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Continuous chemical reaction network versus chemical reaction network - difference?
I'm reading a paper entitled Reachability Problems for Continuous Chemical Reaction Networks and as I was reading it I realized: I have no idea what the difference is between a continuous chemical reaction network (CCRN) and a chemical reaction network (CRN). The paper does distinguish between the two, so they aren't two terms for the same thing. The wikipedia article doesn't distinguish/give a definition for a CCRN. Most of the results when I search google are papers that are way over my head - the one I'm reading right now is to be quite honest over my head.
Any help would be appreciated. Thanks!

I suspect that the distinction is that a CRN considers discrete species concentrations whereas a CCRN considers continuous species concentrations: "The CCRN
model is continuous, dealing with real-valued concentrations of species...". Certainly on a molecular level concentrations must be discrete, but continuous concentrations are easier to work with.

@a-cyclohexane-molecule, I think I understand your meaning, but would you mind turning your comment into an answer (and perhaps giving an example)?

I suspect that the distinction is that a CRN considers discrete species concentrations whereas a CCRN considers continuous species concentrations: "The CCRN model is continuous, dealing with real-valued concentrations of species...". More precisely, whereas a CRN deals with concentrations $x(t), y(t) \in \mathbb{Q}$, a CCRN deals with concentrations $x(t), y(t) \in \mathbb{R}$.
On a microscopic level, it's easy to see why $x(t), y(t)$ should be elements of $\mathbb{Q}$: a microscopic concentration is simply a ratio of the number of one molecule to the total number of molecules in the system, and the numerator and denominator must thus be integral.
For macroscopic systems, however, since the denominator---the total number of molecules in the system---is so large, it becomes a very good approximation to take $x(t), y(t) \in \mathbb{R}$. Moving to $\mathbb{R}$ is convenient, since then by physical considerations $x(t)$ and $y(t)$ are continuous, and we can, under mild assumptions, take derivatives and return to a rate-law formalism involving differential equations rather than difference equations.

Remark. I'm not familiar with this field, and I don't know how common the CRN/CCRN distinction is. Indeed, the Wikipedia article assumes differentiability of $x(t), y(t)$ and hence that $x(t), y(t) \in \mathbb{R}$.


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How to identify if it is a buffer solution

Aqueous solutions of $\ce{HNO3}$, $\ce{KOH}$, $\ce{CH3COOH}$, and $\ce{CH3COONa}$ of identical concentrations are provided. The pair (s) of solutions which form a buffer upon mixing is(are)?
A. $\ce{HNO3}$ and $\ce{CH3COOH}$
B. $\ce{KOH}$ and $\ce{CH3COONa}$
C. $\ce{HNO3}$ and $\ce{CH3COONa}$
D. $\ce{CH3COOH}$ and $\ce{CH3COONa}$

I think that option D is correct as it a solution of a weak base and its a salt; that's why it is a buffer solution. But how do I check the other options?

A screenshot or picture of an exercise is not searchable. Please consider rewriting it, so that it can be of help for future visitors.

Well, it all boils down to the preparation of buffer solutions in the first place. One of the methods of preparation includes the mixing of a salt solution of an weak acid and a strong base with a solution of a strong acid, with the solution of the strong acid as a limiting reagent, which leads us to conclude positively for the third choice.
Alternatively, you can think which of the mixtures will resist the change in pH on mixing a strong acid or a strong base; for instance in the third choice, nitric acid being a strong acid will help to resist the change in pH when a strong base is added, and $\ce{CH3COONa}$ being a weak acid + strong base type of salt resists the change in pH that could've been caused due to the addition of a strong acid.

Why option B is wrong

@samsung Neither of $\ce{KOH}$ and $\ce{CH3COONa}$ would be able to help when you add a strong base.

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Thermal and photochemical excitation of electron in photochemistry
One of the basic rules of photochemistry is that the electronic configuration of a conjugated polyene does not change when excited thermally, but changes when photochemically excited, i.e. an electron moves from HOMO to LUMO.
Can anyone please explain this to me in terms of quantum mechanics?
I have done some calculations, which I am enclosing here.
(P.S.: Please do not provide explanation like thermal energy is very less to excite than photochemical, since it would imply that at very high temp (~1,00,00,000 C) the reaction is possible, which is not really the case).

But what you say in your last sentence is true, and not specific to polyenes; it is quite general.  The energy of a visible photon at $500$ nm is $20000~ \pu{cm^{-1}}$ but at room temperature the average thermal energy is only $\approx 210 \pu{~cm^{-1}}$. The probability of a molecule with $20000~ \pu{cm^{-1}}$ of thermal energy is $\approx e^{-100} = 3.7\cdot 10^{-44}$ so small! Thus there in effectively zero chance that by thermal excitation an excited state at 500 nm will be produced. (By excited state I mean an electron is transferred from the HOMO orbital into the LUMO or another, higher, unoccupied orbital).
In your calculation you cannot use $C_V\Delta T$ as its not an operator but in a collision transferring energy there has to be an operator so you cannot just remove it from the integral. The properties of the operator determine the selection rules for transitions.
It is possible by collisions between molecules to move up and down in energy, but because thermal energy is small usually this normally involves vibrational and rotational levels. (The effect is immediately obvious by cooling or heating samples in the appearance of hot bands and in relative rotational line intensities.) If there are very low lying electronic states then I can see no reason why these also could not be excited thermally.

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What is a topological isomer?
I'm working on a project where I have to understand the concept of topological isomer. I found a review or two containing the idea but I couldn't find a book or a detailed description of what a topological isomer is. I would like a suggestion for a book, article or other resource that gives me a clear definition (Wikipedia doesn't do a good job either). Or I would like someone to give me a clear description of what a topological isomer is.

Wikipedia has a page on topoisomers

"Isomers" are compounds that are equivalent but for a certain change in structural property. There's different types of isomers depending on which sort of structural property you need to change in order to inter-convert the two molecules.

Structural isomers are equivalent in their atomic makeup but differ based on their chemical connectivity. For example, 1-propanol and 2-propanol are structural isomers - they share the same formula ($\ce{C3H8O}$) but differ in how those atoms are connected. You can change one to the other by rearranging how the atoms are connected to each other.
Stereoisomers are identical in chemical makeup and connectivity, but differ in the spatial orientation of their atoms. There's several different subtypes, which vary based on what sort of change to the molecules can interconvert them.

E-Z isomers can be interconverted by a theoretical rotation around a double bond - something that normally doesn't happen.
Optical isomers can be interconverted by mirroring the molecule (again, something that doesn't happen)
Diastereomers can be interconverted by mirroring the arrangement around a certain subset of atoms.

Conformational isomers are like stereoisomers, but with the distinction that they differ based on changes which do happen under normal conditions. For example, by a rotation around a single bond, or by the umbrella-like inversion of a nitrogen. Different conformational isomers are normally not isolatable from each other, as they rapidly interconvert.

There's a subset of these isomers called atropisomers which would otherwise be interconvertable, but local conditions (e.g. steric hinderance) means that the interconversion is slow under standard conditions, such that the different isomeric forms can be isolated from each other.

Okay, that brings us to your question on topological isomers. These are conceptually very closely related to conformational isomers and atropisomers. Like conformational isomers, topological isomers are equivalent in chemical formula, connectivity, and even the normal stereochemical properties like E-Z and R-S designations. However, like atopisomers - and unlike most standard conformational isomers - two different topological isomers cannot interconvert on an appreciable timescale, and thus each isomer can be isolated from the other. But unlike atropisomers, the difference between two topoisomers is not due to rotation around a single bond.
Instead, the difference between two different topoisomer is due to the different topology of the molecules. Topology is a somewhat complex mathematical topic, but for our purposes here, you can think of topology as a description about the invariant way something structured on a global scale, that can't be changed by bending, twisting, folding, etc.
For example, take a rubber band. No matter how you bend, twist, fold or tie the rubber band, it's still topologically the same. (That's part of the definition of "topology".) However, if you cut the rubber band, pass it through itself and reattach it, you can get something that's called a trefoil knot. If you look locally at the rubber band, all the connectivity is the same. There isn't any twists in the trefoil that couldn't also be present in the original circular band form. But on a global scale the trefoil and the band are very different. There's no way you can bend, fold, twist, tie, etc. the two to interconvert them without cutting. Because of this, we say the trefoil and the circular band are topologically distinct, in a way the circular band and a twisted form of a circular band aren't.
This is the difference that's being expressed in the case of topological chemical isomers. All of the local structure between the two is the same, but the large-scale global structure of the molecule is such that you cannot interconvert the two isomers without breaking bonds. You can't make one into the other simply by rotating a bond -- not even by rotating a bond that is energetically disfavorable to rotate.

Do the optical isomers of something like hexahelicene count as topological isomers? Would make a good example if they do.


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Why is the number of carbonyl stretches in the IR spectrum different for solid state samples compared to samples in solution?
I am comparing two IR spectrums of the same molecule, one in the solid state and one in $\ce{CH2Cl2}$ solution. There is only one CO group in the molecule.
Why does the solid state spectrum show two carbonyl stretches, while the solution spectrum has only one carbonyl stretch?

Please provide more information about your molecule and the spectra, and the preparation of samples. In its current state the question is very vague and broad.

In solution, the material to be characterised experiences a different environment than in its own matrix.  In extremis there is nujol as apolar matrix, and KCl as salty matrix, for example. In the pure solid state, the accessibility of vibrational states may be a different one, in particular in the crystalline state where special sites and their local symmetry may allow / disallow vibrations to occur (in case of perovskites, for example).


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Can I claim that a hot dip Aluminum coating in steel provides cathodic protection to a substrate since Al can act as sacrificial anode?
I have a school experiment wherein I will put aluminum coating to a steel substrate, wherein the Al will act as a corrosion protector for the steel. We know that aluminum is also classified as a sacrificial anode. Can the aluminum coating on the steel provides a cathodic protection?

Hello and welcome to StackExchange. Have a look at the FAQ to learn how to post good questions, and hopefully get good answers. http://chemistry.stackexchange.com/help

In practice, no -- at least, not at ambient temperature.  Aluminum passivates and this passive film breaks the potential electrochemical circuit.  The passivated aluminum can't react to release electrons to the steel.  For cathodic protection you need something that will corrode instead of the steel, meaning it can't passivate.
Usually, cathodic protection is offered by zinc which has less driving force, but it does not passivate and thus the electrochemical circuit remains intact.
Addendum:
Aluminizing does offer cathodic protection at high temperature where the passivation is less effective.

