text:
Can superglue (cyanoacrylite) be thinned with acetone and still harden?
I'm looking for something with an extremely low viscosity to use as a binder for cold casting iron oxide with very fine detail (it's to make a slotted winding core for an electric motor with $\pu{1mm}$ slots). My plan is to thin out the superglue with acetone enough to make a slurry for pouring in the mold - possibly also mix in some water just before the pour to speed the setting process.
My question will superglue still harden if it has been thinned with acetone or would this destroy whatever properties that make superglue such a useful substance once it hardens? It's stated in many places on the internet that acetone works as a solvent for CA glue but this is only ever mentioned in the context of dissolving the end result.

Superglue for modelers (hobby/craft stores) can be found in different viscosities.

The Iron oxide is extremely fine and requires a lot of mixing to get any liquid mixed through it - I would expect no matter what viscosity the glue came in it would start to set while I was still mixing it through. With a solvent I can wet it first and then mix the glue through with a longer working time (this is my theory - will put it to the test this morning).

The problem with using superglue for this is that it cures quickly and is likely triggered by the OH groups present in the surface of the iron particles. This makes it unlikely you will get good mixing. You probably need to find some alternative polymer where mixing can be done before adding a curing agent to trigger polymerisation.

As you can see from the reaction of cyanoacrylate from Wikipedia, the carbonyl ($\ce{C=O}$) groups do not react with the cyano acrylate which means that acetone should not react with your super glue.

That does not mean though that any water present in the acetone won't. So you should make sure your acetone is dry. But even if the super glue does cure, it is dissolved in acetone and will still act as a binder for the Iron oxide.

Not just dry, but very very very dry. I'd wonder if you'd get something like a sponge structure at best rather than a "complete" solid structure. I'd also suspect a much longer curing time if the water content of the acetone was low enough to not catalyze the reaction.

Thanks, sounds like it's at least worth an experiment then - will dry the acetone with epsom salt crystals first.

Your iron oxide contains a huge amount of water at the surface. There is no chance you will be able to properly mix this before it sets. Why don't you go for some epoxy resin?


text:
Find the Electronic Configuration of an Iron Atom
I'm confused. Iron has 26 electrons and I wrote down $\ce{1s^2, 2s^2, 2p^6, 3s^2, 3p^6, 4s^2, 3d^6}$ yet my answer booklet says $\ce{1s^2, 2s^2, 2p^6, 3s^2, 3p^6}$ but that is only 18. How can it be 18 if $\ce{Fe}$ (iron) is 26? What am I doing wrong?
Thank You

I'm voting to close this question as off-topic because the answer summarises in you're right and book wrong.

@Mithoron - I disagree. RTStriker made a huge mistake. "What am I doing wrong? " You believed the book was infallible. ;-)

Many books have an erratum list somewhere. You might try to look yours up for other stupid mistakes.

@Mithoron I'd leave it open. It's a good example for an  acceptable homework question: RTStriker describes their assignment, their own solution AND the reason for their doubts. I don't see, why "your book is wrong" was a valid close reason.

@Marzipanherz No, this is example of question, that even if not closed, should get later deleted in "Spring Cleaning"

I agree with Mithoron on that the question is not useful; and therefore I will close it. The configuration can easily looked up on the internet, e.g. here. Without the source (complete quote and full citation) it is also not helpful as a reference to future users of your book.

source: Stack Exchange

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text:
Number of electron for complete Discharge
What is the answer for question 17 and what is mean by "number of electrons for complete discharge?"

Which of the following needs the largest number of electrons for complete discharge during an electrolysis process?
a) 3 mol of $\ce{Al^{3+}}$
b) 4 mol of $\ce{Al^{3+}}$ ions
c) 5 mol of $\ce{OH^-}$ ions
d) 7 mol of $\ce{O^{2-}}$ ions

$$\text{a)} 3\times+3 = 9$$
$$\text{b)} 4\times+3 = 12$$
$$\text{c)} 5\times-1= -5$$
$$\text{d)} 7\times-2 = -14$$
This seems like an ambiguous question.

b) "needs" the most electrons supplied to become neutral
d) "gives up" more electrons than b) to become neutral.

I'd have to take "needs" literally and go with b) as the answer.

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Why density functional theory favors a uniform electron density?
Why density functional theory (DFT) favors a uniform electron density when we use it to calculate ground state electronic structure self-consistently?
For example, when we use DFT for calculating the (ground state) electron density of two $\text{H}$ atoms with a separation somewhat larger then the bond length of an $\text{H}_2$ molecule, there will still be a non-negligible electron density between the two atoms as if there is still a covalent bond. As a result, the calculated electronic ground state is essentially an excited state of the system.
I cannot remember where did I read this, and I don't understand why DFT has such a problem. I think the problem resides in the fact that we use a local exchange-correlation functional $E_{XC}$ like LDA and GGA. But I don't understand why a local $E_{XC}$ will lead to a uniform electron density.

I think I can see the confusion, and I will do my best to explain.
As an initial attempt to use the electron density, the uniform electron gas (UEG) model was developed. This fictitious model assumes: an infinite number of electrons, a constant non-zero density, an infinite volume, and a uniformly distributed positive charge. IMPORTANT: The electron-electron interactions (correlation and exchange) are fundamentally ignored.
Two main theorems comprise modern DFT: Hohenberg-Kohn (HK) and Kohn-Sham (KS).
In principle DFT is exact if we knew the universal HK functional (which contains electron-electron exchange-correlation terms); however, no mathematical description has been developed. As we do not know the exact form of the exchange-correlation term, a number of exchange-correlation functional have been developed. These can be derived from experimental thermochemical date or high-accuracy quantum calculations.
As exchange-correlation functionals are approximations, the electron density is not fully localised to a specific area. I hope this helps.


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What is the strongest oxidising agent?
I searched for the strongest oxidising agent and I found different results: $\ce{ClF3}$, $\ce{HArF}$, $\ce{F2}$ were among them.
Many said $\ce{ClF3}$ is the most powerful as it oxidises everything, even asbestos, sand, concrete, and can set easily fire to anything which can't be stopped; it can only be stored in Teflon.
And $\ce{HArF}$ could be a very powerful oxidant due to high instability as a compound of argon with fluorine, but was it even used as such?
What compound is actually used as oxidising agent and was proven to be stronger then others, by, for example, standard reduction potential?

"not much of it is available" is like understatement of century. HArF wasn't isolated at all, only "detected". Highest std. potential known goes to KrF2 with impressive 3.5 value. FOOF doesn't have one because it decomposes in cryogenic temp.

Colleagues, please refrain from close votes. This is a very natural question, which is likely to come up often in the future. Also, it does have a very certain answer, and there is nothing opinion-based about it. That the answer kinda defeats the purpose of the question is another story. After all, an answer "there is no answer" can be as final and as important as any other possible answer.

@IvanNeretin Well, you two didn't do it much favor by nitpicking and stretching to antimatter and beyond. You should better act as if OP posed it properly, by putting a  criterion like std. potential and agent as compound which is actually used to oxidise  other compounds.

inert gases can make compounds?

@Mithoron Gödel and Cohen's answer to the continuum hypothesis is not nitpicking, nor is it a mere statement "We don't know", though it might look superficially like that. It is very much an answer, large as life, as final as could be, though not the one anticipated by Cantor, and not its opposite either. Same thing here.

@user17915 Yes, at very low temperatures and with a healthy dose of ultraviolet light (which is very good at forcing energetically unfavourable bonds) or in plasma. The bond is very weak, needless to say.

I suspect answers that accepted the constrain of "practically useable" would be better than those that aim for "theoretically possible". Though we might want to allow "practically usable by risk loving lunatics" which would include ClF3 and FOOF.

If anyone wants to write a more conventional answer, I will happily upvote.

I edited the question to be more precise. I suggest OP to check it out.

Platinum hexafluoride can oxidize oxygen into Dioxygenyl hexafluoroplatinate. For 300K reasonably stable compounds with net zero charge this is pretty extreme.

Ivan's answer is indeed thought-provoking. But let's have some fun.
IUPAC defines oxidation as:

The complete, net removal of one or more electrons from a molecular
entity.

My humble query is thus - what better way is there to remove an electron than combining it with a literal anti-electron? Yes, my friends, we shall seek to transcend the problem entirely and swat the fly with a thermonuclear bomb. I submit as the most powerful entry, the positron.
Since 1932, we've known that ordinary matter has a mirror image, which we now call antimatter. The antimatter counterpart of the electron ($\ce{e-}$) is the positron ($\ce{e+}$). To the best of our knowledge, they behave exactly alike, except for their opposite electric charges. I stress that the positron has nothing to do with the proton ($\ce{p+}$), another class of particle entirely.
As you may know, when matter and antimatter meet, they release tremendous amounts of energy, thanks to $E=mc^2$. For an electron and positron with no initial energy other than their individual rest masses of $\pu{511 keV c^-2}$ each, the most common annihilation outcome is:
$$ \ce{e- +\ e+ -> 2\gamma}$$
However, this process is fully reversible in quantum electrodynamics; it is time-symmetric. The opposite reaction is pair production:
$$ \ce{2\gamma -> e- +\ e+ }$$
A reversible reaction? Then there is nothing stopping us from imagining the following chemical equilibrium:
\begin{align}
\ce{e- +\ e+ &<=> 2\gamma} &
\Delta_r G^\circ &= \pu{-1.022 MeV} =\pu{-98 607 810 kJ mol^-1}
\end{align}
The distinction between enthalpy and Gibbs free energy in such subatomic reactions is completely negligible, as the entropic factor is laughably small in comparison, in any reasonable conditions. I am just going to brashly consider the above value as the standard Gibbs free energy change of reaction. This enormous $\Delta_r G^\circ$ corresponds to an equilibrium constant $K_\mathrm{eq} = 3 \times 10^{17276234}$, representing a somewhat product-favoured reaction. Plugging the Nernst equation, the standard electrode potential for the "reduction of a positron" is then $\mathrm{\frac{98\ 607\ 810\ kJ\ mol^{-1}}{96\ 485.33212\ C\ mol^{-1}} = +1\ 021\ 998\ V}$.
Ivan mentions in his answer using an alpha particle as an oxidiser. Let's take that further. According to NIST, a rough estimate for the electron affinity of a completely bare darmstadtium nucleus ($\ce{Ds^{110+}}$) is $\pu{-204.4 keV}$, so even a stripped superheavy atom can't match the oxidising power of a positron!
... that is, until you get to $\ce{Ust^{173+}}$ ...

My black hole will oxidize your vacuum and spit out the positron.

@IvanNeretin I do enjoy an escalation which reaches the most fundamental limits of physics!

Of course the corollary to this is that a sufficiently at sufficiently high concentrations gamma photons can reduce Fe3+ to Fe2+ (and a positron). The experimental conditions here might be a little harsh, though.

