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Ethane and ethene: which is easier to burn? Which one burns hotter? Why?
Anything that burns "easy", has a low activation energy ($E_\mathrm a$) for the burning process. Anything that burns hotter, will have a lesser enthalpy and thus, will have a more aggressive exothermic reaction.
Ethane is a molecule with two carbon atoms and 6 hydrogen atoms; all bonded with covalent single bonds. Ethene has the same number of carbons, but 4 hydrogen atoms; the bond between carbon atoms, in this one, is a double bond.
If you had to choose a fuel, and heat was your prior factor, which one would you have chosen? I imagined that for this, the activation energy and the enthalpy comparison could lead to an answer. I could easily have looked both $E_\mathrm a$ and enthalpies up, but I didn't.
Why and how do enthalpy and $E_\mathrm a$ change when two single bonds and two H atoms are replaced with a double bond?

Why and how do enthalpy and $E_\mathrm a$ change when two single bonds and two H atoms are replaced with a double bond?

Olefins are less stable than their corresponding alkane analogues.  For example, when ethene is hydrogenated to ethane roughly 33 kcal/mol of heat is given off because ethene has a higher energy content than ethane.
Why is this, why is a molecule with a double bond higher in energy than a molecule without a double bond?
There are two equivalent ways to describe the orbitals in ethene.  One is the "double bond" description where a pi and sigma bond exist together side by side.  We know that a pi bond is not as strong as a sigma bond due to poorer overlap with the pi bond compared to the the sigma bond.  It only takes something on the order of 60 kcal/mol to break a pi bond and produce cis-trans isomerization about the double bond, whereas it takes something like 90 kcal/mol to break a carbon-carbon single bond. This poorer overlap with the pi bond causes ethene to be higher in energy than ethane.  Alternatively one can view ethene as a two-membered ring, no pi bond, just 2 sigma bonds between the 2 carbons forming a two-membered ring.  It is easy to see that such a ring system would contain a significant amount of strain.
Whether you use the pi bond or two-membered ring approach to describe olefinic double bonds, both lead to the conclusion that olefins are destabilized (higher in energy content) due to poor overlap or ring strain (which in itself is really a reflection of poor overlap).  Because of this destabilization, alkenes will generally have lower activation energies and more heat will be given off when they react compared to alkanes.
Example:
Let's consider the bromination of ethene and ethane
$$\begin{align}
\ce{C2H4 + Br2 &-> BrH2C-CH2Br}\\
\ce{C2H6 + Br2 &-> 2 CH3Br}
\end{align}$$
The heats of formation of ethene, bromine and 1,2-dibromoethane are 52.5, 0 and -37.8 kJ/mol respectively; the reaction is exothermic by 90.3 kJ/mol.  On the other hand, in the case of ethane, the heats of formation are -84.7, 0 and (2x) -37.4 kJ/mol; this reaction is endothermic by 10.3 kJ/mol.  The olefinic reaction is significantly favored for the reasons discussed above.

Ethane will release more heat per mole but ethene should produce a higher temperature.  The extra heat released by ethane goes into heating up more water which has a high heat capacity.
As for which has lower activation energy, ethyne is dangerous to store because it can explode so I am assuming ethene is faster to react because of some trend.  I also worked with someone who was researching possible fuels for a ramjet and his group chose ethene because it was the best compromise between safety and speed of combustion.  I do not know why the trend ethane>ethene>ethyne for reaction rates is as it is though.

Ethene produce a high temperature. Ethane release more heat per mol. The extra heat released by ethane goes into heating up more water which has a high heat temperature.

Probably true, but why?


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How many electrons are there in the pi-system of cyanogen?
I know that there are 18 electrons available for bonding in the entire molecule, and that 6 of these are used for sigma bonds. That leaves 12 left over. Which molecular orbitals best describe the locations of these electrons?

Here is the structure of cyanogen

Each carbon has 4 valence electrons and each nitrogen has 5, for a total of ((4x2)+(5x2)) 18 electrons as you said.  Again, as you noted, 6 electrons are involved in sigma bonds, the 2 C-N sigma bonds and the one C-C sigma bond. Each nitrogen also has a lone pair of electrons and that accounts for (2x2) 4 more electrons.  That leaves 8 electrons to fill the 4 pi bonds.  Just like acetylene with 2 pi bonds, the two carbon-nitrogen triple bonds in cyanogen also have two pi bonds, with each pi bond holding 2 electrons.

It's also worth noting that the internal C-C bond is significantly shorter than a normal single bond, indicating significant delocalization across it.

To visualize ron's answer, here is the valence orbital scheme of cyanogen. Note, that this is the canonical molecular orbital picture, which does not use hybridisation. From this depiction, the stated fact from Geoff is also clearly visible. Due to delocalisation the carbon-carbon bond is significantly shorter than a normal single bond. The right hand side orbitals are taken from a DF-BP85/def2-SVP calculation. (Since I forgot to add that to the picture itself, the horizontal dotted line separates unoccupied from occupied orbitals.)

source: Stack Exchange

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/

sort: votes


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GC-MS unexpected peaks during 1-octanol quantification
Note: This experiment was performed as part of a teaching lab experiment which has concluded, so I would not be able to alter any experimental conditions. I was also unable to modify any of the variables used in the GC-MS experiment itself, but I may be able to provide these upon request.
I would like help in the interpretation of this spectrum from experienced chromatographers, I expect more information to be required other than the spectra provided, but due to my inexperience I am uncertain what they might be. Please do point them out in the comments if necessary.
This is a spectrogram plot performed in Mathematica of the relevant data for octanol quantification (octanol normally elutes at approximately 5.8-5.9 minutes, and did so in the positive control samples.)
The GC-MS machine was not used by any other group during the 2 days, and the only samples run on it during the 2 days were either ddH2O (with or without octanol) or sterile filtered cultures of E. coli in M9 growth media+glucose (with or without octanol).
All images are thumbnailed, click on the link to open a full resolution image.
Representative samples taken on Wednesday, consistent high levels of tailing. All samples are in ddH2O.
ddH2O

25mg/L octanol

100mg/L octanol

Samples taken on Friday, varying amount of tailing
ddH2O

Blank growth media

100mg/L octanol in ddH2O, not tailed

50mg/L octanol spiked into growth media

100mg/L octanol in ddH2O, tailed

The noise present (at m/z 59 and 78) seemed inconsistent with the column (100% DMPS) nor any of the alternatives present on a list of possible column bleed spectra.
Octanol's solubility seems to be 2300mg/L as seen here, and I don't find it very likely that the low solubility of octanol in ddH2O resulted in the tailing.
What could have caused the noise at m/z 59 and 78, and why was the tailing so different despite near-identical experimental conditions?

