AChE Aging time of organophosphorus compounds containing hydroxyl groups Organophosphorus compounds are known to inhibit the enzyme acetylcholinesterase (AChE). This occurs when the OPC phosphorylates the serine-203 residue of the enzyme. If the enzyme is not reactivated in time, a second reaction occurs, whereby one of the alkyl groups on the OP molecule is hydrolyzed off, giving the alcohol and the phosphorylated serine residue. The hydrolysis reaction also protonates the nearby Histidine-447 at the active site of AChE, forming a salt bridge between the OP and the enzyme. This is called "Aging", and makes the inhibited enzyme resistant to activation even by oximes. Depending on the OPC in question, the aging reaction can take between a few minutes (In the case of Soman) to several hours (In most cases). Here is a figure explaining the reaction:

Sources:

  https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=7&cad=rja&uact=8&ved=2ahUKEwi0r7T4g8jnAhXCK80KHfFCBIEQFjAGegQICRAB&url=https%3A%2F%2Fwww.mdpi.com%2F1420-3049%2F22%2F9%2F1464%2Fpdf&usg=AOvVaw0WTpEABL2j3q-kXjlvKdQM

  https://www.ncbi.nlm.nih.gov/pubmed/8250487

The part which I have circled in red depicts the aging reaction. Firstly, is the information which I have found thus far correct?
Secondly, If an organophosphorus AChE inhibitor which already contained a hydroxyl group, such as EA-2192

or (Presumably) fluoro(methyl)phosphinic acid

were to inhibit AChE, how long would the "Aging" reaction take? My guess is that aging would occur even faster (Seconds to minutes) due to the OPC itself being capable of acting as a proton donor, and thus leading to the formation of the salt bridge without any additional hydrolysis needing to take place beforehand. However, I have not been able to find any information on this particular matter, and so that is just pure speculation. Has any research been conducted into this? Is my speculative guess likely to be correct?

Acetylcholinesterase is important enzyme and its mechanism of function is well known. In active site, it has two pockets: anionic pocket (to attract positively charged Quaternary nitrogen of choline) and esteric pocket. The esteric pocket consists of a serine side chain (a $\ce{-CH2-OH}$ function) as depicted in following cartoon, which pacilitate the deseterification of acetylcholine to choline:

The neurotransmitter, acetylcholine bound to this pockets and $\ce{-CH2-OH}$ function of serine attracts carbonyl group of acetyl group as depicted. As a result, choline group will be released and acetylated serine group would then release acetyl group as acetic acid in a moment by hydrolysis to reactivate the enzyme.

However, if you introduce a competitive inhibitor such as neostigmine (which is a reversible inhibitor), it would bind to some of enzymes and decelerate the enzyme for designed time (e.g., use of this phenomena as a medical treatment). Once the inhibitor was released in an extended time period, enzyme will be reactivated.

Yet, there is a group inhibitors called irreversible inhibitors (and commonly known as suicide inhibitors because they kill the enzyme while get killing themselves), of which do not leave the active site once bound, unless an antidote is introduced.  One group of such inhibitors are called nerve reagents (a.k.a. nerve gases). All most all of nerve reagents are organic phosphorus reagents, which consist of a good leaving group such as fluoride (e.g., Soman and Sarin). The suicide-type inhibition of a nerve reagent is depicted in following diagram (The $\ce{-OR_L}$ group here represents a good leaving group):

For nerve gases such as Soman and Sarin, the leaving group is fluorine atom, which leaves as a fluoride ion ($\ce{F-}$). The strong attraction of phosphorous in nerve reagent to Ser$\ce{-OH}$ group causes $\ce{P=O}$ to break. Reformation of $\ce{P=O}$ bond would have made unacceptable hexavalent phosphorus. To avoid that scenario, one group on attached nerve reagent has to leave, and the best one available to leave is fluorine atom as $\ce{F-}$ (e.g., in Soman and Sarin). That leaves you with irreversibly inhibited acetylcholinesterase, which makes all chaos in central nerve system, until you receive an antidote (you should be given the antidote before infected enzyme goes to the next step, aging). The mechanism of the action of antidote (e.g., 2-PAM) is well documented and been illustrated in following diagram:

2-PAM has high affinity for anionic site (which is free now; see diagram) of the enzyme because it consists of part of nicotine moiety (searh for nicotinic receptor description). Once bound to free anionic site, oxime$\ce{-OH}$ has high attraction for phosphorous atom in infected enzyme site by the proximity. This attraction causes $\ce{P=O}$ to break (see pink arrow). Reformation of $\ce{P=O}$ (see blue arrows) would have again made hexavalent phosphorus. Again, to avoid that scenario, one group has to leave, and the best one to leave is ser$\ce{-OH}$, leaving the free enzyme (note that, all bond making and bond breaking in the active site here are in equilibrium, but once Ser$\ce{-O-P}$ bond is broken, the elimination of 2-PAM bound-phosphorus moiety is faster compared to regeneration of same Ser$\ce{-O-P}$ bond).

The aging depends on how easy to hydrolyze the $\ce{O-R2}$ bond by a water molecule. At least, this bond in Soman, $\ce{R2OH}$ of which is Pinacolyl alcohol (3,3-Dimethyl-2-butanol), can be easily hydrolyze by water in few minutes. This mechanism was studied and following mechanism was given (Ref, 1):

In Sarin, the $\ce{O-R2}$ group is the $\ce{O-CH(CH3)2}$, which would not be hydrolyzed as easy as that in Soman. Thus, for sure, aging of Sarin infected enzyme would take more that minutes, perhaps hours.

The given mechanism is supported by the study of different group on another nerve reagent Tabun, the initial leaving group of which is cyano group (as $\ce{CN-}$ compared to $\ce{F-}$ in Soman) to effect acetylcholinesterase (Ref.2). Aging of the effected enzyme is caused by leaving of ethyl group from $\ce{P-OCH2CH3}$ function ($\ce{-OR2}$) in similar manner to Soman-effected acetylcholinesterase:

References:

Gulseher Sarah Sirin, Yanzi Zhou, Lee Lior-Hoffmann, Shenglong Wang, Yingkai Zhang, "Aging Mechanism of Soman Inhibited Acetylcholinesterase," J Phys Chem B. 2012, 116(40), 12199–12207 (https://doi.org/10.1021/jp307790v).
Eugénie Carletti, Jacques-Philippe Colletier, Florine Dupeux, Marie Trovaslet, Patrick Masson, Florian Nachon, "Structural Evidence That Human Acetylcholinesterase Inhibited by Tabun Ages through O-Dealkylation," J. Med. Chem. 2010, 53(10), 4002-4008 (https://doi.org/10.1021/jm901853b).


Charge, Size, and Movement in Aqeuous Ions as Electrolytes Would a pair of aqueous $+2$ and $-2$ ions necessarily conduct electricity better than the same concentration of $+1$ and $-1$ ions? Furthermore, would more massive ions not conduct as well as less massive ions because they cannot travel as rapidly? It seems to me that ions move in the liquid to conduct electricity, because solid ionic crystals do not conduct electricity. If they move, then size and charge should both be factors affecting electrolyte strength. However, if ions are moving, why don't they ever achieve the most stable position and stop moving?

Would a pair of aqueous $+2$ and $−2$ ions necessarily conduct electricity better than the same concentration of $+1$ and $−1$ ions?

Not necessarily. Taking the concentration of 0.001 aqueous solution of $\ce{KOH}$ and $\ce{MgSO4}$ at $18$ degree Celsius, we find that the current conducted by $\ce{KOH}$ is $2.34 \space mA$ while the current conducted by $\ce{MgSO4}$ is $2.00 \space mA$. Why did this happen?

It happened because of the mass. And therefore here comes the answer to your second question.

  Would more massive ions not conduct as well as less massive ions because they cannot travel as rapidly?

Yup, more massive ion means less current. It's because of simple equation (and I hope you are aware of these physics equations)$$qV=\dfrac{mv^2}{2}$$

Here $q$ is the charge on the ion. $V$ is potential difference between both the solutions. $m$ is the mass of the ion. And $v$ is the drift velocity. We know that $qV$ is constant. So we get $$v \propto \dfrac{1}{\sqrt{m}}$$ We can easily analyse that more the mass we have, less the speed we get.

  If ions are moving, why don't they ever achieve the most stable position and stop moving?

They do achieve the most stable position and that's why batteries die.


Reaction of Hydrogen and Dinitrogen Tetroxide? What might be the possible products of a reaction between hydrogen and dinitrogen tetroxide ($\ce{N2O4}$)? They are fuel and oxidizer respectively in rockets.

A likely reaction between dinitrogen tetroxide and hydrogen would be as follows with nitrogen and water being the products

$\ce{N2O4 + 4H2 → N2 + 4H2O~(I)}$

$\ce{\Delta H_{rxn}=[0+(4 x -241$.$ 8)-9$.$2 - (4 x 0)] ~=~ -976$.$4~ kJ}$

For the record, dinitrogen tetroxide and hydrazine (usually in combination with methylated hydrazines) is currently used as a rocket fuel.  The equation describing this reaction is

$\ce{N2O4 + 2N2H4 → 3N2 + 4H2O~(II)}$

$\ce{\Delta H_{rxn}=[(3 x 0) +(4~ x ~-241$.$ 8) - 9$.$2 - (2 x 50$.$6)] ~=~ -1077$.$6~ kJ}$


Basic question about an expression for Gibbs free energy I'm getting into thermodynamics, and I have a basic question about the following expression for Gibbs free energy:

$G = \sum \mu_i N_i$

Is it correct that this equality only holds for constant temperature and pressure, as follows from integration over N of the expression for $dG$:

$dG = -SdT + VdP + \sum \mu_i dN_i$

In the equation

$$G = \sum \mu_i N_i$$

$G$ is the absolute value of the Gibbs free energy at some point in state space. While not explicitly stated, the values of the chemical potentials $\mu_i$ are in general functions of composition and other state variables (e.g. T and p), rendering $G$ a function of these same state variables.

On the other hand, the equation

$$dG = -SdT + VdP + \sum \mu_i dN_i$$

describes an infinitesimal difference in the free energy between two (infinitesimally close) points in that space.

Neither equation puts constraints on T and p, that is, both are general.


Plastic: Reputable Publications I would just like some reputable textbooks and or publication companies that have extensive research on plastics, as, for example, dealing with life cycles of plastics, degradation factors, effect on life, polymer types, etc.

A quick search in my university's library for Life cycle polymer brings me two reputable looking books:

Green composites: Polymer composites and the environment
Fundamentals of Materials for Energy and Environmental Sustainability

A Google Scholar search for life cycle analysis of plastic yields several good looking hits. A similar result for life cycle analysis of polymers.

This kind of information is out there, and there is a lot of it. My examples hopefully will inspire you to spend the effort (and it's not that much effort) to look for it.


Obtaining deuterium directly from water I know that there is 2:6400 deuterium and hydrogen molecules respectively in water.

When electrolysis is done, hydrogen and oxygen will be obtained. The hydrogen obtained should contain small amounts of d2 and t2.

Is there a way to separate it by diffusion or effusion (or any other way)?

Deuterium is already obtained by electrolysis of water. Deuterium is a little less reactive than Hydrogen, So when we repeatedly electrolyze water, we get heavy water. Then we can obtain deuterium from that.