@NicolauSakerNeto is it possible for you to edit in how you did your calculations? I have been trying to replicate your results but I cannot. Additionally, what do you mean by entropy being negligible? And how did you calculate the equilibrium constant?

Well, there is not much to calculate because I made some coarse approximations to get the point across. The electron and positron both have a rest mass of $\mathrm{511 keV\ c^2}$, and due to conservation of energy all this rest mass is converted into the energy of the combined photons ($\mathrm{E=mc^2}$ on the left side). I then call this the "enthalpy change of the reaction" (exothermic by $\mathrm{2 \times 511 keV = 1.022\ MeV })$.

I then ignore the "entropy change of the reaction". I admit I do not know how to accurately calculate this value. However, when I state "laughably small in reasonable conditions", I mean to say that the product $\mathrm{T\Delta S}$ is expected to be very small in comparison at temperatures outside the LHC. The reference you linked in your question implicitly says this: "Once the temperature drops below the electron mass, the annihilation is strongly favored" (this "electron mass temperature" is $\mathrm{511\ keV\ c^{-2} \times 11604\ K\ eV^{-1} = 6\ 000\ 000\ 000\ K}$, by the way).

@BrianBlumberg So basically I equate the annihilation energy with a reaction change of enthalpy, then approximate that as equal to the free energy of reaction, and from there I can use standard formulae in chemical thermodynamics. In essence, all I'm doing is rephrasing the statement "electron-positron annihilation releases a lot of energy" into a chemical thermodynamics/electrochemistry setting, which makes it seem unusual, especially when compared to things more normally studied in such a setting.

I see. Now, I do get the same numbers everywhere except when calculating the cell potential. I used G=-zFE where E is the cell potential. What equation do you use to calculate the potential?

@BrianBlumberg That one or several other equivalent ones should work. What value are you getting? If it's not in the same ballpark, I may have made an algebra mistake when I wrote this answer.

@NicolauSakerNeto So,
-98,607,978 kJ/mol = -1*96485 C/mol * E
=> 10225 kJ/C = E
=> 10,225,000 J/C = E
So the voltage equals 10.225 million  Volts approx.
My answer seems awfully high...

Let us continue this discussion in chat.

There is no definitive answer; if you think you have one, you are wrong.
See, this is much like asking "what is the northernmost big city". Depending on where you draw the line for being "big", the answer may be Moscow (latitude $55^\circ$N, population 13M), St. Petersburg ($60^\circ$N, 5M), Murmansk ($68^\circ$N, 300K), and quite a few others. There is no natural and universally accepted way to draw that line; it is inherently arbitrary.
How's that similar, might you ask, as you don't have any such line in your question? Yes you do, and here is it: you want the compounds which exist. Really, you don't want any compounds which don't exist, do you?
Now there's a catch: we have quite a few different subtle grades to "exist", and no natural way to draw the line that would make it clearcut black-and-white. In fact, in absence of further context it is about as ill-defined and vague as "being big" for a city. In my personal taste, $\ce{HArF}$ does not exist; if you point me to the papers that claim otherwise, I'll throw at you (figuratively) $\ce{He^2+}$, AKA an $\alpha$ particle, which was known for a good century longer, surely does exist, and will easily oxidize $\ce{HArF}$, I'm pretty confident on that.
There is another dimension to the problem. Oxidative ability of any compound is not measured by one number so that you could compare them. True, there is redox potential, but it is measured in standard conditions, and in different conditions things may turn out other way around. So there is not going to be an answer even if we would unanimously agree on the definition of "exists" (which we wouldn't).
So it goes.

HArF exist , its mentioned on wikipedia with all information with melting point and bond energies . What about F2 can protonate From F+ . I don't think so

Yeah, I've seen that page, and that's what I call "don't exist"; also, the page doesn't claim it has a melting point, even if a cursory glance might have convinced you otherwise. The rest is puzzling to me; I never said a thing about anything protonating anything.

I'm with @IvanNeretin - HArF does not exist in any usable or useful form. When you show me a reaction with it acting as an oxidant I will acknowledge its place in the ranks of powerful oxidising agents.

@Waylander Besides, even if we accept it as existing, why would it be any remarkable as an oxidizer? It's just Ar in oxidation state 0, after all.

There is a subtle difference between exists and has been made in bulk and studied. I'm with those who think candidates should have been made in quantity and studied not just shown to exist in very cold matrices of inert gases.

I think the answer would be a black hole made up of antimatter (specifically positrons) . Here you have the annihilation energy along with immense gravitational energy and immense electromagnetic energy. The gravitational tidal forces are strong coupled with the fact that the electromagnetic force repels the nucleus and attracts the electrons it is even more. I can't think of anything better!

The no-hair theorem for black holes suggests that a black hole "wouldn't know" it was made of positrons, and would behave as any black hole with the same mass and charge, so I don't think you get any annihilation benefits. Also, a black hole can't be generated with just positrons, as the electromagnetic repulsion would vastly outstrip the gravitational attraction. There would have to be a lot of electrically neutral "filler" mass to keep things together. Lastly, all black holes are thought to be overall very close to electrically neutral.

All in all, it's an interesting idea, but you may not get qualitatively greater oxidation power even though it is a much more exotic system (even compared to positrons and regular black holes!).

Yeah! But it isn't completely proved right? I did doubt that the repulsions might not let it happen. But what I am saying is if it did exist, it would be powerfull. By the way wouldn't the no-hair theorem lead to asymmetry in matter and anti-matter? But I must confess I am not knowledgeable enough to make these statements. Probably only after a few years (just about to start my UG course) I can make legible statements.

The no-hair theorem is at least a good approximation of reality which meshes well with current astrophysical theory, so without a very good reason otherwise (i.e. degrees in astrophysics!) you and I do well to regard it as robust. I'm not sure this analysis has much to say with regards to matter/anti-matter asymmetry. I see you're an eager young student! The fun thing in science is putting your imagination in a straitjacket; learn the rules of the Universe as well as you can, then find elegant ways to exploit them to do weird and beautiful things!

The most powerful oxidising particle would be positrion. In term of chemistry the strongest oxidiser would not be HArF and definitely not ClF3. ClF3 is more powerful compared to ClF3, but not as powerful as oxygen fluoride. O2F2 is far more reactive than chlorine fluoride or fluorine, in fact it will decompose into fluorine and oxygen at -163 celsius. Oxygen fluoride have a trend of being more reactive the more oxygen and flurine are in the compound. There are compound that are even more reactive than argon fluorohydride, as mentioned by other users dihelium will oxidse HArF.
Trihelium may be even more reactive compared to dihelium as the Van der Waals bond is even longer. There is an acid that is discovered in 1925 called Helium Hydride which is more adicidc than any man-made superacid, it has a lower porton affinity when compared to helium. Neon compound may be the most more oxidising compound but there is no stable neon compound (this is due to how neon is more inert than helium).


text:
What drives the redox reactions in a standard hydrogen electrode?
In this video, the SHE half cell has two redox reactions occurring i.e.
$$\ce{2H+ (g) + 2e- <=> H2 (g)}\\
\ce{H2 (g) <=> 2H+ (g) + 2e-}$$
But how can a single half cell initiate the two reactions? In the video, e- are going into the platinum wire which react with the H+ in the aqueous solution, but where does the e- come from?
And how can $\ce{H2 (g)}$ randomly release two $\ce{e-}$ into the copper wire? Where did the $\ce{H2 (g)}$ get the energy to release the two $\ce{e-}$?

The short answer is that it doesn't. You need something (an oxidant) to pick up the electrons in such a way that the overall process of transferring electrons is favorable. Also, only the first reaction you wrote is a reduction. The other is an oxidation.

It is an equilibrium reaction, so the two reactions occur at the same rate and the net current is 0 (electrons are flowing in both directions). A net current will flow only if another half cell is connected.


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What is "mechanistic duality"?
I am doing a general overview of different kinds of dualities present in science. I am quite well versed with the dualities in physics and mathematics. I was searching for dualities in chemistry, biology and other sciences. In that process, I encountered "mechanistic duality" in chemistry. For instance, see [1] and [2].
But I am not able to find a good introduction or description about "mechanisitic duality". It would also be nice if someone can briefly explain what this duality mean. Does someone have clue about its history (who coined it first and for what purpose)? Also, please point me to relevant book/paper which discuss this.

Denmark, S. E.; Su, X.; Nishigaichi, Y. The Chemistry of Trichlorosilyl Enolates. 6. Mechanistic Duality in the Lewis Base-Catalyzed Aldol Addition Reaction. J. Am. Chem. Soc. 1998, 120 (49), 12990–12991. DOI: 10.1021/ja982993v.
Ross, S. P.; Baire, B.; Hoye, T. R. Mechanistic Duality in Tertiary Amine Additions to Thermally Generated Hexadehydro-Diels–Alder Benzynes. Org. Lett. 2017, 19 (20), 5705–5708. DOI: 10.1021/acs.orglett.7b02888.

This just means that the same type of reaction may proceed by two different mechanisms depending on the exact chemical being used. It's not a particularly widely used term (it is not a "standard" phrase such as wave-particle duality) and there are many other ways of saying the same thing ("divergence", "pathway", etc.).
As an extremely simple example, alkyl bromides can react with nucleophiles via either an SN1 or SN2 pathway, depending on the degree of substitution of the carbon bonded to bromine. This is an example of how a different substrate may follow a different mechanism, although the end result is the same (replacement of Br with the nucleophile).
The papers you cited are just slightly more sophisticated examples of the same concept.

Thanks! So, if I have got you right, "mechanistic duality" is the presence of two different "mechanisms" through which a given reactants produce the same end product.
Here, I have few clarifications

are there only two pathways? and if so why?
In other words, can there be cases where for a given reactants, there can be n number of pathways to end with the same outcome?
Is there any general theory which decides the number of pathways for the particular kind of inputs?

Excuse! I am completely new to this area. Also, any reference you can point to will be of great help.

(1) It depends on the exact case. Two is not a special number in and of itself. (2) Yes, hydrolysis of esters is a good example, with at least three possible mechanisms that I can think of. (3) Yes, it's called chemistry. Generally there are some mechanisms which are considered plausible and some which are considered implausible. To find and identify these, you need to apply principles of chemistry which you have learnt - not just one principle, but many of them, accrued over years of study. Unfortunately, there is no real shortcut.

Thanks for your (1)  and (2). Regarding your (3), probably I was not clear with my question. Consider for instance the duality in mathematics (in set theory, groups, vector-spaces, etc.). After repetitive occurrence of duality, mathematicians and philosophers of mathematicians have attempted to make sense of duality for mathematics. (See https://doi.org/10.1016/j.shpsb.2015.07.004) Similar to this, I was asking have there been accounts (or theories) which explain why there are two (or n) possible mechanisms for given reactants to give rise to the same products.

