text:
What is the purpose of the substructure section of a mol2 file?
I'm looking at the tripos mol2 format. (This is the most recent mirror of the specification document that I have found so far: http://chemyang.ccnu.edu.cn/ccb/server/AIMMS/mol2.pdf). What is the purpose of the SUBSTRUCTURE section? I notice that obabel does not include it.
It appears to specify connectivity, but that seems a duplicate of the BOND section.

Although I am not sure, I would say (based on a quick search), that it seems to exist for defining smaller units within a macromolecule (or other large unit), so that they can be compared against the whole molecule or other subunits. Think of atoms belonging in to chains in PDB-files.

Yes, it exists to define substructures smaller than the entire molecule - something that is relevant for structural manipulation. Imagine for instance that you have two complex molecules sharing a core; defining a substructure for the core allows you to align the molecules based on the core and then compare the rest of their structures.


text:
What's the difference between isoionic point and isoelectric point?
What's the difference between isoionic point and isoelectric point? My teacher uses them interchangeably, but Wikipedia states that isoionic point is when charges are balanced, while isoelectric point is when there is a net charge of 0. Is this not the same thing?

Well, article about isoionic point answers it...

The difference, according the Journal of Research of the [US] National Bureau of Standards (1940) is:

The isoionic point is defined as the point at which dissociable groups of the substance combine equally and only, with hydrogen and hydroxyl ions [6]. This is identical with the isoelectric point only when the substance does not combine with ions other than hydrogen or hydroxyl. It follows then that although it may be possible to determine the isoelectric point by several methods, as by titration, provided that the substance combines with only hydrogen or hydroxyl ions, only an only an electrokinetic method can determine the isoelectric point in every case. The isoionic and isoelectric points of some soluble proteins have been found to be approximately the same. However, in the case of the insoluble materials, especially those which are more or less crystalline in nature, the isoelectric and isoionic points may be far apart. This is true of crystals of even relatively simple substances, such as amino acids [8]. It has been suggested that in addition to the amphoteric properties at the surface, negative ions are also adsorbed at the surface, which results in a shift of the isoelectric point to lower pH values.

In other words, "isoionic" only applies to a zwitterionic substance in pure water, whereas "isoelectric" can be used when there are ions such as Na+ or Cl- present in the solution along with the zwitterion.

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sort: votes


text:
What is "ölsaurem Alkali"
I read a technical German book that refers to "ölsaurem Alkali" which apparently means "oleic acid alkali". How could it be both an acid and an alkali?

Can you cite the book?

At a guess, I would think it means a salt of oleic acid

Actually if you google "ölsaurem Alkali" there's too many German books referring to this term, so I guess there isn't any need for a citation @Pritt ;-)

@GaurangTandon Well, OP's should support their claims, so there is such need.

They were writing it fancy in old times.
Sodium acetate would have been "essigsaures Natrium", potassium sulfate "schwefelsaures Kalium" and so on.
Therefor, "ölsaures Alkali" is an alkali salt of oleic acid (Ölsäure).

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text:
Is chromium (III) green or violet in solution?
Is chromium (III) green or violet in solution? In terms of oxidation reactions between chromate and $\ce{Cr}$ (III) it forms a green color in solution, but when I searched online in different sources it seems to say that it is actually purple. Which is the right answer?

It can be anything, depending on the ligand environment.

[Cr(H2O)6] is a violet solution. Cr(OH)3 is a grey/green precipitate. [Cr(OH)6] is dark green solution. As Ivan said, it depends on the ligands etc.

Related: How hydrate isomerism results difference in colours?

What if it was water alone though? I'm still curious on how many textbooks say for sure that the chromium (iii) in solution or in minerals such as ruby cause the green color but when I search on the internet most sources say its violet.

Even in water this depends on the hydration process, $\ce{[Cr(H2O)6]Cl3}$ is purple. Also the anhydrous $\ce{CrCl3}$ when you buy it is a purple compound. The chloride-ligands exchange for water ligands like $\ce{[Cr(H2O)5Cl]Cl2 * H2O}$ which is blue-green. And in the end the  $\ce{[Cr(H2O)3Cl3]*(H2O)3}$ is green. Some ions like fluoride are said to influence this process but I have not been able to confirm that yet. For Cr(III) there's this interesting phenomenon, and I think I already asked for that, that this exchange seems to be quite slow. I once prepared a Cr(III) citrate solution and it was blue-green, but after about a week it was deep purple. So this exchange may sometimes take some days.

What if it was water alone though?


text:
Why do the unsaturated ketones have C-O bond weaker than saturated ketones? How it is related to resonance structures?
I am studying Infrared Spectroscopy and the book says that unsaturated ketones have a carbonyl group which has weaker bond because it's more single-bond character than single bond because it has these three resonance structures.

On the other hand, carbonyl group in saturated ketones have more double-bond character because it only has two resonance structures.
I can't really understand - why do the unsaturated ketones have $\ce{C-O}$ bond weaker than saturated ketones? How it is related to resonance structures?

The statement can actually be said like this. In Unsaturated aldehydes/ketones the $\ce{C -O }$ bond has significant single bond character associated with double bond characters(i.e. partial double bond character), whereas, saturated aldehydes/ketones the $\ce{C -O}$ bond has almost pure double bond character.
 In unsaturated carbonyl compounds, as the no. of stable resonating structures increase,( the structure where the $\pi$ electron cloud of $C =C$ double bond delocalises is also very stable) the contribution of  that canonical form where the $C$ and $O$ has double bond between them also decreases . So, that's why, in the resonance hybrid of the molecule( which will give the IR spectra) the bond order of $\ce{C -O}$ bond becomes significantly less, and become close to the midway between double and single bond. But in saturated carbonyls, the other resonating form is relatively very less stable ( not complete Lewis Octet and very close  charge seperation)  than the canonical form of pure double bond. So, contribution of that other form becomes significantly less. So, the resonance hybrid becomes almost close to that consisting of pure double bond.So, in the IR spectra, the streching frequency of saturated carbonyls correspond to almost perfect $\ce{C=O}$ double bond.


text:
Is cyclononatetraenylium antiaromatic?

I was thinking that it might be non aromatic since the ring is very big and it might bend just like COT (cyclo-octatetraene) does to avoid destabilization due to antiaromaticity but I don't know if the given does that or not.

This question may be much more interesting then one could think, as I read about a claim, that this cation shows Möbius aromaticity!

If you call cyclooctatetraene "non-aromatic" (I wouldn't argue with that), then what sort of molecule would you actually call antiaromatic?

https://en.wikipedia.org/wiki/Cyclononatetraene#Cyclononatetraenyl_cation

@Mithoron Indeed, it is the poster boy for Möbius aromaticity: Wikipedia.

@orthocresol By definition, there cannot be stable (i.e. ground state) anti-aromatic compounds; these need either be hypothetical molecules or excited states.

@Martin-マーチン Precisely, and that's why I said I wouldn't argue with COT being called non-aromatic. However, that defeats the purpose of the question, because it makes the answer a trivial "no".

Cycloenneatetraenyl cation, using the Greek root for nine, is certainly not antiaromatic.  A ring that large will in general be nonplanar unless strong aromatic coupling is there to drive a planar shape, and if the planar form is antiaromatic so much the better for nonplanarity.  By classical rules such a nonplanar form would be nonaromatic.
Whether this nonplanar form is Mobius aromatic is open to interpretation, or at least it seems that way given the twisted nature of the ring.  In any event Mobius aromaticity is unlikely to give the strong stabilization we see in the Huckel version because the p orbitals do not overlap perfectly.  Most likely the structure and stability of the ion  are more determined by sigma bond interactions (minimizing steric strain) than pi bond interactions.


text:
What's the density of a frozen ethanol or ethanol solutions?
How can we know what's the density of ice (water), but there's no data on how much is the density of frozen ethanol? (as far as I know)

You've probably seen water ice quite a number of times. Frozen ethanol, on the other hand, requires going well beyond $-100^\circ\rm C$, which you probably don't do every day. That said, I believe the data is somewhere out there, only it's not that easy to come by.

The closest online source I found, http://www.celsius-process.com/pdf/ethanol.pdf , only goes to 4K above ethanol's m.p.  It does seem that not all reference material is readily found online... yet.


text:
Are there any substituents other than halogens which are ortho-para directors but ring deactivators?
I was reading about electrophilic aromatic substitution reactions, and in every book, I found that the only example given for ortho-para directors - which deactivate the ring - were halogens. So are there any other substituents, other than halogens, which are ortho-para directors but ring deactivators? If so, do they do it like halogens?

The nitroso $\ce{-NO}$ group is deactivating as well as ortho para directing just like the halogens in electrophilic aromatic substitution.
For reasons of this behaviour, refer to this question on the website.
Here are more examples from Peter Skyes' Guidebook to Mechanism in Organic Chemistry: (least deactivating to most deactivating)

$\ce{-F}<\ce{-Cl,-Br, -I}<\ce{CH2Cl < -CH=CH-COOH < -CH=CH-NO2}$

For all these groups $\ce{-I}$ effect is more dominating than $\ce{+M}$ effect. Hence they deactivate the ring. However, when the arenium ion is formed they stabilise the ring through the $\ce{+M}$ effect (hyperconjugation in case of $\ce{CH2-Cl}$)
The book doesn't mention the nitroso group in the order but since it has very weak +M we should expect it to be after $\ce{CH2Cl}$
$\ce{Cl-, Br- , I-}$ are almost equally deactivating therefore they have been separated by a comma.


text:
Why do metallic elements have a very small band gap while nonmetallic elements have a large band gap?
I recently was going over solid state bonding and conductivity. I was wondering why metallic elements (such as sodium, iron, etc.) have a small band gap / no band gap (as conductors that have the fermi level in the overlap between the valence and conduction band) while elements such as carbon have such large band gaps. What distinguishes carbon and sodium, for example, such that carbon is an insulator while sodium/iron is a conductor of electricity? If you could explain in terms high school to undergraduate level chemistry, that would be great.

https://chemistry.stackexchange.com/questions/51904/formation-of-bands-in-semi-conductors

https://chemistry.stackexchange.com/questions/34750/does-the-delocalization-of-p-orbitals-ensure-conductivity https://chemistry.stackexchange.com/questions/6513/why-are-there-free-electrons-in-a-metal https://chemistry.stackexchange.com/questions/25003/clarification-for-metallic-bond

The band gaps are what determines whether an element is a metal, a semiconductor or a non-metal. Metals have a low band gap because that is the definition of a metal. Perhaps you mean why do band gaps differ among elements?