So the corrosion protection that Al coating provides for steels cannot be considered a cathodic protection?

That is right, at least under ambient conditions.  Aluminizing is more effective at high temperature where the aluminum finally becomes more reactive.  See http://www.onealsteel.com/carbon-steel-sheet-aluminized.html.

I have read about this concept. So if the aluminum coating on steel passivates, it can provide great corrosion protection for the steel, although this process can't be considered cathodic protection, right?

Not cathodic protection, right.


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Is the cyclopropenyl anion aromatic?
Shouldn't cyclopropenyl anion be aromatic as it will keep its lone pair in an unhybridized orbital? My teacher says that the lone pair will be a part of a hybrid orbital and the compound will be anti-aromatic.

Welcome to chemistry.se! In $\ce{C3H3-}$ (with the hypothetical $D_\mathrm{3h}$ symmetry) you would have 4 electrons in the pi-system (all p orbitals). This is only a minimum on the potential energy surface in a triplet state, which is higher in energy than the distorted singlet state. The anion is not aromatic, there are no stable (ground state) anti-aromatic compounds.

No stable antiaromatic compounds?  Some call that a challenge ... https://cns.utexas.edu/news/new-breed-anti-aromatics.

@Martin-マーチン are there really no stable anti-aromatic compounds? 5,5-difluoro-1,3-cyclopentadienes are antiaromatic and you can buy that one.

@DSVA I don't see how the compound you cite is anti aromatic, there is no circular conjugated pi system. If a compound is destabilised with respect to another, distorted arrangement of the 'same' bonding pattern, it cannot be the ground state. The definition of anti-aromaticity dictates that it is less stable than a distorted geometry, because it cites this the reason for the distortion.

@OscarLanzi that name gets tossed around by far too often. I don't have the time to review that article completely, but going from the very definition of anti-aromaticity, being a less stable arrangement, the claim can't be true, it at least has a huge caveat.

@Martin-マーチン that compound basically has a conjugated 4pi-electron system since the CF2 group can participate "hyperconjugatively" with the ring system (original publication http://pubs.acs.org/doi/abs/10.1021/ol401154r , follow up by Houk: http://onlinelibrary.wiley.com/doi/10.1002/jcc.24191/full ). That compound is planar and shows antiaromaticity in every metric we got (aromatic stabilization energies, NICS, ACID, Diels-Alder reactivity,...)

Why will the carbon atom with the negative charge keep its electrons as a part of the conjugated system?


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Why -I effect of SR2+ >NH3+
I have previously asked this question, but that was in a little ambiguous way.
So am re-asking the question precisely.
As electronegativity of N > S, hence the tendency of N to pull the electrons is supposed to be more than that of S.
Moreover there is two alkyl groups are attached to S (though I don't think alkyl groups will show +I effect as probably no scope for hyperconjugation!).
Then why is this contradictory trend?

I'm not personally convinced that this is a duplicate, but please hold on, a moderator will get back to you regarding this question some time soon. Jan's answer in the linked duplicate did argue that as N is more electronegative than S, NH3+ should be more electron-withdrawing than SMe2+. So, it is not exactly true that there is no comparison there. However, that does not automatically make your question invalid, since you are asking about why your textbook says the opposite.

I am also not at all convinced that this is a duplicate. However, you have asked the same question before. Please do not do that. Instead edit the existing question with the relevant changes, making clear why it is not a duplicate. If you disagree with a close vote, you can discuss it on meta. Re-asking the same question will not help you. It is likely to be closed even faster, and many closed questions will trigger a ban.// I personally can't really follow your reasoning though. Please also include a citation for your screenshot.


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The Boiling or Evaporation of Diethyl Ether
When we heat water, we see bubbles appear as the boiling point approaches. The water then "boils" vigorously as it is converted from the liquid phase to a gas.
I realised recently that during an ether extraction of a product, a lot of ether had evaporated, as was evident from the mass balance. The ether vapours were visible, but no "boiling", or fluid motion that we tend to associate with boiling, was observed. The boiling point of ether is $34.6\ \mathrm{^\circ C}$. The ambient temperature then was around $35\ \mathrm{^\circ C}$.
Question: Why do we not see ether "boiling" near its boiling point?
Let me also put forth my thought process: The most apparent difference between the two cases of boiling water and boiling/evaporating ether is the location of the heat source. The ether was kept in a beaker on a stable platform, so it could exchange heat through the upper surface and the walls of the beaker. Heating water from below might create pockets of 'hotter' and hence less dense water that might lead to the "boiling". In the case of ether, the only possible explanation I found believable was that the walls of the beaker were really bad conductors of heat, as we would have seen some fluid flow in the ether case as well.
We also see puddles of water evaporating off. In that case, the heat exchange is possible only from above. Considering no disturbances, there is no visible fluid motion in the puddle. I felt this was because the rate of evaporation was too low. In the case of ether, it is already at its boiling point. This leads me to believe that in a restricted environment as a beaker, when the ether changes phase, the vapour accumulates near the surface as it is denser than air (almost 2.5 times as much). This further slows down the transfer of heat. Taking this into account, I find it very hard to believe that about 5 g of ether was lost in a mere span of a bit less than 10 minutes.
Is my reasoning right, or is it flawed? Please do tell.
Notes & References:

Diethyl Ether - Wikipedia
weather.com and personal experience
Modified Ideal Gas Law: $\rho=\frac{pM}{RT}$, i.e. $\rho\propto M$, taking $M_\text{air}\approx 28.84\ \mathrm{\frac{g}{mol}}$

I believe the ether was slightly cooler than its surroundings (mostly due to evaporation), and thus below its boiling point.

You can see ether boil if you heat it slightly more strongly - DO NOT do that without supervision, though.

This was an EXTREMELY dangerous experiment, if you actually performed it at an outside temperature of 35°C. You break an ether bottle at that temperature, and huge amounts of it will have evaporated when, almost inevitably, the fumes have found an ignition source. In a cool lab, you get scorched a bit, OK. In yours, you´re toast. No fume hood window will save you.

The issue here, as eluded to in your discussion, is that near the boiling point the rate of heat being removed from the ether (via rapid evaporation) was greater than the rate of heat being transferred into the ether, so that the temperature of the ether never reached it's boiling point. In fact, the ether was likely several degrees below it's boiling point due to the evaporative cooling.
The rate of heat transfer into the solution via conduction from the surrounding air is a function of the difference in temperature between the ether and the air, which in this case was very small. If the ether had been placed in an oven of a sufficient temperature, the ether would have boiled.
The mass loss that was observed is not at all unexpected, as the temperature of the ether was just below it's boiling point. Imagine how quickly water would evaporate from a beaker if the surrounding air temperature was $\pu{99^oC}$ (and assuming we're at sea level so the water doesn't boil). The vapor pressure of water, and ether, is only slightly less at a few degrees below it's boiling point than it is right at it's boiling point. (Of course the physical action of boiling also allows for more rapid evaporation, so long as sufficient heat is applied to keep up with the evaporative cooling).
Summary, TL;DR:
As the ether was close to it's boiling point, it was evaporating rapidly, resulting in rapid mass loss. The evaporative cooling near the boiling point happens at a rate to greater than the conductive heat transfer from the surrounding air, so that it never quite reaches it's boiling point, and actually cools to several degrees below it's boiling point.

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Does the Hofmann Rearrangement extend to Secondary Amides?
I have read where it is explicitly stated that the Hofmann Rearrangement is a reaction whereby primary amides are converted into primary amines with one less carbon. I have tried to find examples where a secondary amide is used, but haven't been able to find it addressed anywhere. I have been able to find examples where a N-hydroxylamide is used, but that's still a primary amide.
What happens when a N-methylamide or an N-methylhydroxylamide is subjected to Hofmann Rearrangement conditions?

Simply put, secondary amides will not react.

I think it's not possible, because if you go through the reaction mechanism, you will see that in the 1st and 3rd step two protons are needed to be abstracted by the base to proceed the reaction. So if you take a secondary amide, i.e., one of the H being substituted by any alkyl/acyl/aryl group, the base will not be able to abstract that group and hence the reaction will not progress to any fruitful product.

Pic reference: http://www.name-reaction.com/hofmann-rearrangement

See you the isocyanate intermediate in the mechanism above? It's impossible the existence of this intermediate from a secondary amide. So, Hofmann, Curtius and Lossen rearrangements only works with PRIMARY amide.

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Should the normal mode of vibration leading to the transition state be removed?
I would like to know whether the normal mode of vibration of the reactant should be deleted when calculating $\Delta ZPE$ and $\delta G_{\text{corr}}(T)$, where $\Delta ZPE$ is the zero-point energy difference between the minimum and the TS and $\delta G_{\text{corr}}(T)$ is the difference between the thermal corrections of the minimum and the TS. In the TS with $N$ atoms there are $3N-7$ degrees of freedom. Should I use the same number of vibration in case of the reactant, as well?

Are you talking about zero-point correcting a TS?  Related: https://chemistry.stackexchange.com/questions/6610/why-are-the-total-energies-of-transition-states-so-commonly-corrected-for-zero-p

Yes. What is your opinion about leaving the vibration coordinate conforming to the imaginary mode?

Include it in the zpve correction. This is what is commonly done in the literature.

Including it provides a negative free energy of activation, because the reactant and the product are two conformers of n-pentane, thus they are drastically similar, but they have huge ΔZPVE.

Other source of errors (applying the harmonic approximation and anharmonic correction) cannot occur, all calculations were performed with state-of-art computations within the focal-point approach (e.g. we used geometries close to the CCSD(T)/cc-pVTZ ones, extrapolated ΔHF, δMP2, δCCSD, δCCSD(T) terms, based on MP2/cc-pVXZ and CCSD(T)/cc-pVXZ single point calculations up to X=5 and X=4, respectively). The free-energy differences of the conformers are in agreement with the values from the litterature.


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How to read electron energy levels abbreviations?
The electron energy transfer diagrams I've been coming across all have energy levels labeled as e.g. $^{2}F_{5/2}$ or $^{4}S_{3/2}$. I've tried to Google "quantum energy level abbreviations" but haven't come across anything too useful, just explanations of electron configurations. Can anyone explain or point to a website or textbook that goes through what these numbers/letters/fractions mean for someone who's taken general chemistry but not quantum mechanics?
Thanks!