So, basically, they could have just said "two mechanisms" and everyone would have known what they meant, but they said "mechanistic duality" and people said, "Uh, what's that?"

@DavidRicherby, there is definitely a general trend towards flowery language in chemistry, but I think this particular case isn't really an issue. Just saying "two mechanisms" isn't enough for a cogent title. It would be something like: "Tertiary amine addition to thermally generated hexahydro-Diels–Alder benzynes occurs via two mechanisms", but that's not really a title, that is more like the first line of an abstract. I also think that most chemists would have no trouble understanding what "mechanistic duality" means.

@orthocresol I agree that most chemists won't ahvea problem and, yeah, it does give a snappier title. (And, e.g., "Two mechanisms for..." doesn't convey that both actually occur and that there's no third, etc.)

Looks like both of you are suggesting that "duality" here is not implying anything more than "two in number". This might be true with regard to specific case at hand.  But I hope both you get what I am pushing for here - "Duality" as a concept is much more than mere invocation of cardinal two. In this spirit, I would be bit cautious and would give a much more patient relook. In any case, see this paper (https://www.cell.com/cell-chemical-biology/pdf/S1074-5521(97)90286-6.pdf) where there seems to be bit more stronger formulation of "duality"

@werunom, I am quite certain that if you’re looking for a deeper meaning to the word ‘duality’, there is none to be found in this context, unfortunately.


text:
What is the difference between carbon in an excited state and with sp³ hybridization?
What is the difference between the carbon in the excited state, $\ce{[(1s)^2] (2s)^1 (2p)^3}$, and $\pu{sp^3}$ hybridization?

The difference is that excited state exists in nature, while sp3 hybridization exists in our heads.

Those are two different (but related) concepts. This might get close as a homework question, please have a look here

This really shouldn't be closed as homework. This is one of the problems you'll have when hybridisation is taught as a law and not as a helpful tool. This is clearly a conceptual query, unfortunately based on a misconception.

Whether hybridisation is real or not is not the point here.

You are confusing different concepts here. The first one ($\ce{[(1s)^2] (2s)^1 (2p)^3}$) is an electron configuration. The other one ($\ce{sp^3}$) is an hybrid orbital.
An electron configuration describes the distribution of electrons in available orbitals (occupation). Hence it depends on a choice of orbitals, but that choice is not unique: One usually uses atomic orbitals for single atoms, and molecular orbitals for more than one atom. But one can also use localized orbitals or hybrid orbitals.
As a side note: An electron configuration is strictly speaking not an (excited) electronic state. You may say so in approximation, but there is more to it.
A hybrid orbital is constructed by a certain linear combination of atomic orbitals. It is a possible choice for the basis in an electron configuration.
As a second side note: Hybridization is a concept used in organic and inorganic chemistry to explain bonding situation. But it is just a model, a crutch, to help with interpretation and imagining things. In theoretical (quantum) chemistry it is rarely used, as there is not much value in using them. You get the very same results, if you just stick with the atomic orbitals (the one the hybrid orbitals are constructed from).


text:
IUPAC name for branched alkane
I'm trying to apply IUPAC rules to name the compound below.
TLDR: Why is this 3-ethyl-2,3,4-trimethylpentane and not 2,3-dimethyl-3-(1-methylethyl)-pentane?

Is this the correct source to use?
According to above, p. 60:

P-15.1.7.1.4
  The longest chain is numbered from one end to the other by arabic numbers, the direction being chosen so as to give the lower locants to the substituent groups (side chains) [see P-14.4(f)]. The lower set of locants is defined as the set that, when compared term by term with other locant sets cited in order of increasing magnitude, has the lower term at the first point of difference (see P-14.3.5).

The longest chain possible here is a pentane chain, and ChemSketch says that this is 3-ethyl-2,3,4-trimethylpentane.

However, if I want to select the lowest set of locants, wouldn't 2,3-dimethyl-3-(1-methylethyl)-pentane be preferred, as its ordered set (2,3,3) is lower than (2,3,3,4)?
Also, is 1-methylethyl preferable to prop-2-yl?

That doesn't affect 3., as it would be 2,3-dimethyl-3-(prop-2-yl)-pentane.

Am I correct in assuming that prop-2-yl is preferable to isopropyl?

Otherwise, that would make it 2,3-dimethyl-3-isopropylpentane.

Yes, 2. yes, 3. no, 4. no, 5. yes

Answers to numbered questions:
1.  Yes (with the errata applied)
3.  Yes, however before the lowest locant set rule (P-45.2.2), maximum number of substituents as prefixes rule (P-45.2.1) is to be used. (See e.g. Alkane nomenclature - choosing the parent chain - maximize or minimize substituents?)
4–5. No, “prop-2-yl” is an incorrect spelling of propan-2-yl (P-29.3.2.2), which is preferred to both isopropyl and 1-methylethyl (P-57.1.1.1).


text:
How to use dispersion correction with TPSSh in Gaussian 16?
A colleague of mine suggested to use TPSSh as a functional. Since I am applying Grimme's D3 dispersion correction by default, I was surprised that it wasn't implemented in the standard density functionals, cf. Gaussian Manual (Copy @ The Internet Archive).
Since I might not be the only one to come across that: How can I use dispersion correction with TPSSh (or other non-implemented functionals) in Gaussian 16?
In principle the following route section is in use:
#P TPSSh/def2TZVP           ! verbose, method/ basis set
EmpiricalDispersion(GD3BJ)  ! Switch on D3 with Becke-Johnson damping
opt(MaxCycle=100,VeryTight) ! Run optimisation
int(ultrafinegrid)          ! use finer grid for dfa

Related Error is if attempted:
 R6DS8: Unable to choose the S8 parameter, IExCor= 2523 IXCFnc= 35 ScaHFX=  0.100000 IDFTD=4
 Error termination via Lnk1e in [...] g16.b01_bin/l301.exe at Thu Oct  4 18:59:24 2018.

But it could also be the S6, a1, or a2 parameter. If zero-damping is used, then it could also be S6R.

The answer is quite simple and already given in the manual:

You can use this empirical dispersion method with other functionals via the IOps(3/174-178) (S6 should be 1.0).

Parameters and the standalone program can be found can be found on the DFTD3 homepage at Mulliken Center for Theoretical Chemistry. Here are links via The Internet archive to the tables for zero-damping and BJ-damping.
The Overlay3 manual can be found on gaussian.com, or The Internet Archive:

IOp(3/174)
S6 scale factor in Grimme’s D2/D3/D3BJ dispersion.
0          Default (see subroutine R6DS6).
-1         Set S6 to 0.
NNNNNNNN   A value of NNNNNNNN/1000000.

IOp(3/175)
S8 scale factor in Grimme’s D2/D3/D3BJ dispersion.
0          Default (see subroutine R6DS8).
-1         Set S8 to 0.
NNNNNNNN   A value of NNNNNNNN/1000000.

IOp(3/176)
SR6 scale factor in Grimme’s D2/D3/D3BJ dispersion.
0          Default (see subroutine R6DSR6).
-1         Set SR6 to 0.
NNNNNNNN   A value of NNNNNNNN/1000000.

IOp(3/177)
A1 parameter in Becke-Johnson damping for D3BJ and XDM.
0          Default (see subroutine R6DABJ/XDMABJ).
-1         Set A1 to 0.
NNNNNNNN   A value of NNNNNN/1000000.

IOp(3/178)
A2 parameter in Becke-Johnson damping for D3BJ and XDM.
0          Default (see subroutine R6DABJ/XDMABJ).
-1         Set A2 to 0.
NNNNNNNN   A value of NNNNNN/1000000 Ang.

For TPSSh the route section should therefore look similar to the following:
#P TPSSh/def2TZVP           ! verbose, method/ basis set
EmpiricalDispersion(GD3BJ)  ! Switch on D3 with Becke-Johnson damping
IOp(3/174=1000000)          ! Set s6 to 1.000
IOp(3/175=2238200)          ! Set s8 to 2.2382
IOp(3/177=452900)           ! Set a1 to 0.4529
IOp(3/178=4655000)          ! Set a2 to 4.6550
opt(MaxCycle=100,VeryTight) ! Run optimisation
int(ultrafinegrid)          ! use finer grid for dfa


text:
How can I get this glass stopper out? Need troubleshooting!
I placed a glass stopper in a $\pu{100 mL}$ round bottom flask while my solution was still warm and let it cool overnight. Now I think I created a vacuum seal and I can’t force the stopper out to filter my product. I’ve tried heating it up again but to no avail. I don’t want to slip oil in there and risk getting it in.
Anyone have a good idea?

Sometimes tapping the joint lightly can help let some air in.

Cooling the stopper helps, LN2, dry ice.

Have you access to an ultrasound bath?

I found gently taping a glass stopper with another glass stopper sometimes loosens it(maybe the shock is allowing air to pass through), but you could always put ice on the stopper and surround the top of the flask with warm water and wait a minute or two. This allows the top of the flask to expand while the stopper shrinks just a bit. Then, just take the stopper out. If you lose all hope (I hope that doesn't happen) and don't care about about your solution I would just use WD-40.

If all else fails (unlikely unless you had a strong base in your flask, but who knows), use a glass cutter and snap off the neck of the flask.
Heating the neck with a heat gun should however be sufficient. Don't heat for too long, so the stopper is still cold. The expansion coefficient of borosilicate is small, you want to use the full effect.

I've had success using a heat gun (or hair dryer) to heat up the joint and then using clamps to turn the stopper while pulling it out.

I was looking for help in releasing a stuck No. 29 glass stopper in a one liter bottle.  I had sprayed a little WD40 at the joint and after several minutes it was still very stuck.  After reading here, I held the bottle in a pail of very hot water up to the neck for 30 seconds.  Then I tapped the stopper with another stopper and presto, the stopper popped out.


text:
Restarting a Gaussian partially constrained relaxed surface scan using GICs
I am Using Gaussian 16 rev D.01, and I want to perform relaxed surface scans where I vary a bond distance. At the same time I constrain seven atom coordinates during the entire optimization (this is an enzyme active site cluster model).
Setting up the initial input is not difficult. However, as most Gaussian users would know, the optimization is likely to fail at some point due to the "Curvilinear step not converged" error. That means restarting the optimization.
The question then becomes how to properly restart the relaxed surface scan automatically from the last point in the scan while at the same time not losing my constraints. The Gaussian manual says that
geom=(CheckPoint, NewDefinition)

should not be used, and that instead
geom=(CheckPoint, GIC)

should be used.
My attempted input looks like this
%chk=mycheckpointfile.chk
# P opt=(tight)
# pbepbe/def2svp
# scf=(MaxCycle=1000)
# int=UltraFine
# EmpiricalDispersion=GD3
# geom=(CheckPoint, GIC)
# NoSymm

Relaxed surface scan searching for transition state

-1 1

Atom 30 Frozen
Atom 21 Frozen
Atom 1 Frozen
Atom 19 Frozen
Atom 56 Frozen
Atom 39 Frozen
Atom 55 Frozen
rSC(NSteps=15, StepSize=0.05) = Bond(15,59)

but this resulted in loss of my constraints. Should I add "AddGIC" as well? What is the easiest way to restart such an optimization?
Also, if this is easier with the old "ModRedundant" keyword then such an answer is also appreciated.