Metals don't just have a low band gap - they have NO band gap

The energy of the orbitals may be the first point to explain it at this level. You can use the example of a simple 1-D LCAO to form the bands(check page 723 of Atkins physical chemistry 8th edition). For instance, the energy difference between the bands formed by s orbitals and p orbitals will be small or large depending on the atom.


text:
Glowing stone - Scarab carving
I have an intriquing question and, after searching whole Google, found almost no information. I simply don't know where else to look for answer and so I am asking you.
When I was a child, my parents travelled with me to Egypt and there, they bought me a little scarab beetle carved of some stone, about 5 cm long, 2 cm width and height. The stone is white-ish and translucent and glows green in dark (for some time). I only remember that it might (just a speculation) have been in Luxor in Alabaster workshop. But, according to online sources, Alabaster is not supposed to glow in dark, so what is it then?
Can you please tell me?
Oh yea, right. Here are some images. The fluorescence is really weak and doesn't last longer than 5-10min so I am sorry for a bad photo, just couldn't make it better.

Daylight 01

Daylight 02

Daylight with flash

Daylight against sun 01

Daylight against sun 02

Fluorescence
Well, it doesn't look like Fluorite to me(according to Google images) but then again, I'm no mineralogist either so, maybe it's some form of it... what do you think?

Your description matches fluorite, you can see many pictures in Internet, and compare with your stone. To determine what it is precisely, you have to post a picture. EDIT. After the pictures are posted, I take back my suggestion: the key difference is that fluorite is not transparent, at least those pieces that I possess, and those which I can see on Internet. But in the dark, it looks exactly of the right color. There are very many fluorescent stones, look for example, here: http://www.galleries.com/minerals/property/fluotabl.htm

Earth Science SE will take rock identifications questions if properly presented.


text:
Raman database of polymer spectra for analyses in R/python
Can anyone reccomend a raman database of polymer spectra that is open source or freely available? I'm analyzing spectra of a number of plastics and looking for reference spectra for identification.
The majority of available databases are stored in proprietary file formats that don't allow them to be pulled in to conventional scripting analyses in R/python, etc.
Thank you.

Format conversion...

That would be awesome, through the spectroscopy software platrforms that open the proprietary databases most often limits the ability to export in to something usable like .csv, .txt, etc.


text:
Order of bond rotation about alkene

How to decide which among these has the greatest bond rotation possible? 1 has the greatest and 2 the least, according to me, I think it is due to the presence of bulky groups. What more factors can effect bond rotation?

How should I ask these question I really dont have the concept? @Mithoron

Well, this thought about bulky groups is already something, but what about them really? I think you should use my hint to try different approach, as there's no bulky group in any of these compounds...

https://chemistry.meta.stackexchange.com/questions/141/how-do-i-ask-homework-questions-on-chemistry-stack-exchange

First, recall that a single bond has the highest free rotation possible (and hence it shows conformational isomerism). A double bond instead does not show any rotation at all (hence, $\ce{abC=Cfg}$ type molecules show geometrical isomerism).
So, it's obvious that, the more double bond character a bond will have, the less likely it is to rotate freely. Here, we're talking about double bond character, because, due to resonance effects, the resonance hybrid in these compounds will only have a partial double bond character.
Thus, you only need to compare the partial double bond character in each compound, and apply the above rule. Bulkiness of the groups isn't really a factor, because the groups actually aren't that bulky, and it isn't a t-butyl group anyway.

Does ether oxygen also have, even if tiny, a resonance share? I know there is even 4-bonded oxygen in some cases but, I don't understand this for ether.

@GürayHatipoğlu In case 2? Why not! The lone pair on oxygen is in conjugation with the pi bond, and can freely resonate with it :-)


text:
Are aliphatic amines SABRE amenable?
Hyperpolarization of molecules via the SABRE-SHEATH technique is a fairly new idea. Does anybody know if hyperpolarizing aliphatic nitrogens is possible? The articles I've read mostly use N-heterocyclic compounds like pyridine or imidazole as substrates. If not, why?

There have been a number of interesting papers just this year on this. Simon Duckett from York is leading the charge, and has recently demonstrated successful hyperpolarisation of a number of amines, including ammonia, as well as a number amides. Previously, a limiting factor in the success of the SABRE method has been the ability of the substrate to coordinate to the catalyst centre. Recent results show that this can be overcome using a relayed polarisation transfer method using a co-ligand.
For a couple of examples, check out:

Chem. Sci. 2018, 9 (15), 3677–3684
Sci. Adv. 2018, 4 (1), eaao6250
anything else by Simon Duckett in recent years

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text:
Effect of resistance (distance) on the voltage of a galvanic cell
I'm doing an experiment at school on the rate of reaction in galvanic cell so I wanted to question how the length of wires and the addition of different resistors could affect the rate of reaction. I'm a little confused with the fact that if I increase the resistance, according to the equation $$V=IR$$ current will increase. Theoretically does this mean that no matter how much resistance I provide, the voltage will be almost constant? Also, how does this explain the rate of reaction, since current isn't proportional to voltage because resistance is not consistent?

Do you mean the current will decrease if the resistance increases? @Jess

Theoretically does this mean that no matter how much resistance I
  provide, the voltage will be almost constant?

Mostly, yes, though the current will decrease as the resistance increase. The voltage of an ideal cell at the terminals is determined by the chemical species in the cell and their concentrations, according to the Nernst equation. The current that flows in response to this voltage is determined by the load (ie. resistor), up until the point that the cell becomes rate limited. The current directly describes the reaction rate because each mole of reaction results in 1 Coulomb of charge (or 2, 3, n, depending on the balanced equation). So if you have n amps of current, that's n coulombs/sec and 1 mol/sec.
In a real battery, there is some (small) resistance in the parts, such as the electrode. This means there will be a V=IR drop between the surface of the electrode, and the place you connect the volt-meter. So the measured voltage at high currents will be slightly lower.
At high rates, there comes a point where the battery cannot react fast enough to supply the current. One example is when the interface between the electrolyte and the electrode consumes ions faster than they can diffuse from the bulk. When this happens, the surface becomes depleted of ions and their effective concentration goes down, so the potential changes. Ultimately it's the concentration at the interface that matters.
Also, many (most) cells do not have a constant voltage over their entire charge states. If the concentrations (activities) change at all as the reaction happens, the potential will change as the cell discharges.


text:
Mechanism for the conversion of dimethyl cyclohexa-1,4-diene-1,2-dicarboxylate to dimethyl benzene-1,3-dicarboxylate using bromine
Question

Convert the first compound into the second using only $\ce{Br2}$

I tried solving it by making a 3-ring with $\ce{Br+}$, and then making one of the oxygens break that ring. Than I used $\ce{Br-}$ to steal an $\ce{H}$ and get a lone pair on one of carbons. Lastly, I used the lone pair to create a double bond. Am I on the right track?

I don't think that's the right path. Probable scenario is rather dehydrogenation by Br radicals.

Firstly, I didn't get which oxygen you are talking about to break the ring. If you are considering the oxygens that are part of esters, that is not possible. Secondly, $\ce{Br^-}$ is the conjugate base of s very strong acid $\ce{HBr}$. So, $\ce{Br^-}$ is a very weak base. Therefore, it is very very difficult for that $\ce{Br^-}$ to abstract the hydrogen atom and act as a base.
Instead what you can simply do, you can first perform bromination i.e.  treating the substrate with $\ce{Br_2/ CHCl_3}$. You will first get a bromonium intermidiate, like this:

The intermidiate will form on the farther double bond as the $\pi$ -electron cloud of the other one is pulled by the electron withdrawing ester groups, and also to form that ring, there will be huge steric crowding. The other $\ce{Br^-}$ will act as nucleophile to break this ring, (normal mechanism) and you will get this product:

Now, the two $\ce{Br}$ are in anti position. So, if you now treat the compound with excess $\ce{EtO^-}$ in $\ce{EtOH}$ and heat it, anti-periplanar E2 elimination will occur, and finally you will get your desired product as the one mentioned above.

The elimination will occur from two different carbons as the bromines are in anti position. So, to follow anti elimination of $\ce{Br}$ with $\ce{H}$, the elimination has to be from two different non adjacent carbons.

Using ethoxide in ethanol for the double elimination will lead to ester exchange. Go with methoxide in methanol. Alternatively, start with free radical bromination and loss of HBr may be spontaneous by an SN1 process.


text:
What really are the units of enthalpy change?
A reasonable definition of standard enthalpy change of formation would be:
'The enthalpy change when one mole of a compound is formed from its constituent elements under standard conditions'
Enthalpy change itself is understood as 'the exchange of energy between a chemical reaction and its surroundings at constant pressure.'
I am a little confused on what the units of enthalpy change should be. For example, for the enthalpy change of formation of water, this is shown: ΔHfo = -285.8 kJ mol-1
Now, if enthalpy change already contains 'one mole of a compound' in its definition, do we really need to include the per mole (mol-1) part in the enthalpy change? If enthalpy change is the change in just energy, how are the units kJ mol-1 and not just kJ ? I am sorry if I am overthinking here.

You grab a shovelful of your compound, it reacts and gives you some kJ, you measure those and write the number down for later use. Next time you have a whole dump truck of the same compound. Will the number of kJ be the same?

I agree with your logic, but enthalpy change of formation itself describes me taking only a mole in my shovelful of compound (as far as I know).


text:
Is there any difference between "autoignition temperature" and "ignition temperature"?
Is there any difference between the "autoignition temperature" and the "ignition temperature" of a substance?

Sure. Autoignition temperature is the temperature at which the substance catches fire, and ignition temperature is not a thing at all. Then again, there is a thing called flash point: it is a temperature above which the substance will burn if ignited.

https://chemistry.stackexchange.com/questions/82810/what-is-the-difference-between-ignition-temperature-and-flash-point

The autoignition temperature is the lowest temperature at which a substance can spontaneously ignite in a normal atmosphere without an external source (spark,flame etc).
"Ignition temperature" is what people call autoignition temperature who don't care about details. As Ivan said, there is no such thing as ignition temperature.


text:
Will ice be cold or not when formed using pressure?
We know that ice can be formed either by lowering the temperature of water or by applying more pressure. If we lower the temperature of water the ice so formed will be cool which is obvious.
If ice is created by applying pressure, will it be cold as well?

Being cold and having a cooling effect are two different things.

Who downvoted without comment? Doesn't help the poster...