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What compounds exist in the bitter peel of specific fruits?
Actually, my original question was: What makes a pith of pomegranate so bitter?
Then I decided to be more generic. You'll hardly find someone that consumes pomegranate and doesn't nag about the bitter taste the white pith has! Additionally, when I was a kid, my parents taught me not to eat the bitter peel of an orange, which I later learned that is basic. So, since the evolution of these kind of fruit trees has unanimously been altogether, the chemical species that cause the bitterness of their fruits' piths can be of the same group.
I have a hunch what I seek is the "family" of phenols, but that is nothing more than a hunch. Answer in detail please.

Related: https://biology.stackexchange.com/questions/57011/why-do-some-vegetables-taste-bitter

The reason behind the bitterness of most fruits is due to presence of Tannin in them.
Tannin is a naturally occurring polyphenol found in plants, seeds, bark, wood, leaves and fruit skins.
Tannin tastes bitter,dry and astringent and you can feel it specifically on the middle of your tongue and the front part of your mouth. Besides the peels of fruit, we can feel bitterness in other food products as well like,
High-Tannin Foods =
 Tea Leaves
Walnuts, Almonds and Nuts with Skins
Dark Chocolate
Cinnamon, Clove and other spices
Pomegranates, Grapes
Quince
Red Beans
The  bitterness and astringency are caused by tannins binding with proteins in our saliva. The chemical constituent of tannins are: gallic acid , ellagic acid and their derivatives(present in hydrolyzable tannins) , catechin and its derivative (present in condensed tannins)

As for your specific question about the the pomegranate fruit pith, it contains ellagitannins, pelargonidin, punicalin, punicalagin, anthocyanins, cyanidin, ellagic acid which makes it bitter.

+1, answer voted up. Wait, can you give me a link of where you found these?

Well I am a pharmacy student . So I have to read these in our course .I extracted these information through my course book called Pharmacognosy by kothari, things that are taught by my teacher and some internet searches.http://en.wikipedia.org/wiki/Tannin  https://www.google.com.np/url?sa=t&rct=j&q=&esrc=s&source=web&cd=10&cad=rja&uact=8&ved=0CFcQFjAJ&url=http%3A%2F%2Fsiba.unipv.it%2Ffarmacia%2Fart%2FMarrubini%2FNat%2520Prod%2520Rep%25202001.pdf&ei=oSSYVOvpDc23uATRq4LYCQ&usg=AFQjCNFFi7b7t1U1HvVdGj-lifdqhpXrDA&bvm=bv.82001339,d.c2E

You can easily get these info from Google search as well.

file:///home/dragon/Downloads/molecules-16-02191-v2.pdf  ,   file:///home/dragon/Downloads/Condensed%20Tannin%20Structural%20Chemistry.pdf   .These two articles and the link that I posted in earlier comment will definitely help you alot for acquiring sound knowledge of tannins and its characters.


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What physical properties are manifested when an electron leaves Na and moves to Cl to form an ionic bond?
When an ionic bond is formed between Na and Cl, the lone electron in Sodium's outer shell leaves, and completes Chlorines outer shell.
Are there any physical characteristics that I can use as an observer to see this movement?
When the reaction begins (as seen in this video https://www.youtube.com/watch?v=oZdQJi-UwYs) a lot of light and heat is created. Is this the electron moving from one atom to the other? If not, what is causing it?
Secondary to this. Why does the reaction need help (by either adding water in the video above, or pre-heating the Sodium before dropping it in the Chlorine gas) before it gets started?

The light and heat are created by the exothermic nature of the reaction.  If you look up "Born Haber cycle sodium chloride" you will see information on why the reaction is exothermic.
The reaction "needs help" because of the fact that the sodium will likely be coated with a layer of sodium oxide or hydroxide that will not allow chlorine to collide with the sodium metal.  Even then, with pure sodium metal the reaction will be faster if heated as any reaction needs activation energy to proceed.  Adding water causes an exothermic reaction that can provide activation energy for the reaction with chlorine.


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Fireplace window: Can one influence how fast soot is building up?
I'm sorry for such a stupid question, but after googling for half an hour I know two things

There are at least a zillion pages, blog-posts and articles about how to keep your fireplace window clean or how to clean it.
There is not a single page which explains what I'm looking for and which doesn't make claims without supporting them with scientific facts.

Therefore, here the short form of my question: Assuming I have a fireplace with a window, which looks like this after a while:

Further assume I'm able to keep all other factors constant (like type of wood, temperature of the fire etc.), is there something I can apply/do to the cleaned window to prevent the building of soot for a longer time?

The reason I'm asking is this article I stumbled over, which claims that normal corn-starch can be used to fill the pores of the glass. This left me with more questions

does each glass (especially my one) have pores with a size that starch particles can fit in? I read somewhere that starch has a diameter between 2-170 µm depending on the type.
I'm making a fire where I burn wood! Why wouldn't the starch simply burn and create itself the soot?
As we know for a while now making a surface really smooth doesn't mean it keeps clean. With a size of 10-20 µm of the lotus papillae, it seems to fall in the same range as starch.

And for the diligent, here two more sub-questions:

How does soot building work? I mean, why does the soot stick so hard at the window? After all it's just little particles flowing around and landing on the window. (I know it's called burning in, but how does the heat work on a already burned particle?)
Is there some connection to the pores (if any) in the glass in this process?

Update
After using my fireplace one winter, I'd like to make recommendation based on my experience so far. Important is that I can easily remove the window from my fireplace for cleaning. In case the window is fixed, cleaning might become a bit messy. Let me bring it down to 3 tips:

My current method is to use a cleaner for ceramic glass cooktops. This works like a charm. In case someone wants to know the specific product: it's a cleaning stone for class ceramic which is probably only available in Germany but I guess something similar can be found all around the world. This thing comes with a sponge and you just have to rub the wet sponge a few times over the cleaner to apply some. Cleaning the window with this is done in less than a minute, even if there is a lot of soot.

If you don't want to invest any money, the most simple and very effective cleaning method is to use the ash itself to scrub the window. Simply take water and a newspaper, if possible make the window itself wet or better rinse it with water. Take a sheet of newspaper, make it wet and apply ash directly from the fireplace itself. It's really surprising how easily the soot can be scrubbed away with this method.

Last tip: Heat! As already said in the accepted answer, soot is basically the deposition of incomplete combustion products from a flame. There is no doubt and I have tested this very often: a dirty window can partially be cleaned by make a hot fire with very dry wood and a lot of air. Therefore, try to fill your fire-room with the maximum possible amount of wood, open all air-valves and let it burn for a while.

I don't have a fireplace or wood stove, so I don't have a practical technique that I know works to prevent soot buildup on glass, but I can answer some of your questions about glass and soot. To start, I should add a caveat here: while many fireplaces use tempered glass windows, wood stoves generally have glass-ceramic windows to withstand the increased heat. As far as surface properties go for this, glass-ceramic isn't dramatically different from normal glass.

does each glass (especially my one) have pores with a size that starch
  particles can fit in? I read somewhere that starch has a diameter
  between 2-170 µm depending on the type.