I assume you use revision B not D as that is the latest out there. Could you also post the input that fails. I think you need the addGIC option to specify for Gaussian to read the extra input. GIC does only specify to run the new algorithm. Otherwise I don't see any issue. How do you know you lose the constraints? Does it simply optimise? Can you find on the input of it read and GIC?


text:
Why does ATP has a lower entropy than ADP
I was confused about why ATP has a lower entropy than ADP when ATP has 1 more phosphate group. According to https://chem.libretexts.org/Textbook_Maps/Biological_Chemistry/Metabolism/ATP%2F%2FADP#Why_is_ATP_hydrolysis_an_exergonic_reaction.3F It states that

The entropy, which is the level of disorder, of ADP is greater than that of ATP.  Therefore, due to thermodynamics, the reaction spontaneously occurs because it wants to be at a higher entropy level. Also, the Gibbs' free energy of ATP is higher than that of ADP. Naturally, molecules want to be at a lower energy state, so equilibrium is shifted towards ADP.

Can someone please explain to me why does ATP has a lower entropy than ADP?

It doesn't $  $.

If you treat the entropy as a "level of disorder" the overall entropy of ADP and phosphate seems to be higher than the entropy of ATP.

Sorry for duplicating your answer! I think we were typing them at the same time.

The key is, I suspect, in the full reaction, assuming that the reaction is the production of ADP and free inorganic phosphate:
$\ce{ATP -> ADP + P_{i} }$
When one molecule disassociates, the total entropy of the products is typically greater, simply because (in general) there are more configurations and conformations available to two molecules than there are to one.

Do you mean that it is the overall entropy of ADP+Pi that is greater than ATP instead of the entropy of ADP larger than ATP?

Yes. If anything, I would imagine that (all other things being equal) the entropy of ADP alone would be lower than that of ATP, as it has slightly fewer possible conformations.


text:
Faraday's Law of Electrolysis problem
Question:

How many moles of iron will by passage of $\pu{4 A}$ of current through $\pu{1 L}$ of $\pu{0.1 M}$ $\ce{Fe^{3+}}$ solution for $1$ hour. Assume $\ce{Fe^{2+}}$ is not present in the solution initially.

Attempt:
Using Faradays' Law, we have:
$\text{mass delivered = } \mathrm {Z\times I\times t} = n\times M$  where $\mathrm Z = \dfrac{\text{Equivalent mass }}{\pu{1F}}$
Equivalent mass of Iron here would be $\dfrac{M}{3}$ where M is molar mass.
So,
$n \times M = \dfrac{M}{3}\times \dfrac{\pu{1 mol}}{\pu{96500 C}}\times \pu{4 \frac{C}{s}} \times \pu{60 \frac{s}{min}} \times \pu{60 \frac{min}{hr} \times \pu{1 hr}}$
$\implies n = 0.049$ moles.
But answer given is $\pu{0.0245 moles}$
Where have I gone wrong? I can't figure out why there's a difference of factor of 2 in the answer.

For one using units will prevent most errors though not in this case.

I think the key part of the answer is

Assume $\ce{Fe^{2+}}$ is not present in the solution $\color{red}{\underline{\text{initially}}}$.

Since iron reacts with $\ce{Fe^3+}$ to form $\ce{Fe^2+}$ via:
$$\ce{Fe^0 + 2 Fe^3+ -> 3 Fe^2+}\tag 1$$
You have to account for this additional reaction. If the reaction did not produce any $\ce{Fe^2+}$ as you assumed then of the $\pu{0.100 mol}$ of $\ce{Fe^3+}$ in the solution then $\pu{0.051 mol}$ would remain after and you would be correct, but it does react. Now, my math says that the answer is more like
$n \times M = \dfrac{M}{3}\times \dfrac{\pu{1 mol}}{\pu{96500 C}}\times \pu{4 \frac{C}{s}} \times \pu{60 \frac{s}{min}} \times \pu{60 \frac{min}{hr} \times \pu{1 hr}}\tag 2$
$\implies n = \pu{0.0497 mol} = \pu{0.050 mol}$
Which means $\pu{0.050 mol}$ of $\ce{Fe^3+}$ remains which must be neutralized by the $\pu{0.050 mol }\ce{Fe^0}$. From equation 1 we can see that it takes one $\ce{Fe^0}$ for every 2 $\ce{Fe^3+}$ which menas $\pu{0.025 mol}$ of $\ce{Fe^0}$ is needed to react with $\pu{0.050 mol } \ce{Fe^3+}$ and your remaining $\ce{Fe^0}$ is $\pu{0.025 mol}$ or $\pu{0.0245 mol}$ in the case of $n = \pu{0.49 mol}$.


text:
What is (if there is) the reaction between manganese chloride and oxygen when dry?
I had a chemistry lab and I did a error. I was supposed to make some $\ce{MnCl2}$ and then used fire to dry it but then some of it turned black, and I am supposed to quantify my error. So I am assuming the reaction to be between $\ce{MnCl2}$ and $\ce{O2}$ but I don't know what they make.

Heating $\ce{MnCl2}$ in presence of air made it to oxidize and form black manganese oxides.
$$\ce{4MnCl2 + 3O2 ->[\Delta] 2Mn2O3 + 4Cl2}$$

You don't get gaseous chlorine this way, only mixture of oxides and chlorides, IMO.


text:
Does concentration remain constant in an electrochemical cell's half cells?
We have from Nernst equation:
$\mathrm{E_{cell} = E^\circ - \dfrac{RT}{nF} \ln Q_c}$
But I was wondering that in a simple cell like Daniel cell if concentration of $\ce{Zn^{2+}}$ keeps increasing and that of $\ce{Cu^{2+}}$ keeps decreasing, won't the value of $\mathrm{Q_c}$ keep increasing?  In that case, won't that lead to a decreasing emf? If it does, then what's the use of such a cell that yields decreasing emf?

Batteries don't last forever. Aren't you just making that claim?

Note that because you have a logarithm, the value of $\ln Q_{c}$ stays relatively constant for larger changes in the value of $Q_{c}$, implying that the battery can maintain a reasonable EMF as you use it up.

But I was wondering that in a simple cell like Daniel cell if concentration of $\ce{Zn^{2+}}$ keeps increasing and that of $\ce{Cu^{2+}}$ keeps decreasing, won't the value of $\mathrm{Q_c}$ keep increasing?  In that case, won't that lead to a decreasing emf?

Have you ever wondered how battery meters on phones, laptops and other electronics can tell how charged your battery is? The question you just asked is the answer. As energy from the battery is consumed, the voltage will drop which can be correlated to a percent charge if known.

If it does, then what's the use of such a cell that yields decreasing emf?

These cell are still useful.  Astute engineers know this and design around it using batteries that provide more voltage than needed when charged. If I have a $\pu{3.3 V}$ circuit, I can use a $\pu{3.7 V}$ Lithium battery (note $\pu{3.7V}$ is the charged voltage) to run the device without damaging it and will continue to work until the battery declines to below $\pu{3.3V}$ (actual/rated voltage). Similarly car batteries are typically rated for $\pu{12V}$ but if you measure it with a galvanometer (voltmeter) you will find the voltage is on the order of $\pu{13-14V}$.

Doesn't decreasing emf cause any problems? Decreasing emf would lead to lower current...


text:
Silver in defect chemistry
When talking about conductivity in combination with lattice defects you can often see that AgX is taken as an example for a Frenkel defect where the $\ce{Ag^+}$ moves to intersticial voids. I remember that my professor once called this a property of silver, that silver likes itself so much that it can be close to other silver cations. And I think we see trends like this in for example $\ce{Hg2}$-dimers. Is this some sort of relativistic effect or can I explain it with certain orbitals overlapping that I can get a quite close contact for two $\ce{Ag^+}$ ions?

(partial answer)
Interesting question.  Silver also forms an unusual "subfluoride", Ag2F.  My understanding is the Ag is NOT in dimers.  But the structure has sorts of "sheets" of silver in that all of the octagonal holes are filled but only in alternating layers.  This allows silver atoms to be very close to each other, similar to the metal.  Presumably there is a lot of silver-silver bonding going on.  But more like sheets of metal than dimers.
The explanation of Frenkel defects you usually hear is that the cation is small enough versus the structure to allow dislodging by radiation (i.e. not a thermodynamics distribution of defects but a provoked one). See the Wikipedia page for instance.  However defect models are idealizations.
It would further be interesting to know how often the Frenkel defect is associated with some higher conductivity.  A purely billiard ball dislodged model would be less prone to conductivity than one where silver-silver bonding is the driver.


text:
Determining R and S configuration for a cyclic compound when the lowest priority group is on the plane of the paper
I am struggling with determining R and S configuration for the following compound

I know how to give R and S configuration  for a cyclic compound like the compound above if the lowest priority group is on a wedge or dash bond but on the above compound the lowest priority group is on the plane of the paper so how do I determine the R and S configuration for the chiral carbon on this  compound?

Build a model and turn it around.

@orthocresol yeah I could simply do that but on the exam we are not allowed to use a model so how could I determine the Rand S conformation without a model?

Just determine the priorities on the nearest atoms (not super easy in this case), then dissect the cycle (possibly redraw it with numbers or letters instead of chemical groups), and swap any two substituents twice in order to get  a comfortable structure orientation, if you have not 3D imagination to rotate the molecule or your point of view in your head. (And note that you're determining configuration, not conformation)

This similar question may help? https://chemistry.stackexchange.com/questions/55676/absolute-configuration-when-hydrogen-lowest-priority-group-is-in-the-plane-of?rq=1

Try this method after assigning priorities: https://chemistry.stackexchange.com/questions/99721/twisting-stereoisomers-with-rings-to-determine-r-s/99750#99750

The only two ways to do this are either:

Move yourself (or imagine your eye) to the right place so that the lowest priority substituent is away from you, or
Rotate the molecule so that the lowest priority substituent is away from you as you look at the page from above.

Either approach takes a bit of mental gymnastics and gets practice to perfect. For (1), I fine it helps to hold the paper up or move my head so I am close to the page, which makes it easier for me to imagine looking along that bond. For (2), it's easier if you abstract away all the detail by replacing the four groups with just the priority numbers. Even if you can't use a model in the exam, I still think models help to get to grips with the mental manipulations.
Here's a quick sketch - hope it helps:  Good luck!

Thanks! Is it possible to use the swap method for this cyclic compound? If yes,  Do I only need to consider the molecules which I gave the priority (1,2,3,4) when doing molecule swap method to determine the R and S configuration of the above compound(4th priority on the plane of the paper)?

