text:
Are our hands really chiral?
Everyone says that our left and right hand are mirror images of each other and are non-superimposable, unlike a sphere whose mirror image is exactly superimposable. But shouldn't our left and right hand be considered achiral?
If I move my hands in front of me such that palm of both the hands face towards me and move my say left hand horizontally such that it can be superimposed on the right hand clearly they can't be because the thumbs point in opposite directions.
But if I rotate my left hand by 180 degrees so that now the palm of my left hand faces away from me then both the left and right hand are superimposable. So shouldn't they be achiral? I know there are two more similar questions akin to this but none of the answers answer my simple question.

Holding your hands in this way merely proves that your hands are mirror images. If you take any object (chiral or not) and hold it up to a mirror, you can always align common features.
Imagine instead placing one of your hands inside the other. You may be able to align the overall thumbs and fingers, but they will be facing opposite directions, thus are not completely superimposable.
If you'd like further confirmation that your hands are chiral, try shaking someone's right hand with your left. Or try putting a left hand glove on your right. The interaction of mirror image forms (enantiomers) of a chiral object (like your hands) is different with the mirror image forms of another chiral object (like a glove or someone else's hand).

But if i rotate my left hand by 180 degrees ie now palm of my left hand faces away from me then both the left and right hand are superimposable .

Turning your hands this way only makes them superimposable if you make the assumption that they are two-dimensional objects, where the normal vectors coming out of both sides of a given hand are indistinguishable.  To note, this sort of assumption is what is typically made when analyzing planar molecules.
From a fully three-dimensional perspective, in this configuration, your fingers and thumbs are indeed aligned but your palms are now facing opposite directions.  If the normal vectors from each hand are taken as distinguishable (the "palm side" is different from the "back side"), you cannot just choose to ignore the misalignment of your palms and declare your hands superimposable, and thus achiral.

So does this mean that a planar molecule can be chiral ?

@RaghavSingal I think not; I'm pretty sure planar molecules are always achiral -- but I'm not completely sure. Planar molecules are treated as something of an idealization, where both normal vectors off of the plane of the system are indistinguishable. Hands are three-dimensional objects, and the palm-side and back-side normal vectors are typically not treated as indistinguishable.

Can you give an example ?

Also if I ignore the thickness of my hand so my hands are one example of a planar chiral object.

@RaghavSingal In that case, yes, if you make the assumption of treating your hands as 2-D objects, then I would say you're correct that they become superimposable.  But, only under that assumption.

@RaghavSingal See edits to answer.

But still my palms are facing opposite directions when I try to superimpose them

Ok. Why do we see according to the normal vector ? How is it different or same in a situation ?

I don't think that assumption is made when analyzing planar molecules, but it does seem to be what the questioner is assuming about hands. Ethylene is a planar molecule. It's mirror image is superimposable on itself, even if you consider it a 3D object, which is good, because it is 3D.

A planar molecule does not have a front- and backside, and therefore cannot be chiral.

Planar molecules possess a mirror plane of symmetry which renders them achiral ($\sigma \equiv S_1$).

@hBy2Py Yeah they cannot, I have explained it the best I can in my answer

"if you make the assumption that they are two-dimensional objects" Even if I consider them 2 dimensional objects (say cardboard-made palms) and colour one side, it becomes chiral. If you tell cardboard is also 3D with some thickness, so an imaginary perfectly-planar cardboard-palm could be chiral if we associate an area vector upon it.

@AlwaysConfused Good point -- one has to make an even stronger assumption than what I did.

"Superposable" means exactly identical in shape. Superposability is a thought-experiment to know 2 shapes are  identical or not.
Our 2 hands are mirror-symmetrical, and of similar size; but no; they are not of identical shape.
Here is a small test. Stand facing the East. Bring your hands at front side of your body. Bring your palms together as shown in FIG2. Seems the shapes are same?  (I'm ignoring tissue-details, just look the overall shape) Look closer. Think. Got anything?
For your left hand, now the palm-side is facing south. And its knuckle+nails side is facing north.
For your right hand, now the palm side is facing North. and knuckle+nail side is facing south.
They are so alike but they are so different!
If you now try to bring the both palm facing same direction, as in fig.3; your finger won't match. you may also change the orientation of wrist, and see any difference. No benifit, they wont "superpose". They are so similar but so different. They are sort of "opposite" shapes!
That's chirality.
Yes our hands (left and right) are chiral and enantiomorphic from external morphology.

FIG 1
Image source: The Chemistry LibreTexts library. 25.7. Chirality in Organic Chemistry
Our 2 hands are not superposable i.e. not identical, because if you try to merge theme as follows the nails do not match.

FIG2
The nails, knuckles, and darker, outer side of your left hand stay towords your left. And the nails, knuckles and the darker side of your right-hand, stay towords your right.

Reference:

This particular example was taught in our college chemistry class.
Solomon & Fryhle's Organic Chemistry, 10th Edition, by T.W. Graham Solomons and Craig B. Fryhle, Wiley Student Edition.

Chapter 5 (stereochemistry)
Here is a google book result of an older-edition;  Ed-8

FIG3

In general, when an object is asymmetrical (reflectional) in n dimensions within n-d euclidean space, it is chiral. Make an object that has a definite 'up' and 'down'. A definite 'forward' and 'back' and a definite 'right' and 'left', and enforce a rule that objects are only permitted to travel through the space in an unbroken line.
Suppose left is 'anticlockwise' to up on one 2d object and we take a copy of this item but swap the position of up and down, now we claim that left is 'clockwise' to up. These two objects are indistinguishable in 3d because we can rotate one 180 degrees to make the other. In 2d they are chiral. But if we introduce an asymmetry in the third dimensions, the act of fixing the chirality in the second dimension will create a different image in the third dimension. A hand is chiral because there is 3d asymmetry of reflection.
It is impossible for the designer of the object to be able to to use some mathematical tool or understanding to 'tell' whether their object has a left or right chirality without having something with an already known chirality to compare with it. The words 'clockwise' and 'anticlockwise' have no meaning without a given chirality in the first place. What I mean is that there is nothing left about a left hand, other than it is simply not a right hand. They can only be told apart from each other.
You may ask why we say that rotated objects are indistinguishable and reflected objects are not. There is really no good mathematical reason for this terminology. From a physical perspective, this only makes sense because of the conservation of angular momentum.
https://en.wikipedia.org/wiki/Noether%27s_theorem#Examples


text:
Energy cycle centered around methane production / consumption
Many car manufacturers have developed projects for hydrogen cars but I've always been wondering why they put so much effort into hydrogen, as it is hard to store, and fuel-cells use to restitute energy back are based on rare-earth-metal not suitable to produce many cars for everyone.
Nature mainly store energy with carbon, I'm not a chemist but is it that hard or not yet enough economically viable for not seing rising projects around simple hydrocarbure synthesis. Easy to store, easy to consume, but maybe hard to produce, right ?
Production : $\ce{CO_2 + 2H_2O}$ (+ Renewable Energy Source) $\ce{-> CH_4 + 2O_2}$
Consumption : $\ce{CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O}$ (+ Energy To Use)
Could even $\ce{CO_2}$ be simply captured from air with limewater in such cycle ?
Sorry if my question may seems too simplistic, but I'd really like to handle (in simple terms) what are the economic/technical constraints to elaborate such apparently simple round-trip.

The economic barrier to this approach is the current situation of cheaply available methane via mining (natural gas wells) compared to the cost of synthesizing methane from carbon dioxide and water.

Hum ... I quite envisage that economic barrier is still higher than technical one :/

Carbon sequestration may be easier growing corn, feeding chickens and using their excrement to produce biogas. No need for any new technology.

Vehicles that run on natural gas already exist, and have done for many years - you can find CNG (compressed natural gas) cars and larger vehicles around the world.
But there's very little synthesising of methane because there are very few places with a sensible carbon price, so it's just cheaper to extract natural gas and burn it.
Research is ongoing on methane synthesis - this is usually done under the name "power to gas", which can refer to hydrogen or methane. Either way, it usually starts with the splitting of water to get hydrogen, and then is sometimes combined with methanation of the hydrogen. The latter is an exothermic reaction, so combining the two processes can give pretty good efficiencies (see e.g. this paper).
Hydrogen / methane for cars is looking like a lost battle: electric cars have almost certainly already won; it's just going to take a few years until they completely dominate. However, there is still the possibility that larger vehicles (heavy goods vehicles, boats, aircraft) might settle on hydrogen or renewable methane as their decarbonisation option.
By the way, just because a fuel-cell might use rare-earth metals, doesn't make it unsuitable for mass market. Most rare-earth metals aren't rare, despite the name. Fuel cells have lots of issues, and fuel-cell vehicles really haven't achieved much yet, but that's to do with lack of reliability and longevity, rather than scarcity of components.

Could then electric cars if they won be powered by methane-cells if it exits (so as to acheive longer distances autonomy / easier storage than pure hydrogen and less polluting than current batteries) ?

That's probably best asked on [engineering.se]; but AFAIK methane fuel-cells have low electrical efficiency, so they get pretty hot, and aren't really yet up to the job.

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text:
Fe with different charges
Taken these reactions:

\begin{align}
\ce{2\overset{0}{Fe} + 3Cl2 &-> 2\overset{+3}{Fe}Cl3} \tag{1}\\
\ce{\overset{0}{Fe} + 2HCl &-> \overset{+2}{Fe}Cl2 + H2} \tag{2}
\end{align}

Why does $\ce{Fe}$ have +3 charge in the first example and +2 charge in the second?

Iron can have two different ions which you have correctly identified as Fe +2 and Fe +3. Other elements are capable of this as well. You may find this link useful, http://www.syvum.com/cgi/online/serve.cgi/squizzes/chem/valency3.html

Related: http://chemistry.stackexchange.com/questions/41095/formation-of-ferrous-chloride-or-ferric-chloride

http://chemistry.stackexchange.com/questions/34132/how-is-the-oxide-of-iron-fe3o4-possible

http://chemistry.stackexchange.com/questions/32996/calculating-valence-of-oxides/33005

Iron exists in the +2 oxidation state (iron(II)), also called ferrous iron, as well as the +3 oxidation state (iron(III)), also called ferric iron.  There are sever other oxidation states of iron that are not pertinent to your reactions.