It's a bit tough to understand where these numbers come from without QM, and I don't really want to write any technical inaccuracies (because this topic is confusing enough for QM students and having another sloppy account online won't help). However, as a general chemistry student, perhaps you are aware of the quantum numbers of an electron: $(n, l, m_l, m_s)$. $m_s$ has to do with the spin of the electron, which is an "intrinsic angular momentum" of the electron. $m_l$ has to do with the orbital the electron is in, which confers a certain amount of angular momentum on the electron as well.

... If you add up the angular momenta of all the electrons in an atom, you can get the angular momentum of the atom. In short, then, the top-left number represents the spin angular momentum of an atom, the middle letter represents the orbital angular momentum of an atom, and the bottom-right number represents the total angular momentum of an atom, which is simply a sum of the previous two angular momenta.

Those are Russel Saunders atomic term symbols and represent the quantum numbers of the many-electron wave function.
The notation scheme is $^{2S+1}L_J$ where $2S+1$ is the spin multiplicity, $L$ is the angular momentum quantum number and $J$ is the total angular momentum due to the coupling between the former two.
There are also molecular term symbols using greek instead of latin letters for the angular momentum. Those are commonly used for diatomic molecules.
You should find the Russel Saunders Coupling scheme in standard physical and quantum chemistry books. Molecular term  symbols are for example explained in detail in Molecular Spectra and Molecular Structure: I. Spectra of Diatomic Molecules by Herzberg and Huber (1950).
This link may give you a preliminary idea : http://wwwchem.uwimona.edu.jm/courses/RScoupling.html


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Why does ice melt slower at higher altitudes?

Why does ice melt slower at higher altitudes and how does Le Chatelier's principle play a role in this?

This question appeared right after this chapter in my text book and I've also read a very rough explanation from another source which cited this principle.

What a poor question. Dumping that statement with its tremendous lack of context and forcing the link to be found leaves it wide open to misinterpretation and speculation, and is likely to either cause confusion or establish wrong connections.

I'm sorry but I've found this question in my book Just as I've wrote and as concerning to context, I'm only asking is there a link to La Chateliers Principle to the phenomenon.

Sorry, I didn't mean to imply you are at fault, it's the book's author who should done a better job. You're fine!

A simplified attempt.
For a simplified answer that attempts at a solution using Le Chatelier's principle, you'll need a constraint: Ice and water are at an equilibrium at $\pu{0^\circ C}$.

$$\ce{Ice <=> Water}$$

Simple enough, now pull up the values of densities of ice and water, $\rho_{ice}= 0.9167~\pu{g/cm3}, \rho_{water}= 0.9998 ~\pu{g/cm3}$. Mass being the same, you can conclude that volume of ice is greater than that of water, so in accordance with Le Chatelier's principle, when you move at higher altitudes, the atmospheric pressure falls, thus the equilibrium is favored towards a side where the species occupy a higher volume, i.e. towards the formation of ice. This works antagonistically for $\ce{Ice->Water}$.

Alternatively, have you heard of phase diagrams? This might help you to understand things a bit more clearly after you go through its definition from other sources: https://chemistry.stackexchange.com/a/72732/43942
The phase diagrams for water are different though.
https://d2gne97vdumgn3.cloudfront.net/api/file/m1GFJVbQSBmwkDXRvglV

This, too indicates that for water, at lower pressures, the formation of solid (i.e. ice) is more favored.

A realistic insight.
To be more realistic, as MaxW pointed in the comments, the pressure difference is rather small. The summit pressure at Mt. Everest is $\pu{33.7 kPa}$, while at the sea level it is $101.3$, which gives us a difference of $\pu{67.6 kPa}$ which converts to a mere $\pu{0.676}$ Bar.
Although this is quite a significant pressure (see Jon's comment) for our perception, it doesn't affect freezing that much. For a comparison, this answer https://chemistry.stackexchange.com/a/8883/43942 states that the pressure required to solidify $\ce{CO2}$ to make it into dry-ice at room temperature, you'll need a pressure of about $\pu{56 bar}$.
Freezing at higher latitudes, requires us to also ponder on the ambient temperature. You probably wouldn't want to talk about latitudes as high as the Everest, because then the temperatures will be extremely low (averaging about $\pu{-19^\circ C}$ in summers!) and you'll have to apply an external pressure to adjust the melting point. Latitudes where the ambient temperature is (say) $\pu{3^\circ C}$ have a less temperature gradient when compared to a place (probably at the sea level) with an ambient temperature of (say) $\pu{25^\circ C}$. A higher magnitude of a gradient means a high rate of heat transfer, which hints at a slower melting of ice at higher latitudes.
Also, airhuff suggested that thermal conductivity of thin air at high latitudes may have a role to play in this. I searched this up, you might want to look at this paper: http://nist.gov/data/PDFfiles/jpcrd269.pdf

Even on Mt. Everest the pressure difference will be small. The main effect on ice melting is temperature not pressure.

The pressure at the summit of Everest is roughly 1/3 of an atmosphere. You would notice it...

@MaxW This has nothing to do with temperature. We're talking about the equilibrium between air and water at 0°C (which I've stated in the first line of my answer). Also, if you're so keen about the temperature of the surroundings, Everest's (summit) temperature averages -19°C in summer and -36°C in winter. Apply Fourier's law (law of heat conduction); it tells you that heat will now flow from ice to the surroundings, thus further lowering the temperature of ice.

@BerryHolmes - I think you constrained the problem beyond how it was stated.  The original problem makes no mention of constraining the consideration to 0°C.

@MaxW Talking about the ice-water equilibrium (probably you're thinking of an elevated melting point) at yet another temperature asks for a changed (lower in your case) pressure in itself (see the phase diagram), which contradicts your first comment.

@BerryHolmes  - Ugh... From your own data... which is more significant on Everest (a) a summer temperature of -19°C or (b) a pressure of 250 torr? // My point about a small difference in pressure is that 250 torr is much closer to 760 than 4.58 torr since the relationship on the phase diagram is logarithmic.

@MaxW Edited, please have a look.

Easy. As you get higher, there is less air, making the area colder. So, the ice is surrounded by colder and colder tempatures.


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effect of temperature on position of equilibrium
I was wondering if there was an intuitive way of explaining how a change in temperature would effect the position of an equilibrium, without using le chateliers principle. Le Chateliers principle doesn't seem to explain why the equilibrium shifts the way it does, I was wondering if there was a way of explaining this.

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Well, For endothermic reaction the Number of moles at the product side increase with increase in temperature.
And for exothermic, it decreases with Increase in temperature..
And exactly vice versa with respect to the reactant side..
It is in Accordance with the Le chaitliers Principle..


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Why do alkenes with more surface contact have greater London forces?
Why is it that alkenes with greater surface contact have greater London forces? I thought greater London forces were dependent on the size of the molecules, or the number of electrons, rather than the surface contact.

You are correct that London forces are dependent on the size of a molecule, but think about why this is. One reason is that a larger molecule tends to have a greater surface area than a smaller molecule, and thus there is more surface area available to interact via London (and other) forces.
Also as stated in this Wikipedia article:

London forces become stronger as the atom in question becomes larger, and to a smaller degree for large molecules. This is due to the increased polarizability of molecules with larger, more dispersed electron clouds.

While yes, London forces do depend on the mass of the molecule, but they also depend on the distance between the two attracting molecules.
London forces are inversely proportional to the sixth power of the separation between them:
$$F_{\text{london}} ∝ \frac{1}{d^6}$$
Now, consider pentane and 2-methylbutane. See how the molecules approach each other and show attractivity:
Pentane:

2-methylbutane:

The former shows attractions of 5 atoms, while the latter shows attractions of only 4 atoms.

London forces are proportional to the sixth power of the separation between them or  proportional to the sixth power of half the separation between them (radius)?

@MelanieShebel Both.

Well yeah, haha. Touché. I mean you state, "London forces are inversely proportional to the sixth power of the separation between them:" and then show an equation with radius. Did you mean to put distance in the equation or radius? Or did you mean to say, "London forces are inversely proportional to the sixth power of half the separation between them."

@MelanieShebel My bad, sorry, I'll fix it

Data to supplement this answer: Isopentane has a boiling point nearly 29degC. Pentane instead has a boiling point nearly 36degC.


text:
Is there an explanation to the atomisation energy trend of transition metals?
I am familiar with the double hump trend for atomisation of transition metals, and I can explain the unusually low atomisation of $\ce{Mn}$ with the exchange principle, specifically that in the solid state the spins are more localised whereas for the gas/vapour the spins are able to match and hence there is a gain in exchange energy (stabilising).
However, what I am struggling to understand is (ignoring the middle drop due to exchange):

Why there is a hump in the first place?

I have come across a few ideas, some suggesting that it is a bonding/antibonding situation, whereby, first you fill a d-band gap up to $\ce{Mn}$ and then an antibonding contribution comes in.
Is there a more eloquent way of explaining this trend?

The hump is apparent because of the dip. To be more clear, it is good to know that all these attributes follow the same trends:

Metallic character
Strength of metallic bonding
Enthalpy of atomisation
Melting point
Density

If you plot a graph with any of these attributes on the $y$ axis, you'll notice that more or less in each of these cases, the graph rises first reaches a maxima just before the middle of the series, undergoes a dip in the middle again rises and then finally decreases till the end of the series.
It can be somewhat explained (these observations are made on an experimental basis) by the strength of metallic bonding. As the number of unpaired electrons increase, the metallic bonding becomes stronger. Note that $\ce{Cr}$ has the highest number of unpaired $e^-$s: $6$. This explains the hump.
$\ce{Mn}$ on the other hand has a half-filled electronic configuration. Thus, it is more stable. Stability order on the grounds of electronic configuration follows the order:

Fully filled > Half filled > Partially filled

A lesser strength of metallic bonding, plus the stability collectively account for the dip.
Also, as a quick roundup of densities, density of $\ce{Os}$ is the highest: $\pu{22.59 gm cm^{-3}}$ and $\ce{Ir}$ stands quite close $\pu{22.56 gm cm^{-3}}$. Another fun fact, gold and mercury are also quite dense: $\pu{\rho_{Au} = 19.32 gm cm^{-3}}$, $\pu{\rho_{Hg} = 13.56 gm cm^{-3}}$. As a comparison (as to how dense I actually mean): $\pu{\rho_{Pb} = 11.34 gm cm^{-3}}$.
The low melting points of $\ce{Mn}$ and $\ce{Tc}$ are also mapped to half-filled stable configurations.


text:
Aluminium, heat and biological safety
I'm currently working on a really simple project, an ashtray that can stop cigarette combustion.
I planned to put an aluminium block into my ashtray, with some kind of curves or holes in it, in order for the cigarette to fit in, then stop combusting by lack of $\ce{O2}$, and maybe by heat transfer, as the cigarette tip and the aluminium will be in contact.
I choose aluminium, because I've got some, because it seems quite inert to me, and because it's not too hard to work with. But I can't solve the safety estimation for this material.
Here is some information I found around :

A cigarette tip being around $\pu{900^oC}$
Aluminium point of boiling at $\pu{2519^oC}$
Aluminium toxicity seems still debated, but aluminium is more or less eatable, i.e. used as pigment in food (E173)

Of course, aluminium won't burn, but I'm afraid that heating it may produce a layer of aluminium oxide, which may scatter on the cigarette tip, or get combined with ash components.
So, how can we estimate the safety of this material under these condition?

