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Which is the major product formed on nitration of benzyl methyl ether?
Which is the major and minor product formed from the reaction of benzyl methyl ether with dinitrogen pentoxide?

I know that $\ce{NO2+}$ is behaving as an electrophile here and attacking the aromatic ring. Furthermore, the substituent directs electrophilic substitution to the ortho- and para-positions. I thought that since the ortho position is more sterically hindered, the para product should be the major product and the ortho  product minor.
However, the answer I was given says the opposite, i.e. that the  ortho product is major and the para product is minor. Where am I going wrong?

The ortho position isn't that hindered and there are two ortho positions compared to only one para so this probably accounts for the observed products.

This isn't a literature reaction, and in general nitrations using N2O5 are not common. There is only one paper which mentions any kind of nitration of benzyl methyl ether, and the authors note that with a similar compound (PhCH2CH2OMe) the regioselectivity of nitration depends on the choice of reagent: https://pubs.acs.org/doi/10.1021/jo9606867.

1) $\ce{NO+2}$ electrophile.

Yes

2) Benzene ring is attached to an activating group, so meta positions
  are enriched and electrophile has chances of attacking ortho/para
  positions.

The effect of the substituent in your example is probably similar to the effect of a methyl group (e.g. replacing a methyl hydrogen with a methoxy group won't have a big effect on the electron donating ability of the substituent).  Look at the relative rates of nitration in the following figure.

(image source)
We see that in toluene the ortho and para positions react 43 and 55 times (these two numbers are rather close) faster than in benzene, while reaction rate at the meta position changes little from benzene.  This is consistent with the pi-donating ability of a methyl group adding electron density to the ortho and para positions making them more attractive to attack by an electrophile.

3) ortho position is more sterically hindered, hence para- major
  product and ortho minor product.

Look at the relative rate for the ortho position in t-butyl-benzene, here steric factors definitely play a role in slowing down attack at the ortho position. By comparison, the methyl group (and your substituent) is not very large and may slightly reduce the reaction rate at the ortho position in toluene (43 vs. 55 rather than both numbers being 55) or your example.
Another factor to consider is statistical in nature, there are two ortho positions, but only one para position. So, if everything else were equal, we would expect twice as much ortho product as para product.
In summary,

the electron donating effect of the substituent increases the rate at the ortho and para positions to a similar degree
steric factors might reduce the rate at the ortho position a small amount
statistical factors favor twice as much attack at the ortho position compared to the para position

This statistical effect is the largest of the three and is consistent with the ortho isomer being the major product.
One final cautionary note. When we discuss reaction rates we are really discussing kinetics - which product will form the fastest. However, electrophilic aromatic substitution is an equilibrium process. If you allow the reaction to run at high temperature, or for a long time, then thermodynamics can take over and we may wind up with a product composition based on relative product stability rather than kinetic factors.

Is there any example of nitration reaction on substituted benzene where there is a clear majority of para substitution?

sure, t-butyl-benzene

If we take statistics, H-bonding would come into play right? Then if we take Intermolecular H-bonding > Intramolecular H-bonding.... ?

@SujithSizon How would H-bonding come into play here?

@ron intramolecular H-bonding between nitrogen attached oxygen(ortho case) and  hydrogen attached to the starting/ending methyl group?

No, there is no hydrogen bonding involving the methyl hydrogen.  In fact in the starting material or the two products, there is no hydrogen bonding.

@ron glad to have that cleared, thank you so much.

@SujithSizon Your welcome!


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How can I find the bond angles for carbamic acid?
I need to find the bond angles for carbamic acid (as a self-given practice). As I understand it, the structure looks like this:
        O
        ||
H - N - C - O - H
    |
    H

I'm trying to find the O-C-O bond angle, but I'm not sure how the - H on the end will factor in to the answer. Everything I read about it is dense and fairly hard to understand.
I've tried finding the number of electron domains around the C atom (I think it's 4), but I'm not sure where to go from there.
What is the general procedure for doing this? How would it look if I were trying to find a different angle (like H-N-H)?

A quick google of 'carbamic acid bond angles' reveals many papers in the literature measuring/computing the bond angles in this and similar molecules. See here, for example.
That said, chemists are often not concerned with the exact bond angle, instead, we often look at the structure and make predictions based upon bonding and the substituents around the atoms.
In carbamic acid, the simplest carbamate, we can consider the central carbonyl to be sp2 hybridised, giving it a planar structure with bond angles of 120. Due to conjugation with the nitrogen lone pair, the N can also be considered to be sp2 hybridised, and also have bond angles of around 120.
Although this is quite a crude analysis, it holds up quite well for sp2 centres, with very little variation (except in cases of exceptional steric bulk forcing groups to move closer/further apart, or in strained ring systems). The paper referenced confirms that in this system (carbamic acid), the deviation is as little as 4 degrees.

Wow, that made a lot of sense. So the O-C-O bond angle would be the same even if the rightmost H atom was (somehow) left out and the rest of the molecule was otherwise the same?

@Kaylee  Not exactly, but anion also has this type of geometry.

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How to calculate concentrations and what is the concept of 'mole'?
Say I have a $50~\mathrm{ml}$ solution of a liquid mixture which is $30~\%$ concentrated. Now if I take $20~\mathrm{ml}$ of that liquid out from that mix, what would be the resulting concentration of both the liquid remaining and the liquid extracted? Will the concentration of both still be $30~\%$?

This seems like a homework question. We have a policy which states that you should show your thoughts and/or efforts into solving the problem. It'll make us certain that we aren't doing your homework for you. Otherwise, this question may get closed.

What do you think? Please provide your thoughts so we can guide you better.

According to me the concentration should remain the same since it is dependent on moles and hence even if we take out 20ml from a concentrated solution the moles should not be affected. However my doubt lies in the fact that does this stand true in all terms of concentration , eg in cases of molarity or maybe volume by volume concentrations etc

@SushmitaNayak yes, if it is a solution all portions will have the same concentration.  If it is merely a "mixture", like orange juice, then not necessarily.

Please guide me by explaining the main concept and not by random values used in the question

@SushmitaNayak the main concept is that all samples, portions, subdivisions, etc. of  a solution have the same concentration.

So that means even it is a 1M solution or a 20% solution(volume by volume) all of them technically on changing the volume do not get affected by the concentration ?

You have to assume that the solvated molecules are uniformly distributed in the liquid for all subdivisions of the liquid to have same concentration. If you have a glass of lemonade, and pour half of it into another glass, you will have two glasses of lemonade of the same strength. See the point?

Concentration of a liquid can be given by the formula C=N/V, where V is the volume of the solution and N is the number of moles present in it.
To help you understand exactly what a 'mole' is, think of it like this: when you ad a certain amount of powder to water, say sugar or a fruit drink, how many spoons/ grams of powder you add is the number of moles. (1 mole is actually 6.02214129 * 10^23 particles of the substance)
The concentration can be determined by how much water you add to the drink: If you add more water, the solution is said to be diluted, since the ratio of solute:solvent is lower. So in a drink with 30% concentration, the ratio would be 3:7, 3 moles of powder for 7 moles of water. The concentration can ONLY BE CHANGED if the amount of water changes. The ratio remains same throughout, so concentration would be 30% if you take 20ml, 10ml, or any measurement of volume.
Hope this train of thought helped your mind to get the reasoning :)

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Magnetic susceptibility trend
So I was recently looking at this website.
http://www.fizika.si/magnetism/MagSusceptibilities.pdf
It lists the susceptibility values for inorganic compounds, and I've noticed that the trends don't appear to follow logic. The stronger the ligand OH, CN etc.., the larger the energy needed to reach the higher orbitals, and thus more paired electrons so a stronger diamagnetic effect, whilst the weaker ligands cause the pairing energy to be higher relative to the energy needed to go up to the higher shells, so less paired electrons so higher magnitude of magnetic susceptibility.
If one takes a look at CuI, CuBr, CuCl, and CuCN we can see that the negative susceptibility values become more positive (from -63, to -49, to -40, to -24) as ligand strength increases. Similarly, some compounds don't follow the trends of stronger ligands like $\ce{NiI2}$, to, $\ce{NiBr2}$, to $\ce{NiCl2}$. How can this be?
Also, I've tried plotting transition metal radius alongside their susceptibility values for bonding w/ Br for same oxidation states. I got a proportional relationship. Apparently, the atomic radius of the transition metal increases the susceptibility. Does this make a bit of sense? I mean, logically, larger radius essentially means larger diamagnetic and hence lower paramagnetic susceptibility values, right?


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Has the exact formula for computing the density of water in terms of temperature ever been determined?
According to a simplified quantum mechanics theory where electrons and nuclei are point charges and there's no nuclear chemistry, and the gravitational constant and cosmological constant are zero, with all physical constants derivable only from the following constants: speed of light; Planch's constant; proton charge; Coulomb's constant; electron mass; and proton mass; has anyone ever determined the exact formula for the density of water in its liquid phase at 1 atm in terms of temperature and those 6 physical constants? The atmosphere unit and the exact celsius scale itself are defined in terms of the triple point of water which in turn can be computed in terms of those 6 physical constants. Is it also true that the density of supercooled water and superheated water can't be determined exactly because they're not infinitely stable and will eventually undergo homogenous nucleation of the stable phase? Is it true that the less supercooled water is, the lower the space-time probability density of homogenous nucleation so the more accurately its density can be determined but its density can only be determined exactly from its freezing point to its boiling point? However, the reciprocal of the space-time probability density of homogenous nucleation probably varies superexponentially as the reciprocal of the amount supercooled the water is. Is it even true that the function that determines the density of water in terms of temperature is analytic at all temperatures above the freezing point and below the boiling point but not analytic at the freezing point or boiling point?

No.  That's not possible.  What about neutron mass?  What about the mass decrease due to the binding of nucleons through the residual strong force?
Beyond that, for even something as simple as $\ce{H_2}$, as considered using the non-relativistic Schrodinger equation, there is no exact solution for even the wavefunction of one molecule.
What NIST uses is this: http://nvlpubs.nist.gov/nistpubs/jres/097/jresv97n3p335_A1b.pdf

A neutron can be made by fusing a proton and an electron together. Can it still not be done because the loss of nuclear potential energy reduces the total mass of a proton and electron and the reduction in nuclear potential energy wasn't given?

@Timothy no, you would need an electron anti-neutrino as well.  Also, the neutron mass is not the sum of the electron mass and proton mass.

A better answer would be if you wrote an answer that starts off with supposing the constants I gave as well as the mass of an oxygen nucleus was given and then answers the question of whether it has been determined in terms of those 7 constants or researchers have the resources that they would immediately be able to figure out how to do so.

@Timothy Then you would have the issue of calculating an infinite number of Feynman diagrams to determine anything http://physics.stackexchange.com/questions/111400/feynmans-infinite-amount-of-logic-for-one-tiny-bit-of-space or even if you stuck with the Schrodinger equation, the three-body-problem prevents calculating even the wavefunction for H2 exactly https://en.wikipedia.org/wiki/Three-body_problem

The 3-body problem is subject to chaos theory. That just means the further in the future, the harder it is to compute the exact state at that time from the exact initial state. It doesn't mean it can't be done. You're correct that at any given time, we can't observe any wave function exactly. However, the bigger a body of water is, the less random fluctuation in its average temperature with time, so an infinitely big body of water of a given average pressure should have zero fluctuation in average temperature and have an exact average density.