In your first reaction, the strongly oxidizing $\ce{Cl}$2
  oxidizes the neutral, metallic iron to it's highest stable oxidation
  state, iron(III).
In your second reaction, $\ce{HCl}$ oxidizes iron to the lower stable
  oxidation state, iron(II).

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text:
In the carbonate anion, which atoms gain the two electrons?
In the carbonate ion ($\ce{CO3^2-}$), is it the oxygen or the carbon that has gained the two additional electrons?

The two additional electrons (no, you cannot tell which electrons are extra because all electrons are the same) are mainly delocalized between the three oxygen atoms (carbon atom should also share a tiny proportion of the extra charge).

One can draw Lewis structures to see where the lone pairs of electrons may be.

Here you can see that you can draw the ion like this, with one of the oxygen atoms having a double bond with the carbon, and the other two having a single bond. This means that the double bonded oxygen atom has no charge. The single bonded oxygen atoms have a net charge of $-1$ each as they have gained a lone pair of electrons.
However this picture does not tell the whole story. In reality, the negative charges are not localised to individual oxygen atoms in the ion. We can draw a total of three structures (called canonical forms):

These can be drawn because in actual fact there is delocalisation of electrons between the carbon and each oxygen atom. The true structure is not changing/oscillating between these three forms, but is actually called a resonance hybrid of all of the canonical forms. This is shown below. Therefore you can see that the negative charges are not localised to single oxygen atoms, but are instead delocalised across the entire ion.

If you want to read more about the carbonate ion and its structure: https://en.wikipedia.org/wiki/Carbonate#Structure_and_bonding


text:
Work done in irreversible isothermal expansion against vacuum
Why is the work done in an irreversible isothermal expansion against vacuum equal to zero?

A system does its maximum amount of expansion work when the external pressure is equal to the pressure of the system all the way through the expansion process.
At constant pressure work done is given by:
$$w=-p_\mathrm{ex}\Delta V$$
Where $p_\mathrm{ex}$ is the external pressure. Therefore the minimum work done is when the external pressure is $0$, which is the case under vacuum. This would mean that the work done under vacuum is $0$.
This is called free expansion. The greater the opposing force then the greater the amount of work done is (as work is equal to force applied over a distance). Therefore when the external pressure is $0$ there is nothing for the piston to push against, because there is no opposing force, no work is done by the system.

If you are interested to see where $w=-p_\mathrm{ex} \Delta V$ comes from then note that
$$w=Fd$$
Because pressure can be given by $P=F/A$ and volume can be given by $V=Ad$ we can see that $$w=\frac{F}{A} Ad = p\Delta V$$
This is work done on the gas. For work done by the gas, we put a negative sign in front of the pressure term.
*Note that this is at constant pressure.

Good answer. It really is intuitive as well, work is defined by a transfer of energy from one system to another - if the other system has vacuum, then no change in its volume will transfer any energy to it. Trying to pump something with vacuum on the low pressure side of the pump will not produce any gainful results.

@PiyushKumar  you wouldn't mind marking my answer as correct then would you? ;)

Yeah sure george!!


text:
Good summary of highschool to university chemistry
DISCLAIMER: I've browsed similar questions, but sadly they refer to books that're either too basic or too specialized.
Hello. I'm studying mechanical engineering (1st year), and we have to do a module in chemistry. It's mostly highschool chemistry (equilibrium, redox reactions, acids and bases, ...) with small snippets from unknown topics, such as thermodynamics and crystal structures. We are now beginning with "the chemistry of non-metals", and will eventually cover "the basics of organic chemistry" and "the chemistry of metals and solids". The problem is, we've gotten a new professor who's awful as an educator: he barely speeks the language, and never prepares the lessons. Instead, he just clicks through powerpoint presentations, slowly reading them out, while facing the screen.
I've found that - although more productive than his "lessons" - just reading the presentations is not enough. I recognize what I already know, but require an unreasonable ammount of time to understand the rest, which usually involves browsing wikipedia.
That's why I'd like to ask if anyone could suggest a book that summerizes highschool chemistry, and ventures into topics above highschool level in a way someone with a highschool education would understand. Obviously, alternative mediums such as videos are also fine. I just want something that more or less complements the lectures, and can always just google whatever details are not covered.
Thank you in advance.

I think Khan Academy is pretty good for chemistry. I am a second year chemical engineering student and I find that the videos are still helpful as a quick refresher for topics I've gotten a little rusty in.

The good news is that I'm writing this text book. The bad news is that it's not going to be ready any time soon... :(


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

No, the bond angles will not be the same.  See this earlier Q&A "Why does bond angle decrease in the order H₂O, H₂S, H₂Se?".  A similar explanation would apply here.

Bond angles and other aspects of molecular geometry don't follow associative rules - you have 3 different molecules, so 3 different structures. You can add a 4th - $\ce{H2O}$ - with a bond angle of 104.5$^{\circ}$ to extend your comparison.

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


text:
How do I select a sub-freezing temperature for my experiment that will allow the formation of frost, but not supercooled liquid water (rime)
I once conducted a set of experiments in which I had to form ice by direct vapor deposition (not condensation of liquid water followed by freezing, aka riming) onto a Peltier-cooled glass-coated copper rod.
The experiments were conducted outside at night within a small wind tunnel, with ambient air temperatures ranging from -2 oC to -25 oC and the glass substrate was cooled from 1 oC to 10 oC below the ambient air temperature.  Occasionally supercooled water droplets would condense, followed by freezing (forming rime ice) rather than direct deposition of ice crystals from the vapor phase.  These events would ruin my time-consuming experiments.
The only relevant parameter under my control was the temperature of the glass-coated copper rod.  The only relevant parameters I could measure during the experiment were the air temperature and humidity. I also introduced trace-levels (single to hundreds of ppb) of several C1 to C4 alcohols in order to later measure their incorporation into ice during ice deposition from the vapor phase.  However, the low levels of these compounds did not affect whether riming or direct vapor-ice deposition would occur, as based on a series of 'blank' experiments without the trace gases.
If I were having trouble with liquid water condensation, in what way would I want to alter the glass substrate temperature so that only the direct deposition of solid ice could occur?

This is hardly a chemstry question. Otherwise, the heat of condensation can easily heat your glass rod locally by a few degrees. Heat conductivity of glass is lousy, and how would you cool a glass rod effectively and uniformly, especially to a different temperautre than the surrounding air? The way you describe it, this experiment cannot work reliably and reproducibly.

As I stated @Karl, I control the temperature of the glass rod.  Obviously it took engineering and proper calibration but it was very doable.  And very publishable.  And if phase transitions are not relevant to chemistry, then I guess I don't know what field I'm in.

Meteorology is physics. And sorry, from your description it sounds like a misguided experiment. Perhaps you can describe it better? Is your "glass rod" actually a pipe with flowing coolant? What temperatures did you measure? What "bulb"? How did you know the actual temperatue of the glass rod, if you didn't measure it?

You've asked at least five questions about my experimental conditions since stating in one post then reaffirming in another that you do not believe my question is related to chemistry.  If I'm off topic, at least it's by accident.  If others also believe it to be so or if there is a general lack of interest then I will without hesitation voluntarily remove this Answer-Your-Own-Question post.

I find your question interesting, but I'm not on SE:physics. ;-) I think you can migrate it with comments and everything.

@Karl. Along those lines, I changed the subject tags and also clarified some aspects of my experimental conditions, changed some terminology and tried to make the real-world analogy in my answer more interesting.  If others down-vote this question-answer, I will probably just take it down as being off-topic.  I don't think it is but I'm trying to listen to the community as a whole, and I appreciate the constructive advice that you had to offer.

Surprisingly, at a glance, I had to run my experiments with the glass substrate at higher temperatures in order to ensure ice would form and not liquid water.  When you think in terms of the difference between the saturation vapor pressure of liquid water and that of ice at the same temperature, this solution is obvious.  The temperature has to be above the dew point but remain below the frost point, which is always higher than the dew point at temperatures below 0 oC.  The reason for this is that at any given temperature, liquid water has a higher vapor pressure than ice.
As a real-world example of the ice-growth mechanisms observed in my trace-gas uptake experiments, what was a problem for my experimental conditions actually has important implications for ice-particle growth in clouds, in which context it’s called the “Wegener-Bergeron-Findeisen Process”.  Here, when there is a mixture of ice and supercooled water droplets, water vapor condenses onto the ice particles so that the humidity with respect to liquid water decreases enough for the liquid water to further evaporate.  Thus there is a net transfer of water mass from liquid droplets to solid particles and the solid particles may then grow large enough to precipitate (as in fall to the ground in this context).


text:
What gives fresh milk a yellow tinge?
Every now and then, when I pop over to my grandparents' house, they make sure to give me a glass (or two) full of fresh milk (or "whole milk" if you will) and then proceed to pamper me for the rest of the day.
Now, it suddenly struck me that the fresh milk I get over there actually has this really faint yellow tinge, as opposed to the skimmed milk I get at the store which is almost pure white.
I did try Googling this up, but I can't seem to find anything on what imparts the yellow tinge to fresh milk.
So,
What substance/pigment is responsible for the yellow tinge associated with fresh milk?

White light scattered in a colloidal medium has slightly different colours depending on the angle to the incident light under which you see it. Mie scattering. If it does not depend on the angle, then it is probably something in the milk.

@Karl " probably something in the milk" Yeah, fat globules ;)

See: Butter ...

You wrote ‘skimmed milk’ but I assume you wanted to write ‘white water’.

The main substances that cause the yellowish color of the milk are carotenoids [1]. The main carotene involved is the beta-carotene coming from the feed that cows eat.

Some studies have been carried on and it has been noticed that the milk with a more yellow tinge was collected during late spring and early summer when carotene levels are at a maximum [2] unfortunately for this reason in many products as mozzarella they add titanium oxide.
The amount of Riboflavin is too little compared to that of beta-carotene and generally gives a slightly green-ish tinge to the milk as reported here and here [3].

P. F. Fox, Fundamentals of Cheese Science,p. 13
Kosikowski, Frank V., and David P. Brown. "Application of titanium dioxide to whiten mozzarella cheese." Journal of Dairy Science 52.7 (1969): 968-970.
Marion Eugene Ensminger,Audrey H. Ensminger,Foods & Nutrition Encyclopedia, 2nd Edition, Volume 1,p. 798

Also worth noting that carotenoids are fat-soluble, so their concentrations are going to be much much less in skim milk than whole milk.