While special techniques exist for making porous glasses, normal glass is not porous and float glass (the standard technique for fabricating flat glass panes) produces quite smooth pieces even without polishing (hundreds of nm peak-to-peak), so probably whole starch particles are not getting stuck in the surface of the glass and the surface roughness probably has little impact on soot adhesion in this case.

I'm making a fire where I burn wood! Why wouldn't the starch simply
  burn and create itself the soot?

Unless the flame is right up to the glass, it probably doesn't get hot enough to really burn starch (if any is even there). That site didn't compare it directly with not applying starch so it's not clear that it makes any difference.

How does soot building work? I mean, why does the soot stick so hard
  at the window? After all it's just little particles flowing around and
  landing on the window. (I know it's called burning in, but how does
  the heat work on a already burned particle?)

Soot is basically the deposition of incomplete combustion products from a flame. A wood flame appears yellow or orange because tiny particles of mostly carbon are incandescing. There is insufficient oxygen to fully combust the carbon in the fuel to $\ce{CO2}$, so the combustion results in carbon molecules in various states of oxidation from small polymers to elemental carbon. These species are really only stable as gases within the heat of the flame itself and will thus readily deposit onto cold surfaces, like the window glass. Many modern wood stoves seem to have "air-wash" systems that flow air past the inside of the window to reduce how much soot can actually reach the surface.
What is actually deposited as soot is a complex mixture, but what is emitted from the flame is related to how much air is available and the temperature of the flame (the two are also connected with each other). If the oxygen supply is increased, more complete combustion is favoured so less carbon species that can deposit as soot are formed. The temperature that is attained influences the composition of soot particles: higher temperatures produce more aliphatic and less aromatic compounds. The practical significance is that reducing the flame intensity by limiting oxygen can greatly increase the production of soot. (and may have some effect on how well it adheres to the glass, though I wasn't able to find anything concrete to that effect)
From the non-scientific sources I glanced through, it seems to be well known that excessive soot buildup can be due to wet fuel or low airflow and that burning a very hot fire can help make soot easier to remove. As for the corn starch, it may indeed be leaving some deposit that hinders soot adhesion (there are commercial cleaners that claim to leave a film behind that makes soot easier to remove), the act of applying it may be removing existing soot that's too small to see but new soot adheres readily too, or it may do nothing. Perhaps you might want to try applying corn starch to half of a freshly-cleaned window to see if there is a noticeable difference.

Thank you for your detailed answer. It clears up a lot of my questions. I cannot withstand to say that I had hoped that it is (scientifically) easily possible to (dis)prove especially what was advertised in the blog-post I linked.

Excellent question. One I've been pondering myself. I moved to Norway 3 years ago. Before then, I'd never had a wood burning stove in any of the places I had lived in. Now, our main sources of heat are wood burning stoves with electrical unit for the coldest times. So, it has been a bit of a learning experience for me and one that I've found fascinating, having had a Biology & Chemistry tertiary education (as well as having slight pyromaniac tendencies!).
To answer your question...
I've found that soot deposition on the glass doors is dependant on wood composition. More resin means more soot on the door. It also makes the soot difficult to remove. Pine is the worst, closely followed by fir. Hard wood is the best. My personal favourites are beech and elm but both can be difficult to come by. Birch and willow are what I use mostly. Although I've had a bit of a windfall - one of our old elms was blown down in this year's autumnal storms. It is a bit of a shame as I really like the tree but I'll be glad to make use of it. I tend to use a mixture of coniferous and hard woods. We have an excess of the former. They burn very easily and quickly so are good at the start. Hard wood comes later once I've got a good collection of embers and the oven has heated up properly. but I digress...
Starch is combustible so, after the coating applied and fires have been burnt the chemical structure will be have changed. The process is called The Maillard Reaction. You'll find plenty of articles about that online. The products of the Maillard reaction tend to be brown in colour so the starch film would need to be very thin or you would see it. They also tend to have a hydrophylic nature. If there are indeed pores then these will end up as pockets of hydrophylic Maillard reaction products. Coniferous resin is mainly composed of hydrophobic compounds. (I found this fascinating paper that gives details of them and their structure. Also discusses high temperature treatment products. See text around Fig. 2.1 in "Chemical Investigations of the Organic Embalming Agents
Employed in Ancient Egyptian Mummification" https://research-information.bristol.ac.uk/files/34498395/396674.pdf) Hydrophobic and hydrophylic repel each other, think oil in water. So, if this theory of pores and starch holds true then it will the mutual repulsion that is most likely to be the factor that influences the soot deposition rate. I have to say though - I do doubt it because starch is combustible. There will be plenty of oxygen and heat to facilitate that.
As stated elsewhere, high airflow / oxygen availability prevents soot build up. That is because the resin combustion reaction is allowed to go to full completion. That is why, as you've noticed, a roaring fire at a later time clears the glass. It basically finishes the job.
Delivering the appropriate airflow is a fine art, one I've yet to master. It dependant on air pressure and wind as well as how wide the stove's vents are. My step father just burns with his vents full open. He has done this all his life so perhaps he is right. However, imho his method wastes a lot fuel and heat as the wood burns very quickly and draught sucks a lot of the air out of the room which is replaced by cold air from outside. So, I try to find the balance.
For cleaning, I use a hoover with plastic scraper to start (reduces mess), then a steam lance (gets most of it off) and finally oven cleaner (removes the most hard stuck deposits). That works a treat.
I like the idea of using the ash. Thank you!
By the way - I got to this forum post via google. It was highly ranked. So thanks for asking and thanks for your update.

Thanks for your long answer. Since my question is from 2014, I have some years of experience now. I can support everything you say and to my astonishment, you found out most what I have learned. Especially, that the appropriate airflow highly depends on the weather outside. It's clear that on windy days, the chimney will create a larger underpressure, sucking the air through the stove. But even without wind, there seems to be a difference between the days and after some time, you know how this will influence the air flow. Additionally, I would disagree with your dad as well, ...

... for the exact same reason you gave. It wastes wood and a lot of heat is lost. When you got the fire burning well with all valves open, I close them and look at the flame. You don't have to forget that at this moment, the chimney is still hot and creates underpressure due to the hot air that is flowing upwards. So I close the valves until the fire is not entirely calm. With this, the stove instantly begins to heat up and I can leave it for like that for 1-3h depending on the wood. Thanks again for sharing your experience and I'm happy that my dumb question is well received :)


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How do I purify the resulting compound after a nitro- to amine-group reduction?
I have an aromatic compound that I reduced with either $\ce{Fe + HCl}$ or $\ce{Sn + HCl}$.
How would you choose to go about extracting the compound?
I'm thinking to dry off all the $\ce{HCl}$, redissolve in water, and react out the iron salts to Iron Hydroxide, which can be filtered out.
Then dry out again, dissolve in something like acetone then filter out the sodium chloride?