The cigarette tip will not vaporize or react with the aluminum in any significant way. Also, aluminum does form an oxide layer pretty much immediately on contact with air under standard conditions, no elevated temperatures required. This stable oxide layer is really what the cigarette tip comes in contact with.

Also, to be honest, I think the biggest safety concern is caused by the cigarette itself. Aluminum oxide is a relatively inert material, which is why  chemists had a real hard time converting it to aluminum in the early days. Also, unlike iron oxide, the aluminum oxide layer does not shed off from the aluminum surface quite easily. Which is why aluminum oxide coatings are popular.

The use of aluminum in the construction of your ashtray as you've described will not result in the ingestion/inhalation of any form of aluminum in any significant way.
All aluminum that has been exposed to air is covered in a chemically and physically stable oxide. This aluminum oxide has an even lower vapor pressure than aluminum metal, such that the tip of the cigarette will actually be more than $\pu{2000^oC}$ below it's boiling point. By physically stable, I mean that aluminum oxide adheres strongly to the bulk aluminum and will not flake off as in the case of many other metal oxides.
Although your question was about potential safety issues, I will make one comment regarding your discussion of the design. You suggested that heat transfer into the aluminum might play a role in extinguishing the cigarette, which is not the case. So long as there is sufficient oxygen available for the combustion, the cigarette will continue to burn.

My english is worst than I though, special thanks for the massive rectification in the question :)


text:
Propose the mechanism of a hypothetical reaction
I was given a hypothetical reaction $\ce{A + 2B + 4C -> 2D + E}$ for which I have to propose the possible mechanism with a requirement, that the third step is rate determining.
From the other given inputs I determined that the rate law for this reaction will be $\ce{B + 2C}$.
I came up with a number of "possible" reaction steps, all of which seem to me equally unrealistic. Examples:
A +  B -> X
C +  X -> Y
B + 2C -> 2D + Z
C +  Z +  Y -> E

This does not make sense to me as $\ce{C}$ reacts with $\ce{X}$, $\ce{B}$ and with the product of its reaction with $\ce{B}$.
A +  B -> X
C +  X -> 2D + Y
B + 2C -> Z
C +  Z +  Y -> E

Same problem here, $\ce{C}$ reacts with just about everything.
A +  B -> 2D + X
C +  X -> Y
B + 2C -> Z
C +  Z +  Y -> E

Needless to say I am skipping the variation where the last step uses 3 reactants and produces 2 products, as it's obviously introduces another multi-step reaction:
A +  B -> X
C +  X -> Y
B + 2C -> Z
C +  Z +  Y -> 2D + E

Of course there can be 4 variants of each of the above approaches, achieved by simply combining the variables in a different order.
The above considerations do not seem to be bothering the author of the course. For the hypothetical reaction $\ce{2A + 3B + 5C -> 2D}$ they consider the following the correct answer:
2A + B -> W
2C + W -> D + X
B + 2C -> Y
C +  X -> Z
B +  Y + Z -> D

Here $\ce{C}$ reacts with $\ce{B}$ and the product of its own reaction with $\ce{W}$. I do not understand how it is possible to come up with a "realistic" mechanism of a reaction where the specific substances are unknown.
I tries google and found a number of videos explaining reaction mechanisms on the example of $\ce{NO2 + CO -> NO + CO2}$. In those videos they propose a mechanism, where $\ce{NO2}$ reacts with itself to form $\ce{NO}$ and $\ce{NO3}$. (Are they kidding? If this was possible, production of nitric acid would be a joke.) If I was to follow this approach, there could be a number of variations of the mechanism above where each of the reactants and intermediate products reacted with itself. I estimate that would introduce several hundreds of variants.
How is an average grade 12 student expected to solve this task?

The third step being rate-determining is hardly a constraint, as any step can be rate-determining, especially since the structures of your reactants and products are arbitrary to begin with. What's stopping you from just writing the first mechanism that you can conjure up? It's not like anybody can say it's wrong. Personally, I think this is a pointless exercise.

Let's not debate what is given. The 3d step as rate determining is a requirement of the assignment.

You don't seem to have used that piece of information in coming up with your mechanisms.

I thought I did - the 3d step of each reaction is rate determining, it matches the rate law of the reaction. I skipped typing "fast" or "slow" indeed, but that is insignificant to this question.

Oh, I see. My bad about that, then. But it still seems very pointless, as there are almost infinite possibilities. Incidentally, when you say "the rate law is B + 2C", do you mean that the rate is of the form $r = k[\ce{B}][\ce{C}]^2$, or do you mean that the rate has some different form but using some other information you deduced that the third step must be of the form $\ce{B + 2C -> ...}$?

Does the problem state that $A \neq E$? One simple answer that comes to mind is a three body reaction. 2C and B form one 2CB. The A collides with the 2CB and forms D. The A loses energy/gains energy. http://elte.prompt.hu/sites/default/files/tananyagok/AtmosphericChemistry/ch03s03.html

In other words, A would be acting sort of like a catalyst.

Honestly, this is a bad question (not the OP's post, but the question presented to him).

Honestly what kind of question is it?


text:
Python package for modelling chemical reactions
I'm looking for a package that is compatible with Python 3 and can actually do work with reactions along with, for instance, referencing values. Ideally, it should also not be a package that will take 20 days to learn how to access one value.
More specifically: there should be a way to figure out all possible reactions a certain state could lead to, a way to manipulate lists of possible reactions and applicable reactions in certain states, and to apply reactions to states to get a resulting new state (this, of course, may require some coding on my part, but just an idea of what it should lead too - the more the package can do the better).
Thanks; I'd be glad to provide any clarification in the comments.

Maybe you want to clarify this question. What does "work with reactions" mean?

@Greg see the second paragraph...I think I gave decent examples there, or are those unclear?

Well maybe it is unclear therefore I am asking to help you formulate better question. What state? What kind of chemistry? You mean fully automated and figures out all chemistry for you? Of course for free, in Python, preferably open source and runs on a laptop, right?

@Greg no need to be sarcastic. I never said "fully automated and figures out all chemistry for [me]" - I said "this of course may require some coding on my part, but [...] the more the package can do the better". Most of the chemistry packages for python I've seen have an emphasis on having values stored for access. I didn't know if there was one that found the result of a reaction or some such problem. "State" means some general state that I input.

I am not sarcastic. I am pointing out that your question makes not much sense so far in this shape and form. What do you mean by a state? What properties/quantities do you want to calculate (if you want to calculate anything) and what info do you already have? Do you want to calculate kinetical equations? Do you want to figure out the equations themselves? " python I've seen have an emphasis on having values stored for access" What packages? What values? To solve what problems? Do you mean kinetical calculations done using pre-determined rate constants for the elemental reactions?

This is not really my area of expertise, but a quick search for "python chemical reactions" revealed three hits I've never seen before that may be of interest, the first one being closest to what you want.
chempy
from chempy import ReactionSystem  # The rate constants below are arbitrary
rsys = ReactionSystem.from_string("""2 Fe+2 + H2O2 -> 2 Fe+3 + 2 OH-; 42
    2 Fe+3 + H2O2 -> 2 Fe+2 + O2 + 2 H+; 17
    H+ + OH- -> H2O; 1e10
    H2O -> H+ + OH-; 1e-4
    Fe+3 + 2 H2O -> FeOOH(s) + 3 H+; 1
    FeOOH(s) + 3 H+ -> Fe+3 + 2 H2O; 2.5""")  # "[H2O]" = 1.0 (actually 55.4 at RT)
from chempy.kinetics.ode import get_odesys
odesys, extra = get_odesys(rsys)
from collections import defaultdict
import numpy as np
tout = sorted(np.concatenate((np.linspace(0, 23), np.logspace(-8, 1))))
c0 = defaultdict(float, {'Fe+2': 0.05, 'H2O2': 0.1, 'H2O': 1.0, 'H+': 1e-7, 'OH-': 1e-7})
result = odesys.integrate(tout, c0, atol=1e-12, rtol=1e-14)
import matplotlib.pyplot as plt
_ = plt.subplot(1, 2, 1)
_ = result.plot(names=[k for k in rsys.substances if k != 'H2O'])
_ = plt.legend(loc='best', prop={'size': 9}); _ = plt.xlabel('Time'); _ = plt.ylabel('Concentration')
_ = plt.subplot(1, 2, 2)
_ = result.plot(names=[k for k in rsys.substances if k != 'H2O'], xscale='log', yscale='log')
_ = plt.legend(loc='best', prop={'size': 9}); _ = plt.xlabel('Time'); _ = plt.ylabel('Concentration')
_ = plt.tight_layout()
plt.show()  # doctest: +SKIP

OEChem
Cantera
There is RDKit and Pybel as well, but those are more traditional cheminformatics tools.

Chempy seems like it'll work very well - thanks!

One suggestion that's missing is Bill Green's group RMG (Reaction Mechanism Generator) software:
http://rmg.mit.edu/
It certainly isn't simple to use, but I'm told it's extremely powerful. One can construct reaction mechanisms using pre-defined rules (class based methods), I believe it supports combinatorial and heuristic pathway network generation, and it has very useful support libraries that can calculate transport and thermochemical properties of reasonable first-guess accuracy.