Water can be formed by applied pressure, but only at certain temperatures. Looking at the phase diagram below, you can see that water must be at 0 C until extreme pressures are reached. This can create ice, but it would only remain solid under these pressure conditions at that temperature. The ice would have a cooling effect on surroundings that have a higher temperature, in fact the rapid change back to liquid would cool the surroundings, but the ice would not be cold. This ice would also quickly revert to another phase if not held under these pressure conditions.
So, if at the bottom of a deep, high gravity sea, the ice will form at temp above 0C. If it starts floating upward, the lower pressure will allow it to melt.
If you create warm ice in a hydraulic ram, it will melt as soon as pressure is relieved.


text:
How to calculate the resonance energy of thiophene?
How to calculate the Resonance energy of Thiophene?
I know how to calculate the resonance energy of Benzene. I am applying same procedure for thiophene, too. (Wikipedia: Resonance energy)
The values are taken from NIST Chemistry WebBook.
\begin{array}{lccc} \hline
  \frac{\text{Heat Of Formations}}{\pu{kcal mol-1}} &
  \text{Thiophene} &
  \text{Tetrahydrothiophene} &
  \text{2,5-Dihydrothiophene}\\
\hline
\text{Experimental}          & 27.82 & -8.03 & 20.86 \\
\text{Using Argus Lab (AM1)} & 29.65 & -13.7 & 15.31 \\
\hline
\end{array}
From moving from tetrahydrothiophene to 2,5-dihydrothiophene, there is a change of $\pu{28.89 kcal/ mol}$, so ideally thiophene should have
$28.89 \times 2 (=57.78) \pu{kcal/mol}$ more energy of formation, but it has only $\pu{35.85 kcal/mol}$ more. This difference should be called resonance energy, i.e. $\pu{35.85 kcal/mol} - \pu{57.78 kcal/mol} = \pu{-21.93 kcal/mol}$.
In the literature, the value is given as $\pu{-29 kcal/mol}$. (1) (2)
I applied the same method on furan, where I got an error within 1%, but in case of thiophene, I am getting an error of about $\pu{7 kcal/mol}$.
Why? When using Argus lab the error even reaches above 50%.
I've posted this question on another forum also. I didn't get any satisfactory answer. Why in the cases of benzene and furan does this method work, but in case of thiophene I'm getting this error?

Resonance energy is not a real measurable physical value anyway, so why bother.

@Ivan Neretin Hahaha, Right. But at least it gives us some insight into the stability of different molecules. Moreover, My teacher gave me this assignment to calculate the resonance energy using Argus Lab, but results are drastic. First, my teacher didn't know how to calculate it :3  , Then I tried to find it then I faced this problem.

Thiophene, 2,3-dihydro-: 21.68 kcal/mol -> R 23,57 kcal/mol. That's pretty good for guessing a random number. It is also clear that AM1 is not able to describe the heats of formation very well; I mean they are 5 kcal/mol off, that is a lot. Please add where you got your reference value of -29.1 kcal /mol from. In what context and how did they measure that?

@Martin I've added two references. Please check them.


text:
What are the IUPAC names for singly- and doubly-protonated piperazine?
If ammonia, $\ce{NH3}$, is protonated, it becomes ammonium: $\ce{NH4+}$.
If piperidine is protonated, it becomes piperidinium:

If piperazine is protonated once, what does it become? By analogy, I would assume piperazinium:

If the resulting monocation is protonated again, what does it then become?  The best I've come up with are piperazine dication and piperazin-dium, but that latter seems really awkward:

The relevant rule in the current (corrected) version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) is:

P-73.1.1.2 General rule for systematically naming cationic centers in parent hydrides
A cation derived formally by adding one or more hydrons to any position of a neutral parent hydride (listed in Chapter P-2), or whose degree of hydrogenation has been modified (see P-31) is named by replacing the final letter ‘e’ of the parent hydride name, if any, by the suffix ‘ium’, preceded by multiplying prefixes ‘di’, ‘tri’, etc. to denote the multiplicity of identical cationic centers. (…)

Therefore, the preferred IUPAC names are piperazin-1-ium and piperazine-1,4-diium, respectively.

"... by replacing the final letter ‘e’ of the parent hydride name, if any, by the suffix ‘ium’, preceded by multiplying prefixes..."  Shouldn't it be 'piperazin-1,4-diium', without the 'e' immediately before the locants?

@hBy2Py The elision of vowels is described in more detail in Subsection P-16.7: “Vowels are systematically elided as follows: (a) the terminal letter ‘e’ in names of parent hydrides (…)
when followed by a suffix or an ending beginning with ‘a’, ‘e’, ‘i’, ‘o’, ‘u’, or ‘y’;” Therefore, it’s “piperazinium” but “piperazinediium”.

Interesting. Obviously not the way I would've initially thought to do it, but I can see the logic. Thanks!

Thanks to AvnishKabaj reminding me of ChemSpider's structure search, the names for these compounds are piperazin-1-ium and piperazinediium.
Interestingly, the IUPAC name for the singly protonated species appears to require the locant for the protonated nitrogen, despite its being nominally superfluous, likely due to logic similar to the methyl group in the hydrocarbon discussed in this question.


text:
How to specify complete active spaces in the Dalton software package?
What do the numbers for specifying the inactive and active space of a Dalton calculation mean?
For example, I do not understand what the following numbers refer to:
.INACTIVE Doubly occupied orbitals
1 0 0 0 1s on oxygen

.CAS SPACE
4 2 2 0

Dalton manual p.14 has explanations for these values.

I am guessing that this is an example taken from the Dalton manual on page 13.
The complete listing is
**DALTON INPUT
.OPTIMIZE
**WAVE FUNCTIONS
.HF
.MP2
.MCSCF
*SCF INPUT
.DOUBLY OCCUPIED
3 1 1 0
*CONFIGURATION INPUT
.SYMMETRY
1
.SPIN MULTIPLICITY
1
.INACTIVE
1 0 0 0
.CAS SPACE
4 2 2 0
.ELECTRONS
8
**START
.SHIELD
.MAGNET
**PROPERTIES
.DIPGRA
.VIBANA
**END OF DALTON INPUT

There is a lot of unimportant stuff for your question, but we do a calculation on a water molecule, represented in the C2v point group, so we have 4 irreps.
We want to keep the 1s electrons on the oxygen doubly occupied, and to do this, we write
.INACTIVE
1 0 0 0

This means that there is one inactive (doubly occupied) orbital in the A1 irrep, while none of the B1, B2 or A2 irreps have any irreps forced to be doubly occupied.
This corresponds to the 1s orbital on the oxygen atom, which is so much lower in energy that is acceptable to not correlate.
A water molecule has 10 electrons, and since we have already used two to fill the doubly occupied 1s orbital on oxygen, we have 8 electrons left.
These are distributed according to the specification
.CAS SPACE
4 2 2 0

This means that we have 4 orbitals with A1 symmetry, 2 with B1, 2 with B2 and 0 with A2 symmetry, which we would like to correlate.

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


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Unstable or radioactive as a property?
What is better term for describing a chemical element; "radioactive" or "unstable"?
I am leaning towards the "unstable", because it seems to me that it better describes nature of the element, as radioactivity is consequence of instability.
However, I am not sure.

"Unstable" is ambiguous in certain contexts. For example, ozone is definitely unstable, but not radioactive. (Also, it is not really an element, but it consists of one element, hence a confusion may arise.)

So, you would actually prefer "radioactive element" than "unstable element" . If you specify what it is, could we call it better specified than if we used word radioactive?

An unstable element could mean that it reacts violently with air (For example caesium or rubidium). It could also mean that it's unstable as in it radioactively decays (for example thorium). Or as Ivan commented with ozone, it's not very stable and turns from O3 to O2. The word unstable for a chemical element (as Ivan commented) is pretty ambiguous. So it helps to have some context

The thing is, we have a word for that - "reactivity". Reactive / highly reactive. I understand unstable tho, but more in the sense of compound. Say explosives are unstable (or metastable).


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Anodisation of aluminium in sulfuric acid - Electrolysis reactions
As an experiment for electrolysis and to show how a metal can be protected of corrosion, I am trying to anodize an aluminium plate, using sulfuric acid as electrolyte.
I know the aluminium plate is the cathode and, thus, connected to the positive terminal of the generator, while for the cathode I am using aluminium foil.
Supposing I have $H_{2}SO_{4}:80ml, 3M$. I have two questions:
$1.$ Knowing:
$$\epsilon^{o}_{Al^{3+}/Al} = -1.67V\\
  \epsilon^{o}_{O_{2}/H_{2}O} = 1.23V, pH = 0\\
  \epsilon^{o}_{2H^{+}/H_{2}} = 0V, pH = 0\\
  \epsilon^{o}_{2H_{2}O/H_{2}} = -0.83V, pH = 14$$
I want an explicit explanation using those potentials as to why the reaction at the cathode is $$2H^{+} + 2e^{-}\rightarrow H_{2}$$
and why the reactions at the anode are ( If i am not mistaken )
$$Al \rightarrow Al^{3+} + 3e^{-}\\
  2Al^{3+} + 3O^{2-}\rightarrow Al_{2}O_{3}\\
  2Al^{3+} + 3OH^{-} \rightarrow Al_{2}O_{3} + 3H^{+}$$
  Especially the second and third reactions confuse me
$2.$ Knowing:
$$\epsilon^{o}_{Cu^{2+}/Cu} = 0.34V\\
\epsilon^{o}_{Pb^{2+}/Pb} = -0.13V\\$$
Could I use a plomb or copper blade instead of the aluminium foil?

Right now, people are flagging this as two broad: I think your two questions should be asked separately, as they seem relatively unrelated. If you edit this question to only contain one, and ask a new question, it may be better received.

As the experiment is described, in the solution we have :
$$H^{+}, H_{2}O, Al_{(s)}, SO_{4}^{2-}, OH^{-}$$
Taking that into account and seeing how the environment is acidic ( $pH<7$ ), we can see that the reaction of reduction reaction that can take place is $H^{+}/H_{2}$
Following the same kind of reasoning, reaction $(1)$ in the anode is caused by the odydation of the aluminium plate when placed in an acidic environment, $O^{-2}$ is present because of the oxygen in the atmosphere and $OH^{-}$ exists in the solution anyway(water autoionization).
As far as the second question is concerned, notice that the aluminium foil in the anode does not participate in any reaction. It is the $H^{+}$ that is reduced, while the aluminium foil serves as a conductor for the current. Knowing that, we could replace the aluminium foil with either $Cu_{(s)}$ or $Pb_{(s)}$, that may also be used as conductors ($Cu^{2+}, Pb^{2+}$ will not be present in the solution and we will have the same reduction reaction in the anode), no matter their standard electrode potentials.
Finally, it is interesting to point out that in this experiment the anode and cathode are in the same solution which is used at the electrolytic reactions (acidic environment ) and at the same time plays the role of the electrolytic bridge constantly providing $H^{+}$ and keeping the solution electrically neutral.


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Extraction of metals using electrolysis
I came across a question that goes as:

The electrolytic method of reduction is employed for the preparation of metals that are:

Less electropositive
Moderately electropositive
Highly electropositive

I was thinking that highly electropositive metals like Na+, Li+ and others which have highly negative reduction potentials should be difficult to reduce as hydrogen would be preferentially reduced. For example, in electrolysis of NaCl, Na+ is left in the solution as NaOH while it is hydrogen that is reduced.
However the answer is highly electropositive metals. I searched for the reason elsewhere but did not find it.

