text:
Less substituted vs more substituted alkene
In general, more substituted alkenes are more thermodynamically stable. However, in certain cases such as enamine alkylation, the less substituted double bond is favored. How do we know that which is favored during a particular reaction? Can we have a generalized rule for it?

Welcome to ChemSE. Try to be more specific in phrasing your question. Specific examples would help. Your question is too broad as it stands.

Thanks for your comment.This question is intended to be general.I mean to say I would like to understand in what contexts more substituted 'ene' is favored and also what is the driving force for the formation of less substituted 'enes' although more substituted 'enes' are stable.

There is no single rule; chemistry can sometimes tend to be a collection of various rules which must be applied at different times. Learning chemistry entails learning when to use which rule.

Arguably, it's also unclear; I don't know what 'enamine alkylation' is referring to. The answer currently refers to relative stabilities of the enamines, which is unrelated to their alkylation. It would be helpful if you could clarify (and also come up with a title that describes the context).

I'll address your concerns about alkene stability in general and enamines in particular. It is useful to examine the acyclic permutations of C6H12. The heat of formation of these isomers is a good measure of their relative stability. As the double bond is increasingly substituted with alkyl groups, the heat of formation becomes more negative, i.e., more stable.  Tetrasubstitution is better than trisubstitution which is better than disubstitution. Clearly, monosubstitution is the least desirable case. The stabilization by alkyl groups in this case is due to the hyperconjugation of the bonding σ-CH orbital and the anti-bonding π*orbital. The more opportunities for this interaction, the more the stabilization. Heats of formation are available at the NIST site.
The pyrrolidine enamine of 2-methylcyclohexanone prefers the trisubstituted over the tetrasubstituted position. Because there is a strong resonance contribution (planarity) in the enamine, the tetrasubstituted isomer has strong steric interactions between the methyl group and the methylene adjacent to the nitrogen in the pyrrolidine ring. The equilibrium is not spontaneous but it is catalyzed by water or traces of acid. In the trisubstituted isomer, the methyl group adopts an axial position to avoid an A1,3 interaction if the methyl group were equatorial.

1) International Institute of Standards and Technology, Dept. of Commerce, USA.
2) F. Johnson and S. K. Malhotra, J. Am. Chem. Soc., 1965, 87, 5492.

I'm not sure where the alkene is during enamine alkylation. I can only see an alkene in the enamine starting material, so I will try to explain the alkene formation selectivity for enamine formation.
The less substituted alkene is favored for enamine formation because in the enamine, the N atom pushes electron towards the alkene by resonance. A less substituted carbon atom accommodates the negative charge better than a more substituted carbon atom, so a less substituted alkene is more stable in this case.
For other reactions such as alcohol dehydration, the more substituted alkene is formed because the intermediate has a carbocation character. The more stable intermediate is the more substituted intermediate, thus giving the more substituted alkene as the major product.

Thanks for answering.So, in general, if there is a probability for negative charge to accommodate through resonance, a less substituted one is favored and vice versa?

It all depends on the mechanism of the specific reaction. No such general statements can be made.


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Why ionisation enthalpy of early lanthanides are expected to be greater than early actinides?
In my book Chemistry Part I (NCERT XII), a statement on p. 232 goes like this:

It is evident from the behaviour of the actinoids that the ionisation enthalpies of the early actinoids, though not accurately known, but are lower than for the early lanthanoids. This is quite reasonable since it is to be expected that when 5f orbitals are beginning to be occupied, they will penetrate less into the inner core of electrons. The 5f electrons, will therefore, be more effectively shielded from the nuclear charge than the 4f electrons of the corresponding lanthanoids.

My thought on this matter is that $Z$ will increase and f-orbitals will shield electrons poorly and so ionisation enthalpy should increase. But I'm not able to understand what my book is trying to say. What is meant by less penetration to inner core of electrons? To be honest I wasn't able to understand all the above mentioned 3-4 lines. So, can anyone explain this properly to me, please?

@HahaHahaha How did I what? Yes, I added a literature source since OP failed to do that, what about it?

If anyone could confirm, if I understood the meaning. 5f orbitals being more on vicinity as compared to 4f, will have more effectively shielded, and since, as mentioned, they penetrate less into nucleus, connotes more ionisation enthalpy. @andselisk what's more accurate to say actinides or actinoids and similarly lanthanides or lanthanoids, I have never seen latter in either case...

@Zenix Nomenclature - Lanthanoids vs Lanthanides (and Actinides vs Actinoids)

@HahaHahaha I don't see how this is relevant. If you search Google Books for the exact phrase, for example, "The 5f electrons, will therefore, be more effectively shielded from the nuclear charge", you only find this very NCERT book. Google still works for everybody regardless the nationality.

I'd suggest you follow the book blindly if you're preparing for JEE, even if the data might be wrong. There have been instances before when JEE Mains has considered this book to be correct despite it having the wrong information.

https://chemistry.stackexchange.com/questions/109178/why-ionisation-enthalpy-of-early-lanthanides-are-expected-to-be-greater-than-ear

This data is incorrct, and I am saying so because the Ionisation energy of the actinoid is a decent bit higher than the corresponding lanthanide
In each case IE 1 is significantly higher for lanthanoids, IE 2 is still higher for lanthanoids and IE 3 is comparable.
$\ce{Ce}$ $534.4$, $1050$, $1949$
$\ce{Th}$ $587$, $1110$, $1930$
These were two examples where i showed all 3 IEs
For, the sake of further clarification, here is a the IE1 of the first few
$\ce{Pr}$ $527$, $\ce{Pa}$ $568$
$\ce{Nd}$ $533.1$, $\ce{U}$ $597.1$
$\ce{Pm}$ $540$, $\ce{Np}$ $604.5$
$\ce{Sm}$ $544.5$, $\ce{Pu}$ $584.7$
(units used are $kJ/mol$)
The reason you stated in your doubt is absolutely correct, as even if we were to conclude at these aren't accurately determined, we see this phenomenon elsewhere in the periodic table also, where the values are accurately determined.

$\ce{Sn}$ is the lowest IE element of GP 14 and not $\ce{Pb}$.
$\ce{Fr}$ has a higher IE than $\ce{Cs}$.
$\ce{Ra}$ has a higher IE than $\ce{Ba}$.

All due to the same reason poor shieling, which increases the effective nuclear charge, which increases nuclear attraction, which in turn leads to a higher IE.
References
Wikipedia data page

@Andselisk I tried to format this in the best way possible, however, are there some improvements that you can help me with?

I find it hard to track what values are taken from where and what do the represent, especially in the first array. When you cite a textbook, it's almost always obligatory to include the page numbers. Also note that symbols for chemical elements must be upright (use \ce{…}).

Please visit this page, this page and this one on how to format your future posts better with MathJax and Markdown. See also: MathJax basic tutorial and quick reference.

@andselisk my stupid laptop does not have a backward slash or curly brackets lol, so idk what to do, as said in the references all values are simply taken from the Wikipedia data page, which has taken its from the two given books which is given in their references, so basically I study NCERT too, i was almost sure this statement was wring, so i just googled data and found this, then i looked up these books and their rewives to see if the wiki editor has taken daya from good books, turns out they are both well reputed books

Switch to US international layout which will give you access to all ASCII symbols including {} and  \\, which are essential if you do slightly advanced posting or any programming. As for the rest, make sure you read the sources you cite and don't throw them in because they look relevant.

Hey thanks @andselisk so iremoved those sources and just left it at wikipedia data page. Now looks okay? What about the answer itself? Did you have a look at it?

i wrote the elements upright also, now the answer seems complete

heyy @Zenix thanks! Actually i was off SE for 2 days because of jee stage 1

How was it? Was it too tough, I saw few questions of maths and chemistry, idk physics much. I felt maths was easy, except for 4-5 questions which I couldn't do. Chemistry was easy too, except I felt that few questions were impossible to solve. What was the point of asking structure of chloramphenicol?

Ahh, there were slots. The chloramphenicol question came in day 1 slot 1. I was in day 2 slot 2. My paper was decent, not too difficult. But i suck at phy, so my phy went bad. Otherwise, maths and chem which i checked with an answer key relaeased by some private intitutes i should be good to go with a roundabout 90-95% score (180-190ish out of 200). So it all comes down to my PHY, which will decide my rank. Maths was in general very easy, and chem they also asked vitamin deficiency diseases in day 1 slot 1, which is so not chemistry lmao pity those dudes

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Why is the carbanion resonance structure being used in the aldol condensation mechanism?

In the above mechanism's first step, what motivates the use of the carbanion resonance structure of the enolate ion if the negative charge is better localized on the more electronegative oxygen?
Edit: I might not have been clear in what motivates my confusion. Maybe I should rephrase what I'm asking. Basically, why would the carbon on the enolate attack the carbonyl carbon as opposed to the oxygen on the enolate attacking the carbonyl carbon? I assume there's a mixture of both products, but why is the aldol the major product?

See this link: https://www.google.com/url?sa=t&source=web&rct=j&url=http://butane.chem.uiuc.edu/jsmoore/chem232/notes_current/The_C-alpha_Nucleophile/NOTES-Enolate_and_Enol_Nucleophiles.pdf&ved=2ahUKEwjp7c2NwqvgAhWTWX0KHcSsD0sQFjABegQIEBAE&usg=AOvVaw2TxLxuBy7LxROabWcLdSJt&cshid=1549611013315

They are two ways of representing the same thing, don’t overthink it.

The negative charge is placed on the carbon atom to signify that it is the nucleophilic center which can attack the aldehyde. Indeed the negative charge is better placed on the oxygen atom, but after performing a nucleophilic attack on the aldehyde, an unstable acetal is formed reversibly.

I understand that. I was looking for a deeper motivation as to why the carbanion resonance structure was the primary structure being shown. Is it possible the solvent choice would have it tending more towards carbanion character?

Maybe I should rephrase what I'm asking. Why would the carbon on the enolate attack the carbonyl carbon as opposed to the oxygen on the enolate attacking the carbonyl carbon? I assume there's a mixture of both products, but why is the major product?

Who is to say that the oxygen of the enolate doesn't add to the carbonyl group? If so, it is a non-productive step that forms a hemiketal which reverses to the reactants in base. So you are back where you started.

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Raman transition and IR transition
In a character table one can find the reducible representation of vibration of a molecule. In such a table we also see things like $xy,z,x...$. If this corresponds with an irrep we say that IR transitions can occur if $x,y,z$ and quadratic terms give Raman transitions. This is because the dipole moment and polarizability are non-zero and can change in that case. My question is: where does this come from? I prefer a physical or mathematical answer involving quantum mechanics since I am from physics.