RMG is a great tool for constructing reaction mechanism if that's up anyone's alley. The documentation is fairly good and should be enough to get you up and running. For solving systems of rate equations, I strongly suggest the use of Cantera. It's very robust and essentially a free alternative to the popular (and costly) Chemkin software program. If the original poster is basically just looking for a simple code that can balance equations and act as basically a glorified ODE solver though, chempy seems suitable. For those interested in surface reactions, CatMAP is a good option too.


text:
Hydrolysis of a salt B2X
An aqueous solution of soluble salt $\ce{B2X}$ has concentration of 0.1 M.
Given data :

$\ce{K_b(BOH) = 2 × 10^{–5}}$
$\ce{K_{a1}(H2X) = 2 × 10^{–7}}$
$\ce{K_{a2}(H2X) = 5 × 10^{–10}}$

What is the pH of the solution?
My attempt,
$$\ce{B+ + X2- -> BOH + HX-  \,\,...      (1)}$$
The hydrolysis constant of $\ce{HX-}$ comes out to be about $10^{-7}$, so I decided to neglect it (is this correct?)
$$Kh=[HX-][BOH]/[X2-][B+]$$
So  the  value  of  $$\ce{K_h=\frac{Kw}{K_{a2}*K_b}}$$
Upon substituting the values I get the Kh=1..(something i have never seen before)
So this means that for the reaction
$$\ce {B+ + X2- -> BOH + HX-}$$
We will end up with 0.05M of $\ce{B+}$ , 0.05M of $\ce{of X2-}$, 0.05M of $\ce{BOH}$ , and 0.05M $\ce{HX-}$ if we start with 0.1 M of $\ce{B+}$ and 0.1 M of $\ce{X2-}$
Now, I do know that
$$\ce{pH=7+\frac{pK_a + pK_b}{2}}$$
I also know that this is true when the hydrolysis constant is smaller than 1. Therefore I cannot use if for this problem.
(Nevertheless, I disregarded this and just used it. I got the answer to be 9.305)
So I am stuck over here.

I got another idea.
When I now look at the sum, the resulting solution looks like a mixture of two buffer solutions,
Buffer 1: $\ce{BOH/B+}$
Buffer 2: $\ce{HX-/H2X}$
I don't know whether this observation is correct. Even if it is correct, I am not sure whether it be used to solve the problem.

Hello Vedant, welcome to chemistry SE. It isn't clear as to what you are asking. It appears that you have to determine the the final pH. However, I am not sure.

I am sorry!...yes the final pH is what I am asking. I have edited the question(added what I was asking for)

I am posting a solution image regarding same question. 2nd hydrolysis of X2- is neglected as you did. important thing is degree of hydrolysis is not small and should not neglected.

A screenshot or picture of an answer is not searchable. Please consider rewriting it, so that it can be of help for future visitors. // You can format mathematical and chemical expressions on Chemistry.SE using MathJax; this post contains further details.

Thanks Ramazan Khan! I agree with your solution. I also agree with orthocresol. For the benefit of other users. I'll repost or edit your answer(in a more visitor friendly manner), considering even the second hydrolysis so that I can truly show that it's impact is negligible.


text:
Why does it hurt in the mouth when drinking carbonated beverages?
The typical explanation for the origin of the "sting" is that $\ce{CO2}$ turns into carbonic acid (reddit, yahoo), however the "explanations" there are   incomplete.
The answer I'm looking for would cover:

Is conversion to carbonic acid is really the case here? And what are the original compounds the same enzyme(s) is used for?
Why is there no "sting" from $\ce{CO2}$ found in air?

http://articles.latimes.com/2013/aug/22/science/la-sci-sn-the-soda-bite-not-bubbles-20130821

"The earliest clues that carbonic acid, not bubbles, were responsible for this sensation came from mountaineers’ accounts of the “champagne blues.” After pouring a bottle of celebratory champagne at the summit, mountaineers who had taken medication for altitude sickness disappointedly remarked on the drink’s flat taste — despite its fizziness. It turns out the medication blocked the enzyme that converts carbon dioxide into carbonic acid."

@QuantumAMERICCINO thanks. They don't mention why this is not happening with air.

Hmm interesting. According to https://chemistry.stackexchange.com/questions/9067/what-is-the-carbon-dioxide-content-of-a-soda-can-or-bottle
, a can has 2.2 g of CO2 in a 355 mL can of Coke. Thus, it would be roughly 6000 ppm. Atmospheric CO2 ppm is around 400 ppm. Thus, it seem like concentration is what causes it. Maybe it has to do something with the medium. Maybe the CO2 has to be solvated in an aqueous medium for the enzyme to reach it.

Isn't it a coincidence that the user "Sparkles" is asking about carbonated beverages, one of them being "sparkling water"?


text:
Is the molecule in the attached image technically an amino acid?
Is the molecule pictured below an amino acid? Not necessarily one of the common (natural) amino acids.

Hello Mike! Welcome to StackExchange! Your molecule isn't an amino acid, since amino acids by definition, must have an amino and carboxylic acid group.

Since you're new to this site, have a look at the FAQ and learn how to use this site better. https://chemistry.stackexchange.com/help

It is a sort of peptide as it has the C-CONH-C grouping.

No it is not, as there is no carboxylic acid functionality present. An amino acid is defined in this Wikipedia article as follows:

Amino acids are organic compounds containing amine ($\ce{-NH2}$) and carboxyl ($\ce{-COOH}$) functional groups, along with a side chain (R group) specific to each amino acid.

No, it is not.

However, the starting structure for it is gamma-Aminobutyric acid, or GABA:

Rather than the carboxylic acid ($\ce{-COOH}$) at one end, there is a peptide bond ($-\ce{C(=O)NH}-$).
Attached to the peptide bond is the most unnatural part, the $\ce{-CH2C(CH3)=CH2}$ organic tail.


text:
What are AE Salts?
I saw this sentence as I was reading Agatha Christie's Five Little Pigs:

Coniine and AE Salts comes under Schedule I of the Poisons Acts.

So I've been wondering what are the mentioned "AE salts"? It's such a little detail but I'm interested to know.

Seems to be fictional.

I find that unlikely, since coniine is real, and also Agatha Christie often put her professional knowledge into her books.

I can only find one reference to "AE salt" and it refers to a mixture of salts and organic materials used to culture plant cells. AE are the initials of the workers who developed the medium (von Arnold and Eriksson). The composition is here and none of these are likely to be found anywhere on a Poisons list.

and the original reference for AE medium: Can. J. Bot. 1981, 59, 870–874, but if anybody knows anything better and writes an answer...

"Five Little Pigs" was published around 1942-43. Agatha Christie also worked in pharmacy around that period of time, and, most likely, referenced Poisons Acts of 1933 by Royal Pharmaceutical Society of Great Britain.
It is also worth noticing that translations of the paragraph vary quite a bit:
English:

"He was very distressed by the whole thing, poor gentleman. As well he might be! Blamed himself for his drug brewing-and the coroner blamed him for it too. Coniine and AE Salts comes under Schedule I of the Poisons Acts. He came in for some pretty sharp censure. He was a friend of both parties, and it hit him very hard-besides being the kind of county gentleman who shrinks from notoriety and being in the public eye."

Russian, missing "and AE Salts":

"Он был очень расстроен случившимся, бедный джентельмен. И правильно! Винил себя за то, что приготовил эту настойку, - и коронер тоже винил его в этом. Кониум входит в список ядовитых веществ №1. Мистеру Блейку было выражено порицание в самой резкой форме. Он дружил и с мистером, и с миссис Крейл, а потому случившееся переживал особенно болезненно, не говоря уж о том, что ему, как человеку, постоянно живущему в деревне, такая популярность была совершенно ни к чему."

German, entirely missing "Coniine and AE Salts comes under Schedule I of the Poisons Acts":

"Er war höchst unglücklich darüber, der arme Mann. Er wurde von Gewissensbissen wegen seiner Giftmischerei geplagt, und der Gerichtsarzt machte ihm auch schwere Vorwürfe."

Which makes me think that "AE salts" is probably not that important, and slightly expands the definition of the poison.
I didn't find original Schedule I of the Poisons Acts 1933, though its brief overview states that it mostly consists of various alkaloids and its salts. At that period of time alkaloids were not synthesized, but obtained from the plants by extraction, so I would suggest AE stands for Alkaloid Extraction.


text:
Thyroid peroxidase - can atomic iodine serve as iodinating agent?
McMurry's Organic Chemistry (7th Ed.) states, that Tyrosine is iodinated by mechanism of electrophilic aromatic substitution and the iodinating agent is $\ce{I+}$ or $\ce{HIO}$ formed by thyroid peroxidase. The mechanism is quite clear in this case.
Wikipedia article about thyroid peroxidase states that:

thyroid peroxidase oxidizes iodide to atomic iodine ($\ce{I}$) or iodinium ($\ce{I+}$).

From this figure from previous wikipedia article it moreover seems like $\ce{I^{0}}$ would be able to function as an iodinating agent.

I would like to ask whether atomic iodine produced by thyroid peroxidase could act as an iodinating agent by itself (if yes, by what mechanism, because I think electrophilic aromatic substitution cannot be used, morover, this reaction is supposed to happen in the colloid without any additional enzymatic support) or if I just didn't understand the article and figure correctly and additional oxidation to $\ce{I^{+}}$ is indeed needed (atomic iodine is just an irreactive intermediate).
I tried to find some references for this, but I was unable to do so.
NOTE: I originally posted this in biology.stackexchange.com but got no answers, so I thought here it might be more suitable.


text:
Reaction of acids with water
Here is the reaction of $\ce{HCl}$ with $\ce{H_{2}O}$ I found on a website
$$\ce{HCl  + H_{2}O -> H^{+}(aq) + Cl^{-}}(aq)$$
However, in my text-book, the reaction is given as
$$\ce{HCl + H_{2}O -> H_{3}O^{+} + Cl^{-}}$$
Can anyone tell me what's going on here ? Which one is correct ?

See https://en.wikipedia.org/wiki/Hydronium

Both are correct. Moreover, in fact both are the same.

@IvanNeretin could you elaborate it a bit?

@IvanNeretin I think I have to disagree. Bare protons are way too much polarizing power to exist without binding to solvent molecules (in this case it's $\ce{H2O}$).

Odd that Community♦ redirected the duplicate to another duplicate.

@PrittBalagopal Exactly. That's why, in my opinion, "being solvated" for a proton in water means almost the same as "being H3O+".