Why is magnesium hydroxide not hygroscopic? I asked this question today and my professor was unable to provide any rationale.

I know that sodium hydroxide is hygroscopic, but why not magnesium hydroxide (at least to the same extent as $\ce{NaOH}$)? Could it be that $\ce{Mg(OH)2}$ has a higher lattice energy; the hydroxide is more closely associated with the magnesium ion and less able to react with carbon dioxide?

Try looking at Inorganic Chemistry in Aqueous Solution by Jack Barrett around page 41 on google books or elsewhere.

Entropy of hydration of Mg2+ is very negative compare to Na+.

According to Barrett, more highly charged ions restrict the movement of water molecules of hydration more.  Also, the smaller the radius of the ion, the more restricted the movement of the hydrating waters is.  Both the number of waters whose movement is restricted, and the degree to which they are restricted, is affected by charge and radius.

A complete analysis would consider both the entropy and enthalpy contributions.


Etymology of "normality" Why the concentration of a solution expressed in gram equivalent weights of solute per litre of solution is termed as "normality" ?

Does it have anything to do with the word "normal" ?

The term "normality" stems from the older German term for equivalent concentration (die Äquivalentkonzentration), namely die Normalität, which was coined by Karl Friedrich Mohr.
According to German Wikipedia which also refers to Straube's Der historische Weg der Chemie [1, p. 220], Mohr introduced the concept of normal solution and normality (as it's known in the modern world) in the first edition of his Lehrbuch der chemisch-analytischen Titrirmethode [2], e.g. the preface contained the following definitions among others:

Die Normaltemperatur der Flüssigkeiten ist 14° R. = 17,5° Cent.
[…]
Normalflüssigkeit bedeutet eine Lösung, welche 1 Atom Substanz, in Grammen ausgedrückt, in 1 Litre enthält.
Zehend-Normalflüssigkeit bedeutet eine Lösung, welche $1/10$ Atom Substanz, in Grammen ausgedrückt, auf 1 Litre enthält oder mit $1/10$ Atom eines darin enthaltenen Körpers (z. B. Sauerstoff) wirkt.

The book also lists numerous "normal" definitions not really in a sense of "adequacy", but rather in a a sense of "standard": Mohr basically invented standardized quantitative analysis and proposed numerous analytical tools and techniques.
Among these definitions there are:

Normalwägung — standard weighing [procedure];
Normalbürette — standard burette (a glass tube $\pu{500 mm}$ long, $\pu{13.5 mm}$ inner diameter, holds $\pu{60 cm3}$ of liquid divided into five parts, each $\pu{cm^3}$ is $\pu{7 mm}$ long);
Normale Stärke — strength of a normal solution of acid/base;
Normalsäure and Normalnatron — standard/normal solutions of acid and soda, which, interestingly enough, Mohr suggested to label with the pieces of red and blue paper, respectively;
Normalkleesäure, Normalsalpetersäure, Normalchlorbaryum, etc. — normal solutions of oxalic, nitric acids and barium(II) chloride and so on.

At the same time, normality was used to describe the normal ratio between stoichiometry of the elements (esp. C, O, N) and the number of found equivalents in complex compounds such as proteins or drugs [3].
References

Wilhelm Strube: Der historische Weg der Chemie. Aulis Verlag Deubner & Co KG, Köln 1989, ISBN 3-7614-1180-4. (In German)
Friedrich Mohr. Lehrbuch der chemisch-analytischen Titrirmethode; Vieweg und Sohn: Braunschweig, 1855. (In German. Available for free at Google Play (accessed 2019-05-25))
Zeising, A. Das Normalverhältnis der chemischen und morphologischen Proportionen; Weigel, 1856. (In German. Available for free at Archive.org (accessed 2019-05-25))


Can an amide nitrogen be a hydrogen bond acceptor? Can $\ce{N}$ (i.e. main chain $\ce{NH}$s of recognised residues in a peptide or protein) be a hydrogen bond acceptor? It is a well known fact that the main chain $\ce{NH}$s can be a hydrogen bond donor (as observed in the α-helix), but can they act as a hydrogen bond acceptor? If yes, help me with an example or even better a reference. And, if no, why can't they act as an acceptor, after all it's an electronegative atom?

If I understood you correctly, you are talking about the peptide bond nitrogens ($\ce{R-C(=O)-\color{red}{N}H-R}$). This is, reduced to its significant chemical functional group an amide, more precisely a carboxylic amide.
The amide nitrogen technically has a lone pair and thus technically could function as a hydrogen bond acceptor when viewed alone. However, this lone pair is actually significantly delocalised towards the carbonyl bond. If we ignore the two $\ce{R}$ residues, we could write this the following way:
$$\ce{O=C-N-H <-> ^{-}O-C=N^{+}-H}$$
The lone pair is not really available as a lone pair on the nitrogen, but forms a π-bond to the neighbouring carbonyl carbon to a significant extent. Thus, these nitrogens cannot be hydrogen bond acceptors.
Further, the fact that a Grignard reagent cannot react with an amide gives great support for the fact that nitrogen’s electrons are delocalized onto the carboxylate.

Finally, thank you Orthocresol for supplying me with the paper given at the end of this post. The authors performed some quantum chemical calculations to extract the energy difference between free formamide and water molecules and various conceivable hydrogen bonds between them. The strongest stabilisation can be found in an amide–amide hydrogen bond of the following type:
$$\ce{H2N-CH=O\bond{...}H-NH-CO-H}$$
At around $-40~\mathrm{kJ \cdot mol^{-1}}$ for a $\ce{H\bond{...}O}$ distance of $285~\mathrm{pm}$ and a $\ce{C=O\bond{...}H}$ angle of $60^\circ$.
$$\ce{H2N-CH=O\bond{...}H-OH}$$
For the interaction of a water molecule donor and the carbonyl oxygen as depicted just above, the stabilisation is given as up to $-27~\mathrm{kJ \cdot mol^{-1}}$ for a distance of $280~\mathrm{pm}$ and a $60^\circ$ $\ce{C=O\bond{...}H}$ angle.
$$\ce{OHC-HN-H\bond{...}OH2}$$
For this interaction where the amide nitrogen is donor and water is acceptor, the stabilisation was calculated to $\Delta E \approx - 30~\mathrm{kJ \cdot mol^{-1}}$ — achieved twice. Once for the geometry $d(\ce{H\bond{...}O}) = 280~\mathrm{pm}$ and $\angle(\ce{H\bond{...}O-H2}) = 30^\circ$ (the angle is measured between the $\ce{H-O-H}$ plane and the amide plane), and once for $d = 280~\mathrm{pm}$ and $\angle = 15^\circ$.
$$\ce{OCH-H2N\bond{...}H-OH}$$
This is the one you are looking for. These energy differences are all given as positive. The lowest is $\Delta E \approx +0.8~\mathrm{kJ \cdot mol^{-1}}$ for a distance $d(\ce{H\bond{...}N}) = 360~\mathrm{pm}$ — $1.3$ times the previous distances. (The angle is not given but from the picture we can assume a $90^\circ$ angle between the amide’s plane and the $\ce{N\bond{...}H}$ axis.) For the $280~\mathrm{pm}$ which brought the most stabilisation in the other cases and which thus seems to be the ideal hydrogen bond distance, they report $\Delta E \approx + 30~\mathrm{kJ \cdot mol^{-1}}$ — this is about the dimension of the stabilisation of the previous hydrogen bond only in a destabilising manner!
These studies let us conclude that it is safe to say the amide nitrogen will never be a hydrogen bond acceptor.
Reference:
A. Johansson, P. Kollman, S. Rothenberg, J. McKelvey J. Am. Chem. Soc. 1974, 96, 3794. DOI: 10.1021/ja00819a013.


Synthesis of Diazepam with (2-amino-5-chlorophenyl)(phenyl)methanone I tried to draw the mechanism of the diazepam synthesis.
Question: Do I really get the tertiary amine or does dimethyl sulfate methylate the amine twice?

While amines may undergo multiuple methylations to form quaternary ammonium salts, the nitrogen atom being akylated in the diazepam synthesis is an amide. Amides are different.

The lone pair on the amide nitrogen atom in diazepam is thoroughly delocalized into the carbonyl, into the aromatic ring, and all the way to the imine (see image below - some resonance contributors not shown). Due to this delocalization, that amide nitrogen atom is not nucleophilic and will not be methylated a second time. I would be much more worried about alkylating the imine nitrogen atom, which is made more nucleophilic by delocalization.

This delocalization is the reason why you need to add a base in order to do the methylation step in the first place. The amide in N-demethyldiazepam is likewise a poor nucleophile due to delocalization.

However, once a base deprotonates the amide, the amide becomes the better nucleophile due to the concentration of negative charge. Even though one of the lone pairs is delocalized, the other is not.


Does 30% hydrogen peroxide eat away at platinum? if I soaked a platinum dish with 30% hydrogen peroxide, will it eventually eat away at it?

Yes, as @Ivan said, hydrogen peroxide will not corrode platinum, instead hydrogen peroxide will get decomposed.

Catalytic decomposition of hydrogen peroxide by metals and alloys of the platinum group:

  The catalytic decomposition of dilute aqueous solutions of hydrogen
  peroxide has been studied on a number of Group VIII metals and binary
  alloys, including palladium-gold. The kinetics were measured at 27 °
  as a function of catalyst composition and pH of the solution. In
  neutral solution, the activity of the individual metals of Group VIII
  was generally greater than those of Group VIII2, the order of specific
  activity (per unit surface area) decreasing in the sequence $\ce{Pt > Os >
 Ir > Pd > Ru > Rh}$. (...)

Other references:

http://ac.els-cdn.com/S0926337306001561/1-s2.0-S0926337306001561-main.pdf?_tid=a114888e-1c29-11e7-9622-00000aacb360&acdnat=1491635258_a3e20984dda9eaaf07e78f4814383b96
http://pubs.acs.org/doi/abs/10.1021/ed080p788
http://jcsp.org.pk/ArticleUpload/2541-11513-1-CE.pdf


Finding chiral carbons in aromatic or cyclic compounds I know that for a carbon to be chiral four different groups should be attached to that particular carbon. But, what about in the case of cyclic compounds? What's the method to check chirality if the carbon is not $\mathrm{sp}^3$ hybridized and only three, or two groups are attached to it (through double or triple bonds)?

If an atom is part of a double or a triple bond it is never a stereogenic center because then this atom and the atoms bound to it form something planar, so that you can always find a symmetry plane.
But an atom that is part of a double bond can be prochiral, if it can be converted from to a stereogenic center in a single step.


Why is the lone pair of pyridine's nitrogen atom not part of the aromatic ring? In pyrrole, the lone pair of electrons belonging to the nitrogen is part of the aromatic ring. However, in pyridine it is part of an sp2-hybridized orbital. Why can't it be in the p-orbital and take part in the aromatic ring?

I mean, why can't the lone pair in pyrrole be in the p-orbital, with the sp2-hybridized orbital (which is not bonded to anything) sporting a single electron?

I'm not talking about the nitrogen atom having two p-orbitals participating in the aromatic ring. That would be strange. I wonder why the existing p-orbital of N cannot carry two electrons, the way it does in pyrrole.

