanks!

@katara Eq 14.4.12 in https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_ChemPRIME_(Moore_et_al.)/14%3A_Ionic_Equilibria_in_Aqueous_Solutions/14.04%3A_The_pH_of_Solutions_of_Weak_Acids


text:
Relative lowering of vapour pressure
Why is lowering of vapour pressure (Po - Ps) temperature dependent? And why is relative lowering of vapour pressure (1 - Ps/Po) independant of temperature?

Why indeed should the vapour pressure NOT be temperature dependent? Please clarify your question, which one are you unsure about?

i want mathematical approach to the terms (1 - Ps/Po) that how they balance the effect of temprature!

also in case of (Po - Ps) why it depends on temp? because when temprature is increased both Po and Ps increase, they should have balance each other? why they don't balance?

In the simplest terms and most convenient definitions, the lowering of vapor pressure is not a colligative property while the relative lowering of vapor pressure is indeed a colligative property.
A bit more explanation: A colligative property is dependent on the moles of solute only. The ratio in the definition you provided eliminates the temperature dependence of formulation (assuming ideal solutions) by making the measurement relative to another quantity, which in this case is the vapor pressure of the pure solution. Otherwise, the influences on vapor pressure are multifactorial. I.e. external pressure may affect it, as well as temperature.
Here is a somewhat hand-wavy out-of-the-pocket calculation to demonstrate. Assume the vapor pressure can be expressed as an ideal gas. Therefore pressure may be expressed using the ideal gas law, $P = \frac{nRT}{V}$. Also know Raoult's law for vapor pressure, $P_s = \chi_s P_o$ where $\chi_s$ is the mole fraction. Using your definitions, we have for absolute pressure difference
$$P_o - P_s = P_o - \chi_s P_o = \left(1-\chi_s\right)P_o = \left(1-\chi_s\right)\frac{n_oRT}{V}$$
This is therefore dependent upon temperature. Now do the same for the relative difference in pressures
$$1-\frac{P_s}{P_o} = 1 - \frac{\chi_sP_o}{P_o} = 1-\chi_s$$
The expression is now dependent upon the moles only, i.e. a colligative property. Of course, in reality this depends on the assumptions we made under ideal Raoult conditions.

is volume same for the term Po and Ps? or if it is different, can we mention Vo and Vs instead of just V?

I wrote this under the assumption that volume does not change. You can create this isochoric process experimentally when adding solute if needed. But for this particular problem, it does not matter if volume is constant and you can say it changes from $V_o$ to $V_s$. I just wanted to explicitly show how the absolute pressure difference can be a function of temperature. For instance, we can leave the math at$$P_o-P_s=(1-\chi_s)P_o$$and make that argument that because the pressure $P_o$ is present, then the difference in pressure is still subject to all the same forces that change it, like temp.

I don't understand how lowering of vapour isn't a colligative property. Based on the equation you provided for lowering of vapour pressure, it is dependent on T, V and n sub o. I don't see any variable that accounts for the nature of solute. Is it necessary for a colligative properties to be ONLY dependent on number moles? If it depends on temperature but its still independent of nature of solute then it's not colligative property?


text:
Formal Charge in Coordinate covalent bonds
I am in an honors chemistry course right now and I would like to know the formal charges of the atoms in CO, my confusion is that because the bond is partially coordinate covalent, are the bonded electrons still counted as #bonds/2 or should I count the number of electrons donated by each atom (making the formal charge 0 for both) Please help me understand and explain WHYnot just how. Thanks a lot.

CO is a more complex molecule and analyzing it as having a dative bond and 'normal' covalent bonds is bound to get you into trouble. https://chemistry.stackexchange.com/a/31003/7448

The coordinate covalent bonds are still counted as 1 electron, since the pair is shared between both atoms. That means the Carbon will have a formal charge of -1 and Oxygen a formal charge of +1 on Carbon Monoxide.
Carbon FC: $\ce{(4 valence electrons on neutral atom) - (1*2e- from lone pair + 3*1e- from covalent bonds) = -1}$
Oxygen FC: $\ce{(6 valence electrons on neutral atom) - (1*2e- from lone pair + 3*1e- from covalent bonds) = +1}$


text:
What atom would be protonated first in a strongly acidic solution?

