text:
Why do alcohols not undergo resonance?
Take for example $\ce{CH3OH}$. The lone pair on the oxygen atom may be donated to form a double bond and giving $\ce{CH3-}$ and $\ce{O+}$ ?
Is it because we cant have + charge on $\ce{O}$ atom?

Octet rule is broken on the carbon.

Please tell me do you the conditions for resonance to take place!!

To see a similar substance that can undergo resonance, look at phenol. (Alcohols like vinyl alcohol really overdo it, they isomerize to aldehydes).

It looks like I misread the question which meant the structure
$\ce{H_3C- =O^+-H}$ rather than $\ce{H_3C^- O^+}$ which I assumed previously. This structure is not possible because the carbon atom's octet is violated, as it has 10 electrons around itself. Carbon should always have an octet structure (if possible) or lesser number of electrons (like in carbocations) in the resonance contributor structures.

O would still have 8 el. but C gets 10 and that's the problem

oh sorry I shall edit.I got confused by the CH3- O+ part in the question

@Mithoron I have edited my answer.

You'd better integrate those parts, You don;t have to use this EDIT convention.


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Why is my rusting metal chainlink fence glowing bright orange and sparking?
This is a really practical question, but I can't find anything on Google.
My metal chainlink fence is rusting and I have noticed that where the metal from one wire makes contact with another wire there is a point of bright orange glowing which sparks if the metal is moved. It does not feel at all warm to the touch, though outside temp is hovering around 0C right now.
Back story: I noticed it tonight when I went out to toss the compost in the leaf bin, and saw a very bright orange glow. I got a light and inspected it a bit closer with my wife and she noticed a second spot on the fence. Both were at junctions in the chain, where there was significant corrosion. Those locations were both heavily corroded such that visible gouges were forming at the junction where the bright orange spot was. I gave the fence a kick and both spots threw off a good number of very bright white and orange sparks. It did this about 5 or six times then stopped, and the metal was no longer orange in those spots. It did not feel hot to the touch either, though I didn't touch it when it was orange. I'm perplexed but figured it seemed like a good questions for some chemists, as I am guessing it has something to do with the corrosion process.

Sounds like thermite

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text:
Converting a cis-alkene to trans-alkene
I was thinking about simple method to convert a cis-alkene alkene like cis-2-butene to trans-2-butene.
I came up with this method (the method can also be used to convert trans alkene to cis alkene):

First use a peracid like m-CPBA to convert the cis alkene to an epoxide:

Then I used the Wittig reagent $\ce{PPh_3}$ to get a betaine. For the $\ce{- O-}$ to be able to attack $\ce{PPh3}$, they should be on same side. So rotation about $\ce{C-C}$ bond:

Finally, the leaving of $\ce{POPh3}$:

I would like to know if my method is correct and if there are any other methods to bring about the conversion.

In cases of cis alkenes heating should suffice, the reverse way is actually quite interesting.

This method will convert trans to cis as well right?

Maybe. The theory is sound I think, not so sure about the praxis. We would have to dig into the literature.

I went through Jerry  March's book. This is from page 1374: https://books.google.co.in/books?id=by05kNKm_xYC&pg=PA1374&lpg=PA1374&dq=jerry+march+betaine&source=bl&ots=DsnZQB2oDQ&sig=nARlAk1_LUTtUCJKIFoSHexUAuU&hl=en&sa=X&ved=0ahUKEwjW_bPtvuDJAhUNJI4KHdxqAWwQ6AEIGTAA#v=onepage&q=jerry%20march%20betaine&f=false

Your method looks fine. There is another simple method (with three steps though).

Convert the the cis alkene to a dibromo derivative using bromine;
Use two equivalents of $\ce{NaNH2}$ to get an alkyne (elimination);
Selective reduction to trans alkene by using sodium in liquid ammonia.
However this requires that Hydrogen be present as a substituent on the double bond, on both carbons.

I know that method. Do you have anything else to add about my method?

For E/Z trisubstituted alkenes capable of E2 elimination the following sequence may apply. Add the elements of water in a syn, anti-Markovnikov fashion to the double bond. Remove the elements of water as p-toluenesulfonic acid via anti E2 elimination. The cycle below illustrates this point.

There are two other methods which I know, if you were to take a cis-1,2-diaryl alkene and expose it to UV light then you will excite the alkene into a 1,2-diradical. The $\ce{C-C}$ single bond in this is free to rotate.
It will then randomly form a $1:1$ mixture of the cis and trans isomers. As the photophysical properties of the two isomers are different (extinction coefficient) it is possible using light of a given wavelength to convert any mixture of cis/trans into a mixture with a given cis/trans ratio. A photostationary mixture. By changing the wavelength the ratio in the mixture changes.
Another method would be to react the cis alkene with a metal hydride which has a vacant coordination site such as $\ce{RuHCl(CO)(PPh3)2}$. Then do a β hydride elimination to form an alkene and the metal hydride again. This relates to the way in which trans fats are formed during the hydrogenation of plant oils (lipids with cis alkenes).

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Tips for maximizing yield, purity and crystal size during recrystallization
I have already asked a similar question on titrations before. In a few months, I have a test on chemical synthesis. It is marked on the percentage yield, purity and crystal size of the product. I do not have a lot experience in doing syntheses and I am looking for any tips that will help maximize the yield, purity and crystal size of my synthesis.
Most of the synthesis that we do, the products are obtained by recrystallization. The solid is dissolved in a solvent at a high temperature and then it is placed in a cool bath. Then the product will crystallize on the bottom of the beaker.
Reading through an analytical textbook, it states a few ways to improve yield and purity but it doesn't actually state how you would do it practically:

allow the solution cool down as slowly as possible. This will maximize crystal size.

maintain a low relative supersaturation of the solution, this will increase crystal size and improve purity

describes the process or digestion or Ostwald ripening where larger crystals grow at the expense of smaller crystals

Does anyone have any suggestions on how I would actually do these 3 above tips? Also does anyone have any other tips?

It's a bit confusing what you mean with yield. With recrystallisation you cannot increase the yield of a reaction, you need to start earlier than that. If you are talking about the yield of the crystal, then just make everything as slowly as possible. Say you have a saturated solution at 100 °C, then let it cool down to room temperature over a day or so. Also your starting crystal should be as pure as possible. Other than that, I have not really a clue. Many repeats though can help grow a crystal larger.

When I grew crystals of aluminium potassium sulphate I filled a jar with boiling water and placed it in a cooking pot that was also full of water. Doing this helps to maintain a constant temperature. I set the cooker to a low heat and began dissolving the alum until the solution was saturated. At this point I removed the jar containing the solution and suspended a piece of string for a crystal to grow on. This would be my seed crystal. I left the jar on a window sill to cool and waited.
A few days later a crystal had formed so I lifted out the string and set it aside. Then I reheated the solution to dissolve any crystals that had formed on the jar as they would compete with the suspended seed crystal, reducing its size. It is also a good idea to filter the solution at this point. Once I had finished this step I placed the string with the seed crystal back into the jar and placed it on the window sill again. After a week I moved the jar to a refrigerator (around 3 degrees C)  to further cool the solution. This left me with a pretty big, well shaped crystal.
I didn't do this but you could try evaporating off the water after the previous step to grow the crystal further.

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Horizons of a bachelor degree in chemistry
If I have a bachelor degree in chemistry, what master programmes can I apply for ? In other words, are the choices limited in chemical engineeirng, biochemistry,etc? Is it possible to apply for a master degree in computer science for example ?

It should be asked rather on Acaemia SE

I'm not sure if this would even be considered a good question there @Mith. Prolly too broad. I'm not sure though.

You would need certain prerequisite courses to get the master's degree in a different field.  For example, Boston University explains:

The Master of Science (MS) in Computer Science program is intended for computer professionals and for people who intend to move into the computer field from other areas of study (emphasis added).

So yes, you can come from a different background.

Applicants to the program are required to have a baccalaureate degree from a regionally accredited institution and proficiency in the areas covered by the following courses:
◾MET CS 201 Introduction to Programming
◾MET CS 231 Programming with C++ or MET CS 232 Programming with Java
◾MET CS 248 Discrete Mathematics
◾MET CS 341 Data Structures with C++ or MET CS 342 Data Structures with Java
◾MET CS 472 Computer Architecture

So it would be fine to switch to physics, biology, biophysics, or many other programs as long as you have the right prerequisite courses.
However, some schools only accept people wanting a PhD, and give a master's degree along the way, which you get to keep if you don't finish the PhD.


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Unknown white precipitate in electrolysis of water
I was electrolysing water using a copper anode and zinc cathode.  The water was from the tap and where I live is very hard.  I was using about 20V between the two electrodes.  As I applied the voltage, a white precipitate formed around the anode.  What is strange is that I can't replicate this using any other type of metal as the anode.  Any ideas on what this might be?   Or, I suppose a better question is: are there any ideas on how I can find out what this was?

What other anode materials have you tried?

I've tried iron, aluminum, and graphite.

The aluminum probably just anodized and stopped the current flow -- I'm guessing there was no bubbling there?  As for iron, was it actually iron, or was it a steel of some kind?

Yes, it was steel

Aha --- stainless steel?

No, it was from a computer case.  Once I left it for a few days after my electrolysis experiments it most certainly stained :-P It's all rusted now.

The slight bluish tinge to that white precipitate makes it look to me more like copper(II) hydroxide than anything to do with the hardness minerals, especially since you haven't observed it with other anode metals thus far.
One way to test this would be to make a whole bunch of the precipitate, by running the current for a long time, and then taking a small sample of the liquid with precipitate in it and adding an equal volume of distilled vinegar (5% acetic acid).  If it is indeed $\ce{Cu(OH)2}$, I think it should dissolve to form a clear blue solution of copper(II) acetate.
The only problem might be if 5% acetic acid isn't able drop the pH far enough, in which case you'd have to track down a more concentrated acid. One choice might be $\ce{HCl}$, which is available at, e.g., aquarium supply stores under its historical name of muriatic acid. You'd then get a solution of copper(II) chloride, the color of which will vary depending on the $\ce{Cl-}$ concentration. Another option, per MaxW's comment, would be pickling vinegar (10% acetic acid) -- in fact, a quick search just revealed that Amazon has several vinegar options available in higher acidities.

How would I determine if it was produced?  I'm a bit new to electrochemistry but does it have something to do with the overvoltage I used?

@MichaelStachowsky As re overvoltage, not really. Anodic electrolysis of copper in neutral aqueous media is likely to make $\ce{Cu(OH)2}$ regardless of the applied potential, as long as it's large enough to get electrons flowing.

Since the water was noted as hard, I'd guess some sort of copper (II) carbonate would be in the ppt as well. // "Picking vinegar" is 10% acetic acid, as opposed to "regular vinegar" which is 5%.