P.S. A related question, with a very interesting answer: sp2-hybpidization in pyridine and pyrrole

As follows from our discussion in the comments to the question, pyridine uses only a single electron in the p-orbital in order to comply with Huckel's rule. Kudos to Orthocresol:

  Huckel's rule demands that there be 4n+2 pi electrons in the ring. Each carbon contributes one, making for a total of 5 from the carbons. If your nitrogen lone pair is in the ring, then you will have seven pi electrons. $7≠4n+27≠4n+2$ for $n ∈Z$

Of course, electrons know nothing of Huckel and his rule, but apparently they position themselves into the most energetically suitable configuration. I am not yet qualified to explain why exactly it is the most suitable one.


Why hydrofluoric acid (HF) has two Raman bands? Hydrofluoric acid ($\ce{HF}$) is a diatomic molecule with only one vibrational coordinate. Therefore I would expect only one Raman band, however there are two. Why?

(source)

I've been doing some digging and I think I have a possible answer so I'm gonna post it and if I'm misinterpreting information, I'm open to being told so!

As I said in my comment, this vibration has to be coming from some dimer which is formed because if one of these modes were the $\ce{HF}$ fundamental it would be up around $4000\ cm^{-1}$.

So, I found this paper which uses FTIR to analyze $\ce{H2O/HF}$ complexes. They find two librational modes (which I believe is just a solvated rotational/rocking mode basically) which they identify as coming from the $\ce{H2O--HF}$ dimer. They assign these two frequencies at $635.5\ cm^{-1}$ and $753.1\ cm^{-1}$. Those two values are extremely similar to the frequencies we're seeing in the spectra you posted which is why I think we might be on to something here. Unfortunately I don't have access to the paper so I can't go into any more detail really.

In answer to your question of how we could even be seeing multiple peaks in an $\ce{HF}$ spectrum, the answer is that we're not looking at just $\ce{HF}$ here!

We know that both $\ce{H2O}$ and $\ce{HF}$ are very strong hydrogen bonding molecules, so it's no surprise that we find peaks which are associated with the various vibrational modes of the dimers present. And, once we have a dimer, we expect to find all sorts of rotational and vibrational modes.

So, to conclude, I'm fairly certain that the modes we're seeing in this spectra are two (and I think the only two because my digging around says this dimer has $C_s$ symmetry) librational modes of the $\ce{H2O--HF}$ dimer, each rotating about a different axis of the water.

Source:

Andrews, L., & Johnson, G. L. (1983). FTIR spectra of water–hydrogen fluoride complexes in solid argon. Evidence for inversion doubling in the HF librational modes of H2O––HF. The Journal of chemical physics, 79(8), 3670-3677.


Confusing definition of mol when considering second order rate equations I am confused by "$\text{mol}$".
I denote the unit of a physical quantity with square brackets (e.g. $[V]=l$), and the value of a physical quantitiy using curly brackets (e.g. $\left\{ V\right\}$ = 0.01).
A simple second order rate equation for number concentrations $n=N/V$ is given by,
$$\dot{n} = k^*_{on}n^2,\;\;[1]$$
where the $[k^*_{on}]=l/s$.
The same second order rate equation but for molar concentration $c=\frac{N}{N_AV}$, is given by
$$\dot{c} = k_{on}c^2,\;\;[2]$$
where $[k_{on}]=\frac{l}{\text{mol}~s}$.
When transforming equation $[1]$ into equation $[2]$, we get the relation
$$k_{on} = \left\{k_{on}\right\}\frac{l}{\text{mol}~s} = k^*_{on}N_A = \left\{k^*_{on}\right\}\left\{N_A\right\}\frac{l}{\frac{1}{\left[N_A\right]}s}$$
Hence $\frac{1}{\left[N_A\right]} = \text{mol}$?
But this would be in contraction to the definition I find on wikipedia where it is written that:

In short, for particles 1 mol = $6.02214076×10^{23}$.

Hence, $1~\text{mol} = \left\{N_A\right\}$.

I think the sentence cited above in the wikipedia article is very confusing.
But the wiki article about the Avogadro constant helped me resolve my confusion, which defines $mol$ via the Avogadro constant:
$$N_A = \left\{N_A\right\}[N_A] = 6.022\times10^{23} \text{mol}^{-1}.$$
Hence
$$k_{on} = k^*_{on}N_A = \left\{k^*_{on}\right\}\left\{N_A\right\}\frac{l}{\frac{1}{[N_A]}s}=\left\{k^*_{on}\right\}\left\{N_A\right\}\frac{l}{mol~s}$$
This definition is also given on the hompege of the National Institute of Standards and Technology (NIST), see here.


Why can dissolution reactions (e.g. dissolution of calcium chloride) be exothermic? Dissolution reactions are normally endothermic, and crystallization reactions are normally exothermic, but there are some exceptions -- such as the dissolution of calcium chloride. Why is this so?

The endothermicity (taking up heat) or exothermicity (giving off heat) depends on a balance of factors including the entropy change, the degree with which the ions become associated with water, and the possibility of actual chemical reactions.

The most notorious example is sulfuric acid, which reacts so strongly with water that it can actually boil the water, and spatter the acid, which is why one always adds the acid to the water(water as solvent) and precautionary measures (gloves and proper clothing) are needed to handle the reaction.


What are the organic reactions that are taking place in this reaction? I know that A is the product of a Knoevenagel condensation between the aldehyde and the diethyl malonate, due to the acidic protons of the latter, but I don't know what the reaction between that product and potassium cyanide would be.

Here is what I think is going on:

$Step$ $1$: As the OP correctly identified is a Knoevenagel condensation to give A diethyl 4-cyclohexyl-benzalmalonate.
$Step$ $2$ is an example of little-used ester hydrolysis using cyanide ion to give initially the acyl cyanide which is unstable in aq EtOH and gives the diacid B
$Step$ $3$ is loss of one carboxy group by acid catalysed decarboxylation to give the cinnamic acid C
$Step$ $4$ formation of the mixed anhydride D
$Step$ $5$ cyclisation through formation of an aluminium-acyl complex literature example here to give the final product 6-cyclohexyl-Inden-1-one E


Is equilibrium potential for K+ always ~-90 mV? Equilibrium potential is the difference in charge relative to the inside of the cell. -90 mV means that the inside of a cell is more negative than the outside. However when you talk about equilibrium potentials, I thought you only consider an individual ion like potassium? Potassium usually has 150 nM inside the cell and 5 nM out. If you allow potassium to freely pass through the membrane, wouldn't it eventually balance out so that the net charge on both sides is 0 (concentration would also be equal)? So why do people say that the equilibrium potential for K+ is -90?

Most ions don't pass freely through the membrane. This is prevented by the hydrophobic (water "fearing") fatty acid tails of the phosholipids that provide the basis of the membrane. The regions outside and inside the membrane are aqueous and attract ions much more than does the hydrophobic fatty region in the interior of the membrane bilayer.

Under usual physilogical conditions, the only way ions can cross the membrane is through protein channels. But these channels are gated and only open in response to certain stimuli. Stimuli can include a chemical substance binding to the gate and opening it, this is how neurotransmitters work. Many gates respond to changes in the overall potential across the membrane. These are termed voltage regulated gates and, among other finctions, explain how action potentials (impulses) travel down neurons and muscle cells.

Regardless of how these gates open, the effect is temporary. Neurotransmitters break down, the membrane potential continues to change in a manner that voltage regulated gates once again close.

During these events only small amounts of ions cross the membrane compared to the vast supplies on whichever side has the surplus. That would be the interior of the cell in the case of potassium. Yes, this might eventually deplete the concentration if allowed to continue without some mediation.

That mediation is the sodium-potassium pump. The Na-K pump, using the energy in ATP, pumps Na+ out and K+ into the cell. But it pumps out 3 Na+ for every 2 K+ it brings in which makes the overall resting membrane potential -70 mV. This pump is busy and important. It is responsible for using about 20% of the ATP your body produces. Even more if you're inactive.

So, after some of the K+ ions are allowed out through gated channels, the potential changes very locally. Then other K+ ions migrate into this area, down their concentration gradient, restoring mmost (but not all) of the normal K potential. Operation of the Na-K pump provides the additional K+ ions to restore it completely


Why is NH2 negatively charged? Whenever I look up the Lewis structure for $\ce{NH2^-}$, it shows 8 electrons. Now I understand the octet rule, but I still don't see how an electron can just appear from nowhere. There are 5 valence electrons with Nitrogen, and 2 more, one from each hydrogen. I know that formal charge is calculated by subtracting the nonbonding electrons and number of bonds (or number of electrons in bonds divided by 2), but why is the formal charge of $\ce{NH2^-}$ "-1", and why the extra electron out of nowhere? Because 5 - 3-2 is still just 5

For atoms in the first 2 rows of the periodic table the octet rule arises as a consequence of filling orbitals four orbitals which each hold 2 electrons - each of which appears as a pair of dots in your Lewis structure.

Were you to construct the Lewis structure for neutral $\ce{NH2}$ you would indeed have 5 electrons from the nitrogen, and two from hydrogen, 7 i.e. 3 pairs and 1 extra electron on its own in a orbital.  Electrons in unpaired orbitals are however very reactive, the pairing of electrons lowers the total energy of the system.  As such, it is more favourable for the $\ce{NH2}$ molecule to remove an electron from some other molecule or atom near by and form ions e.g. $\ce{Na}$ to create $\ce{Na+NH2-}$.

Also by creating the charges, you create an electrostatic attraction between the ions which also lowers the energy and makes the ion more stable than the neutral molecule.


ASTM D2270 Viscosity Index Formula/Table derivation How was the formula (3) for viscosity index $\mathrm{VI}$ derived in section 5. Procedure A—Oils of Viscosity Index Up to and Including 100 [1]?
Specifically with reference to the reference oils considered and the table for it.

$$\mathrm{VI} = \frac{L - U}{L - H} \times 100$$

References

D02 Committee. ASTM D2270 - 04 Practice for Calculating Viscosity Index from Kinematic Viscosity at 40 and 100 °C; ASTM International. DOI: 10.1520/D2270-04. (PDF)

back when it was invented they measured all the oils they could find.  They got the worst (most viscosity change) and the best in each group and ranked these as 0 to 100.  All others were plotted on this "linear" scale.

The unit was used for dome time until the best was bettered, then they invented VIE (E is a subscript, and stands for extension, strictly all VI's above 100 are VIE's).

It is a horrid unit because it is far from having any useful meaning, unfortunately it has some significance.


How to do Lowdin symmetric orthonormalisation? Löwdin symmetric orthonormalisation seems to be a common practice in quantum chemistry. I come from a different background though and have to understand it and possibly implement in a computer code. Here is the definition of the transformation;

$$
|{\phi'}\rangle = S^{-1/2}|\phi\rangle,
$$

where $S$ is the overlap matrix between the non-orthogonal basis,

$$
S = \left< \phi|\phi \right> .
$$

My first question is how to obtain $S^{-1/2}$ analytically?
Are there any parallel algorithms available to compute it numerically?

