text:
Un-denaturing industrial alcohol
I have a question regarding the denaturation of industrial ethanol with methanol to discourage its consumption.
The boiling point of methanol is 64.7 °C, while the boiling point of ethanol is 78.37 °C. Wouldn't it be trivial then, to separate the methanol by simply distilling industrial ethanol at 64.7 °C ?
Is there some technical reason as to why this wouldn't work? Because if it does work, then why isn't it a widespread practice among criminal elements and such.

The mixture simply would not boil at 64 °C. That's not how mixtures work.

Some zeolites can selectively adsorb water and methanol and be regenerated by heating. So passing the denatured ethanol slowly through a column of zeolites should remove water and methanol.

“…why isn't it a widespread practice among criminals and such.” What does having to be a supposed “criminal” have to do with this?

@JakeGould It's 'criminal' because it deprives the government of the booze-taxes which it requires to function :P

A major reason why it isn't done is probably economics. It is cheaper and easier to make your own alcohol by fermentation and then distill it (and it is safer as fermentation doesn't give methanol as a contaminant).

@matt_black Wrongly prepared and distilled ferment leads to poor taste and methanol-rich spirits.

The article "The Chemist's War" http://www.slate.com/articles/health_and_science/medical_examiner/2010/02/the_chemists_war.html goes through the various changes to denaturing agents used after the passage of the 18th amendment (Prohibition). I think the cost/benefit tradeoff still makes it unattractive

I notice you didn't bother comparing freezing points...  The 16 C gap is easier to stay in using expensive commercial freezers.  However, not all denaturants are methanol.

@AaronAbraham Molecular sieve 4A absorbs water and methanol.

@matt_black Almost all fermentation products contain some methanol (eg coming from pectins and other esters), which then present even in high-quality booze.

 (I'm no chemist, but here's my take on it)
Let's see: I’m an alcoholic, and I really hate having to dole out large amounts of cash to purchase your usual "drinking"-ethanol.
So I get this amazing idea to separate the ethanol by distillation (methanol has a lower boiling point than methanol, so what I'll really be doing is distilling out the methanol, leaving behind ethanol and water). Pretty smart, right (for someone who's still inebriated)? Thus invigorated, I sally forth... and return with a dozen gallons  of industrial (denatured) ethanol. However, as I slowly get sober, I begin to see two major obstacles to my plan:

It’s going to be very, very difficult to maintain a constant temperature of exactly 64.7 °C (boiling point of methanol). Now, even though ethanol boils at around 78 °C, trying to maintain the denatured alcohol at under (say) 70 or 75°C isn’t really an option either. Remember, what I’m trying to do here is criminal (where I'm from at least), plus I don’t really have a Nurdrage-worthy laboratory. If say, I were to simply buy a large metal vat with a distillation apparatus clamped on the top and light a fire below, at such a large scale (a dozen gallons in my case) and with the non-professional equipment I use, it’ll be impossible to ensure that the vessel is uniformly heated at a specific temperature. So what’ll happen is, I’ll end up distilling out a mixture of ethanol and methanol, and the ethanol that remains will also have significant amounts of methanol in it (10ml of methanol's enough to make you go blind… just to give you an idea about what I mean by “significant”). Regardless, I wouldn't want to engage in something that involves producing methanol vapors (be it in a proper laboratory or otherwise)
Say I did manage to get hold of the required laboratory apparatus (I, uh, happen to know this cancer-stricken, bald, high-school Chemistry teacher who was willing to lend me them), and I were to distill the alcohol in small batches to ensure I evenly heat the solution at the desired temperature, but the whole process would be so time-consuming and labor intensive that it would probably just be better if I just went and bought "drinking"-ethanol in the first place.
(Should have mentioned this first, but since you were talking about methanol, ah well…) To begin with, I probably wouldn't know what the manufacturers actually used to denature the alcohol. They could have used methanol, pyridine, or some other foul-tasting, foul-smelling additive for all I know, or maybe even mixed all of them together. So trying to get ethanol out of that mess, is going to be a HUGE pain in the neck.

Also you might want to read up on denatured alcohol here.

I wasn't able to find anything on ethanol-methanol azeotropes in this Wiki page, so if anyone knows where I can find something on it, gimme a shout! :-)

10 ml of Methanol may be enough to blind you, but as long as you have about .7 g of Ethanol per kg in your body, due to competitive inhibition of Methanol being decomposed by ADH it will not have any effect. So if you just kept up the buzz, this shouldn't be an issue! (Not recommending it, though... ;) )

Methanol is indeed dangerous

LOL - It is no trouble to identify all the denaturing agents. You just run a sample through your $500,000 GC/MS. Distilling a couple of million quarts of alcohol will yield $500 or $600 on your investment.

"Remember, what I’m trying to do here is criminal ..." - Is it? I assume it's illegal to sell drinkable alcohol without charging/paying the appropriate taxes, but is it illegal to remove methanol from denatured alcohol for your own use? (Honest question, I couldn't find anything on the subject)

@marcelm Well, my CONSCIENCE tells me extracting 'drinkable' ethanol from industrial ethanol in order to avoid paying the taxes I'd normally pay to get a bottle of regular booze; taxes which enable our noble and selfless  government to improve public facilities, create jobs, boost our economy, wage useless proxy wars in Syria, continue screwing around with Russia, manipulate FBI reports on Clinton's ''misdemeanors '', etc, etc is a selfish and criminal act.

@marcelm Depends where you are. In the UK, it's legal to brew alcoholic drinks for personal consumption but illegal to distil, even for personal consumption.

I don't understand how the process in the answer is any different from, say, the distillation of corn whiskey; they boil several dozen gallons of fermented malt, throw away the first products which include most of the methanol, and then either redistill the rest or bottle it.

From what I read, there has been no documented case of moonshine ever causing blindness due to methanol, except for a few cases when they started denaturing industrial alcohol with 50% methanol, and the moonshiners decided to take that as a challenge.

@nexus_2006 you're dealing with a much narrower temperature gap and a much larger percentage of toxins making it more difficult.  If it's European denatured alcohol with Isopropanol and Butanone it's even worse with only 4/2 degree temperature deltas to work with.  Although as a counterbalance they're less toxic than methanol, you'd also have a bitterant to deal with (separately from distillation?).

@AaronAbraham I didn't ask about your conscience; I asked if the behaviour is actually criminal. How you (or I) feel about it has no bearing on that; for something to be criminal, it has to be against the law. So my question still stands: is it actually illegal?

@marcelm "Conscience" is quite a poor argument, but it does seem like it is illegal: https://www.ttb.gov/spirits/home-distilling.shtml

The reason criminals don't undenture alcohol is probably because it's cheap and easy to make your own alcohol at home.  Plus it's legal in the US (click on G1) and most other English-speaking countries.
You can get equipment kits from anywhere between 50 and 200 USD (more expensive kits will get you tools that make things easier or improve quality and add an ingredients kits).  Ingredients kits for beer, wine, cider, etc., run anywhere from 30 to 300 USD.
Now, if you wanted to go cheap, you could make kilju, a cheap homemade Finnish alcoholic beverage pronounced, basically, "kill-you".
For equipment, you'd get a food grade bucket from the hardware store (a 5 US gallon one is about 4 USD locally) and thin towel or piece of cheese cloth (about 2.50 USD locally) for a total of 6.50 USD for equipment.
For ingredients, you'd use 3 lbs of sugar plus 1 oz of raisins (for extra nutrients for the yeast and more reliable results) per US gallon.   We need to leave a gallon or so of headspace, so in that 5 USG bucket, we'd make 4 USG.  That's 12 lbs of sugar and 4 oz of raisins for 4 USG of water.  I can get sugar for about 2.39 USD for 2 lbs, so that's 14.34 USD for the sugar and 1 USG of spring water goes about 0.89 USD here, so that's 3.56 USD for the water.  A bag of 10 oz of raisins goes for about 1.99 USD.  Champagne yeast (you can use baker's yeast which you can find in a grocery store but champagne yeast will give much better results and less yeasty flavors) runs about 0.50 USD which gives us a grand total of 20.39 USD for the ingredients.
Once you have that, you mix the sugar into the water in the bucket then add the raisins then sprinkle the yeast on top and cover the bucket with the cheesecloth.  Once bubbling stops (about a week), syphon it out, leaving the dregs (raisins and dead yeast) behind.  You'll have a little less than 4 USG of kilju with a potential alcohol content of about 16% (assuming I did my math correctly).
Everything I've described up to this point is legal in the US and most other English speaking countries.  The next procedures I'm going to describe is probably illegal in most English speaking countries (except, apparently, New Zealand).
If 32 proof alcohol isn't strong enough for you, you can distill the kilju.  An old fashioned easy way is freeze distillation (or jacking).  Place your kilju in your freezer and when it gets slushy, pour it through a sieve and toss the solid portion.  Repeat a few times until it doesn't freeze very well.  What you'll have is moonshine (although most moonshiners seem to use heat distillation instead of freeze distillation).

Heat is used because you wind up with a purer final product.  Distillation discards all but the gas phase over a certain temperature range.  Generally the very beginning of the distillation is discarded (there goes methanol, acetone, and more), the middle portion is kept (almost exclusively ethanol-water azeotrope), and the end of the distillation is discarded (when other non-ethanol contaminants appear.)  Freezing removes water and some fraction of high freezing point compounds, but leaves the ethanol as a mix of fermentation byproducts and volatile, undesirable compounds.

In Soviet Russia, where alcoholism was widespread and homebrewing illegal, undenaturing alcohol was a pretty common thing - a part of national culture, in a way - along with such practices as:

drinking straight down various non-food grade products containing ethanol
or extracting ethanol from them
or stealing ethanol from industrial processes
or, well, moonshining.

That being said, unlike in some other countries, denaturing was almost never done with methanol, for fear of people drinking themselves to death, which is bad for planned economy (and which they did anyway, when they could lay their hands on methanol in applications where it is genuinely needed). In fact, the authorities went to certain lengths to substitute methanol with ethanol wherever possible, considering illicit drinking a lesser evil than the alternative. Methanol was marked most prominently and guarded most heavily, and still regularly caused poisonings.
Denaturing alcohol usually was done with some additives that gave it bitter taste (or worse) and foul stench; distillation alone would not remove the latter. Undenaturing typically included oxidation of the additives by $\ce{KMnO4}$ and/or absorbing them with activated charcoal (both were dirt cheap in the drugstores), and then probably distillation as an additional step. Mind you, methanol can't be dealt with in this manner.
Moonshining included either distillation or freezing out; the latter was usually done outdoors during the winter, by pouring the raw fermented brew down a sloped U-beam, for domestic fridges in those days were ridiculously weak and barely could get anything to $0^\circ\,\rm C$. BTW, methanol can't be reliably removed this way either.
Consumption of non-food grade products included things so abhorrent I myself have troubles believing. How about stirring a certain kind of industrial glue with a power drill, so as to separate the glue proper from its solvent (which contained some ethanol)? Or putting a thick layer of shoe polish on a lump of bread, thus making alcohol to penetrate into the bread? Drinking cheap cologne and window washer looks pale and innocent in comparison.
Nowadays, with relatively cheap and available legal booze, most of these practices seem to fade into oblivion. Let them be recorded here, if only as a negative example for the generations to come.