@MaxW Agreed, $\ce{CuCO3}$ is certainly possibility, depending on the type of hardness in OP's water. Either way, acidification should solubilize it, I figure?

Metallic copper at the anode works as an electrode producing $\ce{Cu(OH)2}$, the slight blue pricipitate you observed in the course of reaction.  The $\ce{Cu(OH)2}$ gradually releases water creating a black precipitate of, CuO. Please read our work in which we described the homemade reactions to synthesized copper(II) aspirinate and acetate:
[1]
[2]

This may be a matter of language but a sacrificial anode is not what copper is  in this case.


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Particle size of oxides of nitrogen and sulfur pollutants and HEPA filters
I am thinking about buying air purifier as pollution is high in my city.
Going through the specifications of HEPA filters, I found that they block particulate matter of 2.5. PM 2.5 contains dust and other fine particles.
Googling tells me that their size is around 2.5 micrometers.
My question is: Do pollutants such as sulfur oxides and nitrogen oxides (and related compounds) fall into this category, specifically in regards to size?

If the high efficiency particle air (HEPA) filter says it blocks PM 2.5 that's good as pm 2.5 particles are the ones your lungs cannot easily dispel and cause haze.  Unfortunately sulfur oxides and nitrogen oxides are gasses, so no particle filter will remove them.


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Should I include solvation in quantum chemical calculations of a simplified protein-ligand interactions?
I am running a B3LYP-D3/6-31+G geometry optimization of a protein-ligand interaction. Well, not quite, I have simplified the protein to four amino acids surrounding the active site, but I keep the entire ligand (170 atoms total).
I am wondering if I should optmize the geometry in solution, i.e. will including solvation effects make my results (energies) be more realistic? My mind is split.
On the one hand, I am modelling a biological system which usually takes place in aqueous solution. On the other hand, my system is so simplified, that it no longer has much root in reality. By "placing my molecules in water", the water molecules will interact with parts of my system that otherwise would not interact as strongly (there would be much less water inside the protein than outside it, I mean). So, I feel there is no guarantee that my results will improve, and I am not able to find any data indicating one or the other. If I included the entire protein, then the case would be different, perhaps, since at least then that protein would be "solvated correctly".
So, given the simplified system, will solvation improve the resulting geometry, and thus the interaction energy between the protein residues and the ligand, or not?

Solvation corrections at this level of theory will not do you much good. Unfortunately B3LYP/6-31+G is probably far too simple for that system in any case. Calibration is necessary. I am obviously not familiar with your project, so any advice is flawed to begin with. Have a look at this question and you might see how difficult it can be to even selct the appropriate functional, not to speak of basis set or any other shortcomings.

I should add that I use the B3LYP-D3 method. Still, you think solvation will not benefit much? Your answer in the linked question was insightful, thank you for the link.

In case you are interested in binding to protein, the method doesn't mean much, a fair amount of research is done using molecular mechanics of all things. What is more important is to approximate solvation using actual solvent molecules and some stochaistic or systematic method to find the global minimum.

I don't think there is a definitive answer to this.  The best thing is to do it both ways to see if it makes a difference.  See for example this paper http://dx.doi.org/10.1002/jcc.21458

This is difficult to answer without full knowledge of the system, but some tips:
1) The overall energy of the system is arguably irrelevant.  Rather, the relative difference between different conformations, and perhaps, different ligands is of utility.

2) The level of theory and basis set your using is quite low.  If you have the time and the resources, scale it up.  For biomolecules I've found the minnesota functionals to be most accurate (if insisting on DFT) but I still believe MP2 is king.

3) Referring back to #1, if you solvate, then your comparison should too be solvated.  Perhaps your approach would be to run a solvated and unsolvated model for both your ligand, and perhaps, another known ligand. This could too help convince you of the utility of the solvation model.


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What is a dry electron?
I was reading about water photolysis which produces a positive water radical and a dry electron, or an excited water $\ce{H2O^*}$ through the following equation
$$\ce{
H2O ->[$hv$] H2O^{.+} + e_{dry}^-, or~ H2O^*
}$$
(as seen in this link)
By opposition I have also seen $\ce{e_{aq}^-}$ (as seen in pp. 2 of this link).
I can understand the "positive water" and even the "excited water", but what is a "dry electron"? And what is an $\ce{e_{aq}^-}$ (wet electron? I am unsure of how to translate the term I had learned to english). Why not simply say "free electron"?

There are a zillion hits on a google search. It is evidently an electron that retains some of the kinetic energy from whatever interaction created the free electron. https://books.google.com/books?id=W_d6qz2YMzMC&pg=PA157&lpg=PA157&dq=%22dry+electron%22&source=bl&ots=3ViHtmy7Oc&sig=kHnAw9yVJONSovdjPsBDdvI6P4s&hl=en&sa=X&ved=0ahUKEwjazort6uDJAhVM_mMKHS-wCh44ChDoAQgrMAQ#v=onepage&q=%22dry%20electron%22&f=false

@MaxW that doesn't explain the term "dry" which is the focus of my question. By opposition I suppose there is also the $e_{aq}^{-}$ ? (adding this to the question)
Example on pp 2 in https://www3.nd.edu/~ndrlrcdc/Compilations/rxn.pdf

$e_{aq}^-$ is quite a natural thing; that's a hydrated electron. Just rip an electron off something and let it go in the water (not that it would survive there for long, but anyway). But dry electron sounds pretty weird to me.

@Sosi - :-( again - A "dry electron" is a free electron which  still has some of the kinetic energy from the event which created it. When it loses enough kinetic energy so that it becomes solvated then it is $e^{-}_{(aq)} $.

Yes, I got that; I just wanted to note that it still sounds strange to me.

@MaxW, would you like to post that as an answer please?
Thanks everyone for the patience to answer such rookie questions!

A "dry" electron is an electron released from the event of an ionization. Depend on source the electron can either be defined as:

"The $e^-$ is a 'dry electron' that still retains some of the excess
  kinetic energy acquired from the ionizing event."$\ce{^{[1]}}$

or,

"The 'dry-electron' (defined as an electron before solvation)
  ..."$\ce{^{[2]}}$

A dry electron is thus, an electron that is not assiciated with any atoms, molecules or solutions.
Reference
$\ce{[1]}$ Thormod Henriksen, David H. Maillie; Radiation and Health; page 157
$\ce{[2]}$ J.D. Simon; Ultrafast Dynamics of Chemical Systems; page 144


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Does light absorption result in a momentarily weaker bond?
I may have just misunderstood the topic, but light absorption by a molecule is a result of a promotion of an electron into an empty antibonding orbital; hence, does that mean that the strength of the bond is momentarily weakened by absorption of light? What processes account for the actual radiation part (release of heat) associated with the absorption of light?
http://www.chemguide.co.uk/analysis/uvvisible/bonding.html

Well, the electron is in a different orbital (has an excited wave function) - that required an input of energy in to the bound system. The energy difference between the excited state and the ground state may either move to a different electron on the molecule, be re-radiated as a photon, go in to vibration or rotation modes, or some combination of all of those.

Right, but since the excited wave function is antibonding, does that mean that the bond is weakened until the electron can return to ground state?

Kind of by definition, yes. The binding energy of the molecule as a whole is reduced by the energy of the absorbed photon. Of course, certain photon energies will result in the molecule falling apart  (which is a clear indication of the bond being weaker!).

@JohnSmith Depends. Some molecules exists only in excited states.

Woah, I have never heard of an entire molecule falling apart from interactions with light. Could you give me an example?

@JohnSmith https://en.wikipedia.org/wiki/Diazirine#Chemistry . Again, excitation may lead to strengthenning of a bond.

Good as the example is, I'd rather point out that any molecule can fall apart from interactions with light, given that photons have enough energy. Also, the orbital from which an electron is excited is not necessarily bonding, and the one to which it gets promoted is not necessarily antibonding, so yes, in some (relatively rare) cases certain kind of excitation can make a bond stronger rather than weaker.

@IvanNeretin Not entirely correct. One-photon adsorption results in one-electron excitation. If a bond has order more than one, one-electronexcitation can't reduce it's order to zero. There is quite a set of molecules with double and triple bonds. (though gamma-photon interacting with a nuclei of the molecule will surely break it apart, as the energy is extremely high. But it's beyond chemistry)

@permeakra You are right about the multiple bonds. But wait, your very example shows how two separate bonds can be broken simultaneously in a single-photon process.

@IvanNeretin That's because  the example is kind'a special case. Typically, when photolysis is employed, the molecule in question is inherently unstable towards fragmentation, so even slight change in bonding causes fragmentation. The example molecule has 1) high strain and 2) double N=N bond. It is already ready to break and does so from a funny look.

True. So let's add the word "almost" to my statement, since almost all molecules have single bonds somewhere. As for those which do not... well, I'm not quite sure whether or not $\ce{CO2}$ can be broken by a single-electron excitation.

Another thing to consider - don't think just of 'orbitals' - the electronic states of a molecule are more distributed than around a lone atom. By moving an electron to a different state, you have changed the total electron density function, which could impact the molecule as a whole. An analogy is defect levels  (like the As interstitial in GaAs) in semiconductors - if you change the charge state by electron capture, the atomic positions of the interstitial and surrounding lattice atoms changes. Similarly, exciting an electron may result in a different configuration for the entire molecule.

Molecules dissociating with light is very common, Oxygen & Ozone by UV in the atmosphere, for example, or Iodine vapour by blue-green light (<  approx 500nm), or chlorine by shorter wavelengths.  The molecule NO has a shorter bond length in some of its excited states than it has in its ground state, indicating stronger bonding even though it will have absorbed about 5 eV of energy.

I think the most general answer to this question: depends on the system.  The ground and excited-states are characterized by orthogonal eigenfunctions (wavefunctions) of the Hamiltonian.  Each wf has an associated electron density matrix/function/distrobution which describes the density of electrons in 3-dimensional space.  It is the distribution of electrons relative to the nuclei which determines how some bonds may strengthen and others weaken upon excitation. @JonCuster is right to remind us that orbital-to-orbital single electron excitation is a drastic simplificaiton for many systems.

it may happen but it is not always necessary as 1S antibonding molecular orbital is lower in energy as compared to 2S bonding MO If electron jumps from 1S ABMO to 2S BMO  then BOND ORDER willhence bond will become stronger increase σ(1s) <σ∗(1s) < σ(2s) <σ∗(2s) < π(2px) = π(2py)


text:
Carboxylic acid strength due to the inductive effect
Suppose we have a mono carboxylic acid . In that if attach a +I group then sigma electrons will shift towards the compound and O-H bond will become weaker.  Hence H bond can easily break so it should be more acidic but it is not i.e greater the +I effect less the acidic nature

No clear to me what "+I" group is. But yes, the pKa of the carboxylic acid group is affected by the group attached.