@Timothy I was thinking more of within one water molecule, the 3 nuclei and 8 electrons.  That's 11 bodies.   The Schrodinger equation can't be solved exactly  for more than 2 bodies.

@Timothy - you appear to be using a lot of techno-babble. Various things actually can't be done, not because they are hard to do, but that it is actually not possible to do.


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What does equilibrium constant mean?
I really don't understand equilibrium constants: they seem to change nonstop! Can someone clarify the definition of equilibrium constants and give me examples of them?
Also, I looked in my textbook and the explanation for heterogeneous equilibrium used an example of decomposing calcium carbonate.
$$\ce{CaCO3 (s) <=> CaO(s) + CO2(g)}$$
The textbook continued to explain that $K^{'}_{c} = \frac{[\text{CaO}][\text{CO}_{2}]}{[\text{CaCO}_3]}$
There is my first obstacle: What is the meaning of $K^{'}$ (I don't think they're talking about the reverse process equilibrium constant)? And the next part to me is even stranger...
I do get that the concentration of the two solids are basically constants, so we can treat those parts as constants.
But why would you do $K^{'} \times \frac{[\text{CaCO}_{3}]}{[\text{CaO}]}$, which equals $K$, according to the textbook, the actual equilibrium constant of this reaction?

I am sure the book, if it is any decent, should have said something like this already. $K'$ is not a measure of the forward reaction, nor is it a measure of the reverse reaction. It is a measure of the equilibrium position.
Equilibrium occurs when the rate of the forward reaction is equal to the rate of the backward reaction. By the laws of thermodynamics, this occurs at one special point where the instantaneous concentrations of all the compounds satisfy the equation
$$\frac{[\ce{CaO}][\ce{CO2}]}{[\ce{CaCO3}]} = K'$$
$K'$ for a reaction is a constant. Clearly, a large $K'$ indicates that when equilibrium is reached, there is a lot of $\ce{CaO}$ and $\ce{CO2}$, and only a little bit of $\ce{CaCO3}$. We say that the equilibrium position lies to the right. Conversely, a small $K'$ means that the equilibrium position lies to the left, i.e. reactants predominate.

The equilibrium constant is properly defined in terms of activities of the compounds, not concentrations. This means that for the reaction you gave,
$$K = \frac{(a_{\ce{CaO}})(a_{\ce{CO2}})}{a_{\ce{CaCO3}}}$$
The activity of a pure solid is defined to be equal to $1$, which means that you can simplify this by removing all the terms corresponding to pure solids:
$$K = a_{\ce{CO2}}$$
The "equilibrium constant" that you are learning that involves concentrations is a simplification.


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Is this graphic of molecules correct?
First off, huge fan of chemistry, not a chemist – do be nice :D
Would an image like this make sense as an infographic for the molecular composition of the air? I especially wonder if the different sizes of the molecules (or do they all have the same size?) and the constellation make sense in theory.
Many thanks

Differences in sizes would only make sense if smaller ones were further imo. Anyways that's definitely not air

Seconded - that's not air.. personally I would prefer it if you represented them accurately. The different atoms should have appropriately different sizes - look for tables of "atomic radii". As for sizes of whole molecules, that's harder to estimate, but if you get the circles the correct size then they should be about right.

thank you very much! I'll try my best. It should actually show the content of CO or CO2 etc. in the air – it's all about pollution.

To accurately show CO2 in air, you would draw 3,000 molecules and have one be CO2.  For CO you would have to draw on the order of a million other molecules for one CO molecule to be accurate.

No.  Most molecules of air have two atoms.
So if you want to be quantitatively accurate, if you have 100 molecules in a graphic, it would show
78 dinitrogen (N-N) molecules
21 dioxygen (O-O) molecules
1 argon atom (Ar)
http://eesc.columbia.edu/courses/ees/slides/climate/table_1.html
And you could add some water (H-O-H) molecules if you want to depict humidity.

I'd say that even if you don't want to be quantitatively accurate, you should have more two atom molecules. Also, water doesn't have a 90-degree angle, but 104.5°.

Molecules do not have the same size, mass, etc. Air is composed of many different elements and compounds, so I think this image would be suitable for the infographic. But unless we are talking about pollutants, I don't think there are any compounds that exist in bulk in the atmosphere with that complicated chemical structure, but that may be a detail you can overlook.
Also, this topic may be marked as off-topic. If you're new here, please post relevant chemical conceptual questions, not objective imagery questions :P.

Yes we are talking about pollutants and air quality in general (CO, CO2 etc.) Sorry for posting here and thank you very much

authors: محمد بن راكان


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Synthesis of 2-methyl-4-nitrophenol from benzene?
Benzene first reacts with the reagents $\ce{CH3Cl/AlCl3}$, to attach a methyl ($\ce{CH3}$) group to the benzene ring.
Then I suggested reacting with $\ce{OH-}$. Because the methyl group is an activator, it will cause the $\ce{OH-}$ electrophile to attack either in the para or ortho positions. In this case I suggest it will attack the ortho position. Now we have two activators, hydroxyl and methyl, but because hydroxyl is a stronger activator, the position where the next molecule attacks will be controlled by the hydroxyl group.
The next electrophile that attacks is the $\ce{NO2+}$, through the reaction $\ce{HNO3/H2SO4}$, the nitro group will attack the para position of the benzene and we get our final product 2-methyl-4-nitrophenol.
However, I am not 100% sure if the methyl group should be introduced first to form toluene or the hydroxyl group to form a phenol.

OH- is nucleophile not electrophile.

First convert benzene to phenol. Then do nitration since -OH is an Ortho para director. This brings -NO2 to para position. Then use FC alkylation and the only spot available is the Ortho position.

I agree with Aditya Dev. Also, for future reference, when you have a activating group that directs incoming groups to the ortho and para positions the para position is favored due to steric approach control.

There is a lot of good thinking in your proposed synthetic strategy.  The one major shortcoming is, as @Mithoron points out, $\ce{OH^{-}}$ is not an electrophile.
We need to introduce a methyl, hydroxy and nitro group onto a benzene ring.  If we look into our synthetic "toolbox" here are some thoughts that might come to mind:

A methyl group can be readily introduced using the Friedel-Crafts alkylation reaction with$\ce{CH3Cl}$. However, be aware that there are some general issues with the Friedel-Crafts alkylation reaction.  For example, if the alkylating agent can rearrange to a more stable carbocation it will likely do so ($\ce{CH3Cl}$ can't rearrange).  Also, the mono-alkylated product is more reactive than the starting material, so polyalkylation can occur (but in your case, we can use an excess of benzene - it's cheap - and get a good yield of the desired product toluene).  Finally, the Friedel-Crafts alkylation does not work well with deactivated aromatic rings. Therefore, the comment suggesting we run the alkylation after introducing the nitro group is incorrect; we must use the Friedel-Crafts alkylation before we introduce the nitro group.
There aren't any easy ways to electrophilically introduce a hydroxy group into an aromatic ring.  When I need an aromatic hydroxyl substituent, I think of the Sandmeyer reaction.
The nitro group can be introduced via straightforward electrophilic nitration.

There are a lot of ways to pull this synthesis together, but here is a relatively direct proposal.

The first step is our Friedel-Crafts alkylation to prepare toluene.
Next, we'll introduce our ortho hydroxy substituent.  To do this let's first sulfonate using fuming sulfuric acid (oleum) and block the para position. Blocking the para position will simplify our next step - we won't have to separate ortho and para isomers and we will get a higher yield of only the desired ortho (to the methyl group) product. Now the sulfonic acid group (meta-director) and the methyl group reinforce one another and will direct our next electrophile to the position ortho to the methyl group.
Now we nitrate using nitric and sulfuric acid and get only the product where the nitro group is ortho to the methyl group.
Next, we heat the reaction. The sulfonation reaction is readily reversible, by heating our reaction we can readily remove the sulfonic acid substituent since it has served it's purpose and is no longer needed.
Now it's time for the Sandmeyer reaction. We first reduce the nitro group with tin and hydrochloric acid to an amino group.  Then we diazotize it and decompose the diazonium salt in the presence of acidic water to introduce our hydroxyl group.
Finally we nitrate one more time to introduce the nitro group.  The directing effect of the hydroxyl group is much stronger than that of the methyl group and will primarily direct the nitro group into the position para to the hydroxyl.

http://www.mhhe.com/physsci/chemistry/carey/student/olc/ch24phenolpreparations.html . I'd go for benzene-benzenesulfonic acid - phenol - ortho-formylation product - kishner-wolf reduction - nitration (mono-nitration is reasonably easy to achieve on cold with weak reagent and orto- and para- produducts should be separatable by water vapor distillation)

Are we in danger of over alkylating?


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How important is it that geometry be optimized at a high level of theory?
I've been doing research with a computational chemist for a little while now and in one of our projects we are dealing with a rather large system and because I go to a relatively small university, our resources are limited. Thus, when we began to ramp up the calculations in the project, we opted not to re-optimize our structure at CCSD(T)/def2-ATZVPP despite the fact we are doing the rest of our calculations at that level of theory with that basis set. The reason we chose not to optimize is we have a limited amount of time/resources and we estimated the optimization would take around 45-50 days to complete which would basically hog a computer for that long and stall the other projects we are working on.
I'm wondering how significant of an error, if much at all, that will introduce?
To give a few more details, we are studying vibrational properties of the system using a local mode approach. The structure we have is optimized at MP2/def2-ATZVPP.
Regardless, I'm just looking for a general understanding of how important the optimized structure might be. It seems like it would be less important, but could be quite dependent on what one is studying, than other things based on the fact that no system is truly at that optimized structure, but is only in that geometry on average. Is that correct at all?

Method

Most of the time CCSD(T) would indeed be a huge overkill for geometry optimisation. DFT and MP2 have way better performance/cost ratio. Note that I said DFT and MP2 above, not or MP2; this is well known procedure to compare DFT geometries with MP2 ones to be on the safe side: there should be no big differences.
On the DFT side hybrids might be overkill as well. Try simple pure GGA functionals like BP86 and PBE as well meta-GGA functionals like TPSS first. There might be almost no advantages in using hybrids for geometries.
Include empirical dispersion corrections in DFT calculations even if you do not suspect dispersion forces to be decisive. Just in case.
Employ the density fitting (also known as the resolution of identity) for DFT as well as MP2 to significantly accelerate calculations.

Basis set

With respect to the basis set convergence of geometries is typically reached at the triple-zeta level. This is fairly well-known from excessive benchmark studies for correlation consistent1 and polarisation consistent basis sets2, but Turbomole def2 basis sets should have similar behaviour as well. References are exclusively about DFT, but the same is true for WFT as well.
In practice, though, for DFT calculations (but not necessarily for MP2) a triple-zeta basis might also be an overkill. Try double-zeta def2-SVP first (augmented with diffuse and polarisation functions to your taste). Even Turbomole developers themselves describe DFT/SV(P) and MP2/TZVPP levels of theory as "almost quantitative".