@pwcnorthrop And much higher in butter ... which is why we think of butter as being light yellow.


text:
How are bubbles formed when heating water?
My thinking is that bubbles are formed due to the dissolved gases in water. Consequently, upon heating, bubbles are formed and do not remain intact with water molecules, and thus the bubbles are formed. And since the bubbles are less dense than water, they goes up and pop due to air pressure. Water vapor is then formed in the above air. Is this all correct?
I have researched this topic and it's very difficult for me to understand. Please explain it as if you're explaining it to someone who knows very little about chemistry.

Related: http://chemistry.stackexchange.com/questions/38653/vapor-pressure-and-boiling-point-for-solutions-in-open-containers/38654#38654

There are three aspects: Gas dissolved in the water, water that evaporates (at the surface air/water), and boiling water. Your question is a bit hard to understand, btw. Perhaps try to write shorter sentences?

Is it a rule that when gases dissolve in water, bubbles will form?

Saying that the liquid is converted to bubbles is not correct. The liquid simply becomes a gas (so inside the bubble there is the gas phase of your liquid) this is a phase transition and depends on the pressure and temperature of the portion of liquid close to the bottom of your container.
The points where the bubbles appear depends on the local variations of temperature and pressure.

Original Post has been updated. Please tell me if I am correct or not. Thanks!

Also I said that I didn't understand from resources.. Please read the OP. :(


text:
Why do molecular modelling simulations predict that halobromides are more stable than halochlorides?
The C-Cl  bond energy is greater than that of C-Br, yet the minimised energy of say, methyl bromide is larger (more negative) than methyl chloride. I've tried using STO 6G, HF 3-21G, and 6-21G*, all show similar trends.

This sort of comparison is unreliable. "More stable" is a pretty subjective term and should only be used when comparing two otherwise equal situations: for example a reaction that produces two different isomers. One might be more stable and formed preferentially. The minimized energy is related to the heat of formation, and we should expect different heats of formation for the two species. Instead, if we compare the reactions of both compounds with the same reagent to make a common product, then we could talk about "relative stability", but "relative reactivity" is a better term.

What kind of energies are you comparing? Absolute energies as obtained from Hartree-Fock? Do a calculation of a single Br atom and then of a single Cl atom, you might notice that you obtain a much larger energy for the Br atom. Please tell us what energies you are comparing and what methods you are using. Quantum chemical calculations are dangerous as in if you do not know what you're doing, your results are meaningless. Do not interpret anything into your calculation please.

The absolute energy of different molecules is irrelevant because they have a different number of particles (electrons and protons); the one with more particles (C-Br in your case) will typically display a more negative energy.
Thus, one must compute relative energies using systems with the same number of particles. For example, you could compare the energy of CH$_3$Cl + Br$^-$ with that of CH$_3$Br + Cl$^-$.


text:
Why is phenyl shift preferred over methyl shift in cumene hydroperoxide reaction?
In the cumene process, which converts cumene to phenol and acetone, there is a step where cumene hydroperoxide undergoes acid-catalyzed rearrangement:

(Image source: Wikipedia)
Why does the phenyl shift happen preferentially? The methyl group is smaller and should have a higher migratory aptitude.

Despite the mechanism shown in your question, which is commonly used to depict a phenyl shift, phenyl shifts occur by a slightly different mechanism.

Phenyl shifts do not usually happen by the normal 1,2-sigmatropic "hop". Instead, the phenyl group shifts by more of a "walk". It's not really a sigmatropic rearrangement. First, one of the $\pi$-bonds in the phenyl group can attack the electrophilic center. This leads to formation of a spirobicyclic intermediate where the cation is delocalized throughout the rest of the ring. Then, the three-member ring breaks open at the expense of a carbon-carbon bond to rearomatize the phenyl ring and form the new resonance stabilized cation.

That's pretty neat. Is this mode assumed based on kinetics from ring substitution?

I wonder if there is a name for the intermediate carbocation. I recall the carbon analogue is called a phenonium ion and can arise via similar chemistry, however the IUPAC definition strictly disqualifies this one with an oxygen in it.

Hi, thanks for your answer! If I may ask, could you please tell how do we know the phenyl shift occurs by a "walk" and not a "hop"? (nice analogy btw!) Do we have experimental evidences, or some other proof? Many thanks!


text:
How to bleach polyester
I have a challenge. I recently bought a cotton/polyester coat with the intention of bleaching it to near-white. To my surprise, normal bleach with sodium hypochlorite had very little effect. It changed the color from light brown to mustard-yellow-ish, even after two soakings in a 30% mixture using water and 5% sodium hypochlorite for about 10 hours each.
My second attempt was better. I put a test piece of fabric in a solution and boiled it in the microwave. It made it even lighter, but still far away from my desired near-white (white-yellow or offwhite is what I hope to achieve).
I also tried to soak it in white spirit, with a small added extra bleaching effect.
Now, using common household chemicals, how can I achieve the desired bleaching? Can I increase the effect of the chlorine? Can I use peroxide, ammonia, or any other chemical to strip the fabric of its color? I'm willing to risk to have the fabric destroyed in the process.

Forget it. The colour is obviously not very sensitive to oxidation.  I am quite sure you will irreversibly damage the fabric. Soaking for hours in hypochlorite at elevated temperature is not nice. And i hope you DO NOT INTEND TO USE A POLYESTER COAT AS A LAB COAT for chemistry! Sorry for the shouting.

Not at all, it's for a costume party. The coat is made from thick material, almost like felt. As I said, I've had some success soaking a test sample in a bleach solution for a few hours, and more success at an elevated temperature. But since I'm getting to a point where the effect is waning, I've come here for expert advice.

This isn't an entirely chemistry question, given that most cotton-polyester blends are actually cotton threads wrapped around a polyester core for maximum comfort and stability. It could be that the bleach is acting on the cotton surface layer, and has no effect on the core polyester, giving you a mustard yellow color (white cotton wrapped around brown = yellow)

Here's an excellent article about the various chemicals that are commonly used for color removal and bleaching: http://www.pburch.net/dyeing/FAQ/discharge_chemicals.shtml

Why not just paint the coat? Fabric paint or acrylic paint might work. You could use a spray bottle and do very thin layers. After it's dried a good washing would get rid of any stiffness.

I tried painting it with acrylic paint, but to no avail. It doesn't stick, and just falls off as dust.

The yellowish mustard color I would have to say was caused by your efforts to de-color this polyester, polyester is a decently strong polymer and thus is resistant to many forms of chemical tamper, and in all effects the bleach is damaging the fabric and leaving behind that yellow color, I would suggest using Sulfur Dioxide if you can get your hands on it as that can bleach polyester, you should be left with a white translucent fabric if the bleach hasn't already damaged the fabric too terribly.
Sulfur Dioxide part is iffy, you may choose to disregard that.

You can't bleach it entirely since some of the color has actually migrated inside the polyester thread.

But what if I mix common household bleach and ammonia? Then I'll get chlorine gas, which can strip color more or less completely, right?

That won't work and the reaction products are extremely dangerous to your health. Do not try that.

I'm well aware of the dangers, but very interested in any possible results.

You will damage not only your health, but also the coat. One go in the washing machine afterwards and you can use it as a see-through negligee.

That's actually a lot better progress than I've been getting so far :) I'm just skeptical to people saying "it can't be done".

NEVER ever mix chlorine bleach with ammonia. You will get chloramine gas that is deadly if breathed in concentrated amounts, as it eats holes in your lungs that would leave you with permanent lung damage at best or give a very painful death at worst. Don't do it, whatsoever. This coat is not worth your life or wellbeing. https://www.fs.usda.gov/Internet/FSE_DOCUMENTS/stelprd3793292.pdf
That being said, probably you won't be able to get a total white color from bleaching. Polyester is a synthetic fiber, and is resistant to dying and bleaching. One of my chemistry professors tried tie-dying her polyester lab coat, and it came out in pastel colors instead of the brights she wanted. And just a couple days ago, I tried using chlorine bleach with water to create an acid-wash affect on a navy blue dress that was a cotton/polyester blend. I soaked it in the solution for 6 hours, and it came out a sage green. Probably your best bet is just buying a white coat and saving the hassle.

It wasn't me that downvoted, but don't you agree that part of the point of this forum is to actually learn something. Even if it's not worth the hassle, we're all building knowledge in the process ;)


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Finding the binding energy iof methylated miRNA to an mRNA
I'm working on a project in biophysics and I need to find if there is a difference in binding of an miRNA to an mRNA when in one case the miRNA is methylated and in the other it isn't.
What tools should I use to calculate the binding energy in each case?
I have the binding sequence of the miRNA.

Something like free energy perturbations?

Yes, after reading about FEP in wikipedia I believe it might be what i'm looking for.


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Sweet non-toxic salts
I have recently read one post over here in chemistry exchange about tastes of ionic salts, some described lead and beryllium ones to be sweet but toxic, some my point is there any salt that is relatively non-toxic (NaCl for example) but tastes sweet (unlike NaCl)

From this answer, lead and beryllium salts are sweet but they are toxic.

after some searching and a little help from google I have found that Potassium (K) Bromide (Br). In a low concentration of an aqueous solution it will taste sweet. In higher concentrations it will bitter and even salty. wiki
The toxicity of Bromide isn't very high and it can be tolerated by a healthy human body under some doses. Yet it's long half-life in the human body make it hard to avoid intoxication (as written in the wiki page).
A question arises, why does the change in concentration affect the taste? I would guess that in higher conc. the KBr distorts the water structure and thus affect the interaction of the ions and cations with the taste buds.
I wouldn't use it as a sweetner, and wouldn't suggest tasting it to see the effects of concentration change on the taste.
If we don't restrict ourselves to heteroatomic salts then Sucrose Na is a sweet salt as can be seen here: http://dx.doi.org/10.1016/0031-9384(80)90077-3

I would say that sodium saccharin is sweet,  given its main use is as a sweetener. The sodium salt is used as a more water soluble form of saccharin itself.


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Naming ionic compounds of group B
When naming an ionic compound which includes a transition metal, the oxidation state is written between parantheses, yet some metals show an exception. Which elements represent this exception, and what is the reason?

What is ‘group B’?