Look up the Hammett equation https://en.wikipedia.org/wiki/Hammett_equation for some background of what you're trying to say.  Hope that helps.

+I means inductive effect

The strength of the carboxylic acid (and any other Bronsted acid, for that matter), is related to the 'stability' of its conjugate base, the carboxylate anion. The carboxylic acid and the carboxylate anion are in equilibrium with one another, and the relative acidity of carboxylic acids depends upon the position of this equilibrium.
One of the main reasons why carboxylic acids are acidic is due to the ability of the charge to be delocalised around the pi-system:

However, in addition to this delocalisation about the pi system, extra stability can be gained by the presence of electron withdrawing groups adjacent to the carboxylate group (i.e. -I groups attached to the R group in the diagram above).
These electron withdrawing groups draw electron density away from the carboxylate anion via the sigma system (the single bonds). This is advantageous as the stability increases with the number of atoms it can be spread across (i.e. how diffuse the charge is).
Electron releasing groups (+I as you call them) destabilise, by increasing the electron density around the carboxylate group, which of-course makes it less favourable for the carboxylic acid to lose a proton and hence makes it less acidic.
The +I groups aren't involved in breaking the O-H bond to any great extent, as you allude to. This is because when the proton leaves, it transfers its electron density to the oxygen it was attached to, cleaving the bond and allowing it to leave as H+. If the +I groups were donating into this bond, you would have to have the H leaving as H-, which is rarely observed.


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drawing epoxides, the 3d shape
given the image on the left, when you look at it from the side would it look like the image in the right? since the thicker lines means it's coming out of the page?

The oxygen will be going down: try and get access to Avogadro (I'd recommend installing it on your laptop) and draw the molecule.  A bit of googling gave this page, https://www.rollapp.com/app/avogadro it means you can draw the molecule and manipulate it without having to download the software.

From PubChem

So the molecule is sort of boat shaped.

the first illustration I drew is from the textbook. How does that illustration reconcile with the 3d shape you created?

Think of the chair and boat forms of cyclohexane. In this case the boat form would seem to be the more stable. I don't know what the ratio of the two forms is, but the ratio is surely a function of temperature.

The oxygen bridge strains the C-C bond. The "boat" form has the two hydrogens on the carbons with the bridged oxygen pointing away from each other. If the oxygen end were flipped into the "chair" position then the two hydrogens would be much closer together.

If I use chemapp and draw the structure of cyclopentane-1,2-epoxide, it looks like this:

You can see that the the epoxide group is kinda tilted upwards. This is perfectly understandable, since a trans geometry is unacceptable.

@Fl.pf. Thanks for pointing it out, its fixed now.

And thus, I shall remove my downvote


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Why is calcium stearate (soap scum) insoluble in water while potassium/sodium stearate are soluble in water?
So, potassium and sodium containing soaps work because they have a polar end and a non-polar structure for the rest of the molecule - they can thus be dissolved in polar and non-polar solvents. This is why they're used for cleaning, as they can basically clean off anything non-polar like oil while still lathering/rinsing off with water.
Now, my question is, what changes when 'hard soap' is formed? Here, when the stearate ion of soap is released in hard water, an ion of magnesium/calcium bonds with it forming 'hard soap'. The thing I don't get is why it is now insoluble in water. What's changed?
Based on electronegativity values, there's very little difference between atoms of potassium, sodium and calcium/magnesium, not enough to change the polarity. The ion from the hard water bonds with an oxygen atom, and the electronegativity difference between an atom of calcium/magnesium and an atom of oxygen is still large enough for it to be polar at the same end as it was with sodium/potassium.
What's going on?

'Hard soap' isn't totally insoluble in water, it is just less soluble that the sodium or potassium forms. In making lye soap the ancient process was in fact to "salt" out the soap by adding salts (sodium chloride usually...).

@MaxW That makes sense, but why exactly is it it so much less soluble than the sodium/potassium forms? The electronegativity change doesn't account for it whatsoever, is there another reason?

The "hand waving" reason - With a soap that has a monovalent anion, a cation with a +1 charge still has a "polar" end for the "molecule." With a divalent cation the two polar ends are both attracted to the cation, thus the polar part is in the center of large molecule with no polar end.


text:
Why does this E2 reaction occur on the less substituted carbon?

In compound C it eliminates the beta hydrogen attached to the less substituted carbon, rather than attacking the beta hydrogen attached to the tertiary carbon. Why is this the case?

E2 elimination is stereospecific for anti elimination. In both variants of compound B that you have drawn, the OTs and the proton on the tertiary centre are syn to each other, meaning that the elimination is disfavoured.
For this reason, the NaOMe deprotonates at the secondary centre, generating the products as drawn, despite the fact that this goes against the Zeitsev rule which favours the most substituted alkene.


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Equilibrium constant and thermodynamic relationship
Why is Equilibrium constant (Ratio of rate constants) considered as a thermodynamic quantity whereas the rate constant as kinetic quantity. Moreover why is extent of reaction also a thermodynamic quantity.?
Please explain briefly as I'm stuck.

Well first think of thermodynamics as determining what a system wants to do, and kinetics as what the system is allowed to do.  The rate  constant gives how fast something will react (kinetic rate).  The thermodynamic equilibrium constant gives how much of something will react.  Kinetics are dictated by the movements of atoms, Thermodynamics is dictated by the changes in energy and entropy.

But how will you justify your statement
Thermodynamic equilibrium constant gives how much of something will react.
Moreover, what exactly is meant by  extent of reaction, and why  is it a thermodynamic entity?

We have a simple relation between Gibbs free energy change and the equilibrium constant for a given reaction as:
delta G naught= -RTln(K)

There is a relationship between the equilibrium constant and the forward- and reverse kinetics reaction rate constants:  $$K=\frac{k_f}{k_r}$$This relationship can be derived directly by setting the rate of the forward reaction equal to the rate of the reverse reaction at equilibrium.  This determines the extent of the reaction.  You can also derive a relationship between the thermodynamic heat of reaction and the activation energies for the kinetic forward- and reverse reaction rate constants.

So, what exactly does thermodynamics tell about a chemical reaction (Whether feasible or not)?
I didn't get the extent of reaction part in your answer, could you please elaborate it more, as I have an interview.

If you know the initial concentrations of the reactants and the equilibrium constant, you can calculate the final concentrations of reactants and products.  This is what I called the extent of the reaction., and this is what thermodynamics tells you.  If you want to design a reactor to achieve a certain conversion of reactants to products, you need to know the kinetics so that you can determine the size of the reactor needed.

Okay so, does that mean that thermodynamics can only be applied to equilibrium reactions.?

Most people would say that thermodynamics can only be applied to equilibrium situations.  Most would not accept the concept of determining the thermodynamic functions S, U, G, A, H for states of a system that is not at equilibrium (i.e.,  is undergoing an irreversible change).  I'm not one of those people, and I am not alone.  In my judgement, it is valid to determine the thermodynamic functions locally within a system not at equilibrium, and then integrating to get the functions over the entire mass of material.

okay, a small doubt, difference between a thermodynamic variable and kinetic varible

A kinetic variable tells you how fast a reaction is taking place.  A thermodynamic variable does not address rates; for example, the equilibrium constant just tells you the state of your system at final equilibrium (when the net rates are all zero).


text:
Functional Isomer of Formic Acid is?
I have read like ''Functional Isomers of Carboxilic Acids are Esters'', then my doubt is; What about Formic Acid?

Well, if you'll try to make a functional isomer of formic acid in the same manner, you'll end up with formic acid. Guess that makes it an exception.

Note the image of acetic acid (top) and the image of the carboxylic structural isomer methyl formate (middle).  As you can see the bonding location of the proton and the methyl group switch between the structural isomers.  Note the image of formic acid (bottom).  If you switch a hydrogen and a hydrogen you still get formic acid.  Yes, structural isomers of carboxylic acids make esters, but formic acid does not have a structural isomer, so the rule does not apply.
 Acetic Acid
 Methyl Formate
 Formic Acid


text:
What is the most compact way to store Oxygen?
What is the most compact way to store Oxygen? Perhaps I should qualify that as being at room temperature since the most compact form of oxygen is probably liquid oxygen. At room temperature would it be better to store it as rust or compressed air or something else, like hydrogen peroxide?

Degenerated matter ;)

Rust? Why not water, then? Or... wait, your body stores a good deal of oxygen, albeit in a chemically bound form. What exactly do you want?

@IvanNeretin I think the question is pretty clear. If an organic compound is the most compact way to store water, then what is that compound? Water is a bad answer because hydrogen pyroxide (which I already mentioned) stores oxygen more compactly than water. Did you even read the full question.

@ShakaBoom Did you even write the full question? If you are really interested in packing as many as possible atoms of O in a given volume, regardless of the energetic factors (this kinda strikes me as odd, but... none of my business, really), then indeed $\ce{H2O2}$ (about 65 moles of O per 1 liter) is a pretty strong contestant and definitely better than water (55 mol/l). However, $\ce{H2SO4}$ (molar mass 98, density 1.84) is even better than that.

Nitrogen Dioxide would be the best as it is liquid and will spontaneously decompose in the presence of the right catalyst

Solid ozone has a density of 1.728 g/cm3 according to The Density of Solid Ozone J. Am. Chem. Soc., 1959, 81, pp 805–806.
Best I can think of at atmospheric pressure, assuming your goal is to store and recover as much pure oxygen as possible in a given volume.
If your goal is just to pack as many oxygen atoms in a given volume, BeO, which has a density of 3.01 g/cm3 is better.
At high pressure there is metallic oxygen.


text:
how to find valency of calcium using spdf electronic configuration
Electronic configuration of calcium is $\ce{1s^2 2s^2 2p^6 3s^2 3p^6 4s^2}$. The orbitals are fully filled with electrons so would its valency be two or zero?

The configuration $$1s^2 2s^2 2p^6 3s^2 3p^6 4s^2$$ has a valency of zero since there are as many electrons as protons. The configuration $$1s^2 2s^2 2p^6 3s^2 3p^6 $$ is two electrons short, so it has a valency of +2.

Yes, the valency is 2. It's all orbitals are completely filled, but it's outer most orbit is incomplete. Ca can easy donate 2 electrons and can become stable cation.


text:
Is ammonium hydroxide safe to be used for sanitizing food?
At least one big fast food chain uses ammonium hydroxide for making "pink slime" edible by killing off harmful micro organisms: https://en.wikipedia.org/wiki/Ammonium_hydroxide
Is this product from this process safe to consume?

The FDA and local health inspectors do a good job controlling food safety in the US. looking at it backwards, that is why some food contamination problem is such big news in the US now. In contrast to the industrial plants, or restaurants, most home kitchens and food preparation would not pass a health inspection.