@IvanNeretin I guess I misinterpreted your statement then. :)


text:
Quantitative elemental analysis in mixtures with internal standard normalization
I am trying to make a quantitative model for Laser Induced Breakdown Spectroscopy (LIBS) analyzing metal alloy samples. I am using an internal standard normalization (so far it produced the best repeatability results) and would like to some how to keep using it in the quantitative model. My issue is that if the base element of an alloy is Fe and I am trying to measure %w of Ni and Cr, then the ratio of Ni/Fe is not only affected by the %w Ni, but also by the %w of Cr as well, since more Cr means less Fe and a different Ni/Fe value for the same %w Ni.
I have standards with different and more-or-less known amount of Ni and Cr.
If there a standard way to handle this type of problems?
It seems like something people have run into many times and there should be cookie-cutter solutions, but I have a lot of difficulty to find one. So far the only discussed models I've seen is a change in %w of one element, while others are locked constant.
A textbook or a journal paper discussing the solution would do just fine.
Thanks in advance!

Can you define the acronym "LIBS"?

Laser-Induced Breakdown Spectroscopy

Ooooo neat. Can you edit your question to include that in the main body?


text:
How do atoms lose their induced dipole?
There exists the basic chemistry lesson with induced dipoles where the electron density of one atom changes causing a dipole to form and causing a cascading effect where other atoms adjacent to that polarized atom also become polarized.
How do those atoms lose their dipole if their wavefunction changes to "line up" with the dipole? Does a probabilistic event need to happen at the same time for all atoms in order to "lose" the dipole where the electron density goes back to its normal state?

The said lesson tends to oversimplify things to the point of getting them downright wrong. Atom dipoles do not line up with each other. They are way too weak for that. Should it be according to your description, they would indeed line up all in one direction and stay locked forever, much like magnetic particles in a magnet.
Now, I brought up the magnetic analogy because it is much better known than the similar electric phenomenon, called ferroelectricity. That's where your explanation is precisely right. Compounds that exhibit ferroelectric properties, once polarized, never lose their dipoles unless you force them to. But they constitute just a tiniest minority of all compounds.
What happens in the rest of them? Well, the molecules turn this way and that, atoms bounce and wiggle, spontaneous dipoles arise every now and then, only to vanish almost instantly. But the position where two neighboring dipoles are aligned in the same direction is energetically favorable, so they spend a little more time in this position than in the opposite. And that's what we call induced dipoles, dispersion forces, etc.
So it goes.


text:
Is there any way to quantitatively find when a reaction has completed?
So I'm designing an experiment for measuring how the reaction time of copper and ferric chloride is dependent on temperature; here's the reaction:
$$
\ce{FeCl3 + Cu -> FeCl2 + CuCl}
$$
The problem I have right now is how to quantitatively measure when the reaction has ended. I know you can visually see when a small copper rod has completely dissolved, but I wanted there to be a quantitative aspect to it rather than qualitative.
I've thought of using a pH probe to see if the PH changes after the reaction takes places and stays static, which I could leverage as an endpoint of the reaction, but the pH probes my school are damaged, I think, since the reading keeps jumping around by about 3 or 4.
If you have measured reaction time in general before, I'd love to hear how you did it.

You'd probably want to determine the order of your reaction; further calculations down the line, then, are a cakewalk.

UV-vis would make this very simple, you just need to monitor when the absorbance at a certain wavelength (probably that of Fe(III)?) stops changing.

I will definitely look into the order of the reaction.

Option 1: Spectrometry
Easiest method, in most cases accurate enough. If you know that there are no side-reactions, this is a safe bet. Simply measure some characteristic wavelength for one reactant (preferably $\ce{Fe^{3+}}$) and one product (I would go for the $\ce{Cu^{+}}$ there) and then make measurements of your reaction mixture on the fly.
Option 2: Titration
Not that recommendable if you need any reliable results, but if you are low on budget and/or want a didactic effect. Take a little sample from a big flask and do a $\ce{Fe^{2+/3+}}$-redox-titration.
Option 3: Electrolysis
Don't know how to implement that, but basically you would take the sample and measure the amount of electricity needed to convert all $\ce{Fe^{2+}}$ in that sample back to $\ce{Fe^{3+}}$. Probably more accurate than spectrometry as the measurements of currents can be done quite precisely
You could also do some weird stuff like quantitative chlorine-NMR, but I can not think of any benefit that you would have with that.

Possibly relevant reading: (http://faculty.sites.uci.edu/chem1l/files/2013/11/C08aMANCVhydroxylation1.pdf)

I think I will go with the Spectrometer method. My school has one of those, so that will be possible. Looks like I have a potential solution; thanks so much for your input!


text:
Is the term "heteroatom" reserved for exclusive elements?
From my experience, typically in organic chemistry, the implication of "hetero-atom" in some molecule is understood as B, N, O, S, P or a halogen -- or any nonmetallic element, perhaps.
If one describes a nanostructure that contains inorganic and organic ligands (like metal-organic frameworks), is it safe to describe the metals as heteroatoms? Or should the term be reserved for covalent-type atoms only?
For example, is this an acceptable statement?

Snurr and co-workers showed results for simulated H$_2$ sorption in a
  variety of rht-MOFs containing only carbon and hydrogen in the
  linkers, as originally these MOFs did not contain heteroatoms
  other than oxygen and the metal.

I would argue that this is propably a matter of style much more than one of definition or scientific accuracy. An organic scientist may have other views on that than somebody occupied with inorganics. Take heavy atoms for example, which for some people start with nitrogen, other would start somewhere around silicon/phosphorus, while somebody who works a lot with different metal might at the split only at iron.

The IUPAC Goldbook provides no 'official' definition of the term heteroatom,and as such you're somewhat free to use (and abuse) the term as you see fit.
Since no IUPAC definition exists, it may be helpful for you to define what you plan on calling a heteroatom at the beginning of whatever you're writing, to inform the reader (common practice in scientific literature for poorly defined terms). Clearly an organic chemist would have a very different definition to someone involved in purely metallic synthesis.
I would however, point to several other IUPAC documents in which, whilst no 'definition' of a heteroatom is given, they do suggest lists of possible heteroatoms. For example, the following statement from the IUPAC nomenclature guidelines which includes many elements (including metallic ones, possibly going against your suggested definition).  This priority order of heteroatoms is repeated multiple times.

P-22.2.2.1.3 If two or more kinds of heteroatoms occur in the same name, their order of citation follows the sequence: F, Cl, Br, I, O, S, Se, Te, N, P, As, Sb, Bi, Si, Ge, Sn, Pb, B, Al, Ga, In, Tl, Hg.

The most common context is as you've indicated; a heteroatom is typically a non-metallic atom that replaces a part of the hydrocarbon backbone of an organic molecule. However this is not the only context, and a metal atom would also be considered a heteroatom, and the statement that you've made would be acceptable.
Also, here is an example of a somewhat different context from this Wikipedia article:

In the context of zeolites, the term heteroatom refers to partial isomorphous substitution of the typical framework atoms (silicon, aluminium, and phosphorus) by other elements such as beryllium, vanadium, and chromium.


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Hybridisation and the Schrödinger equation
I am slightly confused about hybridisation and how it relates to molecular and atomic orbitals, despite having pored through many sources online. I was hoping someone could verify whether my current understanding is correct, in particular regarding what hybridisation actually is/does because I have not read this explicitly, but am assuming it is the case from what I have read so far:

The Schrödinger equation can be solved to give the atomic orbitals (at least, some simplification involving the effective nuclear charge can be used to find the outer (and inner?) atomic orbitals.
In hybridisation, we first consider the shape of a molecule and then we consider each atom seperately and how we could combine the atomic orbitals so that the geometry about each atom is correct. Now what I am particularly unsure about is: Are the hybridised orbitals entirely a conceptual construct, or are they mathematical solutions to the Schrödinger equation? I was thinking that maybe the hybrid atomic orbitals that are created are a linear combination of the atomic orbitals, and thus this linear combination also solved the Schrödinger equation and can exist but would have a different energy to the individual atomic orbitals? I am finding it difficult to think how a linear combination of the atomic orbitals could produce something with a completely different shape though (although I suppose the orbitals are orientated and essentially vectorial, so I think it could work). Perhaps  the hybrid atomic orbital form could also be, say, the product of the atomic orbital forms? Although then I do not think this would solve the Schrödinger equation anymore. But I am not sure. And anyway, all of these ideas assume that hybridisation has a mathematical basis in the Schrödinger equation, which I am not sure about at all!
Then molecular orbital theory takes the orbitals of each atom (I think theoretically it takes every orbital in every atom?) in the molecule and combines them to form a molecular orbital for the whole molecule - no longer simply an overlap between two atoms. This certainly also solves the Schrödinger equation. From what I understand, theoretically the calculation to form MOs should involve only the AOs of each atom, but to simplify the situation sometimes hybrid atomic orbitals are used in MO construction? If so, surely the hybrid atomic orbitals must be a linear combination of the atomic orbitals that solve the Schrödinger equation, so that this can be used in molecular obrital theory which is based in the solution to the Schrödinger equation? Also, where in the calulcation are the coefficients for each atomic orbital arising? Is it from the boundary conditions including things like molecule geometry, or perhaps that we know the energy of a molecule and when we put this particular energy in to the Schrödinger equation it gives us the coefficients?

I apologise for the long post and I realise there seem to be many questions within here, however they are all linked and about the mathematical underpinnings of hybridisation/valence bond theory and molecular orbitals, so I thought they belonged in one post.

When you say hybridization, are you referring to LCAO (linear combination of atomic orbitals)? Or are you referring to the hybridization taught in organic chemistry classes (where they only deal with the valence orbitals)?

@QuantumAMERICCINO I think the latter (sp/sp2/sp3 mixing etc)- i.e. the hybridisation where you want it to match up with the shape of the molecule

An important point is that the total electron density described by the hybrid orbitals is the same as the density of the unhybridized orbitals. We're just carving that space up differently in terms of names like drawing different boundaries on a map. The land isn't changed by the change in names.