The relevant document including the rules of formal IUPAC nomenclature is the Nomenclature of Inorganic Chemistry — red book from 2005. Section IR-5.4.2.2 (Use of charge and oxidation numbers) is relevant for the cases you discuss. It reads:

[…]
[…]
The oxidation number (see Sections IR-4.6.1 and IR-9.1.2.8) of an element is indicated by a Roman numeral placed in parentheses immediately following the name (modified by the ending ‘ate’ if necessary) of the element to which it refers. The oxidation number may be positive, negative or zero (represented by the numeral $0$). An oxidation number is always non-negative unless the minus sign is explicitly used (the positive sign is never used). Non-integral oxidation numbers are not used for nomenclature purposes.
Several conventions are observed for inferring oxidation numbers, the use of which is particularly common in the names of compounds of transition elements. Hydrogen is considered positive (oxidation number $\mathrm I$) in combination with non-metallic elements and negative (oxidation number $\mathrm{-I}$) in combination with metallic elements. Organic groups combined with metal atoms are treated sometimes as anions (for example, a methyl ligand is usually considered to be a methanide ion, $\ce{CH3}$), sometimes as neutral (e.g. carbon monooxide). Bonds between atoms of the same species make no contribution to oxidation number.
Note that oxidation numbers are no longer recommended when naming homopolyatomic ions. This is to avoid ambiguity. Oxidation numbers refer to the individual atoms of the element in question, even if they are appended to a name containing a multiplicative prefix, cf. Example 12 above. To conform to this practice, dimercury($2+$) (see Section IR-5.3.2.3) would have to be named dimercury($\mathrm{I}$); dioxide($2-$) (see Section IR-5.3.3.3) would be dioxide($\mathrm{-I}$); and ions such as pentabismuth($4+$) (see Section IR-5.3.2.3) and dioxide($1-$) (see Section IR-5.3.3.3), with fractional formal oxidation numbers, could not be named at all.

While the section in itself does not mention the possibility of omitting oxidation numbers, the examples obviously do:

$\ce{Na2[Fe(CO)4]}$   sodium tetracarbonylferrate($\mathrm{-II}$), or sodium tetracarbonylferrate($2-$)

The generally accepted rule is that oxidation numbers can be left out if the element in question is commonly known to exist in only one oxidation state. That, for example, includes groups 1 and 2 but also elements such as zinc of which only one state is common.

Mostly elements with numerous oxidation states have their oxidation state written to specify which compound is reacting. For example:

iron(III) chloride is a brown yellow compound, whereas iron(II) chloride is a green compound
chromium(III) chloride is violet compound, chromium(IV) chloride is red and chromium(II) chloride is green

The general exception is when we are sure that mostly this element exists at one oxidation state. For example with sodium, we usually do not write oxidation state as it is known (group I, II etc).

So, perhaps i should write orbitals (1s2 2s2...) to decide ?

Do you mean which metals or which transition metals will be an exception.  Base metal ionic compounds (groups 1 and 2) do not show a Roman numeral because they have no D electrons and so will lose 1 or 2 electrons corresponding to their group number.  If you mean which other metals, if I remember correctly Silver (+1) Zinc (+2) as well as Aluminum (+3 but in the Boron group) are the three non base metals that do not require a roman number to be shown in the compounds that they form.
So among ALL metals, Base metals (alkali and alkaline earth), silver, zinc and aluminum I believe do not require a roman numeral to be shown in compounds that they form unless maybe they deviate from their usual oxidation number.


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Why do chemistry and physics have different sign convention in thermodynamics?
In physics class, we write the first law of thermodynamics as $\mathrm dU =\mathrm dQ - \mathrm dW$ and in the physical chemistry class, we write the same law as $\mathrm dU =\mathrm dQ + \mathrm dW$.
The reason being the sign convention is different in both the cases.
In physics we take work done by the system as positive and in chemistry work done on the system.
I realize that this does not cause any change in the actual law of nature but I just want to know why we have different sign conventions. Wouldn't just one convention make life easier?
Is there a historical reason? Or is this is to differentiate between subjects?

The standardized definition according to ISO 80000-5 Thermodynamics of internal energy $U$ for a closed thermodynamic system is $\Delta U=Q+W$, where $Q$ is amount of heat transferred to the system and $W$ is work done on the system, provided that no chemical reactions occur.

Personally I would describe the version with the "-" as the engineering convention. When you are building engines (as were the people who invented thermodynamics) the process you care about putting heat in and getting work out, and you define both those quantities as positive. Pure physics thermodynamics often (mostly?) uses the "+" version.

Even in engineering, there are two conventions... Mechanical engineers think work leaving the system is positive work, chemical engineers think work done on the system is positive... Other engineers just fake it.

It has got to do with IUPAC's ruling then.

This is not a simple physics versus chemistry distinction. I taught Physics for 25 years and saw many examples of either usage in multiple textbooks. In fact, at some point in my tenure, the AP Physics committee swapped conventions on the equation sheet for the AP Exam.
Just my take here: I've always attributed the work-done-by-the-system camp as being more prone to be used by engineering types who want to know "what the system can do for us" in practical applications. On the other hand, work-done-on-the-system seems to foster the view of an experimenter or theoretician operating on a system from without.

"Ask not what your system can do for you..." :-P Also, for what it's worth, only the second version makes sense to me. In the first version you're considering positive heat to be something that is supplied by the surroundings, and positive work to be something that is supplied by the system. That's obviously inconsistent, and by fixing the inconsistency you get the second equation. Just my (correct) opinion... ;)

@Mehrdad Agreed. That's the same argument I gave my students.

As I said above, only mechanical engineers are greedy and care about what the system can do for them :D

@B.Kelly Is that why there is no difference in sign convention of heat?? Since heat released by system is a loss of energy that could have done work for engineers and heat loss is decrease in system energy for theoretician it is negative.  While heat gain is positive . For engineers it is input energy and for experimenter ìt is increase in system energy

There is no difference and the equation will yield the same answer. The difference is by deciding the point of reference for the work done.
Where we have $W = \pm P\Delta{V}$, you just have to decide where do you look from. Am I the system and work is done on me, or am I outside and doing work on the system? anyhow, once you choose your point of reference you should stick to it and do all your calculations according to it. I guess, as in most of these cases it's historical.
Another example is different notation, chemists use $A$ to symbolize the free Helmholtz energy while in physics it's $T$.

Seems like using T for Helmholtz energy would be a nightmare, T is already an overloaded symbol even just sticking to chemists notation (temperature, kinetic energy, certain tensor operators).

I've never seen "T" being used for Helmholtz energy. I think you mean "F".

Physicists are interested in getting work done by the system. In physics we always look for what system is giving to us (surrounding). So, in physics we study work done by the system.
Whereas chemists are interested to know what work have the surroundings done on the system, in order to get the reaction completed. So, in chemistry we study work done on the system.

Chemistry made the switch over to the IUPAC convention back in the early 90's which dictates the first law as the sum of $Q$ and $W$ and in turn requires $W$ to be defined as the negative integral of $P(V)\mathrm{d}V$. See the recommendations of the International Union of Pure and Applied Chemistry.
As pointed out, physicists use $Q-W$ but define $W$ as the positive integral. Same final results though.

This is because....
In physics , we are concerned about the surroundings
And in chemistry, we concerned about the system.
That's why ,
In physics , work done by the system is positive because change in energy (due to the work done) of the surroundings is positive ! (Energy is released from the system to the surroundings)
Whereas In chemistry , work done by the system is negative because the system loses some amount of energy to do work !
That's it!

Take it simple; it's nothing but a reference in different field for different uses that changes the formulae. If system is doing work on surrounding, U will decrease if W is taken positive – and vice versa.

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


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Physical chemical property that determines redox potential?
Can someone explain (on an advanced level) what phys-chem property of compounds/elements determines what their half-cell redox potential will be? For example, Zn and Cu are structurally very similar and their ionization energies, electronegativities, etc. won't be all that different. So, why is the 2-electron redox potential for each so different? ie Zn is -0.76 V (SHE) and Cu is +0.34 V (SHE), which is a significant 1.1 V different.
Isn't electric potential the amount of "stored" energy? So, wouldn't removing or adding 2 electrons to Zn or Cu be very similar since they are next to each other on the periodic table?
Edit: I've added some sources and clarity to the discussion in the answer below.

I wonder, what you mean by "advanced level"? Potentials are connected with Gibbs' enthalpy in simple way, but don't arise from electron configuration in any simple way, neither do other common properties.

I meant at a graduate level, preferably from a theoretical standpoint. For example, if I were going to computationally derive the redox potential, what properties would go into that model? Can you elaborate or provide a link that would help?

Also, I found NIST enthalpy for ionization energies and Cu and Zn 2 electron ionizations were very similar, but their redox is not?

From a theoretical standpoint you will have to  account for solvation energies.

As I understand it, the potential is related to the difference in energy between the oxidized and reduced state (via the Gibb's energy as Mithoron noted). While their individual electron orbital energies are probably similar given the similar nuclear charges, the oxidized and reduced states are differently occupied and so have different energies.
In zinc for example, you have the reduced state of 4s2, 3d10 and oxidized state of 3d10. Copper on the other hand goes from 4s1, 3d10 to 3d9. $\ce{Zn^2+}$ has a fully occupied d-shell while $\ce{Cu^2+}$ does not, meaning that the reduced state for zinc is lower energy and more stable. As a result, zinc has a stronger tendency to oxidize and consequently a lower reduction potential.

Okay, that makes sense about not having a complete shell, but 1.1 V of a difference? I still can't wrap my head around going from a full d shell to 1 e- off equating to that huge of potential difference. Why isn't iron or cobalt or silver similar in difference? Why doesn't redox potential follow a periodic trend like other phenomena (assuming they are all eg 1 electron reaction). The explanation is probably complex, but perhaps someone can point me toward a good source? I couldn't find anything in my chemistry and EC texts.