To simplify slightly both Raman and IR spectroscopy show the vibrational modes of a molecule (though the techniques used to reveal these are very different).
IR spectroscopy relies on coupling between the electromagnetic field of light passing through a sample and the electric dipole of the molecule. But that absorption is only possible is the vibration in the molecule changes the molecules dipole. Consider, for example, carbon dioxide, a linear molecule with no net dipole moment. Symmetric stretches (ie both oxygens moving the same amount in opposite directions) don't change the dipole so can't be detected by IR. But asymmetric stretches or bending will change the dipole and can be detected by IR.
Raman spectroscopy (simplifying a little) relies on detecting changes to the molecules polarisability not the dipole. So the symmetric vibrations of carbon dioxide are detectable because they do change the polarisability of the molecule.
In both cases the symmetry tables tell you whether the symmetry of the vibrational mode changes the electric dipole or the polarisability of the molecule. In simple molecules physical intuition can usually tell the same thing but symmetry tables are more reliable.
A full understanding would involve a quantum mechanical description of the interaction of EM radiation with the possible vibrational modes of a molecule considering the allowed and forbidden transitions. But the physical intuition above captures the essentials without getting too mathematical.

It might be worth mentioning in addition that polarisability involves a change in the shape or size of the 'electron cloud' and so the em field see the projection of this i.e. a 2D shape change hence quadratic terms in the point group.

This may be unnecessary clarification, but to connect to the character table, the $x$ associated with a specific character means that the $x$-axis transforms with this symmetry. Since the $x$-axis is asymmetric about the origin (+ vs -), any vibration that has the same character also involves asymmetry along the $x$-axis. Same applies to $y$ and $z$ axis of course. Since asymmetric movement creates a changing dipole, those vibrations give IR absorbance as described above.

Edit: my comment above should say "antisymmetric about the $x$ axis" rather than "asymmetric. . .." Apologies for sloppy terminology.


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I am looking for any substance that would be unstable or dangerous when it is cooler or not on heat, and is stable when it is being boiled
I am a first responder and one of our protocols when entering any unknown lab environment is to turn off any burners. Most of the time, we try to not disturb the environment as much as possible so a colleague of mine and myself were curious as to any substance that would actually be more dangerous after turning off said burner. Thank you.

Substance? No. Device? With some far-fetched assumptions, maybe yes.

Can you please explain what exactly a "first responder" is?

@Karl In case of a medical emergency, a first responder is a designated trained person providing first help prior to the arrival of an ambulance.  Typically, these have more repeated training "under the belt" than "just" the red cross course you pass when applying for a driver's licence. (In some schools / universities, you are not allowed to enter the chemistry lab until at least one first responder is in proximity.)

@Buttonwood In Germany, we have "Ersthelfer", employees with a professional or academic degree, and a one week first aid course (+ regular refresher courses). Every "workplace" building must have a certain number of them registered per floor, depending, and their names and office numbers are printed on the evacuation plans you find on every corridor. But: Those people know what is going on where they might be called for help, and in a chemistry building, they are generally chemists. This seems not the case with our OP, if I understand correctly.

Well, in theory, a few compounds could become explosive when crystallizing out of a solution (which is promoted by low temperatures) though I do not think that "turning off burners" would be enough. The "best" candidate I can imagine would be "azidoazide azide" which can detonate for... just any reason! But nobody uses it, just because of that.

It seems as azidoazide azide is pretty unstable with most things. I do like where you are going with something crystallizing as time passes. Sometimes we are at a scene for days. I guess the question would be, is there something that is air reactive in a crystalline state that would be stable as a solution.

Another mechanism would be the solvent evaporating over days. So the solution of the compound is stable, but the dry compound is not.


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Lewis structure of a compound
What is the correct lewis structure of ammonium cyanate  ion? I do have a structure but it seems partially correct. If I'm wrong, then please inform that also.

$\ce{NH3 }$ has a lone pair, $\ce{NH4^+ }$ doesn't.

What lone pair? NH4+ has none.

First thing is that there will be no lone pair in nitrogen as the  lone pair is donated to the 4th H+ ion. This forms the ammonium ion.
Now, cyanate ion is formed by [OCN]-. The negative charge is due to the three lone pairs in O (instead of two).


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Simulation of Miller-Urey experience
The Miller-Urey experiment shows that it easy in a primordial atmosphere
similar to early earth to generate amino acids,my question: is it theoretically possible to simulate this experience on any computer ?

It would be pretty hard to do on an abacus, but that is a time constraint. Any Turing machine is, theoretically, capable of implementing a chemistry computation that simulates something.

Good question! There are two attempts that I know of: https://doi.org/10.1038/nchem.2099 and https://doi.org/10.26434/chemrxiv.7660532.v1 The primary motivation in both papers is to showcase a theoretical method rather than exploring the origin of life.

Update: https://doi.org/10.1073/pnas.1402894111 HT @fxcourder on Twitter

Theoretically you can simulate any reaction an a computer, provided it is big enough of course.
Practically however, this should be close to impossible. Here are some thoughts on how to approach that and their related problems.
First, one would have to model a complex reaction mixture including the educts $\ce{H2O}$, $\ce{CH4}$, $\ce{NH3}$, and $\ce{H2}$. Simulating those alone, without considering reactions, should be possible using molecular mechanics. This should scale up to a couple of thousands of these simple molecules.
Modelling all the possible reactions requires quantum mechanical approaches. But this does not scale anywhere close to macroscopic quantities.
One could model all the individual possible reactions one at a time. Then it is possible to compare how fast they are and how much energy is required. From that one could deduce how likely a certain reaction is, and feed this into the molecular mechanics calculation mentioned above.
First one would have to figure out all possible product of the reaction mixture, including all intermediates. This alone should yield a huge number of candidate compounds. But maybe one could try to focus just on a single product and one path there. Based on experimental results, pick a target compound of interest and a possible reaction path/mechanism. Experiments should give information on both, products and intermediates.
But even simulating a predefined reaction path (i.e. the educts, intermediates and final product) is a challenge. Different relative positions of the atoms in a reaction need to be considered (the energy landscape along the reaction coordinate). Ab initio molecular dynamics (AIMD) is an approach to do that. But there are still a number of approximations involved. This is costly and of limited accuracy.
We also need to consider the spark from the electrodes. This will lead to electronically excited state and ionic intermediates. This adds quite a lot of complexity to the already difficult task.
In summary, tackling the whole thing at once is impossible. One would have to break it down into many small steps, maybe then it becomes feasible to focus only on very specific aspects.


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Lewis structure of ammonium ion
What would be the lewis structure for ammonium ion?
Is there a coordinate covalent bond as well in the structure?
My confusuion would be dealth with if i get the structure of ammonium cyanate as well.

Will there be any lone pair left to nitrogen after the compound is formed?

Coordinate covalent bonds do not exist.  They are all plain old covalent bonds.  We use "coordinate covalent bonds" as a means of counting electrons from atoms that go into bonds; here nitrogen contributes five electrons to four bonds so we figure that the nitrogen must have contributed two electrons to one of the bonds.  But the actual bonds formed between the nitrogen and each hydrogen atom are all alike.  Ammonium ion has the same tetrahedral structure as methane, and for methane we do not need coordinate covalent bonds to count the electrons.


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Wouldn't adding a proton acceptor still make a solution just as acidic?
Say if you add NH3 to a solution, and it accepts a proton, it still becomes NH4+ in a solution, which is also acidic. A base should decrease the hydrogen ion concentration in a solution, but how does a solution become more basic if it now has an increased concentration of NH4+?

When $NH_3$ accepts a proton, what molecule does the proton come from, and what does this molecule look like after losing a proton? Are there any other acids or bases in the solution, and how will their concentrations change?

Indeed, H+ is acidic, and so is NH4+, but there is a catch. Indeed, Mike Tyson is strong, and so is an elephant, and so is a nuclear blast, and so is the black hole's gravity. One may be tempted to say they are all the same, but there is a catch.

Severe misconception. Hydrolysis reaction is $\ce{NH3 + H2O ⇌ NH4+ + OH-}$
Hence, dissolving $\ce{NH3}$ in water produces $\ce{OH-}$ ions, making the solution more basic, as $\ce{[OH^{-}] > [H+]}$

Just to elaborate more on what Mr Organic has said.
Basically what made a solution acidic is when $\ce{[H3O+]>[OH^{-}]}$ and what made a solution alkaline is when $\ce{[H3O+]<[OH^{-}]}$.
You are right about $\ce{NH4^{+}}$ is acidic, but why it is acidic? Because you need to remember the following happen when $\ce{NH4^{+}}$ is dissolved water and caused $\ce{[H3O+]}$ to increase $$\ce{NH4^{+} + H2O ⇌ NH3 + H3O+}$$
But then you would say, the $\ce{NH3}$ will accept a proton again and decrease the $\ce{[H3O+]}$. The thing is the reaction is a reversible reaction and will eventually become equilibrium which means the concentration of all reactants and products will not changed. And at that moment, the increase of $\ce{[H3O+]}$ caused by $\ce{NH4^{+}}$ is far more than the decrease of $\ce{[H3O+]}$ caused by $\ce{NH3}$. Thus $\ce{[H3O+]>[OH^{-}]}$ and make the solution acidic.
Now back to your question. First the $\ce{NH3}$ will increase the $\ce{[OH^{-}]}$ by the following reaction $$\ce{NH3 + H2O ⇌ NH4^{+} + OH^{-}}$$.
But the $\ce{NH4^{+}}$ will release proton and increase $\ce{[H3O+]}$. However, the increase of $\ce{[OH^{-}]}$ caused by $\ce{NH3}$ is far more than the increase of $\ce{[H3O+]}$ caused by $\ce{NH4^{+}}$ . Thus $\ce{[H3O+]<[OH^{-}]}$ and make the solution alkaline


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Perpendicular orbitals in aryl and vinyl carbocations?

This point (iv) has been given under reasons for extremely low reactivity of aryl and vinyl halides. I cannot understand it. First of all, isn't the positive carbon atom in phenyl as well as vinyl carbocation sp-hybridized. But in the point, the positively charged carbon atom of the phenyl carbocation is referred to as sp2-hybridized. I cannot understand why. Also I cannot understand the "perpendicular part". That the orbitals are perpendicular and so the carbocation is not formed. But how are they perpendicular and how does this hinder the formation of carbocation. I cannot understand the last diagram (for vinyl carbocation). Please help.

No, I do not think my question is a duplicate. My concerns are a bit different. If you read my question properly, you can understand. My question is mainly about why the perpendicularity of the orbitals destabilizes the carbocation, the hybridization of the carbon atoms and the meaning of the last diagram.

Well here is a simple answer to your question.
Consider the resonating structures of aryl halide and vinyl halides. You will find that the C-X(carbon-halide) bond will have partial double bond character. Now double bonds are more powerful than single bonds. Hence they will not break easily and hence aryl and vinyl halides are less reactive towards neucleophillic substitution reaction.