In pure form, aqueous solutions of ammonia ($\ce{NH3}$) cannot be consumed. Solutions containing 5 % ammonia or more are considered “corrosive” (causing severe skin burns and eye damage).
However, in reasonable quantities or in dilute form, aqueous solutions of ammonia can be generally used safely in food. They are sometimes named “ammonium hydroxide”, “ammonia aqueous”, or “aqua ammonia”. The food additive code (the E number, which is commonly found on food labels) of aqueous solutions of ammonia is E 527.
This food additive is approved in the European Union as well as in the United States.
For example, aqueous solutions of ammonia are used in food processing as pH control agent (acidity regulator), leavening agent, surface-finishing agent, or additive to boiling water.

Generally one should not consume ammonium hydroxide, however I do believe food grade ammonium hydroxide can be purchased and is safe in small quantities.


text:
What is the pressure belonging to in Boyle's Law?
Just a hour ago, I was reading the great and highly interesting, Boyle's Law. In my textbook, there is given Boyle's Law and leaving some pages, there is stated Dalton's Law of Partial Pressure.
Boyle's Law is stated as, "As so long as the mass of the gas and temperature is constant, the product of pressure by volume of the gas remains constant."
Experiment of U-tube is well known. I am not going to copy it here but I would directly explain my problem. From U-tube experiment, I grasped that the pressure, mentioned there, is one that is applied and not of the gas. However, the difference stays vague in the law stated earlier that whether this is the pressure applied by external agency or the pressure applied by gas molecules or gas.
Later, I read there the Dalton's Law of partial pressure. There is a primary equation concerning to partial pressure of a gas viz.,
$$\frac {P_{gas}}{P_{mixture}}=\frac{m_{gas}}{M_{mixture}}$$
It is clearly mentioned there that this pressure either $P_{gas}$ or $P_{mixture}$ is the pressure applied by gas/mixture and not the pressure applied by external agency. This is even more clear from the relevant experiment.
The problems start when I begin to analyze the derivation for the equation above. Author of my textbook hasn't given any derivation for it but my teacher let us did that.
Derivation:
Consider  the mixture of gases A, B and C.
Let the number of moles of gas    $A=n_A$
Similarly, number of moles of gas $B=n_B$
And that of C is given by $n_C$
Total number of moles=$n_T=n_A+n_B+n_C$
We know that, $$P=\frac{nRT}{V}$$
Therefore pressure applied by gas A is given by, $\color{red}{P_A=\frac{n_ART}{V}}$
For gas B, $\color{red}{P_B=\frac{n_BRT}{V}}$
And for gas C, $\color{red}{P_C=\frac{n_CRT}{V}}$
For mixture of A, B and C, $\color{red}{P_T=\frac{n_TRT}{V}}$
Diving $P_A$, $P_B$ and $P_C$ individually by $P_T$ we get,
$$\frac{P_A}{P_T}=\frac{n_A}{n_T}$$
$$\frac{P_B}{P_T}=\frac{n_B}{n_T}$$
$$\frac{P_C}{P_T}=\frac{n_C}{n_T}$$

The red marked statements are where my previous concepts were denounced. I was thinking that the derivation maybe wrong but I am not sure so I want you people to judge my concepts and my teacher as well and inform me.

Thanks.

Your derivation of Dalton's Law of partial pressure is fine.
I think you are confused about how pressure is "applied" to a gas, and what that does to the internal pressure of a gas.
Imagine that you have a "sphere" of gas surrounded by a "force field" (like in Star Trek) in empty space, then the force field containing the gas is dropped. Since there is nothing to confine the gas it just expands into infinite space.
A lot of ideal gas models, like the Wikipedia article on Boyle's Law, show the gas as being contained in a cylinder with a piston pushing on the gas to create the volume and pressure of the system. The external force on the piston is counterbalanced by the pressure of the gas times the surface area of the piston. The pressure of the gas is the same on all surfaces of the container. The gas doesn't just "press back" on the piston.
Pressure and volume are inversely related for the same system. So at the same temperature,
$PV = k$
Thus if the pressure increases, then for the same amount of gas the volume must decrease to keep $k$ a constant.

Suppose I put a gas in a container and put a piston over it and moreover put a weight on piston so that the pressure applied on gas becomes $x$. The gas pressure would balance it and we would say gas pressure is also $x$. If I do it with an other gas, it would have $x$ pressure too. Now put these gases in a common container with same external pressure $x$ and now mixture would have same pressure $x$.  That would mean that the mixture of non reactive gases and component gases (individually) would always exert 760 torr pressure when kept under standard pressure i.e. 760.

Right.  But in that case, you would end up  twice the volume of gas, and the partial pressure of each gas would be 380 torr.


text:
order of basic strength
2 is the least basic as the lone pair is involved in aromatizing the ring
1 is the maximum basic as one of the lone pairs is used to aromatize the ring and the other lone pair is available
but how do we distinguish between 3 and 4
the options are:
a.1>3>4>2
b.3>2>4>1
c.4>3>2>1
d.3>4>2>1
by the above discussion we can make out that the answer is a.
but how to distinguish between option 3 and 4. how can we say that 3 is more basic than 4.

All your answers are wrong, and that's particularly bad question. 3 and 4 are more basic than 2.

As Mithoron points out this is a bad question.  The basicities of 3 and 4 are close and they flip-flop depending upon whether we are discussing gas phase or solution basicity.  In solution (you can google the pKas) the correct answer would be 3>4>1>2. You are correct that 2 is least basic because the lone pair is involved in aromaticity.  2 is more basic because the most available of the 2 lone pairs is in an sp2 orbital.  Finally 3 and 4 are the most basic because their lone pairs are in an sp3 orbital.

I think 4 should me more basic than 3 since $ -CH_3 $ has $ +I $ effect due to which electron density on $N$ increases.

but according to the answer it is 3>4

How can you say 1 is most basic? both 3 and 4 too have localized lone pair and since in 1 Nitrogen is $sp^2$ hybridized its basic strength decreases with respect to 3 and 4.

1 is the most basic because the lone pair on nitrogen(the one that is at the bottom)will delocalise to make the ring aromatic, and the other lone pair is always available(a stronger base is one which can donate lone pairs).

In both 3 and 4 lone pairs are available too and 3°N is more basic than 2°N

authors: Юлия Курочкина


text:
Use of combinations to determine number of isomers
Our teacher asked us how many geometrical isomers are possible with the formula $\ce{c2ClBrIf}$.
I answered with the logic $^4C_2=6$ and it turned out to be correct.
Is the use of combinations this way generally helpful in determining the number of possible isomers? For example, if I have a rigid molecule with three carbon atoms and six substituents (perhaps a cyclopropane), is the correct number of possible isomers 20?
$$^6 C_3 =\dfrac{6!}{(6-3)!\cdot 3!}= 20$$

LOL - The reason?!? - You got lucky since you didn't understand the concept of isomers.

@MaxW - Actually, the approach described in your answer is equivalent to the statement $^4C_2$, which is the number of distinct combinations of a set of four taken in pairs - in this case a set of four substituents divided over two carbon atoms. It is evaluated equivalently mathematically as well: $$^k C_j = \dfrac{k!}{(k-j)!j!}$$ $4!=24$, and you determined that only one in four structures is a different isomer (as every isomer can be drawn in four degenerate ways), and $(4-2)!\cdot 2!=4$. $\frac{24}{4}=6$

@BenNorris - Yes and no. You need to look for the general principle not just a particular example. // Your edit greatly changed and expanded the question...

What is $\ce{c2ClBrIf}$? Do you mean $\ce{C2ClBrIF}$?

For the following I've numbered the positions on ethylene where the four halogen atoms could go. (The Carbon-Carbon double bond doesn't "spin.") It doesn't matter which halogen goes to 1, so let's make it Fluorine. If Fluorine were in any other position then you could flip or rotate the molecule to get Fluorine to position 1. Now there are three choices left for position 2, which leaves 2 choices for position 3. So the number of possible isomers is:
$3 \times 2 = 6$
Note that if you just willy nilly drew molecules then you could draw 4! or 24 permutations. However when you rotate and flip the molecules, then only 6 unique molecules exist.

Okay I get what MaxW was trying to explain there. But here with another logic can we just not take in account the case of circular permutation and get the total number of geometrical isomers as 3!=6?


text:
Refrigeration for Cryochemistry −35 °C
I need to cool a tube containing liquid to −35 °C for a cold chemical experiment that will last about a week. The glass vessel I need cooled is 500 mL in volume and will be giving off $\ce{H2S}$ gas. I currently only have a fume hood. Most freezers do not seem capable of this low temperature. Most cryochemistry refrigeration seems designed for much lower temperatures, and is needlessly expensive. I realize that refrigeration is relatively simple: 1 compressor, 1 expansion valve, two heat exchangers, and electronic control/monitoring. My first impulse is to modify a freezer, but I am completely new to this. At this point I would be very happy with some general advice and perhaps a reference for custom designing these sorts of cooling devices.

Please note that you should never store flammable liquids or gases in a normal household refrigerator or freezer, which have various ignition sources (lights, switches, compressor motors) that could ignite vapours in the unventilated cooling compartment. The flash point of hydrogen sulfide (the lowest temperature at which it can vaporize to form an ignitable mixture in air) is approximately −82 °C.

I would suggest a Frigorific Mixture of calcium chloride (often sold as DampRid) and Ice in a cooler.  You will need quite a bit of each and to moniter at least twice a day, but it wont break the bank and can reach the desired temperature.

That or a Cooling Bath seem ideal.

If your going to use dry ice I would suggest a [Dewar condenser]  [1] as is typically used in such cryogenic applications, noteably Birtch reduction
[1]: http://ivyscientific.com/images/categories/CD-0054.jpg


text:
Can you write (with and ink pen) on hydrophobic paper?
Would the ink be repelled from the paper or prone to smearing?

http://chemistry.stackexchange.com/questions/27384/what-forces-hold-ink-on-paper

This depends on what your ink is based on. If it can be washed off ordinary paper using water, it will be repelled on hydrophobic paper. However, if it can't, it has a chance of staying on.
Note that the paper in particular might repel all or most liquids, not just water. In this case, the ink has a very low chance of staying on the paper.

Most inks are hydrophobic I believe if we're talking printer inks or pen inks.


text:
Writing equilibrium expressions with aqueous species
If we have the reaction: $$\ce{A +BC_{(aq)} <=> D}$$ where $\ce{A}$ and $\ce{D}$ are generic compounds (neither is pure solid nor pure liquid) and $\ce{BC}$ is an ionic compound dissociated in water (consisting of $\ce{B+}$ and $\ce{C-}$ ions), would the $K_{\mathrm{c}}$ expression be written as
$$K_{\mathrm{c}}=\frac{[\ce{D}]}{[\ce{A}][\ce{BC}]}$$
or
$$K_{\mathrm{c}}=\frac{[\ce{D}]}{[\ce{A}][\ce{B+}][\ce{C-}]}?$$

The fact is that BC(aq) does not exist, so the equation is not really "correct" - we just use that as shorthand for B+(aq) + C-(aq).