The answer is A. I thought it would be D since the ester group has the highest pka and the more basic functional group would be more prone to protonation. Also, I don't really get how to rank C in terms of atoms that would be protonated first as I don't know the pka of the ether group.

Atom C is in nowheresville.  By donating one of its "lone pairs" to the ring this oxygen atom gets a positive formal charge.  Not good for then accepting a proton.

What is "protonated first" even supposed to mean. Does author imagine they would get a dication? Something like this would break down much earlier.

Also if you, say, added conc. H2SO4 to this, there would be protonation equilibria for all these groups, so I suppose properly formulated question would be here which Keq would be highest.

An interesting question would be: What happens were this compound refluxed in dilute sulfuric acid for 24 hours?

In short, none. As Mithoron said, the oxygen atoms can all act as Lewis bases and will all be protonated to an extent. If you rephrase the question to "Which of the atoms has the highest Kb?", well that's another story. Atom D has the lowest Kb, because the carbon in a carbonyl has a slight positive charge. Atom C would be difficult to protonate, the oxygen is already donating a pair to the furan ring to make it aromatic. So the debate is between a carbonyl and a primary alcohol. After consulting some pKa tables I found that protonated ketones have a pKa of around -7.3 while protonated primary alcohols of around -2.4 thus the alcohol is a stronger base (higher pKa means weaker acid thus stronger conjugate base). Finally we can conclude that the primary alcohol -OH group will be the first to accept a proton.
P.S. Fun fact, the quickest to be protonated was none of the ones marked with letter, it would have been the ester's carbonyl group!

As far as D is concerned, I think that mesomeric effect may more important then inductive,  Not bad for a first answer, though ;)

I looked at the pka chart my instructor provided me. On that pka chart, it said the pka of R-OH (A in the molecule) is 16. B (the ketone group) is 20. D (the ester group) is 25. Nothing was given for the ether. A higher pka means its more basic, and thus a higher kb. If it is the highest kb, shouldn't it be D (ester) then, as the pka is higher than the rest.

As I said, those I listed were the pKa values of PROTONATED species. The protonated species with the highest pKa forms the quickest.

In the given molecule

D's lone pair on oxygen is in conjugation with C=O group. Consequently it acquires a positive charge, its Kb is less for protonation to take place (as shown in scheme 1, as pointed by Mithoron).
At C lone pair on this ring (furan) are delocalised and is AROMATIC (as shown in Scheme 2). Comparatively is kb value is less for protonation to take place.
At B carbon of carbonyl carbon is sp2 hybridised and is more withdrawing then sp3 hybridised carbon (hybridisation is useful here). Hence due to greater inductive polarisation by carbonyl carbon lone pair on oxygen are donated less.
At A, lone pair on oxygen is not in conjugation as in C and D. Further carbon bonded to alcohol at A is sp3 hybridised and is less withdrawing inductively then carbon at B which is sp2 hybridised.

Summing, OH group at A has highest Kb value. Its tendency to donate lone pair is comparatively more then B, C or D.


text:
Determining identity of unknown compound with NMR data

I have the following IR spec, C NMR and H NMR data. The boiling points determined for the unknown compound were 116.8 and 119.1°C (two trials). Given the possible compounds, the unknown can be one of:
3-methyl-2-butanol
1-butanol
3-pentanol
2-pentanol
2-methyl-1-butanol. (These are based on boiling point)
I have eliminated some of these choices based on the number of different types of carbons (4, as shown on C NMR) but am not sure where the H of the OH is indicated in H NMR. There is no singlet present among the 3 groups of peaks, so where would it be?

Peaks of acidic protons in an NMR spectrum can change (in position, form and intensity) due to exchange with other volatile protons.