Mathematically, atomic/molecular orbitals are 1-electron wavefunctions (hydrogen-like wavefunctions) that are used as a basis with-which the total N-election wavefunction is expanded.  The N-election wavefunction is a determinant (or a linear combination of determinants) built from these 1-electron wavefunctions: an anti-symmetric (Fermi statistics) linear combination of Hartree products (usually a product of MOs expanded in a basis of AOs).
For simplicity's sake, lets assume we are interested only in 1-determinant wavefunctions.  The Hartree-Fock method will (when used where HF is appropriate) result in a variationally-optimal 1-determinant wavefunction that is an energy eigenfunction of the applicable Hamiltonian operator.  Not only is the final, self-consistently optimized HF total N-electron wavefunction an energy eigenfunction, but the individual HF canoncial MOs are also eigenfunctions of the Hamiltonian/Fock operator, with eigenvalues that are the so-called 'orbital energies' (canonical orbitals are defined as the optimzed orbitals which diagonalized the Fock matrix).
However, there is nothing special about these canonical orbitals in the context of the total system energy.  Any linear, norm-preserving transformation of the occupied orbital space will result in an N-electron wavefunction that is still an eigenfunction of the Fock operator, while the individual MOs are no longer eigenfunctions of the Fock operator (for example, localized MOs, LMOs).
So, once we have a HF wavefunction and optimized MOs, we can linearly transform the MOs however we want.  Such a linear transformation might have the effect of producing orbitals that resemble hybridized orbitals.  Or maybe we are interested in generating localized MOs.  The point is, a linear transformation within the occupied orbital space results in a wavefunction that has the exact same electon density and energy eigenvalue as the wavefunction HF originally provides us in terms of Canonical orbitals.
Orbitals transformed to look like hybrid-orbitals are on the exact same footing as any other choice of orbital.


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Non-bonding molecular orbital in triatomic hydrogen
I am slightly confused about how non-bonding MOs form, for example in H3+  ion as shown below.

Why does the central atom not contribute an orbital?
The source mentions symmetry, but I can only see that the central orbital has a positive or negative phase, so it will inevitably cancel somewhat on one side and add with the wavefunction on the other side. This orbital will have a distinct appearance.
Could someone explain why there is no contribution, despite us adding s orbitals whose symmetry always allows them to be added (and even if we had p orbitals, surely the p orbital which can contribute will contribute?)

Could the down-voter please let me know what is wrong with my question and how I could edit it?

H3 cation is triangular not linear, also @Not.

I know Mithoron — its quite a common undergrad exercise to compare linear and triangular however (cf Flemings MO book )

In order to draw the MO diagram for the H3+ system, it helps to first thing about the MO diagram for a simple diatomic case, and then mix in the additional s-orbital.
The diagram below hopefully illustrates this technique (and can be found in many introductory textbooks).

First consider two s-orbitals (the right hand side of the diagram). They can either mix in a symmetric (in phase) or anti-symmetric (out of phase) combination.
The two atomic orbitals we initially combine can give rise to two molecular orbitals, as shown on the right hand side.
We can give them symmetry labels to help us identify what kinds of things these molecular orbitals can then combine with to form the new molecular orbitals for the triatomic system.
The labels g and u are related to whether the molecular orbital is symmetric or anti-symmetric with respect to an inversion centre.
Once our two molecular orbitals for the diatomic system are formed, we can mix these orbitals with another hydrogen s-orbital in order to generate the final MO diagram as per your question.
Only the g molecular orbital from the diatomic system is of the correct symmetry to interact. This mixing creates two molecular orbitals (the top and bottom ones) for the triatomic system. The middle, non-bonding molecular orbital is simply the orbital of u symmetry from the initial diatomic system, which is unable to interact since there is nothing of the correct symmetry for it to do so.


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How to obtain reactant lifetime using transition state theory?
I've read this paper and the authors calculate life times of an intermediate using Quasi-Classical-Trajectories (QCT), RRKM and transition state theory (TST). Table 3 on page 4755 show the results.
If I understand correctly for the QCT calculations they take all (several hundred in that case) trajectories and just do statistical analysis of the lifetime of that intermediate. That perfectly makes sense and they come up with a value +/- error.
For TST they also calculate a lifetime of that intermediate, assuming it is in equilibrium with it's surrounding. This makes me think that they used the reaction barrier from intermediate to product for their calculation of the lifetime. How to do this? Up to now I've only used TST in combination with Eyring equation for calculation of rate constants.

As far as I understand, there are two things that I should answer for you.
Calculating rate constants with any trajectory methods is different than using TST. In fact, the whole point of using QCT is so we can account for nonstatistical effects. These nonstatistical effects are usually extremely important, as it is also mentioned in the article

... This dynamic matching leads to pronounced nonstatistical effects
  on the lifetimes...

Any time you are using TST, you are assuming there are no nonstatistical effects whatsoever. This approximation is probably never true, but it leads to surprisingly good numerical results. The way you calculate reaction rates in QCT is by integrating the reaction probability using a Monte Carlo integration formula, cleverly sampling the initial conditions of the reactants. I usually recommend a good book of W. Miller. The book itself is a bit outdated, but the derivation by Porter and Raff in the first chapter is beautiful.
Now you asked about the determination of the lifetime. The article, as I understand it, talks mostly about the lifetime of the activated complex. This can be done because in QCT we know the coordinates of atoms in every single timestep, so we can compare them to some given values to decide whether we are in the activated complex region or not. From the article:

Transition zones were defined for the C–C forming (3‡) and the C–N breaking (5‡) steps (see Supporting Information for details). [...] The lifetime of
   intermediate 4b was defined as the elapsed time between departure from the transition zone 3b‡ and entrance in transition zones 5b‡

So, long story short: QCT determines lifetimes differently from TST, this is because TST is inherently statistical, while QCT also considers nonstatistical effects. Instead, QCT calculations rely on defining geometries for different states, and simply calculating how many integration steps the system spends there.

Thank you for the answer. I understand and/or knew the things you explained here and it doesn't really answers my question. I know how to calculate a rate constant using TST (eyring equation) and I know how to determine the lifetime of this intermediate using QCT. What I don't know is how to calculate the lifetime using TST (and how to calculate rate constants using a trajectory - if that's even possible). How do I get to the lifetime of this intermediate using TST?

As for calculating the rate constant by QCT, I'd still refer you to the Raff-Porter chapter I linked earlier, it is well explained there. As I mentioned in my comment, you do a Monte Carlo sampling of the initial condition, and if you choose the right distributions, then the reaction rate calculation reduces to counting the ratio of reactive collisions to nonreactive collisions. Note that this is true only for a well-derived Monte Carlo sampling, that is what is found in the book I mentioned earlier. Another reference can be found at Gaussan help page: http://gaussian.com/glossary/hase96/

I am not quite sure about determining the lifetime of the activated complex by TST calculations, as I have no real experience in TST methods. I have a hard time imagining an activated complex with a non-zero lifetime tho. It is a local energy maximum, thus a metastable configuration. I'd make a wild guess that it can have no lifetime whatsoever. Someone who is more experienced than me should correct me tho.

thanks, this sounds very interesting I will look into that. Let's see if someone else knows how to do this. I'm able to calculate the lifetime of the activated complex but not the reactant.

If you can calculate the reaction rate, isn't the reactant lifetime just the reciprocal of the rate coefficient?

yeah that's what I figured out in the last hour, just similar to lifetime of radioactive particles. It does make sense, but somehow the numbers don't add upvery nicely. For whatever reason I'm always a little bit off (7600 instead of 7000 for example). I have to check with their actual numbers from the SI to confirm. But thanks for your help!

Are you off compared  to what? Compared to QCT? That's no wonder, because as I have already tried to point out, QCT and TST are inherently different methods. Another source of confusion might be with lifetime and half-life. Lifetime is just the reciprocal of the rate coefficient, while half-life is ln(2) divided by the coefficient.

no, off compared to their TST calculated values if I take their barriers and calculate the lifetime as reciprocal of the rate constant using eyring equation. For example the anti pathway for the reaction of tetrazine with cyclopropene they got a barrier of 2.3kcal/mol according to that paper and a lifetime of 6794 fs. Now with 2.3kcal/mol I get around 7600 fs. I looked at the supporting information and calculated the barrier from the calculated energies and I get a barrier of 2.218 kcal/mol and with that a lifetime of 6805 fs (the difference has to be just accuracy of the used constants)

Just dropping in a question here: is it common practice in theoretical organic chem to use the Eyring equation and call it transition state theory? I myself have some research experience in reaction dynamics of small systems (like atom + diatom reactions), and when I say TST, I  do not mean Eyring's equation, which is - as far as I know - correctly termed as activated complex theory. This might seem nuisance, but I've been thinking about picking up reactions like this. I have always been fascinated by Diels-Alder, and I think it is not well understood - so I am asking it for my clarity.

yes it is.  Activated complex theory is basically just another name for TST and Eyring equation is the way to go ( IUPAC goldbook shows the eyring euqation in their entry of TST: http://goldbook.iupac.org/html/T/T06470.html ). There are very small differences in the definition of "activated complex" and "transition state" but practically they are the same. But to keep it clean here I'm happy to discuss this further in chat.

Let us continue this discussion in chat.

One thing to consider is that the TS can, if the reaction is indeed elementary, go back to the reactants as well as products (which is contary to the assumptions in rigorous TST). Make sure you consider the rate constant for the reverse process in your lifetime calculation (ie take the inverse of the forward and reverse rate constants) - it might well account for the ~800 fs...


text:
Why would a reaction favor one isotope over another?
In my instrumentation class last semester, we were asked to read a short paper. The paper described a technique for determining whether dinosaurs were cold-blooded or warm-blooded based on the amount of a certain isotope in a sample of dinosaur bone.
What would cause a reaction to favor one isotope over another? In my general chemistry course, I learned that isotopes act similarly in chemical reactions. Of course, I have found over the years that pretty much everything taught in general chemistry was approximations probably so students don't feel inundated.

https://en.wikipedia.org/wiki/Kinetic_isotope_effect however, context doesn't sound like KIEs

Is the wikipedia page here any use to you? It describes the isotope ratio idea and links to the original research

Good to know: https://en.wikipedia.org/wiki/%CE%9418O

This Wikipedia entry for isotopic fractionation, and it's links to equilibrium and kinetic fractionation might be helpful.