Your reasoning is not wrong, in that it takes additional energy to reduce metals like Na and K that are highly electropositive. This is the preferred method to extract these metals, though, because alternative methods are far more expensive.
For example, the first attempts to extract aluminum (or aluminium, if you prefer) by Ørsted, Berzelius and Wöhler between the 1820's and 1845 reacted potassium with aluminum compounds. The metal thus produced was costly, valued more than gold, even though it was neither pure nor useful. King Christian X of Denmark had a crown made of aluminum! In 1854, Deville brought down the cost to that of silver by using sodium instead of potassium... still rather pricy.
Paul Héroult and Charles Hall made innovations in 1886 that made electrolytic production of aluminum practical, so that now the cost is low enough that aluminum foil is considered a disposable (though better recycled) commodity.
With plentiful electricity, the cost of producing aluminum has dropped to ~US\$2/kg, and electrolytically produced sodium is ~US\$4,000/tonne.
Less electropoisitive metals are produced electrolytically where there is an economic advantage or to provide high-purity metals. Copper is a good example, because high-purity copper is a better conductor of electricity, so less metal is needed to make a wire of sufficient capacity, and because small amounts of precious metals such as gold and silver are recovered from the electrolytic sludge, helping to defray the cost of re-refining.

Your explanation was good. But one last doubt remains: what makes electrolysis 'better' for highly electropositive metals as compared to those that are moderately or less electropositive? Those with less electropositivity would rather be more easily reduced when electrolysed.

Economics: it is cheaper to use readily available coal, for example, to reduce iron oxide to iron, than to use electricity. That said, many less-electropositive metals of high purity are produced by electrolysis.


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Ionic liquids and deep eutectic solvent cation vs anion shape
If we consider an ionic liquid of a deep eutectic such as the choline chloride type DESs, why are ionic liquids (and deep eutectics) with not very symmetric cations and very symmetric anions more common than ionic liquids which are the other way around.
Consider for example Choline chloride (plus ethylene glycol) and butyl-methylimidazonium hexafluorophosphate

Complex anions are labile - exchange ligands till equilibrium. You just can't get most of mixed ligand inorganic anions.


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Is there a simple way to predict the molecular pi orbitals in conjugated pi systems?
Particularly for longer alkyl chains. Or is predicting these molecular orbitals from quantum mechanics too complex to be able to be translated easily as a geometric pattern? If Frost Circles exist, I just figured something could for this as well. The only pattern I've been able to identify is that each molecular orbital is symmetrical, starts with no nodes and goes up by one with each higher energy orbital but for systems with many carbon atoms, this isn't a specific enough set of rules.

How many carbon atoms are you intending to look at? From looking at this figure there does seem to be a pretty clear pattern in the same vein as a frost circle (at least for the even numbered case). @Elmer

It all boils down to eigenvalues of adjacency matrices.

@IvanNeretin That isn't really useful. Even if the questioner knows what those words are, they wouldn't know that they should look for the Huckel method.

@pentavalentcarbon That's why this was a comment and not an answer. Anyway, thank you for making it more useful.

@Tyberius Let's say we're looking at 10 carbons. And  what I'm looking for are the positions of the node(s) at each orbital. All I know now is that they go up by one each time starting at zero and ending between each pi orbital and that the positions of the nodes are arranged symmetrically, but this doesn't help me predict the orbitals for molecules beyond a certain length.

@IvanNeretin Can you refer me to a different resource that elaborates on this?

What's wrong with the mentioned Wikipedia page?

@IvanNeretin From my organic chemistry course I figured all Huckel's Method was just the 4n+2 rule restated, my bad. The math for it seems to be beyond me unfortunately but would you say its difficult to learn the Ritz method for example? Looks like linear algebra, something I haven't really been exposed to

This is linear algebra, and there is no going around it.

@Tyberius You answered it perfectly! Thanks for the help.

This should work for any even number of conjugated carbons (and with slight modification, for cations/radicals/anions of uneven numbers of carbons). You can follow a simple procedure to generate all the $\pi$ orbitals.

Generate the lowest orbital with no nodes
Generate next orbital by creating a node through the middle bond
For all subsequent orbitals, generate by taking the previous orbital, placing nodes to the left and right of the original node, and then removing the original node.
End when all bonds are separated by nodes

We can write these as binary strings where a 1 is one orientation and 0 is the other. So for example, you could write the lowest orbital of N=10 chain as 0000000000, the next as 0000011111, the next as 0000110000, and so on.
The Huckel method mentioned in the comments is a more formal way of obtaining the orbitals by solving for the eigenvalues and eigenvectors of an adjacency matrix. The adjacency matrix is formed by numbering the carbon chain and forming an NxN matrix with the rows and columns numbered like the carbons. In each row, put a 1 if the carbon of the row and column are adjacent and put an x if they are the same. A simple online calculator with a few examples and instructions for this can found here (this only gives the eigenvalues, you would probably have to download or code your own version to give the eigenvectors). You only really need these eigenvectors to get an equation for the orbitals, but for your case you can do that just as well with the simple procedure by simply writing the MOs as linear combinations of the AOs.


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What computational methods are available for predicting nanoparticle structure?
I'm a quantum chemist, and nearly every system I've ever cared about has been studied by DFT. However, I am currently trying to model the surface interactions on a nanoparticle around 100 nm in diameter. I know the composition of the nanoparticle, but I do not know which atoms are at the surface versus in the core. It's too large to really be reasonable in my regular DFT packages. Any ideas on where to start?

For big nanoparticles, you're probably looking at something like ReaxFF. For 100 nm, if this is a million or more atoms, classical MD is not really there. You would need to perform finite element analysis and lose all atomistic information.

Much appreciated @pentavalentcarbon! I think that's what I've been leaning toward and avoiding. Looks like I can't avoid it anymore...

I agree with the ReaxFF comment. However, are you sure that you need to model surface interactions on a 100 nm NP? I can't think of many reasons why you'd need to look at a 100 nm surface for a surface interaction (but it's not my work after all!). Why would a periodic cell containing dozens, or even hundreds, of atoms not be sufficient?

@Argon in this case, I am looking at a hetero-nanoparticle and the surface interactions are the second half of a project plan. Initially, I just need to figure out what atoms preferentially appear at the surface. From there, I can simplify my system and look at smaller sections of the surface.

@AlyChem Depending on the problem at hand, perhaps you can get away with a smaller cross-section of the NP surface by doing an ab initio thermodynamic approach to predict the most favorable surface structures (e.g. see here http://www.diss.fu-berlin.de/diss/servlets/MCRFileNodeServlet/FUDISS_derivate_000000002356/04_chap4.pdf?hosts or check out Chapter 7 of David Sholl's DFT text).


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Why is a spin change favourable in intersystem crossing?
All phosphorescent molecules go through the following transitions:
 $$\text{excited singlet state}$$
$$\Bigg\downarrow$$
$$\text{[intersystem crossing]}$$
$$\Bigg\downarrow$$
$$\text{ excited triplet state}$$
Why does that spin change always occur, given it’s a “forbidden” transition and, thus, unfavourable and much less likely to take place?

Well, if there's no easy path, unfavourable  needs to be taken.

The triplet phosphorescence  is also spin forbidden as is $S_1\to T$.  As it actually occurs there is an allowed spin changing mechanism that lets this happen.  This is spin-orbit coupling. See this answer for an explanation of radiationless transitions in general of which intersystem crossing is an example. https://chemistry.stackexchange.com/questions/28883/non-radiative-transitions-or-what-features-of-rhodamine-result-in-it-being-so-hi/55287

It doesn’t always occur,nor favorable.

There are two things going on here. You have conflated with "forbidden" with unfavorable. In many systems, the first excited triplet is lower in energy than the first excited singlet. Hence, the order of energies excited singlet > triplet > ground state singlet is correct and sensible. If your question is "why is this the order?" that should be a separate question.
The forbidden part of the transition has to do with the fact that, because of conservation of (electron spin) angular momentum, you can't excite directly from a singlet ground state to a triplet excited state with a photon. The forbiddenness is a statement about not being able to get there, not unfavorability. The triplet is still lower in energy and the molecule has to do some other shenanigans (often involving spin orbit coupling) to do the intersystem crossing into the lower energy triplet state. If your question is "what is the mechanism of intersystem crossing?", that should be a separate question.

Pardon my limited knowledge of quantum physics, for I really thought the first excited triplet is actually lower in energy. On spin orbit coupling, it appears it’s just the thing I needed, so I’m going to read about it, thanks a lot!


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Is there a simple way to memorize/reason out the solubility rules?
Normally, to determine if a compound is soluble, I would just apply the solubility rules, which can be found here.
The problem is, these are not fun to memorize as there are often loads of exceptions for each ion. Also, they are easy to forget. Is there an alternative way to reason out if something is soluble or not?
I was thinking that lattice energy might be a factor in determining whether or not something is soluble or not because if the lattice energy is low, the ionic compound can easily break apart and form ions. Thus, if the charges are low, it should be soluble. However, this hypothesis doesn't really explain why fluoride and hydroxide salts are insoluble.

Lattice energy surely is a factor, but there are no easy ways to memorize or derive it either.


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How is different flame color of alkali metals related to nuclear charge?
Why do different alkali metals produce different flame colors ?
According to my teacher,different atoms have different nuclear charge and due to difference in effective nuclear charge,the orbit to which an electron jumps differ even though energy provided from Bunsen flame is same.And on coming back it emits energy in the form of light.
This is very different from the reason I found when I googled my doubt. I wanted to know whether the explanation teacher me gave was correct.

There are different ways of explaining why electrons have different energy levels, but ultimately, the nuclear charge does determine the orbitals. What contradictory information did you find?

They produce different colours because different types of atoms have different and discrete energy levels. The photon is emitted when an electron moves from a higher level to a lower one. If you look up the (simple)  Bohr model of the atom you will see that the quantised (discrete) energy levels depend on $Z^2$ where $Z$ is the nuclear charge. You teachers explanation is correct.


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What is the major product of bromobutane and sodium ethoxide?

I know that the it is SN2 reaction and $\ce{Br}$ is the leaving group. Second Carbon(from right side) is the beta carbon. As the $\ce{Br}$ group leaves, $\ce{OH}$ will create a bond with one of the beta hydrogen and will lead to formation of a double bond between the second and third carbon.
Can someone please check and explain why the structure i got for major product is wrong ?