Then again, some guys just seem to stick to the old ways. http://www.bbc.com/news/world-europe-38363441

As @matt_black and @Aaron Abraham suggested, cleaning the denaturated ethanol is difficult and expensive. You don't know, what chemicals other than ethanol and methanol are in the solution and in what concentrations.
Selective absorbtion using zeolites, as @SteffX suggested, is also blank. You don't know what to filter out. Maybe using regenerative filter that will absorb ethanol and water only and then regenerate it untouched.
Much cheaper is preparation of your own ferment and distilling it in distillery or on your own. This takes time and care (collecting and fermentation) and knowledge and skill (distilling). Otherwise you will end up with poor-taste methanol-rich liquid. Hard alcohlic, thinking of de-denaturation of denaturated alcohol is unable to do it for sure.

Another way is to focus on very cheap drinks. Czech final-stage alcoholics, homeless mostly, drink Okena™, some of them, the rich ones, drink it through a bread.
Playing with the devil is quite common in black market in post-communist countries though, see: Czech methanol scandal and Pärnu methanol tragedy.

Distilling your own spirits is fine and legal as long as you make no more than five gallons of finished ethanol. Anyways, you just need to throw out the first ounce or two of distillate if small scale, up two 4oz if large scale. That will be mostly methanol, other than that just be careful and God bless chemistry!

Don't give legal advice... Especially, if you do not state the jurisdiction it is not useful, and often misleading.


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Mechanism for the conversion of 4-bromobenzyl bromide to 4-bromobenzoic acid using Oxone (potassium peroxymonosulfate)?
Oxone is a mild, inexpensive and easy to handle and environmentally benign oxidation agent that has many application in organic synthesis.

4-Bromobenzyl bromide can be converted to 4-bromobenzoic acid using Oxone. I am curious what would be the mechanism for the oxidation.
The proposed mechanism seems like that, but how the carbocation is formed in the first step?

This is the simple $\mathrm{S_N1}$ mechanism.

@Jan, how it is SN1 Reaction?

Please look it up in your textbook.

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equivalents for products?

I do not understand how to use the equivalents to find the millimoles of the theoretical product. This is the part that I'm struggling with the most. My TA says that after I find the millimoles using the equivalents, then I can solve for grams and the mL of water.

First, you have to find out the limiting reactant in the reaction, in your case it's seems that cyclohexanone is limiting reactant. Theoretically, one equivalent of cyclohexanone (27mmol) react completely with 2 equivalent of benzaldehye (54mmol),but benzaldehye is used in excess (73mmol) to gave one equivalent of product (27mmol) and two equivalent of water (54mmol).
When you multiply 27mmol of product with mol. weight (274.36g/mol) which is equal to =7408mg/1000= 7.4g
Similar you can do for water as well..
I hope this would be helpful, right now i don't have pencil with me, otherwise i will post a picture to this problem.
anyway if you didn't understand let me know..
sincerely,

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Interstitial Sites in unit cell
I wanted to calculate how many interstitial sites (tetrahedral and octahedral) are present in a simple unit cell and their location and radius.
I am guessing they are at the center of unit cell and at the center of each edge.
But can someone please confirm it and explain this in detail? Thanks

Since a 'unit cell' is not a uniquely defined thing, you should be careful. For example, a Wigner-Seitz unit cell will have a lattice atom at the center, which means that can't be an interstitial site as well.


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Calculating the energy used during electrolysis of NaCl
Suppose we are getting the necessary $\ce{Na}$ for $\ce{HCl}$ out of $\ce{NaCl}$. We need to end up with $V = 100~\mathrm{m^3}$ of $\ce{HCl}$ at standard conditions. Assuming that $\ce{HCl}$ is an ideal gas, and using the ideal gas law, I found there are $$ n = 4272~\mathrm{mol}$$ of matter in those $100~\mathrm{m^3}$. The applied voltage to the cell is $$U = 3.6~\mathrm{V}$$ and the current utilization factor is $\mu=88\%$.
Then we also have Faraday's law of electrolysis:
$$\frac{m}{M}=\frac{q}{F}$$(the oxidation state of chlorine is $z=1$).
Since there is a constant voltage being applied to the cell, the energy used in this reaction is
$$E = \frac{UI \Delta t}{0.88} = \frac{Uq}{0.88} = 470~\mathrm{kWh}$$
However the correct answer is $$E=48.9~\mathrm{kWh}.$$ Where did I go wrong? Perhaps it's something blatantly obvious.

How did you calculate the number of moles of HCl gas?

I used the equation $pV=nRT$

I think you did that wrong. $$\dfrac{1*10^5}{22.414} = 4461$$

I used $T=298K$ and $p=101 325 Pa$

But still, we are off by a factor of 10.

The question needs to be clarified. You are using Faraday's law but I am not seeing a redox reaction. Are you oxidizing Na to Na+? Where is the 3.6V coming from?

We have a solution of NaCl,  and a dc voltage of 3.6V is applied across the two electrodes dipped into that solution. I have an electrical engineering background, and perhaps I have trouble communicating myself in the chemsitry world.

@EmirŠemšić This is probably not related to your stoichiometric and electrochemical calculations, but how are you supposed to produce gaseous HCl from the electrolysis of NaCl?

Let's reformulate your equations a little bit.
The ideal gas law was just fine:
$$n = \frac{p V}{RT}$$
A problem lies in your formulation of Faradays law.
We are not interested in masses, but in moles. Of course it is interchangeable by using the molar mass, but why make things complicated. As a rule of thumb you can remember for such tasks: Convert from mass in the beginning and convert to mass in the end (if needed). Nearly all equations in "the middle" are based on the concept of fixed relationsships between the number of involved particles and all equations are much easier to understand if you use $n$.
Faradays law means that the overall charge is just the product of charge per particle times the number of particles:
$$
Q = nzF
$$
The energy is just the product of $Q$ and $U$ divided by $\mu$. You don't need time and amperage.
This gives:
$$
E = \frac{QU}{\mu} = \frac{zFpVU}{RT\mu}
$$
Unfortunately when putting values into these equations you get another result. The following python code will give 448.4 kWh as result. I think that your result is correct and our difference results from differently used constants...:
constants = {}
constants['F'] = 96485.33289
constants['R'] = 8.314
conversion = {}
conversion['to kWh'] = 2.7777e-07

n = lambda p, V, T : p * V / (constants['R'] * T)
Q = lambda n, z : n * z * constants['F']
E = lambda Q, U, mu : Q * U / mu
# Using p=1.01325*10^5 Pa, V=100m^3, T=298K, z=1, U=3.6V, mu=0.88
print(E(Q(n(1.01325e5, 100, 298), 1), 3.6, 0.88) * conversion['to kWh'], 'kWh')

So, what result do you get when you run the above code?

48.9 kWh. Your desired result

I get a different result see here: http://tinypic.com/r/x1isgm/9

I've also run the code - the answer generated by the code is 448.39. While the answer is wrong, I am not voting to delete this answer (who nominated it for deletion, btw???) because it needs to be debugged and is otherwise not "low quality."

Perhaps it's all just a typo in the textbook, and that really is the answer. I hope we can come to a conclusion before the bounty runs out.

I definetely made a typo when reading your desired result and put the decimal point one back while reading. :/ Nevertheless I am pretty sure, that the equations and the python code are correct. I have edited the answer accordingly


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Are any bonds broken when NO2 becomes N2O4?
This reversible reaction occurs in room condition:
$$\ce{2NO2 <=> N2O4}\quad(\Delta H=-57.23\mbox{ kJ/mol)}$$
This is the Lewis structure of $\ce{NO2}$ (courtesy Ben Mills via Wikipedia):

This is the Lewis structure of $\ce{N2O4}$ (courtesy ChemSpider):

To me, it seems that no bonds need to be broken for the forward reaction ($\ce{2NO2 -> N2O4}$) to occur, which is strange indeed.

No bonds are broken, but there are several isomers of N2O4.  The unpaired electron can be on either the N or the O, for each NO2 molecule. The long bond between the monomers involves these two unpaired electrons.
See sections 7-1 to 7-3 of Bonding in Electron-Rich Molecules: Qualitative Valence-Bond Approach via Increased-valence Structures for more information.
According to First-Principles Study of the Role of Interconversion Between NO2, N2O4, cis-ONO-NO2, and trans-ONO-NO2 in Chemical Processes J. Am. Chem. Soc., 2012, 134 (31), pp 12970–12978:
there is no barrier to the bonding of two monomers to form the planar isomer with the N-N bond.

So the energy diagram (enthalpy) would have no uprise; only downfall?

@user34388  There is no barrier for the formation of the isomer with the N-N bond, see the reference I just added to the answer

As stated by @DavePhD no bonds are broken. The $\ce{NO2}$ is a brown paramagnetic gas with the single electron mainly located on the nitrogen, and $\ce{N2O4}$ a colourless and diamagnetic gas. This molecule has a very long N-N bond, 0.178 nm, the O-N-O angle is 134 degrees the same as in $\ce{NO2}$. (The single bond in $\ce{H2NNH2}$ is 0.147 nm). The bond appears to be mainly of a $\sigma$ nature but delocalised over the molecule. (ref Ahlrichs & Keil J.A.C.S. 1974,96,7615) Although there are 3 isomeric forms the most stable by far in the solid and liquid has the geometry you show in the question. The structure $\ce{O-N-O-NO2}$ is observed at 4 k in infra-red spectra. This structure may be important in some reactions even in the gas phase but this is speculative. The $\ce{N2O4}$ is planar at room temperature but twisted in an inert matrix at low temperature; this is the third isomeric form.


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Acrylic paint on polypropylene PP
I have bought a set of polypropylene (PP) test tubes that I wish to paint with acrylic to achieve the same effect as the following image:

Can anyone recommend a process I can apply to the test tubes (in a home setting) or type of paint (acrylic not necessary) that will bind and last and allow me to achieve a similar effect as is pictured above. My main worry is that it may bind but a year or two later will begin to flake off.
Any primer must be completely transparent.
Apologies if this is not the typical question asked here I initially posted this in arts and crafts but then realised this is less about art than it is to do with the properties of polypropylene and someone here is likely to have a greater expertise in that area.
EDIT: As far as I am aware the reason acrylics don't bond well to PP is due to the low surface energy (LSE) of PP. I have found some products that purport to raise the surface energy of PP to be used with adhesives but before I go ahead and purchase them I would love for someone with more knowledge to confirm that this is likely to work. For example:
Techsil® Clear Cyanoacrylate Primer 77 or LOCTITE® 770™

Paint the insides. Acrylic paint does not adhere well to polypropylene, and would soon flake off the outside. Stopper the tubes to last longer. If there is much thermal cycling (e.g. left outdoors), it may still flake somewhat.