Benchmark

Compare the calculated geometries with experimental data if such are available for your (or at least similar) compounds. At the end of the day, how well does a particular level of theory reproduces experimental geometries is the main criteria of choice.

1) Wang, N. X., & Wilson, A. K. (2004). The behavior of density functionals with respect to basis set. I. The correlation consistent basis sets. The Journal of Chemical Physics, 121(16), 7632–46. http://doi.org/10.1063/1.1792071
2) Wang, N. X., & Wilson, A. K. (2005). Behaviour of density functionals with respect to basis set: II. Polarization consistent basis sets. Molecular Physics, 103(2-3), 345–358. http://doi.org/10.1080/00268970512331317264


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Is it possible to make plastic out of fibres?
Can we make plastic, especially bio-plastic from vegetable and fruit fibres?

We kinda can, but the result is not called "plastic".

Sure; it is one of the oldest "plastics": cellophane and rayon.
It is not difficult to demonstrate, though care must be taken with the somewhat poisonous cuprammonium solvent.
Other plastics can be made from substituted cellulose, such as cellulose acetate... what could be more "organic" than vinegar and wood pulp?

On a way one can see paper as a polymer hybrid material, too.

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Are the collisions between the real gas particles perfectly elastic?
Well my question is simple, if two real gas particles are colliding (head on collision) then will the kinetic energy will be conserved i.e. will it be a perfectly elastic collisions?

Well, what is an inelastic collision, really? Suppose you have two balls made of steel; they collide, then fly away with some lasting deformation, so some energy is lost. With molecules, it is not quite like that. You can't leave a dent on a molecule. It has certain discrete energy levels, and that's it. You either excite the molecule to one of these levels, or you don't excite it at all.
To sum it up, some collisions of molecules are perfectly elastic, and others are combined with excitation of some rotational or (more probable at higher temperatures) vibrational mode in one of the molecules, or maybe in both.
Noble gases which have no molecules and hence neither rotational nor vibrational modes may enjoy perfectly elastic collisions up to pretty high temperatures.

" It has certain discrete energy levels, and that's it. " are you referring to Molecular orbital theory? Because, why would otherwise a molecule have energy levels?

A molecule has discrete energy levels because it is a finite quantum system. Whether you describe it in terms of molecular orbital theory or otherwise is irrelevant.

authors: مرتضى الطائي


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Continuous flow mass spectrometer for isotope analysis
If i took thousands of discrete samples of mineral fragments in a field study using a specialized drill, what mass-spectrometry techniques/equipment would allow me to analyse the samples as fast as possible?
Some machines can take 5000 readings per second, how can that be applied to a continuous flow of mineral fragments?
The premise is a feasibility-study of a robot archaeology tool:
Cartesian Axis Robots have an accuracy of about 0.01mm, so it would be possible to excavate sensitive sites to sample submillimeter 3d slices of the soil.
The technical challenge of the machine would be how to prepare and record tens of thousands of chemistry results every hour.

To find out about current spectrometer sampling speeds, research the term: "spectra per second"
There is alot of precis informations i.e.:

"The standard interface supplied with all nTM detectors is
12 bit USB2.0. With this interface and our SpectraArray SL software,
up to 30 spectra per second can be captured.
For even faster acquisition, a PCI bus interface is available as an
option. This  allows up to 160 spectra/second to be acquired. "
50 to 80 micrograms of mineral is sufficient for a typical analysis, although small samples on the order of 12-20 micrograms (single foraminifers) can also be analyzed

So it seems that a major challenge is the data bandwidth between the spectrometer and the SSD/HDD.
Spectra per second returned by the machine are not going to be the limiting factor, because they can go up to 10 000 spectra per second. A graph with 250 floating point humbers is 4bytes times 250, so about 1kb per graph, research is necessary to understand why USB can only handle 12 samples per second, it should be in the 1000ds.
In spectroscopy, the key is to constantly calibrate the sample against an empty control sample. If the sample is drilled, it can be ground into a find dust and then it should be possible to process it in a homogenous way similar to liquid or gas spectroscopy.
All the different spectroscopy methods can be considered, they are all very fast, and have different detection strengths. IR and Xray have different advantages, IR is easier to deal with. The selection of appropriate IR or other wavelengths to use for a specific type of study is an advanced topic.


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Computing accurate vibrational and rotational contributions to the free energy of transition states and loosely bound complexes
Recently I have been dealing with a lot of transition states and relatively loosely bound ion-dipole complexes and I have some trouble figuring out how to make sure that the rotational and vibrational contributions to the Gibbs free energy are as accurate as they reasonably can. (I am using Gaussian 09 rev C)
I am using MP2/aug-cc-pVTZ, based on literature results it should be in close agreement with even better methods for the type of molecules I am working with.
First I have optimized the geometry to verytight convergence, then reoptimized it with verytight, CalcAll to make sure the geometry is as converged as possible, and get the thermochemistry output at the same time.
I have once read in a CCL post, that tightening SCF convergence can increase the accuracy of analytic derivatives, so I am performing all calculations with SCF=Conver=12.
My concerns are mostly related to having a number of low frequency modes, and the possibility of coupling between vibrations and rotations.
For example, in the case of the transition state between $\ce{MeCl + F-}$ and $\ce{MeF + Cl-}$, I have three low frequency modes: 258.4867 $cm^{-1}$, 258.4867 $cm^{-1}$ 303.2302 $cm^{-1}$ (and of course an imaginary frequency). Sure enough, Gaussian warns that:

Warning -- explicit consideration of   3 degrees of freedom as vibrations may cause significant error

Based from what I gathered from various sources, this means that these low frequencies have significant uncertainties in them, and they may or may not be internal/hindered rotations. I have ran a calculation with Freq=hindrot but it did not identify any of the modes as rotations. However, the Gaussian manual says that the hindered rotor analysis is not always successful at finding rotations when transition states are involved.
My other concern is the validity of the rigid rotor approximation for molecules having low frequency modes (ie. low force constants). My intuitions tell me, that the lower the force constants in a molecule, the worse are the errors caused by the centrifugal distortions of the molecule.
So my actual questions are:
1.What is the proper way to accurately handle low frequency modes? How does one make sure that they are not some sort of rotation, when hindered rotor analysis does nothing? If they legit low frequency vibrations, are they reliable? If they are not reliable how can they be corrected?
2.Do I have to be concerned about low frequencies also implying that the rotational contributions are also affected due to centrifugal distortion?

I forget.. does Gaussian "automatically" remove the 6 rotational and translational modes for you? (Some programs do.) The mode energies you gave don't seem especially low.

You can also convert between the rotational constants in the Gaussian output (usually in GHz) and wavenumbers to compare with the frequencies you mention.

Yes, the translational and rottational modes are handled automatically.

@GeoffHutchison I think I saw somewhere that both the T+R unprojected and projected frequencies are output. Dunno if there's a flag you have to set for that, though.

@Brian Just looked it up - it won't output the XYZ of the modes, but does give the frequencies. The Gaussian manual recommends "Opt=Tight Int=UltraFine" for nailing down these frequencies accurately.

"I have optimized the geometry to verytight convergence, then reoptimized it with verytight, CalcAll to make sure the geometry is as converged as possible" I don't see how re-optimization or CalcAll would any way improve the "tightness" of your convergence. Would you elaborate it? It seems like a hell a lot of cost for nothing.

If you are trying to be that accurate, you may have significant error from sources you have not considered: e.g. anharmonicity of the potential energy surface, esp for the soft nodes.

The reoptimization with CalcAll was not expensive at all, as the geometry was already very near optimum. In fact in most cases it only did a single iteration (and found that the geometry was already converged), so since I needed the frequencies anyways it was practically no extra cost(compared to say an equivalent OPT FREQ job).

@uLoop Cites added to my answer. Sorry it took so long.

Given the accuracy it sounds like you're looking for, and the systems you're studying, this indeed appears to be a Very Hard Problem. I think your intuitions are well founded.
Briefly, to your questions:
1) How to handle the low-frequency modes?
A lot of manual exploration of the PES, doing things like running normal mode scans and optimizing intermediate geometries along suspected internal rotations. This gives you an idea how the various energetic barriers to non-harmonic motions compare to a Boltzmann distribution of states.

If the barriers are high, the harmonic approximation is probably decent and you can just use Gaussian's results
If barriers are low, then you have ~free internal rotation.
If intermediate, then you're facing hindered internal rotation.

There's literature on evaluating enthalpy/entropy contributions of free and hindered internal rotations.  Some can be found in my answer to a different question, here.  Other potentially relevant resources include:

Czaszar (WIREs Comput Molec Sci 2 273, 2012)
Overview of anharmonic force fields and their computation.
Bloino (J Chem Theor Comput 8(3) 1015, 2012)
Overview of the VPT2 anharmonic computational system implemented in recent versions of Gaussian.
Carbonniere (Chem Phys Lett 392 365, 2004)
Description of the vibrational-rotational (Coriolis) coupling anharmonic correction and its potential significance for some systems. More recent references are likely available.
Meal (J Chem Phys 24(6) 1119, 1956)
Meal (J Chem Phys 24(6) 1126, 1956)
Classic discussions of Coriolis coupling.

You should be able to turn up quite a bit more information with some quick further searches, if need be.
2) Do I need to be worried about contributions from centrifugal distortion?
Very likely. Unfortunately, I don't know of very many examples where people have attempted corrections such as this. Certainly you're looking at manually developing a suitable analysis; I can't imagine any software packages out there at present are capable of automatically chugging through this for you.


text:
What is the relation between partial pressure and concentration?
So, one of my books said that partial pressure plays more significance in determining the gradient than concentration. So, it said it was possible for a gas to go from high partial pressure at point A to low partial pressure at point B, even though there is low concentration at point A and high concentration at point B.
How is this even possibly? I thought partial pressure and concentration were always related.

because of temperature. Temperature at A must be more than temperature at B

This does not really answer my question. :(

MATHEMATICALLY
At point A
The pressure according to ideal gas equation is $P_A = C_ART_A$. Here $C_A (= n/V)$ is concentration at A.
At point B
Similarly, the pressure according to ideal gas equation is $P_B = C_BRT_B$. Here again $C_B$ is concentration at B.
Now, $P_A$ > $P_B$
So, $C_ART_A$ > $C_BRT_B$
i.e. $C_A/C_B$  > $T_B/T_A$  .....equation 1
Now as the condition is that concentration at A is lower than the the concentration at B. Therefore we can say that
$C_A/C_B$ < $1$ .....equation 2
Now from equation 1 and 2 we have
$T_B/T_A$ < $1$
which gives us $T_A$ > $T_B$
So in the end, yes, your book was right. It is possible if the temperature at A is more than that at B.
"HOW TO FIND IT IN SECONDS" or "I HATE TO DERIVE BY USING EQUATIONS"
Think of two containers A and B which are connected by a pipe. Both are connected with the help of a pipe which is closed (i.e. it doesn't let the gases exchange). Now we know that concentration of gas in A is lower than the the concentration of gas in B. If we want more pressure of gas in A then the velocity of its particles should be more. Now it is possible if it has more thermal energy. So the temperature should be more in A.


text:
Does the potential energy increase when temperature is raised?
The potential energy of a material is its ability to do work. So I would suspect the potential energy to increase during phase changes that increase entropy (solid $\rightarrow$ liquid and liquid $\rightarrow$ gas), and decrease in the opposite direction.
However, what about when the temperature is raised for a solid, up until the melting point - will the potential energy increase? The heat from a 1500 K gold plate could do work on liquid water (evaporate) to a higher extent than a plate at 300 K, so I guess the potential energy has increased. So would the potential energy as a function of temperature, of an arbitrary material, look something like this ($K$ should be $T$)?