As depicted, it is unlikely that the halobenzene splits to yield the phenyl cation, because the phenyl cation is of very high energy.  It is of very high enery since the then empty C(sp2) orbital is not stabilized by the neighbouring C(p) orbitals (which otherwise create the continous cyclic cloud of electron density above / below the plane of the benzene molecule). Since the empty orbital (formal positive charge) is pointing away from the molecule -- think as pointing into the direction of the formerly bond halogen atom -- there is no stabilization of this orbital possible.  A better visualization is the following one:

(credit:  Grossmann book, 2nd edition, p. 109.  To match your case, think substitutent R equally stands for a hydrogen .)
While the dissociation drawn in your reference for halogen is unlikely, I would however object the idea that aryl cations can not exist.  The intermediate generation of the aryl diazonium ion at low temperatures, typically followed by release of nitrogen gas (triple bond $\ce{N#N}$) is one of rare cases where this high energy situation may be compensated, offering a wide array of subsequent chemistry.

(credit: source)


text:
most frequently used organic (name) reactions
The situation: Beside the relevant inventory in the Merck index, organic-chemistry.org lists 243 name reactions (as counted on 08-Feb-2018), the compilation by Kurti and Czako includes "250 selections" (to quote K.C. Nicolaou's introduction), Li's general reference on the topic "covers over 320 classical as well as contemporary name reactions" yet he equally compiled name reactions relevant to heterocyclic compounds to mention a few possible sources.  There may be a significant overlap of content, though.  The discern of nucleophile and electrophile, as pointed out here, or sometimes anthromorphisms like
"What would you do if *you* were an electron"

(Grossmann book) to ease drawing mechanisms are ackowledged, too.
The question: But, isn't there a more condensed list consisting of maybe 50 to 100 organic name reactions most often used reliably in small molecule organic chemistry?  As one possibility for "reliable" I think of reactions potentially used on larger scale (kg-lab, or journals like Org. Process Res. Dev.) or the ones suggested by computers in the course of an automated retrosynthetic analysis.

Even though I admit I like the question, I doubt there is a "Leeloo" for "Leeloominaï Lekatariba Lamina-Tchaï Ekbat De Sebat" of all name reactions. Probably defining a field, or how small the "small molecules" are (I suspect you use a definition used by crystallographers) might be helpful.

@andselisk Frankly, "Leeloo" is new for me and quite to the contrary, rather than collecting all reactions (as if Diderot and d'Alembert were re-born chemists ...) I speculated e.g. Org. Process Rev. Dev. as primary literature already were surveyed like "a typical issue / volume of reactions published consists of 20% reductive amination, 10% Michael/Knoevenagel/Perkin, 5% esterification, 2% Finkelstein, etc.".  Org. Syn. seems to me less suitable here.  Eventually, "small organic molecule", correct, influenced by crystallography, excludes here typical polymers, proteïn / DNA / RNA.


text:
product of the reaction iodine with alcohol
I have a multiple choice question.
The product from the reaction of iodine monochloride, ICl, with pent-1-ene is:
A CH3CH2CHICHClCH3
B CH3CH2CH2CHICH2Cl
C CH3CH2CH2CHClCH2I
D CH3CH2CHClCHICH3
the correct answer is C but I really don´t get it at all. Does this have to do with electrophile and nucleophiles?

The answer will be (C) option.
As we know Cl is more electronegative than I hence ICl will break as I+, Cl-.
Now the double bond will break and positive charge will appear on the second carbon ( this is due to higher inductive effect of 2 alkyl groups), whereas the negative charge will appear on the first carbon. You may think of rearrangement but that will not happen here because the degree of the carbocation will not change (rearrangement occurs only in case of increase in degree of carbocation).
Now join the positive and negative charges and you will get your answer.
HOPE THIS HELPS.(IF I AM WRONG PLEASE DO INFORM ME. BUT I AM 99% SURE OF THIS ANSWER)


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How do I use the priority list of substituent groups to decide the parent chain for this compound?

I know that the -OH group has a higher priority than the -Cl group, but does that mean I should favour including it in my parent chain even if there's another chain that has a lower locant on the first substituent group?
My intuition is that the chain with the lower locant should be favoured. Is my reasoning in line with IUPAC rules?

You misnumbered the image on the right. There are 4 carbons in that chain.

The most important simplified criteria for the choice of a principal chain are:

greater number of substituents corresponding to the suffix (principal characteristic group)
longest chain
greater number of multiple bonds
lower locants for suffixes
lower locants for multiple bonds
greater number of prefixes
lower locants for prefixes
lower locants for substituents cited first as a prefix in the name

The corresponding wording of the rules taken from Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) is as follows.

P-44.1 SENIORITY ORDER FOR PARENT STRUCTURES
When there is a choice, the senior parent structure is chosen by applying the following criteria, in order, until a decision is reached. These criteria must always be applied before those applicable to rings and ring systems (see P-44.2) and to chains (see P-44.3). Then criteria applicable to both chains and rings or ring systems given in P-44.4 are considered.
P-44.1.1 The senior parent structure has the maximum number of substituents corresponding to the principal characteristic group (suffix) or senior parent hydride in accord with the seniority of classes (P-41) and the seniority of suffixes (P-43).
(…)
P-44.3.2 The principal chain has the greater number of skeletal atoms [criterion (b) in P-44.3].
(…)
P-44.4.1 If the criteria of P-44.1 through P-44.3, where applicable, do not effect a choice of a senior parent structure, the following criteria are applied successively until there are no alternatives remaining. These criteria are illustrated in P-44.4.1.1 through P-44.4.1.12.
The senior ring, ring system, or principal chain:
(a) has the greater number of multiple bonds (P-44.4.1.1);
(b) has the greater number of double bonds (P-44.4.1.2);
(…)
(h) has the lower locant for an attached group expressed as a suffix (P-44.4.1.8);
(…)
(j) has the lower locant(s) for endings or prefixes that express changes in the level of hydrogenation, i.e., for ‘ene’ and ‘yne’ endings and ‘hydro/dehydro’ prefixes (P-44.4.1.10);
(…)
P-45.2.1 The preferred IUPAC name is based on the senior parent structure that has the maximum number of substituents cited as prefixes (other than ‘hydro/dehydro’) to the parent structure.
P-45.2.2 The preferred IUPAC name is based on the senior parent structure that has the lower locant or set of locants for substituents cited as prefixes (other than ‘hydro/dehydro’) to the parent structure.
P-45.2.3 The preferred IUPAC name is based on the senior parent structure that has the lower locant or set of locants for substituents cited as prefixes to the parent structure (other than ‘hydro/dehydro’ prefixes) in their order of citation in the name.
(…)

So, yes, the priority list of substituent groups has to be considered first in order to decide the parent chain for this compound according to Rule P-44.1.1. However, you have already done that in both of your suggested examples: The principal characteristic group that is expressed as a suffix is the acid group ($\ce{-COOH}$). Therefore, the parent structure is a butanoic acid.
Looking at your two examples, your actual question is about which of the two possible butanoic acids is the principal chain: 3-chloro-2-(2-hydroxyethyl)butanoic acid or 2-(1-chloroethyl)-4-hydroxybutanoic acid. Thus, your actual question is about how to identify the principal chain after you have already considered the priority order of functional groups.
Both possible chains have the same length; therefore, the next rule (P-44.3.2) cannot be used to make a decision.
Both possible chains don’t have any multiple bonds; therefore, also Rule P-44.4.1 (a and b) doesn’t help.
In both possible chains, the principal characteristic group ($\ce{-COOH}$) is located at the start of the chain; thus, also Rule P-44.4.1 (h) doesn’t make any difference.
The number of substituents expressed as prefixes is the same (two) in both cases: chloro and 2-hydroxyethyl compared to chloroethyl and hydroxy; therefore, Rule P-45.2.1 cannot decide which chain is the principal one.
The decision is finally made using Rule P-45.2.2 since the locants et ‘2,3’ in 3-chloro-2-(2-hydroxyethyl)butanoic acid is lower than ‘2,4’ in 2-(1-chloroethyl)-4-hydroxybutanoic acid.
Therefore, the preferred IUPAC name for the compound that is given in the question is 3-chloro-2-(2-hydroxyethyl)butanoic acid.


text:
What does x mean in (1-x)A – xB composite materials, molar ratio, weight ratio or volume fraction?
I've been collecting data about ceramic materials and often I get to see composite materials are described as $(1-x)\ce{A} - x\ce{B}$. I am confused what this $x$ means. And I haven't seen any of the papers mention what $x$ is unless %wt is given specifically.
For an example in $(1-x)\ce{MgO} - x\ce{LiF}$ ceramic material, does $x$ mean molar ratio or weight ratio? For me it doesn't make sense to think of $x$ as a volume fraction as most of the times powders of these materials are taken for sintering.

If nothing is mentioned, is it all right to assume $x$ as the molar ratio?
Can we assume volume fraction is equal to molar ratio even in solids or liquids?

Here, $x$ is a variable $(x\in[0;1])$ denoting mole fraction of a given constituent in a phase of variable composition (non-stoichiometric compound).
This is an algebraic way of denoting a given compound from the phase diagram for a solid solution: $x$ corresponds to its abscissa axis.
And no, volume fraction is not equal to mole fraction and is overall a very inconvenient variable to use, so it's usually being avoided.
For example, volume fraction would deviate for the different morphologies and allotropes of the same solid, which is not useful for the description of chemical composition, not to mention it's also temperature-dependent.

Since the OP refers to sintering I don't think that mole fraction can be assumed. In fact the OP mentions that some articles explicitly specify wt %.

@MaxW Mole fraction is pretty much a default in the absence of additional remarks.  Of course, if there is %wt notation, then it's mass fraction.

@MaxW I believe almost all the ceramic compounds are made by sintering. Do you believe 0.8MgO - 0.2LiF could mean weight fractions? Most of the related papers don't mention what x is, although when they do mention about 'weights', they make sure to present it as a percentage, like MgO - 20% LiF. But I'm sure there should be some convention among scientists to understand what x really is unless explicitly specified?

@avin Again, unless there are any specific notes, $(1-x)\ce{MgO} - x\ce{LiF} ≡ \ce{Li_xMg_{(1-x)}O_{(1-x)}F_x}$, so by default $x$ implicates mole fraction.


text:
Does ammonium react with water in an ammonia solution?
In the following equilibrium:
$$\ce{NH3 + H2O <=> NH4+ + OH-}$$
Is it possible for the ammonium to simultaneously establish another reaction with water?
$$\ce{NH4+ + H2O <=> NH3 + H3O+}$$
If so, we would enter a cycle where the additional concentration of ammonia resulting from the second reaction would, in turn, shift the first equilibrium to the right according to Le Châtelier's principle. I surmise that this system will eventually stabilize yielding hydroxide ions (with a negligible concentration of hydronium ions), hence the basic character of ammonia. Are these valid conclusions? Thanks in advance.