@DrMoishePippik Thanks I intend to achieve the effect by dripping it inside the tubes and then stopper the ends and seal them in polyurethane. I wonder would the amount of paint affect how quickly it flakes?


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How are clathrate hydrates formed in the lab?
How do chemists create clathrate hydrates in the lab (for example, those of $\ce{CO2}$, $\ce{O2}$, $\ce{H2}$ or $\ce{CH4}$)? I'm assuming that they just place a pressurized container of water and the gas in question into a freezer (or liquid nitrogen). However, I'm probably somewhat incorrect and don't have the resources currently to perform such an experiment.

This might be interesting: http://pubs.acs.org/doi/abs/10.1021/ie900679m

I don't know about some of the clathrate hydrates you mentioned, but I've recently been reading some papers about both bromine and chlorine clathrate hydrates, and I imagine the procedure would be very similar for the systems you've mentioned.
I'll summarize from Spectroscopic Signatures of Halogens in Clathrate Hydrate Cages. 1. Bromine.
Basically, they're procedure is as follows. Get some really pure water and really pure bromine solution. Make a solution of bromine and water and place it in a quartz cell (cuvette). Place dry ice above the solution and the solution will begin to crystallize on the quartz cell wall. After it is clear that the solution has condensed/frozen onto the the side of the wall, place the cell into around $3^\circ C$ water. This is because at this temperature, both water and bromine will be liquid, but a bromine clathrate hydrate will be solid.
Thus, one can isolate the crystal by simply melting off the water and bromine which are not in a clathrate hydrate cage-type structure.
I believe this same sort of procedure could easily be applied to many of the other guest molecules you mention, however, because in general clathrate hydrates are going to have a higher melting point than ice. For instance, carbon dioxide clathrate melts at about $10^\circ C$ according to wikipedia.
I would suspect that methane clathrates and other clathrates with more weakly interacting guests would melt at a temperature very close to ice, so the procedure might be more complicated or sensitive. Indeed, that paper I cited above make some bromine-doped clathrates of dichloromethane, which is fairly insoluble in water, so a slightly different procedure is chosen.
One thing I find is that you can make these clathrate hydrates at around the melting point of ice, but then people frequently cool the clathrates down to very low temperatures to conserve the structures and keep the gases trapped in the cage, not necessarily because the this is needed in the formation of the clathrates.

After reading a bit more, I believe that a nearly identical procedure can be used for weakly interacting gases like methane, but the process is done under high pressure of the gas desired inside the cage which obviously shifts around the temperatures but doesn't change the idea presented above.


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Incense Stick vs Candle
*Not sure whether to put this on Phy or chem.. *
What exactly happens when you light an incense stick?

How is this different from a candle, or embers for that matter?
From what I understand, a candle-flame is..... well.... basically a flame and embers are pieces of glowing hot coal or wood, but then where does the lit portion of an incense stick fit in?

https://en.wikipedia.org/wiki/Smouldering

Yes, smouldering the process. It could be that I misunderstood the question.

In a candle, the wax first melts, then turns into a vapor above the wick and the oxidation region, where the vaporized fuel mixes with air, surrounds the central, cooler region. You can demonstrate this by extinguishing a candle and then relighting the smoke above the candle... the flame will flash back to the wick. This "flash-back" can be hazardous in a lab using ether, since a nearby flame can ignite the fumes which will burn back to the ether can.
Incense, smoldering cotton and similar combustion relies on thermal degradation of a solid, producing a narrow flame-front on the surface of the solid fuel. Since oxygen cannot penetrate the solid, the rate of combustion is limited. For that reason, cotton bales burn slowly but are very difficult to extinguish, since the interior remains hot and water cannot quickly penetrate. On the other hand, stretching a cosmetic cotton ball changes the burning from smoldering to an intense flame.

Wow! never knew about flashback. So basically inscense is like a candle withless exposed surface area which causes it to burn slowly?        Or is it a decomposition reaction?

Again, candles have solid fuel converting to gaseous fuel, so the solid wax is not burning, just the mixture of fuel in air. Incense burns at the surface of the solid fuel.

An application of reducing the temperature under the combustion threshold: https://diy.stackexchange.com/questions/201121/stopping-an-incense-stick-from-burning/201122#201122

Temperature is the chief difference. 500 to 800C vs 1500+ Celsius. smouldering is defined as slow, low temperature incomplete combustion, the ash layer and design of the stick depletes oxygen resulting in lack of open flame. a candle flame is hot gasses and incandescent soot particles, an incense stick has an ember which is just the wood itself becoming slightly incandescent due to temperatures in excess of 500C thus unlike metal wood cannot glow without slowly burning in the process as it’s a combustible material, you’ll notice an incense ember is not as bright as a candle flame due to a much lower temperature gradient in degrees.

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Why impurities of the solute are not separated in the process of crystallization?
In evaporation the impurities of the solute are separated with it but in case of crystallization only pure crystals are formed and the impurities of solute remains dissolved in it.why?why does it happen so?

First, in evaporation, the solvent is separated into its components, leaving the impurities (solute) behind. Fractional distillation can be used to separate the components of the solvent, though azeotropes may make separation difficult.
Second, in crystallization, those components that fit into a regular structure precipitate out preferentially, since the regular pattern is a lower energy condition. However, "pure crystals" do not always form. For example, various alums co-crystallize without separating, so that a non-stoichiometric mixture may result.


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Periodic table and p block elements
I have two questions:

It's said that in p block there's gradation in physical and chemical properties of elements. What is meant by this statement?
Why are the different blocks named s, p, d and f?

First question
From P.248, Rudiments of Chemistry:

'p' block elements contain both metal and non-metal and we observe a gradation of properties (such as electropositivity, electronegativity, reducing or oxidising property etc.) as we move from left to right in a period.

Second question
From the Wikipedia article on atomic orbital:

They are derived from the description by early spectroscopists of certain series of alkali metal spectroscopic lines as sharp, principal, diffuse, and fundamental. Orbitals for ℓ > 3 continue alphabetically, omitting j (g, h, i, k, …)

Aren't trends followed in every block?If yes,why did my teacher tell that point only for that specific block?

@AbhijithDileepKumar That would be something you should ask your teacher.

I am sorry.But could you tell me what "complexes"are in chemistry?

@AbhijithDileepKumar Make that a separate question with sufficient context (where do you see this word? can you quote the whole sentence?)

https://en.wikipedia.org/wiki/Coordination_complex


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Formation of a bubble which is not a sphere
Recently a friend of mine received a project from her college which was like -  "Bring a bubble which isn't spherical in class. ".  Could anyone please suggest how to prepare such a bubble sufficiently big and which can sustain enough. So that she could bring in class.  I confirmed that a soap film(like a cylinder) could also do.
Please help me in suggesting a method and the liquid we should use to develop this bubble

A lava lamp might do...

Is it invalid to merely sustain a bubble between two bubble wands?

How do you define bubble?

Surface tension will relax most or all bubbles into a
spherical shape around internal pressure.  So, one solution is to create a situation where surface tension is nonuniform:a bubble adhering to the wall of a glass. Because the glass is not part of the liquid's surface, that
face of the bubble needn't be spherical.
Another is to modulate the pressure the film contains.  Two bubbles in collision is the
easiest way: if the bubbles are the same internal pressure, they
will form a flat wall where they meet.  If they do NOT have the same
internal pressure, the wall curvature is concave toward the
center of the smaller bubble.


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Is Fe(OH)3 a salt or an alkali?
I know that a neutralization reaction (i.e. a reaction between an acid and a base) results in the formation of a salt and water, but when I came to know that $\ce{Fe(OH)3}$ is salt, I began to have doubts:
If bases are ionic compounds that break apart to form a negatively charged hydroxide ion ($\ce{OH-}$) in water, then why is $\ce{Fe(OH)3}$ is salt and not a base?

Who said that $\ce{Fe(OH)3}$ is not a base?

Also, it does not break apart to form hydroxide ion in water.

Compounds do not class themselves as either one thing or another. Looking at a specific compound, there will always be classes that it is not, but save very few very simple compounds all will fit at least two classes.
$\ce{Fe(OH)3}$ is both a base and a salt. And a transition metal compound and a hydroxide and …
Let’s take an analogy. To some, I am a friend. To others I am a student. To others again, I am a teacher. I can be all of those, depending on who looks at me.


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Why is Promethium the only radioactive Lanthanide?
Why is it that only Promethium (Pm) is the only radioactive lanthanide?
I was trying to figure out what percentage of all radioactive elements in the Periodic Table the f-block accounts for, when it struck me that Promethium was the only radioactive lanthanide shown. I found this a bit odd, since I expected to see a couple of other radioactive elements there as well, especially since Promethium is (somewhat) near the middle of the series.

Nuclear properties have absolutely nothing to do with chemistry (f-block and everything). Promethium being radioactive is just bad luck. Technetium is another such case.

Glen Seaborg might disagree a little bit on it being chemistry or not ;), but @IvanNeretin is generally correct. The issue isn't f-electrons so much as nuclear structure, where it turns out that the Samarium nucleus is more tightly bound (see nuclear shell model), so flipping one of Promethium's neutrons to a proton results in a lower-energy nucleus (not that it stops there in general, but...).

It is not the only radioactive lanthanide. It is the only one with no stable isotopes. Other stable lanthanides also have naturally occurring radioactive isotopes (samarium is a good example).

@IvanNeretin  I would advise against making such a sweeping absolutist statement.  It is well-established that nuclear properties can affect chemical properties.  Granted, it's not typical, but it's nevertheless incorrect to take the hard-line stance that it doesn't happen at all.  See, for instance:  https://www.tandfonline.com/doi/abs/10.1080/00222348808212315?src=recsys

@theorist See, in natural sciences there is always an exception, or a subtlety, or a tiny deviation in nearly any rule and law. If we start with exceptions, the students never get to know the rules in the first place. Here I stand, and I can do no other.

@IvanNeretin while I agree this one would be better in the physics forum there are number of things in chemistry that depend on nuclear properties. Can we no longer discuss NMR here? A few others that come to mind with a few seconds thought are the kinetic isotope effect, Mossbauer spectroscopy, isotope shifts in vibrational spectroscopy and neutron activation analysis

Lanthanoids are not generally supposed to be radioactive, with exception of mentioned very long lived radioisotopes.
The longest half-life radioactive nuclides of lanthanoids ( in years )

$\ce{^{150}Nd}:\ \pu{6.7e18}$
$\ce{^{151}Eu}:\  \pu{5e18}$
$\ce{^{148}Sm}:\  \pu{7e15}$
$\ce{^{144}Nd}:\   \pu{2.29e15}$
$\ce{^{152}Gd}:\  \pu{ 108e12}$
$\ce{^{147}Sm}:\ \pu{106e9}$
$\ce{^{138}La}:\ \pu{ 102e9}$
$\ce{^{176}Lu}:\ \pu{ 38.5e9}$

Promethium is rather a victim of circumstances there was no free position to have at least one stable isotope. It is related also to the Liquid drop model.
There are few semi-empirical rules, related to proton and neutron fermionic pairing within the nucleus model:
Elements with the odd proton number have maximally 2 stable isotopes.
2 Isotopes of adjacent elements with the same nucleon number, like $\ce{^N_{A}X, ^N_{A+1}Y}$, are not both stable. ( But some of these unstable ones  have very long half-life as seen above.)
Isotopes not close enough to the Valley of stability are not stable, as they have too little or too many neutrons.
The stability of isotopes of similar nucleon number generally decreases in order

both proton/neutron numbers are even
one number is odd, the other is even.
both numbers are odd.