I was able to get some clarity on my own question, which may lead to further discussion. From "Inorganic Chemistry in Aqueous Solution by Jack Barrett," Section 7.4.2. The portion below is paraphrased for brevity.
The two-electron reduction potential related to the enthalpy of oxidation of half a mole of dihydrogen to give one mole of hydrated protons and ionization energies is as follows:
$\ce{E(M^2+/M)=-[-\Delta _{hyd}H(M^2+) + 840] + [I_2 + I_1 + \Delta _aH(M)]/2F}$
The term [$\ce{-\Delta_{hyd}H(M^2+) + 840}$] tends to make the value negative, while the latter term has the opposite effect (paraphrased). For Ca, V, Cr, Mn, Fe, Co, Ni, and Zn, these two terms combine to give a negative SRP, while Cu has a positive value.
The unique behavior of Cu is attributed to its inability to liberate $\ce{H_2}$ from dilute acids (i.e., those with molar concentrations of 1 M). The reason for this behaviour is the high energy needed to transform Cu(s) to $\ce{Cu^2+ (g)}$, which is not exceeded by the hydration enthalpy of the ion. (as @porphyrin suggested with regards to solvation energies)


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Anti-bonding molecular orbitals
The anti-bonding molecular orbitals are higher in energy than the constituent atoms in a diatomic molecule. That means it is less stable. Why does it form? Why can't all the molecular orbitals be bonding molecular orbitals?

There ain't no such thing as a free lunch. You got some stabilisation into the bonding MO, so the antibonding MO has to get destabilised.

Molecular orbitals form when you apply the "Linear Combination of atomic orbitals" (LCAO) model to atoms bonding. Where you take a linear combination of the relevant atomic orbitals to create an MO. In each case of combination a bonding and an anti bonding (ab) combination forms. The ab MO form as a result of applying the Huckel model which results in a square root in the expression for the energy, thus giving a negative and a positive contribution to the energy. One of the expressions E1 will be lower in energy and will form the bonding and the second E2 will form the higher energy MO - anti bonding.
Sometimes a non-bonding orbital forms, which doesn't contribute to the total bond energy, where the squared root expression equals zero.
To understand it more thoroughly one needs to understand the hydrogen atom quantum model and it's solution which give the orbitals we know as: s,p,d,f etc. In the case of MO we apply say that each atom "brings" it's set of occupied orbitals from it's valence electronic shell, combining these gives MO diagrams.
It's an approximation but the results it gives describe to some extent what can be measured in the lab.


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History of the word copperas
I searched for copperas on google and it says ferrous sulphate. My dictionary shows that the word copperas is derived from copper.
I was just wondering what has ferrous sulphate to do with copper

You are on the right track.
The Brockhaus Konversationslexikon (Brockhaus Encyclopedia, 14th edition, 1894-1894)
explains the German term Kupferwasser (copper water) as

veraltete Bezeichnung für unreinen (kupferhaltigen) Eisenvitriol

which means outdated name for impure (copper-containing) iron vitriol.
Note that iron vitriol is an old term for iron(II) sulfate,  such as vitriolic oil is an outdated name for sulfuric acid.
If you take a look at English wikipedia page on iron(II)sulfate, you'll see that one of the common minerals is a heptahydrate, melanterite. If you follow the link, you will recognize that one of the associated minerals is chalcanthite, which is $\ce{CuSO4·5H2O}$.
To cut the long story short, the term copperas stems from the fact that natural iron(II)sulfate was often found to be associated and maybe contaminated with copper(II)sulfate.


text:
Can anions be reduced at the cathode in an electrochemical cell?
Two related questions:
(1) It's my understanding that for a galvanic cell, as electrons flow from the anode to the cathode, the cathode (+) becomes negatively charged due to the accumulation of electrons. If I have a sodium chloride electrolyte, the sodium ions will move toward the cathode. However, does that mean chloride will not be reduced at the cathode?
(2) Is the same true for an electrolytic cell? The cathode is (-), so I assume it stays negative. But, what if I wanted to force the reduction of an anion like perchlorate - is it possible or will there be too great of an electrostatic barrier?
I see people reporting on the electrocatalytic reduction of anions, but I don't understand how this is possible if the ionic current is the opposite direction.

If the anion is the species that gets reduced, it will do so at the cathode.

Reduction should happen exclusively at the cathode. Oxidation at the anode. By definition.

Yes, of course anode is oxidation, cathode is reduction. I'm asking about movement of ions toward electrodes. So, ionic movement is driven by potential difference of the ionic species?

Theoretically, it is possible for anions to get reduced more. According to wikipedia's reduction potential table page, this reaction occurs at +0.08V vs SHE:
$\ce{S4O6^2− + 2 e− ⇌ 2 S2O3^2−}$
In your chloride example, though, you're trying to reduce $\ce{Cl^-}$ to (at least) $\ce{Cl^2-}$ and the potential for that reaction would be extremely negative. In most real-world scenarios, you would start reducing other things way before you hit that potential. Most likely you'd start reducing water at -0.8V.
$\ce{2H2O + 2 e− ⇌ H2(g) + 2 OH−}$
As an aside, be careful with your terminology. Reduction always happens at the cathode, that's what "cathode" means. Sometimes we get sloppy and don't switch the terms when we go from galvanic to electrolytic mode, but that doesn't change the chemistry. If electrons are flowing into an electrode, it's the cathode. I find the (+)/(-) labels on terminals more confusing than anything else.
Also, you said that

for a galvanic cell, as electrons flow from the anode to the cathode, the cathode (+) becomes negatively charged due to the accumulation of electrons

This is backwards. The cathode does not become negatively charged because electrons have flowed into it. Rather, reduction occurs spontaneously at the cathode, leaving it electron-deficient. Electrons then flow from the anode to the cathode to compensate this charge imbalance and make both sides neutral again.

Okay, I think we are referring to two different states of a galvanic cell. I'm referring to the point just before oxidants diffuse to the electrode surface. At that point, the cathode would be negatively charged as no reaction has occurred, not positive as you have stated, which would be at equilibrium. However, if the oxidant never reaches the electrode due to electrostatics then no reduction will occur. That's what I'm really asking here: what's the state of electrostatics? Will anions be able to diffuse (effectively) toward a negatively charged cathode?

Wait, or are the ions already adsorbed to the electrode before the electrons arrive? This is what caused the flow of electrons in the first place...this created potential difference? So, I can ask this another way, if assuming the cathode is net positively charged throughout the experiment - will cationic oxidants effectively diffuse to the electrode surface or will there be electrostatic repulsion preventing this?

Ok, let's say the cell is not connected to start with. Electrodes are submerged in the electrolyte. The oxidant diffuses through the solid-electrolyte interface and adsorbs to the surface. An electron transfers from the electrode's conduction band to the oxidant, which gets reduced, leaving the electrode with a net positive charge. The quickly stops as electrostatics repels more oxidant. Now we connect the external circuit and electrons flow from the anode to the cathode to balance the built up charge, which relieves the electrostatic repulsion and let's redox continue.

Okay, so that brings me back to my original thought - can anions or cations be reduced at the cathode, irrespective of its charge? ie there is no electrostatic barrier?

Sure. It's just really hard to do for most anions and neutral metals.


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How to tell the needed doping element for semiconductors?
I was reading about a crystal that is a semiconductor, galena. I am confused on how to tell what elements are needed to dope it to make a p or n type semiconductor, as well as how to know which elements to use for doping other semiconductors.

A priori, you might have a really hard time. One would need to look at it experimentally, or perhaps with modeling like density functional theory (DFT). While you are, perhaps, lulled into a false sense of security through thinking you understand silicon, remember that even there boron is a great dopant while nitrogen is  absolutely not. And don't try to compare silicon with, say, diamond...


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Why is 2-methyloxirane not named 1-methyloxirane?

The image above depicts 2-methyloxirane. Shouldn't the numbering be so that the methyl substituent receives the lowest number, being 1?
So shouldn't the name should be 1-methyloxirane?

Ethyl substituent gets the lowest number.

According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), heteromonocyclic parent structures with three through ten ring members and containing one or more heteroatoms are generally named by using the extended Hantzsch-Widman system; however, retained names are preferred for several heteromonocyclic compounds. When a single heteroatom is present in the ring, it is assigned the locant ‘1’.

P-22.2.2.1 Constructing and numbering Hantzsch-Widman names
P-22.2.2.1.1 (…) The presence of a single heteroatom determines the numbering in a monocyclic compound; the heteroatom has the locant ‘1’ and the numbering usually proceeds clockwise, when unsubstituted.
  (…)

Therefore, the oxygen atom in oxirane has the locant ‘1’.

This numbering is also used in substituted heteromonocyclic compounds. The heteroatom keeps the locant ‘1’; then low locants are given to the substituent prefixes. If there is still a choice, the lowest locant is given to the prefix cited first in the name.

P-14.4 NUMBERING
When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority:
(a) fixed numbering in chains, rings, or ring systems, i.e., when the numbering of a system is fixed, for example in purine, anthracene, and phenanthrene, this numbering must be used, both in PINs and in general nomenclature;
(…)
(f) detachable alphabetized prefixes, all considered together in a series of increasing numerical order;
(g) lowest locants for the substituent cited first as a prefix in the name;
(…)

Therefore, the oxygen atom in 2-methyloxirane has the locant ‘1’ according to (a), and then the lowest possible locant ‘2’ is given to the methyl substituent according to (f).

Accordingly, the oxygen atom in 2-ethyl-3-methyloxirane has the locant ‘1’ according to (a), which leaves the locants ‘2’ and ‘3’ for the substituents. According to (g), the compound is named 2-ethyl-3-methyloxirane rather than 3-ethyl-2-methyloxirane since the prefix ‘ethyl’ is cited first in the name.

The name goes by the highest priority carbon 2,3.
ethyl side=>  ring carbon connected to a carbon has more priority C,H,H
methyl side=> ring carbon connected to a carbon has less priority H,H,H
That's why you should give number 2 to the ethyl.

No, the Cahn-Ingold-Prelog system is not used to determine the locant numbers. Ethyl comes before methyl simply because of alphanumerical ordering: E comes before M in the alphabet


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Where did amino acids get their one-letter codes?
Some amino acids have a one-letter code that's just the first letter of the name of the amino acid. This makes sense and obviously, since there is more than one amino acid that begins with the same letter, other letters had to be used.
But why were these other letters chosen? For instance, is there a reason 'W' specifically was chosen for tryptophan (other than the fact that 'T' was taken)?

http://www.biology.arizona.edu/biochemistry/problem_sets/aa/Dayhoff.html

Tryptophan

For instance, is there a reason 'W' specifically was chosen for tryptophan (other than the fact that 'T' was taken)?