Welcome to chemistry.se! If you have questions about how to beautify your posts, have a look at the [help]. Do you want to know more about this site, please take the [tour].

What is the reaction you have just done and the compound you are trying to purify. This may help to determine how you purify it.

I downvoted. In this form we cannot possible help you, aside from very general statements. Please refer to the links @Martin kindly gave you to learn about how this Q&A site operates and what constitutes to a good question.

It is very easy to purify the amino compounds from the reaction mixture of reduction of nitro- to amine- group. Firstly, we will do  workup after completion of the reaction by extraction method with organic solvents  either with DCM or with ehtylacetate, then organic layer wlii be separated, and then dried with anhydrous sodium sulphate and the solvent will be evaporated under reduced pressure by applying rotary evaporator. Then after making slurry by adding little amount of silica gel we will pack slurry in Column-chromatography silica gel (100-200 mesh) by using eluent as Ethylacetate:Hexane (20% to 60%),  the amino- group compond will be purify. This is so because nitro- group is less polar compared to anino- group. So, nitro- compound will come first from column chromatograhy and amino- compound will come letter in pure form.


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Why is tetracyanonickelate square planar?

Why is tetracyanonickelate, $\ce{[Ni(CN)4]^{2-}}$ square planar?

$\ce{CN}$, for being a strong field, would not follow Hund's rule.
the central atom $\ce{Ni}$ would have a +2 charge so its configuration would change from $\ce{3d^8 4s^2}$ to $\ce{3d^8 4s^0}$
Using the statements above and using crystal field theory I get its hybridization as $\ce{dsp^3}$, which suggests that its shape must be trigonal bipyramidal while it is not.

Crystal field theory and hybridisation does not really mix well.

Alright i made a mistake.. 3d8 4s2 would change to 3d8 4s0 .But even after this correction and on applying crystal field i get the hybridisation as dsp3 and not dsp2

You should not apply hybridisation. The splitting in CFT refers only to the d-orbitals.

So you are saying that CFT cannot be applied to this compound (to predict its shape)?

No, I am saying hybridisation should not be applied to this compound.

This is the square planar splitting diagram

Many d8 metal complexes are usually square planar because a very stable electron configuration is achieved if you put the electrons on this splitting diagram -6 electrons stabilizing the compound and only two slightly destabilizing electrons-
Now you could ask, why then not d6 and d7 adopt this cinfiguration? Because for 6 or 7 electrons, an even more stable state is achieved with octahedral fields.
About hybridization... Forget about it when talking about coordination complexes. It's completely useless. It can't explain or predict anything that CFT/LFT can't

$\ce{[NiCl_4]^{2-}}$, $\ce{[NiBr_4]^{2-}}$, and $\ce{[NiI_4]^{2-}}$ are tetrahedral

Yeah, of course they are not all sqp

You can rationalize this in terms of the spectrochemical series; the halogens are very weak field ligands so they encourage the formation of high spin tetrahedral complexes.

The use of hybridisation for transition metal complexes is extremely outdated and often leads to incorrect predictions and rationalisations of transition metal chemistry. If you are using it routinely, please stop using it.
You may wish to first refer to this other answer of mine for a discussion of a very similar case:
Why is [PdCl4]2- square planar whereas [NiCl4]2- is tetrahedral?
There, the comparison is between a 3d and a 4d metal. However, the factors that affect the geometry of a $\mathrm{d^8}$ complex are the same. They are described in much more detail in that link. To summarise:

Adoption of a square planar geometry is favoured due to occupation of four low-energy orbitals formed.
Adoption of a tetrahedral geometry is favoured due to (a) relief of steric repulsions; and (b) the relief of electron-electron repulsions due to adoption of a high-spin configuration.

When the ligand is $\ce{CN-}$, a strong sigma donor and pi acceptor, the corresponding splitting of the orbitals in $D_\mathrm{4h}$ symmetry will be very large. Consequently, the electronic stabilisation derived upon adoption of a square planar geometry in $\ce{[Ni(CN)4]^2-}$ is much larger than for other complexes, e.g. $\ce{[NiCl4]^2-}$.
It's quite possible that the linear ligand $\ce{CN-}$ also introduces less steric repulsions in a square planar geometry than a halide, like $\ce{Cl-}$ or $\ce{Br-}$, would. However, the main factor is likely to simply be the large electronic stabilisation, which in turn derives from the fact that $\ce{CN-}$ is very high in the spectrochemical series.

It's tetrahedral; as not 6 lone pairs are coming from $\ce{CN}$,  only $\ce{CN}$ are attached. Therefore, only 4 lone pairs will affect the configuration.
So by now applying CFT you will get its hybridization to be $\ce{dsp^2}$ which will make the shape of tetrahedral.

Please make use of the punctuation marks. It is difficult to understand what you've stated.

Sorry, this is incorrect. It is square planar as is common for d8 complexes for the reasons given above, see e.g. https://en.wikipedia.org/wiki/Cyanonickelate . Also hybridization is one of those tools you learn earlyish in your chemical career and then spend a lot of time trying to avoid using it.


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Current in Battery
When a battery is performing work on a charge and moves it to the higher potential, why doesn't the charge move through the battery and return to the low potential? Why does it go through the circuit?

Batteries are constructed with an internal barrier between the two electrolytes that does not conduct electricity (i.e., electrons can't pass), but that does allow ions to pass through.  See the figure below (taken from Wikimedia, open source), of a simple alkaline dry cell (e.g., an AA battery), in particular the part labeled "ION CONDUCTING SEPARATOR."

The reason batteries are made this way is because in order for the anode electrolyte and the cathode electrolyte to come to thermochemical equilibrium, there has to be a transfer of both electrons and ions.  By putting the separator in there that only conducts ions, you block equilibration of electrons except in the case where you have it connected to an external circuit.  If you conduct an imaginary experiment where you * SNAP * your fingers and make that ion conducting separator disappear, the electrons would indeed surge from one electrolyte to the other as you describe, as fast as mass transfer would allow.  (You would then having to imagine the battery either exploding or melting, because this would make a LOT of heat.)

The charge doesn't "move back through the battery" because in order to do that, it would have to pass directly through the chemical reactions that put it there in the first place. It moves through the external circuit, because that's the only path available to it.
To take it to the next level of detail, the half-reactions that occur at the two plates inside the battery are actually occurring in both directions all of the time. When the battery is open-circuit, both reactions are occurring at the same rate. The forward reactions are slowed down by the electric field created by the separation of charge, and the reverse reactions are accelerated by it, and the battery maintains equilibrium.
When an external path for current is created across the battery terminals, some of the charge flows through it reducing the voltage and the E-field slightly, and this allows the forward reactions within the battery to occur at a higher rate than the reverse reactions, and this is what eventually dicharges the battery as the reactants are consumed.