For Pm and Tc, both have odd proton numbers what gives them disadvantage.
All their possible stable isotope candidates are already taken by their  neighbors as energetically preferred variants. Therefore all isotopes of given 2 elements beta decay to nuclei of adjacent elements with less energy.

We may also seek odd neutron counts that fail to give stable nuclei, like 43 or 61 with protons. The first "unlucky" number with neutrons is only 19 (meaning there are stable nuclei with 1,3,5,...,17 neutrons but not 19). Trace amounts of sulfur-35 ($\ce{^_{16}^{35}S}$) do occur as a cosmogenic nuclide, generated by cosmic-ray spallation of argon.

Additional fact: The heaviest stable "odd-odd" nucleus is nitrogen-14. For heavier ones is instability due odd-odd parity bigger factor than the deviation from the "valley of stability". (there are heavier ones, that are kinetically almost stable like Tantalum-180m)

@Poutnik potassium-40. Had this not been so kinetically stubborn potassium would be monoisotopic and 19, unlucky with neutrons, would be almost so with protons. We  nearly don't have liquid soap (often made with KOH).

But this is still quite radioactive, halflife comparable to uranium-238 or thorium-232.


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How to calculate the amount of substances product produces from the masses and amounts of starting materials?

Calculate the amount of carbon dioxide formed when we burn 1 mole of carbon in the presence of 16 g of dioxygen.
  $$\ce{C + O2 -> CO2}$$

1 mole of carbon means 12 g. Therefore amount of carbon dioxide is 12 + 16 = 28. Am I correct?

From the equation,
$$\ce{C + O2 -> CO2},$$
we can see that one mole of $\ce{C}$ reacts with one mole of $\ce{O2}$.
Therefore, let us calculate the amount of each reactant present.
The amount of $\ce{C}$ is already given: $n(\ce{C}) = \pu{1 mol}$.
The molar mass of $\ce{O2}$ is $M(\ce{O2}) = \pu{32 g//mol}$, so $\pu{16 g}$ of $\ce{O2}$ are $\pu{0.5 mol}$.
Since one mole of $\ce{C}$ only reacts with one mole of $\ce{O2}$, there remain $\pu{0.5 mol}$ of unreacted $\ce{C}$, while $\pu{0.5 mol}$ of $\ce{C}$ react with $\pu{0.5 mol}$ of $\ce{O2}$.
In formal terms, oxygen is called the limiting reactant as it limits the amount of product formed.
From the equation, one mole of $\ce{CO2}$ is produced per each mole of $\ce{C}$. Therefore, $\pu{0.5 mol}$ of $\ce{CO2}$ is produced.

The name of the quantity $n$, which is usually expressed in the SI unit mol, is “amount of substance”. It shall not be called “number of moles”, just as the quantity “mass” shall not be called “number of kilograms”. According to SI, ISO, and IUPAC, the words “of substance” can be replaced to specify the substance, so that one may talk of “amount of carbon dioxide”. Although the word “amount” has a more general dictionary definition, this abbreviation may be used for brevity. Therefore, “amount of carbon dioxide” is the amount of substance $n(\ce{CO2})$, which is usually expressed in mol.

@Loong Wow. My school teaches me "number of moles".

Yes, the name “number of moles” is often used for “amount of
substance”, but this is deprecated.

Oh ok.Since 1 mole of oxygen required for  producing 1 mole carbon dioxide.16g of oxygen=0.5 moles of oxygen are there.So,0.5 moles of carbon dioxide is required,right?

@Abhijith It would be more accurate to say that only 0.5 moles of carbon dioxide reacted.

The one mole of carbon would react with one mole of dioxygen to give one mole of carbon dioxide.
$\pu{12 g}$ of carbon would therefore react with $\pu{32 g}$ of $\ce{O2}$ to give $\ce{44 g}$ of $\ce{CO2}$.
As someone else has pointed out, it would be tricky to get absolutely all the carbon to burn in oxygen unless you set it up in a particular way to do so. Probably in real life some would not burn efficiently and you might end up with carbon monoxide, which is poisonous if you breath it in. Some carbon might also stay as carbon (soot).

Did you read in the question that the available amount of dioxygen is limited?

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


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Color shifting reaction in black sealcoat
When I have my driveway seal coated, it turns blue a few days after it is done and then turns back to black. I've heard it called "blue flash." What actually happens chemically to cause this?

It's not a chemical reaction, AFAIK, but due to the Tyndall effect or Rayleigh scattering from the fine particles of the asphalt-water emulsion as water evaporates and the size and/or dispersion of the asphalt particles change.
It is also possible that interference of light in an adhered water layer plays some part. This is the same effect that gives soap bubble their rainbow colors, which change as the bubbles dry -- note that just before it pops from drying, one area of a bubble  may go colorless due to thinning of the film beyond the range of visible light interference.

Thanks for your insight. Curious Fun question.. Why does it turn blue and not green or purple or pink?

Most likely because it is the scattering, rather than interference. The longer wavelengths pass through, and shorter are scattered more, but both scattered and transmitted rays are mixtures of various colors. Put a drop or two of milk in a glass of water, and shine a flashlight through. Looking into the beam looks orange-red, from the side, blue.

Chemistry is so much fun. I will do the milk experiment with my son. Thanks :)

It's easy to see the effect with a white emulsion. Milk, especially skim milk when diluted with water will look blue to scattered light (and orange to transmitted light (looking thru milk into light source)). This is the Tyndall effect, aka Rayleigh Scattering. I have some doubt that this is what you're seeing. Without seeing it myself, or getting a better idea of what you mean by "blue" (and black), my first guess is that a BLACK asphalt emulsion will absorb too much light to ever turn blue. So, if the material is black initially, it should stay black (assuming the surface didn't get wet and that the coating didn't separate as it dried, squeezing the white-ish latext up to its surface - either of these situations could produce Tyndall effect reflection) So, if neither of those things happened then the only explanation I have for a black coating to change to blue then back to black would be interference colors. Interference colors (like many butterfly wing colors) are very sensitive to thickness so generally you get more than just one color - AND the color you get depends on the viewing angle compared to where the Sun is. This would be the same effect as that you would see if you put a drop of oil onto the surface of a pool or container of water -in general the colors aren't uniform. So, I have no 'good' answer to your question, but hope that the above helps YOU figure out what it is you saw.
I should add that it is barely conceivable that there were light stabilizers (UV stabilizers) in the mixture (although most asphalt emulsions won't contain such expensive ingredients) which might have separated out and turned the surface to a different color until they fully deteriorated or were washed away. This would be a real chemical reaction. I mention it to be thorough, I'd be quite surprised to ever encounter such a product...and frankly I'd worry about what I'd bought if I did see it.


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vapour pressure above 2 separate beakers containing liquids which form ideal solution
Suppose we have A moles of liquid 1 in one beaker and B moles of liquid 2 in the other. Beakers are placed together in a biger closed container. Both have different vapour pressure when isloated. Now both components can't have their individual vapour pressures as it would disrupt the equilibrium of the other. It turns out the compensated vapour pressure would be the same as both liquids paced in one beaker. Why so? Is liquid from one beaker moving to the other and making the concentration equal in both?

Well the title leads to a wrong impression for the mixture in one beaker. Ideal liquids in this case doesn't just mean that the individual liquids that are ideal, but rather that their mixture that ideal also.

Liquid might or might not be moving, but not quite for that reason. Say, you have a beaker of water and a beaker of mercury. Inside your container there would be some amount of water vapor and some amount of mercury vapor (and probably some air too, unless you've bothered to vacuumize the container beforehand, which is not important anyway). Just why and how would they disrupt each other?

I just know they would disrupt each other, don't know much in detail how they do, although I would love to know that too. But my real question is that why is the vapour pressure same as a mixture of both liquids in the same beaker? Does having them in different beakers change anything?

Define what is an ideal liquid. (This is a much less universal concept than an "ideal gas" or "ideal solution"). Chances are that your question is hinged on a flawed premise.

Wanted to make my title shorter which maybe lead to the confusion. I was wrongly calling the liquids which would supposedly form ideal solution as "ideal liquids".

Then the vapor pressure over two different beakers is not the same as that over a mixture.

What would the vapour pressure be then? Let's say the beaker 1 has 2 moles of A, beaker 2 has 3 moles of B. Both beakers are placed together in a closed container. Let isolated vapour pressures be P1 and P2. If the same contents formed a mixture in a single beaker, vapour pressure can be given by Raoult's law. What happens when we have separated beakers?


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Does dilution of a buffer affect pH?
The Henderson-Hasselbalch equation for the $\mathrm{pH}$ of a buffer solution of the monoprotic acid $\ce{HA}$ is given by
$$\mathrm{pH}=\mathrm pK_\mathrm a+\log{\frac{[\ce{A-}]}{[\ce{HA}]}}$$
Since concentration appears in both the numerator and denominator of the fraction $\frac{[\ce{A-}]}{[\ce{HA}]}$ and $\mathrm pK_\mathrm a$ is constant (at a fixed temperature), it appears that dilution of the solution with pure $\ce{H2O}$ would not change the $\mathrm{pH}$. However, since
$$\mathrm{pH}=-\log{[\ce{H+}]}$$
 the amount of substance of $\ce{H+}$ must increase in order for $\mathrm{pH}$ to stay constant upon dilution.
Where is this additional $\ce{H+}$ coming from? I know that diluting an acid causes it to dissociate to a greater extent. But at the same time, you would be diluting its conjugate base and causing it to associate more, cancelling the dissociation of the acid.

Your intuition is right. Strictly speaking, dilution does affect the pH of the buffer because it affects the position of the equilibrium $\ce{HA + H2O <=> A- + H3O+}$. However the effect is really very small which is why it is commonly said that the pH is unchanged.

Related-http://chemistry.stackexchange.com/questions/32176/is-the-henderson-hasselbalch-equation-volume-independent?rq=1

In the Henderson-Hasselbalch equation, $K_\mathrm{a}$ is a product of concentrations and considered a constant.
In reality, $K_\mathrm{a}$, when defined as a product of concentrations, is not a constant:

Upon dilution (decrease in ionic strength) the $\mathrm{p}K_\mathrm{a}$ will change, and therefore the pH of the solution will change.
In addition to the above reason, pH will always approach 7 at extreme dilution as it approaches being pure water.