I will say the latter form is more correct, and in fact an important way to consider equilibrium with solubility and the common ion effect.  You should not count any solids in equilibrium expressions though as they do not determine the final equilibrium.


text:
Experimental determination of pKa
I have $20~\mathrm{cm^3}$ of a $0.1\ \mathrm{M}$ ethanoic acid solution and have to find out the $\mathrm{p}K_\mathrm{a}$.
In experiment this solution had a $\mathrm{pH}$ of $2.91$.
Do I use the formula $\mathrm{p}K_\mathrm{a}= 2 \mathrm{pH} + \log(c (\mathrm{acid}))$ ?
What I do not understand is that my $\mathrm{p}K_\mathrm{a}$ value varies. I measured many solutions with 18:2, 15:5, 10:10 sodium ethanoate/ethanoic acid composition. Which I calculated with the formula $\mathrm{p}K_\mathrm{a} = \mathrm{pH} - \log \frac{c(\mathrm{salt})}{c(\mathrm{acid})}$
I get close values but not the equal.  So the one above would be $\mathrm{p}K_\mathrm{a}=4.82$   and another value I obtained is $4.62$ (for the 15:5 solution with $\mathrm{pH}\ 5.11$).
Is this deviation purely due to experimental inaccuracy or does the composition matter.   Would it make more sense to use the literature $\mathrm{p}K_\mathrm{a}$ value for further calculation rather than those experimental ones?

A pH of 2.91 is far more acidic than a 0.1 molar solution should be. How did you measure pH - I assume a pH meter. So how was the meter calibrated?!?

@MaxW no it's not... ! Adam, can you be a bit more precise in what you did, I don't really understand what you did during your experiment :/

@Shadock - You're correct. I actually did the calculation and the solution would be more acidic than I thought.

Do I use the formula $\mathrm{p}K_\mathrm{a}= 2 \mathrm{pH} + \log(c (\mathrm{acid}))$ ?

No, for the pure acide use:
$$K_\mathrm{a} = \frac{([\ce{H+}])^2}{[\ce{HA}]_\mathrm{initial}-[\ce{H+}]}$$
Then, if you are still having problems, see this question and its answer:
Difference in calculated pH and the real pH of a phosphate buffer?


text:
By which mechanism would azide react with alcohols or alkyl halides?

$\ce{NaN3}$ produces $\ce{N3-}$ ions. So the reaction with an alkyl halide will either be $\mathrm{S_N1}$ or $\mathrm{S_N2}$.  After some reading, I found that $\ce{N3-}$ is a good nucleophile. Hence the $\ce{S_N2}$ mechanism is followed.
But if I use $\ce{HN3}$ on an alcohol, the $\ce{H+}$ ions produced will help in removing $\ce{OH-}$ as water forming a carbocation. So either $\mathrm{E1}$ (if the alcohol is tertiary) or $\mathrm{S_N1}$ (if the alcohol is primary or secondary) can take place.

Is this correct?

SN1 on a primary alcohol? Why can't you just have SN2 of azide ion with water as the leaving group?

This is what my Teacher told. I have seen reaction in SN2 where H2O leaves since OH- is a weak leaving group.

The first point seems correct though.

No primary carbocations!!! I really want to shout this at every first year organic chemistry student. You could get $\mathrm{S}{N}1$ or $\mathrm{E}1$ for a tertiary alcohol. The primary alcohol might undergo $\mathrm{S}{N}2$, but there's a good chance that it will be a bad reaction. Your best bet would be to convert the primary alcohol into a tosylate.

After some reading, I found that $\ce{N3-}$ is a good nucleophile. Hence the $\mathrm{S_N2}$ mechanism is followed.

Unfortunately, it is not as simple as that. Whether $\mathrm{S_N1}$ or $\mathrm{S_N2}$ is followed, depends on both partners but is more a function of the electrophile than one of the nucleophile. Typically, primary halides will undergo $\mathrm{S_N2}$ substitutions while tertiary ones will undergo $\mathrm{S_N1}$. Secondary ones can do both. However, note that secondary halides that are not 2-halopropanes will undergo Wagner-Meerwein rearrangements if a carbocation is formed.
In an $\mathrm{S_N1}$ reaction, the rate-determining step is the formation of the carbocation, hence weaker nucleophiles can react equally rapidly as stronger ones. In the $\mathrm{S_N2}$ reaction, the nucleophilicity of the nucleophile directly determines the rate of reaction, i.e. both partners are important. Hence these tend to prefer better nucleophiles.

But if I use $\ce{HN3}$ on an alcohol, the $\ce{H+}$ ions produced will help in removing $\ce{OH-}$ as water forming a carbocation. So either $\mathrm{E1}$ (if the alcohol is tertiary) or $\mathrm{S_N1}$ (if the alcohol is primary or secondary) can take place.

Again, primary alcohols will never react by an $\mathrm{S_N1}$ mechanism and secondary ones are very unlikely to do so. The $\mathrm{p}K_\mathrm{a}$ value of $\ce{HN3}$ is $4.72$ — which means that alcohols will practically not be protonated (you need strong acids for that, e.g. $\ce{HBr}$).
I would assume that $\ce{HN3}$ will not react with alcohols at all since it is too weak to protonate them. There will also be no elimination side reactions neither by $\mathrm{E1}$ nor by $\mathrm{E2}$ — a hydroxide ion will never be a leaving group in itself.


text:
How to determine the mole fraction of the liquid phase from a given mole fraction of the vapour phase for a benzene/toluene mixture?

A quantity of benzene, $\ce{C6H6}$, and toluene, $\ce{C6H5CH3}$, is placed in a $\pu{1 L}$ evacuated vessel at $\pu{25 ^\circ C}$. At equilibrium, a small volume of liquid is visible at the bottom of the container. A sample of the vapour phase is analysed and found to contain $\pu{53 mol-\%}$ benzene.
  What is the mole fraction of benzene in the liquid phase?
Vapour pressures at $\pu{25 ^\circ C}$: $P_\mathrm{vap}(\text{benzene})= \pu{0.125 atm}$, $P_\mathrm{vap}(\text{toluene}) = \pu{0.037 atm}$

First of all, I am not even certain if this a Raoult's law problem. I just assumed so, since Henry's constant was not given in the problem.
First I'm getting the total liquid pressure using Raoult's law.
\begin{align}
P(\text{benzene}) &= X(\text{benzene}) \cdot P_\mathrm{vap}(\text{benzene}) &
                  &= 0.53 \times 0.125 &&= 0.06625~\mathrm{atm}\\
P(\text{toluene}) &= X(\text{toluene}) \cdot P_\mathrm{vap}(\text{toluene}) &
                  &= 0.47 \times 0.037 &&= 0.01739~\mathrm{atm}\\
P_\mathrm{total}  &&&= 0.06625 + 0.01739 &&= 0.08364~\mathrm{atm}\\
\end{align}
Then using Dalton's law, I am getting the mole fraction of Benzene in the liquid phase.
\begin{align}
P(\text{benzene}) &= Y(\text{benzene}) \cdot P_\mathrm{total}\\
Y(\text{benzene}) &= \frac{P(\text{benzene})}{P_\mathrm{total}}\\
                  &= \frac{0.06625}{0.08364} = 0.792.
\end{align}
However, my answer is incorrect according to the answer key that has $0.25$ as the answer. I have a feeling my whole approach may be wrong since my answer is off by a lot. What am I doing wrong?

The question gives you the mol percentage of benzene ($53 \%$) in the vapour phase, and not in the liquid phase.
It's always good to write down what you have first. (note: I use $y$ to denote mole fractions in the vapour phase and $x$ to denote mole fractions in the liquid state)
First of all, we have $y_\text{benzene} = 0.53$. Additionally, we have been given the vapour pressure of pure benzene and toluene (denoted with a superscript $*$):
$$p^*_\text{benzene} = 0.125\ \mathrm{atm}  \qquad p^*_\text{toluene} = 0.037\ \mathrm{atm} $$
and we require $x_\text{benzene}$
We also, know $p_\text{benzene} = x_\text{benzene}\ p^*_\text{benzene}$, and the total vapor pressure is given by $$P = p_\text{benzene} + p_\text{toluene} = x_\text{benzene}\ p^*_\text{benzene} + x_\text{toluene}\ p^*_\text{toluene} = x_\text{benzene}\ p^*_\text{benzene}+ (1- x_\text{benzene})\ p^*_\text{toluene}$$
Finally,
$$y_\text{benzene} = \frac{p_\text{benzene}}{P} = \frac{x_\text{benzene}\ p^*_\text{benzene}}{x_\text{benzene}\ p^*_\text{benzene}+ (1- x_\text{benzene})\ p^*_\text{toluene}}$$
It is now, trivial to solve this equation. Since I am lazy, and was not in the mood to write down two lines of algebra and reach for a calculator, I chucked it into Mathematica, and the result is:

Thus, $x_\text{benzene} = 0.25$, which is indeed the result given in your answer key.

You solved the problem assuming that 0.53 is the mole fraction of benzene in the liquid phase.  However, the problem statement says that "the vapor phase is analyzed and found to contain 53 mole percent benzene."  Call P the total pressure. What is the partial pressure of benzene in the vapor?  What is the partial pressure of toluene in the vapor?  In terms of P, what is the mole fraction of benzene in the liquid?  In terms of P, what is the mole fraction toluene in the liquid.  In terms of P, what is the sum of the mole fractions of benzene and toluene in the liquid?  From this equation, what is the total pressure P?

Okay, so I retried again and was able to arrive at the correct answer with your steps, but I don't quite understand the chemistry behind it. Partial pressure of benzene in vapour is 0.53P, 0.47P for toluene. Mole fraction of benzene in liquid state is 0.53P/0.125 and for toluene it is 0.47P/0.037. This part, I am confused by why the partial pressure in vapour is used in Raoult's law when I am calculating mole fraction in liquid.

Raoult's law says that the partial pressure in the vapor is equal to the equilibrium vapor pressure of the pure substance at the given temperature times the mole fraction in the liquid.  So, the mole fraction in the liquid is equal to the partial pressure in the vapor divided by the equilibrium vapor pressure of the pure substance.

Thanks a lot! To clarify on the states, is it always the case that Raoult's law says partial pressure in vapor equals equilibrium vapor pressure times mole fraction of liquid? The partial pressure part is always in vapor right? However the mole fraction part can be liquid/liquid solution or liquid/solid solution?

Yes, for vapor-liquid equilibrium.  For liquid/liquid or liquid/solid solution, a different law applies.