You obtain the diagonalization of $S$ as defined
$$
S = U \cdot s  \cdot U^{\dagger}
$$
(which holds for the case of symmetric $S$, which it is) and apply the power needed
$$
S^{-\frac{1}{2}} = U \cdot s^{-\frac{1}{2}}  \cdot U^{\dagger}
$$

For more background, try the application section of Wikipedia's article on diagonalizable matrices.
As far as understanding it goes, I'm afraid not much beyond doing some model calculations on 2D or 3D vectors can be done.

I do not have a strong background in parallel computing, but all operations involved are well-known and your linear algebra framework (BLAS, anyone?) will document how to run any of this in parallel and what storage may be needed for it. Note that, typically, diagonalization is not the most expensive step of a self-consistent field calculation (Hartree-Fock or DFT) anyway, so parallelization may not be critical for this step.


Transistors: How does the gate work? Two questions:

First, how does the gate provide a positive charge by running a current through it? I really just don't see how this positive charge could be achieved, unless you had some sort of battery in the gate itself and attached a wire to donate some of the electrons flowing through to the cathode of some other battery. Is it possible to get electrons out of the gate by some other means?

The gate of a FET works much like the grid of a radio tube (valve). The gate is insulated from source and drain; the only current flow (if the insulator were perfect) is to charge the intrinsic gate capacitance. Even a MOSFET has some leakage, but it's pico- or femtoamperes. See this discussion on capacitance and leakage.


Why is there syn-addition in this bromination of an alkene? In my book I read that halogenation of alkenes results in anti-addition. Is this example showing syn-addition?

The example:

In this example, it is impossible to tell from the product whether the mechanism was syn- or anti-addition. Let's look at the two possible products. After anti- or syn-addition of bromine, two compounds are produced that look different. However, facile bond rotation around the C1-C2 bond shows that they are in fact the same compound.

How would one go about proving that anti-addition is actually occurring in 4-methyl-1-pentene. One solution would be to use an isotopically labeled variant as shown below. Starting from the deuterated version of 4-methyl-1-pentene, we can draw the anti- and syn-products again. Similar rotation around the C1-C2 bond makes it clear that the products are not the same but instead are diastereomers. It would be possible to distinguish these products by some analytical technique (1H NMR) and show that the anti-product is formed selectively.


Can someone explain the equation for entropy? Here's the equation:

$\Delta S = \frac{\Delta q}{T}$

The confusion I have with this equation lies in the variable for temperature, $T$.  Why is the temperature constant in this equation? Why can't the equation include $\Delta T$ instead of just $T$ and, surely, if $\Delta q$ is negative or positive the temperature must decrease or increase, respectively.

Your relation is not correct, so you must not understand something. If I were you I will asked the question in Physics SE but nevermind.

The good relation is : $$\mathrm{d}S=\frac{\delta Q}{T}$$

I let Richard Feynman explain it to you here (when you click on the link wait few seconds)

I hope it will help you, have fun, live long and do chemistry !


Why do electronegative atoms depress coupling constants? Related to this previous question about detection of small couplings between protons.

In a recent conversation with our labs NMR technician, it was stated that coupling constants decrease in the presence of electronegative atoms:

  ...and you have a lot of electronegative atoms around, so thats going to make those coupling constants drop off significantly...

  Source: 'Personal communications', aka the labs NMR-tech

Thinking about it, this is something I've always known but possibly never thought about to any great deal, despite the fact it shows up in even the most basic organic chemistry textbooks:

  ...electronegative atoms, which decrease 2J and 3J couplings between protons

  Source: Organic Chemistry, Clayden & Warren

Digging a little deeper, this idea comes up a lot in NMR textbooks too, and a few trends are discussed, for instance in Günther's NMR spectroscopy:

  For substituted ethanes the relation between the change of electronegativity, caused by the replacement of a hydrogen atom with a group X, and the coupling constant is given by:

  3J = 9.41 – 0.80 (EX – EH).

  Source: NMR spectroscopy, Günther

What I can't find, in any great detail, is why the coupling constant depends on electronegativity the way it does.

From what I've read, it has something to do with the polarisability of the bond, but again, its maybe not completely intuitive how this effects coupling between two protons separated by 2/3 bonds.

Let's set some starting parameters so we are hopefully working from the same page. Interestingly, the summary that an increase in electronegative substitution leads to a decrease in coupling constant is not strictly correct per se.

For geminal coupling, electronegative substitution at the alpha position leads to a more positive coupling constant. H-H Geminal couplings are negative, and hence we observe a coupling that becomes smaller in magnitude. (There are a few examples where they actually go the other way, but still become more positive).
 On the other hand, H-F geminal couplings are positive, and so an increase in 2JHF coupling is observed with an increase in electronegative substitution.
Electronegative substitution at the beta position has the opposite effect, leading to a more negative coupling; this actually gives a coupling that is larger in magnitude. Effects from substituents at the beta position are usually smaller than for the alpha position, but the effect is nevertheless opposite.

For vicinal couplings, electronegative substitution leads to a reduction in magnitude of the coupling constant.

The magnitude of scalar spin coupling depends on a few parameters:

the angles between overlapping bonds between coupled nuclei
electronegativity of substituents
bond length

The traditional accounts for the mechanism of spin-spin coupling might go something like...

In the simplest terms, spin coupling comes about because the magnetic moment of nucleus A interacts with the electrons surrounding it, and these electrons produce a local magnetic field at the neighbouring nucleus B. Spin coupling is dominated by the Fermi contact over a short number of bonds. The Fermi contact describes the magnetic interaction of an electron within the nucleus. Only s-orbitals can contribute electron density at the nucleus (all other orbitals have nodes at the nucleus), and so bonding to entities that will promote s-character of molecular will strongly influence the coupling constants.

The more stable orientation of for the nucleus and electron are antiparallel (because they have opposite sign magnetogyric ratios), and for electrons in a bond also to be antiparallel. For the most stable arrangement we get something like (lower case = electron spin, upper case = nuclear spin, u=up, d=down, - are bonds):

Ud-uDu-dUd-uDu-.....

Essentially, the observable nuclear transitions depend on the molecular orbitals, and how well the molecular orbitals stabilise the antiparallel, or spin-polarised configuration.  There are some comprehensive explanations of the molecular orbital model given in Gunther, as you have already encountered, and first published by Pople and Bothner-by. You will also find an excellent account, as always, on Reich's NMR page.

A coupling constant is positive, by definition, if it stabilises the anti-parallel electron spins, as shown above. So, a 1J will be positive, 2J negative (usually) and 3J positive. If the probability of finding an electron in the up state is the same as for the down state, the coupling constant will be zero. For a simple 2 nucleus, 2 electron bond system, if there is an up-electron at one nucleus, then there is a very high chance that there will be a down electron at the other nucleus, and so the coupling will be large. For 2 nuclei across multiple bonds, the spin state of an electron at nucleus A will have a negligible effect on the spin state of electron at nucleus B, and so the coupling will be very small.

...Pause for a breath....

So, to the crux of your question...
If a substitution element can help stabilise the anti-parallel arrangement of the electron spins, by reducing the energy gap for the HOMO-LUMO of the molecular orbitals then the coupling constant will become more positive. Electron-withdrawing groups serve to reduce the energy gap between these orbitals (arguments based on symmetry-allowed transitions - read the above references). σ-acceptors will reduce the LUMO, while π-donors raise the HOMO; both have the same outcome of making the coupling more positive.

This explains the geminal trend, but what about vicinal? There are some theories about alternating polarisation along the bond pathway, so that stabilising the anti-parallel alignment in one molecular orbital will destabilise the adjoining orbitals, hence the trend towards reducing J. It all gets a bit vague about here, and not very convincing. There was some good work by Altona et al in the 80s/90s (see for example) extending the Karplus equation to account for electronegativities, and improved Karplus equations were published, but nothing universal. At this point, we begin to accept that the ability to describe the physical nature of what is going on here is not really going to work using an emperical model.

More recent work by Cremera and Gräfenstein, for example, pretty well put these old 'textbook' explanations back in their boxes, and describe the mechanisms for spin coupling as composites of 4 contributing factors: Fermi contact, spin dipole, diamagnetic spin orbit and paramagnetic spin orbit. It all gets very Hamiltonian early on, but essentially there are co-dependent interactions of all these terms in the electronics. Fermi contact coupling is propagated over multiple bonds and occurs through space via a molecular orbital vector involving the tails of the orbitals, and these tail interactions are strongly influenced by the substituents. An electronegative substituent on a carbon will increase the local Fermi contact, but lead to a decrease in the tail orbital overlap, leading to a reduction of vicinal coupling, for instance.

Some very useful references for you might include:

Phys. Chem. Chem. Phys., 2007, 9, 2791–2816. Calculation and analysis of NMR spin–spin coupling constants
Annu. Rep. NMR Spectrosc. 2000, 41, 55–184. Advances in theoretical and physical aspects of spin-spin coupling constants


Can you separate CO2 into carbon and oxygen using pyrolysis? I recently read about a pyrolysis process that separates methane into Carbon and Hydrogen. The basic technology/process is you have a vertical column of super-heated liquid metal and you inject methane into the bottom of the column. The methane breaks down into hydrogen and solid carbon. Are there any similar processes that separate CO2 into solid carbon and oxygen using liquid metals and/or salts etc.?

Standard enthalpy of formation of carbon dioxide is -393.5 kJ/mol. Standard enthalpy of formation of both oxygen and graphite is 0 J/mol. Standard entropy of carbon dioxide is 213.8 J/mol-K (NIST). Standard entropy of oxygen is 205 J/mol-K (Chem LibreTexts). Standard entropy of graphite is 5.69 J/mol-K (ChemLibreTexts).
Everything below is a lot of fairly inexact approximations/assumptions because I didn't want to use a better model or search literature for empirical adjustments and/or constants.
Assuming constant T,P (which I doubt would be true under reactor conditions) G = H - TS. Approximating $\Delta H, \Delta S$ as constant with T, $\Delta G = (0 - (-393500)) - (T \cdot (5.69 + 205 - 213.8)) = 393500 + T \cdot (3.11)$. Then $\Delta G < 0$ regardless of temperature. Catalysts also do not affect thermodynamics, so this process can never occur spontaneously. Increasing pressure of the reaction chamber would not help appreciably either, as the reaction has 1 mol gas on each side. While assuming $\Delta H, \Delta S$ are constant with T is quite incorrect as conditions become more extreme, the approximation should suffice to show that the reaction does not occur easily. Additionally, even if one were to account for the perturbations in $\Delta H, \Delta S$ with T and P, these conditions would be expensive to maintain in a reactor. (Note that this analysis does not even account for the activation energy/kinetics problem! I did not look up the activation energy of the problem, but it is entirely possible that the extreme conditions would still not allow the process to occur at a reasonable rate.)
As a result, this reaction is (probably) practically only driven forward by direct energy input (aka a coupled reaction/external energy). This often, if not always, involves some energy input, such as light or a coupled chemical reaction, that directly impacts the electron states of the participant compounds. These changes in electron states change the identity/properties and stability of compounds, and, when applied effectively, they can help drive the reaction forward. However, relevant changes to electron states using heat not only often involve very high temperatures, but the effect of heat on electron states is also a highly statistical process. Thus, it may be relatively inefficient. Thus, pyrolysis, while it may be theoretically possible, would be incredibly impractical at achieving separation of carbon dioxide. (Also, to compare to your methane example, methane combusts readily, carbon dioxide does not. Methane has a standard enthalpy of formation of -74.87 kJ/mol.)
Finally, if you still need more reasons to look at a different way to drive this reaction, I have not mentioned: energy/cost to continuously scrape/collect the produced carbon, purify/collect/pressurize an inlet carbon dioxide stream, (if you actually want to run the reaction, you'll probably want) catalytic surfaces to encourage carbon deposition, possible packed/fluidized bed catalysts, heat exchangers, and much, much more.
ELI5: Methane combusts/reacts readily and is an "unstable" compound. Carbon dioxide is very stable and does not react easily. Higher temperatures can make some reactions that do not happen at a lower temperatures occur, but this particular reaction receives no such benefit. Heat and catalysis can also drive forward the speed of a reaction, but the reaction has to be able to occur first. Pyrolysis, as you described it, involves catalysts and heat. As a result, pyrolysis cannot be used to separate carbon dioxide into carbon and oxygen.
tldr: Excepting conditions extreme enough to perturb $\Delta H_{rxn}, \Delta S_{rxn}$, conditions of constant T,P never favor breakdown of carbon dioxide into elemental carbon and diatomic oxygen. Pyrolysis cannot drive this reaction.