Chemically speaking compounds containing isotopes are very similar but not exactly similar
Broadly when compounds contain different isotopes of the same element (say different isotopes of oxygen) their reactivity is the same. But this is a little over broad. There are subtle and (usually) small differences and these can show up as small differences in chemical reactivity.
To get an idea why consider the simple physical properties of water and heavy water. In heavy water the two hydrogens are replaced by deuterium (hydrogen with an extra neutron in the nucleus). A heavy water molecule is about 10% heavier than a normal water molecule (molecular mass ~20 rather than ~18). As a result the physical properties are a little different: bp is about 2°C higher, mp about 4°C higher and the liquid is about 10% denser. So careful techniques can separate the two (by repeated fractional distillation, for example).
The slight differences also appear in chemical reactions. The reasons why are more complex than the explanation of simple physical effects. But one hint that they exist is simply that the vibrational frequency of a bond depends on the mass of the atoms making up the bond. Isotopes have different masses so there will be some difference with different isotopes. And bonds vibrating are part of the mechanism of reactions involving that bond so the reactions will have different rates. The detailed theory is pretty complex (see this) and the size of the effect depends on the detailed mechanism of the reaction.
Reaction rates for some reactions involving bonds to hydrogen can differ by large factors (because deuterium is twice as heavy as hydrogen) and are sometimes 10 fold. Most other isotopes have only small relative mass differences compared to their parent element and the net rate differences are often only small single-digit percentages. This results in some reactions concentrating some isotopes over others (eg 18O vs 16O. But these can be large enough to detect by mass spectroscopy.
So, despite their chemical similarity, some reactions and some physical processes will favour one isotope over another (though often only very slightly). If you understand the key reaction mechanism, the small differences in the concentrations can tell something about the conditions of the reaction (temperature dependence, for example).


text:
nM or μM, which is more concentrated?
A study found that a particular herb fueled cancer at nM concentrations but actually inhibited cancer in μM concentrations. Which is the higher concentration, nM or μM?

Although molar is the standardised SI unit for concentration, metric prefixes also exist in order to make the numbers more manageable (i.e. to allow small fractions of units to be discussed easily).
The Wikipedia page for molar concentration gives a table of commonly used prefixes, you can clearly see that a micromolar solution is 1000 times more concentrated than a nanomolar solution (which itself is 1000 times more concentrated than a picomolar solution).

Source: Wikipedia

Perfect. Thank you very much! I could not understand without your interpretation of the concentration.


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Removing aluminum oxide from impure silica
We have an assay for a mine. Our goal is to have only silica. Is there a cost effective way to extract the contaminants? The product is a fine powder called diatomaceous earth, or fossil shell flour. Our goal is to use the product to help poor pacific islanders increase their silica uptake and assist in heavy metal detox. Many members of the pacific island nations suffer from heavy metal poisoning. The composition of the powder is:

> 84.7% $\ce{SiO2}$
< 4% $\ce{Al2O3}$
< 2% $\ce{Fe2O3}$
< 0.33% $\ce{CaO}$
< 0.36% $\ce{MgO}$
< 0.5% "other"

We aren't concerned with the calcium, iron or magnesium oxides, just the aluminum oxide.

Is there a simple process to reduce the aluminum oxide levels?
Also, what is the easiest process to convert the silica to orthosilicic acid for increased bio-availability?

I believe this paper tackled the exact same problem. I'd give you more details, but I don't have access from my location at the moment. http://onlinelibrary.wiley.com/doi/10.1034/j.1600-0692.2002.310205.x/abstract

a) silica deficiency is a thing? b) why do you need to lose the aluminium?

You may try the famous Bayer process, which is used for extracting alumina form bauxite. When treated at $160°$C and high pressure with $\ce{NaOH}$, a mixture like yours dissolves $\ce{SiO2}$ and $\ce{Al2O3}$, and lets the other compounds insoluble. This is a first purification. $$\ce{SiO2 + 2 NaOH ->  Na2SiO3  +  H2O}$$ $$\ce{Al2O3  +  2 NaOH + 3 H2O -> 2 Na[Al(OH)4]}$$Later on, bubbling $\ce{CO2}$ in the solution makes $\ce{Al(OH)3}$ to precipitate. $$\ce{Na[Al(OH)4] + CO2  -> Al(OH)3(s) + NaHCO3}$$ Of course $\ce{NaHCO3}$ is immediately transformed into $\ce{Na2CO3}$ by $\ce{NaOH}$. $\ce{Al(OH)3}$ can be eliminated by filtration. And the final solution contains only sodium silicate, apart from the excess of $\ce{NaOH}$. Silica can then be obtained pure by adding  enough $\ce{HCl}$ to destroy $\ce{Na2SiO3}$ and the NaOH in excess.$$\ce{Na2SiO3 + 2 HCl + H2O ->  Si(OH)4 + 2 NaCl}$$ It is advisable to highly dilute the $\ce{NaOH + Na2SiO3}$ solution before adding $\ce{HCl}$. Because the precipitate of $\ce{Si(OH)4}$ is extremely voluminous, and fills up the entire solution.

Treat it with concentrated $\ce{HCl}$, it will react with the aluminium oxide and turn it into aluminium chloride. You can separate it from the silica powder by a simple filtration.
$$\ce{Al2O3 + 6HCl -> 2AlCl3 + 3H2O}$$

I doubt this will work when the alumina is part of some complex clay mineral (which it is in this context).

ore or mineral should be treated with some flux like CaO, which react with silica to form Ca-silicate. and which can be separated after filtration process.


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How many significant digits are appropriate when one or more digits are eliminated (during subtraction)?
Say, for example, $50.0-49.6=0.4$. Does this result have $1$ significant figure, or $3$ (as in the data: $50.0$ and $49.6$)?
Had it been $50.0-4.6$, it is understood that the answer is $45.4$, by the rules of significant figures. How do I apply them in the "$50.0-49.6$" case?

This question deals with $50-49.6$; so a related but not the same question.

This might help: https://chemistry.stackexchange.com/a/74496/43942

Had checked that already. It justifies the rules-of-significant-digits[-specially-concerned-about-addition], So, doesn't answer my query; at least, I don't see it, if it does.

There's an easy way to look at this.
Lets say the value $50.0$ refers to $\pu{50.0 cm}$ measured accurately to $\pu{0.1 cm}$, and that $49.6$ refers to $\pu{49.6 cm}$ measured accurately to $\pu{0.1 cm}$. The difference would be, as you've said, $\pu{0.4 cm}$ measured to $\pu{0.1 cm}$ accuracy.
So, yes, the answer has only one significant digit.
Your initial measurements aren't more accurate than $0.1$, so adding two extra significant digits is incorrect.
I hope this helps.


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How do I define the stereochemical configuration (R/S nomenclature) for this case?
I am aware of the R/S configuration of chiral centers in which the molecule is depicted in dash wedge form.
I wanted to know how the configuration would be mentioned if a chiral center is attached with:

two atoms/groups in wedge position
one atom at dash position, and
only one straight line bond

Hmm, I'm not sure what do you exactly mean by the representation, but I'm guessing that it'd be something similar to the one I've drawn. Such a case is not possible because when we represent something by a wedge/dash configuration we plan to represent a $\text{3D}$ molecule in a $\text{2D}$ drawing sheet of paper.
In reality the chiral molecule with four different groups is $\pu{sp3}$ hybrid, which means that it exists as a tetrahedron.  The wedge actually mean that the group points towards you, and the dash means it is going away from you. And by going towards or going away, we mean along a line that's perpendicular to your drawing sheet.
Now that you can't have two groups approaching you perpendicular to the drawing sheet, talking about the configuration makes no sense here.

I agree with you . And what you've drawn is exactly what i was looking for..


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Dibenzoyl-L-tartaric acid won't show up on HPLC chromatogram
I'm trying to develop a method for HPLC to detect dibenzoyl-L-tartaric acid. Concentration doesn't matter at this point; I'm only concerned with detection. The problem I'm having is the compound won't show up at any wavelength I've tried. I think I have the appropriate column (one specifically designed for organic acids). The run time is more than enough, the sample is essentially saturated and my injection volume is 25 mL.
The compound seems like it would have a chromophore, so I've been trying wavelengths between 210 and 270, but nothing works. Literature also turns up nothing.
Can anyone steer me in the right direction?

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Are you sure it's coming off the column at all? (Try and TLC) the waste stream. Carboxylic acids (especially multiple in one molecule) can be quite difficult to move on a HPLC column— adding some modifier to keep them proton ated often helps (0.1% TFA or similar).

What are you using as the mobile phase?

Is this normal or reversed phase? I'd also guess that the injection volume is not 25 mL. Maybe 25 uL or 2.5 uL?

Test the method with something else, for example, benzoic acid.  If that doesn't show up either, there is a problem with your system.  What's the column packing, back pressure, flow rate and mobile phase?


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Is Gibbs free energy the sole determiner for spontaneity?
If a reaction has negative $\Delta G$, can we then say with certainty that the reaction will be spontaneous?
Is negative $\Delta G$ the sole determinant for spontaneity?

please ignore kinetics of reaction in above quo

What do you mean by reaction? Are we talking about whole conversion of product to reactants? Or are we talking about whether one molecule of reactant converts at all to product?

@Zhe. I would like to know how does that matter? Well,take my previous comment casually: so that some spontaneous reaction taking too long may not become point of argument.

Because the microscopic view is governed by likelihood. Likelihood compounds in samples with large numbers. If you're talking about 1 molecule, maybe it's not completely unlikely. If you're talking about a reaction, it's quite possible that given the numbers, you'll never observe an appreciable amount of the product in the sense. Though spontaneity has a very specific definition, so I guess you're right that my previous comment is moot.

For a spontaneous process at constant temperature and pressure, $\Delta G$ must (emphasis on must) be negative.
Spontaneity of a process , explicitly, depends on the enthalpy change and the entropy change. The Gibbs free energy change includes both these factors to predict the spontaneity of a process.
So, yes it can be said that Gibbs free energy change is the main factor that decides the spontaneity of a process.
Putting into better words, $\Delta G$ being negative is a consequence of a process being spontaneous.

Technically,  $\Delta G <0$ isn't really necessary.  It's just most probable.  But for any macroscopic sample the huge number of atoms/molecules stacks the odds very heavily.  If this were sports, I would all $\Delta G <0$ a prohibitive favorite.

To add to what @OscarLanzi said, the requirement that $\Delta G$ be negative for spontaneity is just a Rule of Thumb.

Then again, you could just argue that "spontaneity" itself is a probabilistic label.

@orthocresol , you are welcome to edit my answer.

It's not really important, I was just tacking on to the two previous comments. The entire Second Law is probabilistic in nature, in that a process with $\Delta G < 0$ is much more likely to occur than one with $\Delta G > 0$, but it is a point that is only barely related to the question being asked. Your answer is quite fine as it is.

You might want to mention that different state variables are used to determine spontaneity when "at constant temperature and pressure" does not hold; for instance, Helmholtz free energy for reactions taking place at constant temperature and volume but not constant pressure.

@zwol. What if pressure and temperature aren't constant, how do we evaulate it then?

@altosanta I'm sorrry, I don't remember anymore.