A very fast exchange makes the peaks coalesce, you get one singlet between the two "original" (expected) chemical shifts, the exact position depends on the respective amount of both.
A slower exchange, one "on the NMR timescale", broadens that peak considerably, typical case is residual water and a COH proton. The peak might in some cases be so broad it becomes invisible.
Even slower, this peak splits into two, which become sharper and further removed from each other.
A very slow exchange makes no change to the positions, but an exchange with deuterium (i.e. from the solvent) changes the relative intensity of the respective peaks over time.

Is the greatest possibility in this case then a slower exchange where peak is invisible? I did not do the experiment myself so I am unsure.

No, in case of a slow exchange, the proton would instead appear as a large residual solvent peak. But the chloroform peak at 7.2 is nice and small. I believe that there is an exchange with an approximately equivalent amount of residual water, happening with a rate of a hundred or a thousand Hertz, broading the peak so much it vanishes into the baseline. My best guess.

The proton on O or N atoms can be acidic enough to exchange with the deuterium atoms of the NMR solvent, thus "hydrogen-bonding protons" are often absent from H NMR spectra, or may show up "anywhere".

Alcohol protons don't exchange with CDCl3. (notice the peak at 7.1ppm). What can make them "vanish" is the exchange with residual water.

The chloroform peak would esp. be much larger if there was an exchange.


text:
Michaelis menten constant motivation behind 1/2?
So the michaelis constant is defined as the substrate concentration at half maximal „reaction-speed“.
I was wondering why 1/2 ?
I guess for higher values you need more substrate, which could be a problem. But would 1/3 also work ? Or does 1/2 optimize some other parameter?

1/2 is the least arbitrary choice of a number between 0 and 1. To me, that's good enough.

The „least arbitrary choice“ is defined in which way? Smalles sum of denominator and nominator ? Seems kinda odd to me, that there is no further considerations defining such a important constant

If you need a formal definition, then yeah, this one will do.

in the derivation of the M-M equation using rate constants (lower case k's), the k's are grouped in such a way as to make the entire expression most simple, in particular to isolate [S]. When you then define Km as the other term in the denominator, the coincidental result of that denominator (Km + [S]) is that when [S]=Km, we can replace Km with [S] and get rate = Vmax/2. This definition of Km is also conceptually useful because as $k_{-1}/k_2$ goes to infinity, Km becomes equal to Kd for the ES complex in the simple case.

@Andrew I think that's an answer.

So i see the connection to the mm equation but i guess that is the same problem: where does the mm equation comes from? Can you provide more information about the limit of Km with the small ks?

You could define a new constant $K_\mathrm{Paul}$ that is twice as large as $K_\mathrm{M}$. In that case, the rate expression would be $\displaystyle v_0 = v_\mathrm{max} \frac{[S]}{0.5 K_\mathrm{Paul} + [S]}$, and the rate would be two-thirds of the maximum for $K_\mathrm{Paul} = [S]$.