So if I did dilute a buffer solution with water, in a 10 to 1 ratio, would that affect the pH or not? I'm still unable to understand... My book says that the pH will remain unchanged, but as per your answer, the pH should change, and approach 7

@Abhihyan it won't charge much, but it could change on the order of 0.1 pH units

So basically, there is a pH change, but the change is negligible... That's what I understand... Is that right? (Also, it's Abhigyan :) )

@Abhigyan if you only dilute by a factor of 10, and you consider changes on the order of 0.1 units negligible, then it's negligible.

thanks so much... that really clears things up for me, and hope it does the same for many people to come...

I did the experiment.  My buffer was a commercial product a simple packet of salts probably phthalate based. I made it up in deionised water to the right volume then measured its pH with a simple all-in-one probe-meter I measured the buffer neat and then again after 1/5 serial dilutions. I rinsed the probe with deionised water between readings.
The pH fell from 4 to 3.45 at dilution number 4 before it climbed again.
See the table below. I was not expecting this. The change appears significant. I am inclined to agree that the Ka is dependent on ionic strength and to a  greater level than I previously thought.
Relative
Concn
 Buffer pH measured
1           4.0
0.2         4.0
0.04        3.7
0.008       3.5
0.0016      3.45
0.00032     3.75
0.000064    4.25


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Which conformations of 2,3-dimethylbutane are more stable?

I am not sure what the proper arrangement of these conformations are from least to most stable. I got the wrong answer, but my thinking was this:
 First, I thought that because 2 and 4 are in eclipsed conformations they are least stable.  From there I thought that 4 was less stable than 2 because of the steric interactions of the methyl groups.  Then of the two staggered conformations 1 and 3, it appears to me that 1 has more steric interaction of the methyl groups than 3.  So my ordering from least to most stable was 4,2,1,3.  Is there an error in my line of reasoning?

I followed your reasoning and actually, I would have reached the same answer as you did. I suppose the only possible area up for debate is between 1 and 2 - it basically comes down to whether 3x Me-Me gauche interactions are worse than 1x Me-Me syn - without running some calculations it is difficult to answer that question so I would not fault you (and myself) for getting the wrong answer.

..so I just checked the assignment for the millionth time.. because for the past 2 days I've been sure I'm right about this....just discovered the mc answer option I selected was 4,2,3,1 (swapped 1 and 3) and 4,2,1,3 was not listed..the correct answer was none of the above.  So I'm pretty sure that my reasoning is correct here. Unless anyone can point me to an error still?

Nah, I figure you and I are correct. Sucks to have questions that are like that, sometimes feels more like a test of reading carefulness than chemistry.

The correct arrangement (from least stable to most stable) is probably: $4 > 2 > 3 \gtrapprox 1$ based on my calculations for the four gas phase rotamers you can see in the figure below.

Your reasoning for 4 over 2 sounds good, that would also be my point. For the two minima 1 and 3 (with Boltzmann weights of 46%:54% at the highest calc. niveau (red), 33%:66% at the lowest (blue)) you might also be right. But from the really small energy difference I don’t want to argue too much about it.

My calculation methods seem a bit random, but at first I only wanted to make that fast wB97X-D3/def2-TZVP//HF-3c scan and then decided to calculate the bigger one.

I found a question in Chem 220 - Organic Chemistry
Problem Set 2.
The question is such:

Draw Newman projections for the eclipsed and staggered conformations of 2,3-dimethylbutane viewed along the $C_2-C_3$ axis. Calculate the energy of each conformation, both staggered and eclipsed.
Eclipsed (kcal/mol): H/H, 1.0; CH3/H, 1.3; CH3/CH3, 3.0 kcal/mol. Staggered (kcal/mol): H/H, 0; CH3/H, 0; CH3/CH3, 0.9.

Solution given:

This solution seems reasonable,your book might be wrong.

New link to user16347's answer. http://ursula.chem.yale.edu/~chem220/chem220js/PROBSETS/PS13spr/PS2-S13-ans/PS2-S13-ans.html. Problem #6.

The solution given by user16347 seems logical but is not really the correct solution. According to the answer the anti form is more stable since it has the least energy, this is actually false. Experimentally, the gauche form with 3 gauche interactions ( that is structure with energy 2.7 ) is more stable than the anti form. The reason being that in the anti form the geminal repulsions due to the methyl on adjacent carbon hakes the system more unstable than the gauche interactions. This for this particular molecule the gauche conformer is more stable.
For details into this concept of geminal repulsion, refered Chapter 2 - Strain and Staibilty from the book Physical Organic Chemistry by anslyn doherthy


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Understanding group theory easily and quickly
Please can you explain me the group theory especially the reducible and irreducible representation and class and subclass and how to tell they are in the same group or not. I know it is a tedious task but help will be very much appreciated.

"There is no royal road to geometry", they say. You have no other way but to consume this body of knowledge, one thing at a time. Read the textbooks, do the exercises. What is a group. What is an element. What is a conjugation class... $$\vdots$$
Also, welcome to Chem.SE.

I can't understand how to tell that the set of point groups belong to the same class or not. Atleast that you can explain

This is a huge subject, too big for an answer here.  The first thing to do is to understand the notation in a point group table, the jargon is obscure/arcane but its what we all use. The next thing is to practice with molecules, which is where the web site  www.molecule-viewer.com  really is useful. One of the best places to start from scratch is 'Molecular Symmetry and Group Theory', by A. Vincent, (example based so v good starting point) and by R. Carter with the same title.

@user34577 Could you be yet more specific? Like, here is such-and-such group, what class it belongs to (whatever that might mean)?

I second porphyrin's book recommendation, it is very good. Don't try FA Cotton's book unless you are particularly masochistic.

Consider a search "group theory" (including the quotation marks to indicate the complete string as such has to be present) in the Chemistry LibreTexts.  This leads, for example to books like this.

The figures below show you how to navigate your way round point group tables. The irreducible representations (irreps) are shown as the row of characters. A reducible representation is a collection of the irreps that can be reduced to a number of irreps. Some reminders about definitions, important but tedious. The web page molecule-viewer.com gives many example structures that you can manipulate and to which you can add planes and axes.
some definitions :
A symmetry operation is an operation that moves the molecule about a point into a new orientation indistinguishable to its original one.
A symmetry element is a line point or plane to which the symmetry operation is applied.
(This can be confusing as operations and elements tend to get merged as for example $\ce{C_2}$ can be an element, a 2-fold axis or operation, rotation by 180 degrees)
The first thing to determine is the principal axis, this is the axis with highest rotational symmetry (choose one if there are two or more that are equal). Make this the z axis and choose x and y on your molecule. Make sure that the x and y correspond to the same order of axes as used by the set of point group tables you are using. (There is no standard way). Do this by checking with a simple molecule such as water with $\ce{C_{2v}} $ point group.
There are five types of symmetry operations (1) The identity always present and means 'do nothing' (2) rotations about a symmetry axis $\ce{C_1, C_2, C_3...}$ where subscript 2 means 180 degree rotation, 3 by 120 degree and so on. (3) mirror planes $\ce{\sigma}$ (three types, parallel to principal axis, called vertical, $\ce{\sigma_v}$, perpendicular to principal axis, called horizontal, $\ce{\sigma_h}$ and dihedral $\ce{\sigma_d}$ which bisect two 2 fold axes) Note that mirror planes are defined by their relationship to rotation axes. (4) Centre of inversion i,(5) Rotation reflection $\ce{S_1, S_2, S_3...}$
The figure shows the principal axis and one 2 fold horizontal axis. The point group is $\ce{D_{5h}}$.(from www.molecule-viewer.com) Here there  are approx 500 molecules for you to work out the point group. Some axes and planes can be added as hints. The molecules can be rotated etc. All commonly seen point groups have examples.

Common groups are $\ce{C_n}$ and $\ce{D_n}$ types, $\ce{T_d}$ tetrahedral,and $\ce{O_h}$ octahedral. The C groups have rotation axes, and NO perpendicular 2 fold axis, the D groups have a rotation axes but do have a perpendicular 2 fold rotation axes.

The table below shows the Mulliken labels.

The table shows the symmetry elements for several point groups.It is usually only necessary to identify as few symmetry elements to work out the point group.

How to find irreps from reducible representations.
There are several ways to do this, most books give a formula but this can be tricky as its easy to make little sign errors so I always use a tabular method as explained in the figure below taken from my lecture notes. The reduced representation in $\ce{C_{2v}}$ is shown at the top. You also need this point group show below

To make the table to work out the irreps an extra row is added at the top of the point group table to hold the reducible representation and then the entries in the table are calculated as shown below. The values in the rows are added up and divided by the order of the group. The answer must be a whole number or zero, if not a mistake has been made. (the order is the total number of symmetry operations (columns in table) each first multiplied by number in its class, 4 for $\ce{C_{2v}}$ but 6 for $\ce{C_{3v}}$)
The method is quite quick as you once you have one value worked out others in each column are either the same or multiplied by an integer or zero depending on the point group. Don't forget that not all the characters are 1 or -1, in some tables they can be 2 or 3 etc, so you must always multiply by these values.

It is comforting to know that not everyone around is as lazy as myself.

My instructor taught it using rotational vibrational and translational degrees of freedom. Standard books don't go that way that's why I was so confused

@user34577 Then start a new question to that effect.

@user34577, yes this is the way I usually teach it also. By starting with vibrations etc, one puts three orthogonal vectors on each atom and generate the reducible representation that way. Here I started with this and worked towards the irreps.

@porphyrin how did you find the sigma R row?

@porphyrin the one with 9, -1, 3, 1

This is the reducible representation, and this depends on the problem you are studying; for example, if is the vibrations in a molecule you put xyz, arrow vectors on each atom and count 1 for a mirror plane or inversion operation for each vector unchanged, -1 if inverted and 0 otherwise. Its more complicated for C3 rotations etc so you will need to look in a textbook.


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Coordination dilemma, change of energy related to ligand distance

In octahedral field, the baricenter remains conserved and there is lowering of energy of $\mathrm{t_{2g}}$ orbitals.

What is a baricenter and why must it be conserved?
Why is there lowering of energy of $\mathrm{t_{2g}}$ as ligands approach?
Also when ligands are far there an increase in energy but not when they are nearer. What is going on?
I know energy must be conserved but they are many ways to release/utilise energy like sound, heat, light, kinetic energy.

@Melanie I disagree with this edit, thus I will roll it back in a second. Until OP states otherwise, I am willing to assume that baricenter was a misspelling in OP’s book. Searching for it on Wikipedia gives no result while using a y does.

The baricentre should probably be spelt barycentre. It is something analogous to a centre of mass. Simple models such as some flavours of the crystal field theory will assume a destabilisation only of the metal orbitals. However, for an energetic destabilisation to occur, something else must be stabilised — this is the conservation of energy. Thus, the logic says that the raise in energy of the $\mathrm{e_g}$ orbitals must be offset by a lowering of the $\mathrm{t_{2g}}$ orbitals.
When constructing a proper MO scheme of an octahedral coordination complex, this argument is revealed as bogus. As can be seen in this answer, the $\mathrm{e_g}$ orbitals are actually $\mathrm{e_g^*}$ orbitals, i.e. antibonding with respect to the $\ce{M\bond{<-}L}$ bonds which is the reason why they are raised. Disregarding π interactions, the $\mathrm{t_{2g}}$ orbitals are nonbonding and thus their energy is not affected.