What are the components of a labware setup for short-path distillation? I've recently become interested in short-path distillation and have found that most kits are in the thousands of dollars. So I thought that I would compile the labware myself.

I understand the theory behind short path distillation, but do not know the actual names of the components. I do know I will be using a digital hotplate and labware grade pyrex labware.

So what is the (cheapest) "shopping list" needed to compile and assemble all the labware necessary to construct a short-path distillation apparatus?

Of course the answer depends a bit on the quantity to distill and the conditions (temperature and vacuum applied, difference in the boiling points of the components to separate).

There are about palm size short-path distillation heads, which may be connected easily directly to your round bottom flask and receivers like the following example:

(source)

At chemglass.com, this device (CG-1239) is listed for USS 178.95 and allows you to mount as usual thermometer, water-in and -out for the condenser, and the vacuum by an oil pump.  The efficiency of separation may be increased by mounting a vigreux column below the head, of course, which are available in different (shortened) lengths, too.  There are "jacketed versions" (e.g., CG-1240) which are more suitable if the head temperature to reach is quite elevated (say, 150 $^\circ\mathrm{C}$) already while applying vacuum, at a bit more elevated price tag, too.

Of course, the advantage of the Kugelrohr shortpath's design is that you may moderate the heat gradient better, because the bulb outside the oven may be cooled with a bath of acetone / dry ice, and the rotation reducing sudden evaporation.  The one by Buchi B-585 is more suitable for smaller quantities, while the ones by Aldrich obviously allow distillations out of a 250 mL flask, which truly are both considerably more expensive.  (The larger Aldrich ovens require more bench place, too.)

Do not forget that there are e-bay-like companies specialized in selling used but still working lab equipment, too, e.g. Richmond, LabX to mention two.


Does bonding happen naturally? For example, how do the bonds in NaCl form?

I understand that the elements need to obtain an octet structure to become more stable, by either accepting or losing electrons.

In NaCl, from my understanding, the sodium atom loses an electron and gives it to the chlorine atom, in turn gaining a +1 charge and then causing chlorine to have a -1 charge.

The new compound formed is NaCl. How does it form ? Do i need to place the respective atoms close to each other? What is needed to them to bond? Also, what is needed for bonding for other types such as covalent and metalic? Does it happen naturally ?

The collision theory aims to explain how reactions happen.

In the theory, when the particles of the suitable reactant collide with each other, the particles with the right conditions (orientation, velocity, etc.) will react and form new chemical elements.

It is quite difficult to form $\ce{NaCl}$ from sodium metal and chlorine gas alone. If you have to produce $\ce{NaCl}$ from sodium metal and chlorine gas (which you would not), you would do it in an aqeuous environment:

$$\ce{Na(s) + H2O(l) -> NaOH(aq) + H2(g) ^}$$
$$\ce{Cl2(g) + H2O(l) -> HCl(aq) + HOCl(aq)}$$
$$\ce{NaOH(aq) + HCl(aq) -> NaCl(aq) + H2O}$$

But in this method $\ce{NaOCl}$ is produced as a side product, so I do not recommend using this method.


Are there any solids that undergo diffusion? Are there any solids that undergo diffusion?

Introduction:

In school, my teacher is only talking about diffusion happening in liquids and I though that it may be able to happen to solids also.

Though I did think about how the concentration of the molecules in a solid was too concentrated for the molecules to pass through a cell membrane. What I am mainly wondering and would answer my question, is there a way that the molecules of a solid would break up into less concentration, in which they can undergo the process of diffusion.

Diffusion also occurs in solids. For example, (intentional or undesirable) movement of impurity atoms or other crystal defects can be very important for the fabrication of semiconductors.

The classical experiment to demonstrate diffusion (and to measure the diffusion coefficient) in solids is the diffusion of gold in lead. A thin layer of gold that is labelled with $\ce{^198Au}$ is applied to the surface of a lead sample using electrolysis. The sample is heated, and the penetration of gold atoms into the lead sample is observed using the β radiation from $\ce{^198Au}$.

You could use a similar experiment to measure self-diffusion (e.g. of gold atoms in gold).


Can we relate approximate Hartree-Fock orbitals to true solutions in the basis-set limit? Assume that we've "solved" the RHF Hartree-Fock equations in some finite basis, that is, we have arrived at a self-consistent set of coefficients $c_{\mu{}i}$ such that each spatial molecular orbital is expanded as $\chi_i = \sum_{\mu=1}^K c_{\mu{}i}\varphi_i$. By construction, this set of orbitals is orthonormal, and forms a collection of eigenfunctions for the Fock operators in the given basis, so $f_i |\chi_i\rangle = \varepsilon_i |\chi_i\rangle$ where $\varepsilon_i$ is the $i$th orbital energy.
Question: what, if anything, can we say about these orbitals in terms of their relationship to the "true" orbitals in the basis set limit, denoted $\chi_i^\mathrm{BS}$? Here, I mean those orbitals that would be obtained if the calculation were to be extended to an (infinite) complete basis set. (I realise that the assumed existence of a single unique set of true orbitals may not be justified...)
For example, is it possible to estimate, bound, or otherwise describe the overlap $\langle \chi_i | \chi_i^\mathrm{BS} \rangle$ between an approximate solution and the true solution? How about matrix elements such as $\langle \chi_i | \mathcal{H} | \chi_i^\mathrm{BS} \rangle$, where $\mathcal{H}$ is the electronic Hamiltonian, or indeed any self-adjoint operator?
I suspect/hope that some kind of a statement towards the first part might be possible, given that the Roothan-Hall equations are effectively a collection of Galerkin equations. One of the standard hand-waving statements about solutions obtained using Galerkin methods is that they are in some sense projections of the true solution onto finite-dimensional subspaces, and have residual vectors that are "orthogonal" to that subspace(for a potentially complicated definition of "orthogonal").
(I'm also interested in answers that would apply in any of the HF settings, i.e. UHF, ROHF, etc.)

In quantum chemistry numerical experimentation is the norm and rigorous mathematical proof the rare exception. As a result, not many rigorous results are know, not even for the good-old Hartree-Fock method. For a review of mathematical results I recommend you have a look at these references:

Claude Le Bris, Computational chemistry from the perspective of numerical analysis, Acta Numerica (2005), pp. 363–444
Pablo Echenique and J. L. Alonso, A mathematical and computational review of Hartree–Fock SCF methods in quantum chemistry, Molecular Physics, Vol. 105, Nos. 23–24, 10 December–20 December 2007, 3057–3098

In particular, section 3.3 of the first reference is relevant, but don't expect to find mathematically strong results. For example, here's a quote:

The choice of an AO basis for solving
a given problem mostly relies upon some practical know-how. The lack of
rigorous understanding is a pity, because the output of the calculations
(typically some molecular property) might be very sensitive to the choice
of the basis set. The only available measures of the quality of the basis
set are obtained, in the chemistry literature, by choosing test cases, i.e.,
reference systems, where the solution of the exact Schrödinger equation may
be computed, mostly through numerical computations and, when possible,
with the help of an analytic calculation.

From a practical point of view I don't see why occupied Hartree-Fock orbitals expanded in the usual Gaussian basis sets should not converge (in the integral norm sense, ie leaving aside the lack of cusp at the nucleus) to the exact solutions of the Hartree-Fock equations. I have little idea as to what happens to the virtual orbitals.


Entropy change in non-ideal solutions We know that some solutions show positive deviations and some show negative deviation from Raoult's law.The former is endothermic and the latter is exothermic. And again, the volume change show positive and negative terms.  In which case entropy change is more? And why is that?

The detailed calculation for $\Delta G$ is here,  Derive expression for internal energy of mixing and entropy of mixing using statistical thermodynamics  which gives $$\Delta G = RT[n_1\ln(x_1)+n_2\ln(x_2)] +(n_1+n_2)x_1x_2w$$

and as $\Delta S= -d\Delta G/dT$ then

$$\Delta S = -R[n_1\ln(x_1)+n_2\ln(x_2)] - (n_1+n_2)x_1x_2\frac{dw}{dT}$$

where $w \approx (2E_{12}- E_{11} - E_{22})$ and $E_{12}$ is the interaction energy between molecules of type $1$ and $2$ when they are adjacent and similarly for $E_{11}$ and $E_{22}$ all of which are negative. From a consideration of the energies involved $w$ is expected to be positive.

The first term is the  Raoult's law expression for an ideal solution and the second term the deviation, which is sometimes also called the excess  entropy of mixing. For benzene cyclohexane when $n_1=n_2=x_1=x_2=1/2$  the excess entropy  $\displaystyle -\frac{1}{4}\frac{dw}{dT}\approx 1.7$ J/mol/K at $298$ K and $1.2$ at $343$ K.


Why does dimethyl dichlorosilane undergo hydrolysis to form siloxane polymers, but its carbon analog does not? Dimethyl dichlorosilane will undergo hydrolysis to form siloxane polymers. But why would its carbon analog dimethyl dichlorocarbon not be able to undergo a similar hydrolysis reaction to form polymers?

My guess is that $\ce{Si-Cl}$ bond being much weaker than $\ce{C-Cl}$ bond would be easier to break and undergo hydrolysis, and that the geminal diol is unstable, hence it is less likely to form.

I'm unsure of my answer though. Could anyone please explain and elaborate?

Actually a $\ce{Si-Cl}$ bond is much stronger than a $\ce{C-Cl}$ bond.  It takes about 90 kcal/mole to break a $\ce{Si-Cl}$ bond, but only around 81 kcal/mole to break a $\ce{C-Cl}$ bond.  The strengths of the corresponding $\ce{Si-O}$ (110 kcal/mole) and $\ce{C-O}$ (85 kcal/mole) bonds also need to be considered.

In the silicon reaction we need 90 kcal/mole to break the $\ce{Si-Cl}$ bond, but we get 110 kcal/mole back when we form the $\ce{Si-O}$; the reaction is exothermic by roughly 20 kcal/mole - a large amount.  In the analogous carbon case, we need 81 kcal/mole to break a $\ce{C-Cl}$ bond, and we only get 85 kcal/mole back when we form the $\ce{C-O}$ bond; the reaction is exothermic by only 4 kcal/mole - not much.  There is quite a difference in the overall driving force for the two reactions with the silicon case being energetically much more favorable than the carbon case!