The answer is: it will! When no salt bridge or similar conducting element is present between the two half cells, the reaction will quickly run to equilibrium. The state in which the potential force of the difference in electron affinity between the two electrodes more or less equals the Coulomb force of the opposite charges that accumulate in them. However, if you make use of a salt bridge this charge difference accumulation can be removed, since an electrolyte, which sort of completes the circuit. Usually salt bridges are constructed in a manner that prevents intermixing of the ions in both half cells, so that there is some certainty about what redox reaction is actually going on, and how. An example of a common salt bridge used in high school chemistry experiments is a filter paper soaked in a solution of potassium chloride.


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Reaction Rates & Equilibrium
So I have a quick question about some concept I have watched in a video, it is illustrated as follows.
$$\ce{A + B <=> C + D}$$
So what the video is saying is that if you increase $\ce{A}$ then you get the products to increase and the $\ce{C + D}$ to increase as well.
My question is that shouldn't the products also depend on the $\ce{B}$ and not only just the $\ce{A}$. So to increase the products don't we need to increase both the reactants?
Here's the video for reference.


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Hydrogen peroxide from water
Is it possible that $\ce{2 OH-}$ from water could react to form $\ce{H2O2}$?
I mean most autoionization forms $\ce{H3O+}$ and $\ce{OH-}$.
However I think it is possible that $\ce{H-}$ and $\ce{OH+}$ forms, though the amount formed of these 2 is small. This is from water acting as an acid and H+ acting as a base.

The only "hydroxyl"-type species whose recombination has been observed to yield  hydrogen peroxide are hydroxyl radicals.
$$\ce{HO^. + ^.OH -> H2O2}$$
In order to generate these hydroxyl radicals from water, a lot of energy has to be provided, typically by radiolysis (short wave uv, $\gamma$, electron beam).
The conceivable steps yielding $\ce{HO^.}$ are either
$$\ce{H2O  -> H^. + ^.OH}$$
or
$$\ce{H2O  -> H2O^{+.} + e-}$$
$$\ce{H2O^{+.} -> H+ + ^.OH}$$

Is it possible that $\ce{2 OH-}$ from water could react to form
  $\ce{H2O2}$?

There would be two electrons left unaccounted for.  There needs to be a balanced chemical equation, then we could consider the equillbrium constant for the reaction.
We could write $\ce{2H2O -> H2 + H2O2}$, but it is extremely unfavorable, like water decomposing to hydrogen and oxygen gas. You could calculate the Gibbs free energy of the reaction and an equilibrium constant, but the equilibrium constant will be extremely small.

This is true but to make it a better answer I'd do the actual calculation and show the real reason why this doesn't happen.


text:
Does hydrogen bonding strength correspond to Brønsted basicity in a given medium?
It does seem that there is a correlation. Both are measures of proton affinity. Heck, isn't hydrogen bonding present in the transition states of most acid base reactions (excluding those involving proton tunneling?) Also, it seems that hydrogen bonding would be subject to the same limitations as Brønsted basicity; i.e. it seems that steric hindrance in some molecule would inhibit both proton transfer and limit hydrogen bonding capability. At least it seems that way to me prima facie.
Also why do we depict acid base reactions as concerted? Is it because the time scale is extremely small for proton transfer reactions, so who cares if it's actually stepwise or concerted?

@Dissenter, you really need to stop asking questions like this.  Once again you point out that the emperor has no clothes!  :)
It turns out that the mechanism of simple proton transfer is an active area of investigation.  There is not a single yes or no answer that would adequately address your questions.  But if this truth were told in the introductory texts who would continue on when the answer to seemingly simple questions leads to such a morass?  Instead, particularly in introductory circumstances, there is a gentlemen's agreement of sorts that we will simply say that the emperor's clothes are the finest we've seen.
Let's look at your two questions in reverse order.

why do we depict acid base reactions as concerted?

In many introductory texts, and even some that are for advanced courses, proton transfer is presented as a concerted process based largely on Occam's razor (the simplest explanation is often the correct explanation).  However, enough experiments in this area have been run that I think it would be fair to say that a stepwise mechanism is now favored, but by no means mandated for all proton transfers (we'll see why some proton transfers proceed in a concerted fashion below).  The stepwise mechanism proceeds as follows:

step 1, formation of the encounter complex $\ce{A-H}$ and $\ce{B-}$ diffuse through solution until they are close enough to form a hydrogen bond

$$\ce{A-H + B- -> A-H..B-}$$

step 2, the $\ce{A-H}$ bond elongates while the $\ce{B-H}$ bond contracts and a transition state is reached.

$$\ce{A-H..B -> [A..H..B]}$$
The transition state is not necessarily linear or symmetric.  There could also possibly be an intermediate accompanying the transition state.  This depends upon your definition of an intermediate.  How long does a species need to survive in order to be called an intermediate; is it 10 vibrations, 1 vibration or what?  Obviously it is difficult to detect such short-lived molecules, but ultra-fast spectroscopy has made great strides and we may soon know whether an intermediate exists along this multi-step reaction coordinate.

step 3, the transition state elongates in the opposite direction

$$\ce{[A..H..B] -> A..H-B}$$

step 4, the hydrogen bond is broken and the complex diffuses apart

$$\ce{A..H-B -> A- + B-H}$$

Does hydrogen bonding strength correlate with Brønsted basicity in a
  given medium?

If we throw out cases where geometry interferes with hydrogen bonding (e.g. one example would be where steric factors preclude close approach or proper alignment of the species involved), then it would not be unreasonable to expect some correlation between hydrogen bond strength and the pKa difference between the two species involved.  There is certainly experimental evidence to support this view.  However, it has been found that if we step back and examine a larger range of pKa differences, the correlation breaks down.
Here is a diagram showing how the rate of proton transfers might change as a function of the pKa difference between the two species involved.

If we look at a small portion of the plot, the part at low pKa, we could fairly easily convince ourselves that there is a linear correlation.  But eventually as the pKa becomes large, the rate of proton transfer becomes diffusion controlled and insensitive to any further increase in the pKa.  The same concept applies to hydrogen bonding.  There could be reactions (acid-base proton transfers for example) where there is no activated process, the reaction is diffusion controlled; as a consequence we cannot expect to accurately measure or compare hydrogen bond strengths as a function of Brønsted basicity differences in such a regime.
Summary Answers

why do we depict acid base reactions as concerted?

To keep it simple in introductory classes.  It turns out that many acid-base proton transfers are not concerted, but rather involve a multi-step process.  Some other proton transfers, for example those involved in diffusion controlled reactions, are probably concerted.  This is still an active area of research.