According to Raoults law, it should be 0,25.
Lets say, you have the molar fraction of each component in vapor, and lets say you have the saturated pressure at some temperature, than according to Raoults law you can say:
0,53(molar fraction of benzene in vapor)Pi(which is equal to Xbpb(saturated, benzene)+Xtpt(saturated, toluene) = Xbpb
than:
Xb*0,125=0,53*(Xb*0,125+(1-Xb)*0,037)
Xb=0,25


text:
Does resorcinol give positive iodoform test?
It's given in my book that other than the usual compounds that answer iodoform test, resorcinol (benzene-1,3-diol) also gives positive iodoform test.
Is this because of tautomerization which gives a diketone with an active methylene group? But why should a ring lose its aromaticity by doing so?

There is an equilibrium between the keto enol forms, however small the keto form may be as a contributor. The rest is as you say. The precipitation of iodoform drives the equilibrium further to the products, as per Le Chatelier's principle.Of course, it is not in all cases where the equilibrium is controlled by such factors. The net gibbs energy change for the reaction must be negative.

Speculation aside, neither resorcinol nor phloroglucinol give positive iodoform tests. Fuson, R.C. and  Tullock, C.W., J. Am. Chem. Soc., 1934, 56, 1638.

@user55119 Note that the conditions used by the experimenters in the study are in fact quite mild, i.e. 2 min at 60 deg C. The following study which was also not done under vigorous conditons, at temperature of 20 deg C. However, the experimenters waited long enough for iodoform reaction to occur for resorcinol to give a positive test. Hence, they concluded that resorcinol does give a "slow test". Seelye, R. N., & Turney, T. A. (1959). The iodoform reaction. Journal of Chemical Education, 36(11), 572. doi:10.1021/ed036p572

Tan Yong Boon: Yes, if you pound on the substrate, I'm not surprised. Marrian (1932) treated estrone with KOH/I2 to obtain the dicarboxylic acid from the D-ring. Heer and Miescher (Helv. Chim. Acta, 1945, 28, 156) treated estrone benzyl ether with methanolic KOH/I2 at 80 oC to obtain O-benzyl marrianolic acid. Although the reaction presumably goes through the diiodoketone, I would be reluctant to call this reaction or that of resorcinol "haloform tests".

Yes, resorcinol gives iodoform test.
This is due to the fact that a keto- form of resorcinol exists (via tautomerism as suggested by the original poster), which contains an active methylene group in a cyclic compound, this being a positive condition for the iodoform test.
However, the reaction will be slower than expected and is only driven forward by Le Chatelier's principle which states that precipitation drives a reaction in the forward direction.
Any question in theoretical organic chemistry asking whether Resorcinol gives the Iodoform test will always specify the rate (eg. A question with the statement: Which of the following gives iodoform test instantly? has Resorcinol being a wrong answer).


text:
Using UV-C (254nm) light to cleave the amino acid off of vyvanse
Vyvanse:

Desired product:

Amide bonds absorb in the region of about 190 to 240 nm, so could using UV-C light conceivably cleave the amide bond?
Also I don't think this reaction is as simple as putting vyvanase powder under a UV-C lamp because the reaction intermediates probably need to be stabilized by a proper solvent or something, correct? What would the intermediates likely be, and what solvent would likely help?
Obviously the easiest way to get the desired product is probably simple hydrolysis, as there are not really any other hydrolyzable function groups on the starting material apart from the amide C-N bond. So this is mainly a theoretical question.

there seems to be something to the notion. http://home.physics.wisc.edu/~himpsel/456_jcp.pdf

Honestly, I wouldn't bet on a cleavage here.
In principle, excited state amides and imides can undergo electron transfer reactions. We did work a bit on that (DOI) and an early example on intramolecular electron transfer between an aromatic amide and a tertiary amine, resulting in amide cleavage, was given by J.D. Coyle (PDF).
In your case, however, the situation is a bit different. You don't have an aromatic amide but an arene and an aliphatic amide. The latter aren't supposed to exhibit a strong absoption at 254 nm, see an older article by Nielsen and Schellman (DOI).
Even if you'd manage to excite the amide, I'd exclude electron transfer from one of the primary amines in your starting material and hence the cleavage of the amide.


text:
Calculating molarity of a gas using ideal gas equation
Is hydrogen gas at 1 atm pressure and 298 K temperature considered as a 1 molar solution? If yes then can I use ideal gas equation $$PV=nRT$$ to find molarity as $$\frac nV = \frac P{RT}?$$

I get confused after reading the following paragraph .The standard hydrogen electrode consists of a platinum electrode coated with platinum black. The electrode is dipped in an acidic solution and pure hydrogen gas is bubbled through it. The concentration of both the reduced and oxidised forms of hydrogen is maintained at unity (Fig. 3.3). This implies that the pressure of hydrogen gas is one bar and the concentration of hydrogen ion in the solution is one molar. so I asked the question. It is Said that concentration of redox couple should be one for finding standard potential

Oh sorry, that homework part was not supposed to be there. But if you have questions that tend to go in that direction, then it's good if you know about the policy anyway.

Hydrogen at one atmosphere pressure and 298 K is certainly not a one molar solution. However, you can still calculate the concentration with the ideal gas. Note that this is a gaseous solution and since it is a pure gas, the solute is also the solvent.
$$
c(\ce{H2}) = \frac{n(\ce{H2})}{V(\ce{H2})}
           = \frac{p(\ce{H2})}{\mathcal{R}T}
           \approx 40.9~\mathrm{mol\,m^{-3}}
           \approx 0.04~\mathrm{mol\,L^{-1}}
$$

So it means that 0.04 molL-1 of hydrogen gas at equilibrium with one molar H+ ion solution will give standard electrode potential

Not quite, see wikipedia for the set-up of the standard hydrogen electrode. If you have any more questions about that, you should consider asking a new question: [ask].

I am unable to understand the folowing statement of Wikipedia :The concentration of both the reduced form and oxidised form is maintained at unity. That implies that the pressure of hydrogen gas is 1 bar

@JawadMirza you should ask a new question, the SE system works better this way. Anything we resolve in the comments won't benefit other users.


text:
Elimination reaction on 2-fluorobutane using alc.KOH
Question: Find the major product formed when 2-fluorobutane reacts with alcoholic $\ce{KOH}$
My answer: Elimination of $\ce{HF}$ takes place and a more substituted alkene should have formed. (Zaitsev product). I think the mechanism is E2 since strong base is used.
but answer given is a less substituted 1-butene. Is the answer wrong or am I making any mistake?

E1CB elimination takes place because of a poor leaving group. This is because of the high $\ce{C - F}$ bond strength. Stronger bonds are more difficult to break, making them less reactive.
If the reaction was E2, then zaitsev product would have been formed. But since the mechanism is E1CB, the reaction proceeds via carbanion.

From wikipedia: The first step of an E1cB mechanism is the deprotonation of the of β-carbon, resulting in the formation of an anionic transition state, such as a carbanion. The greater the stability of this transition state, the more the mechanism will favor an E1cB mechanism.

I hope you know that for carbanions, the stability order is primary > secondary > Tertiary.
Now that you know the stability order of carbanions, the preferred position for carbanion is a primary carbon. So base abstracts a proton from a primary carbon.

This answer is, unfortunately, not correct. There is no reliable experimental evidence to demonstrate that a primary carbanion is more stable than a secondary carbanion. In fact, the opposite is true (JACS 1989, 1968). And neither carbanion would be stable enough to be formed.
So, what's the answer? I would honestly say that no-one knows for sure. High level gas phase calculations reproduce the experimental outcome, so it's unlikely to be due to solvation.
It's a little unfortunate that this has become a textbook example, since the textbook explanations do not withstand scrutiny. Please understand, I am not criticising the person who responded. It just seems that this reaction is a bit more complicated than it seems at first, and (in my opinion) it needs a bit more work to understand it.
Sorry if this response is not very helpful, but it's better to tell the unpalatable truth.


text:
Standard Enthalpy of Formation for Branched Alkanes
I'm looking at the heat of formation for branched and straight alkanes and I've noticed a few odd trends I'm having trouble explaining. For straight-chained alkanes, the addition of carbon atoms to the main chain seems to consistently decrease the value of $\Delta H_\mathrm f$. This seemed logical as the compounds become more stable. However, for branched alkanes, I'm having a bit of trouble explaining them. The more bent it is the more stable it seems to be, as $\Delta H_\mathrm f$ decreases. But I'm not quite sure why.
Does it have something to do with geometry? This is a bit confusing because the boiling points of branched alkanes are less than that of their straight-chained isomers, implying that they're more stable. The location of the methyl groups seems to also have a part to play as if they're in the center (most spherical shape) they'd be least stable, but still, have higher $\Delta H_\mathrm f$ than straight-chained ones.

Why would you think the boiling point has anything to do with ΔHf?


text:
Can you tell me which one of these substances are highly or easily flammable?
List is next:
ethylene glycol (EG),
diethylene glycol (DEG),
triethy-lenglycol (TEG)
basicly liquids from polyol groups,
I know alcohols are flammable but what about polyols? and how easily

http://www.sigmaaldrich.com/ Look up the safety data sheets for the compounds and see whether they say "flammable".

they dont have all of them

They have all the ones you listed. Check your spelling of triethylene glycol. In any case I am fairly sure all of those are rather flammable.

The USA legal definition per 29 CFR 1910.106 of "flammable" is:

Flammable liquid: any liquid having a flash point below 100ºF (37.8ºC)

None of the three substances is flammable by that definition.
The respective flash points by the Penskey-Martens Closed Cup method are:
Ethylene Glycol: 241ºF (116ºC)
Diethylene Glycol: 310ºF (154ºC)
Triethylene Glycol: 350ºF (177ºC)

Update: Now the rule has been changed to:

Flammable liquid means any liquid having a flashpoint at or below 199.4 °F (93 °C)

How can I find flash point of these substances?

@SSimon I added flash points and reference to answer

thank you very much @DavePhD do you know if sparkling wire ( electrical current) above liquid can cause substance to flame? ( Current spark is around 500 C)

@SSimon if the temperature of the liquid is below the flash point, a hot spark above the liquid will not ignite the liquid or its vapor.

@DavePhD Is this hazard classification still valid in the US since the new OSHA Hazard Communication Standard (HCS) has adopted the Globally Harmonized System of Classification and Labeling of Chemicals (GHS)?

@Loong now the definition has been changed to "Flammable liquid means any liquid having a flashpoint at or below 199.4 °F (93 °C)" http://www.ecfr.gov/cgi-bin/retrieveECFR?gp=1&SID=6a2be1c9d6b4f12de646eba495c8ee3f&ty=HTML&h=L&mc=true&n=pt29.5.1910&r=PART#se29.5.1910_1106

@DavePhD do you remember about you question on academia that you asked about some author that stole your idea from Sceptics SE?