B-b-but the Wikipedia article spells it barycenter.


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Copper-free Click Chemistry?
Classic click chemistry, also known as Huisgen 1,3-Dipolar Cycloaddition, is the azide-alkyne cycloaddition in the presence of copper as catalyst.

The toxic metal used as catalyst precluded its application in medicine until the recent introduction of copper-free click chemistry.
It involves the reaction between cyclooctyne (azadibenzocyclooctyne) and azide, which proceeds like an explosion to give a single product without the presence of a toxic metal as catalyst.

Why is a catalyst unnecessary for this reaction?

Because it's faster? Asking why it's faster is rather a complicated question which I suspect nobody will be able to answer.

@bon It's mean there is no answer to that question,

The short answer to your question is that instead of increasing the reaction rate by providing an alternate reaction pathway (lowering the reaction's activation energy) with a catalyst, the catalyst-free version uses a higher energy starting material, due to the ring strain from the triple bond and electron-withdrawing groups, which lowers the activation energy by increasing the energy of the starting point. I'm not familiar with the precise mechanisms involved and the design of different starting materials, so I will leave it to others to give a more complete answer.

@MichaelD.M.Dryden It mean due to strain from the triple bond lowers the activation energy, and  reaction happened even in the absence of catalyst.

@N.ALodhi: yes, the activation energy is lowered by increasing the energy of the starting material (i.e. it is less stable). If you use a catalyst, you lower the energy of the transition state(s).


text:
Chemical shifts seen during XPS
When in X-ray Photoelectron spectroscopy, an Si $2p$ peak is defined or a C $1s$ peak is defined at a certain binding energy, how is it done? I presume it is done from mono-crystalline Si but even in that case the Si peak might not only be experiencing chemical shifts due to the $sp^3$ hybridized bulk state, the case for a C 1s peak I guess would be more complicated due to its allotropic distribution, graphene, CNTs, Fullerenes and Diamonds would in my opinion show separate chemical shifts due to the outer hybridized orbitals being differently hybridized (oriented) and strained (in the case of CNTs, Fullerenes and Graphene since they are all $sp^2$ hybridized). The best way in my opinion to get an idea of the binding energy of the samples, would be to perform the tests in a vacuum tube under very high temperature so that all species are gasseous and are not bonded in any way. I would really like to know how these discrepancies are managed.

As noted, peak positions have been experimentally determined for a number of clean surfaces of reference materials. In 'real' measurements, chemical shifts are observed. I'm not clear on what your question really is, though. For, say, crystalline silicon, one would expect that there would be a chemical shift relative to mono-atomic gaseous silicon. Further, the absolute binding energy is not that important - it is the differences in the electron energy levels that comes in to play.

So, my point is that we don't have an absolute binding energy for an $Si 2p$ peak, so the $Si 2p$ peak we see is in bulk $Si$ or in $SiC$, especially true i would guess for $C 1s$ peaks then, since we would not have an absolute $C 1s$ peak , but a $C 1s$ graphene or a $C 1s$ diamond or a $C 1s$ fullerene

And that is a strength of XPS - it is sensitive to the chemical environment.


text:
Distribution of reaction times in (typical organic) chemistry
Is there any research on how reported reaction times are distributed (in typical organic chemistry)?
I feel like I remember that there was a paper/talk discussing this, finding that there was a steep drop around 10-14 h (since one would then have to setup the reaction very late or come in very early for workup - to avoid staying in the lab in the midst of the night), but right now I cannot find anything on the topic.

16 hours probably means "I set it up before leaving and found that it was done when I came back the next morning.", doesn't it? ;)

I guess so! I remember a total synthesis a few years ago, where one of the steps would have 200+ hours... Probably setup before going on vacation for Christmas/New Year's ;)

Also, another view is that if it isn't done overnight, the reaction does not work.  Your question needs a better specification.

I do most of reaction overnight, sometime i wake up late in the morning, and time exceed more than 12 hours but i always mention it as overnight. This doesn't make any difference, I have seen same time of reaction in different papers having different reaction times.

I'm voting to close this question as off-topic because it isn't really about a chemistry concept. It's really asking about studies on the behaviour of organic chemists.

@bon, in my opinion the question is really important, I am also curious to know how reaction time is defined ,

Why is it important? Reaction time is presumably defined as the time between putting the reactants in the reaction vessel and stopping the reaction by quenching it/removing the contents etc.

@bon, how you define, stirred overnight ?

What do you mean?

A professor in Munich once said: ‘There are only three reaction times. 5 minutes, i.e. you set it up and then you work it up; 2 hours, i.e. over lunch break and 16 hours i.e. over night.’ However, his analysis is skewed: It forgot 70 hours (over the week-end).


text:
Entropy change when indistinguishable particles suddenly become distinguishable
We have that
$$
S = \frac{U}{T} + k\ln Q
$$
Using that for indistinguishable particles in the canonical ensemble
$$
\ln Q_{\mathrm{indis}} = N\ln \frac{q_{\mathrm{tr}}e}{N} + N\ln q_{\mathrm{rot}} + N\ln q_{\mathrm{vib}} + N\ln q_{\mathrm{el}}
$$
Similarly for distinguishable particles we have that
$$
\ln Q_{\mathrm{dis}} = N\ln q_{\mathrm{tr}} + N\ln q_{\mathrm{rot}} + N\ln q_{\mathrm{vib}} + N\ln q_{\mathrm{el}}
$$
Then we assume the indistinguishable particles suddenly become distinguishable. By how much does the entropy of the system change?
$$
\Delta S = S_{\mathrm{dis}} - S_{\mathrm{indis}} = k(\ln Q_\mathrm{dis} - \ln Q_\mathrm{indis}) = Nk[\ln(N)-1]
$$
In the special case that the system contains only one particle ($N=1$), the entropy changes by $-k$.
Intuitively, it does not matter whether one particle becomes distinguishable from the rest if there is only one particle; Assigning a label to just one particle does not really modify the information within the system.
So how can this be understood?

How can an indistinguishable particle suddenly become distinguishable without any energy?

Whether the particles are indistinguishable or distinguishable does not affect the energy.

Just out of curiosity does this describe a physical process? It's very interesting regardless, but just wondering.

Very interesting question. The issue is that your formula for $\ln Q_\mathrm{indis}$ does not hold for $N = 1$.
Since the rotational, vibrational and electronic degrees of freedom do not come into play I will just ignore them. The way you derived the term $\ln (q_\mathrm{tr}e/N)$ comes from the use of Stirling's approximation
$$\begin{align}
Q_\mathrm{tr,indis} &= \frac{q_\mathrm{tr}^N}{N!} \\
\ln Q_\mathrm{tr,indis} &= N \ln q_\mathrm{tr} - \ln(N!) \\
&\approx N \ln q_\mathrm{tr} - N \ln N + N \\
&= N\ln\left(\frac{q_\mathrm{tr}e}{N}\right)
\end{align}$$
Plugging $N = 1$ into Stirling's approximation however tells you
$$\ln (1!) \approx 1\ln 1 - 1$$
or $0 \approx -1$, which is obviously a problem. Stirling's approximation is really meant to be used for large $N$ only!
Now, you could try working it out without using the approximation: plug $N = 1$ into the line before the approximation $\ln Q_\mathrm{tr,indis} = N \ln q_\mathrm{tr} - \ln(N!)$ and you will get the result
$$\ln Q_\mathrm{tr,indis} = \ln q_\mathrm{tr}$$
which matches the expression for the distinguishable case, so all your terms will cancel out and you will get the desired answer of $\Delta S = 0$.


text:
What is the physical basis for Hund's first rule?
According to Hund's first rule, a set of degenerate orbitals are singly occupied first, before the second slot in any of the orbitals are populated. This is quite intuitive because electron-electron repulsions would make an atom more unstable if the electrons start filling two at a time in a single orbital.
However, the rule also states that in the ground state, electrons that singly fill the orbitals will adopt the same spin. Why is it necessary for the electrons to have the same spin?

Related: Why is up spin in the px orbital and down spin in the py orbital considered an excited state?

The lowest energy state has parallel spins to maximize the exchange energy.
As you say, there's a Coulomb repulsion between two electrons to put them in the same orbital.
There's also a quantum mechanical effect. The exchange energy (which is favorable) increases with the number of possible exchanges between electrons with the same spin and energy.

Going from the top state to the middle state, we remove the Coulomb repulsion between electrons in the same orbital.
Going between the middle to the bottom (most stable state and that predicted by Hund's rule), we gain the exchange energy, because these two electrons are indistinguishable.

Disclaimer
My following answer is the "traditional" explanation of Hund's first rule, which is based on a smaller value of $V_\mathrm{ee}$ (electron-electron repulsions) in the triplet state arising from Fermi holes. According to Levine's Quantum Chemistry 7th ed.:

This traditional explanation turns out to be wrong in most cases. It is true that the probability that the two electrons are very close together is smaller for the helium $^3S$ $1s2s$ term than for the $^1S$ $1s2s$ term. However, calculations with accurate wave functions show that the probability that the two electrons are very far apart is also less for the $^3S$ term. The net result is that the average distance between the two electrons is slightly less for the $^3S$ term than for the $^1S$ term, and the interelectronic repulsion is slightly greater for the $^3S$ term.
The calculations show that the $^3S$ term lies below the $^1S$ term because of a substantially greater electron nucleus-attraction in the $^3S$ term as compared with the $^1S$ term. Similar results are found for terms of the atoms beryllium and carbon.

Wildcat has an answer which briefly mentions this. Aside from that Levine gives several references:

Katriel, J.; Pauncz, R. Theoretical Interpretation of Hund's Rule. Adv. Quantum Chem. 1977, 10, 143–185. DOI: 10.1016/S0065-3276(08)60580-8.
Shim, I.; Dahl, J. P. Theor. Chim. Acta 1978, 48 (2), 165. The DOI doesn't resolve properly, but here's the link to the publisher's website.
Boyd, R. J. A quantum mechanical explanation for Hund's multiplicity rule. Nature 1984, 310, 480–481. DOI: 10.1038/310480a0.
Oyamada, T.; Hongo, K.; Kawazoe, Y.; Yasuhara, H. Unified interpretation of Hund's first and second rules for 2p and 3p atoms. J. Chem. Phys. 2010, 133, 164113. DOI: 10.1063/1.3488099.

With that said, here is the old-fashioned argument for why parallel spins are favoured over paired spins (all else being equal).