Possible duplicate of Is there a difference between ‘ammonium hydroxide’ and ‘aqueous ammonia’?

Everything is in equilibrium at the same time. I think you are confusing what happens to a single molecule, but the equilibrium process happens in bulk.

Oh, ok. I see. When the equilibrium is reached in the first reaction, there is no such thing as a second reaction because the system is already in equilibrium? Am I correct?

To be complete, you should also include the reaction $\ce{H3O+ + HO- <=> 2H2O}$. Perhaps then it will be more clear how everything equilibrates.

@Andrew If you add up the two reactions given by the OP, you get the auto dissociation of water. It is a bit hidden, but already there. Alternatively, you could keep the auto dissociation reaction and drop the 2nd reaction given by the OP to remove the redundancy.

@Karsten Theis I realize it's already implicity included, but I thought that writing it out explicitly might help the OP see more clearly all of the reactions that are going on simultaneously. But it's good to point out that it's redundant mathematically.

@Karsten Theis Thank you for your answer! Then, is it true that the acid dissociation constant of ammonia factors in, so to speak, the secondary equilibrium reaction that takes place simultaneously in the system? That is, its value corresponds to the equilibrium of the system as a whole.

@Andrew Understood, thanks for writing it out for me. Would it be correct to say that the first reaction is representative of the state of the whole system? If the first reaction has reached equilibrium, this must also be the case for all the other secondary reactions that occur in the system.

Yes, as Karsten Theis noted in his comment below your answer, if this reaction has reached equilibrium, the others must have as well.

Is it possible for the ammonium to simultaneously establish another reaction with water?

It might help to add these two reactions to see what the net effect is:
$$\ce{NH3 + H2O <=> NH4+ + OH-}$$
$$\ce{NH4+ + H2O <=> NH3 + H3O+}$$
$$+\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ $$
$$\ce{2H2O <=> H3O+ + OH-}$$
So yes, this is the autodissociation of water, and this reaction goes on at the same time. The three equilibrium constants are linked (K1 x K2 = Kw), so there is never a situation where two of these reactions are at equilibrium and the third one isn't, causing an endless change in concentrations. In fact, for reactions in solution, there is always a state where all of the reactions are at equilibrium, and if all reactions proceed, they will all reach equilibrium eventually.
You can take any reaction that includes hydronium (or hydroxide) ions and formulate another reaction that includes hydroxide (or hydronium) ions instead by combining it with the autodissociation reaction.

If so, we would enter a cycle where the additional concentration of ammonia resulting from the second reaction would, in turn, shift the first equilibrium to the right according to Le Châtelier's principle.

There is no cycle, unless you call reactants making products and products making reactants at the same time a cycle. The latter is a feature of all reactions that are at equilibrium. If you label a hydrogen, it will forever cycle between being part of ammonia, ammonium, hydroxide, water, and hydronium. There are some more complicated systems of reactions where the path toward equilibrium involves oscillations of concentrations, but in most cases, concentrations approach the equilibrium concentration in a monotonic way.

Since the equilibrium has been reached in the first reaction, the concentration of ammonium no longer varies and therefore there is no reaction with water. I was confusing this with salt hydrolysis, where the ammonium resulting from a salt does react with water because salts are strong electrolytes and there is no equilibrium involved.

That's a misconception. If you have multiple equilibrium reactions in the same system, all have to reach equilibrium before changes in concentrations cease. Otherwise, the first equilibrium will be disturbed as the second equilibrium process changes concentrations to reach its equilibrium.


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How can I determine the highest boiling point given a list of molecules?

I know that the highest boiling point has to do with which has the strongest intermolecular force. I also know that the strongest would be ionic, then hydrogen bonding, then dipole-dipole, then london dispersion.
Can someone show me how I am supposed to solve these questions step by step though?
Am I supposed to create the Lewis dot structure for all of them and then see which is nonpolar vs. polar?
Also, once I come up with the Lewis structure, is that all I need to determine polarity? I just look to see if it's symmetric or not?
I know the 3D VSEPR shapes, so am I supposed to come up with that and then see which is nonpolar or polar?
Can someone provide me with a step by step solution as to how I can solve any of these problems from start to finish so that I have a good methodology?
Thank you!

:/ It's impossible to do without some prior knowledge. If you have it, drawing anything would be usually a waste of time. Without it you might run out of time on exam and still put wrong answer.

Q1 and Q2 have the same approach. One notices that they are homologous alkanes and picks the biggest and the smallest, having the most London dispersion/van der Waals force and least, respectively. Other forces are known not to be present between alkanes.$^1$ (There is a small difference between branched and linear alkanes, but that is negligible compared to another $\ce{CH2}$ group.)
For Q3, one sees that they are combinations of hydrogen and halogen. While van der Waals forces are also present here, they are likely to be dominated by dipole-dipole interactions and hydrogen bonding (where applicable).$^2$ Which of the two is actually determining here, does not matter, since both point to a high boiling point in $\ce{HF}$.
In Q4, one notices that the question deals in homo-atomic molecules, so there can be neither hydrogen bonding nor dipole-dipole. Thus, the more electrons, the easier it is to polarize the molecule, and the heaviest molecule wins.
To summarize: To me, the questions are more about test-taking than actual chemistry. I cannot say whether the boiling point of $\ce{C4H10}$ or of $\ce{HF}$ is higher without looking up the values or knowing them and I do not expect most chemists to. Look for which forces may apply and reason about the trends. To do so, one may have to derive the molecular structure, but I would expect that in a test, one either does not need to or has the time to do so.

$^1$ This is where practice and knowledge are indispensable, which is unfortunate for the learner.
$^2$ In this case, this is trivial, but there are cases where exact quantitative calculations would be necessary. One can construct systems in which van der Waals forces outweigh ionic repulsion.

Your rule of thumb that the smallest (lowest MW) alkane will have the lowest BP and the largest (highest MW) will have the largest is just wrong. It works in this case because the molecules are all very small, "essentially" linear. Branching in higher weight alkanes makes it impossible to predict BP solely on the basis of MW. For $\ce{C5H12}$, neopentane has a BP of 9.5 °C, but n-pentane has a BP 35.9 to 36.3 °C. N-butane has  a BP of −1 to 1 °C, but isobutane has a BP of  −11.7 °C.

Yet the pentanes all boil higher than the butanes according to MaxW's list.

My point was that the "rule" doesn't work for higher alkanes because there is a significant spread in the BP due to branching.

It is true that Q1 and Q2 can be addressed having the same approach since all of them are hydrocarbons, which do not have any other forces acting on them other than the London dispersion and van der Waals forces. Also, they are homologous alkanes, which increase those forces according to their size. Yet, these forces also depend on how branched their molecular structures are. Therefore, I have to agree with MaxW's concern that isomerization on boiling points has an upper hand over increasing $-\ce{CH2}-$ units. For example, compare boiling point of n-decane ($\ce{C10H22}$), which is $\pu{174.1 ^{\mathrm{o}}C}$ with boiling point of 2,2,4-trimethyl-3-isopropylpentane ($\ce{C11H24}$) that is $\pu{173.3 ^{\mathrm{o}}C}$ (Ref.1). Even lower molecular weight examples are given in Ref.1: Easily, if you look at $\ce{C10H20}$ vs $\ce{C9H20}$ isomers, you find the boiling points of 2,2,5,5-hexamethylhexane, 2,2,4,5-hexamethylhexane, and 2,2,3,5-hexamethylhexane, all of which are $\ce{C10H22}$ isomers, are $137.5, 147.9,$ and $\pu{148.4 ^{\mathrm{o}}C}$, respectively. Yet, the boiling point of n-nonane, which has  one less $\ce{CH2}$ group ($\ce{C9H20}$ isomer), is $\pu{150.8 ^{\mathrm{o}}C}$. Another isomer of $\ce{C10H22}$, 2,4,6-trimethylheptane has boiling point of $\pu{144.8 ^{\mathrm{o}}C}$, which is also lower than that of n-nonane.
All of these examples means, in order to predict boiling point simply by molecular weights, we may need additional information such as molecules structural features, etc. We could have used the trend of lower atomic (molar) mass having a lower boiling point on Q1 and Q2, only if it states that answers (C) and (D) are n-$\ce{C5H12}$ and n-$\ce{C4H10}$, respectively (or i-$\ce{C5H12}$ and i-$\ce{C4H10}$, respectively). Simply, needs to say all have similar structural features.
The trend of lower atomic (molar) mass having a lower boiling point can be applied only to Q4 because all answers are homo diatomic molecules.
In the case of Q3, there is another force acting of $\ce{HF}$ other than London dispersion and van der Waals forces: H-bonding, which would overcome effects on molecular mass and other forces in this case. The simplest reason is $\ce{F}$ the highest electrnegative atom and capable of having strongest hydrogen bonding. If you Googgled, you may find boiling point of $\ce{HF}$ is $\pu{19.5 ^{\mathrm{o}}C}$, while that of $\ce{HI}$ and $\ce{HBr}$ are $\pu{−35.4 ^{\mathrm{o}}C}$ and $\pu{−66 ^{\mathrm{o}}C}$, respectively.
Reference:

S. W. Ferris, Handbook of Hydrocarbons; Academic Press, Inc. Publishers: New York, NY, 1955, pp. 187.


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Where does the potential energy of capillary action come from?
Capillary action is when water pulls itself up a paper towel, glass tube or a tree. It is going up, which results in an increase in potential energy. According to the law of conversation of energy, energy cannot be created or destroyed. Where does this potential energy of capillary action come from, or is this an exception or something? My question is different from How does water defy gravity in paper towels?, because the other question is asking why it moves up, but mine is asking where the energy comes from.

The increase in gravitational potential energy is countered by a reduction in the chemical potential of the substances due to a change in the areas of mutual contact.

"Water at the surface of a droplet is unhappy" (in a state of high surface tension)

Wetting a surface is an endothermic process. The ultimate energy source is the thermal energy in the components.

please check J Chem Ed 2004, vol 81, p. 150

I believe The Young-Dupre equation describes this. You could rationalize the climb of the liquid as being caused by the release of energy that is associated with a phenomenon known as wetting (see image).

You can relate the wetting angle of a droplet on a solid material to the surface energies of the different interfaces (gas-liquid, liquid solid, and solid-gas). These energies (labeled as sigmas) are listed as energies per unit area. This means that for every unit of area I increase the droplet by, I have to decrease the other areas by a similar amount.
This often releases some energy (from the newly wetted solid-liquid area and the increase in liquid-gas area), but it costs some amount of energy to disrupt the solid-gas interaction too. We balance these energy factors with terms of sigma × the respective area.
Now, if we want to get an idea as to whether a droplet will climb, we can look at the droplet's wetting behavior. If the droplet wets the solid (say water and paper) with an angle less than 90 degrees, then we can conclude that wetting is favorable.
Accordingly, favorable wetting releases an amount of energy (usually as heat, usually small). However if the liquid must climb up the walls of a tube in order continue to wet, it can do so spontaneously, to a point. I am personally not sure what that point is, but surely involves balancing the surface energies and their respective areas, given some knowledge about the geometry of the surfaces.
To answer your question explicitly though, the source of that increased potential energy comes from the energy released upon wetting.