Does hydrogen bonding strength correlate with Brønsted basicity in a
  given medium?

The answer is "yes" if we 1) examine only a small range of basicities and 2) look in the range of basicities where the hydrogen bonding is described as an activated process.

There are several methods which can be used to compare the Brønsted-Lowry (proton-transfer) basicity and the hydrogen bond basicity.
One of the most common defines a parameter $ \beta_{2} =  (log K_{B}^{H} + 1.1)/4.636 $, where $logK_{B}$ is calculated via the solution of  $log K^i = L^ilogK_{B}^{H} + D^i$ ($K$ being the experimental equilibirum constant, $L$ and $D$ acid-dependent parameters) over 34 reference compounds. The numerical correction results in $ \beta_{2} = 0 $ and $ \beta_{2} = 1 $ being the lower and upper limits for hydrogen bonding strength.
$ \beta_{2} $ values are not well correlated with measures of Brønsted-Lowry basicity (e.g. proton affinity); they are well correlated when comparing very divergent systems such as water vesus phenols but not when comparing members of the same family; e.g. different halo-subsituted phenols.
We can rationalize this by considering that hydrogen bonding strength is dependent on the electronegativity of the donor atom as well as the donor lone pair LUMO/H $\sigma*$ HOMO energy-match. Increasing the electronegativity of the donor atom at constant $pK_{a}$ results in greater hydrogen bond basicity.
As for your second question, proton transfer is not going to occur step-wise because an unsolvated proton is very unstable; loss of a proton is always associated with formation of a charged complex in solution
Sources: 1, 2, 3


text:
Terminals Of An Electrochemical Cell
Does the positivity and negativity of terminals of a battery depend on the terminal where the electrons enter the external circuit?

The positive terminal of a battery is where 'conventional current' comes out of the battery and flows through an external circuit, returning to the battery via the negative terminal. Since electrons are (by convention) negatively charged, the negative terminal of a battery is where electrons are 'released' by the battery.
NOTE: Of course, if it is a secondary battery which is 'recharging', electrons will 'enter' the battery at the terminal marked negative.


text:
How to explain the increase in current with the increase in the concentration of responsible ions?
I have conducted an experiment wherein I have used different concentrations of $\ce{H2SO4}$, $\ce{HNO3}$ and $\ce{H3PO4}$ as the electrolytes. I filled the acid in glass tube and used a 9V battery, wires and two graphite electrodes to complete the circuit. I also attached an ammeter to check the current variation during a time period of 1 minute. I observed that the current was higher with more concentrated solutions, although my query is that when I add more of these acids, the concentration of the $\ce{OH-}$ ions reduces; it is the $\ce{OH-}$ ions and the $\ce{H+}$ ions that dissociate. How exactly do I explain the increase in current due to increase in concentration?

The electrical current is carried by all ions in the solution, including the $\ce{SO_4^{2-}}$/ $\ce{HSO_4^-}$, $\ce{NO_3^-}$, and $\ce{PO_4^{3-}}$/ $\ce{HPO_4^{2-}}$/ $\ce{H_2PO_4^-}$ ions, and not just by $\ce{H+}$ and $\ce{OH-}$.  So, even though the $\ce{OH-}$ concentration is decreasing with increasing acid concentration, like you say, it's much more than made up for by the increased concentration of the anions from the acids.
EDIT (in response to a comment): The contribution of ions that don't participate in the electrochemistry differs depending on the system.  For example, in $\ce{NaOH}$ the $\ce{Na+}$ effectively just distributes to balance the charge from the concentration gradient in $\ce{OH-}$:

$\ce{H2O}$ is reduced to $\ce{H2}$ and $\ce{OH-}$ at the cathode, increasing the $\ce{OH-}$ concentration there
$\ce{OH-}$ is oxidized to $\ce{O2}$ and $\ce{H+}$ at the anode, decreasing the $\ce{OH-}$ concentration there
Thus, $[\ce{OH-}]$ changes from high at the cathode to low at the anode
In order for the solution to maintain electroneutrality, something has to balance the higher concentration of negative charge at the cathode, and in this system $\ce{Na+}$ is the major contributor.  (Water self-ionization can't provide enough $\ce{H+}$ to maintain the charge balance.)

In other systems such as $\ce{H3PO4}$, the acid-base equilibration behavior of an electrochemically inert species can contribute to the charge carrying function:

Adjusted to pH 7 with, say, NaOH, phosphoric acid exists mainly as $\ce{H2PO4-}$ and $\ce{HPO4^{2-}}$
Water is oxidized at the anode, producing $\ce{H+}$, which reacts with $\ce{HPO4^{2-}}$ to form $\ce{H2PO4-}$
Water is reduced at the cathode, producing $\ce{OH-}$, which reacts with $\ce{H2PO4^{-}}$ to form $\ce{HPO4^{2-}}$ and $\ce{H2O}$
So, $\ce{HPO4^{2-}}$ has a higher concentration around the cathode than the anode, and $\ce{H2PO4^{-}}$ has a higher concentration around the anode than the cathode.  These concentration differences result in a net transfer of charge between the two electrodes.
All of the negatively charged phosphate ions are balanced out by a corresponding distribution of positively-charged $\ce{Na+}$, similar to the $\ce{NaOH}$ case.

thank you! but say i did not use acids and used salts instead, could the same explanation be used?

@nick yes. current is the rate of flow of charge. whether the charge comes from a salt or from and acid is irrelevant

thank you! but at the end only particular ions go through the redox reactions at the electrodes. so how do the other ions that do not go through the redox reactions at the elctrodes contribute?


text:
What type of Organic Substance is used in LCD
I know that the liquid crystal displays contain some class of organic compound, which is somehow used to polarise light to make contrast variations(In calculator displays).
My Question is What type of Organic Compound is used in this case and How does it polarise light only when stimulated by external agents such as current in LCD's

Liquid crystals are usually rigid rod-like molecules that tend to show some degree of intermolecular order even when notionally liquids (there is a good summary on Wikipedia). This happens because the rigid polarisable parts of the molecules can have relatively strong directional interactions even in the liquid phase.
Some of the structures they form can rotate the plane of polarised light You can get some intuition about why knowing that polarising lenses consist of highly ordered molecules whose interaction with light is highly direction dependent. In normal liquids the molecular orientations are effectively random so, even in each molecule interacts with light depending on its orientation, this effect washes out due to the randomness of the overall structure. Liquid crystals form structures where there is the possibility or large-scale order allowing macro-effects on polarisation (just like the rigid structure in a polaroid film).
The other thing that makes liquid crystals useful is that the details of the ordering in the liquid can be affected by external electric fields, especially in thin films of the compounds. In principle this is possible when any molecule is polar (that is it has a net electrical dipole). Of course, making an effective display requires a lot of fine tuning of the electrical, mechanical and chemical structure, but the basic details are simply that the molecules are affected by electrical fields and this allows some control of the macro-structure which, in turn, alters the way the liquid crystal interacts with light. the effect is not "turned on" by the field, just altered and that is enough to make a viable display technology.
Liquid crystals have a long history. The first compounds known to show liquid crystal behaviour is cholesterol benzoate, the first molecule shown below, which was dicoveredn in 1888. The other structures show some actual molecules from the early days of LCD displays to modern developments. For more examples of the history there is a great summary from Merck (one of the major manufacturers) in this slideshow.


text:
Why do gases only transmit sound waves whose wavelength is long relative to the mean free path?
Also, when is this fact useful in practice?