Introduction
The origin of Hund's first rule lies in "exchange energy", which is a way of saying that electrons with like spin repel each other less than electrons with unlike spin.
To see why, let's consider an excited state of the helium atom: $\mathrm{1s^1 2s^1}$. (The ground state configuration $\mathrm{1s^2}$ is of no use to us because those electrons have to be paired with unlike spin.) In the excited state, the electrons can either have parallel spin (the triplet case) or paired spin (the singlet case).
The total electronic wavefunction comprises a spatial part, which describes which orbitals the electrons are in (i.e. it specifies $n,l,m_l$), and a spin part, which describes the, well, spin ($m_s$). According to the Pauli exclusion principle, the total wavefunction must be antisymmetric with respect to interchange of the electron labels: $\Psi(1,2) = -\Psi(2,1)$.
Spin wavefunctions
The spin wavefunctions are simultaneous eigenfunctions of the $\hat{S}^2$ and $\hat{S}_z$ operators. The total spin quantum number, $S$, can be found from the eigenvalue of the operator $\hat{S}^2$:
$$\hat{S}^2\psi = \hbar^2 S(S+1)\psi$$
For more detailed discussion, refer to any QM text on the coupling of angular momenta.
There are three possible triplet spin wavefunctions, which are degenerate.
$$\psi_{\mathrm{spin,triplet}}(1,2) = \begin{cases}
\alpha(1)\alpha(2) \\
\beta(1)\beta(2) \\
\frac{1}{\sqrt{2}}[\alpha(1)\beta(2) + \alpha(2)\beta(1)]
\end{cases}$$
It turns out that $\hat{S}^2\psi_\mathrm{triplet} = 2\hbar^2\psi_{\mathrm{triplet}}$, which means that these have the quantum number $S = 1$. This corresponds to a multiplicity of $M = 2S + 1 = 3$, hence "triplet". These wavefunctions represent the case where the electrons have parallel spin. (Loosely speaking, each electron has a spin of $1/2$, and if these are aligned parallel the total spin is $1/2 + 1/2 = 1$.)
The singlet state is
$$\psi_{\mathrm{spin,singlet}}(1,2) = \frac{1}{\sqrt{2}}[\alpha(1)\beta(2) - \alpha(2)\beta(1)]$$
and as you might guess, $\hat{S}^2\psi_\mathrm{singlet} = 0$, meaning that $S = 0$ and $M = 2S+1 = 1$. This represents the case where the electrons are paired - one can think of it as the two $1/2$ electrons spins “cancelling each other out”.
The important thing to notice is that the triplet spin wavefunctions are symmetric with respect to permutation of the labels, whereas the singlet spin wavefunction is antisymmetric. For example:
$$\begin{align}
\psi_\mathrm{singlet}(2,1) &= \frac{1}{\sqrt{2}}[\alpha(2)\beta(1) - \alpha(1)\beta(2)] \\
&= -\frac{1}{\sqrt{2}}[\alpha(1)\beta(2) - \alpha(2)\beta(1)] \\
&= -\psi_\mathrm{singlet}(1,2)
\end{align}$$
Spatial wavefunctions
The only way of constructing these are the linear combinations below.
$$\begin{align}
\psi_\mathrm{space,symm} &= \frac{1}{\sqrt{2}}[\mathrm{1s}(1)\mathrm{2s}(2) + \mathrm{1s}(2)\mathrm{2s}(1)] \\
\psi_\mathrm{space,antisymm} &= \frac{1}{\sqrt{2}}[\mathrm{1s}(1)\mathrm{2s}(2) - \mathrm{1s}(2)\mathrm{2s}(1)]
\end{align}$$
Remember that the atomic orbitals above are functions of the coordinates of the electron: $\mathrm{1s}(1) = \mathrm{1s}(\vec{r}_1) = \mathrm{1s}(r_1,\theta_1, \phi_1)$. Now, the antisymmetric spatial wavefunction has an interesting property. Let's explore what happens when the electrons get very close together, i.e. $\vec{r}_1 = \vec{r}_2 = \vec{r}$.
$$\begin{align}
\psi_\mathrm{space,antisymm} &= \frac{1}{\sqrt{2}}[\mathrm{1s}(\vec{r})\mathrm{2s}(\vec{r}) - \mathrm{1s}(\vec{r})\mathrm{2s}(\vec{r})] \\
&= 0
\end{align}$$
If the wavefunction is 0 when the two electrons are on top of each other, that means that there is no probability of this happening. The average distance between the two electrons $\left\langle\left|\vec{r}_2 - \vec{r}_1\right|\right\rangle$ will therefore be larger, and on average the electrons will repel each other less, since they are on average further away from each other. This is known as a Fermi hole.
In fact, one can also show that the symmetric wavefunction has a larger probability amplitude when $\vec{r}_1 = \vec{r}_2$. This is known as a Fermi heap.
The total wavefunction
As I mentioned earlier, the Pauli exclusion principle dictates that the total wavefunction be antisymmetric. This means that the triplet spin wavefunctions, which are symmetric, must be paired with the antisymmetric spatial wavefunction. Likewise, the singlet spin wavefunction, which is antisymmetric, must be paired with the symmetric spatial wavefunction.
$$\begin{array}{c|c|c}
\text{Spin} & \text{Spatial} & \text{Total}\\
\hline
\text{Triplet (S)} & \text{AS} & \text{S } \times \text{ AS} = \text{AS} \\
\text{Singlet (AS)} & \text{S} & \text{AS } \times \text{ S} = \text{AS}
\end{array}$$
So, there is a decrease in electron-electron repulsion when the electrons are parallel - because this means they have a triplet spin wavefunction, and hence an antisymmetric spatial wavefunction, which has a Fermi hole.

Exchange integrals
There is another, perhaps more mathematical, way of looking at it, and that is to calculate the expectation value of the electronic energy for both wavefunctions. For the helium atom, under the assumption of an infinitely heavy nucleus (essentially an atomic version of the Born-Oppenheimer approximation), the total Hamiltonian (in atomic units) is
$$\hat{H} = -\frac{1}{2}(\nabla_1^2+\nabla_2^2) - \frac{2}{r_1} - \frac{2}{r_2} + \frac{1}{r_{12}}$$
where $r_i$ is the distance of electron $i$ from the nucleus, and $r_{12}$ is the distance between the two electrons. The expectation value of the energy is simply $\left\langle\Psi\middle|\hat{H}\middle|\Psi\right\rangle$ (assuming our wavefunction is already normalised). It turns out that the only important term in our present discussion is the electron-electron repulsion term $1/r_{12}$.
Let's look at the triplet spin wavefunction (of course paired with the antisymmetric spatial wavefunction) first. The operator doesn't depend on spin degrees of freedom, so we can separate those, and the inner product of the spin wavefunction with itself is just unity.
$$\begin{align}
\left\langle\Psi_\mathrm{triplet}\middle|\frac{1}{r_{12}}\middle|\Psi_\mathrm{triplet}\right\rangle
&= \left\langle\psi_\mathrm{spin,triplet}\middle|\psi_\mathrm{spin,triplet}\right\rangle\left\langle\psi_\mathrm{space,antisymm}\middle|\frac{1}{r_{12}}\middle|\psi_\mathrm{space,antisymm}\right\rangle \\
&= \frac{1}{2} \left\langle \mathrm{1s}(1)\mathrm{2s}(2) - \mathrm{1s}(2)\mathrm{2s}(1) \middle| \frac{1}{r_{12}} \middle| \mathrm{1s}(1)\mathrm{2s}(2) - \mathrm{1s}(2)\mathrm{2s}(1)  \right\rangle \\
&=
\frac{1}{2} \Biggr\{ \left\langle \mathrm{1s}(1)\mathrm{2s}(2) \middle|\frac{1}{r_{12}}\middle| \mathrm{1s}(1)\mathrm{2s}(2) \right\rangle  \\
&\quad\quad\quad - \left\langle \mathrm{1s}(2)\mathrm{2s}(1) \middle|\frac{1}{r_{12}}\middle| \mathrm{1s}(1)\mathrm{2s}(2) \right\rangle \\
&\quad\quad\quad - \left\langle \mathrm{1s}(1)\mathrm{2s}(2) \middle|\frac{1}{r_{12}}\middle| \mathrm{1s}(2)\mathrm{2s}(1) \right\rangle \\
&\quad\quad\quad + \left\langle \mathrm{1s}(2)\mathrm{2s}(1) \middle|\frac{1}{r_{12}}\middle| \mathrm{1s}(2)\mathrm{2s}(1) \right\rangle\Biggr\}
\end{align}$$
Ugly, but that's what it is.
We can simplify it however by noting that the first and the fourth brackets are the same (since the variables in the integral are just dummy variables, interchanging the electron labels does not affect the value of the integral). This integral is traditionally called a Coulomb integral, since it simply reflects the average Coulombic repulsion between two independent (uncorrelated) electrons, one in the 1s orbital and the second in the 2s orbital. It is given the symbol $J_\mathrm{1s2s}$ (the subscripts indicating the orbitals that the electrons are in). Because this is an expectation value for a potential energy arising from repulsion, it is necessarily a positive quantity.
The second and third terms are actually also the same, again because the variables are just dummy variables and you can interchange the labels. This integral, however, does not have a physical interpretation; it is a purely quantum mechanical effect. It is called the exchange integral, and is given the symbol $K_\mathrm{1s2s}$. It's a bit harder to show, but the exchange integral is also a positive quantity (there is a question on it here). However, its magnitude is not as large as $J_\mathrm{1s2s}$.
Putting it all together (leaving out subscripts for clarity) we have
$$\left\langle\Psi_\mathrm{triplet}\middle|\frac{1}{r_{12}}\middle|\Psi_\mathrm{triplet}\right\rangle = \frac{1}{2}(J-K-K+J) = J-K$$
and if you were to perform the same calculation on the singlet spin wavefunction (not very difficult, just replace the minus signs with a plus!) you would get
$$\left\langle\Psi_\mathrm{singlet}\middle|\frac{1}{r_{12}}\middle|\Psi_\mathrm{singlet}\right\rangle = \frac{1}{2}(J+K+K+J) = J+K$$
Since $J > K > 0$, the expectation value of the energy for the singlet state is higher.
But I have more than two electrons. What now?
Well, you just have to do a ton more ugly maths and perhaps a few hundred more integrals. But the idea is the same. For every pair of electrons with parallel spins, there is a corresponding exchange integral that reduces the expectation value of the energy.
Therefore, in a $\mathrm{3d^5}$ manganese atom (we ignore the closed shells) there are
$$\frac{5!}{2!(5-2)!} = 10$$
different pairs of d-d electrons with parallel spins. Therefore if you evaluate the expectation value of the energy, there will be a $-10K_\mathrm{3d3d}$ term, which is stabilising. This is precisely the exchange energy effect that Geoff describes in his answer.

"since the variables in the integral are just dummy variables" is this your way of saying that electrons are indistinguishable?