Once you have assigned the other 19 amino acids, there are only 7 letters of the alphabet left: B, J, O, U, W, X, and Z. (Certainly not a nice Scrabble hand to have!)
If one wants to use a letter found within the name of the amino acid, the only available letter would be O. However, the usage of U and O was historically discouraged because these letters could be easily confused with other letters (U with V; O with G, Q, C, D, and the number 0).
It turns out that the choice was made because W is a very fat letter and was reminiscent of the indole ring system present in tryptophan (the only amino acid to contain a bicyclic system). (See below for source.)

What happened to the other six letters, then?
All 26 letters of the alphabet now find use as a one-letter code for amino acids or various combinations thereof.
After their discoveries, selenocysteine and pyrrolysine (the latter only found in bacteria) were assigned U and O respectively. Furthermore, B is used to represent aspartic acid OR asparagine; J is used to represent leucine OR isoleucine; Z is used to represent glutamic acid OR glutamine; and X is used to represent an unknown amino acid.
I believe B and Z find use because, in protein sequencing, acid hydrolysis is often used to break peptide bonds. This has the undesirable side-effect of hydrolysing the amide groups in asparagine/glutamine, leading to the formation of aspartic/glutamic acids, which means that one cannot tell exactly which amino acid it was at the start. J is used in NMR spectroscopy where isoleucine and leucine are difficult to distinguish.

Why did they choose the letters they did?
As far as I am aware the usage of one-letter symbols is adopted by both IUPAC and IUB (since 1991, IUBMB) in their joint 1983 recommendations on "Nomenclature and Symbolism for Amino Acids and Peptides".1 ankit7540's concise summary of the historical development already mentioned these recommendations.
In particular, Section 3AA-21.2 "The Code Symbols" has a description of why the letters were chosen. This document is probably the most authoritative stance on the matter. The rationale is mostly in line with Jan's answer:

Initial letters of the names of the amino acids were chosen where there was no ambiguity. There are six such cases: cysteine, histidine. isoleucine, methionine, serine, and valine. All the other amino acids share the initial letters A, G, L, P or T, so arbitrary assignments were made. These letters were assigned to the most frequently occurring and structurally most simple of the amino acids with these initials, alanine (A), glycine (G), leucine (L), proline (P) and threonine (T).
Other assignments were made on the basis of associations that might be helpful in remembering the code, e.g. the phonetic associations of F for phenylalanine and R for arginine. For tryptophan the double ring of the molecule is associated with the bulky letter W. The letters N and Q were assigned to asparagine and glutamine respectively; D and E to aspartic and glutamic acids respectively. K and Y were chosen for the two remaining amino acids, lysine and tyrosine, because, of the few remaining letters, they were close alphabetically to the initial letters of the names. U and O were avoided because U is easily confused with V in handwritten material, and O with G, Q, C and D in imperfect computer print-outs, and also with zero. J was avoided because it is absent from several languages.
Two other symbols are often necessary for partly determined sequences, so B was assigned to aspartic acid or asparagine when these have not been distinguished; Z was similarly assigned to glutamic acid or glutamine. X means that the identity of an amino acid is undetermined, or that the amino acid is atypical.

One can only hypothesise what they meant by "associations that might be helpful in remembering the code" in the case of N/Q/D/E. My best guess is:

D and E were possibly chosen for aspartic and glutamic acids because they were the only consecutive pair of letters left, emphasising their chemical similarity. Aspartic acid is shorter than glutamic acid by one methylene group (CH2), so it gets the earlier letter D.
Glutamine sounds like Q-tamine. If you don't think it sounds similar, repeat it 50 times until you do.
AsparagiNe was assigned N.

Reference

IUPAC-IUB Joint Commission on Biochemical Nomenclature. Nomenclature and Symbolism for Amino Acids and Peptides: Recommendations 1983. FEBS J. 1984, 138 (1), 9–37. DOI: 10.1111/j.1432-1033.1984.tb07877.x. A HTML version (perhaps more user-friendly) can be found at this address.

I've always used the mnemonic "asparDic acid", though how much that was in the minds of the biochemists who originally assigned it, I don't know.

It's anybody's guess!

For D and E the oxygen atoms of the side chains are delta and epsilon atoms, respectively: Oδ1,2 and Oɛ1,2.

Some single letter codes that aren’t the amino acid’s starting letter actually make sense when viewed from certain angles. Here’s the list starting with the bloomin’ obvious:

G — Glycine
A — Alanine
V — Valine
L — Leucine
I — Isoleucine
P — Proline
S — Serine
T — Threonine
C — Cysteine
H — Histidine
M — Methionine

Some amino acids have a letter that alludes to how the name is spoken rather than written:

F — Phenylalanine (starts with /f/)
R — Arginine (the first syllable could be are or the pronunciation of the letter R)
N — Asparagine[1]

One amino acid is labelled by its second letter (two if you count arginine above):

Y — Tyrosine

A few were labelled with letters close to their starting letter because the actual starting letter is taken:

K — Lysine (L is already leucine)
E — Glutamate (F and G were taken; but so were H and I. E is at least a letter of the name)
D — Aspartate (Basically a shorthand version of E, glutamate)

Which leaves us with tryptophan and glutamine. Glutamine is another really stretchy one:

Q — Glutamine (C and G was taken but Q is round on the left just like a G. Totally obvious?)

And finally, for tryptophan, they decided to take the actual shape of the molecule. You don’t see it? Here it is:

W — Tryptophan

Actually, perhaps it’s a mere coincidence as in they tossed a coin with the letters left; for both glutamine and tryptophan. But I like this explanation.

I always remembered tryptophan by mentally pronouncing it trip-TOW-fan, thus injecting the needed W.

See IUPAC-IUB Commission on Biochemical Nomenclature
A One-Letter Notation for Amino Acid Sequences The Journal of Biological Chemistry Vol. 243, No. 13, pp. 3557-3559, 10 July 1968. :

The possibility of using one-letter symbols was mentioned by
  Gamow and Ycas (2) in 1958. The idea was systematized by
  Šorm et al. (3) in 1961. It was used by this group (4-10) and
  also by Fitch (11) in several papers on the structure of proteins.
  In extensive compilations of protein structures, Eck and Dayhoff (see References 12 to 14) systematically used one-letter symbols
  derived partly from the code of Šorm and Keil. Independent
  proposals were made by Wiswesser (15) and by Braunstein.

...

3.2 Initial letters are used where there is no ambiguity. There
  are six such cases: cysteine, histidine, isoleucine, methionine,
  serine, and valine. All the other amino acids share the initial
  letters A, G, L, P, or T; therefore, assignments of them must be
  somewhat arbitrary. These letters are assigned to the most frequently
  occurring and structurally most simple amino acids.
  On this basis, the letters A, G, L, P, and T are assigned to alanine,
  glycine, leucine, proline, and threonine, respectively.
3.3 The assignment of the other abbreviations is more arbitrary. However, certain clues are helpful. Two are phonetically
  suggestive, F for phenylalanine, and R for arginine. For tryptophan,
  the double ring in the molecule is associated with the bulky
  letter W. The letters N and Q are assigned to asparagine and
  glutamine, respectively; D and E are assigned to aspartic acid
  and glutamic acid, respectively. This leaves lysine and tyrosine,
  to which K and Y are assigned. These are chosen rather
  than any of the few other remaining letters because they are
  alphabetically nearest the initial letters L and T. U and O are
  avoided because U is easily confused with V in handwritten
  work and O is confused with G, Q, C, and D in imperfect computer
  print-outs and also with zero. J is avoided for linguistic
  reasons.
3.4 Two other abbreviations are necessary in order to avoid
  ambiguity. B is assigned to aspartic acid or asparagine when
  this distinction has not been determined. Z is assigned when
  glutamic acid and glutamine have not been distinguished. X
  means that the identity of an amino acid is undetermined or that
  the amino acid is atypical.

The three-letter designation came before the single letter code. The three-letter codes themselves went several refinements and revision finally accepted by IUPAC and International Union of Biochemistry under the leadership of HBF Dixon. Later single letter codes were proposed. (apparently, because they were tiresome to write). This proposal came from Czech biochemists. (check history of the nomenclature of amino acids)
Later, the single letter codes were recommended and accepted in 1994 (in a similar meeting of IUPAC and IUB). The recommendation can be found here Amino acid nomenclature recommendation, also available here NCBI. After acceptance of this new nomenclature, several major publications published the new guidelines.


text:
What happens to acids/bases left out overnight?
I conducted a titration experiment a few days back and was unable to neutralize the solution in time, so I had to leave the solutions of HCl and NaOH out overnight in order to use them the next day. Also, one of my partners blew into the buret to get a drop of NaOH out. At the end of the experiment, we found that we required 6-7 mL more base to neutralize the HCl solution; another group who was forced to leave their solutions overnight used 3 mL more even than we did.
My hypothesis is that CO2 from the air formed carbonic acid in both solutions, mainly due to the instance where my partner blew into the buret, and thus the acidity of both solutions required that we use more base than we would have originally. As far as I've researched, this seems to be the case. Just to be sure, though: is this a correct assumption? If not, what could be the cause of this discrepancy?

Well, sodium alkali draws CO2 from air (carbonates not carbonic acid are created) but it depends on concentration of NaOH and water used, as normally there's CO2 in there from the start.

http://www.middleschoolchemistry.com/lessonplans/chapter6/lesson10 - How would you explain this, then? Is this property of CO2 not applicable to the solutions present, or what?

This is what is called an analytical BLUNDER. It is like making a wrong turn and the GPS keeps yelling at you. Reagents must be used properly and procedures exacted and data carefully recorded for results to be meaningful. You have failed in all three. Repeat the experiment properly. Then if you really want to know what goes on over night set up a controlled experiment.

First of all, blowing into the buret seems like a less-than-safe thing to do. $\ce{NaOH}$ could easily fly out into his/her mouth and nose.
I would expect that your acid and base solutions were already at the equilibrium concentration of $\ce{CO_2}$ when you got them, since they were exposed to air. I think that dissolving large amount of $\ce{CO2}$ into solution at atmospheric pressure is no easy task. I just don't think small amounts of $\ce{CO2}$ would account for the difference.
When you say "left them out" do you mean uncovered? If so, there's probably a fair amount of evaporation happening. The analyte solution could be affected depending on your experiment, and the concentration of your titrant would go up, which means you'd actually need less of it to reach the end-point.