Yes, I shouldn't have been so hasty in saying there's only two ways to do this! Swapping two substituents always inverts the configuration (eg. S -> R), so doing one swap, followed by a second swap gets back to the original configuration (eg. S -> R -> S). So, rather than rotating like I did in (2), you could draw out the chiral centre with 1, 2, 3, 4, then do two sequential swaps that end up putting the lowest priority group going away from you (ie. with the hashed bond).


text:
Seeming discrepancy between pKa and corresponding Ka of methylamine
The first entry in PubChem for the $\mathrm pK_\mathrm a$ of methylamine is $10.657$ at $25\ \mathrm{^\circ C}$. My elementary understanding is that this would be found as the negative log of the $K_\mathrm a,$ but the $K_\mathrm a$ is given as $2.70\times10^{-11}$ for which the negative log is about $10.568$.
My guess was that the two are determined independently, but can anyone shed some light on the seeming discrepancy?

A discrepancy of 0.1 in pKa is as good as no discrepancy at all.

@IvanNeretin: Oh. They are each found experimentally...?

Maybe.$\mathstrut$

Humans create all that data. I'd guess that the 5 and 6 got transposed.


text:
Question about the mole
Why is  it to get from x amu---->xgram you need a mole of the substance.
I mean I underdstand this to be the case for Hydrogen. Why does a mole of a substance  CONVERT the substances amu into grams?

Well, the moles of all elements have different weights too, so what's the problem?

How does a mole of a substance equal its amu in grams?

I cant see the answer on that lini.

I often find units help me understand things like this. So, if:
1 g = mass of Avogadro's number of hydrogen atoms / protons / neutrons, and
1 amu = mass of a hydrogen atom / proton / neutron, and
1 mole = Avogadro's number of atoms or molecules of a substance, then:
if I make a substance heavier than hydrogen, the atomic or molecular mass will go up by one amu for every proton or neutron, and the mass of a mole of that substance will go up by 1 gram for each of those extra protons or neutrons, since a mole of protons or neutrons has a mass of 1 g.

Why with a different amu its still 1 mole in grams?

Both AMU and grams are units for expressing mass. AMU is used for weighing single protons, neutrons ore atoms while grams are used for more convenient amounts.
So: 1 AMU = mass of one proton = mass of one H-atom (neglecting the electron) = 1,7*10^-24 g.
But if you want to do some calculations in gram, that figure is rather unhandy, and instead you use the mass of one mole. One mole is, as you probably know, 6.23*10^23 pieces of whatever we are talking about (Avogados number).
The mass of one mole of H is therefore: 1 AMU * 1 mole = 1 gram (round figures). But as Ivan said: other elements weights more.
An example: carbon (6 protons + 6neutrons) weigh 12 AMU and one mole (6.23*10^23 C-atoms) weighs: 12 AMU * 1 mole = 12 grams.
Ore to be more precise: 12 grams per mole

So can we say that 1 mole =g/amu

Fundamentally yes, but what is your point with this? One mole is a fixed number You can look up, so there is no reason for calculating it?

There are two different ways to measure a quantity. Think about bananas. You can weigh the bananas or you can count them.
Now let's think about a reaction
$$\ce{2Na + Cl_2 -> 2NaCl}$$
It takes one atom of sodium to react with one atom of chlorine to get one molecule of sodium chloride. Now we can't really count individual atoms in a typical chem lab, so we need some way to covert between a count of atoms and a mass of atoms. The way to convert is through the use of "moles".
A mole is just a big counting number like a dozen or a million. But a mole is a much much bigger number. The value is about $6.022\times10^{23}$. So if you "count out" $6.022\times10^{23}$ atoms of sodium you get a mass 22.99 grams of sodium atoms that corresponds to the atomic mass of sodium.
The same notion applies to the chlorine gas. $6.022\times10^{23}$ molecules of chlorine gas have a mass of 70.90 grams of chlorine gas. But each chlorine molecule has two atoms of chlorine. So we'd only need 35.45 grams of chlorine gas to react with 22.99 grams of sodium.

You say "if you "count out"  atoms of sodium you get a mass 22.99 grams of sodium" , how do you know that ?

Better example than bananas would perhaps be water. A gallon of water weighs 8.34 pounds. So I can measure water in pounds or gallons. If I measure gallons, I can convert to pounds by knowing the conversion constant. The primary units of measure form the International System of Units.


text:
Confusion about the relation between gibbs free energy and electromotive force
$$\Delta G = -nFE$$
My textbook says that we can measure that this $\Delta G$  is measured when the work is done reversibly. But my confusion is how will we work reversibly when we have an electrochemical cell? And what guarantee do we have that while measuring $\Delta G$, E will remain constant? Isn't it possible that after 0.25 moles of electrons have reacted, E would change because it depends on concentration.

As the reaction progresses toward completion, it becomes less spontaneous, so that is exactly correct.

What that means is that $\Delta G$ is the maximum work available and therefore is only a thermodynamic limit. That's why $E$ is measured with a voltmeter where no current is flowing (actually extremely low current, i.e. the voltmeter has very high impedance). An analogy would be measuring gas pressure in a pressurized vessel, only in e-chem this is "electron pressure". Once you do work with an e-chem cell, the current will encounter many resistances which translate into voltage drop and in turn are energy losses and E will change.

The limited current will also ensure that the concentrations of reactant and product species don't change noticeably so that $E$ is constant during the time it takes to measure the voltage.


text:
emulsion - why does it make a differnce if you have have more/less water/oil
So I have been wondering why you can see the following:

Using 10 ml oil and 1 ml water with and emulgator leads to a way better emulsion than if you use 10 ml water with 1 ml oil (if you use the same amount of emulgator)

btw: as an emulgator we used the yellow part of an egg

Could you clarify your question?

My question is why oil in water is  makes a more stable emulsion than water in oil using the same amount of emulgator.

The two combinations (10/1 and 1/10) of oil and water will give different emulsions: the first will be water-in-oil and the second will be oil-in-water. The difference lies in the continuous phase; the minor phase will be tiny droplets, with a size depending on how much energy you put into shearing it. Neither emulsion will be particularly stable; that's why we need emulsifiers: to stabilize the emulsion, and also to make it easier in the first place.
The yolk of an egg is primarily fatty (that's why it tastes so good). It's so fatty that it does not absorb the egg white, which is 90% water and about 10% protein. The yolk does contain some hydrophilic groups - but not enough to emulsify the albumen. But the yolk can emulsify/stabilize a small amount of water, and the yolk will disperse easily in (some) oils. So the yolk, spread out in oil, can stabilize a smaller amount of water. But you can't get the yolk well dispersed in water: it agglomerates quickly. And extra oil just makes the destabilization quicker and more obvious.

Thanks for the answer! Do you also know the mechanism behind it, why oil in water makes the destabilization quicker?

While egg yolk has some hydrophilic groups (not enough to be really stable in water), the oil probably has much less hydrophilicity (depending on the oil, of course). The benefit of the oil is to disperse the egg yolk and allow it to coat tiny globules of water (which have to be made tiny by shear energy from the mixer blade - the process is not spontaneous). After the water is sheared into tiny droplets and coated, the emulsion is viscous enough to be stable for a while. Oil alone doesn't stabilize water droplets well, so they are larger and settle quicker.

So the emulgator is the important part right? It decides whether you get oil in water or the other way around, something explained by the Bancroft-rule.

Pretty much. Emulsifier is a more common term. It is a combination of hydrophilic and hydrophobic parts in the same molecule; it is uniform and homogeneous, but has the ability to coat tiny droplets and make them seem opposite to their real nature. The chemical magic is that emulsifiers (also called surfactants) can be synthesized with different balances of hydrophilic and -phobic groups. This allows formulating with different oils and different amounts and different stabilities. And yes to Bancroft's Rule.


text:
Can two atoms with the same amount of electrons but differing electronic configurations be considered isoelectronic?
Can two atoms with the same amount of electrons but differing electronic configurations be considered isoelectronic?
For example, would we consider yttrium(I) $(\ce{Y+})$ with configuration: $\ce{[Kr] 5s^1 4d^1}$ and strontium $(\ce{Sr^0})$ with the configuration $\ce{[Kr] 5s^2}$ to be isoelectronic?

$(\ce{Y+})$ with configuration: $\ce{[Kr] 5s^1 4d^1}$ and strontium $(\ce{Sr^0})$ with the configuration $\ce{[Kr] 5s^2}$ are not isoelectronic. In order to be isoelectronic the two atoms would have to have the same electronic structure as well as the same number of valence electrons.
IUPAC Goldbook

isoelectronic - Two or more molecular entities are described as isoelectronic if they have the same number of valence electrons and the same structure, i.e. number and connectivity of atoms, but differ in some of the elements involved.


text:
Can the ions of dissolved salts be physically separated with a strong enough external electric field?
Say you had a arbitrary amount of sodium chloride dissolved in water. Could an external electric field (e.g. the electrically charged plates in Millikan's oil drop experiment) physically pull apart the dissolved ions in the water if it was strong enough? What forces would it need to overcome?
If it is possible, how powerful would the electric field have to be? What would be the outcome if the $\ce{Na+}$ and $\ce{Cl-}$ ions were then physically separated into two different containers?

I think you went too far with the follow up question.  Less the struck out part the question should be okay.

Rather than close this question I think it could easily be edited.

Could an external electric field (e.g. the electrically charged plates in Millikan's oil drop experiment) physically pull apart the dissolved ions in the water if it was strong enough?
What forces would it need to overcome?

Yes, this is the principle behind electrophoresis and electric double layer supercapacitors where charge is separated by an electrostatic field shown below.

Diagram of Double Layer Capacitor charging and Discharging

What would be the outcome if the $\ce{Na+}$ and $\ce{Cl-}$ ions were then physically separated into two different containers?

Assuming you could so this without any interaction with the container, you would have 2 Leyden Jars that were opposite in charged and storing a lot of energy.

Is water usually not the electrolyte in these cases?

That just won't happen. You can't put more than a couple of volts on the electrodes without electrolyzing the water.  Yes you can polarize the electrodes but that will create a double layer. So a layer of $\ce{Cl^-}$ ions will attract a layer of $\ce{Na+}$ ions. Thus you can't separate $\ce{Cl^-}$ ions in one half of a cell and  $\ce{Na+}$ ions in the other half.


text:
Is a hue indistinguishable from its equivalent frequency of light?
A hue is the color characteristic of a pigment, a particular chemical molecule, when illuminated by sunlight.
Of course such a molecule will not emit a single frequency of light, but will emit a spectrum of light at different frequencies and intensities. This depends on the absorption characteristics of the molecule and its emission characteristics. In some cases, for example, light energy can be absorbed at one frequency and re-emitted at different frequency. In any case, what results is a spectrum of color.
The eye perceives this spectrum as being of a single color, called the pigment's hue.
https://physics.stackexchange.com/questions/ask
My question is whether this hue is truly indistinguishable from energy emitted purely at that frequency. For example, the pigment vermillion, mercuric sulfide, has a nominal wavelength of 607nm. If we create a pure light source of that frequency only, is it indistinguishable from the light of the pigment to the eye?

related: https://chemistry.stackexchange.com/a/102492/23561

Yes, and no... The gist is that the hue of a pigment depends on the light source which is shining on the pigment. So with different light sources a pigment can have different hues.