@Martin I just meant that mathematically the integral is the same - in the sense that e.g. $$\iint 3x^2+4y,\mathrm{d}x\mathrm{d}y = \iint 3y^2+4x,\mathrm{d}y\mathrm{d}x$$ (as long as the limits of integration are the same, which they are). If I am not wrong, this is a manifestation of indistinguishability, because it means you cannot assign a particular label 1 or 2 to a particular electron, which means $r_1$ and $r_2$ are interchangeable in the integral

@orthocresol, we have had this discussion long before....sometime in october. I am just revising for exams. I need a clarification on this. The electron in $1s^1$ and $2s^1$ are in different shells right? why should they be thought of as identical particles? Are their wavefunctions overlapping? They already have the principal quantum number different from each other. Do we NEED to apply the condition of antisymmetric wavefunction to them?

@PrasadMani My reply is late - sorry - but yes, the wavefunction must be antisymmetrised. It is impossible to place a label on an electron, and it is impossible to say that "this electron is in the 1s orbital" or "that electron is in the 2s orbital". The topic is covered, very nicely IMO, in Griffiths' book Introduction to Quantum Mechanics 2nd ed. Chapter 5 is devoted to identical particles.

Some understanding can be gained by looking at the symmetry of the orbital parts of the wavefunctions involved.
The total wavefunction for electrons must be anti-symmetric with respect to exchanging electrons, the Pauli Principle. The total wavefunction has spin and spatial parts, $\psi =\psi _{space}\psi _{spin}$.
When electrons are paired with total spin = 0, (a singlet state) the spatial part of the wavefunction is symmetric and the spin part therefore antisymmetric. Its the other way round for the triplet state, spins parallel and total spin = 1,  $\psi_{triplet} =\psi _{space}(antisym)\psi _{spin}(sym)$.
The effect is that the electrons occupy different regions of space in triplet states; they have a zero probability of being in the same place. In singlets the electrons crowd together;  formally there is a non-zero probability of being in the same region of space. This is why the electrostatic repulsion, which gives energy $e^2/r_{1,2}$ and is large when the electrons are near to one another, raises the energy of the singlet. Thus the degeneracy between singlet (spin paired) and triplet (spins 'parallel') is split.
Therefore the energy difference between singlet and triplet ( and more generally paired and unpaired electrons) is a consequence of the wavefunction symmetry and so the value of the total spin.
The picture shows the effect on the spatial part of wavefunctions illustrated with 2D 'particle in a box' wavefunctions.

With respect to the coulomb and exchange integrals and energies, K and J respectively, both are positive, the total energy works out as $K+J$ and $K-J$ which shows the energy splitting.
(For completeness both Coulomb and Exchange integrals are forms of electrostatic repulsion. If W is the potential energy, which may contain several terms positive and negative of the form $e^2/r$ , then the Coulomb integral with electrons 1 and 2 and orbitals a and b, is $\int \psi(1,a)\psi(2,b)~W~ \psi(1,a)\psi(2,b)$  and exchange integral is  $\int \psi(2,a)\psi(1,b)~W~ \psi(1,a)\psi(2,b)$  where the electrons are 'exchanged')


text:
Why is 1-ethylidene-4-methylcyclohexane chiral?

As far as I can tell, there doesn't seem to be a chiral carbon in the compound, 1-ethylidene-4-methylcyclohexane. The C-4 of the cyclohexane ring has two groups with exactly the same connectivity, and the exocyclic double bond can't give rise to optical isomerism. What am I missing?

I'd say it's an interesting case of a compound with two stereoisomers, that are in the same time enantiomers (optical isomers) and “cis/trans” (E/Z) geometrical isomers.

The strict criterion for a compound to display chirality is that it must not be superimposable upon its mirror image. Let's ignore the chair conformation of the ring for a while, and assume it adopts a planar conformation. You could draw a side-on view of the ring like this:

We can see that if we reflect the molecule (in the plane of the ring + double bond), its mirror image is not superimposable on itself, which makes it chiral:

In practice, the cyclohexane ring does adopt a chair conformation. That does not fundamentally affect the fact that the compound is chiral, as the presence of a chair conformation cannot "remove" this existing source of chirality (for example, there is no way for the hydrogen and methyl groups to swap places by virtue of a cyclohexane ring flip or a similar process).
Actually, this is very similar to the case of an allene, which you may or may not be familiar with. Notice how the conformation drawn above looks almost like this allene:

with the planar ring taking the place of the second double bond. These are examples of axial chirality (Wikipedia; IUPAC), where the chirality stems from the disposition of groups about an axis (in both cases, the axis in question is the C=C bond axis). Further explanation of axial chirality in allenes can be found in this question.

A note on nomenclature
The tetrahedral chiral centre can be named with the usual (R) and (S) stereodescriptors, as described in user55119's answer.
When it comes to the double bond, ChemDraw (and likely some other software) suggest that they can be named with the familiar stereodescriptors (E) and (Z) (the process is also described in user55119's answer). However, this is not fully appropriate.

The exact rules are complicated, but it boils down to the fact that the double bond in question is enantiomorphic: that means that when it is reflected, the configuration of the double bond is reversed. We can see that from the names generated above: the two mirror images have supposedly different configurations at the C=C double bond.
The descriptors (E) and (Z) are supposed to be used only for diastereomorphic double bonds, where the configuration is invariant (i.e. does not change) upon reflection. The large majority of stereogenic double bonds fall into this category, but this one does not.
In place of (E) and (Z), the preferred descriptors are respectively seqTrans and seqCis (see mykhal's answer, and P-92.1.1 (f) of the 2013 IUPAC Blue Book) .

The analogy to an allene is a good one.  An equivalent description of a double bond views it as a two-membered ring (see here for example).  In the OP's molecule the two-membered ring has simply been replaced with a 6-membered ring.

In a sense, you do have a chiral center at the ring carbon opposite the double bond.  If you go around the ring in one direction you pass by the CH3 group at the far end of the double bond just before reaching the double bond itself.  Go around the other way and you pass a hydrogen atom at that point in the sequence instead.

@OscarLanzi You are right, and not just in a sense: that carbon is actually a proper chiral centre (this is reflected in the naming: it gets R/S configurations). I didn't spot it when I reread through.

Did you ask this question 5 years ago and answer it yourself immediately after or something???

@JoshMitchell It's not particularly unusual, actually: I prepared the question and answer together. If you look through my profile you'll see I've done that on several occasions.

This question has been answered in elegant fashion by @orthocresol. I'd like to expand the discussion somewhat by detailing how digraphs and Cahn-Ingold-Prelog-(Helmchen) rules account for the ZR and ES assignments for 1-ethylidene-4-methylcyclohexane. To determine the configuration at C4, the double bond in ZR-1 is separated into two double bonds as a function of the pathway to C4. Digraph ZR-2 assigns the R-configuration to the chiral center (CH2CH2Z>CH2CH2E>CH3>H).
 To assign the geometry of the double bond (ZR-3), the double bond is kept intact and the methyl group is separated into two pathways to the double bond. The red dots terminating the two phantom chains are copies of the sp2 carbon of the double bond. Temporary Ro/So assignments are made with the phantom chains having the second highest priority. The double bond has the Z-configuration because Ro has priority over So and CH3 has higher priority than hydrogen.  The ES enantiomer can be analyzed in a similar fashion.

I would add that from the organic chemistry nomenclature standpoint, the double bond is enantiomorphic. In the current IUPAC Nomenclature of Organic chemistry, special stereodescriptors seqCis/seqTrans instead of Z/E are used in similar cases, here's an example

P-93.5.1.4.2.2 Specification of exo-cyclic double bonds by  stereodescriptors other than ‘E’ and ‘Z’.
(…)
Example 2:

(1seqCis,3S)-1-(bromomethylidene)-3-propylcyclobutane (PIN)
  (3P)-1-(bromomethylidene)-3-propylcyclobutane

(Note: I think there is a typo in the latter M/P axial chirality stereodescriptor locant, should be 1, no 3)
So the preferred names of the two enantiomers are1

(1seqTrans,4S)-1-ethylidene-4-methylcyclohexane

(1seqCis,4R)-1-ethylidene-4-methylcyclohexane

(1) as provided by ACD ChemSketch 2018.1.1


text:
Free database of polymers with Hansen parameters
I am developing a product that needs a component to expand when in contact with crude oil.
Found the wikipedia article about Hanses solubility parameters and thought it would be a good idea to run the parameters of crude oil against a list of polymers to find maybe the five closest polymers.
Where can I find a database of solvents and polymers online?
It should be noted that I know very little about chemistry, but enough about computer programming to be able to run the test against any database format.

I'm afraid that due to copyright claims the only database available is HSPiP, plus there is the most recent handbook Hansen, C. M. Hansen solubility parameters: a user’s handbook, 2nd ed.; CRC Press: Boca Raton, 2007. The book costs ten times less than a database and contains tables for solvents on pp. 345 -- 485, and for polymers on pp. 487 -- 500.


text:
Fructofuranose vs fructopyranose
What's the difference between the two molecules? I can always read about β-D-fructofuranose in the chemistry books, but unfortunately, I've never found any info concerning β-D-fructopyranose. Is there a big difference between them (apart from their ring structure)?
And where can the furanose and the pyranose form be found in nature?

While I don't have time to write up a full answer, I've always found it interesting that the reason you don't use honey as a sweeter in baking is because of the poor control of sweetness which is caused by the furanose-pyranose interconversion at high temperatures. You can read more about it at https://biochemislife.wordpress.com/2013/03/15/honey-what-happened-you-were-once-so-sweet/

@FelipeS.S.Schneider I think you removed a part of the original post..? The extra question was in the edit, which was rejected anyway. You may check the revision history :)

@WilliamR.Ebenezer You're absolutely right, my fault. I just added the question again. Thanks for spotting that one out :)

They are isomers. Both of them have the same molecular formula ($\ce{C6H12O6}$) but different structures.

Structure of β-D-fructopyranose

Structure of β-D-fructofuranose

Could do with some elaboration, I suppose.

Fructopyranose is the free form of fructose. When fructose is attached to another glucose forming sucrose or attached to other fructose making fructans, it does so in its furanose form.
When invertase breaks down sucrose into glucose and fructose, the furanose form of fructose that occurs in sucrose, is now a pyranose form that can be metabolized in the liver by fructolysis.
I hope this helped you!


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bio-compatible thermoset resin
I am looking for something rather odd and specific.
I need a chemical resin that is liquid up to 250 deg. F,  Is a thermoset (it cures at temperatures above 250 F), and is biocompatible in that if the resin is consumed or enters the blood stream it is not toxic to humans in quantities up to 2 cm^3.  If anyone has insight it would be extremely helpful!


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Rate constants using Arrhenius equation
If I'm trying to use the Arrhenius equation in this form:
$$\ln\left(\frac{k_2}{k_1}\right) = -\frac{E_\mathrm a}R\cdot\left(\frac1{T_2}-\frac1{T_1}\right)$$
And I am given the activation energy ($E_\mathrm a$) in kJ, do I need to convert it into J before I plug in my other givens?

What are your units for R?

You just need to make sure the units cancel out with those of the gas constant $R$. If $R$ is in $\mathrm{J~K^{-1}~mol^{-1}}$, then your $E_\mathrm{a}$ should be in $\mathrm{J~mol^{-1}}$ and your temperatures in $\mathrm{K}$.