Where does this potential energy of capillary action come from, or is this an exception or something?

First, the laws of thermodynamics still apply. Second, you have to account for the energy balance when you go from dry to wet on the one hand, and continuous flow on the other hand.

Initial wetting of the paper towel or glass tube: Quoting from one of the comments, "the increase in gravitational potential energy is countered by a reduction in the chemical potential of the substances due to a change in the areas of mutual contact". That explains why the capillary fills up in the first place, and in this case it is fine to say "water pulls itself up".
Continuous flow. If you are asking about a tree, where there is continual flow, you are not increasing the water-capillary contact area anymore, it remains constant. The energy has to come from somewhere else.
It takes some energy to remove a water molecule from the top of the
capillary because as you do, you are pulling all the water molecules
up a tiny bit. As explained in the excellent video by Veritasium, water can have negative pressure in the capillary, so trees can grow higher than atmospheric pressure might suggest (thanks, A.K., for the pointer to the video). This is not a problem for the tree because the top of
the tree is the site of photosynthesis, so there is free energy
available to do the "pulling". So you can think of the water-filled
capillary serving both in the role of the rope and the bucket, if
you compare it to fetching water from a deep well. In cases where the water simply evaporates, the energy comes from cooling down the surrounding as the water molecule pulls up another water molecule and evaporates. In the case of continuous flow, water does not "pull itself up", something on the top (or bottom) has to provide the energy. For example, if you put a towel in a bucket of water with the towel reaching over the rim a bit, the towel will get wet. However, if the end of the towel outside of the bucket reaches below the water level, it will continuously siphon water out of the bucket, with water dripping from the towel.

As an aside, it does not take much work to lift a water molecule up a tree (1.8 J/mol for a 10 m tree). Hydrolysis of a single ATP affords about 50 kJ/mol under physiological conditions, so it is sufficient to get 25,000 water molecules up the tree.Gravity is weak compared to electromagnetic interactions, which dominate in chemistry.

"Electromagnetic interactions" doesn't do the nature of those interactions (Van der Waals, Hydrogen bonding, and Dipole-Dipole interactions) any justice. While accurate on an molecular level, it's misleadingly sparse and overly general.

I was talking about the four physical forces: gravity, electromagnetism, weak force and strong force. If you had to pick from those four to explain chemistry, which one would you pick?

Chemistry is all electromagnetic forces, the attractions of electrons to nuclei. the question is: "Which electrons and which nuclei?" Gravity is just a complication Think of it as an invisible container. Then transfer your experiment to the Space Station where its effects are minimal to vanishing.

None of the answers answer your question, they just layout equations of how to calculate the force/energy. To understand where the force comes from, you have to understand a surface  conceptually. Think of the energy input required to make a surface (break something). It takes energy input to break the bonds between atoms to make a surface, thus a surface has an inherent energy. In fact image breaking the object in pieces until you only had atoms. Think: how much energy would that take? If we recognize that bunch of unbonded atoms are in a gas state, then we can know that this energy is the same as the heat of sublimation. Unless your material is a gas at equilibrium conditions, it will want to reduce surface area to release the energy inputed to break apart the material. This is why water in space forms a sphere, it is trying to minimize energy which means minimizing the amount of surface.
Now consider a solid-liquid interface. If the surface energy (enthalpy of sublimation) is higher for the solid than the liquid, then when the liquid covers the solid energy is released by the destruction of the solid-gas (or solid-vacuum) interface, but some of that energy must go into creating more surface area of the liquid-gas interface. This will cause the liquid to sprawl out and is referred to as wetting. If the surface energy of the solid is less than the liquid, then energy will be released by breaking the solid-liquid interface which will cause the liquid to contract forming a bead which is non-wetting behavior (a very creative term).

The driving force is the difference in waters chemical potential in the capillary and outside it. At equilibrium they must be equal, so water will move up the capillary until chemical potentials are equal. It is the same in osmosis. No need for equations, it is thermodynamics. If equations are desired
$$\mu_{\rm water, capillary} = \mu_{\rm water, bulk}$$
For more on the chemical potential: wiki/chemical potential
Chemical potential is a form of work. In my opinion, the most important form of work in chemistry and chemical engineering.


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Solubility of dimethyl ether in aliphatic hydrocarbons
I'm having trouble finding information on the solubility of dimethyl ether in a hydrocarbon like butane, propane, or isobutane.
I've read that its soluble in methanol, ethanol, isoproanol, chlorinated hydrocarbons, and toluene but cant find much else.
Assuming 1 atmosphere of pressure and both chemicals are at a temperature of -11°F, would the then liquid isobutane be misicible with liquid Dimethyl ether?

DME is a polar aprotic compound; hydrocarbons like butane or propane are for the major part, non-polar; like dissolves like.

Your choice for the example for the miscibility with butane and propane is strange as both are gases at STP. Are you asking about solubility at higher pressures?

@andselisk : And dimethyl ether  is a gas at STP as well. Yet, if OP is asking solubility of gases, it's okay.

I was asking about solubility of these solvents as liquids at higher pressure but I've got no specific pressure value to give you for that question. So in an attempt to simplify this question, lets use Isobutane and Dimethyl ether as the example gasses since they both have a boiling point of -11°F assuming it's cold enough for the gasses to be liquid would one dissolve in the other in significant amounts?

Would be better to take hexane and diethylether so you focus to the core of your curiosity. @YUSUF HASAN true but common ethers go smoothly with hexane, cyclohexane, toluene, petroleum ether(s), ...

Since dimethyl ether is a very uncommon solvent, it's not a big surprise that you have hard time finding its information just using Google. However, I found two relatively important parameters of dimethyl ether: Its Hildebrand Solubility Parameter is 7.3 and Kauri-Butanol index is 90 (INVENTEC). For comparision, Hildebrand Solubility Parameter of diethyl ether is 7.62 (Solubility Parameters: Theory and Application). According to this reference, if two solvents have about the same Hildebrand values, theoretically, such solvents should have solubility behaviors similar to each other.
Thus, dimethyl ether should behave similar to diethyl ether (in liquid phase, I believe).  Keep in mind that most of hydrocarbons dissolve in diethyl ether in all ratios. For example, we run column chromatography using such solvent systems (diethyl ether/hexanes) to isolate close nonpolar organic compounds from complex mixtures.


text:
When a piston cylinder system that evolves isothermally and quasistatically is irreversible?
Suppose we have two states: $\ce{A}$ and $\ce{B}$.
Considering a quasi-static isothermal process of an ideal gas, (and another condition, that is $T = T_\mathrm{surr}$) that join those states.
Then the Poisson's relation $P_\ce{A}V_\ce{A} = P_\ce{B}V_\ce{B}$ is fulfilled.
that means after integrating
$$W = nRT\ln\left(\frac{V_\ce{B}}{V_\ce{A}}\right)$$
$$\mathrm{d}S_\mathrm{s} = -\mathrm{d}S_\mathrm{surr}$$
because $T = T_\mathrm{surr}$. Then
$$\mathrm{d}S_\mathrm{u} = 0.$$
So my guess is that no, that irreversibilities on those conditions come only on finite temperature difference with surroundings.
But then I remember friction. If piston with friction has $x$ joules lost by friction, how does that change everything?
I need someone to "shed light" on this issue. There is always a lot of confusion when we consider a quasi-static but irreversible process.

With piston friction present, the situation does indeed change for both the quasi static and the non-quasistatic cases.  @BobD and I have been working on problems of this type in another forum:  https://www.physicsforums.com/threads/thermodynamics-gas-expansion-with-piston-friction.963282/  Maybe you would like to see what we have done so far, and join in?

Thanks I'll take a look, kinda confusing when friction appears.

So, when friction appears it seems that work cannot be calculated using Psystem=Pexternal ... But it is OK for internally reversible processes?

I don’t quite follow the question, but you can see some actual problems analyzed in detail in the link I referenced.  And it addresses the issue of precisely what one identifies as the “system” (which goes along with what one identifies as Pexternal).  Please feel free to post in that thread.

If it is done quasi statically with friction involved and, if the system is in contact with a constant temperature bath at the initial system temperature T, then, for the same volume change, the change in entropy of the system will be the same.  That means that all the entropy generated as a result of frictional irreversibility will be transferred to the surroundings (the constant temperature bath).  This will be equal to the frictional work divided by the temperature T.  So the entropy of the universe will increase by this amount.

Got it, thanks. An external irreversibility.

Yes, if you want to call it that.  I've never heard this term used before.  Also, if you define your system as the gas, cylinder, and piston (rather than just the gas), then the irreversibility is internal to the system.  Of course, that doesn't change the answer to the problem.


text:
What library/program can draw molecules in 3D programmatically?
Molecular graphics software is usually interactive.
But I am looking for a library or program that would be able to render a given molecule based on the commands from another program. I would like to give it a molecule description, the vantage point, options, etc, and it should produce the picture without opening any windows.
The input is the same data as contained in an XYZ file (or the file itself), the expected output is a 3D molecule drawing.
What is the easiest way to do this?

From the top of my head: Open Babel; chemfig (LaTeX package); RDKit (Python). Is that what you are after?

Try Jmol, many of the  3D rendered molecules you see on wikipedia use this.

@porphyrin Thank you, Jmol works. Their "jmoldata" command (JmolData.jar based) is able to convert the xyz file to nicely rendered 3D image based on the provided Jmol script.

The "jmoldata" command (using JmolData.jar from Jmol) converts xyz/pdb and many other molecule formats to nicely rendered 3D images based on the user-provided Jmol scripts.
(answer based on the hint from the user @porphyrin)

Before monitors had graphics capabilities, all graphics programs worked that way. For macromolecules, you could use molscript, and for smaller molecules, you could use ORTEP. Both are still available, and are script-driven.
Molscript is open source now and lives on Github. You write a script describing how to show the molecule (there is a generic script you can use, or write one specifically for each structure), and choices of output formats (PostScript, Raster3D and VRML).
For small molecules, a good option would be Jmol in its java standalone version. Using command line options, you can load a file with coordinates, a generic script, turn of the graphics display and choose an output file format.

Could you probably add a brief description/links/examples to make it look more like an answer and not a comment? Also, from what I remember, ORTEP is like half-century old and some implementations of it are proprietary and is overall a pain to work with on both Windows and Linux. I'm not familiar with molscript though.