Do you mean E2…

This is an SN2 mechanism, not an elimination mechanism. The $\ce{NaOEt}$ is not bulky, and the leaving group is primary. Because of this $\ce{^-OEt}$ will attack the carbon where $\ce{Br}$ is attached via an SN2 backside attack. This will replace the $\ce{ Br}$ with $\ce{^-OCH2CH3}$.
I remember organic was a bit of a rough time. Here is a flow chart I used back in organic from Clutch Prep.

Isn't $\ce{EtO^-}$ a hard nucleophile? Wouldn't that favor the elimination product isnteas of substitution?

@AlyChem: What about solvent effect?

There are always exceptions, but typically even a hard nucleophile would undergo SN2 because of the primary leaving group. Solvent effects from what I've seen are definitely important, but not usually the driving force behind determining the major product. I'm definitely not an organic chemist, though.

You may be knowing that  SN2 do not undergo $carbocation$ $rearrangement$. Therefore your product is incorrect. Now The $\ce{NaOEt}$ is not a bulky group . Because of this $\ce{^-OEt}$ will attack the carbon where $\ce{Br}$ is attached via an SN2 backside attack. This will replace the $\ce{ Br}$ with $\ce{^-OCH2CH3}$.


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Why does the order of adsorption of hydrated alkali metals on an ion exchange resin decrease down the group?

Why does the order of adsorption of hydrated alkali metals on an ion exchange resin decrease down the group?

I recently did a question on this topic and they say adsorption order is $$\ce{ Li+ > Na+ > K+ > Rb+}$$
I answered it the other way around.  The reason that made me choose my answer was that  $\ce{Li+}$  has a small atomic radius and hence it should have high hydration enthalpy. It should be forming $\ce{[Li(H2O)6]}$ complex which is large enough to show low adsorptivity.
So same trend should be followed as we go down the group. But, it is not the correct answer. So, please point out my mistake.

Consider charge concentration as well.

I think the order should be the other way around. The size of the hydrated Li ion will be the greatest, followed by Na, K and Rb. Refer
This will make the charge to size ratio smallest for the hydrated Li ion, giving it the lowest adsorptivity. That is confirmed by the relative selectivity coefficients of these ions with certain ion exchange resins. Refer


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What is the identity of the unknown compound using GCMS and NMR?
I've been asked to identify an unknown compound by performing several analytical techniques on it.

The compound was a white powder and had a boiling point of 65 degrees celsius.
I have the GC mass spectrum as well as both Carbon-13 and Proton NMR spectra (note that the compound was dissolved in chloroform for NMR analysis).
So far I think it contains an aromatic ring because of the chemical shifts for the NMR spectra but that's all I have so far.

These are the images of the spectra (Fig 1. GC and Mass spectrum, Fig 2. Proton NMR spectrum, Fig 3. Zoomed in Proton NMR spectrum, Fig 4. Carbon NMR spectrum, Fig. 5 Zoomed in Carbon NMR spectrum):

Hints:

If you counted carbon signals in $\ce{^{13}C}\mathrm{~NMR}$, you'd find the compound has at least 10 $\ce{C}$ atoms.  Thus, it's indicated boiling point is incorrect, unless it's determined at reduced pressure. If so, you need to give that pressure.
You are correct about saying your compound should be aromatic. Did you count how many aromatic hydrogens are there? At least, by doing that, you may able to speculate some framework for your compound, as a beginning point.
Now, if you are careful, you see no aliphatic carbon signals in your $\ce{^{13}C}\mathrm{~NMR}$. Then, what is the signal at $\mathrm{\delta~1.6}$ in your $\ce{^{1}H}\mathrm{~NMR}$ doing there?
Now, go to $\mathrm{GC/MS}$. In there, your parent peak is almost looks like a doublet. Do you know why? Can you recall what element have the two stable isotopes about 50:50 natural abundance with $2~m/z$ units? If you don't remember, then subtract each of your parent peaks from your base peak in your $\mathrm{MS}$. Now, you almost have the empirical formula of your compound. Now it's matter of analyzing peak patterns in $\ce{^{1}H}\mathrm{~NMR}$ and fragmentation pattern in $\mathrm{MS}$.

Thanks for your response!
I meant the melting point was 65. Apologies for any confusion.
So far I think there are 10 carbons but I’m struggling to count the hydrogen atoms and determine whether the peaks at 7.733 and 7.717 are singlets or one doublet…
As for the signal at 1.6 in the H NMR I have a feeling it's water in the chloroform solvent?
I believe the element you're referring to is bromine? Does that then mean the compound contains two bromine atoms? If so a possibility could be C10H12Br2. But I still need to have a better look at the H NMR to try to count the hydrogen atoms

Emma, you are correct about water in the chloroform and bromine. So, I felt you are on right track. Now, you have confirmed there is no aliphatic portion in your molecule. Thus, it's a substituted aromatic compound.

PS: Also, you are wrong about having two bromine atom. It is a characteristic feature that if you have one bromine in the molecule, you's see equally intense two peaks ($\mathrm{2~m/z}$ apart for ebvery bromine containing fragment in $\mathrm{MS}$, each representing $\ce{^{79}Br}$ and $\ce{^{81}Br}$. Also, if you do close inspection on $\mathrm{^{1}H~NMR}$, you'd see only 7 proton signals. Now tell me what is the compound? :-)

I have a feeling it might be the solid form of bromonaphthalene as that would be consistent with the formula C10H7Br and the carbon and proton NMR peaks. However, I'm not sure the proton NMR spectrum above reflects the proton coupling in bromonapthalene (I think the spectrum should show more triplets). Thanks very much for your help so far though, you've been very helpful!

Bromonaphthalene has two positional isomers: 1-bromonaphthalene and 2-bromonaphthalene. Which is the solid close to mp of $65~\mathrm{^\circ C}$? And remember, each isomer has different $\ce{^1H}~\mathrm {NMR}$ patterns.

Ah yes that makes sense. 2-bromonapthalene has a melting point just below 60 degrees C, whereas 1-bromonapthalene has a much lower melting point and so is a liquid at room temperature. Thanks for your help.


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Why does potassium react more violently with water than lithium?
Recently, I was telling my friends about the violent reaction that takes place when you throw potassium into water. Soon after, a friend of mine claimed that lithium would react more violently than potassium.
I disagreed with him, because potassium is more electropositive than lithium and thus more reactive.
My friend claimed lithium to be more reactive than potassium due to its position in the reactivity series of metals:
$$\ce{Li > Cs > Rb > K > Ba > Sr > Ca > Na > Mg}$$
Then we found out that potassium reacts indeed more violently in water.
But what about his argument? Why isn't he right?

related: https://chemistry.stackexchange.com/q/2606/23561

For the reaction,
$$\ce{M -> M+ + e-}$$
the heat liberated is highest for lithium owing to its high negative $E^\circ$ value so one would think that the reaction must be most vigorous.
The reason behind the more violent reactivity of potassium rather than lithium lies in kinetics and not in thermodynamics.
No doubt, maximum energy is evolved with lithium but the vaporization and ionization will also consume maximum energy (the melting point and ionization energy of lithium are the highest) and so the reaction proceeds gently. On the other hand, potassium has a lower melting point and ionization enthalpy. The heat of reaction is sufficient to melt it. The molten metal spreads over the water and exposes a larger surface to water. Also, the hydrated radius of lithium is the greatest out of all alkali metals. This reduces the ionic mobility which in turn reduces the speed of the molten metal.
That's why potassium gives a more violent reaction with water.

Reference:

Kumar, Prabhat Conceptual Inorganic Chemistry; Shri Balaji Publications: Muzaffarnagar, U.P., 2014.

Could you please clarify this line, and its relevance with the question? "This reduces the ionic mobility which in turn reduces the speed of the molten metal."

More than 30 years ago in high school chemistry, our teacher put a flake of sodium in a beaker filled halfway with water and the sodium flitted around the beaker like a little motorboat letting out small wisps of smoke.
Once that was done he put in a small flake of potassium.  It reacted much more violently.  The potassium shot around the beaker with smoke and sparks.
The sparks came as a result of the reaction being so energetic that the heat of the reaction ignited the free hydrogen:
$$\ce{2 K + 2 H2O -> 2 KOH + H2}$$
The way he explained it was that the single electron in the outer shell of Potassium was further from the nucleus than Sodium (and of course even further from the nucleus than lithium) and was thus freed from the nucleus' electrostatic bond much easier.
This might not be the technical answer but it was easy for me to understand and remember all these years later.

The rapid kinetics of the reaction of sodium and potassium with water are well explained by the spikes formed and the "Coulomb explosion". But the unexpectedly slow reaction of lithium in water is still a puzzle.
I have watched a small piece of lithium wire (~2 grams) reacting in about 100 mL of water. The reaction is quite rapid at first, then slows down, and then almost stops. The major changes in the system (besides the evolution of hydrogen) are a moderate temperature rise and a pH increase to about 14. Lithium hydroxide is not very soluble (17%, hot). The temperature could not possibly get hot enough to melt lithium (mp = 180 C), whereas sodium melts at 98 C and potassium at 63 C. I watched an irregular piece of sodium (~20 grams) bubble in water (initially cold) until it melted and became globular - just before it exploded!
Lithium is reminiscent of magnesium in water, where magnesium develops a protective oxide/hydroxide coat which inhibits further reaction with water unless the water is very hot. It is worth noting that metals that corrode in water may be inhibited in alkaline media (like iron in concrete). Sodium and potassium are excepted from this inhibition because of the great solubility of their hydroxides and the possibility of melting.
Lithium may start out moderately fast, but slow down as the pH rises, and overall seem a lot slower than sodium or potassium because it will not melt in water and has a solid surface that becomes inhibited by the high pH after some reaction.

Lithium is very reactive, but reactions with water tend to proceed at a somewhat consistent rate.  Sodium and potassium are more prone to concentrate the reaction exclusively on outward projections in such fashion as to make a lump become more and more spherical, until there are no more outward projections, whereupon they will suddenly shoot out spikes in what is called a "coulomb explosion" which causes a massive increase in surface area and consequently reaction rate.  Search "sodium coulomb explosion" on YouTube for more information.
I don't know the exact details of why lithium doesn't go "bang", but I suspect that's because--unlike sodium and potassium--the will reaction proceeds in some measure on the entire surface of a lump.  The reaction may be concentrated on outward projections, but not exclusively as is the case with sodium or potassium, and thus the conditions don't arise that would force the lump to shoot out spikes.

Potassium is higher up in the reactivity series. Reactivity of metals increases as more shells are added. It becomes easier for electrons to be lost, which actually influences the reactivity of the metal.


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Do two Miller planes need to have similar arrangement of atoms?
For two Miller planes, do I need to have the same arrangement of atoms in both of them?
Does it need to look like the same plane translated a little?
I have this question because if this is not the case, then how do you do the calculations for interference, etc, using Bragg's law, where we assume the atom of the bottom plane is directly below the atom of the top plane?