Can you please provide a precise definition of the potential energy of a macroscopic system like the ones you are describing?  In what ways is it related to the internal energy?

As was once pointed out, most introductory chemistry textbooks define energy as "the ability to do work," then turn around and say that free energy is "the part of energy in a system that is available to do work." They never note that their two definitions are inherently contradictory: if some of the energy can't do work, then you can't define energy as the ability to do work! Defining the potential energy of a system as "its ability to do work" strikes me as similarly nebulous---is there a more precise definition you could give?

Yes, potential energy increases with increasing temperature for at least the following three reasons:

At a higher temperature, more atoms/molecules are in excited electronic states.  Higher electronic states correspond to greater potential energy. Potential Energy is -2 times Kinetic Energy. So actually, at higher temperature, when more atoms are in higher electronic states, there is more potential energy and less kinetic energy (just considering electronic energy).
At higher temperature, more molecules are in excited vibrational states.  Higher vibrational states correspond to greater potential energy. Half the energy of each vibrational state is kinetic, half potential.
At higher temperature, more molecules are in excited rotational states.  While for a rigid rotor, this would not involve increased potential energy, real molecules are not rigid.  The rotation of a diatomic molecule about an axis perpendicular to the bond stretches the bond as a centrifugal distortion, which represents an increase in potential energy.

Additionally, for intermolecular forces, to the extent that they involve potentials of the form r^x, the Virial Theorem requires that total energy is divided into potential and kinetic energy in a specific proportion, so total energy can not be increased without increasing potential energy.
Free translational motion is not subject to a potential, nor is rigid rotation.

I wanted to ask that we read that temperature is due to random internal kinetic energy. Is temperature also due to the internal potential energy or the total internal energy or only because of kinetic energy

Links [1] and [3] appear to be malfunctioning

Lets talk about a heating curve
As heat is constantly added to a substance in the solid phase, the KE increases but the PE remains the same. As the substance reaches its melting point, commonly known as fusion (border between melting and freezing), the KE remains the same while the PE increases. After a while the substance becomes completely a liquid, and heat is still being added constantly and the KE increases while the PE remains the same. The substance now reaches its boiling point, and starts to change into the gas phase, in this process the KE remains the same while the PE increases.
KE = kinetic energy
PE = potential energy

Welcome to Chemistry.SE! Take the [tour] to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax. For more information in general have a look at the [help]. This seems to be much too simple to be true. It would be appreciated if you could add a reason to your statements, i.e. why does the KE increase while the PE stays the same.


text:
Why can't phosphorus trifluoride be used as a Lewis acid catalyst in electrophilic substitution of benzene?

Which of the following cannot be used as Lewis acid in the electrophilic substitution of benzene?

$\ce{BF3}$
$\ce{SbCl3}$
$\ce{FeCl3}$
$\ce{PF3}$

My Attempt:

$\ce{BF3}$ and $\ce{FeCl3}$ are Lewis acids.
$\ce{PF3}$ though it has a filled s and p orbitals it can accept electrons from donors and expand its coordinate number up to six due to vacant d orbitals. It acts as a mild Lewis acid.
$\ce{SbCl3}$ has a completely filled s,p,d orbitals so it can no longer accept electrons, hence doesn't act as a Lewis acid.

My answer is option (2) but the answer provided is (4), where am I going wrong?

This is indeed confusing. I expect both B and D not to be Lewis acids (they are both Lewis bases).

@Jan They can both act as acids or bases depending on situation. https://en.wikipedia.org/wiki/Antimony_trichloride, https://en.wikipedia.org/wiki/Phosphorus_trifluoride

@Jan perhaps the real question might be, will all Lewis acids catalyse EAS reactions?

Yes, probably PF3 is simply too weak to do it.

The best answer I can give is that there is enough space on $\ce{Sb}$ in $\ce{SbCl3}$ for an additional Lewis Base to interact with a nonbonding orbital on the $\ce{Sb}$ atom. While $\ce{P}$ can be hypervalent, the LUMO in this case would be $\ce{P-F}$ $\sigma^\ast$, which is really high in energy (meaning that it doesn't easily accept electron density from other species in solution).

Do you mean the back bonding due to fluorine, and hence filling up of the the empty d orbitals orbitals, making it pretty stable, and it refusing to add any more electrons?

@SujithSizon They don't really fill any d-orbitals and there's no such backbonding

@Mithoron But phosphorous is electropositive and lone pair of fluorine is hence more closer to the central phosphorous atom, partially occupying d orbitals, right?

@SujithSizon http://chemistry.stackexchange.com/a/5242/9961


text:
Precipitation with NaI
I dissolved $\ce{[Pt(NH3)4](NO3)2}$ in water then I added small amount of $\ce{NaI}$ ($1~\mathrm{M}$) to it. Precipitation occurred. I want to know what happened?

Hi and welcome to chemistry.stackexchange.com. Feel free to take a [tour] of the site. I improved the formatting of your post by using MathJax; you can learn more about that in the [help], this meta-post or this one.

You precipitated platinum(II) iodide.

Just out of curiosity, did you see any gas evolution?

I thought it is PtI2. but when I took xrd for  the precipitation, I saw too many peaks which I got confused on assigning  the peaks to PtI2.

I didn't see any gas evolution.

The original solution contains the square planar $\ce{Pt(II)}$ complex ion $\ce{[Pt(NH3)4]^2+}$. When sodium iodide is added to the solution, one or more $\ce{NH3}$ molecules are replaced by $\ce{I-}$ ions in a ligand exchange reaction. In particular, replacing two $\ce{NH3}$ molecules with $\ce{I-}$ ions yields the neutral complex $\ce{[PtI2(NH3)2]}$, which is poorly soluble in water, and therefore a precipitate is formed.
Remarkably, two isomers of square planar $\ce{[PtI2(NH3)2]}$ exist: $\ce{$cis${-}[PtI2(NH3)2]}$ and $\ce{$trans${-}[PtI2(NH3)2]}$.
When starting from $\ce{[Pt(NH3)4]^2+}$, the first $\ce{I-}$ can replace any of the four equivalent $\ce{NH3}$ ligands. Because of the larger trans effect of $\ce{I-}$ compared to $\ce{NH3}$, the second $\ce{NH3}$ is preferably replaced trans to the first $\ce{I-}$ ligand and therefore the trans product $\ce{$trans${-}[PtI2(NH3)2]}$ is obtained.
(Note that, if one starts from $\ce{[PtI4]^2- }$ and replaces two $\ce{I-}$ ligands with $\ce{NH3}$, the cis product $\ce{$cis${-}[PtI2(NH3)2]}$ is obtained instead.)

But when I solved pt(NH3)4(OH)2 with NaI in water, precipitation didn't happen. Why it didn't give precipitation?
Thanks


text:
How to prepare 2-bromobutane from 1-butanol?

So my answer (g then b) was incorrect. I thought that the bulky strong base of b would do E2 on 2-bromobutane, making it but-1-ene, then just add $\ce{HBr}$. What is wrong with my way of thinking?

That you’re starting with butan-1-ol. Alcohols don’t eliminate easily.

$\ce{OH-}\mathbf{\neq good~leaving~group.}$

The hydroxyl fragment isn't a good leaving group, unless the medium is acidic. Using just HBr, you can convert the hydroxyl group into an oxonium ion (the H2O fragment would be a good leaving group). Upon heterolysis of the C-O bond, a hydride shift occurs simultaneously to create a secondary carbocation. The exogenous bromide can act as a nucleophile towards the secondary carbocation to form 2-bromobutane.

Apart from a few weird references and those that include spacy catalysts, reacting an alcohol with $\ce{HBr}$ gives the corresponding bromide from the $\mathrm{S_N2}$ mechanism thus in this case 1-bromobutane. That corresponds to my experience using $\ce{HI}$ where I got 1-iodobutane exclusively in near-quantitative yield.

when I was a graduate student, I TA'ed organic lab where the students had to do this exact experiment. If my memory serves me correctly, they got a mixture of both.

Scratch the previous comment, the experiment the students had to do was very similar, but not exactly the same.

right. So the first step would be using HBr to form an oxonium ion. Can you describe the hydride shift afterwards? I'm a little confused on that. Would the H2O just leave since it's a good LG and then one of the beta hydrogens going over to where the H2O left? Then there will be a secondary carbocation and Br- attacks there?

https://www.dropbox.com/s/c7dnaqgmhac2iuc/bromo%20ex.pdf?dl=0. Copy and paste the url to see what I meant.

$\ce{(CH3)3COK}$ reacts with $\ce{R-X}$ to give a Hoffman alkene.
Elimination of the OH group usually requires protonation of oxygen, which requires an acidic medium.
A good solution to the given problem would be to add HBr without peroxide, which will give SN$_2$ (no rearrangement, i.e. no cation shift) and give us $\ce{CH3CH2CH2CH2Br}$.
Then, add $\ce{(CH3)3COK/(CH3)3COH}$ to get but-1-ene. Then, add HBr to get 2-bromobutane.
Also a useful sidenote is that methanol, ethanol, n-propanol, and n-butanol all undergo SN$_2$ when reacted with $\ce{HX}$. Other alcohols undergo SN$_1$.


text:
Why does sulfonation of alcohols even work?
I am now near the end of organic chemistry 2, and I'm trying to understand the patterns of reactions so that I can predict them rather than brute force memorizing all of them. Looking at the sulfonation of alcohols using mesyl chloride, I don't think I would ever predict that. I don't understand why the oxygen should attack the sulfur in mesyl chloride. To me it looks like a nucleophilic substitution reaction, but alcohols aren't even powerful nucleophiles. Not to mention that the end of this reaction creates HCl a very powerful acid. I thought when reactions occur they like to create relatively weaker acids and bases? What is driving the reaction, and what patterns should I be noticing here for future predictions?