@MaxW As it says in the question, the light source is assumed to be the midday sun.

The colour of most objects does not come from their fluorescence or phosphorescence, which is often weak or absent,  but from light reflected off them by sunlight. The reflected light and that observed by the eye are those frequencies that the object does not absorb, i.e. made up of many wavelengths. How we perceive this is another matter and is how of brain interprets what the eye detects. The 'colour wheel' describes what absorbing wavelengths produce the colours we perceive.

Can you please provide references for "The eye perceives this spectrum as being of a single color"? Since human beings have three type of cone cells https://en.wikipedia.org/wiki/Cone_cell , it is not evident that it perceives a spectrum as being of a single color.

I suspect an answer may be closer to biology than physics. You don't see colours directly, your brain simply processes the relative excitations of your colour cones. As long as the colour cones are excited in the same proportions, it shouldn't matter how they were excited; a polychromatic light source is fine, a combination of monochromatic light sources is fine, even direct electrical stimulation is fine.

Certainly not
The problem is not, strictly, one of physics but one of perception. The eye doesn't detect individual wavelengths of light but the relative amount of red, green and blue light (but the sensors in the eye cover a relatively wide and overlapping range of wavelengths centred on what we might measure as "pure" blue, green and red. If a "pure" yellow light (say a wavelength of around 580nm) is shone into the eye it will excite both the green and red sensors and the brain will interpret the result as yellow. So the actual colours we perceive will match a pure light of that colour in that case. But we could get the same effect by shining two lights into the eye of pure green and pure red. The brain can't tell the difference and you would still perceive yellow.
But this system is no perfect and we can perceive colours that don't appear anywhere on the spectrum of "pure" colours. An example of this is magenta which is colour that does not appear in an spectrum of pure, single wavelength, light (or on the rainbow). The brain manufactures this colour when red and blue sensors are activated but green sensors are not. A very simple illustration and explanation of this is given in this short Royal Institution video.
This simple examples shows that colour perception does not relate to or match the perceived colour of a "pure" wavelength of light. Perceived colour is more complex that simply matching the wavelength in a light source and there are colours that we perceive that simply don't have a matching wavelength.

Continuing the physiology, the green receptor comes in two wavelength flavours and some women are heterozygous for it. Unfortunately the wavelengths are rather close and as far as I know, experiments to test for tetrachromatic vision have been inconclusive. Because the two kinds of green receptors produce such very similar signals in every natural case, the brain doesn’t seem to bother to wire itself up to distinguish between them.

The colour we perceive for any compound, such as vermillion is not associated with a single wavelength, but rather with all of the wavelengths reflected by the compound that are in the visible portion of the spectrum.
The reflectance spectra of vermillion under different conditions is show below. Source

We can see that most visible light with wavelengths below ~600 nm is absorbed by the compound.   A laser beam with a wavelength of 607 nm would have this colour (HEX: #FFC800, RGB: 255, 200, 0).Source

Vermillion is a brilliant red/scarlet colour (HEX: #e34234, RGB: 89, 26, 20). Source

The particular hue that we perceive for vermillion is due to intensity of the all the wavelengths of light reflected by the compound.

It seems that the premise "the eye perceives this spectrum as being of a single color, called the pigment's hue" is wrong.
Consider, for example the CIE-1976 colorspace. Its chromaticity diagram is a diagram in two dimensions. Just the boundary of it is the result of a pure frequency light source. The colours in the middle of the diagram are the result of at least two frequencies.
[Edited after Shaka Boom's comments.]

First of all, this is a comment, not an answer. Second of all, the hue is the angle on such charts.

It may be a wrong answer, if you think so, but it is an answer. It says that a hue does not have necessarily an equivalent frequency of light because some hues require more than one frequency.

Isn't the purple at (u',v') = (0.3,0.1) one hue and the bluish green at (u',v') = (0.1,0.5) another hue?

Indeed, you are right, the hue is the angle on such charts. So two different colours can have the same hue.


text:
Does the field of organic chemistry cover all group 14 elements or only carbon?
Does the field of organic chemistry cover all the other group 14 elements or does it strictly deal with Carbon chemistry?  Since the other group 14 elements have similar chemical properties, such as silicon, wouldn't they fall under the same discipline? Why or why not?

They aren't similar at all.

This is pretty much reposting! https://chemistry.stackexchange.com/questions/82318/does-organic-chemistry-cover-all-group-14-elements-compounds

Yes, but my question was closed because it my question was "unclear." You're welcome to vote to have it removed.

The other group 14 elements don't have similar chemistry. In general, the elements in the 2p block are "unusual" compared to the 3p/4p/5p blocks, just as how the elements in the 3d block are quite different compared to the 4d/5d blocks, and just as how hydrogen ("1s block") is very different from the alkali metals.
The obvious difference in chemistry comes from the unparalleled ability of carbon to catenate, or in other words, to form chains of single bonds. Silicon does not do this to any appreciable extent, mainly due to bond energetics (the Si–Si bond, for example, is much weaker than C–C).
Many of the functional groups familiar to organic chemistry have little parallel in silicon chemistry. Carbonyl groups are ubiquitous, but Si=O compounds are barely stable (see Wikipedia), which is a result of the poor π overlap between Si (valence 3p orbitals) and O (valence 2p). The same applies for C=C and C≡C bonds. Silicon analogues are known, but are very difficult to make and generally need to be stabilised using a variety of tricks (see: How come there aren't any silicon analogs of alkenes or alkynes? and links within).
So, traditionally, the chemistry of Si/Ge/Sn/Pb is within the realm of inorganic chemistry. Of course, it is never a clear-cut distinction, since these elements can coexist with carbon in the same molecule. But that is a story for another time.

Or rather, they don't have similar chemistry under room conditions. Since we mostly care about stuff that exists in the same temperatures and pressures as us, carbon utterly dominates.

@Luaan, a fair amount of organic chemistry deals with high or low temperatures or pressures. Although biological systems are indeed important, organic chemistry isn't solely about them or how close we can get to them. Furthermore, I don't think that C/Si chemistry is any more similar at 4 K than it is at 298 K, although I am open to being corrected on this.

It deals with those temperatures and pressures mainly to increase the production yield of compounds stable at normal temperatures and pressures, doesn't it? And I've read a few papers suggesting conditions under which silicon behaves similar to carbon (while carbon no longer does, of course), but I can't find the references right now. I'll see if I can find something.

Yeah, fair enough, that makes sense!

Curious and sort of related.  Do silicon based amines exist?

@user148298, yes, these are known. Si–O and Si–N single bonds are pretty strong and don't simply fall apart, so stuff such as HMDS are pretty stable. Silyl ethers are very stable too.

The idea of organic chemistry being confined to a single element and its compounds, makes it seem like a lonely field. I would hate to be told as an organic chemist that I'm not permitted to sit with the inorganic chemists at the cafeteria.

On the other hand, I would hate to be a inorganic chemist and told that I am not permitted to party with the organic chemists who have fun and exciting compounds at their parties like graphene.


text:
why fluorine and oxygen have different behaviour towards electron gain enthalapy and electronegativity?
I have seen following definitions  in my textbook:

Electron gain enthalapy  may be defined as enthalpy change taking place when an isolated gaseous atom of the element accepts an electron to form a monovalent gaseous anion.
Electronegativity  may be defined as the tendency of an atom in molecule to attract towards itself the shared pair of an electron.

I know that fluorine and Oxygen have low ${\Delta H_{eg}}$ than  Chlorine and Sulpher respectively due to the fact when an electron is added to these atoms, the electron goes to the relatively compact second energy level which results in significant repulsion from other electrons present in the shell.
But I have seen that Fluorine and Oxygen have very high electronegativity.
(i)I can't understand how can an atom attract shared pair of electron very well when it is unwilling to add elctron in its valence shell?

Very unreactive? Flourine still has the second highest electron affinity in the periodic system. Flourine and oxygen are by far smaller than chlorine and sulfur, so the repulsion is higher. Etc. A very similar question is why HI is the strongest of the HHal acids, and HF the weakest. Stability of the ion has no direct relation to electronegativity in a covalent bond. It's a different electronic configuration, you cannot compare the two just like that.

@Karl : thanks for the answer, but the (i) question is still unclear to me.

The one electron is added to an atom orbital. The other phenomenon is the pull on  a pair of electrons that are already in a molecular orbital. It's incomparabale. And both oxygen and fluorine have a negative electron affinity, it's just slightly weaker than of sulphur and chlorine. ?!

The reaction between atomic hydrogen and atomic halogen can be considered as a two-step reaction: single electron transfer furnished by  the formation of bond between halogen anion and hydrogen cation. Even if the real mechanism is different, it is of no importance as long as we consider thermodynamics.
The second step, from the point of view of pure electrostatics, should release more energy in the case of fluorine than in the case of chlorine because the shorter bond is formed.

Thanks for the answer but I more interested in understanding (i) question as the (ii) question is simply  a part of that.


text:
What is the intuition of using "dimer method" for searching transition states?
I learned dimer method for searching transition state in this paper:
A dimer method for finding saddle points on high dimensional potential surfaces using only first derivatives.
However, I still cannot build an intuition for making sense of the method.  By reading section 2 of the paper, here is my attempt:
The constructed dimer (the two ends of which are two images of the system) is used for approximating the eigenvector with the lowest eigenvalue of the Hessian matrix.
Since the direction (the eigenvector) with the smallest eigenvalue is the direction along which the curvature of PES is the smallest and the energy of the dimer directly determines the curvature of the PES, we just need to rotate the dimer in order to minimize the energy and thus the curvature.
My confusions:

The rotation of the dimer is for aligning the dimer with the desired eigenvector of the Hessian matrix, but what is the purpose of translating the dimer?
Why the eigenvector of the Hessian matrix with the smallest eigenvalue is so important? How the eigenvalue of which is associated with the curvature of the PES?


text:
Does the translational part of internal energy mean the kinetic energy of the gas?
Internal energy of an ideal gas consists of energy due to translational, rotational , vibrational etc.
This line from WP says:

In thermodynamics, the internal energy of a system is the energy contained within the system, excluding the kinetic energy of motion of the system as a whole and the potential energy of the system as a whole due to external force fields

At the same time it says:

The ideal gas is a gas of particles considered as point objects ... Such systems are approximated by the monatomicgases, helium and the other noble gases. Here the kinetic energy consists only of the translationalenergy of the individual atoms. Monatomic particles do not rotate or vibrate, and are not electronically excited to higher energies except at very high temperatures.
  Therefore, internal energy changes in an ideal gas may be described solely by changes in its kinetic energy. Kinetic energy is simply the internal energy of the perfect gas and depends entirely on its pressure, volume and thermodynamic temperature.