Chemical composition of seawater Is it true that the sea water is composed of about $86\%$ oxygen, $11\%$ hydrogen and $3\%$ of minerals? The chemical formula of water is $\ce{H2O}$ (two hydrogen and one oxgen) that shows that the number of hydrogen is greater than that of oxygen.

If the number of hydrogen is greater, then why does the sea water consist of $11\%$ hydrogen and $86\%$ oxygen, which is lesser than the oxygen?

The book which I am reading says which is confusing me:

  ... Seawater is composed of about $86\%$ oxygen, $11\%$ hydrogen and $3\%$ of minerals, consisting mainly of sodium and chlorine.

The book that you're reading is measuring by mass.

If you have pure water then you would expect oxygen to make up $\frac{16}{16 + 2}\times 100\% \approx 89 \% $ by mass. Likewise, hydrogen would make up $\frac{2}{16 + 2}\times 100\% \approx 11 \% $ by mass.


Why are the bond angles in sulfur dichloride and oxygen dichloride not the same? Why are the bond angles represented below for each molecule not the same? they have the same form: one central atom and two others in the sides, so the bond angle has to be the same for the three, doesn't it?

Why should the bond angles be the same? You have different atoms every time leading to different molecules. Different molecules have different properties; that goes both for macroscopic properties (they will have different melting and boiling points, their liquid phases will have different densities and refraction indices) and microscopic ones (i.e. their molecular structure will be different). Indeed, there are three-atom molecules with a $90^\circ$ bond angle and those with $180^\circ$. I’m not sure whether it ever gets any smaller than $90^\circ$ but I suspect ‘no.’


What is the temperature needed for a reflux using a mix of solvents as the reaction medium? I had a quick query regarding the setup of a reaction. The procedure that I am working from states that I need to put my reaction mixture under reflux for 3 hours. The solvents used are diethyl ether (30 mL) and THF (10 mL). How do you determine the temperature needed? Do I go via the boiling point of diethyl ether or THF?
If more details are needed regarding the reaction I am carrying out then please let me know. But my query leans more towards the determination of mixed solvent systems in general and how one would determine the reflux temperature needed.

Trial and error, but most often the boiling point of the lower-boiling solvent.
The absolutely correct answer would be to look at a phase diagram, determine whether you have a positive or negative azeotope (or non at all), find out where on the phase diagram you are and then derive the temperature. That would tell you approximately at what temperature your mixture will boil – after you factored in boiling point increase due to dissolved compounds.
In practice, you need to set your oil bath to some 5 or 10 degrees higher than the boiling point of the lower-boiling solvent anyway. So you can just ballpark it, selecting a temperature that would keep the lower-boiling solvent refluxing. Come back after 15 minutes, if it is not refluxing, increase by 5 or 10 degrees.


How is the change in Gibb's free energy a condition for spontaneity for reactions not happening under constant pressure conditions? According to what I have studied, the change in Gibb's free energy is only a condition for spontaneity for processes happening at constant pressure and temperature conditions.

Consider a reversible reaction happening in a closed vessel at a constant temperature.
We check the feasibility of the reaction at any instant using the following equation:

$\Delta G = \Delta G^{.} + RT\ln(Q)  $

As the reaction happens, the partial pressures of each gaseous component will keep on changing as well as the total pressure. Then how can we use Gibb's free energy for checking spontaneity?

Gibbs energy have an another equation which is:-

So according to this equation spontaneity of a reaction depends both on the enthalpy change as well as entropy change.
We all know that a system attains stability as the system's energy is lowered that is it has to release energy. If in a reaction energy is released enthalpy change is negative.
We also know that a reaction is favoured in the direction of increased chaos or a positive entropy change.

So from this we can see that Enthalpy change is negative and entropy change is positive for a spontaneous reaction to take place and the overall equation of

becomes negative that is Gibbs free energy must always be negative for a spontaneous reaction.....Can be applied to any reactions.
Hope this helps.


what's the ratio of atoms of these two isotopes? here is a question says
:

I know that atom's number is found by =  $6.022*10^{23} * mol$

but how this formula is used here is what i dont know

what's the $idea$ of this question?

You need more information.  They probably expect you to look in a book and see that the atomic weight (which is a weighted average over all isotopes) of $\ce{Tl}$ is 204.383.  Now if x is the amount of $\ce{^203Tl}$ with atomic weight 202.97 and (1-x) is....

Does that get you started?

$$\ce{(x*202_{.}97) + ((1-x)*204_{.}97)~=~?}$$

Can you get to the answer from here?


Leaving capacity of halogens Why is F a bad leaving group? Say in HF and HI, F is more electronegative than I, then shouldn’t HF be a better acid ?

When HF loses proton the conjugate base formed is F- Fluoride anion,Since fluorine is small and compact atom having lower affinity to keep electrons because of lack of d orbitals thus fluoride ion is unstable conjugate base compared to iodide ion which is stable.Happy conjugate base(stable), more reactive acid. Hence HI is more acidic than HF.

Even though fluorine is highly electronegative(tendency to drag shared electrons in a covalant bond),it has lower affinity to keep this electrons(lower electron affinity) due to its smaller size .


Can silver metal be used to electrochemically reduce copper? I have a solution of copper(II) ions and several cations.  I am trying to reduce the copper ions to copper metal with electricity.  My Ideas was to use silver as the anode and a copper wire as the cathode.  I have seen people say to do this using platinum as the anode, however this is expensive.  I was wondering if silver could be used as an alternative anode because it has a higher reduction potential than copper.  If not, could any metals besides gold and platinum work.  I have attached a picture of my proposed cell.  I have tried the cell, but the silver began to corrode and form a silvery deposit, probably a silver salt.

In principle, electrically depositing copper(II) ions onto a silver electrode should be possible. As it should also be possible to deposit those ions on just about every other cathode in existence as long as there are no more noble cations in solution.

However, depending on what you want to do with the copper, you may run into problems. Do you want copper-plated silver? That should be fine. Do you want to redissolve the copper to get a pure solution by acidifying? In that case, you cannot use the typical $\ce{HNO3}$ method because that would oxidise the silver, too, and you would end up with a mixture of copper(II) and silver(I) nitrates in solution.


What makes diazo compounds so unstable and explosive? I once had an Orgo TA refer to a diazo compound as "diazo-boom-boom" (the technical term).  I have always been curious as to the reason behind the instability and reactivity.

According to Wikipedia

  Some of the most stable diazo compounds are α-diazoketones and α-diazoesters since the negative charge is delocalized into the carbonyl.  In contrast, most alkyldiazo compounds are explosive

What is it about the alkyldiazos that makes them so much more unstable?  There doesn't appear to be any bond strain or other factors.  Naively, I can say that having a resonance structure should make it marginally more stable.

Well, your question is equivalent to “what is it about α-diazoketones that makes them so much more stable?”, which is easier to see. Compared to an alkyldiazo, the α-diazoketone has a resonance structure in which the negative charge goes to the ketone’s oxygen (and far away from the positively-charged nitrogen atom):

Because the oxygen is a quite electronegative element, the resonance form is quite stable and explains the extra stability of α-diazoketones. It is for the same reason that the protons in position α to the ketones are always more acidic than alky chain protons.

Coming back to alkyldiazo compounds, you have to realize that merely being able to write a resonance structure does not intrinsically imply stabilization: the resonance structure has to have some intrinsic stability factor. In the alkyldiazo, the resonance form you wrote is a carbanion, which is considered quite unfavourable unless it has a further stabilizing factor. Moreover, the most common reactions gives N2, which is a very stable compound… the reaction is thermodynamically very favourable.


Is this reaction with H+ and OH− a possible reaction? I am wondering about the reaction $\ce{H+ + O^2- -> OH-}$. Is it possible?

I think this is not a possible reaction, because here a strong acid plus another particle reacts to a strong base. If it is possible, what would this mean for $\ce{O^2-}$ in the acid-base system? Would it be a very, very strong base?

You can try this in the lab: Take some $\ce{CaO}$, add $\ce{HCl}$ and see what happens. Actually, no don’t. Don’t try this unless you know exactly what you are doing.

A single proton does not exist, especially not in aquaeous solution. Using $\ce{H3O+}$ as a proton-donating species is a much better way to formulate reactions. A professor once said, the smallest actually proven proton-equivalent be $\ce{H9O4+}$, so four water molecules that shift the additional proton between them, but since it was first-year, he didn’t go into details. Also in organic solutions or others, protons will always be bound to something, however weakly.

Now that I got that out of the way, let’s take a look at the $\ce{H3O+}$ particle I just postulated. It has three hydrogens connected to an oxygen and in a very, very, very broad way of arguing, every hydrogen can be considered a potentially acidic proton. Therefore, we can think of hydronium as a triprotic acid with the following conjugate bases:

$$\ce{H3O+ ->C[- H+] H2O ->C[- H+] HO- ->C[- H+] O^2-}$$

Aha! Oxide ions can be considered as fully deprotonated water. Of course, the entire reaction series is reversible, so add ‘protons’ to oxide ions (whatever the source) and find that they get protonated to hydroxide and finally to water. You knew of the second step (hydroxide to water). And here’s the big surprise: The first one is identical, only the equilibrium is much, much further on the hydroxide’s side.

Adding excess acid to oxides would just create water. However, the acid-base neutralisation reaction is exothermic. Very exothermic. Which is precisely why I wrote do not try this unless you know exactly what you are doing at the top. If you have ever neutralised semi-concentrated hydrochloric acid and sodium hydroxide, you know how much heat can be created. Think more than twice that.

In fact, water has what is termed a nivelling effect: Water itself is too strong an acid to allow oxide bases to exist in its vicinity; so $\ce{O^2-}$ cannot be dissolved as is, it will immediately get protonated to hydroxide. Find some $\ce{CaO}$, dissolve it in water, and you would immediately get $\ce{Ca(OH)2}$. Oxides are only able to exist in aquaeous solutions if they are either complexed entirely by cations ($\ce{Al}$ can form oxido-hydroxido multi-core complexes at $\ce{pH 7}$ which include internal oxide ions) or if they are in a bound state, so not technically oxides, such as in the vanadyl cation $\ce{(VO)^2+}$.

Finally, you stated something in your question saying that a strong acid would react with something to a strong base which you think is impossible. That is not the case; if the product of the reaction of something with a strong acid is a strong base, then ‘something’ just needs to be an even stronger base. Which is precisely the case of oxide ions. Just to recall for a second that strong acids and even stronger bases react very, very exothermicly. I feel I have not stressed it enough: Do not try this unless you know exactly what you are doing!


What is the difference between the KLMN and SPDF methods of finding electronic configuration? I've always been puzzled by this because my teachers happen to use only the KLMN method, but what is the difference between the KLMN and SPDF methods of finding electronic configuration?

The KLMN(OP) method is based on electron shells, with the labels KLMN(OP) being derived from an experiment in which the spectroscopist wanted to leave room for lower energy transitions in case there were any.