To see why the $K_m$ is equivalent to the substrate concentration at which the rate is equal to $V_{max}/2$, we need to look at the derivation of the Michaelis-Menten, which is usually included in detail in any introductory biochemistry book. Here is a brief version.
We will consider an enzyme-catalyzed reaction described by the following scheme:
$$\ce{E + S <=>[k_1][k_{-1}] ES ->[k_2] E + P},$$
where "E" is enzyme, "S" is substrate and "P" is product.
To get an expression for the rate, we start by noting that the rate of formation of product is $k_2[\ce{ES}]$, so we need to find an expression for $[\ce{ES}]$ using known quantities. Next, we make a steady-state approximation, setting $\frac{d[\ce{ES}]}{dt}=0$, so we have
$\displaystyle k_1[\ce{E}][\ce{S}]-(k_{-1}+k_2)[\ce{ES}]=0$
$\displaystyle \implies [\ce{ES}]=\frac{k_1}{k_{-1}+k_2}[\ce{E}][\ce{S}].$
To finish, we note that when the reaction has not consumed very much substrate, we can approximate $[\ce{S}]$ as equal to the starting substrate concentration $[\ce{S}]_0$. We also know that the total amount of enzyme in the reaction must be equal to the sum of the concentrations of E and ES, ie $[\ce{E}]+[\ce{ES}]=[\ce{E}]_{Total}$. We already have an equation for $[\ce{ES}]$ in terms of $[\ce{E}]$, so with some algebra, we can find that
$\displaystyle [\ce{ES}]=\frac{k_1[\ce{E}]_{Tot}[\ce{S}]}{k_{-1}+k_2+k_1[\ce{S}]}$.
Combining everything, we get
rate of formation of $\displaystyle \mathrm{P} =k_2[\ce{ES}]=\frac{k_2k_1[\ce{E}]_{Tot}[\ce{S}]_0}{k_{-1}+k_2+k_1[\ce{S}]_0}$.
In order to isolate $[\ce{S}]_0$ in the denominator, we divide the numerator and denominator by $k_1$:
rate $\displaystyle =\frac{k_2[\ce{E}]_{Tot}[\ce{S}]_0}{\frac{k_{-1}+k_2}{k_1}+[\ce{S}]_0}$,
and our final simplification is to define the unwieldy $\frac{k_{-1}+k_2}{k_1}$ as $K_m$ and $k_2[\ce{E}]_{Tot}$ as $V_{max}$:
rate $\displaystyle =\frac{V_{max}[\ce{S}]_0}{K_m+[\ce{S}]_0}$.
Getting back to your original question: Note that we never explicitly defined $K_m$ to be the substrate concentration at half-maximal rate. It simply resulted from our choice to isolate the substrate term in the denominator. However, that was not a completely arbitrary choice.
If we look at $K_m=\frac{k_{-1}+k_2}{k_1}$, we can see that if $k_{-1}$ is large and $k_2$ is small, $K_m$ approaches $\frac{k_{-1}}{k_1}$, which is the equilibrium constant for dissociation of the substrate from ES. For that reason, $K_m$ is often treated as an apparent measure of the affinity of the enzyme for the substrate, and that is an additional reason why we choose to define $K_m$ this way. Always remember that it is not actually the dissociation constant, and in some reactions will be quite far off from the actual $K_d$.

Now i am satisfied, thank you very much

"quite far off from the actual Kd": You can't have a Km and a Kd at the same time. The Kd is for a molecule that binds but does not react (i.e. inhibitor bound to enzyme, or substrate bound to inactivated enzyme) while the Km is for a molecule that binds and reacts. So there is no "actual Kd", it is a though experiment.

@KarstenTheis - I disagree. A substrate molecule can react or it can fall back off. There is no reason why we cannot limit our focus to the elementary step of reversible binding, just as we often break down kinetic schemes into elementary steps.  In the most basic M-M scheme, we have $\ce{E + S <=>[k_1][k_{-1}] ES ->[k_2] E + P}$. The $K_d$ is given by $\frac{k_{-1}}{k_1}$, while $K_m$ is $\frac{k_{-1} + k_2}{k_1}$.

I'm just saying that you can't measure Kd if you can measure Km. In a thought experiment, you can disregard k2, but in reality, the enzyme substrate complex has two fates, not one. You are right that in cases where k(-1) is much larger than k2, Km approaches what Kd would be if k2 were zero. If you simply define Kd as the ratio of k(-1) and k1 (and not as a measurable quantity), then you are absolutely right and the adjective "actual" is appropriate.