So my question is:
Does the translational internal energy of the diatomic or polyatomic molecules also refer to their kinetic energies?

The words "as a whole" are the key.

Thanks to @IvanNeretin for pointing me in the right direction.
Imagine a box full of an ideal gas. Let the box and gas together be of a mass $m$.
Now imagine that the box is moving with a velocity $v$. Thus the system will have a kinetic energy $\frac{1}{2}mv^2$. Now this energy is of the system as a whole and is therefore not counted as internal energy.
Similarly imagine the box placed at height $h$. It will have a potential energy $mgh$. Again this will not be counted as internal energy.
When talking about diatomic or triatomic gases, we will say that that the total kinetic energy is the sum of all the possible ways in which the particle can move in space. This includes motion like translational motion, rotational motion and vibrational motion. The number of dimensions of such motion are called the degrees of freedom($f$) of the molecule and the total kinetic energy of the system is given by
$$U = \frac{f}{2}nRT + \phi$$
where the first term represents the total kinetic energy and the last one represents all the other energies associated with the gas molecules
For example for a diatomic gas, $f=5$ because it can move in $3$ dimensions and rotate in $2$. So $$U = \frac{5}{2}nRT + \phi$$
For gases in high temperature, the degree of freedom would increase by due to the vibrational motion of the gas particles.
Wikipedia

thanks for the quick reply. That's what I want to stress on...while internal energy doesn't involve the KE due to the motion of the system in total...it involves the KE of the gases in case of ideal gases...this is what the kinetic equation of ideal gas  also says.But my question is if I were asked to calculate the KE of diatomic or polyatomic gases (which are not deemed ideal) then is it Ok to say that the translational part of internal energy is the Kinetic energy of the gas molecules?

The kinetic energy would also depend on the rotational motion and the vibrations. So only the translational part of the internal energy cannot be called the total kinetic energy.

Kinetic energy is energy due to "motion", in this case there more than one ways of doing "motion", as rotating is also a form of motion. And, so does vibration.
Energy either stay (as potential energy) or move (as kinetic energy).

The internal energy of a substance is consists of the kinetic energy and potential energy associated with its constituent particles (gas molecules in case of gases). It has nothing to do with the kinetic energy or potential energy of the substance itself. Also the kinetic energy of constituent particle is because of the translational, rotational and vibrational motion of the constituent particles. While potential energy for e.g. may be because of the chemical bonds, such as ionic or covalent bonds.
When we are not dealing with the chemical reactions we can consider the change in internal energy of the system, as change in the total kinetic energy. When heat energy is supplied to the gases the translational rotational and at higher temperature vibrational motion of gases molecules increases (i.e. the gases molecules get excited to the higher energy levels of their respective motions) as a result total kinetic energy increases and thus the temperature of the gas increases. Since temperature is the average kinetic energy of the constituent particles.
However in the case of an  ideal gas the constituent particles are monoatomic hence they do not posses rotational as well as vibrational motions as a result their total kinetic energy is their translational energy.


text:
Is oxepin aromatic?
In my book the resonance structures drawn for oxepin include the lone pair electrons of oxygen. If we include the lone pair electrons the total number of π electrons become 8, so it should not be aromatic. Can we even use Hückel's rule everywhere? Where should we not use it? When should we consider lone pair electrons of heteroatom and when should we not?

Oxepin is certainly not aromatic. If it were flat and the oxygen atom was sp2 hybridized, then it would be an 8π-electron anti-aromatic system, but to avoid being anti-aromatic, it exists in a boat-like conformation, with incomplete conjugation of the π-system.[1] Interestingly, in solution, oxepin is in equilibrium with benzene oxide - the two interconvert via electrocyclic reactions.

In terms of the broader questions about Hückel's rule - it's important to look at whether the system is cyclic, planar and has (or could have) good overlap of p-orbitals on each atom within the ring. The 4n+2 π electron part of the rule is only relevant if these prior requirements are met.
In terms of heteroatoms, they can "choose" to be hybridized in such a way that a lone pair in a p-orbital overlaps with the rest of the π-system and leads to aromaticity, eg. furan, or to be hybridized in such a way as to avoid or minimize overlap, eg. oxepin. If geometrical constraints mean that a heteroatom's lone pairs can't overlap effectively with the rest of the π-system, then that breaks the conjugation and the 4n+2 π-electron rule is not relevant.

Vogel, E.; Günther, H. Benzene Oxide-Oxepin Valence Tautomerism. Angew. Chem. Int. Ed. 1967, 6 (5), 385–401. DOI: 10.1002/anie.196703851.


text:
Eigenvalues of Hessian matrix when the molecule is not at its equilibrium configuration
Given the potential energy surface (PES) of a system, we can calculate the Hessian matrix $H$ of each point $x$ on the PES.
$$H_{ij}(x) = \frac{\partial^2 E}{\partial R_i \partial R_j}|_{R = x}$$
When the structure of a molecule is optimized, the eigenvectors of $H$ are displacement vectors, the corresponding eigenvalues are the 'force constants' and they are all positive. On the other hand, when the structure corresponds to a transition state, some of the eigenvalues will be negative.
When a structure is neither the optimized one nor a transition state, the first derivate of the PES is not zero. However, we can still calculate the Hessian matrix for this structure. Moreover, as long as the PES has continuous second partial derivatives, the Hessian matrix is symmetric and thus diagonalizable (the proof). Can we interpret the eigenvalues here?

I am quite sure a much more thorough answer can be provided, but I will give a brief answer.
The answer is both yes and no. The answer is no in the sense that a harmonic analysis should always be performed at the true minimum on the potential energy surface (PES) because the eigenvalues have a physical interpretation as the force constant, and this must be a positive value. This is true because if one wishes to calculate a quadratic approximation to the PES, one will Taylor expand the potential about the minimum. In this case, the leading term is quadratic in the displacement and the coefficient is the second derivative with respect to the potential. This form of the potential is identical to the usual one for a harmonic oscillator, so the coefficient is a force constant and it is positive because the PES is strictly concave up at the minimum.
On the other hand, the answer is yes both approximately and exactly. First, I have seen people sometimes perform a harmonic analysis away from the minimum and simply trust the frequencies which they get out. This can be justified by noting that if the displacement is mostly along an eigenvector of the Hessian, then one can expect to basically still be at the minimum with respect to all other coordinates because these eigenvectors are mutually orthogonal. You will see this if you do a harmonic analysis slightly away from the minimum. That is, the modes whose eigenvectors do not describe a motion similar to the displacement you performed will have basically the same frequency as at the true minimum. However, if you make many displacements (i.e. randomly jiggle the atom positions) then you have no guarantees the answer you get out means anything.
On the other hand, these frequencies can be made rigorous by projecting out the eigenvectors which correspond to negative eigenvalues and re-diagonalizing the new Hessian. This guarantees all positive frequencies and can be interpreted physically as the instantaneous frequency when the system reaches a particular point on the PES. I think people have tried to use this type of procedure to determine linewidths of vibrational transitions. I am not sure how effective of a method it is. This kind of thing can at least give you some insight into how sensitive a particular frequency is to motion in other vibrational coordinates.


text:
Are there any fluids which are both viscous and volatile?
I'm looking for a viscous fluid which will evaporate (completely) rather quickly.
If propylene glycol evaporated like acetone, that would be the dream ticket, but neither the viscosity nor the volatilty need be quite that high.
Non-toxic would be a plus!

It may be possible to exploit variation of viscosity with temperature, depending on what you're planning on doing. For example, isopropanol is relatively volatile but non-viscous at ambient temperature. However, at -78 °C it becomes very syrupy.

2-propanol is the closest I've found (boiling point 82.6, viscosity 2.038 at room temperature). Not terribly viscous, but possibly viscous enough.

Application: want to apply a paste made from metal powder and a suspending medium to a surface which can be mildly heated (50-60C). Heating should leave the powder only, evenly distributed, with no residue of suspending medium.

Which metal is it? Some solvents may not be compatible. Can you place the surface in a vacuum chamber? That would open up more options. Have you looked into commercial metallic pastes and their compositions? This may be a partially solved problem. For example, silver paste is fairly viscous and often uses terpineol. Also, a sufficiently high weight percentage dispersion of the metal in any liquid will likely make it quite viscous, but it's a matter of knowing whether such a suspension will be stable, which will also depend on e.g. particle size.

Moderators: How does putting this on hold help anyone? The question is exactly as specific as it was intended to be, not an iota more, not an iota less. There's nothing I can change.

Volatility and viscosity are 2 properties going each against the other.
A volatile liquid has rather small molecules a/o  rather weak intermolecular bonds.
A viscous liquid has rather big molecules a/o  rather strong ( often multiple ) intermolecular bonds.
A partial solution to the task is to have composite liquid, a volatile liquid mixed with either viscous liquid ( e.g  glycerine),
either with dissolved thickener ( like polyvinylalcohol or vaseline/wax )

Wouldn't the volatile liquid just evaporate first, leaving the viscous liquid?

It would. But what else would you expect ?

As stated, the liquid needs to evaporate fully - no residue - which I imagine rules out using a dissolved thickener.

That is why I have written partial.

Full solution may not exist, It is hard to have something small and big, strong and weak at the same time. Viscosity means strong bonding. Volatility means weak bonding.


text:
Reducible representation of planar molecule N2H2 with bond lengths as a basis
I'm studying about molecular symmetry and its representations. Today I got a little confused about the planar $\ce{N2H2}$ molecule.
It looks like this

The basis in the example was chosen as $\Delta r_1, \Delta r_2$, i.e. lengths of N-H bonds. Then, it was stated, that the molecule belongs to $\mathrm C_{2h}$ point group, which I agree with.
The problem is, the example stated, that the reducible representation is
\begin{array}{|l|l|l|l|}
\hline
{} & \mathrm E & \mathrm C_2 & i & \sigma_h\\ \hline
\Gamma & 1+1=2 & 0+0=0 & 0+0=0 & 1+1=2\\
\hline
\end{array}
And here I don't understand, what's going on. I understand, that $\mathrm E$ and $\sigma_h$ are 2, as both $\Delta r_1$ and $\Delta r_2$ stay the same after those operations, i.e. bond length doesn't change.
But, bond lengths don't change after $\mathrm C_2$ and $i$ either, so why are there zeros in the table?
I'd understand it if we used vectors of Cartesian coordinates on every atom as a basis, but in this case, I'm quite lost.
Could you, please, explain it to me?

The example is taken from this presentation (slide 18).