I used MolScript from 1999 through 2006 to generate figures. It's solid and the requirements are minimal.

@ToddMinehardt MolScript isn't supported any more, and doesn't build on most modern systems.

Back in the computational chemistry course we used wxMacMolPlt to write the basic input for our calculations. When my output file had some errors I would simply copy paste the optimized geometry coordinates back into the software. It won't generate bonds but you can get the 3D model from the coordinates for example. Depending on the software you can also plot the output file using Avogadro if the file format is accepted.


text:
Stability of methoxy-methyl and tropylium carbocations
Which is expected to be more stable: methoxy-methyl carbocation
$(\ce{CH3-O-CH2+})$ or tropylium $(\ce{C7H7+})$ cation?
I think because oxygen would donate its lone pair so there will be complete octet formed that would make it more stable. Is this argument true and can it be applied everywhere against resonance?
Is it wrong to say tropylium cation is exceptionally stable?
What would be its (tropylium cation) comparison against cyclopropyl cation?

Tropylium cation is aromatic and aromaticity is generally a greater stabilisation than most other effects, I guess.

It would be better to think of the first species as a methyl-activated carbonyl than a stabilized carbocation.

Tropilium salts are a real thing, very empirically.


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Should I plate cast iron with copper first before nickel plating?
I have some antique handplane lever caps I would like to retore back to their former glory. They are made of nickel plated cast iron. The plating  pin holed and flaked in areas, creating rust spots.
I am planning on stripping the nickel using hydrochloric acid and electrolysis. I've heard that nickel has a much better affinity for copper than iron. Would it be a good idea to first plate the lever caps with copper and then plate them with nickel for a more durable finish?  Any other suggestions for improving the quality if the finish?

Will you be doing the electroplating yourself? If so please take note that copper plating on iron/cast iron will require a special type of copper plating solution because the normal CuSO4 solution will not work, that is because Copper is more noble than Iron and it will form a flash coat of copper on iron/steel, but that coat is spongy and comes off real easy!

Can you tell us what is in the special solution?

The standard technique is to copper plate then nickel ,then chromium, if you want chrome. The copper protects the steel then the nickel provides a silver surface color ,normally significantly thinner than the copper.

@blacksmith37 Interesting.. I have thought eventual chromium goes directly after copper. Perhaps nickel forms foundation for more smooth and shiny final chromium layer? Or chromium sticks better to nickel than to copper?


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How to apply the Principal of Atom Conservation (POAC)
27.6 g $\ce{K2CO3}$ was treated by a series of reagents so as to convert all of its carbon to $\ce{K2Zn3(Fe(CN)6)2}$. Calculate the weight of the product.
I am asking here is that, why we are  applying POAC only on Carbon atom not on Potassium Atom.

It's very unclear what you are asking. What is "POAC"? Potassium acetate? What do you mean by "convert all of its carbon to K2Zn3[Fe(CN)6]2"? A nuclear fusion? Why did you mention single C and K atoms?

POAC means Principal of atom Conservation

re: Why only to carbon atom? --  Because that is how the problem is worded.

Because in the question it is specifically written
"All of carbon of K2CO3 has been converted ".
Whereas it has not mentioned anything about potassium.
You have to understand that K2CO3 is conberted to K2Zn3(Fe(CN)6)2 by a series of reactions therefore there might be other potassium compounds which had been formed by the series of reactions.
But as it is given all of carbon converted therefore out of the series of reactions there was only one carbon containing compound K2Zn3(Fe(CN)6)2 .
Hope this helps..


text:
Determining the formula of zirconium oxide
I am doing the following question (from the 1988 HSC Chemistry paper):

In the 19th Century the atomic weight of zirconium (Zr) was known but there was considerable doubt about the formula of zirconium oxide. Experiments determined the relative weight of the volatile chloride to be about 233.
Assuming that zirconium in the oxide has the same valency, this evidence suggests that the formula for the oxide is:
(a) $\ce{ZrO}$
(b) $\ce{Zr2O3}$
(c) $\ce{ZrO2}$
(d) $\ce{ZrO4}$

I am not sure what is meant by the sentence 'Experiments determined the relative weight of the volatile chloride to be about 233'. By 'volatile chloride', are they referring to zirconium oxide or something else? It seems that they are referring to something else since a chloride is a compound of chlorine with another element or group.
Also, by the statement 'assuming that zirconium in the oxide has the same valency' are they saying that zirconium has the same valency as chlorine (i.e. -1)?
Any insights are appreciated.

What is the formula for water and thus what is the valency of oxygen? // The salt with an atomic weight is either an oxide or a chloride. It can't be both in this case.

The formula of water is H2O, so the valency of oxygen is 2- since the valency of hydrogen is 1+. Is the salt with an atomic weight a chloride (as that is what it says in the question)?

Why would the question state that the salt with a mw of 233 is a chloride if it isn't?

@ceno980 1. "the chloride" would be whatever chloride compound is formed with zirconium. 2. "the same valency" refers to the valency of zirconium in the zirconium chloride compound.

First, let's solve the problem.
Both relative atomic weight $A_\mathrm{r}$ and relative molecular weight $M_\mathrm{r}$ are a historical dimensionless terms denoting relative atomic mass and molecular mass, respectively:
$$A_\mathrm{r} = \frac{m_\mathrm{a}}{m_\mathrm{u}}$$
$$M_\mathrm{r} = \frac{M_\mathrm{B}}{M_\mathrm{u}}$$
where $m$ corresponds to atomic mass, $M$ – to molecular mass, and index $\mathrm{u}$ refers to unified atomic mass unit ($= m_\mathrm{a}(\ce{^{12}C})/12$).
Since $\ce{Cl}$ in chlorides is monovalent, then formula of unknown zirconium chloride is $\ce{ZrCl_x}$, where $x$ coincides with the valency of $\ce{Zr}$.
From this the relative weight of $\ce{ZrCl_x}$ is
$$M_\mathrm{r}(\ce{\ce{ZrCl_x}}) = A_\mathrm{r}(\ce{Zr}) + xA_\mathrm{r}(\ce{Cl})$$
relative atomic masses of both elements were known ($A_\mathrm{r}(\ce{Zr})\approx 91$, $A_\mathrm{r}(\ce{Cl})\approx 36$), so that the valency of zirconium can be found via $x$:
$$91 + 36x = 233 \implies x \approx 4$$
Since oxygen in oxides has valency of $2$, then the formula of unknown zirconium oxide is $\ce{ZrO2}$.

By 'volatile chloride', are they referring to zirconium oxide or something else?

No, they refer to zirconium chloride $\ce{ZrCl4}$.
They mention volatility probably to underline how zirconium and hafnium are separated — by differences in sublimation of their volatile compounds.

Also, by the statement 'assuming that zirconium in the oxide has the same valency' are they saying that zirconium has the same valency as chlorine (i.e. -1)?

No, this means that zirconium in both oxide $\ce{ZrO2}$ and chloride $\ce{ZrCl4}$ has the same valency.

I'll point out that $\ce{^{12}C}$ wasn't chosen as the standard until the 1960's in the 20th century.

@MaxW You are correct, I didn't want to overload the answer with obsolete standards and took the modern one. Even if you take former $\ce{^{16}O}$ standard, it wouldn't change the answer, but would cause unnecessary confusion IMO.

It is always helpful if you also cite the reference from which you are quoting the question.  The "relative weight" is old terminology which you will find in 19th-century chemistry textbooks which is the same as molecular weight. Since nobody could determine absolute weights (and the practice still continues), the term relative is added. All atomic weights quoted today are also relative but the word relative is often omitted today and perhaps taken for granted. I am adding a screenshot of an early 20th-century book, Introduction to inorganic chemistry
By Alexander Smith
The second part "Assuming that zirconium in the oxide has the same valency" simply implies that the oxidation state of Zr in the chloride is the same as the oxide. This helped the chemists in the laws of definite proportions.


text:
How does increasing temperature reduce the efficiency of solar cells?
I've seen several answers and trying to confirm.
1) Does temperature increase the ground state energy of electrons once they reach the positively doped semiconductor, thus reducing the voltage between excited electrons from negatively doped semiconductor and ground state they will reach on positive side.
-This was the most compelling and easy to understand answer for me.
-However I'm not sure if it makes sense that "ground states" would rise with increasing temp.
2) Does temperature increase vibrations in wires and semiconductors, such that excited electrons following voltage difference lose energy in collisions along their path from N-doped to P-doped?
Combination of the two? Am I way off?
I am a molecular biology student with some basics in photochemistry. Somewhat familiar with solar cells and looking to dive deeper on the subject so feel free to reference.

While perhaps better suited to Physics SE, I will quote from Sze's Physics of Semiconductor Devices:

As the temperature increases, the diffusion lengths in Si and GaAs will increase ... and the minority lifetime increases with temperature. The increase in minority-carrier diffusion length causes an increase in $J_{L}$. However, $V_{oc}$ will rapidly decrease because of the exponential dependence of the saturation current on temperature. The increase in the "softness" (roundness) in the knee of the $I-V$ curve as temperature increases will also degrade the fill factor. Therefore, the overall effect causes a reduction of efficiency as the temperature increases.

Here, $J_{L}$ is the current caused by light generation of electron-hole pairs. So, it goes up. But, the device is still a diode and must follow the standard diode equation, so the open circuit voltage ($V_{oc}$) drops. This then also impacts the power output.
In brief, neither of your proposed reasons is the basis for it. Instead, you have to probe deeper into the physics of semiconductor devices.

Oh great a third approach XP.
Hmm can you translate this information into plainer English (/ describe the concept) to help me get into the subject?

However,in semiconductors, mobility decreases when temperature increases. The second part of the Q suggest a correct answer. Though the final output of the cell is dictated by more factors as addressed in the answer. Additional note/curiosity: organic solar cells (not yet mature) generally show opposite behaviour as for in disordered material conduction is hopping or hopping-assisted, ie scattering by vibrations is now a positive factor.

@Alchimista - The increases in lifetime and diffusion lengths more than compensate for a decrease in mobility, since the light-induced current contribution increases. But, you are absolutely correct that it is the diode equation (through Shockley-Read-Hall) and saturation current that kills the efficiency. For organic cells, I think you get to through SRH out the window and start with a different theory of carrier dynamics.

Ok SRH got to fame for several reasons but the core is kind of Huckel calculation. Not that I wanted to enter the realm of organic solar cells.  The fact that semiconductor physics took over more basic consideration is making even a bigger mess. Those are still speaking of dumpling bonds in conjugated polymers, to give you an idea.  Of course any IV curve can be modelled using concept at work in Si, but the facts behinds can be totally different. Other point: yes my comment was more on mobility alone rather than the specific of the question, ie a traditional solar cell.


text:
3D printing of alumina
I want to print a structure of alumina, and the 3D printer that I am using is an extrusion-based printer. So I need your help in knowing that what composition (binders and deflocculant) of alumina will be suitable to print the 3D structure in such a manner that it holds its structure (as it is printed)? And to what extent should rheological properties be maintained so that structure doesn't fail?