There are always two sides of a reaction. In the case of an alcohol attacking a sulphonic acid chloride, we can look closely at the alcohol and say: ‘Oh my, this is a bad nucleophile and we cannot deprotonate it with the conditions we have. Probably, this reaction won’t work’. But we can also look at the sulphonic acid chloride and say: ‘Oh my, this is a very strong electrophile and it has a very good leaving group. This reaction will work like a charm!’
So it is important, that you compare the reactants extensively. In this case, the electrophilicity of the sulphur(VI) is high enough to even allow rather bad nucleophiles to work. On the other hand, oxygen atoms aren’t even that bad a nucleophile. Give them enough time and alcohols will attack via $\mathrm{S_N2}$ — it might just take forever.
We can also take electrostatics into consideration. Oxygen is rather negatively polarised since it has a very high electronegativity. The sulphur atom, on the other hand, is strongly positively polarisied, since it is connected to two formal oxide anions and a chlorine atom — all three are more electronegative than sulphur by a noticeable margin. A rather negative particle such as the alcohol oxygen, will happily attack a rather positive particle such as the sulphonic acid chloride sulphur.
Mechanistically, this attack forms an intermediate pentavalent sulphur which will then displace the best leaving group: The chloride atom. Therefore, even the hydrochloric acid that is produced isn’t as bad as it may look. We loose the chloride first and the proton much later. We aren’t immediately removing a strong acid but rather a good leaving group and the proton just has to deal with it.
However, I will allow the objection that a significant concentration of $\ce{HCl}$ would favour the reverse reaction. Thus, when employed synthetically, a base is usually added that serves no purpose other than to capture the proton and keep it somewhere where everybody can be happy.
Too-long-didn’t-read-version:

Alcohol might be a bad nucleophile, but oxygen is negative.
Sulphonic acid chloride is a very good electrophile and the sulphur atom is positive.
Positive and negative like each other, thus alcohol will attack sulphur.
Chloride is displaced because it is a pretty good leaving group.
The proton just has to do what the others force it to do.
Driving force is the liberation of the chloride and turning a weak $\ce{S-Cl}$ bond into a strong $\ce{S-O}$ one.
Add a base to capture the proton for further enhancement.

I love the theory behind this stuff, keep it coming! Also, when you said that HCl is not so bad compared to pentavalent sulfur, I'm not sure I know how to compare those two. From this post, I'm going to assume more than 4 substituents is highly unstable? Does this hold on every atom, or certain types?

@AbidRizvi No, that was not what I was trying to say. You really can’t compare the two. However, the pentavalent sulphur is only a transition state: It has an electron dectet on sulphur which wants to abide by the octet rule as do all main group elements. You can probably better think of the pentavalent sulphur as a halted transition state of an $\mathrm{S_N2}$ reaction. That said, $\ce{HCl}$ is ‘good’ in this context not because it is a good acid or whatever. What we are initally loosing is a chloride $\ce{Cl-}$. Chloride is very stable by itself — see all the chloride salts. (tbc)

(cont’d) As I tried to say, the proton only leaves later and it doesn’t have to end up as $\ce{HCl}$ although it could. And thermodynamically, forming $\ce{HCl}$ molecules isn’t that bad either; hydrogen and chlorine gas react strongly exothermicly to form it.


text:
Szabo/Ostlund exercise on spin orbital orthonormality
I have a question about exercise 2.1 in Szabo and Ostlund's Modern Quantum Chemistry, which concerns the inner product between different spin orbitals: $\require{begingroup}\begingroup\newcommand{\br}{\mathbf{r}}\newcommand{\bx}{\mathbf{x}}$

Exercise 2.1. Given a set of $K$ orthonormal spatial functions, $\{\psi_i^\alpha(\br)\}$, and another set of $K$ orthonormal functions, $\{\psi_j^\beta(\br)\}$, such that the first set is not orthogonal to the second set, i.e.,
$$\int d\br \, \psi_i^{\alpha*}(\br)\psi_j^\beta(\br) = S_{ij}$$
where $\mathbf{S}$ is an overlap matrix, show that the set $\{\chi_i\}$ of $2K$ spin orbitals, formed by multiplying $\psi_i^\alpha(\br)$ by the $\alpha$ spin function and $\psi_j^\beta(\br)$ by the $\beta$ spin function, i.e.,
\begin{align}
\chi_{2i-1}(\bx) &= \psi_i^\alpha(\br)\alpha(\omega) \\
\chi_{2i}(\bx) &= \psi_j^\beta(\br)\beta(\omega)
\end{align}
with $i = 1, 2, \ldots, k$ is an orthonormal set.

I feel quite confused about this question. Consider the spin-orbital integral
$$\langle \psi_i^{\alpha} \alpha (\omega) | \psi_j^{\beta } \alpha (\omega) \rangle  = \langle \psi_i^{\alpha}  |  \psi_j^{\beta } \rangle  $$
which is not necessarily orthogonal.
The result can only be orthogonal if both ket and bra vectors are in the $i$-indexed set. If so, why does the exercise mention a $j$-indexed set in the beginning?

Consider the spin-orbital integral
$$ \langle \psi_i^{\alpha} \alpha
(\omega) | \psi_j^{\beta } \alpha (\omega) \rangle  = \langle\psi_i^{\alpha}  |  \psi_j^{\beta } \rangle
$$
which is not necessarily orthogonal.

Yes, it is not necessarily orthogonal, but the thing is that you don't need to consider such integrals in the exercise. Read more carefully: spatial orbitals from the first "alpha" set can be multiplied only by the $\alpha$ spin function while spatial orbitals from the second "beta" set can be multiplied only by the $\beta$ spin function. Thus, there are only three types of integrals you need to consider:
$$
\langle \psi_i^{\alpha} \alpha | \psi_j^{\alpha } \alpha \rangle \, ,
\quad
\langle \psi_i^{\alpha} \alpha | \psi_j^{\beta } \beta \rangle \, ,
\quad
\langle \psi_i^{\beta} \beta | \psi_j^{\beta } \beta \rangle \, .
$$

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


text:
Controlling the reaction of alcohols with acid to either ether synthesis or elimination
Let’s say you take ethanol: $\ce{CH3-CH2-OH}$ and you add an acid and heat. You can get water, the acid, and $\ce{CH2=CH2}$ (a dehydration reaction). However, you can also get $\ce{CH3-CH2-O-CH2-CH3}$, water, and an acid (symmetric ether synthesis). How would you make one reaction preferable over another?

By modifying the conditions, in this case: the concentration.
The first and rapid step of the mechanism is the protonation of the alcohol:
$$\ce{HA + CH3CH2-OH <=> A- + CH3CH2-OH2+}$$
From then on, it depends what happens:

If the molecule is left alone, i.e. its concentration is low, something that has a just enough basicity might at the correct time be close enough to a methyl hydrogen to allow the elimination to happen. $\mathrm{E1}$ is not an option since the primary carbocation is extremely unstable and Wagner-Meerwein rearrangements for enhanced stability are not possible, thus the elimination must preceed by the $\mathrm{E2}$ mechanism. If $\ce{A-}$ is the only basic compound present, it must accept the proton.
$$\ce{A- + H-CH2-CH2-OH2+ -> A-H + H2C=CH2 + H2O}$$
If the molecule happens to find one if its kind before the former can take place, the two of them could engage in a substitution reaction by the $\mathrm{S_N2}$ mechanism. If the molecule’s concentration is high, this gets more likely. And if it happens, the second molecule’s hydroxy group is nucleophilic enough (even though alcohols are generally bad nucleophiles) to displace the extremely good leaving group $\ce{-OH2+}$ and liberate water. The extraneous proton can later be removed as the ether sees fit.
$$\ce{CH3CH2-OH2+ + HO-CH2CH3 -> CH3CH2-(OH+)-CH2CH3 + H2O}$$
$$\ce{CH3CH2-(OH+)-CH2CH3 + A- <=> CH3CH2-O-CH2CH3 + HA}$$

Thus, the first reaction will always happen but be rather slow. Its rate is close enough to being independent of the molecule’s concentration. The second reaction’s rate is directly dependent on the molecule’s concentration and will get faster the higher the concentration is. High concentrations (and low acid concentrations) can thus favour the substitution mechanism over the elimination one.
This will hold true more or less for all cases where two reactions compete against one another. It is always possible to modify the conditions to prefer one or the other. However, it is often not as simple as in this simple case.


text:
Can dissolved salts (or any soluble chemicals, for that matter) 'bump' each other out of solubility?
Will a more soluble substance dissociate, causing a less soluble substance to precipitate out?
in equation:
AB(s)+ C(aq)+ D(aq)--> A(aq)+ B(aq)+ CD(s)
Does that equation go through, under the assumptions that AB, as a whole, is more soluble than CD, and saturation point is achieved with both substances?
Thanks in advance! This question has been bothering me for a while...

Yes, this effect is termed salting out for non-ionic compounds and the common ion effect for ionic ones.
Note that in specific cases, the converse may hold true. For example, "the solubility of silver chloride in water is reduced if a solution of sodium chloride is added to a suspension of silver chloride in water (op cit)." However, adding more chloride re-dissolves the precipitate due to the formation of complex ions:
$\ce{AgCl(s) + Cl- (aq) -> AgCl2- (aq)}$

I don't think the OP was asking about the common ion effect. More like $$\ce{KNO3_{(s)} + Ag^+{(aq)} + Cl^{-}{(aq)} -> K^+{(aq)} + NO3^{-}{(aq)} + AgCl{(s)}}$$ which I think would have the opposite effect. I'd expect $\ce{AgCl}$ to be more soluble in a $\ce{KNO3}$ solution.

Yeah that's what I was thinking. So if I was to put sugar into saturation point salt water, the salt would precipitate out?


text:
Is this definition of mole correct?
Chemical engineers define one mole as the amount of a substance which possess as many entities as $12\ \mathrm g$ of $\ce{^{12}C}$.
The number of atoms in $12\ \mathrm g$ of $\ce{^{12}C}$ is $6.022 \times 10^{23}$ which is a constant by its definition.
Now come to the relevant definition which is given in my textbook that defines one mole of a substance as the atomic mass, molecular mass or formula mass in grams.
Is this definition correct?

It is the right definition for now. I remember reading about a sphere of silicon which was being proposed for a new standard. The twist here that that the definition of a kilogram is being considered for replacement too.

@MaxW can you explain me this chemical equation? $Zn + 2HCl = ZnCl_2 + H_2$? I think, I can better explain my problem then.

I don't understand what there is to explain about the reaction.

Ok listen.. The equation is balanced. What is the $"2"$ there before $HCl$? If it tells that on the right hand side there are two atoms of Hydrogen and two of Chlorine as well . And two are there on left hand side of Hydrogen and chlorine. Then equation is said to balanced. What my textbook writes just after it is that the $"2"$ before $HCl$ tells us that there are two moles of $HCl$ included in the reaction.  So my problem solicits that if $2$ is there to show no. of molecules then how it is defining the no. of moles at same time?

@SufyanNaeem Wow, I just flagged your comment above 'not constructive' because it has nothing to do with the question but now I realise how it might actually … You are badly misunderstanding things, and your question is a victim of the X-Y problem.

Chiefly the problem is with the definition given in textbook. Above problem is a result of that problem.

Please clarify what your actual problem is.

My actual problem is as stated in question. That problem is rising more problems when dealing with stoichiometry. Read my comment first and then question. I am sure, you will get my problem then.