$r_{1} \rightarrow r_{2}$ and $r_{2} \rightarrow r_{1}$, so there's no $\pm 1$. It's a different object.

@Zhe Would you mind to look at the presentation and explain it in a more detail? It would help me tremendously.

I think you are confused because you're trying to just look at the bond length as a number, but the bond is an object. When rotate it, it goes somewhere else, whereas if you look at the identity or the reflect, it stays in the same place.

@Zhe Ok, so they think about it as bond itself, not only its length, if I understand you correctly?

If that's the reducible representation you're showing me, then yes.

@Zhe Ok. Could you, please, cite the relevant part from the presentation and write a full answer, so I could accept it?

Instead of using the $\Delta r$ as in your figure redraw it with a vector  shown as an arrow parallel to each NH bond and label them 1 and 2. You know from the point group that the NH bonds are the same length and at the same angle so these vectors are similar but point in different directions.
Operate on the whole diagram including arrows (with labels) according to the operations in the point group. If, after an operation the figure is indistinguishable from your starting point, then count $1$ for each vector otherwise count $0$.
In your figure the point in the point group is at the centre of the NN bond. The $C_2$ axis points out of the page centred at this point. The $\sigma_h$ plane is the plane of the figure.
The  identity, which is an 'I exist' or 'do nothing' operation count 2. The $C_2$ operation is rotation by 180 degrees and so interchanges the vectors 1 and 2 so is not indistinguishable  and so each vector counts 0. Similarly for the inversion $i$ through the centre. Reflection in the mirror plane (the plane of the image) leaves the molecule indistinguishable so counts 2. Thus you have the reducible representation you give in your question.


text:
What is the diagram of the roto-inversion of $\bar 6$?
Using the standard way of representation:

Here we have a rotation followed by an inversion (technically called 'rotational inversion'). So as to represent it, what I have done is:
1) Both rotation and inversion occur for the first time. The circle changes of plane:

2) Both rotation and inversion occur for the second time. The circle changes of plane again:

My issue is that I am not convinced of this, because in theory there is rotation followed by an inversion and they do not happen simultaneously. So if this is wrong, what is the correct representation and why?
Diagram for $\bar 3$

Diagram for $\bar 6$

METHOD:

NOTE: I have used comma to stress there was an inversion
Thank you

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

The image below gives some example of different rotation inversion operations. For the $\bar 6$ operation rotate by 60 degrees (black dot to open circle in the direction towards 3) then invert through the centre of coordinates to the opposite side. Inversion changes a point at (x, y, z) to (-x, -y, -z) thus, in the diagram point 1 becomes point 2 after the whole operation, then 2 becomes 3 and so on. Two $\bar 6$ operations moves 1 to 3 and 6 operations places it back where it started.

[The image is from 'Crystallography, An introduction ' by W. Borchardt-Ott, 3rd ed publ Springer.]


text:
Why is the equilibrium constant unaffected by a change of the initial concentration?
In my class, it was taught that the equilibrium constant does not vary with
the intial concentration of the reactants? Why is it so?

Clearly the math tells you that the equilibrium constant is independent of the initial compositions:
For $a\mathrm{A}+b\mathrm{B}\rightarrow c\mathrm{C}+d\mathrm{D}$,
$$K_c=\frac{[\mathrm{C}]^c[\mathrm{D}]^d}{[\mathrm{A}]^a[\mathrm{B}]^b}$$
where $[\mathrm{C}]$, $[\mathrm{D}]$, $[\mathrm{A}]$, and $[\mathrm{B}]$ are concentrations at equilibrium.
This makes some sense intuitively: the rate of the reaction is dependent only on the present concentrations of compounds (and some other stuff like pressure and temperature), not previous concentrations. And equilibrium is defined as a position where the rate of forward and backward reactions are the same.

I have an another question too..Why equilibrium constant is proportional to concentration alone???

@anilbabu What else would you expect it to be proportional to?

temperature, pressure,etc

@anilbabu Exactly. Those are the two other predominant factors. Increasing the pressure moves the equilibrium position towards the side with a smaller number of molecules (assuming all gaseous). Increasing the temperature pushes towards the endothermic direction. Look at Le Chatelier’s
principle for some more stuff.

This is a first order approximation for ideal dilutions only. If the initial concentration becomes large, $K_c\propto K$ breaks down.

@Martin-マーチン Sounds interesting; I just searched online quickly, and I didn't find anything about it (largely because I was unsure what to search for). Do you have any links to stuff about that?

@Martin - マーチン I never knew that $\ce{Kc}$ was an approximation. Can you please elaborate on the actual result then?

@YUSUFHASAN The statement that $K_c$ or $K$ doesn't vary with initial concentration is an approximation; and $K_c$ is an approximation to $K$. Maybe my answer here helps What kind of equilibrium constant we use for Gibbs free energy and Van't Hoff equation? (But I have no time to explain this further.)


text:
Why does polymerization reaction between amino acids produce different products as compared to the reaction between amine and carboxylic acid?
Elaborate:
Amino Acid A + Amino Acid B -> Peptide chain + H20
Amine + Carboxylic Acid -> Salt
Why do they have different outcomes when breaking it down they are both the reaction between the 2 function groups: carboxyl(-COOH) and amino (-NH2)

These two cases are not all that different.
In mild conditions amines + carboxylic acids end up producing salts. So do amino acids. Mind you, there is no need to use two of them, for each amino acid already has $\ce{-COOH}$ and $\ce{-NH2}$. In a way, each amino acid is a salt in itself.
If you remove water, then amino acids produce compounds with $\ce{-CONH}-$ link. So do amines with carboxylic acids. You must have heard of nylon;
indeed, it is a polymer, not unlike peptides.

The issue here is that $\rightarrow$ means two different things in the reactions you've written.
You can't just add two amino acids together and get a peptide. That's the overall transformation, but it takes place via several steps and requires other reagents or catalysts.
If you just mix two amino acids together, e.g., a sample of a single amino acid, you get the second reaction. The case of an amino acid, it doesn't quite make a salt because the acid and base are on the same molecule. Instead, you get a zwitterion. And indeed, amino acids do exists as zwitterions in solution at physiological pH.


text:
How does Nickel harden steel?
Nickel :

Does not harden austenitic stainless steel
Ease the tempering of martensitic stainless steel
Harden maraging steel by forming precipitates

I also read people saying that it harden ferritic steel. Is it true ?
The chemicals reasons behind all of this are quite unclear for me. Could you explain how it works?
Thanks

Hardness is hardly a chemical property.

@IvanNeretin - Hardness is a characteristic of chemical compounds. It is used extensively in mineral identification. // The gist is that think of different alloys as different chemical compounds. The how and why that creates hardness in compounds is the subject for a book.

Unless we want to salami slice part of Chemistry.SE into a MaterialSci.SE, we are going to have to accept materials questions.

Material science is both part of the program of chemistry and mechanical engineering courses. But while mechanical engineers only know what is happening, chemists know why is it happening. This is why I would rather ask this type of question in the chemistry forum

I have the following comments about nickel steels:

Tensile strength is different than hardness. I assume you are specifically asking about hardness, not strength.
Hardness has two different aspects: resisting wear and machinability. Different steels will exhibit different degrees of the two aspects. Ideally, you want a steel that is hard wearing, yet highly machinable.
There are different alloys involving nickel, notably the nickel alloys and chromium-nickel alloys. These two types of steel are different.
Different alloys have very different hardening characteristics. Even very minor changes to the amount of carbon or nickel in an alloy can have dramatic effects on its hardening characteristics.
The main effect of nickel when present in a ferritic or pearlite matrix is to increase the fineness of the grain structure.
Nickel slows the growth in size of ferritic/pearlite grains. This makes it easier to obtain a consistent carburization on the surface of a workpiece. For this reason, small amounts of nickel are valuable in alloys which will be case hardened.
Nickel preferentially dissolves into the ferritic crystals of certain steel phases, therefore it can potentially increase the hardness of a steel without harming its ductility.

Metals are hardened by changing the grain structure to make it more difficult for crystals to move. Adding a substance that precipitates or forms a solid solution, e.g. nickel, and locks crystals in place, makes the metal harder but more brittle.


text:
Will a reverse osmosis membrane exclude perfluorobutanoic acid?
I recently did a water quality test and it showed a detectable level of perfluorobutanoic acid, a potentially dangerous carboxylate PFC.
Will a reverse osmosis water purification system exclude this chemical?

It will depend on the membrane used for reverse osmosis and what your goals are. Are you trying to completely remove the acid or bring it down to an acceptable level?
The study here states that

...reverse osmosis (RO) treatment technologies have the ability to effectively reduce perfluorochemicals (PFCs) in water...

The take away is that it may not eliminate the perfluorobutanoic acid completely, but may reduce it to a safe drinking level.

All RO devices removed PFCs to below testing
  reporting limits of 0.2 µg/L.

The next sentence is probably what you're really after.

The effectiveness of RO devices is due, in part, to these devices
  having both AC and RO components.

So you may want to look into a system that offers more than just RO. This article summarizes some research that was done on RO effectiveness for removing perfluorobutanoic acid which may be of some use for you.
At the end of the day, I would suggest talking with sales representatives of specific RO devices. Tell them what you're after and they should be able to recommend a product for you.


text:
Finding the percent by mass of an unknown substance
I'm doing some chemistry homework and a the problem in question goes as follows:
"A student performs an experiment to determine the percentage by mass of $\ce{MgCl2}$ in a $\pu{2.94g}$ sample of a mixture containing anhydrous $\ce{MgCl2}$ and $\ce{KNO3}$. The student decides to precipitate all of the chloride ion, $\ce{Cl-}$, as $\ce{AgCl}$, by adding excess queous silver nitrate, $\ce{AgNO3_{(aq)}}$ to the mixture sample." $\pu{5.48g}$ of $\ce{AgNO3}$ are produced, so what is the percent by mass of $\ce{MgCl2}$ in the original mixture? I initially just did some basic stoichiometry to find the amount of moles used, but I got $\pu{0.0439mol}\; \ce{MgCl2}$ which produces $\pu{4.17g}$ which is greater than the $\pu{2.94g}$ sample
$$\pu{5.48g}\; \ce{AgCl} \times \pu{1mol}\; \ce{AgCl}\div \pu{64.43g}\; \ce{AgCl} \times \ce{1MgCl2 \div 2AgCl}$$
So I'm simply wondering if I went about solving it wrong or if there's just missing information.

There is one error in the question: the precipitate must be AgCl, not AgNO3.
I can't quite follow your calculation, but the proper way to do it is:
5.48g AgCl/143 g/mol = 0,04 mol AgCl
1 mol AgCl eqals 0.5 mol MgCl2
0.04/2 = 0.02 mol MgCl2 * 95 g/mol = 1.8g MgCl2
The numbers at rounded off, so look up the exact molecular weights if you want a more precise result.