K denotes the first shell (or energy level), L the second shell, M, the third shell, and so on. In other words, the KLMN(OP) notation only indicates the number of electrons an atom has with each principal quantum number ($n$).

The SPDF notation subdivides each shell into its subshells. For further information about the other quantum numbers, especially $l$, which defines the subshell, see the accepted answer to this question: What do the quantum numbers actually signify?

  The K shell can hold two electrons: $n=1,\ l=0$

When $l=0$, we have an s subshell, which has one orbital $m_l=0$, with room for two electrons.

  The L shell can hold 8 elections: $n=2,\ l=0,1$

When $l=1$, we have a p subshell, which has three orbitals $m_l=-1,0,+1$, with room for 6 electrons. The L shell also has an s subshell.

  The M shell can hold 18 electrons $n=3,\ l=0,1,2$

When $l=2$, we have a d subshell, which has 5 orbitals $m_l=-2,-1,0,+1,+2$, with room for 10 electrons. The M shell also has s and p subshells.

  The N shell can hold 32 electrons! $n=4,\ l=0,1,2,3$

When $l=3$, we have an f subshell, which has 7 orbitals $m_l=-3,-2,-1,0,+1,+2,+3$, with room for 14 electrons. The N shell also has s, p, and d subshells.

This table here summarizes the relationship between the two notations for all elements. Here is an example for scandium, $\ce{Sc}$:

Scandium has 21 electrons ($Z=21$). Its electron configuration in KLMN notation is

$$\begin{array}{cccc}
\mathrm{K} & \mathrm{L} & \mathrm{M} & \mathrm{N} \\
2 & 8 & 9 & 2
\end{array}$$

Its SPDF notation (based on the aufbau principle) is $\mathrm{1s^2 2s^2 2p^6 3s^2 3p^6 3d^1 4s^2}$.


How can it be certain that some bonds are 180 degrees? Take acetylene, for example.

It is defined that the two outer bonds are at 180 degrees.

How is it possible to theoretically define all $\ce{C-H}$ bonds in acetylene as being 180 degrees?

If we look at the molecule we see it is symmetrical. The forces acting upon the 2 hydrogen atoms are identical. As you know, when hydrogen forms a covalent bond with carbon, it's electron is drawn more strongly by the carbon than by the hydrogen, which leaves the hydrogen with a net positive charge (carbon is more electronegative than hydrogen).

Also, the 3 covalent bonds between the carbons are kind of bowed outward in the middle. The 3 bonds between the carbons constitute the most negative part of the molecule.

Naturally, the positive and negative charged moieties are drawn toward each other as much as possible. But, given that the force of attraction is equal on all sides of the hydrogen, it will be aligned equidistant on all sides to the negative triple bond.

I don't know if I am painting a good word picture here, but the forces exerted on the hydrogens place them exactly on a straight line, i.e. their bonds are 180 degrees.

Of course there are other forces in play, namely the forces of neighboring molecules. However, they will also align themselves as much as possible so as to minimize the repulsion of having hydrogen atoms close to one another and to maximize proximity of hydrogen atoms to the triple bonds of neighboring molecules. This arrangement will be preserved as much as possible as the molecules move about. So, the net forces upon the hydrogen-carbon bonds will always force those bonds to remain 180 degrees.


Explain the function of this Galvanic Cell in detail So I failed my test because of this question. It was titled Galvanic Cell although it looks more like an electrolytic cell with the external voltage and everything.

What was asked: explain the function of this cell in detail.
some keywords were also presented (not mentioning them all) but there were Gibbs, Faraday, K, and -nFE

I got too confused with the left side while Ni2+ + 2e- -> Ni (s) on the right side looks like a normal reduction in the cathode. What is going on in here?

Standard Reduction Potentials at 25 degrees Celcius:

**Cr2O7- (aq) + 14 H+ (aq) + 6 e- -> 2 Cr3+ (aq) + 7 H20 (l)  +1.33 V
                              Ni2+ (aq) + 2e- --> Ni (s)  -0.25 V**

in the case of electrochemical cells/Galvanic cells: Oxidation at the anode (the A and O of electrochemistry, german people will know what I mean), reduction at the cathode (in electrolysis its the other way round)
$EMF = E_{red} - E_{ox} = E_{cat} - E_{an} > 0$ for the reaction to run without external voltage (<0 in the case of electrolysis) "the higher redox potential always oxidizes the lower one"

In your case this means running the cell without applying any voltage gives an $EMF = 1.33 - -0.25 V = 1.58 V$

reduction at cathode:

Cr$_2$O$_7^{2-}$$_{(aq)}$ + 14 H$^+$$_{(aq)}$ + 6 e$^-$ $\rightarrow$ 2Cr$^{3+}$$_{(aq)}$ + 7 H$_2$0$_{(l)}$
(in the electrolyte -6e charges)

oxidation at anode:

3 Ni$_s$ $\rightarrow$  3 Ni$^{2+}_{(aq)}$+ 6e$^-$
(in the electrolyte +6e charges)
(already multiplied by 3 for the total reaction)

total:

Cr$_2$O$_7^{2-}$$_{(aq)}$ + 14 H$^+$$_{(aq)}$ + 3 Ni$_s$  $\rightarrow$ 2Cr$^{3+}$$_{(aq)}$ + 7 H$_2$0$_{(l)}$ + 3 Ni$^{2+}_{(aq)}$

Now there is however a clear symbol in the electric cycle standing for a voltage source "V". This is indeed confusing and I would ask what was the meaning of that if the figure was actually titled "Galvanic Cell" which is the other way round. So if it would't be for the title and we would consider running the cell with external voltage, this would then turn around all reactions which I wrote before.

Other things that might be relevant:

K$_2$SO$_4$ is responsible for the charge transfer between the cells. Since one cell produces positive charges in solution and the other one negative ones the electrolyte has to establish that the overall charge in both cells stays neutral. The salt bridge connects the electric cycle, without it, the cell won't run for long
the Gibb's energy is related to the EMF. $\Delta G = -n\cdot F\cdot EMF$, so again a positive $EMF$ gives a negative $\Delta G$ which means the reaction runs "alone" without external voltage, the Gibb's energy provides the driving force. n is the number of electrons involved in the redox reaction (6) (F is the Faraday constant, a natural constant)


How can I control the oxygen in an erlenmeyer flask in a bioprocess? I need to control oxygen in a erlenmeyer flask in a bioprocess.  I need to remove overall oxygen in a erlenmeyer flask, and after I need to add oxygen in controlled amounts to a culture in a shaken flask.  Someone knows how can I do it?

Thanks for advance for your ideas.

Controlling oxygen in a flask can be achieved by bubbling a purge gas of some description through it. This will entrain and remove oxygen from the solution as well as creating an oxygen-depleted headspace.

If you have a regulated nitrogen supply, a rubber stopper with a hole in it, an open-ended pasteur pipette and a length of hose, you can bubble nitrogen into the solution to purge oxygen. This requires the vessel to be open somewhere else to equalise pressure, which is achievable with a multi-neck flask but not with a standard Erlenmeyer flask. You can get around this with a stopper with two holes. Continual positive pressure should keep the external atmosphere out.

Depending on how vigorously you want to shake your flask, this may or may not be reasonable. If you are using a stirrer bar you will obviously want to give the pipette sufficient clearance.

As far as controlling addition of oxygen, this would seem to be a bit trickier as you are probably not going to be providing continual positive pressure. You might be able to get around this, depending on the specifics of what you are doing, by pumping in oxygen in the same way and using a balloon or a gas syringe or something to deal with the overpressure.


What decides -1/2 and +1/2 spin values of subatomic particles? This question is in context of nmr. I know that prior to application of magnetic field and RF energy, protons are found in -1/2 and 1/2 spin values. What decides whethere a particular proton has positive or negative values. I know these values are not intrinsic to a proton since applying energy can cause them to change their spins so what decides whether proton has one value over other in first place.

Spin is an intrinsic property of any nucleus, and can be 0, 1/2,1, 3/2, 2........ A nucleus of spin I can adopt 2n+1 stable states or orientations. For 1H, I=1/2, and this means it can adopt two possible stable spin states; +1/2 or -1/2. The energy levels of these two spin states are different (and proportional to the strength of the magnetic field); one is slightly higher than the other, and so there is a population difference between the two states. However, the energy difference is incredibly small, and difference in populations (called the Bolztzmann distribution) is only about one in a million. The individual nuclei can hop back and forth between the two states quite readily, provided there is sufficient energy transfer (into or out of) the system. Indeed, this is the basis of the NMR experiment - we can manipulate spins between the two spin states with relative ease.

For a nucleus with a spin greater than 1/2, there are more than 2 stable spin states that the nucleus can exist in. Deuterium (2H) has I=1, and can adopt one of 3 stable spin states. The distribution of spins amongst these states is approximately equal (less the BD of approx one in a million). Any individual nucleus will switch between these states. Lutetium176 has a spin of 7, and so can adopt 15 different stable spin states.


Why polymorphs form? We always hear about the different arrangements of molecules that lead to the formation of polymorphs, my question here is why would a polymorph form in the first place? The material was given enough time to solidify in  a proper arrangement, why did it choose to solidify in a form that isn't the most stable?

This is a very broad and very interesting question, but the basic point is: polymorphs exist either because

their formation is controlled by kinetics (rather than thermodynamics, which would favor the most stable phase), or
the thermodynamic equilibrium between the various polymorphs is influenced by the synthesis conditions (temperature, pressure, presence of templating agents and other species in solutions…), and once the material has crystallized in a given form, it does not spontaneously undergo a transition to the most stable phase (again, because of kinetics)


How to derive the molar fundamental relation We have the fundamental relation:
$$dS = \frac{1}{T}dU+\frac{P}{T}dV-\frac{\mu}{T}dN$$
how do we get now the following:
$$ds = \frac{1}{T}du+\frac{P}{T}dv$$ where $s = \frac{S}{N}, u = \frac{U}{N}, v = \frac{V}{N}$.
I tried substitution but it got me nowhere..

You have  $$ds=\frac{du}{T}+\frac{P}{T}dv+\left(\frac{u}{T}-s+\frac{Pv}{T}-\frac{\mu}{T}\right)\frac{dN}{N}$$
But, $$\mu=u+Pv-Ts$$
That is, $\mu$ is the Gibbs free energy per mole.  So the term in parenthesis drops out.


Is an acid a salt or not? In our office, we discussed today what the exact definition of a salt is and whether an acid itself can be classified as a salt.

Our first problem was that we couldn't get a definitive definition for what exactly a salt is. Wikipedia states:

  In chemistry, a salt is an ionic compound that can be formed by the neutralization reaction of an acid and a base.

Let's take, for example, carbonic acid: it is a molecule comprised of the ions $\ce{H+}$ and $\ce{HCO3-}$. We could say that this molecule forms due to the reaction of a base ($\ce{HCO3-}$) and an acid ($\ce{H_3O+}$):

$$\ce{H3O+} + \ce{HCO3-} \rightarrow \ce{H2CO3} + \ce{H2O}$$

Thus, it seems that the acid fulfills the definition above. We are pretty sure that acids are not considered salts. So where is the error in our reasoning?

It is a semantics question with an open ended discussion. If you recall the old problem "How many angels can dance on a pinhead? Medieval problem, this issue here is similar.