Since sound waves are longitudinal waves (generally in air) there needs to be some kind of interaction between neighboring "air" molecules (a mixture consisting mostly of $\ce{N2}$, $\ce{O2}$, $\ce{CO2}$ and $\ce{Ar}$). You can see that this is just not possible if the mean free path of the "air" molecules is larger than the desired wavelength. How would the "air" molecules interact on such distance if there are no "air" molecules to interact with? This interaction (i.e. sound of a particular wavelength) will only be possible if there are sufficiently many "air" molecules nearby to interact with, or formulated otherwise: if the mean free path is sufficiently much smaller than the desired wavelength of the sound. Of course, it is possible to make sound waves with a larger wavelength than the length of the mean free path, just not the converse.
One application I can think of lies more in Physics but whatever. In the days before Einsteins theory of special relativity was developed/accepted, there was a common consensus, which was supported by Maxwell himself, that light was travelling in a special reference frame called "The Aether". In this reference frame the speed of light equaled $c$. We now know of course that light can travel through vacuum and that there exists no such special reference frame as The Aether. There were various problems with this Aether theory, most famously the Michelson-Morley experiment that suggested that either the Aether did not exist or was moving for a weird reason with the exact same trajectory and speed as earth in space. Another problem, related to your question, was that the speed of waves in a medium is dependent on the particle density of that medium. And since (at least some) electromagnetic waves have an incredible small wavelength (the $\lambda$ of visible light for example was known), the density of this Aether medium had to be extremely high, yet it had to also weight almost nothing and be able to penetrate trough everything!
From this and other facts Einstein must have concluded that the Aether theory was in fact invalid. This idea ultimately lead to the theory of special and general relativity, both of which have many spectacular and far reaching consequences.


text:
Is the smallest or largest diameter important for the color of a molecule?
I learned that phenol absorbs longer wavelengths than benzene because the electron cloud is bigger.
But what is about Penta-1,2,3,4-tetrien (1) and 2,2,-Dimethylpropane (2)?
(1) has the largest diameter but (2) has the smallest. Which one absorbs the longer waves?

Edit: (1) has the larger largest diameter but (2) the largest smallest.

The size of a molecule does not correlate with the color of a molecule.
Color depends upon which wavelengths of light are absorbed and which are transmitted by a molecule.  Visible light typically ranges from around 390 to 700 nm.  Light with a wavelength between approximately 635 to 700 nm appears red to the human eye.  If we have a molecule that absorbs all visible light below 635 nm, the molecule will appear red; it absorbs all visible light except for the red wavelengths, hence only the red wavelengths will be reflected or transmitted and be available for the eye to detect.
What determines which wavelengths of light are absorbed by a molecule?
This is controlled by the energy separation (energy gap) between the highest energy orbital that contains an electron (the HOMO or highest occupied molecular orbital) and the lowest energy orbital that an electron can be promoted to (the LUMO or lowest unoccupied molecular orbital). If this energy gap falls in the visible range, then the molecule will appear colored to the eye when the molecule absorbs light at the wavelength that corresponds to the energy gap (e.g. when a photon is absorbed that promotes an electron from the HOMO to the LUMO) .
Molecules like 2,2-dimethylpropane contain only sigma bonds.  The energy gap in typical sigma bonds is very large meaning that it takes a lot of energy to promote an electron from the HOMO to the LUMO.  The amount of energy required typically falls in the ultra-violet region, outside the visible range; therefore these molecules are usually colorless since they transmit all visible wavelengths.
Molecules containing pi bonds have smaller HOMO-LUMO energy gaps than molecules with just sigma bonds.  If there are enough pi bonds and they are conjugated then the gap can be small enough that light in the visible range can be absorbed and make the molecule appear colored.  Here is a diagram showing how the energy gap decreases as we increase the length of our conjugated pi system.

Even with 1,3,5-hexatriene the light absorption at 258 nm is still outside the visible  range, so the molecule will not be colored.  But as we continue to extend the conjugation, eventually the molecule will absorb light around 400-500 nm, the blue light range.  As the molecule absorbs blue light it will appear yellow. In addition to pi conjugation, certain substituents such as the nitro group and the carbonyl, to just name two, can also serve to extend the conjugation path in a molecule.
The molecule you asked about, penta-1,2,3,4-tetraene, looks roughly like this.

If we examine it closely, we'll notice that the orbitals really look like those in two distinct (orthogonal) 1,3-butadiene molecules.  As the chart above shows, 1,3-butadiene does not absorb in the visible range.  Nonetheless, because penta-1,2,3,4-tetraene does have an extended pi system it will have a smaller HOMO-LUMO gap than 2,2-dimethylpropane and will absorb light at longer ultra-violet wavelengths (lower energy) than 2,2-dimethylpropane.

Ok I think I understood it but I've one more question: Why is the gap between the two energetic niveaus of the orbitals smaller when there are more orbitals?

@MarvinMeller  Excellent question!  Here is a link to an  answer I provided to a similar, earlier question.  It is based on the physics behind the description of a particle in a box. Hope it helps!

Ok thank you very much for your answer(s) you really helped me out :)


text:
How does Melatonin put you to sleep? How is this compound radio-protective?
Melatonin (IUPAC semi-formal name: N-acetyl-5-methoxytryptamine) is a hormone produced in humans, animals and some of plants. It's mainly famous for being the drug suggested to people with sleeping problems, jet lags and even insomnia (in trivial cases, still experimental). Recent studies have shown this compound's role as an anti-oxidant and a protection for mitochondrial DNA.
Melatonin interacts with the immune system; and is clinically proved to have anti-inflammatory effects. It can act as a radio-protective agent...
Hold it there. This is chemistry, not biology. First of all, how is Melatonin radio-protective? And what is in that structure that makes it radio-protective? 2nd, what reactions occur in the body, in presence of Melatonin, that cause sleepiness (resources have claimed the chemical function to be a mystery but I doubt that's true)?
 Most of the info above is available at the wikipedia: Melatonin.image credit: http://cenblog.org/the-haystack/files/2011/06/melatonin.png