When acquiring both the Acid $\ce{HCl}$ and the base $\ce{NaOH}$, your teacher should have given you a certain amount of either one to start with. Using the data of the given quantity you can acquire the necessary amount of base/acid needed to neutralize the other using the concentration and volume. This is known as the theoretical value needed to carry out the neutralization reaction $\cdots$ and this is probably where from my understanding you misunderstood why you needed more base. Your titration was not wrong as the experimental value (value gotten when experiment is carried out) is usually higher than the theoretical value meaning if 23 ml was required but you used 28 ml, its fine. However the real answer to your question lies in paralax errors (known as reading errors), human errors and friction along the inside of the burret, which causes the discrepancy. Like for example if you wanted to get 200 mL of $\ce{HCl}$ .. you might of gotten 200.5 mL where as the other group got 201 or 202 mL etc.

Yes that's what happened. From my experience, if you want accuracy you need to use the NaOH solution within an hour. Depends on the [NaOH] and the volume/[] of HCl you're titrating. The smaller [NaOH], the more impactful the CO2 has.
To overcome this, it's necessary to do a reference titration using the same NaOH solution to titrate equal amounts of HCl, if you're using the titration analytically to find the [] of another substance in the solution, etc. For example, if you're doing a back titration that combines a solution with an unknown [] of base to a known quantity of HCl, it's important to do a reference to obtain accurate results. Regardless if you've left out the solution or not, you should do this since NaOH and HCl solutions are prepared by hand and are never exactly the concentration you make them as. If you're concerned about the concentration, always do a reference titration before and after the back titration. If it changes slightly, use your judgment as to the reference [NaOH] at the times of the titrations.
I hope this helps you!


text:
How to convert the D'Arcy and Watt model from activity to concentration in mol/m^3
The D'Arcy and Watt 1970 model in "Analysis of sorption isotherms of non-homogeneous sorbents" for water adsorption is as follows:
$$W = \frac{KK'a}{1+Ka} + ca + \frac{kk'a}{1-ka},$$
where:

$W$ is weight of vapour adsorbed per $1~\pu{g}$ of substrate per $1~\pu{g}$ sorbent ($\pu{g/g}$)
$a$ is the water activity or relative humidity (no units, number between $0$ and $1$ or if relative humidity is used - is percentage)
rest are constants with no units

I want to use this model to consider water concentration in $\pu{mol m^-3}$ instead of water activity.
How do I convert this?

The general relation between activity (effective concentration) and concentration is as follows (see Wikipedia for more detail):
$$a_i = \gamma_i \frac{c_i}{c^\circ}$$
The standard concentration is defined as $c^\circ=1~\pu{mol dm^-3}$. For extremely dilute systems it is possible to just use the concentration instead of the activity (often used as a first order approximation, too):
$$\lim_{c_i\to0}\gamma_i\approx1 \implies a_i\to c_i$$
Depending on the accuracy necessary for your model, you might have to look up the specific values.
In any case you should again look up the definition of activity (or relative humidity) in D'Arcy and Watt's paper, because the activity defined for the above relation could be larger than 1.

Ok thanks thats great! Also, you mentioned that $a$ could be larger than 1. They have defined $a$ as $p/p_0$. How much further can that go above 1? I know relative humidity in extreme cases can go higher than 100% to say 102%, is that what you mean?

@Eloise It seems that your paper is not using the relative activity which is defined as $$a=\exp\left{\frac{\mu-\mu^\circ}{RT}\right}.$$ There is no (mathematical) reason, why there should be an upper limit. If they define it as $a=p/p_0$ and $p_0$ being the total pressure, it can indeed be 1 at maximum. Your question does not provide enough context to answer that. From the definition it is related to food chemistry, but I can't help you there.


text:
Effect of pH on electrolysis
I understand (more or less) that acidification of water helps in the dissociation of water into ions. Acids easily dissociate into ions in aqueous solution. So, accordingly, the hydrogen from the acid goes to the cathode and the hydrogen from the water makes a compound with the negative radical from the acid. In the process, oxygen is left alone and hence goes to the anode. I want to confirm whether this is the correct explanation.
I have been trying to find out why is copper sulphate acidified during its electrolytic refining.  I interpret  that acidifying copper sulphate makes it dissociate easily into ions but I want to know what the exact process is. It doesn't seem to me that copper sulphate may behave in the same way as water does above. Please help.
In general, I would like to understand the effect of decreasing or increasing pH on the process of electrolysis.

I'd rather not put it in such terms. Dissociation of water is determined by the constant $\ce{K_w=[H+][OH-]=10^{-14}}$, which you can't change simply by adding acid (or anything else, for that matter). Acids dissociate easily into ions, just like you said, therefore adding acid to pure water (which is a poor conductor by itself) makes the latter more conductive and thus more prone to electrolysis.
Electrolytic refining of copper is a whole different story. Indeed, once dissolved in water, $\ce{CuSO4}$ would dissociate into ions completely, no matter whether or not you acidify it. The role of acid is just to prevent hydrolysis.
In general, as you change pH, some of your ions may undergo certain transformations (like $\ce{CO3^2- -> HCO3- -> CO2\uparrow}$, or $\ce{Fe^2+ -> Fe(OH)2\downarrow}$), which in turn might or might not affect the reactions on one or both electrodes. As it happens, being too general prevents us from drawing more specific conclusions.

Thanks for your response. Would you please also mention how does copper sulphate hydrolyze in water.

Honestly I don't know much about hydrolysis of copper sulphate. So i want to ask whether something with happen if copper sulphate is left alone with water. Will calcium sulphate and water react to form other compounds

$\ce{CuSO4}$ by itself hydrolyzes only partially; however, with local variations of pH created by electrolysis, it may go all the way to $\ce{Cu(OH)2}$ or some basic salts which would precipitate from the solution, and you don't want that to happen.

You are right in that adding acid increases the concentration of dissociated ions in water, and that the hydrogen goes to the cathode but you have assumed too much with the oxygen part.
In electrolysis/electroplating, the positive ions move towards the negative cathode and are reduced, and vice-versa, the negative ions move towards the positive anode and are oxidized. For a pure acid solution (no salts), hydrogen is the only postivie ion so it migrates towards the cathode, however not only do the hydroxide ions move towards the anode, but so do the conjugate bases of the acid. These conjugated bases may or may not be oxidized before the hydroxide depending on potentials and available mechanisms for reaction to occur.
For example chlorine can oxidize before oxygen and electrolysis with chlorides forms aqueous chlorine as shown $\ce{2Cl-_{(aq)} -> 2e- + Cl2_{(aq)}}$ and sulfate can form persulfates at the anode: $\ce{2SO4^2- -> (SO3OOSO3)^2- + 2e- }$. These bases compete with the $\ce{OH-}$ in solution to be oxidized. These species also tend to be very reactive and will often oxidize the anode unless an inert material such as graphite or platinum is used (sometimes nickel will work).
In electroplating, the metal ions compete with the hydrogen ions to be reduced and indeed hydrogen gas is formed at the same time that electroplating occurs.  for ever $\ce{H+}$ reduced there is a $\ce{OH-}$ left in solution. in electrolysis the electron to reduce the $\ce{H+}$ would come from the $\ce{OH-}$, but in electroplating it comes from the metal (i.e. $\ce{Cu -> Cu2+ + 2e-}$). NOw the elctroplating process has created $\ce{Cu2+}$ ions and left $\ce{OH-}$ ions in excess which creates $\ce{Cu(OH)2}$ which precipitates creating anode mud.
This is where the acid comes in to keep the bath clean, I'll use sulfuric acid as an example. The sulfuric acid prevents the formation of $\ce{Cu(OH)2}$ by providing $\ce{SO4^2-}$ and $\ce{H+}$ ions to react with the $\ce{Cu(OH)2}$ and keep the copper in solution ($\ce{Cu(OH)2 + H2SO4 -> CuSO4 + 2H2O}$).


text:
Why does IR spectroscopy specifically require a change in dipole moment, and why doesn't this apply to other types of spectroscopy?
I would really appreciate any amount of detail/depth. I do not have a background in quantum, which is probably why I don't know the answer to this. However, any answer referencing quantum concepts is welcome and appreciated; I will follow up on any resources provided.
My question is basically as follows: Why does IR absorption require a change in dipole moment, and why is it that other types of spectroscopy (UV Vis, microwave) don't have this requirement? What causes this requirement specifically for IR?
Thank you!

All the types of spectroscopy you mention have a similar condition for a spectrum to be observed.  This is that the molecule has to be able to interact with the radiation in the first place and then the radiation has to be of the correct energy to change the energy of the molecule from one level to another.
A plain (classical) explanation is as follows. (We implicitly assume that the radiation has the correct energy). As radiation is electric dipole in nature (it consists of oscillating electric and magnetic fields) if the molecule also generates an oscillating electric field it can interact with the radiation; energy is taken from the radiation field and enters the molecule. Thus a molecule may gain a quantum of vibrational or rotational energy or an excited state may be produced.
In the case of a vibrating diatomic molecule only if it is hetero-nuclear e.g. HCl, will it have a dipole and thus an oscillating dipole and oscillating electric field as it vibrates. Thus hetero-nuclear diatomics show an IR spectrum, homo-nuclear diatomics do not. In a polyatomic molecule vibrations can distort the molecule, e.g. benzene is distorted so that it is no longer 6 fold symmetric, so that a  dipole is produced and an IR spectrum may result. What form this takes depends on the molecule's symmetry and group theory point groups are used to determine this.
In a rotating diatomic molecule we usually use a rigid-rotor model. If the molecule has a dipole then this clearly changes as the molecule rotates, produced an oscillating electric field, and so a microwave (rotational) spectrum is produced. A homo-nuclear diatomic e.g. $\ce{O2}$ has no microwave spectrum.
In a UV/visible transition an excited state may be produced but only if the electron distribution in the excited state is different from that of the ground state, i.e. a dipole is produced that can interact with the radiation. A molecule such a benzene has a forbidden first electronically excited state due to its symmetry (which is however observed weakly due to other effects) and so absorbs very little compared to aniline (amino-benzene) or many dye molecules where symmetry does not limit any electronic dipole formation.
Thus although the details vary, the underlying ideas are the same. You should understand now that the statements about spectra in your question are not all correct. The rules used to determine what transitions are allowed or not allowed (called forbidden) are called 'selection rules'.
You can read about this in more detail, as a quantum version, in most undergraduate phys. chem. textbooks.
You may realise that the vibrational frequencies of homo-nuclear molecules are known even though they have no ir spectrum and this is because they can be measured by Raman spectroscopy. This depends on the change in shape/size of the 'electron cloud' (polarisability) a molecule possesses. Electro-magnetic radiation also possesses an oscillating magnetic field but this interacts far more weakly with molecules than its electric counterpart and can often but not always be ignored, the exception being NMR spectroscopy.