A balanced chemical; equation shows the stoichiometry as you pointed out. The stoichiometry is a just the "count". A mole is just a large counting number - sort of like a million, a billion, or a trillion. It is useful as a counting number in chemistry because it gets a large enough number of atoms or molecules so that that many atoms has a "useful" weight. You can weigh 12 grams of carbon on a "regular balance" but you can't weigh one atom of carbon. You can't just tie the mole to the molecular weight of carbon since carbon has multiple isotopes.

Alright @volunteers  Please tell me about this. "A balanced equation MUST have EQUAL numbers of EACH type of atom on BOTH sides of the arrow." Now My book at a place says that the $2$ prefixed with $HCl$ is telling us that there are two atoms of hydrogen and Chlorine each included in reaction. Next to it, it says that that $2$ is defining no. of moles (of $HCl$) included in reaction. But according to first definition given, $2$ moles of $HCl$ contains $2 \times 6.022 \times 10^{23}$. This really mixes up.

@SufyanNaeem You should read that $2$ as meaning $2x$ and read any (unwritten) $1$’s as $x$. The $2$ just says ‘double the number of where there is no $2$’. In no way at all is the meaning of that $2$ tied to the definition of a mole. And in no way at all does it say ‘exactly two of type $x$’. Otherwise the reaction would stop after two molecules of $\ce{HCl}$ have reacted.

@MaxW, I am close. Just give me your decision whether the definition given in my textbook of mole is correct or not?

The first line of your question is the correct definition of a mole.

And other one???

Is practically always equivalent within experimental error.

‘Yes (unless you ask a theoretical physicist)’

@Jan you say that $2$ is a sort of or a coefficient (simply a number) which doesn't tell us the no. of moles but atoms however. $2H_2O$ doesn't mean $2$ moles of water but it calculate us the total number of atoms however included in reaction. There are 4 hydrogen atoms and 2 oxygen.

Sufyan I don't know exactly how your book's statements are worded to explain the confusion which we are dancing about. In chemistry we slip between talking about individual molecules in a chemical reaction and the number of moles involved rather frequently. So one mole of zinc reacts with two moles of HCl in the reaction, or one molecule of Zn reactions with two molecules of HCl. The real point here is that mass (or weight) isn't such a count. So 1 kilogram of Zinc and two kilograms of HCl are not the correct proportions for the reaction. That is why chemists need "moles."

$2$ is a coefficient. And it only tells us the ratios of different entities.

The mole is a base unit as specified in the Système international d’unités (SI) by the bureau international des poids et mesures. Its decisive definition is that published in French:

La mole est la quantité de matière d’un système contenant autant d’entités élémentaires qu’il y a d’atomes dans 0,012 kilogramme de carbone 12 ; son symbole est « mol ».
Lorsqu’on emploie la mole, les entités élémentaires doivent être spécifiées et peuvent être des atomes, des molécules, des ions, des électrons, d’autres particules ou des groupements spécifiés de telles particules.
La mole est une unité de base du Système international d’unités.

The proposal was brought forth by the International Union of Pure and Applied Physics (IUPAP), the International Union of Pure and Applied Chemistry (IUPAC) and the International Organisation for Standardization (ISO). As with all SI texts, the decisive French version has a semi-official English translation:

The mole is the amount of substance of a system which contains as many
  elementary entities as there are atoms in $0.012$ kilogram of carbon 12; its
  symbol is “mol”.
When the mole is used, the elementary entities must be specified and may be
  atoms, molecules, ions, electrons, other particles, or specified groups of
  such particles.

(The third point is not translated.)
This definition is more or less identical with the one in your first paragraph.
In practice and well within experimental error, this means that your later definition will hold true for any substance. I.e. take $1~\mathrm{mol}$ of an entity and the combined mass of that mole will be the same numerical value in grams as a single entity has in atomic mass units ($\mathrm{u}$). It is not the correct definition, and for any entity that is not carbon-12 the masses will differ slightly (but well within margin of your macroscopic experimental error) but it is good enough for most contexts.

A redefinition of the SI units is being discussed and will likely be adopted at the 26th General Conference of Weights and Measures in autumn 2018. This would redefine the mole in a way that the Avogadro constant is defined to be numerically exactly $6.02214 \cdot 10^{23}~\mathrm{mol^{-1}}$ (with a few further digits appended to the end of the number that yet need agreement). This would mean that the new definition of the mole would be along the lines of:

The mole, mol, is the unit of amount of substance of a specified elementary entity, which may be an atom, molecule, ion, electron, any other particle or a specified group of such particles; its magnitude is set by fixing the numerical value of the Avogadro constant to be equal to exactly $6.02214X \cdot 10^{23}$ when it is expressed in the unit $\mathrm{mol^{-1}}$.

Currently, the Avogadro constant must be measured experimentally giving a value of $6.022140857(74)~\mathrm{mol^{-1}}$; the digits in brackets express the numerical uncertainty.
This will mean that $1~\mathrm{mol}$ of carbon-12 atoms will no longer have the mass of exactly $12~\mathrm{g}$ (but again, it will be well within experimental error for everybody not practising theoretical physics).

To answer the follow-up question you asked in the comments: The coefficients in chemical equations such as
$$\ce{Zn + 2 HCl -> ZnCl2 + H2}$$
are always and exclusively to be understood as ratio coefficients. Thus, instead of thinking one atom or one mole of zinc, think amount $n$ of zinc and amount $2n$ of $\ce{HCl}$.


text:
What is more acidic: D3O+ in D2O or H3O+ in H2O and why?
What is more acidic:
$\ce{D3O+}$ in $\ce{D2O}$
or
$\ce{H3O+}$ in $\ce{H2O}$
and why?
I think it's $\ce{D3O+}$ in $\ce{D2O}$ as I saw somewhere that this property is used in mechanistic studies (the inverse isotope effect), but I need a proper explanation.
EDIT: I found a statement about this in Clayden:  Water $\ce{H2O}$ is a better solvating agent for $\ce{H_3O^+}$ than $\ce{D2O}$ is for $\ce{D_3O^+}$ because O–H bonds are longer than O–D bonds hence $\ce{D_3O^+}$ in $\ce{D2O}$ is stronger than $\ce{H_3O^+}$ in $\ce{H2O}$. That is the origin of the inverse solvent isotope effect.
This is in contrast with all the answers given yet, so I am pretty confused.

related http://chemistry.stackexchange.com/questions/31587/are-there-any-known-chemical-properties-of-tritium-water-that-make-it-unusually

https://en.wikipedia.org/wiki/Heavy_water#Physical_properties_.28with_comparison_to_light_water.29
here you can see that pH of $\ce{D2O}$ is 7.43 which is more than that of water...therefore light water is more acidic. but I dont have the explaination.

the pH of heavy water should be more than that of normal H2O bcoz it is difficult to break the DO bond

But the pKw of D2O isn't 14 right?

For the reasons explained in New point of view on the meaning and on the values of $K_\mathrm{a}(\ce{H3O+, H2O})$ and $K_\mathrm{b}(\ce{H2O, OH-})$ pairs in water Analyst, February 1998, Vol. 123 (409–410), the $\mathrm{p}K_\mathrm{a}$ of $\ce{H3O+}$ in $\ce{H2O}$ and the $\mathrm{p}K_\mathrm{a}$ of $\ce{D3O+, D2O}$ are undefined.
The entire point of the above reference is that
$\ce{H3O+ + H2O <=> H2O + H3O+}$
(which would correspond to an equilibrium constant of 1) is not a genuine thermodynamic process because the products and reactants are the same.
$\ce{D3O+ + D2O <=> D2O + D3O+}$
would also correspond to an equilibrium constant of 1
So when Clayden and the OP write

$\ce{D3O+}$ in $\ce{D2O}$ is stronger than $\ce{H3O+}$ in $\ce{H2O}$

it is wrong for the above reason.
Two genuine thermodynamic equilibriums are
$\ce{2H2O <=> H3O+ + HO-}$ and $\ce{2D2O <=> D3O+ + DO-}$
Experimentally, the self-dissociation constants of $\ce{H2O}$ to $\ce{H3O+}$ and $\ce{OH-}$ and $\ce{D2O}$ to $\ce{D3O+}$ and $\ce{OD-}$ can be measured as in The Ionization Constant of Deuterium Oxide from 5 to 50 [degrees] J. Phys. Chem., 1966, 70, pp 3820–3824 and it is found that $\ce{H2O}$ is about 8 times more dissociated (equilibrium constant is 8 times greater).
But using the above data to say $\ce{D3O+}$ is stronger is misleading, because this corresponds to a reaction with $\ce{OD-}$, not $\ce{D2O}$.  $\ce{D3O+}$ simply has a lower concentration in heavy water than $\ce{H3O+}$ has in light water.
As for why the $\ce{D2O}$ is less dissociated than $\ce{H2O}$, The ionization constant of heavy water ($\ce{D2O}$) in the temperature range 298 to 523 K Canadian Journal of Chemistry, 1976, 54(22): 3553-3558 breaks the differences down in to enthalpy and entropy components, which both favor ionization of $\ce{H2O}$ and states that $\ce{D2O}$ is a more structured liquid than $\ce{H2O}$. Not only the bonds of each product and reactant molecule need to be considered, but also the intermolecular forces: the number and strength of intermolecular hydrogen bonds for each species.  See Quantum Differences between Heavy and Light Water Physical Review Letters 101, 065502 for recent (2008) experimental data.
Numerous references characterized $\ce{D2O}$ as "more structured" than $\ce{H2O}$, meaning more hydrogen bonds, and a more narrow distribution of hydrogen bond lengths and angles.  According to Effect of Ions on the Structure of Water: Structure Making and Breaking Chem. Rev. 2009, 109, 1346–1370 "It is indeed generally agreed that heavy water, $\ce{D2O}$, is more strongly hydrogen bonded (structured) than light water, $\ce{H2O}$."  My explanation would therefore be that there is a greater penalty for placing ions in $\ce{D2O}$ than $\ce{H2O}$ as far as disruption of a hydrogen bonding network.
Also the equilibrium constant for
$\ce{H2O + H2DO+ <=> HDO + H3O+}$
can be measured and it is 0.96 according to Isotopic Fractionation of Hydrogen between Water and the Aqueous Hydrogen Ion J. Phys. Chem., 1964, 68 (4), pp 744–751

Explanation of Normal/Inverse Solvent Isotope Effect
For a kinetic normal/inverse solvent isotope effect
there will be a reactant and transition state.  If (for example) there is a single solvent exchangeable proton that is the same group in the reactant and transition state, for example, $\ce{ROH}$ in the reactant and $\ce{R'OH}$ in the transition state, switching solvents from $\ce{H2O}$ to $\ce{D2O}$ will either favor the reactant or the transition state relative to each other (considering the respective $\ce{OH}$ bond strengths as well and intermolecular hydrogen bonds to solvent). IF $\ce{D2O}$ favors the reactant relative to the transition state (activation energy is increased), this is a "normal kinetic solvent isotope effect".  Oppositely, if $\ce{D2O}$ favors the transition state relative to the reactant this is an "inverse kinetic solvent isotope effect." More complex scenarios involving more exchangeable sites can of course occur.
Similarly there can be equilibrium normal/inverse solvent isotope effect, if there is an equilibrium reaction and then it is reactant vs. product (rather than reactant vs. transition state) that matters.