In answer to you first question, no if hkl differ, yes if they do not. I try to explain below.
In a crystal the unit cell defines the repeating unit. Inside the unit cell the atoms are arranged as they are in the molecular structure with the molecule being at the same angle and position within each of the unit cells. Each atom scatters the x-ray radiation depending on how many electrons the atom has and on the position the atom has in the unit cell. The many repeating identical unit cell structures cause constructive interference and so spots are observed on the x-ray detector. Thus the structure is determined from the intensity of the spots on the detector not their position.
The hkl planes are a construction we devise to allow determination of the x-ray amplitudes because the constructive interference that causes scattering has to occur from atoms in the repeated positions as determined by the unit cell. (There are thousands to millions of unit cells in a crystal). Thus we define planes starting at  hkl = (0 0 0) and continue to larger values such as  (1 2 0 ) etc. etc. (100 50 1) etc. (Negative values have a line above the number $\bar 3$ for example.)
Suppose we choose the plane (1 2 3), this is  parallel to (2 4 6) to (4 8 12) and so on and these planes are separated by a distance we will call $d_{hkl}$. We can reduce all these  planes to (1 2 3) and the reason for this is that the phase of the x-ray wave from plane (2 4 6) is $2\pi$ from that of (1 2 3) and that of (4 8 12) is $4\pi$ shifted. But as the phase repeats itself over $2\pi$ all these larger phases are simply the same as $2\pi$. A phase change of $2\pi$ is equivalent to a path difference of $\lambda$ or a whole wavelength, which might be easier to envisage as then waves add together exactly.
Now, if an atom is found bang on the plane (1 2 3) it will have a phase of $2\pi$ (or path-length change of $\lambda$) with respect to the origin so has the maximum scattering intensity possible for that plane. However, consider that the atom in not on the (1 2 3) plane, but at a position $x$ away (and so between two (1 2 3) planes) then the phase it has is now $\displaystyle \frac{x}{d_{hkl}}2\pi $ and so the scattered x-ray waves are out of phase by this amount and so the scattered amplitude is reduced. Now, of course, every the atom in the molecule will be at some distance from one of the parallel (1 2 3) planes and so will have its own particular scattering amplitude to add to the total for that plane.
The same reasoning applies to each of the unique hkl planes and to each atom. It is the un-ravelling of these amplitudes that produces the x-ray diffraction structure and it is made more complicated because the x-ray detectors measure the intensity of the scattered light (which is  amplitude squared (as $I^*I$)) which means that the phase information is lost. This causes great difficulties in solving the x-ray structure.


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What is Vorlander's rule?
What is Vorlander's rule? I have got it from a book all of a sudden, but without any further information about it; could you please explain it a bit?

Did you googled? https://books.google.pl/books?id=CRgSBQAAQBAJ&pg=PA135&lpg=PA135&dq=Vorlander%27s+rule&source=bl&ots=1NEobHpute&sig=MWubdrWgiH4yDfviMgZq8tS9K14&hl=pl&sa=X&ved=0ahUKEwiTqYWx2cbaAhXRsaQKHTzpD7cQ6AEINDAJ#v=onepage&q=Vorlander's%20rule&f=false

Also such "rule" seems rather part of history of chemistry.

Vorlander's rule discuss about the directing powers of substituent present in benzene ring. Taken from here:

Saturated groups (containing single bonds) direct predominately 2- and 4-positions.
Unsaturated groups (containing multiple bonds) direct predominately at 3-positions.

Here are some examples:


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Reaction of carbonyl cyclic diene with Grignard reagent

Predict the major product of this reaction:

The two compounds on the right are the possibilities that I thought of (weren't given in the question). According to me:

attack on carbonyl carbon by $\ce{MeMgBr}$ and then hydrolysis with $\ce{aq. HCl}$
first tautomerise to form phenol. $\ce{MeMgBr}$ will remove acidic proton of phenol. The $\ce{aq. HCl}$ workup will then add the hydrogen back to phenol.

I an confused as to which product would be the major product. I thought phenol would be the major product. But to my surprise, the answer given is 1-methylcyclohex-2-enol. I am not able to understand how it is formed, and need help here.

Source: Joint Entrance Exam (JEE) 1997 India

Have you looked at grignard reagents?

Yes sir.. I know how grignards react. And drew the products based on that knowledge

That's a pretty blatant mistake. Tautomers of phenol enolise so quickly they can't be isolated.

you may enolise and then form the product.

I am sure that phenol will react with a Grignard as a Bronsted acid, thus forming a magnesium phenolate.

Upon further review, maybe they misdrew the substrate structure?  The given answer makes sense if we have only one carbon-carbon double bond conjugate to the carbonyl group.

Hmm, I wonder how easily that alcohol decomposes to toluene on acidic workup...

The given answer makes no sense with the substrate actually presented (but see "alternative possibility" below). You will get phenol.
If you start with phenol then, of course, the Grignard reagent just yanks off the hydroxyl hydrogen, only to have that hydrogen restored by the acid workup.  Now suppose you were to have the keto form cyclohexa-2,4-dien-1-one.  This tautomerizes to phenol precisely because the methylene hydrogen is more strongly acidic than it would be as a hydroxyl hydrogen in phenol.  So the Grignard reagent would if anything, go even faster for the methylene hydrogen in cyclohexa-2,4-dien-1-one than it would go the hydroxyl hydrogen in phenol - unless the ketone enolizes first.
That latter issue is academic.  Either way you get the same phenolate ion product, which upon acid workup gives phenol.
An alternative possibility
If the substrate structure in the question was misdrawn, then the given answer could be correct. Cyclohexa-2-en-1-one, with only one double bond in the ring conjugated to the carbonyl group, does not have any (relatively) strongly acidic hydrogens and so the Grignard reagent will add to the carbonyl group.  After acid workup the ketone is thereby converted to the tertiary alcohol in the given answer.
Grignard reagents do not react with carbon-carbon double bonds, so after adding to the carbonyl group or extracting an acidic proton that's all there is.

Thank you sir, yes maybe they misdrew the answer. I was thinking MeMgBr was in excess.. will the answer be still phenol/phenoxide??

Methyl Grignard will never add to phenoxide

And it does not in my answer.  The "alternative scenario" does not proceed via phenoxide.


text:
What is the mechanism for iodoform test given by 2-aminoalkanes?
I found 2-aminoalkanes in the list of compounds that give positive iodoform test. I was surprised as they have no keto methyl or methyl carbinol group.
Can someone help with the mechanism?

https://en.m.wikipedia.org/wiki/Haloform_reaction

https://www.google.co.in/url?sa=t&source=web&rct=j&url=https://chemistry.stackexchange.com/questions/73044/iodoform-reaction-of-alkyl-halide/73071&ved=2ahUKEwiqlof858baAhXGuI8KHdcTCy8QjjgwJ3oECAEQAQ&usg=AOvVaw3mGM4lY1x8_G6iOt5jDudF

See these two links.. you would be able to make out the mechanism.. just replace double bond O with imine everywhere and hydroxyl with amine

I require the mechanism involving the amino alkane ,if you could. :)

Ive seen the other mechanisms in my textbook but i have never seen the involvement of this amine compounds.I even googled and I couldnt find any mention of aminoalkane in iodoform reactions.

@NihalN.L Can you post the list here? And mention the source of the list

@Rishi I previously posted same comment and deleted because OP is no longer active.

There is a comment elsewhere in this post saying:

"(The mechanism for iodoform test given by 2-Aminoalkanes might be) similar to the iodoform reaction of secondary alcohols. It is oxidized to imine, maybe then similar procedure."

I won't necessarily agree with that argument. To have a positive iodoform test, it is necessary to acitivate the methyl group (making its three α-hydrogens acidic). That means $\ce{-CH(NH2)\!-}$ of $\ce{RCH(NH2)CH3}$ should be oxidized to $\ce{-C(=NH)\!-}$ group by the available oxidizer, $\ce{I2/NaOH}$. Even if this oxidation happens, without transforming it to the corresponding keto group, those $\alpha$-hydrogens are not active enough to remove by $\ce{NaOH}$.
The reality is it is not easy to oxidize amines compared to alcohol in this situation. In their 1969 publication (Ref.1), E. J. Corey and coworkers have said it the best:

In contrast to the generation of carbonyl compounds by oxidation of primary and secondary alcohols, which is used with great frequency in synthesis and for which many selective reagents are known, the efficient oxidation of primary amines to carbonyl compounds has only been realized by the fore mentioned process.

One of the oxidation method used in the past has been $\ce{NBS/base}$ followed by adjusting the resultant solution to $\mathrm{pH} \ 2$ at $\pu{0 ^\circ C}$ to convert a primary amine to corresponding ketone (Ref.2):

At least, this example shows that the conditions needed to make a ketone from the corresponding imine is to make the solution acidify. Even if the conditions used in the iodoform test $(\ce{I2/NaOH})$ could have been able to oxidize the 2-aminoalkane to the corresponding imine (I doubt it though; vide supra), you still needed to acidify the imine to convert it to the corresponding ketone. Therefore, this point alone, it is safe to conclude that this iodoform reaction is not happening with any of 2-aminoalkanes.
Nonetheless, I am not able to find any reference showing iodine (a milder oxidizer) or $\ce{NaIO}$ $(\ce{I2/NaOH})$ oxidizing amine in literature. Even, $\ce{ICl}$ gives an adduct with amines such as 1-aminotetraline or 1-amino-1-phenylethane, which is used to prepare lactones from corresponding alkene carboxylic acid (Ref.3). Thus, I can say that my reasoning is good enough to convince that this reaction is not possible.
References:

Elias J. Corey and Kazuo Achiwa, "Oxidation of primary amines to ketones," J. Am. Chem. Soc. 1969, 91(6), 1429–1432 (DOI: https://doi.org/10.1021/ja01034a027).
Elias J. Corey, Niels H. Andersen, Robert M. Carlson, Joachim Paust, Edwin Vedejs, Isidoros Vlattas, and Rudolph E. K. Winter, "Total synthesis of prostaglandins. Synthesis of the pure $dl$-$\mathrm{E_1},$ -$\mathrm{F_{1\alpha}},$ -$\mathrm{F_{1\beta}},$ -$\mathrm{A_1},$ and -$\mathrm{B_1}$ hormones," J. Am. Chem. Soc. 1968, 90(12), 3245–3247 (DOI: https://doi.org/10.1021/ja01014a053).
Jürgen Haas, Stewart Bissmire, and Thomas Wirth, "Iodine Monochloride–Amine Complexes: An Experimental and Computational Approach to New Chiral Electrophiles," Chemistry, A European Journal 2005, 11(19), 5777-5785 (DOI: https://doi.org/10.1002/chem.200500507).