@KarstenTheis - I've never heard the claim that equilibrium constants must be easily measurable in order to be real. Furthermore, even if one uses applies that constraint, it is incorrect to say categorically that "you can't measure Kd if you can measure Km". It simply requires more and different data so that you can accurately fit to a model that separates k1, k-1 and k2. Pre-steady state kinetics are often used for this. Here's a good review of the subject: https://www.molbiolcell.org/doi/full/10.1091/mbc.e13-01-0030.


text:
In the Lewis structure for barium fluoride, why does barium lose its valence electrons?
I searched for $\ce{BaF2}$ Lewis structure, and found out that the Barium atom loses its 2 valence electrons to form $\ce{F- Ba^2+F-}$.
Before seeing it I thought each of $\ce{Ba}$'s valence electrons form a single bond with each of the fluoride atoms $\ce{F-B-F}$ which also completes the octet for F atoms,  and the formal charge in this case will be $7-6-\frac{1}{2}(2) =0$ for F atoms and $2-0-\frac{1}{2}[2\cdot 2] = 0$ for Ba, which works out.
I'm looking for an explanation on why Ba loses its electrons to form $\ce{F- Ba^2+F-}$ instead of $\ce{F-Ba-F}$.

You've just stumbled upon an ionic compound. Prepare to see them every now and then, for they make up a sizeable portion of chemical universe. Come to think of it, how could you not notice NaCl, another example of those?

@IvanNeretin I am totally new to this subject, I'll search to understand what ionic compound is.

First, a little bit of background. I assume you're already familiar with atomic orbitals — functions that describe the wave-like behavior of an electron in an atom. There's also molecular orbitals (MO), that describe its behavior in a molecule. You can ignore the odd-looking labels in images and such, since they're not entirely relevant to this question.
Atomic orbitals interact to form molecular orbitals (so called LCAO). The number of resulting molecular orbitals is the same as the number of the atomic orbitals that interact in the first place. To visualize this in simple interactions, you can use a molecular orbital diagram, in which orbitals of lower energy are drawn as lines placed lower than the others. There are three types of resulting MOs:

Bonding MO, which is usually much lower in energy than the starting atomic orbitals
Anti-bonding MO, which is usually much higher in energy
Nonbonding MO, which would have the same energy as the starting atomic orbitals

You can imagine that electrons are rearranged when a molecule is formed, and they fill the orbitals with the lowest energies first, so the order is bonding first and anti-bonding last. In a symmetric molecule, like the boilerplate example $\ce{H2}$, this is what the diagram looks like:

Image taken from this helpful page on Libretexts
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Back to your question, the energy difference between the atomic orbitals and the bonding orbital (presented as the space between the lower bonding MO and the higher atomic orbitals of each H atom) is the same, so they're said to 'contribute the same to the bonding MO'. Bonds between homonuclear diatomic molecules are therefore considered to be purely covalent (whoops, almost!). Covalent compounds are the ones you're taught to draw Lewis structures for, apply formal charges, etc., but it's only one extreme of the spectrum.
The next example would be the MO diagram of HCl.

Cl's 3p atomic orbitals are lower in energy (check the provided link for the reasoning); this time round, the atomic orbitals do not contribute equally to the molecular orbital, and Cl's orbital's contribution to the bonding orbital is greater. Electronegative atoms are generally said to have atomic orbitals of lower energy. This is probably more simply and familiarly stated as "the more electronegative atom pulls/attracts the bonding electrons". Note that, now, there's also an energy gap between the initial atomic orbitals. The $\ce{H-Cl}$ bond is, indeed, somewhere between covalent and the other extreme.
Now to the punchline, what happens when the gap is too large? This would be what an MO diagram of LiF would look like:

Thanks Geoff!
The gap is so large that, essentially, the energy difference between molecular orbitals filled with electrons, and fluorine's atomic orbitals would be very small compared to the energy difference between F's 2p orbitals and Li's 2s. Put in simpler terms, the very electronegative fluorine has such a stronger pull than Li that the electrons are almost fully 'localized' on the fluorine, and Li has approximately lost its electron. This is what we call an ionic bond, the other end of the spectrum.
Most bonds between very electronegative atoms and very electropositive ones are considered ionic, and there's the guideline that metal-nonmetal bonds are often ionic. Don't let it trip you up though, since it's a vague general rule, with plenty of exceptions. As you might have suspected, there's no truly, purely ionic bond either, and it's a spectrum.