How are you going to turn the slurry into a solid? Existing processes for similar things (eg Saffil fibres) require rapidly heating the slurry in a furnace.

Use methylcellulose(MC) as a binder and Ammonium Polyacrylate (NHPA) as your dispersant or deflocculant. The concentration of methylcellulose can be set according to your requirement as its non-ionic (if you want liquid slurry use MC in large amount and if you don't use less amount). For NHPA you have to make different samples of fixed alumina concentration and NHPA of different concentrations. Check viscosity of each sample and the one which gives you least viscosity means the best-dispersed sample and hence your operating concentration of NHPA.


text:
How do the def2-SV, def2-SV(P), and def2-SVP basis sets differ?
I am trying to calculate the energy of complex molecules with DFT and have a question about the Ahlrichs (Karlsruhe) def2 basis sets.
What is the difference  between the def2-SV, def2-SV(P), and def2-SVP?
Do they only differ in whether the basis set includes split valence polarization or not?

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If you want quantifyable results, you should not use SVP basis sets. They have a very broad error distribution and are generally not really usable for accurate results.

Def2-SVP has the polarisation functions on all atoms, -SV(P) does not have these functions on the hydrogen atoms and -SV does without them.
In future, you can check on the EMSL Basis Set Exchange It wouldn't give you an answer explicitly, but you can view the basis set for different atoms and see how the different sets vary.
A side note - If your 'complex molecules' aren't too large, consider the larger Def2-TZVP basis set. Errors at the Def2-SVP level can be several kcal which may significantly impact your study. As an alternative, consider the PC-n family of basis sets from Frank Jensen. If the Def2-TZVP set is too large, the double-zeta PC-1 might work well.


text:
How could the ideal gas law be discovered from experiments on real gases?
The gas laws, namely Boyle's Law, Charles' Law, Avogadro's Law and Gay-Lussac's Law, are all experimental laws. Combining these laws, we get the ideal gas law $pV=nRT$. Also, "real life" gases do not exactly follow this law, so there are more laws for "real life" gases: van der Waals' law, Dieterici equation, etc., which approximately describe these laws within certain boundaries of the gas parameters: pressure $p$, volume $V$ and temperature $T$.
But there seems to be an apparent logical flaw: the ideal gas law $pV=nRT$  was found by experimenting on "real life" gases, but these "real life" gases do not follow the ideal gas law. How could this be the case?

The simple answer is: because while real matter may not be perfectly ideal, it comes pretty dang close.

The same way Newton and Copernicus discovered simple laws of gravitation, even though real gravity is very slightly different (general relativity, and we still don't have a theory of quantum gravity).  The difference is negligible for systems they could observe, so their model did fit the data.  For gases it's easier to create systems there the difference is non-negligible, but still.

The experiments done used to derive the gas laws (boyle, charles, etc) were done in conditons where real gases behave approximately ideally (very low pressures, for example). Thus, they were able to approximate the behavior of ideal gases.

You also have to understand the historical progression towards the ideal gas law. Boyle published his work in 1662. Charles formulated the original law in his unpublished work from the 1780s. Avogadro hypothesized in 1811 that two given samples of an ideal gas, of the same volume and at the same temperature and pressure, contain the same number of molecules.

Nice that @MaxW made us think of 200 y in between. Sometimes we do not get the real flow of things.  Not chemistry, but a genial intuition such as Olbert paradix seems a stupid assertion if state to someone born in the second half of the last century, for instance.

The ideal gas law is a very good approximation of how gases behave most of the time
There is no logical flaw in the laws. Most gases most of the time behave in a way that is close to the ideal gas equation. And, as long as you recognise the times they don't, the equation is good description of the way they behave.
The ideal gas equations assume that the molecules making up the gas occupy no volume; they have no attractive forces between them and their interactions consists entirely of elastic collisions.
These rules can't explain, for example, why gases ever liquefy (this requires attractive forces). But most of the common gases that were used to develop the laws in the first place (normal atmospheric gases like oxygen or nitrogen) are usually observed far from the point where they do liquefy.
As for the volume taken up by the molecules of the gas, consider this. A mole of liquid nitrogen occupies about 35mL and this is a fair approximation of the volume occupied by the molecules. at STP that same mole of gas occupies a volume of about 22,701mL or about 650 times as much. So, at least to a couple of decimal places, the volume occupied by the nitrogen molecules is negligible.
The point is that for gases not close to the point where they liquefy (and few components of the atmosphere are), the ideal gas laws are a very good approximation for how gases behave and that is what we observe in experiments on them. The fancy and more sophisticated equations describing them are only really required when the gas gets close to liquefaction.

Veering slightly into history-of-science with this one, but were measurements back in those days even accurate enough to detect deviations from ideal behaviour?

@orthocresol I'm not sure it is a good generalisation to say that in the "old days" scientists were less accurate in their measurements. More importantly, just how big are the deviations from ideality for typical atmospheric gases in room-temperature laboratories?

I don't know, and am rather too lazy to calculate a ballpark figure; but I was just curious as to whether technology several centuries ago was good enough to detect them. Maybe 'accurate' was a poor choice of word; substitute 'precise' for it if you prefer. (I'm not disagreeing with anything you said!)

Please note that, at STP according to current IUPAC recommendations, one mole of nitrogen actually has a volume of 22.701 l. The mentioned value of 22.4 l corresponds to the old definition of STP.

@Loong Sorry. I've lost the will to keep up. But I will alter the text to match.

I think it's worth mentioning that the "degree of ideality" is a literal experimental parameter called the compressibility factor, which is equal to exactly 1 for an ideal gas and deviates from unity for a real gas. However, it is clear that the value is typically quite close to 1 for a wide range of common gasses in reasonable conditions.

In the last paragraph, the parenthetical note is a "garden path sentence".  Why are most components of the atmosphere not at room temperature?  Are they hotter, colder, or what?  Oh--few of the components of the atmosphere are close to the point of liquefaction at room temperature.  Perhaps the note would be better as simply "(i.e. most of the components of the atmosphere, at commonly available temperatures)".

You must consider this:

The question whether a physical system follows a particular law is not a "yes or no" question. There is always an error when you compare what you measure with what the law predicts. The error can be at the 17th digit, but it's still there. Let me quote a very insightful passage by H. Jeffreys about this:

It is not true that observed results agree exactly with the predictions made by the laws actually used. The most that the laws do is to predict a variation that accounts for the greater part of the observed variation; it never accounts
  for the whole. The balance is called 'error' and usually quickly  forgotten or altogether disregarded in physical writings, but its existence compels us to say that the laws of applied mathematics do not express the whole of the variation. Their justification cannot be exact mathematical agreement, but only a partial one depending on what fraction of the observed variation in one quantity is accounted for by the variations of the others. [...] A physical law, for practical use, cannot be merely a statement of exact predictions; if it was it would invariably be wrong and would be rejected at the next trial. Quantitative prediction must always be prediction within a margin of uncertainty; the amount of this margin will be different in different cases[...]
The existence of errors of observation seems to have escaped the attention of many philosophers that have discussed the uncertainty principle; this is perhaps because they tend to get their notions of physics from popular writings, and not from works on the combination of observations.

(Theory of Probability, 3rd ed. 1983, §I.1.1, pp. 13–14).

The error can be different in different ranges of the quantities you're measuring.
When trying to guess a physical law, scientists often consider the simplest mathematical expression that's not too far from the data.

The "ideal gas law" is very accurate in the range where the temperature $T$ is high, the pressure $p$ is low, and the mass density $1/V$ (inverse volume) is low. In these ranges the law is very realistic. If you 3D-plot various measurements of $(p,V,T)$ for a fixed amount of a real gas, you'll see that a curved surface given by $pV/T=\text{const}$ fits the points having large $T$, low $p$, low $1/V$ very accurately. "Real" gases are very "ideal" in those ranges.
Moreover, the law involves simple multiplication and division: $pV/T$, so it's one of the first a scientist would try to fit the points.
Finally, the law wasn't just suggested by fitting, but also by more general physical and philosophical principles and beliefs. Take a look at
S. G. Brush: The Kind of Motion We Call Heat (2 volumes, North-Holland 1986)
for a very insightful presentation of the history of this and other laws.

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I like @pglpm  answer, but can't comment on it.
I would like to mention that the technology of those years was not precise enough for very exact measurements, and not good enough for measurements at "boundary" conditions (where differences from the ideal gas law increase greatly in magnitude), so the experimental/measurements errors would hide any deviation from the "ideal gas law".

Suppose you have a crate, and when you put one apple in the crate, the crate + apple weighs 1100 grams. The crate plus two apples weighs 1200 grams. The crate plus three apples weighs 1300 grams. Do you have any guesses as to what one, two, and three apples would weigh without the crate?
The gas laws are more complicated, but it's the same basic principle. If you hypothesize that $pV$ is equal to some interaction between an "ideal" function of $nRT$ and some complicating factor from real gasses, it's not too difficult to take a bunch of data points of different $p$, $V$, $n$, and $T$ for different gasses, and isolate out the "ideal gas component" from the rest of the effects, especially since you can take extreme cases where the ideal gas component dominates.
It's made easier by simply applying intuition to the situation. It's logical that the volume should be proportional to number of molecules. Increasing the temperature should also increase the volume. Increasing the pressure should decrease the volume. Much of the the ideal gas law can be worked out from first principles, and experiments then verify the models (as well as providing the value of $R$).


text:
Why ammonium hydroxide precipitates metal salts?
When I was going through the chapter on pnicotgens, I got to know that ammonia acts as a Lewis base and it dissociates in water giving out hydroxide ion and the further hydroxide of ammonia, which we refer to as ammonium hydroxide, precipitates metal salts in a form of metal hydroxides.
I want to know the reason why this thing happens. According to me, ammonium hydroxide should form a complex with the metal being a base after donating a pair of electrons. Correct me if I am wrong. A reaction is given below as an example of a reaction between ammonium hydroxide and zinc sulfate:
$$\ce{ZnSO4 + 2 (NH3 * H2O) → Zn(OH)2 + (NH4)2SO4}$$

Unfortunately, solubility is nowhere near that simple.

Zinc forms ammine complex $\ce{[Zn(NH3)4]^2+}$ all right in liquid ammonia and when exposed to ammonia gas. It just so happens that in aqueous solution zincate $\ce{[Zn(OH)4]^2−}$ formation is favored.

Not so much $\ce{Zn(OH)_4^{2-}}$ as precipitated $\ce{Zn(OH)_2(s)}$.  Getting the hydroxy-complex anion in large amounts requires a stronger basic solution than that formed by ammonia in water.