But "why" is this the case?

@ron I don't think anyone knows the complete reasons.  In “Quantum Differences between Heavy and Light Water.” Physical Review Letters 101 065502 (2008) it is said that D2O is stabilized by more hydrogen bonds per molecule (3.76 vs. 3.62) http://phys.org/news/2008-08-scientists-quantum-differences-heavy.html

So what's then the explanation of the inverse isotope effect?

@Marko section 8.1.6 here: https://books.google.com/books?id=gY-Sxijk_tMC&pg=PA437&dq=%22solvent+isotope+effect%22&hl=en&sa=X&ved=0ahUKEwjrqcfVqLHJAhUJw2MKHRCyBI4Q6AEIKjAB#v=onepage&q=%22solvent%20isotope%20effect%22&f=false is  better than anything I can think of at the moment

This early paper reports that acidic compounds (phenols, carboxylic acids, and others) are noticeably more dissociated in $\ce{H2O}$ than $\ce{D2O}$. Since water (protio or deuterio) is just another example of a weak acid, it is not unreasonable to expect $\ce{H2O}$ and $\ce{D2O}$ to follow this pattern. This would suggest that the equilibrium constant in eqn I would be larger than the equilibrium constant in eqn II. That is, $\ce{H2O}$ in $\ce{H2O}$ should be more ionized than $\ce{D2O}$ in $\ce{D2O}$
\begin{aligned}
\ce{H2O + H2O &~<=> H3O^+ + OH^{-}&&(I)}\\
\ce{D2O + D2O &~<=> D3O^+ + OD^{-}&&(II)}\\
\end{aligned}
The reason why this is the case is the interesting question.
Since equilibria are governed by thermodynamic (rather than kinetic) effects, we must examine the relative stabilities of the reactants and products in these two equations in order to determine the relative positions of these two equilibria.

On the reactant side, $\ce{D2O}$ is more stable than $\ce{H2O}$ since an $\ce{O-D}$ bond will have a lower zero-point energy than the corresponding $\ce{O-H}$ bond. Hence the deuterated molecule will have a lower overall energy compared to its protonated analogue.
On the product side, the deuterated ion must again be more stable than the protonated ion for the same zero-point energy reasons. However, since the bonds in the $\ce{X3O^+}$ ions are weaker (longer) than the bonds in the neutral reactants,  the zero-point energy differences would be compressed or lessened.

The reactant effects discussed in the first bullet will tend to push equilibrium II further to the left than equilibrium I. The product effects discussed in the second bullet will tend to push equilibrium II further to the right than equilibrium I. In other words, the two effects are in opposition, but as explained in the second bullet, the product effect is likely smaller than the reactant effect so that overall the reactant effects win out and equilibrium II lies further to the left than equilibrium I.

Could you elaborate a little this conclusion: Generally, acidic compounds (phenols, carboxylic acids, etc.) are more dissociated in D2O than H2O (ref). This would suggest that the equilibrium constant in eqn (I) would be larger than the equilibrium constant in eqn (II)

@ron are you sure the bonds of the ions are longer?  In liquid, H2O has O-H bond length of 1.01  Angstroms by diffraction and D2O has O-D = 0.98 Angstroms. For hydronium I see 1.002 A calculated in liquid http://www1.lsbu.ac.uk/water/hydrogen_ions.html and smaller lengths for hydroxide  http://www1.lsbu.ac.uk/water/ionisoh.html but not clear if this is in liquid.

No, there's really not much that I'm sure about. By "ion" I meant the $\ce{X3O^+}$ species, not the hydroxide, and would expect the weaker bonds in those $\ce{X3O^+}$ ions to also be longer. I've edited the answer to clarify that point.

Therefore D3O+ in D2O is stronger, right?

No, equilibrium II lies further to the left than equilibrium I, therefore the $\ce{D2O}$ case is less acidic.

If it lies to the left more, that means that D3O+ is a stronger acid as it is on the right side of the equation. I'm confused now.

No, look at equilibrium II.  If it lies more to the left, then there is less $\ce{D^+}$, so the $\ce{D2O}$ case is less acidic than the $\ce{H2O}$ case.  Do you see that?

Equilibrium II lies further to the left than equilibrium I.  Therefore, equilibrium II has** less** $\ce{D^+}$ present, than equilibrium I has $\ce{H^+}$ present.  Therefore, equilibrium II is less acidic than equilibrium I.

Okay, I understand your reasoning that D2O is less acidic than H2O, but Clayden et al. it is stated that: Water H2O is a better solvating agent for H3O+ than D2O is for D3O+ because O–H bonds are longer than O–D bonds hence D3O+ in D2O is stronger than H3O+ in H2O. That is the origin of the inverse solvent isotope effect. I will expand my question.

When comparing $\ce{H2O}$ and $\ce{D2O}$, I can't help but think that the biggest different between the two molecules is their vibrational properties. Specifically, $\ce{H2O}$ is going to vibrate at a frequency which is $\sqrt2$ times larger than that of $\ce{D2O}$ based on$$\nu=\frac{1}{2\pi}\sqrt\frac{{\kappa}}{{\mu}}$$ because they will have virtually the same bond force constant , but the reduced mass of a $\ce{DO}$ bond is (basically) twice that of $\ce{H2O}$.
Thus, I would speculate that an explanation for why $\ce{H2O}$ might be more easily deprotonated lies in the fact that it vibrates at a substantially higher frequency and so it is instantaneously in higher energy states and thus closer to overcoming the energy barrier intrinsic to breaking a bond.
This of course says nothing about the relative stabilities of $\ce{H3O+}$ and $\ce{D3O+}$, but ron deals with that issue in his answer quite well and his analysis corresponds with the claim of my answer too.
It would be interesting to know if tritium is less acidic than deuterium (assuming it is) at a proportion which is smaller than that same proportion between deuterium and protium because that idea would align with the vibration explanation I believe. I can't find that data anywhere though.

This website has some very accessible information on liquid water vibrations if anyone is interested.
Also, feel free to disagree. I'm not completely sold on this suggestion, but it seems like a valid explanation.

a DO bond has a "reduced mass"?

Ya. That's what $\mu$ represents. It's a way of reducing a two body problem into a one body problem which is commonly used for vibrating systems. Essentially you're looking at the vibration of the center of mass rather than the vibration of the two atoms. So, all bonds have a reduced mass not just OD. Sorry if that was unclear.

Nice, I think we are describing the same thing on the "reactant" side. The zero-point energy I refer to is what you are describing using the reduced mass and vibrational energy, see here for example

Well, if you are interested in organic chemistry, you might be aware of the fact that the hyperconjugation effect of -CH3 is greater than -CD3. The reason for this is that the atoms in the molecules keep vibrating and the bonds of H being slightly weaker than D are easily delocalised. (due to heavier weight of D than H, the vibrations are weaker and thus the bonds are stable). The same is the case with water molecules where the bonds between O and H are weaker than O and D bond. Hence H2O is slightly acidic than D2O.


text:
Supercooling, freezing point depression, and finding the right freezing point
Pure liquid can be supercooled below its freezing point. Conversely, impurities in the liquid will cause a freezing point depression. How then is the freezing point accurate?

And what you mean by accurate?

@user11111 The impurities in your second proposition are meant in a different sense than "pure" in the first one. They are homogeneous (i.e. dissolved, like salt in water), while the impurities that prevent supercooling are heterogeneous, like sand, otherwise they won't be able to start crystallization.

Melting (and boiling) points are only defined for pure substances. Rather than defining them as ‘the temperature when pure substance $\ce{X}$ melts/boils’ they are better defined as:

The melting point is the temperature at which for a certain defined pressure (usually $1~\mathrm{bar} = 10^5~\mathrm{Pa}$) the liquid and solid phases of a pure substance are at equilibrium.
The boiling point is the temperature at which for a certain defined pressure (usually $1~\mathrm{bar} = 10^5~\mathrm{Pa}$) the liquid and gaseous phases of a pure substance are at equilibrium.

(I didn’t copy that from anywhere, I made it up. Apologies to everyone who I was unintentionally quoting without attribuition.)
Thus, you can supercool pure liquids but if you allow the supercooled liquid to reach equilibrium at its supercool temperature (and standard pressure), it will solidify. Similarly, you can overheat solids but if you let the overheated solid reach equilibrium at the overheated temperature, it will liquidify. Thus, neither the supercool liquid’s nor the overheated solid’s temperature can be considered the melting point, because neither phase is stable at equilibrium.

Melting (and boiling) points are ideally defined for the "pure" substances. But "pure" is a rather relative concept. With impurities the melting (and boiling) points will change slightly. However if the change is 1% per percent of impurity then for a substance with 1% impurity that actually melts at 50 degrees Celsius, (273+50 = 323 Kelvin) the observed melting would be slighly lower. Consider the table below:
  Purity     M.P. (Kelvin)
    99%        319.770
    99.9%      322.677
    99.99%     322.968
    99.999%    322.997

So the purity required for the melting point determination depends partly on the precision required for the melting point. If you just need +/- 1 degree Celsius, then 99% purity would be adequate for most chemicals. (Remember I made the 1% per percent impurity change up...)
For supercooling a liquid the idea is to observe the solid and liquid phases together. Remember that supercooling a liquid has a slight heat capacity compared to the amount of heat liberated when the solid forms. So the liquid can get below the melting point with a loss of heat and become supercooled, but when the solid phases starts to form then the temperature will increase to the melting point. With further loss of heat the temperature then stays at the melting point until all of the liquid phase has solidified, then the temperature of the whole solid phase can drop.


text:
Change in Gibbs free energy of vaporization reaction
In my chemistry book, I see the equation
$$\Delta G_{rxn} = \Delta G_{rxn}^0 + RT \ln(Q)$$
where $\Delta G_{rxn}^0$ is the change in Gibbs free energy under standard conditions (1 atm pressure in this case), R is the gas constant, T is the absolute temperature, and Q is the reaction quotient. Consider the reaction
$$\ce{H2O (l) -> H2O (g)}$$
Clearly in this example, $Q = P_{\ce{H_2 O}}$, the pressure of water vapor.
Here is my confusion. The earlier expression for $\Delta G_{rxn}$ implies that anytime $P_{\ce{H_2 O}} = 1\ \mathrm{atm}$, we have $\Delta G_{rxn} = \Delta G_{rxn}^0$, which is $8.59 \mathrm{kJ/mol}$ for this reaction. We should see this regardless of temperature, according to this expression.
However, when $T = 100\ \mathrm{^\circ C}$, liquid water and water vapor are in equilibrium at $P_{\ce{H_2 O}} = 1\ \mathrm{atm}$! And if they are in equilibrium, it stands to reason that we should see $\Delta G_{rxn}=0$, which clearly we do not.
Could anyone help me understand this apparent contradiction? My only guess at the moment is that it arises because $\Delta G_{rxn}^0$ is not temperature-independent and at the higher temperatures, perhaps the value of $8.59\ \mathrm{kJ/mol}$ becomes inapplicable. If anyone could explain this to me I would be very grateful!