@Safdar Faisal: you'd miss nothing if you read my conclusion carefully. I'll boldface that statement to be clear.

@MathewMahindaratne I am no expert but doesn’t your answer state iodoform through oxidation is most likely not possible, which doesn’t necessarily rule out the iodoform reaction of the compound through some other pathway?(Knowing that it wouldn’t take place through oxidation is still, though an important piece of information)

@ Rishi: To have a positive iodoform test, you need to acitivate the methyl group ($\alpha$-hydrogens acidic). Without having a keto group those hydrogens are not active enough to remove by $\ce{NaOH}$. If it is not clear for every reader I can make a edit or you may as well.

Addendum to Mathew Mahindaratne's answer (though not appropriate to edit his post for this purpose). As Rishi pointed out, let's try to find a new path for the reaction to occur.
$\ce{RNH2 + NaOH <=> ROH + NaNH2}$
That is a secondary amine forming $2^\circ$ alcohol through $\mathrm{S_N}$2 path. But $\ce{NH2^-}$ is a very poor leaving group as compared to $\ce{OH-}$, so the equilibrium lies largely to the left and hence the above reaction practically does not occur.
Source
Another path is the reaction below, but even that cannot occur because amines such as isopropyl amine are very weak acids, with $\mathrm{p}K_\mathrm{a}$ around 35, so you will need a much stronger base than $\ce{OH-}$ to deprotonate them.
$\ce{RNH2 + OH- -> RNH- + H2O }$
Source

IMO citing Quora and some site that looks like it's been made by toddlers for toddlers as sources isn't a good idea — any intro-level OC textbook would do way better.

Note that using a degree symbol as ordinal indicator in STEM might be confusing and hence this notation is not recommended. Also, $\mathrm{S_N}$2 is poor formatting since 2 may easily become an orphan and there are two different typesetting in a single abbreviation: switch to either SN2 or $\mathrm{S_N2}$.

@andselisk I will work for 1st comment after my exams. Also, this facts are much simple to be known to a person reading this article, thought of taking these reactions as a possible path was more important. $\mathrm{S_N}$2 was done by Mathew hence I don't thought about it.


text:
How can HCl be liquid if it has a boiling point at -85°C?
Shouldn't it at least be boiling? When water reaches its boiling point it makes bubbles, but that doesn't happen with HCl, why?
Did I misunderstand some basic chemistry?

Hydrochloric acid is hydrogen chloride gas dissolved in water.  We often render the solution as HCl when in reality, only the solute is actually HCl.

Thanks, side question; how many acids have this feature?

All acids are essentially some compound dissolved in water.

Not really @Frank.  There are fluorinated superacids (and even at least one chlorinated one, the "carborane superacid") that do not contain oxygen and can't be water plus anything else.

Search "Magic acid", that should be fun

All halogen compounds with hydrogen ($\ce{HX}$) are gases at STP except for $\ce{HF}$, which is a liquid (b.p.: $\mathrm{19~^\circ C}$). When dissolved in water, $\ce{HX}$ dissociates to ions ($\ce{H+}$ & $\ce{X-}$), which are readily hydrated by water molecules as $\ce{H+ (aq)}$ and $\ce{X- (aq)}$, hence not exists as a gas any more.
To answer your side question, a lot of gases have this feature. For instance, sulfur trioxide ($\ce{SO3}$) is a gas but gives sulfuric acid ($\ce{H2SO4}$) when it come to contact with water. Similarly, nitric acid ($\ce{HNO3}$) can be considered as an aqueous solution of nitrogen dioxide ($\ce{NO2}$). In reality, during an industrial process, $\ce{NO2}$ react with $\ce{H2O}$ in air to give $\ce{HNO3}$.
$$\ce{4NO2(g) + O2(g) + 2H2O(l) -> 4HNO3 (aq)}$$
Exception to this trend is carbonic acid ($\ce{H2CO3}$), which decomposes under ambient conditions:
$$\ce{H2CO3(aq) -> CO2(g) + H2O(l)}$$

$\ce{HNO3}$ is NOT a solution of $\ce{NO2}$ in water, because the nitrogen oxidation number is not the same in $\ce{NO2}$ and in $\ce{HNO3}$. It needs oxygen to produce $\ce{HNO3}$ in water. If oxygen is missing $\ce{NO2}$ gets dissolved in water but it produces a mixture of $\ce{HNO2}$ and $\ce{HNO3}$ according to the equation : $$\ce{2 NO2 + H2O -> HNO2  +  HNO3}$$

You are correct. I already has mentioned it in my answer.


text:
How much oxygen will we need to combust 100 grams of diamond
How much oxygen will we need to fully combust 100 grams of diamond given in weight and volume at standard pressure and temperature

diamond is pure C, so just take the mass, look up how many atoms ("mole") per mass.
Realize that the result of combustion is CO2, meaning 2 oxygen atoms per atom C, look up the mass per mole of oxygen, et voilá.
Then look up what gases at one atmosphere pressure take up in terms of volume and you have that, also.

With your answer, OP can stop thinking and just follow the recipe instead. Not good.

@Karl There's so many fully calculated answers to such homework dumps then this relatively good, and, from new user, surprisingly good.

@Mithoron Agreed. I still doubt OP is going to learn anything. Well, you can't force people, can you? ;-)


text:
How do I visualise excited state MOs?
I did an excited state calculation (TD-SCF)in Gaussian and wanted to see the MOs of the first excited state. But when I opened the checkpoint file in GaussView5 and clicked edit -> MOs, I could only see the configuration and MOs of the ground state. How do I see the excited state MOs in GaussView or other visualisers that can be downloaded for free?

Please let it open, as in my eyes there need some misunderstandings be cleared.

I have overruled the community decision of closing; if any of the close-voters are unhappy about that, please open a question on [meta.chemistry.se]. To the original poster I do however strongly recommend to add more context to the question. In any case it is a good idea to specify the versions of the programs you are using. It would also be helpful to know what route section you use. (cc @ph13)

Hm. I am not a Gaussian user but this is almost certainly because one does not need the time-dependent density to calculate the time-dependent energy. Adding density=current should do it.

@pentavalentcarbon That keyword doesn't work.

What you see, when you open your checkpoint file, is all you can see. I assume you use TDDFT and with this method you can't obtain other orbitals than the ones from the ground state. Why is this so?
Because in LR-TDDFT you calculate your ground state and apply a perturbation to it. (You can look up on the behind theory in good books like Introduction to Computational Chemistry from Frank Jensen.) The result from this calculation is presented from Gaussian like this:
Excited State   1:      Singlet-A      1.0662 eV 1162.87 nm  f=0.0007  <S**2>=0.000
110 ->113         0.60155  (0.72)
110 ->115        -0.23139  (0.11)
110 ->116         0.22569  (0.10)
Excited State   2:      Singlet-A      1.2549 eV  988.02 nm  f=0.0002  <S**2>=0.000
109 ->113        -0.39288  (0.31)
110 ->114         0.51726  (0.54)
Excited State   3:      Singlet-A      1.2929 eV  958.99 nm  f=0.0027  <S**2>=0.000
109 ->113         0.50172  (0.50)
110 ->114         0.29522  (0.17)
(The weights in spaces are added by me.)

These are the configurations of three excited states. The first excited state, for example, consists mainly (72%) from an excitation from MO 110 (HOMO) to MO 113 (LUMO+2). There are some contributions with weights of about 10% into the MOs 115 and 116. (On how to get the weights, have a look at an answer by me on this site here.)
You ask, how you can see the excited state in GaussView. You can't ... directly. And this question is also a little bit tricky, because what is it that you want to see? Here are some options:
You could

calculate the electron density of the excited state TD(Root=1) Density(Current) and subtract the electron density from the ground state. Then this difference shows you the "hole" where now is less density and the "hole" where there is more density.
calculate Charge Density Differences (CDD) with Multiwfn or Electronic Difference Density Maps (EDDM) with GaussSum which show you the same but a little bit easier as you do not need an additional calculation because they are based on the excitations and the CI coefficients.
calculate the transition density for the excited state TD(Root=1) Transition=1
calculate Natural Transition Orbitals with, e.g., Gaussian or Multiwfn, for very diffuse excitated states as NTOs reduces/optimizes the number of contributing excitations for each root with Density=(Transition=1) Pop=(Minimal,NTO,SaveNTO)

Just have a look on the manual for TDDFT, Density, FAQ and Pop.

my "Pop"-Link is wrong, I'll change it when I extend my answer soon

What I want to see is the MO of the excited state. In your example, in the first excited state, an electron transfers from MO 110 to MO 113, so I hope to see MO 110 and MO 113. Of course I can see the ground state version of them in GaussView, but I can't see the first excited state version of the two MOs.

That is what I said, you will not get different orbitals. If you optimize your geometry for this excited state, you'll get the orbitals you want. But still you would have to look at the "ground state" orbitals in the geometry of the optimized excited state.

Ok thanks. I'm not trying to visualise it any more. By the way, I don't think the weights you added are useful. We just look at the coefficients and they're helpful enough.


text:
Why are some smells so hard to get rid of?
What may seem to be a silly question could actually have roots in some core chemistry principles. The other day, I was cooking fish when I noticed that I'd forgotten to use a fork, and touched the fish with my bare hands. When I tried to get rid of the fishy smell, it just wouldn't go away.
Why are some smells so hard to get rid of?
My Hypothesis:
I think that the phrase "hard to get rid of" really means "resistant to washing," because that's what we're actually thinking of when we say you can't get rid of a smell. And when we say "wash" we really mean "attempt to dissolve in water using soap emulsification".
I'd conclude that "a smell being hard to get rid of" is caused by chemicals that are not soluble in water, and for some reason, possibly because the molecules are too large to fit in a stable emulsive bubble, refuse to be emulsified into aqueous suspension by soaps. But that's just my theory.

related https://chemistry.stackexchange.com/questions/2384/why-are-smoke-odors-persistent-in-fabrics

Turns out noses can be almost part per trillion detectors in certain cases: https://en.wikipedia.org/wiki/Odor_detection_threshold

Food flavors and odors are comprised of complex mixtures of hundreds of chemicals, which act as stimuli to the human sensory organs, for example, human nose. The odor of food may be attributed to the odor of a single chemical or an integrated response by nose to a mixture of chemicals. This reference to a single chemical or a mixture of chemicals allows trained analysts to develop a common terminology in describing odor, one of which called odor thresholds. Threshold is defined as the degree of stimulation of a nerve or nerve center that just produces a response. If the concentration of the odorous chemical (usually given in parts per million or ppm) is below this point, it will be judged not odorous. Hence, accepted definition of threshold is the minimum
identifiable odor (MIO) or recognition threshold. MIO of some chemicals is very high while some is extremely low. The latter chemicals are the ones you said,  "hard to get rid of." For example, the oder of dimethylformamide is described as 'fishy' and pungent, but you won't smell it if its concentration in air (or on your skin) is below $\pu{100 ppm}$. On the other hand, tar-like, pungent smell of p-cresol can be detected if its concentration in air (or on your skin) is greater than $\pu{0.001 ppm}$.
Now considering the fish smell, it was described short and sweet in following website from American Society for Nutrition: Read
Accordingly, it says:

Water in the open ocean is about 3% salt by weight, but the optimal levels of dissolved minerals inside an animal cell is less than 1%. In order to maintain fluid balance, ocean creatures must fill their cells with amino acids and amines to counter the saltiness of seawater. Ocean fish tend to rely on trimethylamine oxide (TMAO) for this purpose.
The problem is that when fish are killed, bacteria and fish enzymes convert TMAO into trimethylamine (TMA), which gives off the characteristic “fishy” odor.