Not sure what you are looking for. Ammonia is a ligand that can help solubilize metal ions. Exactly when that happens or doesn't happen isn't that easy to predict from first principles. In this case, you have several equilibria in play which further complicates the situation.

I just want to know that why Ammonium hydroxide precipitate out the metal salt what's the reason behind that because in some metals Ammonia form complexes whereas in others it precipitate them out why this difference arises is it due to the configuration of different metals I mean is it due to the different nature of metals or something else thanks in advance.

Sorry, I didn't realize you replied because you didn't @ me in that comment. The reason is because a metal (really any element) is different from another metal (or element). There's no reason why atoms of different elements should behave the same. Some times, opposing factors are on a knife edge. Tip it one way or the other, and you get different outcomes for two very similar systems. Like I said, solubility is pretty complicated.

To my knowledge, nobody knows why Ni, Co, and others, are making complexes with NH3, and why Fe does not make such complexes with the same NH3.

It's simply because of the concentration. In aqueous solution, the formation of a precipitate is favored over the formation of a complex ion, and it can be shown using solubility equilibria and $K_\mathrm{f}$ (but it won't be pretty).
This is just like the QA test for copper(II). After a small amount of aqueous ammonia is added, $\ce{Cu(OH)2}$ is observed. Add more, and you get the deep blue $\ce{[Cu(NH3)4]^2+}$ complex.

Zinc has amphoteric properties like aluminum. It can dissolve in bases as well as acids. In bases it forms zincate anions. The "why" part of science often has no answer because the reason is that nobody fully understands such a complex phenomenon. It is relatively easier to talk about the observational part.
An aqueous solution of ammonium is very basic, which means that there are free hydroxide ions there. Writing NH3.H2O is deceptive because many general chemistry texts emphasize that NH4OH does not exist and they advocate NH3.H2O. One way to rationalize the observation of the Ksp values for zinc hydroxide i.e. if we have free zinc ions and free hydroxide ions, a solid Zn(OH)2 is favored at that particular pH.


text:
Ab initio MD simulations in VASP
I'm not sure if this is an appropriate question for Chemistry.SE, but I'm taking a stab. I'm running ab initio MD calculations in VASP and I'm trying to optimize different input tags. On the VASP website, it says that the MAXMIX tag should be set to at least 3 times as large as the number of electronic iterations required to converge the first step. I tried doing this, but I didn't see any speed up and instead a significant slow down. I'm wondering if anyone else has encountered this issue.
The system that I'm simulating is a silicon supercell in contact with water.

This site seems to be more up to date and detailed. https://cms.mpi.univie.ac.at/wiki/index.php/MAXMIX . Based on what they say here, it seems like getting a speed up from this setting is relatively system dependent (I have never used VASP so take this with a big grain of salt). It may help users who know more about VASP to have a better sense of what your system is.

@Tyberius Thank you. I suppose you are right, I can't think of any other reason. Thanks!


text:
Nuclear equations for the decay of radon-222 to lead-206
I have to answer this on my worksheet. I understand nuclear chemistry equations a bit, as well as the process of alpha decay and somewhat about what happens when an atom changes its number of protons and neutrons, I understand the gamma symbol and the m symbol meaning the element is high-energy state but this has me stumped. It's an 8 step question. It reads:

The inhalation of radon-222 and its decay to form other isotopes poses a health hazard.
  Write balanced nuclear equations for the decay of radon-222 to lead-206 in eight steps.
  Show step 1 below, and show steps 2-8 are on the next page.
a. Step 1: radon-222 decays by alpha emission.
b. Step 2: the daughter product in part a decays by alpha emission and is in a high energy state.
c. Step 3: the high energy daughter product in part b decays by beta and gamma emissions and is in a high energy state.
d. Step 4: the daughter product in part c decays by beta and gamma emissions and is in a high energy state.
e. Step 5: the high energy daughter product in part d decays by beta emission.
f. Step 6: the daughter product in part e decays by alpha emission and is in a high energy state.
g. Step 7: the high energy daughter product in part f decays by beta and gamma emissions and is in a high energy state.
h. Step 8: the high energy daughter product in part g decays by alpha and gamma emissions.

Did you forget to type out part a) of the question? According to https://en.wikipedia.org/wiki/File:Decay_chain(4n%2B2,_Uranium_series).svg, it should be an alpha decay.

My bad. Part 1 a. Step 1: Radon-222 decays by alpha emission. (Radon has the element symbol Rn.) yes it is an alpha decay. I just dont understand how to explain all this through equation as this is literally my first problem like this ever.

Im asusming it starts as 222m/86 Rn----->0/-1 B+(gamma symbol)

You seem to be missing the essential difference between alpha and beta.

Believe it or not, that last comment helped A LOT. Thanks

So what is your first step equation now?

basically 222/86 Rn--->4/2alpha+218/84Po. then for B. I got 218/84 Po--->4/2alpha+214/82Pb. Then for c, I got  214Pb--->0/-1Beta+214/83 Bi+0/0y. For d, I got 214/83 Bi+0/0y--->0/-1Beta+214/84Po+0/0y.

e. 214/84 Po+0/0y--->0/-1Beta+214/85 At  . F. 214/85 At+0/0y--->4/2 alpha+210/83Pb. G. 210/83 Bi--->4/2 Alpha+206/81 Hg+0/0y. H. 206/81Hg+0/0y--->4/2 Alpha+202/79....thats where Im lost because this doesnt seem to add up to 206 lead

In any nuclear reactions the sum of the mass numbers $A_i$ (the sum of the indices on the top left) of the reagents and products is always the same. The same applies to nuclear charges $Z_i$ (indices in the lower left, which are often not indicated). Also, keep in mind the notation for metastable state:
$$
\begin{align}
\ce{^{222}_{86}Rn &→ ^{218}_{84}Po + ^{4}_{2}He}\tag{a} \\
\ce{^{218}_{84}Po &→ ^{214\mathrm{m}}_{82}Pb + ^{4}_{2}He}\tag{b} \\
\ce{^{214\mathrm{m}}_{82}Pb &→ ^{214\mathrm{m}}_{83}Bi + ^{0}_{-1}e + γ}\tag{c} \\
\ce{^{214\mathrm{m}}_{83}Bi &→ ^{214\mathrm{m}}_{84}Po + ^{0}_{-1}e + γ}\tag{d} \\
\ce{^{214\mathrm{m}}_{84}Po &→ ^{214}_{85}At + ^{0}_{-1}e}\tag{e} \\
\ce{^{214}_{85}At &→ ^{210\mathrm{m}}_{83}Bi + ^{4}_{2}He}\tag{f} \\
\ce{^{210\mathrm{m}}_{83}Bi &→ ^{210\mathrm{m}}_{84}Po + ^{0}_{-1}e + γ}\tag{g} \\
\ce{^{210\mathrm{m}}_{84}Po &→ ^{206}_{82}Pb + ^{4}_{2}He + γ}\tag{h} \\
\end{align}
$$

"High energy state" indicates that the nucleus will first decay with emission of a gamma ray, then a particle emission. $$\ce{^{210m}{84}Po -> ^{210}{84}Po + \gamma -> ^{206}_82Pb + ^4_2He }$$

@MaxW That's right, but OP was asked to right down the equations for each "step", and those didn't differentiate gamma emission from the decay acts, unfortunately. Thanks for your comment though!

I understand this is a site where there are some people who are very serious about Chemistry. TBH this is my fourth week of this ever. I am also on Calculus one rn and I'm definitely struggling. I plan to explore as much chemistry as I can take. Im a veteran and a total meat head lol. It WILL take some getting used to. However, You guys are pretty awesome. I hope to exceed and study more sciences in the future and I appreciate any help I can get.

@andselisk: Step (h) in your answer should be: $\ce{^{210\mathrm{m}}{84}Po -> ^{206}{82}Pb + ^{4}{2}He + \gamma}$ (not $\ce{^{210\mathrm{m}}{84}Po -> ^{206}{82}Po + ^{4}{2}He + \gamma}$ as stated). :-)

@MathewMahindaratne Thank you very much, fixed. I knew I would miss something as being lazy I iteratively copy-pasted equations:)


text:
MO diagram for pentaamminechlorocobalt(III)?

I was considering the MO diagram for [Co(NH3)5Cl]2+, and figured it’d be as follows:
5 sigma interactions from the NH3 ligands; by reducing a reducible representation one obtains 2A1 + B1 + E
2 pi interactions (pi donation from Cl- px and pz orbitals), which have symmetry E
And the cation atomic orbitals, which we just read off the character table.

Unfortunately this results in a lot of confusing interactions in the MO diagram.
What would the MO diagram look like, and are my group orbitals correct?
Thanks!

One approach you may find helpful is to break this down into steps. First figure out the MOs for a square planar $\ce{[Co(NH3)4]^3+}$, then add in the axial ligands and see how that changes things.

Since no one's followed up on this - a couple more comments. 1) It appears you're missing the Cl- sigma interaction. 2) One your lower image, you have all the group orbitals at the same energy. It might help you with overall ordering to order those by energy first. I may have time to write out a full answer in a little bit.

Let's start at the bottom. There will be three a1 bonding orbitals. First, we can have the metal $4s$ orbital interact with all six ligands in phase. Let's call this $1a_1$. Next, the metal $d_{z^2}$ can interact with all six ligands as well, but the axial ligands are now opposite phase from the equatorial. Let's call that $2a_1$. Both will of course have antibonding counterparts. The third $a_1$ orbital is an interaction of the metal $p_z$ with out of phase axial ligand orbitals. The equatorial ligands do not contribute to this orbital because they are on the nodal plane. Let's call this $3a_1$.
Mixed in with these a1 orbitals will be the $b_1$ orbital comprised of the equatorial ligand orbitals interacting with $d_{x^2-y^2}$. That will be $1b_1$. It is lower in energy than $3a_1$ but likely higher than $1a_1$ and $2a_1$. (This $1b_1$ and the $2a_1$ correspond to the $1e_g$ orbitals of an $O_h$ complex.)
Near $3a_1$ we have the two $e$ orbitals comprised of the metal $p_x$ and $p_y$, each interacting with a pair of equatorial ligands. Let's call these $1e$. (These along with $3a_1$ correspond to the $1t_{1u}$ orbitals of an $O_h$ complex).
If we are only considering sigma interactions, we next have the $e$ and $b_1$ orbitals representing metal $d_{xy}$, $d_{xz}$, and $d_{yz}$, all of which are nonbonding for sigma. However, the pi interactions with Cl mean that the $e$ orbitals are stabilized and have antibonding destabilized counterparts. Let's call those $2e$. The $b_1$ orbital ($d_{xy}$) remains nonbonding.
Getting the ordering exactly correct is tough in this sort of qualitative analysis, but that should get you a pretty good start.