Hi and welcome to chemistry.stackexchange.com. Feel free to take a [tour] of the site. I improved your usage of MathJax with the \ce{...} command for chemical formulae. It has the nice benefit of being intuitive, i.e. $\ce{2 H2 + O2 -> 2 H2O}$ gives $\ce{2 H2 + O2 -> 2 H2O}$. Learn more about this in the [help] (where you will also find helpful other stuff), this meta post or this one.

Yes. You are correct. $\Delta G^0_{rxn}$ is a function of temperature, and is equal to zero at 100 C for this reaction.

The change in state equation of water is a “half reaction”, but your equation is a phase change reaction.
SI units are confusing with gases:
Temperature is actually 298K in this case (it’s a trick because standard conditions mean temperature is set at 298K and pressure at 1 atm)
Reaction quotant = [products ÷ reactants] = as a  ratio
  $$\rm R = 8.314~J K^{-1} mol^{-1}$$
$$\rm \Delta G = \Delta G^\circ~of~8.59~\frac{kJ}{mol}~ – RTln1 = 8.59~\frac{kJ}{mol}$$
[the little degree symbol means heat of formation / formation energy [Δ= products - reactants] (in this instance I'm guessing its the evaporation energy for the change in state ? anyway) because evaporation of $\ce{H2O}$ is an exothermic reaction and
Other useful Eqs:
$$\Delta G = \Delta H – T\Delta S$$
$$pV=n{\rm R}T$$

You are right $\Delta G^0_{rxn}$ is not temperature independent.
$$\Delta G (T,P)=\Delta G^0 (T) + RTln(Q)$$
When we say standard most of the time we mean a static point, but in this case it's a dynamic line along P=1bar.


text:
Why do some old masters paintings deteriorate faster than others?
Specifically, Wikipedia states that Sir Joshua Reynolds' "The Age of Innocence" was painted by the artist over another one of his paintings because the latter was "had suffered some paint losses," and goes on to say that "Innocence itself has deteriorated since 1859." Why?

The materials used today could have a greater integrity than the ones used in 1859. Also other techniques may be used nowadays in order to preserve the work which my not have been practiced back then.

@Joel: Thank you. That's not what I asked, though. Paintings before and after Reynolds' fare better. What did Reynolds use, or refrain from using, in his paints, that made his work fragile?

I don't know, the paints may not have binded to the material properly or they've denatured somehow..

@Joel: Well, duh. Why do you think I posed the question? Exactly: in order to find out. I can make up theories myself. I'm looking for facts here.

That's why I added a comment not an answer. I was only trying to help sorry I can't provide answer for you.

@Joel: Understood and much appreciated.

Some pictures where lead-based paint was used tend to deteriorate due to formation of black PbS. However, I have found that the cause of this is the artist's 'defective' technique, which caused flaking of paint. It's the matter of technique, not chemistry.

So how exactly was his technique defective? What did he do wrong?


text:
What is the reaction between diethyl malonate and hydrazine?
For the reaction of diethyl malonate with $\ce{NH2NH2}$ in aqueous medium in presence of $\ce{CH3CH2OH}$, I found two types of product with SciFinder:

In presence of $\ce{H2SO4}$, the reaction changes to:

$\ce{EtOH}$ is a good leaving group, but can you explain why there are two different types of products? With what mechanism can they be explained?

The second reaction looks similar to the Nazarov cyclization. The sulfuric acid must dehydrate and provide the protic properties required for the ring formation.

Can you please explain in detail about Nazarov cyclization?

I can't provide a proper answer in the acceptable format for this site on my cheap phone sorry but type it in to Google there's plenty of credible information about it (:

Try @ Nazarov cyclization reaction Wikipedia.


text:
Dealing with different units (psi) in ideal gas equation
My assignment is as follows:

A sample of air is trapped in a container at $\pu{16.2 psi}$. If the initial temperature is increased from $\pu{10 ^\circ C}$ to $\pu{32 ^\circ C}$, what is the final pressure?

I am really confused as to what $\pu{psi}$ is and how am I supposed to convert it to figure out the final pressure.

Strictly speaking, you don't even need to know what psi is. Just apply the ideal gas law and return your answer in the same units.

Be careful here, the constant R has many values, some of which works with psi. But yea, apply ideal gas law with the appropriate R and you don't need to concern yourself with psi. R is approximately 5.961 ft3 psi K^-1 lb-mol^-1

Another question is "What does the number 16.2 mean?" Is this the reading on a really accurate (automobile) tire pressure gauge, and is this supposed to help you predict your tire pressure when the temperature rises from 50F to 90F? If so, add 15 (psi) to the 16.2 psi in order to get the absolute pressure from the gauge pressure; then do the calculation. I love it when teachers connect the theoretical world to the practical world.

In the given experiment, there is no indication that the volume of the container changes and/or a reaction takes place. Both the volume and the amount of substance can therefore assumed to be constant.
Let's assume that the enclosed air can be treated as an ideal gas.
Under these conditions, the pressure is proportional to the absolute temperature (in Kelvin):
\begin{align}
p &\propto T & \frac{p_1}{p_2} &= \frac{T_1}{T_2}
\end{align}
This relation is known as Amonton's Law.

I am really confused as to what psi is and how am I supposed to convert it to figure out the final pressure.

Using the equation above, you would only have to convert the temperatures to Kelvin and could give the final pressure as $\pu{psi}$ too.
If you insist on conversion to Pascal ($\pu{Pa}$) as your favourite SI unit for pressure, remember that a pressure is defined as a force acting on a surface (area):
$$\pu{1 Pa} = \pu{1 N//m^2} = \pu{1 kg//m s^2}$$
The $\pu{psi}$ is a still rather common unit and it stands for pounds per square inch.
This is double weird, since both the force and the area are expressed in non-SI units.
Note that the pound is not the German Pfund ($= \pu{0.500 kg}$), but the angloamerican pound with $\pu{1 lb} = \pu{0.453 kg}$.
The area part in $\pu{psi}$ is given in square inches with $\pu{1 in} = \pu{0.0254 m}$.
In summary, $\pu{1 psi} \approx \pu{6895 Pa}$.

The pound in psi is not pound-mass, but pound-force, the gravitational force on a sample with a mass of a pound on earth at sea level. As long as you are using the Kelvin scale, everything but the psi will cancel out so it does not matter which units everything else is expressed in.

@KarstenTheis:(being nitpicky here, but I think it's important to mention) "..expressed in".. until everything else is the same unit (${}^\circ C$ or K)


text:
Why tungsten does not have stable configuration?
Whenever there are 4 unpaired electrons in d-orbital, elements try to attain stable electronic configuration. For example copper.
Electronic configuration of copper should be [Ar]3d4 4s2, but it tries to attain stable state and so it changes to [Ar]3d5 4s1.
I have seen it happens with most of the elements but not with tungsten. Electronic configuration of tungsten is [Xe]4f14 5d4  6s2 and not [Xe]4f14 5d5  6s1(more stable).
What is the reason behind that?

If it was more stable, it would be adopted, so it isn't more stable. Anyway, the electronic configurations of the transition metals are often quite hard to rationalise, there are many factors that will tip the balance in favour of one configuration or another.

@orthocresol yeah that was my question why 5d4 is more stable then 5d5. Can you sate few factors that... I think you can add it as answer.

@jan not really. There is no where mention of tungsten configuration (at one place there is but that is not what I am asking,right?)

The basis of that question tells you that the electron configuration you termed ‘stable’ is, in fact, ‘abnormal’. So tungsten actually assumes the normal configuration. Thus, I consider it a duplicate.

@Freddy Aesin's answer does give some mention to tungsten at the end. In general, the principles behind the effect are given in the linked question so the same answer applies here.


text:
Reversible and irreversible processes of thermodynamics
My text book states the following:

(Edit:It should be p1/p2, sorry.)
My understanding of a reversible process is that it us a process that is executed so slowly that no energy would be dissipated at all and it is possible to bring it back to the original state again due to this.
(The classic jar and piston example)
An irreversible process is every other process.
I understood the derivation for work for reversible processes but for the irreversible process I am unable to relate to the formula.
Take the example of the jar again.
If we suddenly compress the gas, we are changing the external pressure, right? What would we substitute for p ext ?
Also, how did this formula come and how does it accommodate the extra work we had to do that gets dissipated to get to the same state?
Please correct my understating if it
Is wrong.

it will be $P_1 / P_2$, not $P_2 / P_1$

For both reversible and irreversible processes, the pressure of the gas at the interface with the surroundings exactly matches the external pressure imposed by the surroundings $P_{ext}$, and the work done on the surroundings is always equal to $\int{P_{ext}}dV$.  (The equation you wrote for an irreversible process applies only if $P_{ext}$ is held constant).  However, for an irreversible process, the pressure of the gas at the interface is not given by the ideal gas law, because (a) there are viscous (dissipative) stresses that contribute to the pressure at the interface and (b) the pressure and temperature of the gas is not uniform within the cylinder.  So we need to specify how $P_{ext}$ varies during the process (say, as a function of the volume V) manually in order to analyze an irreversible process.
For more details on this, see the following link:  Calculate Work Done for Reversible and Irreversible Adiabatic process


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Understanding observations of dissolving surfactants in methanol, water and hexane combinations
I added various solvents to a shower gel product to see what would happen. The main surfactant ingredients in it are Methyl Gluceth 20,  Sodium Laureth Sulfate, and Lauryl Glucoside.
First I diluted the sample in water, and added it to a separating funnel along with hexane. After shaking and leaving overnight, the lower water layer was clear and hexane layer was a viscous cloudy white. I extracted the water layer (which was a bit soapy but clear), leaving the hexane layer in the funnel.
I then added methanol to the separating funnel. Adding methanol immediately caused the white soap to clump together at the interface, and the hexane layer now appeared clear. (I don't think I shook the separating funnel in this step)
I then added fresh water to the separating funnel again. This caused the white to diffuse into the lower methanol/water layer so that it could be extracted. The hexane layer was now clear.
I am trying to understand why these things happened. Here is what I think:

The surfactant molecules (e.g. SLS) have long hydrophobic tails, and hydrophobic heads. In water they will arrange themselves into micelles etc with the tails pointing in so they aren't exposed to the water. In hexane they will arrange tails pointing out so the heads aren't exposed. The majority of the molecule being hydrophobic, it will be more favourable to arrange tails-out in the hexane, so this is why the hexane layer went cloudy and water was clear (but still soapy).
When methanol was added, the surfactants clumped together at the interface, and this might be the surfactants arranging with their heads pointing to the hydrophilic methanol and the tails pointing to the hexane at the interface. But why didn't they do this with the water/hexane layers? Water is more polar than methanol (polarity index 9 vs 5.1) so maybe water is too polar even for the heads to be pointing to in that way. It probably isn't because of density because the density of SLS and others is a bit over 1 g/ml.
Adding water to the methanol layer maybe adjusted the polarity just right so that surfactant could dissolve in it, but I'm not sure how that works. Maybe since methanol has a slightly non polar region, that would help solvate the tails while the water solvates the heads?

Let me know if my thoughts sound correct or if there's something I haven't considered.