Basically in chemistry, all the terminology is controlled or endorsed by the International Union of Pure and Applied Chemistry (IUPAC). They have a very open ended definition of a salt: "A chemical compound consisting of an assembly of cations and anions."

If we follow some really old chemistry texts (>1850s), yes an acid was defined as a salt of hydrogen ion, i.e., an acid is a salt whose electropositive element is hydrogen Definition of an acid.

Did this definition by a few chemists become popular? The simple answer is no. Not in a general sense.

The modern interpretation of an acid, again from IUPAC is, "A molecular entity or chemical species capable of donating a hydron (proton) or capable of forming a covalent bond with an electron pair."
If you have heard of ionic liquids, they consist of a large organic cation and an anion, this is a salt as well.

Then you have so-called acid salts, such as sodium hydrogen sulfate e.g.,  $\ce{NaHSO4}$. Their pH is very acidic and these "salts" will neutralize bases like a typical acid.

Thus one may loosely say, that all acids may be considered as salts but not all salts are acids.


Electrolysis of water with aluminum at the negative pole I am doing electrolysis of water to collect the hydrogen and I'm running into a problem: on the positive side I'm using graphite from a pencil (I cannot use any other material because it corrodes too fast) and on the negative side I was using copper. Copper was working really fine, the water was clear and no problem. But I don't have copper with a big surface area so I decided using aluminum (from coca cans) and my solution started to get dirty because of the aluminum and the aluminum is slowly disappearing.

Why? Any other eletrode that I used for the negative pole was not corroded (only on the positive pole everything corrodes). Why is aluminum making my water dirty? Do you think if I use iron this problem will not happen?

If you are using salt to make the solution more conductive the process will cause the creation of Sodium Hydroxide, which will react with aluminium.


Is (Zn(OH)4)2- acidic or alkaline? My chemistry teacher told me that $\ce{Al(OH)4^-}$, $\ce{Pb(OH)4^2-}$, $\ce{Zn(OH)4^2-}$ are called complex ions. So will they break down in some conditions and give out $\ce{OH-}$ to show alkaline property or just remain neutral?

Let's first imagine that one of these complexes is placed in neutral (pH 7) water. If the resulting solution possesses a pH of > 7, we can conclude that the anions show an "alkaline character" as you describe. If the resulting pH is < 7, we'll conclude that the material is contributing an "acidic character."

The way one might go about answering this question without dunking one of these salts in water is by starting with the a more well know complex such as the hexa-aqua complex of each metal and observing the behavior.

$\ce{AlCl3 + 6H2O -> Al^3+.6H2O + 3Cl^-}$ Chloride becomes a spectator anion

This $\ce{Al}$ water complex ($\ce{Al^3+.6H2O}$) possesses acidic protons, just like any polyprotic acid (eg. phosphoric acid.) Thus, each proton also has an acid dissociation constant or pKa. The first pKa for $\ce{Al^3+.6H2O}$ is approximately 5 and would thus acidify the solution. The pKa's of each of the subsequent protons donated to solution is higher than the last and eventually, no more protons would be donated to the solution.

With this basic understanding out of the way, let's look at the approximate pKa's of each of the neutral hydroxide salts of each of these metals.

For $\ce{Al(OH)3}$ the pKa is > 7. This means that by adding another hydroxide to the complex to make it a tetrakis hydroxide complex and would result in an even higher pKa and a more alkaline solution.

$\ce{Pb(OH)2}$ and $\ce{Zn(OH)2}$ are both weakly basic (pKa > 7) and possessing low solubilites. Adding more hydroxide ligands will make each of them more alkaline and should also make each of them more soluble. The zincate anion $\ce{Zn(OH)4^2-}$ is often formed by dissolving Zinc metal in a strongly alkaline solution.$\ce{^1}$

I hope this helps a little.

ref 1. http://depts.washington.edu/chem/facilserv/lecturedemo/BrassPenny-UWDept.ofChemistry.html


Why isn't nylon as strong as kevlar? Why is Kevlar stronger than nylon? They both contain Hydrogen bonding and should have equal strength but this is not the case instead Kevlar is fire resistant. What makes it stronger. Greater hydrogen bonds?!

Kevlar is an A,B co-polymer where the monomers are terephthalic acid and 1,4-diaminobenzene. The amide linkage, together with the aromatic rings in the polymer, makes a very rigid polymeric structure that stacks the chains in an organized way, maximizing the effectiveness of hydrogen bonding between amide linkages, together with pi-stacking of the rings:

http://pubs.acs.org/cen/coverstory/8005/8005olympics.html

The chains form sheets that stack together along the axis of the fiber. In order for projectiles to break these fibers, these strong, inter-strand attractions must be overcome.

Standard Nylon is a similar polymer, where the aromatic moieties are replaced by aliphatic chains. These are much more flexible and give more degrees of freedom to the polymer chains. This results in a more random arrangement of chains that still hydrogen bond, but not in such an organized way, giving a weaker fiber.

https://www.researchgate.net/publication/305313239_Nylon_66_Nonwoven_Fabric_Separates_Oil_Contaminates_from_Oil-in-Water_Emulsions


What is chemical formula of Bauxite? My book says

  Aluminium is extracted from its ore bauxite $\ce{Al2O3.nH2O}$

What does "n" stands for?
Does it mean that there is no fixed amount of water in bauxite?

$\ce{Al2O3.nH2O}$ is just a general (and slightly incorrect, see below) way of writing the formulae of the aluminium oxy-hydroxides that occur in bauxites. These are solid minerals, so there is no "$\ce{Al2O3}$" molecule in them. Instead, it is an (almost) infinite array of Al and O ions in a three dimensional structure. Therefore, what it basically means is that you have three cations of aluminium for every two anions of oxygen. $\ce{Al2O3}$ means exactly the same as $\ce{AlO_{1.5}}$ or $\ce{Al4O6}$ in this context. We just use $\ce{Al2O3}$ because it's simpler: smallest numbers with no fractions.

In the case of the hydroxides, they are also crystalline minerals. You can see some nice pictures of gibbsite, böhmite and diaspore when they are well crystallised. Their formula are:

Gibbsite: $\ce{Al(OH)3}$ which is the same as $\ce{AlO_{1.5}.1.5H2O}$
or $\ce{Al2O3.3H2O}$.
Böhmite and diaspore: $\ce{AlO(OH)}$ which is the same as $\ce{AlO_{1.5}.0.5H2O}$ or $\ce{Al2O3.H2O}$

Bauxites usually contain mostly gibbsite, which is unfortunate economically because it contains the most water in it. When you dig the stuff out, you want the most aluminium per tonne of material, and dilution by water is undesirable.

It's also important to note that even though we write the formulae of the minerals as $\ce{Al2O3.nH2O}$, the water is not hosted in molecular form in the crystal. You do not have proper $\ce{H2O}$ molecules in the structure of the mineral. Instead, you have hydrogen cations attached to some of the oxygen anions. There are, however, minerals where the water is in molecular form. One example is gypsum: $\ce{CaSO4.2H2O}$.

Usually, as a rule of thumb, the way a formula is written can give you an idea on how the water is hosted. If the formula is written with $\ce{(OH)}$ in it, it's probably in the form of hydrogen bonded to oxygen. If it's $\ce{.H2O}$ then you probably have molecular water. This is also important for the physical properties of the crystals. Minerals with molecular water are usually very soft and will lose their molecular water with gentle heating (up to 150 °C or so). Minerals with hydroxide groups in them require much higher temperatures to de-volatilise, around several hundreds of °C.

So, to sum it up, writing the formula as $\ce{Al2O3.nH2O}$ gives the impression that the aluminium hydroxides have molecular water in them, while they actually do not. However, it's a convenient "catch-all" way to refer to all of them together.

Also related:
What is the meaning of the "dot" notation in chemical formulas?


How to index molecules in computer programs? Is there a standard way how can the index over a molecular structure be kept in the computer?

A molecule is a graph, where nodes have labels (chemical element type), and edges are chemical bonds.

There is SMILES representation that determines the molecular structure, but SMILES representation isn't unique.

One can take the set of all possible SMILES representations for one molecule, sort them alphanumerically, and pick the first element. This would be a unique string that is in one-to-one relationship with a molecule, and would satisfy my request.

But is there a standard way how to have a value that is in one-to-one relationship with a molecule that can be used as an index in the database that has many molecules in it?

By extension, how to efficiently search a set of molecules and, given a particular molecule, find a set of the closest structural matches? What characteristics of the molecules should be memorized that would allow such search to be efficient?

A molecule is a graph, where nodes have labels (chemical element type), and edges are chemical bonds.

while fairly common and acceptable for an initial approach, this models fails to catch many types of isomers and tautomers. There is no well-established general approach because the basic definitions are kinda wonky and the subjects itself is extremely complex. For example, monomethylbulvalene would for all intents and purposes work as a single compound yet a set of different molecules would be present in all reasonably analyzed samples.

  By extension, how to efficiently search a set of molecules and, given a particular molecule, find a set of the closest structural matches?

That's a part of QSAR big problem: to identify "similar" molecules. It doesn't work reliably.

  What characteristics of the molecules should be memorized that would allow such search to be efficient?

There are specialized graph databased that aim to solve the problem of efficient search on graph-like structures, but see above.


How to control and vanish termiticide/pesticide fumes from room I had some wood work done in my room and termiticide (UNIBAN-20 (Chlorpyriphos-20% EC) was used so that there is no termites attack in future. The stupid carpenter was using this product for the first time and it had to be diluted 10 times in kerosene oil before use which he didn't do. He poured the concentrated solution of this termiticide.

Now we have very strong smell and poisonous fumes (the head starts spinning within seconds) in the room.

How can I neutralize the smell and get rid of the poisonous fumes? It has been 13 days now.

I have exhaust fan switched on but the smell is still there in the room.

As stated by bon in the comments, this is probably something you should contact someone who is actually supposed to be able to answer this question about. The fumes might just be an annoyance, or they could be highly and immediately dangerous.

One place to start is with the manufacturer. Indeed, the manufacturer is probably the best place to go in general with chemical safety concerns. This chemical seems to be produced by a number of firms, so I was not able to locate a specific one, but a google search shows a number of possible contacts. This MSDS provides contact information for a Dow subsidiary. I suspect they will be willing to help, but if they refuse to because they were not the manufacturers of your specific product, you will need to ask the carpenter for the documentation. In a number of locations, the law specifies he must be able to provide an MSDS.

Alternatively, you might wish to contact your local fire department (non-emergency line). They can probably point you to someone local whose job it is to handle potential chemical contamination.

I will note that the MSDS I cited does not suggest extreme danger. However, it is still best to contact a producer or specialist of some sort.


Why is honey more viscous than water? Why is honey more viscous than water? And, if we boil honey to just below its boiling point, will it become equally viscous as water at normal temperature? How does changing temperature affect viscosity?

Viscosity generally decreases with increasing temperature so honey will become less viscous at higher temperatures. Regarding the actual values, look at this article. It appears that even at quite high temperatures honey remains a couple of times more viscous than water.

I think the reason for higher viscosity in honey is the high sugar () content - hydroxyl groups in sugars can form strong, hydrogen-bonded interactions with water and each other, and the sugar molecules being quite long, they are difficult to shear.