Now, it is nice to know MIO of trimethylamine (TMA) is $\pu{0.00021 ppm}$ or $\pu{0.21 ppb}$ (parts per billion). That's why it is so "hard to get rid of" fishy smell whenever you touch or cook dead fish. But the good news is TMA is a organic base, which would react with any acid to form water soluble ammonium salts. Citric acid from lemon (or lime) or acetic acid from vinegar would do the job.


text:
Is Reverse Osmosis Sanitizer safe to use?
The Sanitizer is called Sani-System (For Reverse Osmosis). The purpose is to get rid of the bacteria within a Reverse Osmosis system (Priority is the tank: that holds the water)
-- edit
Safe meaning that most of the chemicals will wash out of the water tank, after a few flushes (dump & fill). I am worried if it will just make a layer within the tank, and it won't wash out completely. It is important since I am going to take care of 'more fragile' people.
-- edit 2
I don't work on behalf of the company. I am just figuring of if this is 'safe to use.'
-- Final Edit...
I will replace the tank periodically instead of using chemicals to clean the system. :D

https://www.youtube.com/watch?v=JzBlL_w_Afg
I want you guys to look at this sheet and tell me if this looks safe to use.
Safety Data Sheet
Thank you.

"Safe" is a relative term.  The mere fact that this product is produced with the explicit intent for its use in RO units (written on the package) is strong evidence that it is, indeed, safe as defined by society.  If you can be specific in what your personal definition of "safe" is, it will be much easier to help you determine if this product meets your standards or not.

I entered what safe means in my situation.

The first thing I would recommend is to learn any legal liabilities you may have.  Do not take the advice of anyone on the internet when it comes to legal matters.
That being said, the first thing I'd ask is whether you are operating as part of a company which may have company policies to follow.  This sort of thing is certainly common, and they may already have an answer.
Failing that, the next thing you should do is acquire the Material Safety Data Sheet (MSDS) for your product.  This datasheet should be made available by the company who produces the product.  It should be the first place to go.  I could not find a MSDS for the particular product you linked, but I did find one from the same company for their liquid sanitizer, which appears to use the same ingredients.
Now I am not in a position of authority on the precise risks with your "more fragile" people.  I cannot say how they will react differently than a normal person.  You will have to do your own research on that.  However, to your question of whether a film might form, we can address that and provide some evidence that the product will not form the layer you are concerned with.

The MSDS states that the product's solubility in water is "complete."  This means that it will dissolve in water with substantial ease, and is unlikely to come out of solution afterwards.
The MSDS lists hazardous polymerization as "Will not occur."  A polymerization reaction would be the most likely way a product could dissolve into water and then become insoluble later.
The MSDS lists hazardous decomposition as yielding nitrogen oxides and ammonia.  You can do your own research on this, but these are not compounds that are known to deposit in solid form.

Between these, it is unlikely that this particular product will form a layer in your tank.  I can't state it as an authority, but I myself do not see any indicators that would lead me to such a concern.
The instructions for use for this product will certainly call for some amount of flushing of the system after use.  You may choose to flush it more times to provide an additional margin, but without being a medical doctor who is aware of the particular conditions of your "more fragile" people, I cannot say whether this will meet their needs.
Another MSDS for one of the active components in your sanitizer lists the LD50 as 240mg/kg (rat).  For a small 50kg person, that suggests that a lethal does is on the order of 12g.  Obviously that may or may not scale to your people in particular, but it does go to provide evidence regarding how dangerous or benign these compounds are.

I don't work on behalf of the company. I am just figuring out if this product is safe to use.

I am just going to replace the water tank every 3 or so years. Then I don't have to worry about using chemicals.


text:
Break down hemoglobin into heme
Is there chemical that can break down the red blood cells/hemoglobin into smaller protein sub units/globins?
I would like to extract the heme molecules from the blood but not necessarily purify it,or separate it.I would like to get solution containing liberated heme units,its not problem if other stuff,like proteins and globins are floating in that solution,I just want to detach the heme part from the protein.Heme is iron containing coordination complex that binds to oxygen and give blood its red color.

I believe the chemical you are talking about that can break down the red blood cells/hemoglobin into heme and globins is carboxymethyl cellulose (CMC; see Ref). There is a US patent claiming a method for separating blood hemoglobin into heme and globin from blood cells. The abstract states that:

The invention relates to a method of dividing blood hemoglobin into heme and globin by suspending blood cells, separated from blood, in an acid aqueous solution, by precipitating the heme by means of carboxymethyl cellulose, and by separating the precipitated heme-CMC complex out from the aqueous solution of globin.

Title of the patent: Method for dividing blood hemoglobin into heme and globin
Patent: Karin Autio, Martti Kiesvaara, and Yrjö Malkki, US Patent 4,518,525, 1985 (11 pages)


text:
Why doesn't KI(aq) react with HCl(aq)?
When concentrated sulfuric acid is added to anhydrous potassium chloride and the fumes produced are bubbled into aqueous potassium iodide solution, the observed solution would be colourless solution.
I think the first reaction is:
$$\ce{2KCl(aq) + H2SO4(aq) -> K2SO4 + 2HCl(g)}$$
I assumed the second reaction would be:
$$\ce{HCl + KI -> KCl + \frac{1}{2}I2 + \frac{1}{2}H2 }$$
However the second reaction is wrong according to the answer book as $\ce{HCl}$ won't react with $\ce{KI}$. From what I know, $\ce{Cl}$ is a stronger oxidising agent than $\ce{I}$, so shouldn't $\ce{I-}$ in $\ce{KI}$ be oxidized to $\ce{I2}$? I am an A-levels student so would appreciate a simpler answer.

The first reaction is essentially anhydrous and should generate HCl gas; sulfuric acid doesn't oxidize Cl-. The first equation is not relevant. The purpose of bubbling the HCl thru KI solution is unclear. It is stated I2 is not formed, it should not be; aqueous hydriodic acid is exothermic while gaseous HI is slightly endothermic.  Finally chloride ion or Cl in HCl has no oxidative power. Looking at energies it is possible that a saturated solution of HCl and KI at high T and P might produce H2 and I3-, but the oxidizing agent is H3O+ not Cl-.

In your first reaction, you added concentrated sulfuric acid to anhydrous potassium chloride. In your equation, you wrote $\ce{KCl (aq)}$, which is incorrect. Under dilute conditions, that reaction would not take place because all reagents and products would be in aqueous ionic state, meaning there is no reaction.
In your second reaction, the equation is completely wrong. It is true that elemental $\ce{Cl2}$ a stronger oxidizing agent than elemental $\ce{Br2}$ or elemental $\ce{I2}$. But in current situation is ionic $\ce{Cl-}$ versus ionic $\ce{I-}$ where both are aqueous. If all reagents are in ionic form and expected products are also ionic, there is no reaction. See your reaction below:
$$\ce{H+(aq) + Cl-(aq) +K+(aq) + I-(aq) <=> H+(aq) + I-(aq) +K+(aq) + Cl-(aq)}$$
There is no gas, liquid or solid formed, but all ions. Therefore, there is no reaction.

As said by @Mathew that the reactants are in aqueous form and hence the solution only contains ions and thus no reaction takes place. But, if the conditions are changed slightly and if some energy is applied, then the reaction should take place.
So, if the reaction takes place in the presence of oxygen at room temperature and if light is applied as a form of energy, then the reaction takes place. Light as energy helps bonds to break and join to form products and oxygen is applied as an additional oxidising agent as hydrochloric acid is not enough to oxidize $\ce{KI}$. Here is the reaction given:
$$\ce{4KI + 4HCl + O2 ->[h\nu] 2I2 + 4KCl + 2H2O}$$

What happens in the second reaction solution will depend on the concentration of the reagents. When the hydrogen iodide concentration is high, it can form iodine. This is not a reaction of HI with an oxidant. What happens is
2HI ---> H2 + I2
If the HCl concentration in your experiment was very low in the second aqueous solution then I would not expect the reaction to occur. But if you look at an old bottle of concentrated HI, it will always be very dark due to the fact that it contains HI3. I know this as I have used HI solution to convert alcohols into alkyl iodides in lab.
Let us think for a moment about solvent extraction of metals from nitric acid, one of the problems is that the nitrate concentration is not equal to the anayltical nitric acid concentration (by titration) this is due to the fact that the Ka of nitric acid is not infinity. In the same way HI does not have an infinite Ka value.
As the anayltical concentration of HI in a solution increases, as the solvent properties will change and as Ka even in a very dilute solution is not infinity then the concentration of HI (intact HI) will increase. Also the concentration of the HI in the head space above the liquid will also increase.
The thermodynamics of formation of HI from I2 and H2 suggests that it should break down in to its elements. While Max Bondenstein did his work in the gas phase at a much higher temperture, it may be worth looking at.
Sulfuric acid will not form chlorine gas when it reacts with potassium chloride, it will form hydrogen chloride only. I know this as I have done this type of reaction myself.
When I was a teenaged boy I heated sodium chloride and sodium hydrogen sulfate in a test tube, it formed hydrogen chloride gas and not chlorine gas.

Thanks. Why does it form H2 and I2 when concentrated but not when dilute? I would appreciate even if you told me what to look for regarding this.

HI in aqueous solution is very slightly exothermic at high concentrations so it can possibly decompose slightly. At low concentrations it is more exothermic. If O2 is present Iodide ion will oxidize. In the gas phase it is an endothermic compound. One must consider energy and equilibrium constants to determine the direction of reactions.


text:
Is primary valency always ionizable?
In my textbook it is written that primary valences are ionizable. I saw a case of complex $\ce{[Co(NH3)5Cl]Cl2}$  Here it is written that primary valency is 3. The $\ce{Cl}$ inside coordination sphere serves as both primary and secondary valency. Since the constituents inside are non ionizable, will it be right to day that primary valences are always ionizable?

Primary valencies need not always be ionisable.It more correctly refers to the oxidation number of central metal atom.