@SSimon yes, but he agreed to mention Skeptics in the paper, so everything is ok

@DavePhD https://chemistry.meta.stackexchange.com/questions/4354/i-need-permission-to-reuse-photo-from-user-porphyrin-answer-on-question-about?noredirect=1#comment7822_4354 can you help me with this, since you are most into it?


text:
VT graph thermodynamics
If we have one mole monoatomic ideal gas was taken through process AB as shown in the figure then ∆ S is positive and work is also positive but how can we find the value of work and heat absorbed as pressure is changing .k

From what you learned in middle school algebra, what is the equation for the straight line between points A and B in terms of V and T?  If you solve this equation for T and substitute it into the ideal gas law, what do you get for P in terms of V? What is the equation for the work in terms of P and V?  If you know the work and the change in internal energy, do you know the heat?

Yes but I forget that

Do you remember it now, or do you need to go back to your old algebra text book?  This is not really a chemistry question, it's a math question.

But can you help me

Can you help yourself? He literally spelt out everything you need to do. Start by finding $p$ in terms of $V$. Google "equation of a straight line through two points" if you need maths help.

The graph is of volume and temperature no $y= mx+b$  equation is going to help him.

@A.K., are you saying that you can't use the graph to determine the equation of the straight line V=mT+b, where m is the slope and b is the V intercept.  Or, better yet, as orthocresol indicates, you can't determine the equation using the point-slope version of the equation?

He's looking for pressure and heat, which are not graphed.

Since you know the number of moles of ideal gas, you can find the inital and final pressures with the ideal gas law: $PV = nRT$.  For entropy you may first take that $U = \frac{3}{2} nRT$, and that $dU = \delta Q + \delta W$. $ $  $\delta Q $ and $ \delta W$ may be substituted by $\delta Q = T \mathrm{d}S$, and $\delta W = -p\mathrm{d}V$. This should be enough for you to solvew the problem.

Why you have equate Q = Tds and in this how will we find ds

You find d$S$ because you know everything else.

W = -pdV but pressure is changing

Do it in two steps where you change pressure, then volume and add the two results.


text:
Is there a relationship between pH indicator colors and the visible light spectrum?
When I examined the colors showed on a pH paper which is red for acids and violet for bases, greenish yellow for neutral and other ranges in between, I noticed that the sequence is exactly same to that of visible light spectrum that is violet, indigo, blue, green, yellow, orange, and red.
Is there a connection between the two, and if yes what it is?

Yes there is a connection. pH indicators are typically chosen so that the human eye can observe the change in the visible part of the electromagnetic spectrum.  ph indicators https://en.wikipedia.org/wiki/PH_indicator

There is a connection, but not in the way that you think. The fact they happen to appear in the exact same sequence of the visible light spectrum is purely a coincidence.

It has to do with the way pH indicators and particularly large-range pH indicators work.
In simplest terms, a pH indicator is simply a weak acid/base which has one color in its acidic form, and one color in its basic form. When the solution is highly acidic, all of the indicator molecules are protonated, and the color of the indicator is the color of the protonated form. When the solution is highly basic, all of the indicator molecules are deprotonated, and the color of the indicator is the color of the deprotonated form.
Where it gets interesting is where the pH of the solution is close to the pKa of the indicator. In those situations, you have some of the indicator molecules in the protonated form, and some of them in the deprotonated form. You don't ever get "fractionally protonated" forms - each individual molecule is either one or the other. Since light absorption happens on a molecular level, you get
You can see this with a titration of something like Bromothymol Blue [image]. The protonated form is yellow, so at low pH, everything is yellow. The deprotonated form is blue, so at high pH, everything is blue. At around the pKa (~7), there's near equal amounts of the two forms, so it acts like you mixed equal amounts of a yellow pigment/dye and a blue pigment/dye. That is, you get a green solution due to the subtractive color mixing model. As you titrate the pH, you get a range of colors as you add and remove the amount of protonated and deprotonated forms, as would be indicated by the Henderson–Hasselbalch equation.
Okay, so that's a single pH indicator, which is normally limited to a small range. (Because more than ±1 pH unit to either side of the pKa you typically have only a single species present.) To get broad range pH indicators, you have to mix multiple different pH indicators, each with their own different colors and changes. But you have to do this intelligently, as the indicators are always present, and will always contribute to the color.
Here's a chart of some of the major pH indicators, their pH ranges and their colors. You can see what might happen with various indicator combinations. For example, with an ethyl violet/ethyl orange mixture, at low pH you'd have an orange solution (yellow+red), which would transition to be purple (blue+red) at around pH 3 and then to be green at high pH (blue+yellow). But those intermediate colors are going to be pretty muddy. At around 1.5 you'll have an orangy-purple, and at around 4.5 you'll have a greeny-purple.  Imagine what things would look like if you mixed orange paint with purple paint, or purple paint with orange paint.
What we need is a layout of color that can gradually transition one to the other and look good all the way through the approximately 7-ish transitons you'd need to segment up a 14 unit pH scale ... right, the rainbow spectrum has that property.
So there's no real chemical reason for the connection. It's more a practical human consideration, where the people combining the indicator molecules to make a large-range pH indicator need to pick particular ones which make a clear, unambiguous color sequence, and the visual color spectrum has the same useful properties they need.


text:
What is meant by viscous liquid?
I think all the liquids are viscous. So I was wondering what is meant by a viscous liquid as it is written in the first line of page on Oil at Wikipedia.

An oil is any neutral, nonpolar chemical substance that is a viscous liquid at ambient temperatures and is both hydrophobic (immiscible with water, literally "water fearing") and lipophilic (miscible with other oils, literally "fat loving"). ...

Yes, fluids are viscous (well, superfluids...). They will have a viscosity. The liquid we are most familiar with, water, is viscous. But in general use, if something is viscous, it is more viscous than water, and usually noticeably more viscous. I"m not sure there is an official viscosity where things are labeled 'viscous'.

This is somewhat an elementary discussion, but viscosity is typically measured in some sort of viscometer. A typical one is the Oswald viscometer shown below.

Fluid is poured into the apparatus through tube at $f$ to fill neck and bulb at the bottom. The fluid is then sucked up above line $c$. The fluid is then released and drains down the tube to line $c$ at which time the timing for the flow rate starts. When the fluid falls to the line at $d$ is reached then a flow rate can be measured. By using samples of known viscosity, the flow time can be calibrated to the viscosity of the sample.
Now such measurements are in units of pascal-seconds. For reference a short list will be provided.
     Substance           mPa-s
     acetone                0.0306
     water                  0.0894
     sulfuric acid          0.242
     olive oil             81.
     motor oil SAE 40     319.
     glycerol           1,200
     corn syrup         1,380.6
     pitch              2.3×10^8

Wikipedia also lists tables for Viscosity of selected substances
Now let's defined three ranges of viscosity:

Low Viscosity -    0.1 mPa-s >= sample
Medium Viscosity - 0.1 mPa-s <= sample <=1.0 mPa-s
High Viscosity - sample > 1.0 mPa-s

Of course the "exact" ranges are arbitrary. So a viscous liquid is a High Viscosity liquid on our scale.

A viscous fluid is simply defined as a fluid that resists flow due to internal friction for example if you poured oil into a cup with a hole in it ,it certainly would flow less faster than the same quantity of water.

Generally viscosity is an ability to resist flow. More viscous liquid will flow slower. Oil and honey are viscous liquids i.e. liquids with high viscosity.


text:
Molarity of hydrogen gas in standard hydrogen electrode
I am unable to understand the folowing statement of Wikipedia it is about standard hydrogen electrode :

The concentration of both the reduced form and oxidised form is maintained at unity. That implies that the pressure of hydrogen gas is 1 bar.

But why if I apply ideal gas equation then I am getting molarity of nearly 0.004M .this is condradiction to the statement in Wikipedia. N/V=P/RT THEREFORE M=N/V~0.004

The standard hydrogen electrode is a "reference electrode." It is supposed to be used with ultra-low currents.
In operation, hydrogen gas is bubbled through a 1 N acid solution, but there is little hydrogen gas in the solution. In order to get the hydrogen gas to bubble through the solution, a pressure of a bit more than 1 bar must be used.

We know that 1bar pressure is less than atmospheric pressure. The reduced form ie. Gaseous form must have concentration unity. So how then?

In order for the hydrogen gas to bubble out in the solution the pressure MUST be a bit above atmospheric pressure. With constant bubbling the solution is saturated with hydrogen gas.

The $[H^+]$ ion concentration is unity, i.e. 1M. What you are obtaining by ideal gas equation is number of moles of Hydrogen gas per unit Volume.
Hope this helps!

Yes the H+ ion concentration is unity which is the oxidised firm of hydrogen in the redox couple but as from statement both oxidised and reduced forms of redox couple must be a one molar concentration but after applying ideal gas equation iam getting less concentration of reduced form of redox couple

The reduced form is $H_2$ whose concentration or we would rather say Pressure is 1bar (unity) and the oxidised form that is $H^+$ has concentration 1M. The term concentration for gas here implies its pressure and not the actual concentration.

So one bar pressure of a gas is equivalent to one molar solution?

Its just the way of writing. Pressure and concentrations are different things. These is no equivalence between them


text:
Why absorption/transmission spectroscopic data is normalized?
A typical data processing step for acquired absorption/transmission data is normalizing, i.e. stretching the curve such that it is bounded between 0 and 100%.
However, the absorption/transmission curve changes with the density and thickness of the sample. Moreover, absorption is a non linear phenomena (for example Beer–Lambert law), so I would expect that the relative height of the various peak would not remain constant across several samples of the same material.
Why, therefore, is it justified to normalize acquired spectroscopic data? Why not present data with original abs./trans. values, while providing details about the sample itself (for example thickness)?
In addition, how can spectroscopy be quantitative if the relative height of the peaks changes with illumination/thickness/density/etc.?

The Beer Lambert Law is explicitly a linear relation.  I'm some confused about this question.  At constant thickness, the Law is linear with chemical concentration.  At constant concentration, the Law is linear with pathlength (thickness)

@Lighthart how is BL linear? it is exponentially decaying with penetration depth.

Absorbance equals molar absorptivity times concentration times path length? http://life.nthu.edu.tw/~labcjw/BioPhyChem/Spectroscopy/beerslaw.htm

@Lighthart yeah, but intensity is A = -log T = - log (I / Io). Aren't all measurements intensity based?

There is confusion here about absorption at one wavelength and the absorption spectrum. At each wavelength the Beer-Lamber law applies $$I_{trans_{\lambda}}=I_{0_{\lambda}}\exp(-\epsilon_{\lambda} c l)$$ for a compound with extinction coefficient $\epsilon$ at wavelength $\lambda$ and concentration c and path length l.
When an absorption spectrum is produced it is normal to show it as optical density vs wavelength, i.e. $OD_{\lambda}=\log(I/I_0)_{\lambda}=-\epsilon_{\lambda} c l $ and in this case the ratio of peaks for the same compound is constant no matter what the concentration. (Naturally assuming that it does not absorb so much that the instrument fails).