Good explanation.

Lets begin with the absorption.
The probability of ANY absorption taking place is given by the the Einstein's B coefficient, which is \begin{equation}
    B_{n^{\prime}n^{\prime\prime}}=\frac{8 \pi^3}{3h^2}[|\int \psi_{n^\prime}\mu_{X} \psi_{n^{\prime\prime}}d \tau|^2 + |\int \psi_{n^\prime}\mu_{Y} \psi_{n^{\prime\prime}}d \tau|^2 + |\int \psi_{n^\prime}\mu_{Z} \psi_{n^{\prime\prime}}d \tau|^2]
\end{equation}
Above equation is in complete generality, and the wave function $\psi$ is the complete wave function. $\mu_{X}$ is the dipole moment component (in X, using Cartesian coordinates). Integration is over all space.
RULE: if the dipole moment components for a given transition are zero, the absorption will not take place. (By components I mean the integral. If the integral goes to zero then no absorption for that component, if all three integrals are zero then no absorption at all) This determines the selection rules for all absorption processes (and also induced emission, but not discussing here !).
Back to absorption, the wave-function for the molecule  can be expressed as
\begin{equation}
    \psi= \psi_{E} \psi_{V} \psi_{R} \psi_{T}
\end{equation}
where $\psi_{E}, \psi_{V}, \psi_{R}$ and $\psi_{T}$ are the electronic, vibrational, rotational and translational wave-functions. For vibrational analysis the electronic wave function does not matter and we can safely exclude it. (This is the wave function needed for the calculation of the integral. For general understanding consider the wave function as a mathematical function whose integral gives you information about the molecule.)
The translational selection rule is that the change in the quantum number should be $\Delta n =0$ that is the quantum number does not change (while translation in space).
[At the stage we have separated the electronic and translational wave functions, and we deal with only vibrations, hence $\psi_{v}$.]
Coming to rotational and vibrational part,
If $X_{\alpha}$, $Y_{\alpha}$ and $Z_{\alpha}$ is the coordinate which moves with the molecule, then a new coordinate can be introduced which moves and also rotates with the molecule. This is say $x_{\alpha}$, $y_{\alpha}$ and $z_{\alpha}$.
 The relation between $X_{\alpha}$ and $x_{\alpha}$ are governed by angles expressed as trigonometric relations, also known as direction cosines.
 Advantage of this transformation is that we can separate vibration from rotation and discuss vibration.
The electric moment of the full system of molecule has three components in Cartesian coordinates, \begin{equation}
    \mu_{x}=\sum_{\alpha}e_{\alpha}x_{\alpha}
\end{equation}
\begin{equation}
    \mu_{y}=\sum_{\alpha}e_{\alpha}y_{\alpha}
\end{equation}
\begin{equation}
    \mu_{z}=\sum_{\alpha}e_{\alpha}z_{\alpha}
\end{equation}
Now, during vibration the charged entities are moving. To be more clear I would say the nuclei are moving. The dipole moment is dynamic and changing with the vibration. The changing dipole moment can be expressed a Taylor expansion so as to include the change from equilibrium position. This expansion is carried out using the normal coordinates( normal coordinates are coordinates of the atoms which change with the vibration, when all atoms move with same frequency but different amplitudes).\begin{equation}
    \mu_{x}=\mu_{x}^{0}+\mu_{x}^{1}Q_{1}+\mu_{x}^{2}Q_{2}+...
\end{equation}
where $\mu_{x}^{n}$ is the nth derivative. Generally, only the first derivative is taken and hence, \begin{equation}
    \mu_{x}=\mu_{x}^{0}+  \frac{d \mu_{x}}{dQ_{1_{0}}}\hspace{1ex}Q_{1}
\end{equation}
Above result is now used with the vibrational wave function.
\begin{equation}
    \int \psi_{n^\prime}\mu_{x} \psi_{n^{\prime\prime}}d\tau= \int \psi_{n^\prime} [ \mu_{x}^{0}+  \frac{d \mu_{x}}{dQ_{1_{0}}}\hspace{1ex}Q_{1} ]\psi_{n^{\prime\prime}}d\tau
\end{equation}
\begin{equation}
    \int \psi_{n^\prime}\mu_{x} \psi_{n^{\prime\prime}}d\tau=\mu_{x}^{0} \int \psi_{n^\prime}  \psi_{n^{\prime\prime}} d \tau + \int \psi_{n^\prime} [ \frac{d \mu_{x}}{dQ_{1_{0}}}\hspace{1ex}Q_{1} ]\psi_{n^{\prime\prime}}d \tau
\end{equation}
The first term would go to zero since the vibrational wave functions are orthogonal. In the second term, the condition of non-zero derivative of dipole moment is required to have this term being non-zero. Hence the change if dipole moment with vibration is utmost necessary.
Other comments: With similar approach one can derive, for UV and visible absorption that the dipole is required; and for microwave absorption we need a permanent dipole moment.
References:
Physical Chemistry by Atkins, Paula.;
Modern Spectroscopy by Hollas ;
(More advanced reference. Molecular vibrations by Wilson, Decius, Cross.)

I'll improve it further. Thanks for the 'wow'.


text:
Why is the ester bond stable under Reformatsky reaction conditions?
The following reaction is a Reformatsky reaction.

Why is the ester bond not attacked in the last step? Couldn’t $\ce{OR3}$ be replaced by $\ce{OH}$ or $\ce{O(molecule)}$?

You ask why the ester doesn't react with the H3O to form an actic acid and release an alcohol?

@AvishaiBarnoy yes , I want to know that

It probably occurs as it's an equilibrium step.maybe the base adding is under cold condition and prevents further reaction.

Yes, technically you have a zinc alcoholate which could display the basicity and nucleophilicity of any other alcoholate and you know that you can transesterificate with e.g. sodium methanolate.
Aside from the oxyanion here being rather closely associated to the zinc cation which has a larger positive charge than e.g. sodium and thus exercises a greater attractive force, there is also the fact that the alcohol in question is tertiary and tertiary alcohols are bad nucleophiles. In most of the cases, $\ce{R3}$ will be primary or at most secondary, thus the transesterification reverse reaction will be more rapid than the forward reaction.
If, however, $\ce{R3}$ is tertiary, too, you would be dealing with a huge steric bulk on that carbonyl carbon, attempting to have two tertiary $\ce{OR}$ groups on the tetrahedric intermediate. That is unlikely to happen for steric reasons.

If you are talking about the absolute final step, i.e. the aquaeous workup, then you will likely be using ammonium chloride solution or something else mildly acidic. In mildly acidic or otherwise buffered solutions, there is not a high enough concentration of hydroxide ions to attack the ester carbonyl sufficiently for an observable reaction to occur.


text:
Why can't a galvanic cell be a single cell?
I get that galvanic cells require a salt bridge to maintain neutrality so the cathode doesn't become saturated with electrons, but why is a two-cell setup required? Wouldn't the spontaneous reaction occur and neutral conditions be met if Cu and Zn electrodes were placed in a solution of NaCl?
And vice versa, couldn't an electrolytic cell work just fine with a salt bridge?
This website (and many others) leads me to believe that galvanic cells must have a salt bridge and electrolytic cells must be a single cell. My intuition says that's not right though. Is it because you don't want undesirable compounds reacting at the electrodes? I'd think you'd want the same for both EC cell types if that was the case and electrolytic cells would use salt bridges as well (eg proton exchange membranes).
http://chem.libretexts.org/Core/Analytical_Chemistry/Electrochemistry/Electrolytic_Cells

Sure, galvanic cell would work for a while in a single cell setup, but you'll get a lot of side reactions which you don't want. On the other hand, electrolytic cell would work just fine in a two-cell setup with a salt bridge, but then half of your solution would be effectively excluded from the reaction, which we didn't ask for. Also, welcome to Chem.SE.

Thanks. Wouldn't separating the cells in an electrolytic cell be more desirable than galvanic to reduce the back (spontaneous) reaction of by-products? For example, in electrolytic water splitting, the chamber are separated with a proton exchange membrane to reduce back reaction of hydrogen and oxygen to water. Why do introductory texts then claim electrolytic and galvanic cells must be 1 and 2 celled, respectively? Or are they just simplifying it?

That's it: they are just simplifying.

Side note: this whole industry is pretty much based on single-cell electrochemistry for that matter too...

Perhaps this isn't the full answer, but it probably plays a role. It is critical for galvanic cells that there not be any electronic conductivity through the cell between the two electrodes. If there is a path for electrons, they will take it which means the cell will spontaneously discharge and not do any useful work through the external circuit. Pure water is a very poor conductor but becomes conductive once ions are dissolved.
By comparison, the construction of a lithium ion cell is close to what you describe for a single chamber galvanic cell. There is a separator involved but mostly to keep the two electrodes from coming into direct contact. One reason this is possible is that the electrolyte is an organic solvent (usually ethylene carbonate, dimethyl carbonate, etc) and doesn't conduct electrons, even when the appropriate salt (eg. $\ce{LiPF6}$) is dissolved in it. There is no salt bridge in this cell, though it is clearly galvanic.
Regarding the electrolytic cell, I don't see any particular reason that you couldn't build one with a salt bridge, provided the salt-bridge let you balance the charges generated on each side. Maybe they're all presented as one beaker just because it's easier to make and there's not really any advantage to a salt bridge in this case.

What would stop electrons moving through an electrically conductive salt bridge then? For example, fuel cells and batteries don't really use a "salt bridge" as we see it drawn often, they use an ionic membrane. Can't electrons pass through that? But even then, why would they want to - aren't they "forced" to go through the electrode via the potential difference?

Nothing would stop them, that's why salt-bridges aren't electron conductive. I specifically mentioned why Li-ion battery cells don't have a salt bridge, and I think the same applies for fuel cells: the electrolyte is chosen to be non-electronically-conductive. In a galvanic cell, the potential difference is because the electrons can't move between the electrodes internally. In the extreme, imagine a wire connecting the two sides through the electrolyte: the cell would immediately self-discharge and be useless.