You normalize as the emmision strength of machines varies and is not controlable and the precision of spectroscopy is not such that you can justify such high resolution. Franky absolute data would not be very informative.  Further  You have a good observation that thickness can affect results. This is explained by the fact that absorbance is not constant across all wavelengths.


text:
Alcohol Evaporation
If I were to leave an alcoholic spirit out to evaporate naturally (e.g. Campari) until it was just a powder, would that powder be non-alcoholic when rehydrated? In other words would 100% of the alcohol be removed?

Welcome to Chemistry.SE. You might want to look at http://chemistry.stackexchange.com/a/10161/5017

The goo that would be left would probably contain minute traces of alcohol.

Ethyl alcohol is a liquid at room temperature, and in fact doesn't turn into a solid unless cooled below about -114$^o$C! So, whatever solid you might have after drying your spirits to a powder would would be thoroughly alcohol-free.

Vapour-liquid equilibrium (VLE) of water-ethanol mixtures is very well known. We can predict that the vapour in equilibrium with the liquid contains more ethanol than water for solutions with less than 95.6% ethanol (weight basis). In particular, ethanol is much more volatile than water at concentrations around 1% ethanol (in weight basis).
To know the evolution of the water-ethanol solution, you must solve Rayleigh's equation (e.g. here).
If your alcoholic spirit contained only water and ethanol, as it evaporates it would become a solution with an increasingly lower amount of ethanol but, strictly, it would be not possible to obtain an ethanol free solution. However, for any practical purpose, you would obtain at some point concentrations which can be considered in effect ethanol free.
Now, your solution (e.g. Campari) contains carbohydrates, oils... It is very difficult to predict the evolution of the concentration of the remaining liquid as it becomes a very concentrated solution of these components. For instance, we can expect traceshat part of the ethanol will be retained by oil droplets
PS: Mathematically, Rayleigh's equation gives place to a "first-order system" similar to, e.g., a cup of tea that cools down at room temperature. The cup will never reach room temperature in finite time, but after a while, the cup temperature becomes room temperature for any practical purpose. In reality, after a long time has passed, the temperature difference (or the concentration difference) is so low that the hypotheses of continuum that were used to deduce the equations do not hold any longer.


text:
What does nutella add to the Coke+Mentos reaction?
Inspired by this question on SFF.SE.
The Diet Coke + Mentos experiment has been a popular one for a while now, and I've read a little bit about the science behind it.
I recently discovered this video, where an Italian man adds Nutella to the mix1, to his great excitement.
However, I'm not clear on why he did this; I couldn't see a significant difference in the reaction (although my view was somewhat obscured by the condom). What does the Nutella add to this reaction?

1 Do not watch this video in a public place without headphones on (or the sound turned way down); he gets very, very excited when his abomination experiment (literally) blows up in his face

Nutella would be very greasy and not dissolve  readily in aqueous solution. I'd speculate that it just makes a bigger mess. // He only added a couple of tablespoons which would have been initially stuck in the neck of the bottle.

From watching the video I lost 2 minutes of my life. I also think that he just made a non-sense video with Nutella to pormote Nutella or get views for his video from Nutella lovers.  I can find no rationale behind the addition of Nutella.

Nutella was used just to protect cola from incidental reaction with Mentos. And it makes more effective insertion - all mentos tabs are inside at the same time.
The reaction is completely the same with simple Mentos. (this video also is quite noisy)
This is not chemical reaction, is carbon dioxide evaporation.

Fair enough; thanks for taking the time to look into this strange, strange question


text:
Do L-sugars (enantiomers of natural sugars) have a sweet taste?
I'm wondering about how L-sugars taste. They are obviously not digestible by humans, but what do they taste like? If they are sweet, too, why isn't sucrose with the unnatural chirality used as a zero-calorie table sugar substitute?

They're too expensive. https://en.wikipedia.org/wiki/L-Glucose

Good article referenced in Wikipedia article http://spinoff.nasa.gov/Spinoff2004/ch_4.html

Take a look at tagatose. It is a 6-carbon sugar (a ketohexose like fructose) but is inefficiently metabolized and has about 30% the caloric content of sucrose. It is on the FDA GRAS (generally recognized as safe) list. Unlike common artificial sweeteners, you can cook with it like sugar. It has been industrially produced as a sugar alternative, but it is not clear if it commercially available to consumers at this time.

There's an old Wired article from 2003

Unfortunately, I don't think this answers the question at all. The tagatose you've mentioned is actually D-tagatose.


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Which half cell equation do you switch when calculating E° cell?

Given the following standard potentials calculate the overall $E^\circ_\mathrm{cell}$:
$$
\begin{align}
\ce{Pb(SO4)(s) + 2 e- &→ Pb(s) +SO4^2-(aq)} &\quad E^\circ &= \pu{-0.356 V} \\
\ce{Pb^2+(aq) + 2 e- &→ Pb(s)} &\quad E^\circ &= \pu{–0.125V}
\end{align}
$$

I know the answer should be $\pu{0.231 V}$. I was told that $E^\circ$ cell always has to be positive. I tried doing this:
$$E^\circ_\mathrm{cell} = E^\circ_\mathrm{cathode} - E^\circ_\mathrm{anode} = \pu{0.231 V}$$
The only way I get the correct answer is if I do not switch the negative sign of the standard reduction potential of the first equation to a positive sign. Is that correct? I thought you always had to switch signs if you reversed the equation?

Since you flipped the equation from the "reduction form" you must also flip the sign of the potential. Thus: $$\ce{Pb(SO4){(s)}+ 2e- → Pb{(s)} +SO4^{2–}{(aq)}} ~~~~  –0.356~\mathrm{V}$$ becomes: $$\ce{Pb{(s)} + SO4^{2–}{(aq)} → Pb(SO4){(s)} + 2e-}~~~~  +0.356~\mathrm{V}$$ since the reaction is reversible.

Now, lets first clear this up. $E_\mathrm{cell}^\circ$ can be negative, it just means that the reaction won't spontaneously occur. This means that an external current is required to be applied to push the electrons in that direction for the reaction to occur. When $E_\mathrm{cell}^\circ > 0$, that just means that reaction occurs spontaneously. So in voltaic or galvanic cells, $E_\mathrm{cell}^\circ > 0$.
Now to calculate $E_\mathrm{cell}^\circ$, the formula is:
$$E_\mathrm{cell}^\circ = E_\mathrm{red}^\circ \mathrm{(cathode)} + E_\mathrm{ox}^\circ \mathrm{(anode)}$$
As you correctly stated, if you switch the direction of the reaction (so it goes from an oxidising to reducing half equation), the sign of $E^\circ$ changes. Hence the equation for $E_\mathrm{cell}^\circ$ can also be written as:
$$E_\mathrm{cell}^\circ = E_\mathrm{red}^\circ \mathrm{(cathode)} - E_\mathrm{ox}^\circ \mathrm{(anode)}$$
This is the formula that you have used and hence this is why you get the correct answer when you do this.


text:
NMR coupling pattern for hydrocarbon

What does the 1H NMR spectroscopy coupling pattern look like for the hydrogen attached to the carbon that the blue arrow is pointing to? There is a $\ce{CH2}$ attached to ether side of the carbon, but each of those carbons is in a different chemical environment from the other. Does that mean that it is a doublet of doublets? Or just a doublet?

The two $\ce{CH2}$ groups are, as you mentioned, in chemically different environments. Thus, each one will couple separately. But within each of the two groups, there are two protons. So you will not get a doublet of doublets, but a triplet of triplets.
Note that for each $\ce{CH2}$ group the hydrogens are chemically equivalent but not magnetically. Therefore, if you include a chiral environment, the pattern will further complexify itself into a higher-order signal that cannot easily be analysed. But since you’re probably measuring in chloroform you need not worry — there is no chiral information present and the couplings will be degenerate.
If I had to label a spin system here, I would probably choose $\ce{A3BB'CC'MM'X3}$.
By the way, it can be assumed that both coupling constants to the two different $\ce{CH2}$ groups be reasonably similar that the overall pattern looks like a quintet without actually being one.

If I replaced the oxygen and the methyl group to its left with, lets say a single methyl group, would the two CH2 groups be in chemically identical environments, thus giving me a lone triple?

@Lahav If you replace the $\ce{OMe}$ with $\ce{CH3}$, then not only are you creating two chemically equivalent but also two magnetically equivalent environments (okay, it is slightly more complicated when dealing with the magnetic equivalence). The resulting pattern should be a near-perfect quintet (4 nuclei to couple to). However, the actual NMR spectrum of pentane is not easy to analyse because the signals all overlap. Pentane’s spin system should be $\ce{A6B4C2}$.


text:
Synthesize acyl chloride from aldehyde
I know it is usually easier to go from an acyl chloride to an aldehyde,
but is there a way to synthesize acyl chlorides from an aldehyde?
Edit:
Upon some textbook reading (Wiley: Organic Chemistry 11th Ed, Solomons), I realized that aldehydes could also be oxidized to carboxylic acids with chromic acid in aqueous acetone (the Jones Oxidation) and then you could react the carboxylic acid with thionyl chloride or phosphorus trichloride to yield the acyl chloride.
Another method was to convert the aldehyde into a cyanohydrin and through hydrolysis the -CN group could be then converted into $\ce{RCOOH}$. Then react $\ce{RCOOH}$ with $\ce{SOCl2}$ to get the acyl chloride.

If what you want is to convert RCHO to RCOCl, then I think your cyanohydrin method won't quite work.

There are many ways to affect the transformation, as you have realised, and many of which Loong has detailed below in his post.
This answer gives one method which I would be confident of working practically, in the lab, using common reagents. A full search of Reaxys of SciFinder would reveal conditions/methods more suitable for a specific aldehyde.
A Pinnick oxidation will convert an aldehyde into its corresponding carboxylic acid, which may then be treated with a reagent such as oxalyl chloride to give the desired acyl chloride.

In some cases, the direct reaction with chlorine is possible:

Clarke, H. T.; Taylor, E. R. o-Chlorobenzoyl chloride. Org. Synth. 1929, Vol. 9, 34; 1941, Coll. Vol. 1, 155.

In other reactions, radical chlorinating agents are used:

Ginsburg, D. The Action of t-Butyl Hypochlorite on Organic Compounds. II. Aromatic Aldehydes. J. Am. Chem. Soc. 1951, 73 (2), 702–704.

Hebbelynck, M. F.; Martin, R. H. Bull. Soc. Chim. Belg. 1951, 60, 54.

Arai, M. Bull. Chem. Soc. Jpn. 1962, 35, 1272; 1964, 37, 1280; 1965, 38, 252.

Winstein, S.; Seubold, F. H. Jr. Free radical reactions of aldehydes. J. Am. Chem. Soc. 1947, 69 (11), 2916–2917.