Melatonin is not "radio-protective" in the sense that it blocks radiation or anything. As you note, it's an antioxidant. One of the ways in which ionizing radiation can interact with the body is by providing enough energy to break chemical bonds, producing highly reactive radical species that can cause biological damage by reacting with various things in the body. Melatonin can easily be oxidized by many of these reactive species, preventing them from reacting with anything else. This paper shows the products for the reaction with radio-generated hydroxyl radicals.
The role melatonin plays in sleep is fairly well known (though the effects of exogenous melatonin, less so). Without going into too much detail, there are melatonin receptors in the body, found in cell membranes in the brain and a few other places (G protein-coupled receptors). When these bind melatonin on the outside of the cell, their conformation changes, further activating signalling processes on the inside of the cell. The effects of almost all signalling molecules in the body work in this way. The molecules activate some kind of receptor, which then activates a sequence of other signals, eventually resulting in biological effects.


text:
Cannizzaro reaction versus Nitroaldol reaction
In Clayden et al. on p. 696 the following nitroaldol reaction is listed:

Later in the same chapter they explain the Cannizzaro reaction. Why is the yield of this nitroaldol reaction particularly good, while a Cannizzaro reaction is also competing for the benzaldehyde? Is the nitroaldol reaction that much faster?
I looked in some literature and they mostly seem to use nitrogen bases (cyclohexylamine, $\ce{Et3N}$, $\ce{n-BuNH2}$, methylamine, etc.) for these kind of reactions (nitroaldol reaction with aromatic electrophile). The reported yields are also typically lower. See for example the "Literature method" section of the synthesis of 1-phenyl-2-nitropropene on Erowid (which seems to be some kind of precursor to amphetamines, so there is plentiful information about it in comparison to the 1-phenyl-2-nitroethene synthesis listed in Clayden. Can the yields be so different due to the steric hindrance of the additional methyl group?). What is the benefit of these nitrogen bases over $\ce{NaOH}$? Clayden et al. only mentions them with the comment that $\ce{NaOH}$ is not really necessary since nitrogen bases are sufficiently strong for nitroaldol reaction, because the typical pKa value of short chain nitro alkyl compounds is ~10.

Is the nitroaldol reaction that much faster?

My guess would be that the nitroaldol is faster than the Cannizzaro.  The Cannizzaro reaction involves

attack by hydroxide on the carbonyl to reversibly form the tetrahedral intermediate
alignment of the tetrahedral intermediate with a second aldehyde molecule (entropically costly)
hydride transfer in a redox reaction (relatively high energy, even complexed hydride is not particularly stable) to form molecules (alcohol and carboxylic acid) that are not that different in stability than the starting material

The nitroaldol involves

generation of the nitroalkyl carbanion
attack by the carbanion on the carbonyl to form a tetrahedral intermediate
capture of a proton by the negatively charged oxygen to produce a neutral molecule
elimination of water to produce a more stable (more conjugated) molecule than we started with (e.g. energetically downhill)

The nitroaldol doesn't require any special molecular alignment after the tetrahedral intermediate is formed and is energetically downhill.

Can the yields be so different due to the steric hindrance of the
  additional methyl group?

Further alkylation at the nitroalkyl carbanion center will not help the reaction in a steric sense just as you suggest, but also a secondary carbanion will be less stable (higher energy, more difficult to form) than a primary carbanion.

What is the benefit of these nitrogen bases over $\ce{NaOH}$?

They are somewhat milder reagents than hydroxide and therefore less likely to produce side reactions.

One small addition to Ron's excellent answer. Sodium hydroxide has limited solubility in organic solvents, while the amine bases will be very soluble in most organic solvents.

Although it is some time ago already, seeing this question active today made me decide to include the answer given by Prof. Clayden himself over e-mail to me.

The reaction you ask about appears in the classic text on practical
  organic chemistry by Vogel: in fact this preparation is online at
  http://www.erowid.org/archive/rhodium/chemistry/nitrostyrene.vogel5.html
It’s not clear to me why amines would be preferable bases in general,
  although you are right that they avoid the Cannizzaro reaction if the
  nitro compound is hindered. They could also be preferable if the
  alcohol were the desired product, rather than the eliminated alkene.


text:
Density of water and temperature
Is there a formula that shows how the density of water is affected by temperature?
For example, with the speed of sound it can be shown how temperature will affect the speed:
$$V = 331\ \mathrm{m/s} + \left(0.6\ \mathrm{\frac{m/s}{°C}}\right)T$$
Is there a similar formula that relates the density of a liquid to temperature? (in this case, water)

yes, see http://nvlpubs.nist.gov/nistpubs/jres/097/jresv97n3p335_A1b.pdf

In general, we can derive the expression for the density of any substance as a function of temperature using the definition of thermal expansion.
$$\frac{\mathrm dV}{\mathrm dT}=\alpha V\tag1$$
By differentiating $\rho=m/V$, we can show that $\mathrm d\rho/\rho = -\mathrm dV/V$. So using this and equation $\text{(1)}$,
$$\frac{\mathrm d\rho}{\mathrm dT}=-\alpha V\tag2$$
Integrating this above expression and approximating the expression,
$$\rho = \rho^\circ\cdot\left(1-\alpha\Delta T\right)\tag3$$
where $\rho^\circ$ is the initial density at the initial temperature, $\alpha$ is the volumetric thermal expansion coefficient and $\Delta T$ is the temperature.
For water, $\alpha$ is typically $207\times10^{-6}\ \mathrm{K^{-1}}$ at $20\ \mathrm{^\circ C}$. Using this, we can observe how temperature affects the density of water.

Good answer.  It's worth mentioning that $\alpha$ is a function of temperature, which gives rise to nonlinear behavior.  For example, liquid water (at atmospheric pressure) has a density maximum at 4 °C, not 0 °C.

Equation 2 is incorrect (V should be $\rho$). This leads to an exponential function as solution. Equation 3 is a linearized version for small $\Delta T$


text:
Boil to dry 50 liters water
I would like to find out how long it would take to completely turn 50 litres of water to steam. Let's assume power of 2500 W and the water is already at boil. I am not very scientific so please forgive any omissions. Thanks

What actually happens in real life depends on a lot of things. Factors like the shape of the pot can make a big difference to how much of that 2500 W actually goes into vaporizing the water and how much of it is lost to the surroundings. If we assume that all 2500 W is going into the water, it makes things a lot simpler. If the water is already at the boiling point, then all we have to do is calculate the total amount of energy needed to vaporize all the water:
The enthalpy of vaporization of water is 40.65 $\mathrm{kJ/mol}$, which means 40.65 kJ of heat needs to be added to turn one mole of liquid water to steam. 50 L of water is 2775 moles:
$$\mathrm{\frac{(50\ L)\left(1000\frac{g}{L}\right)}{\left(18.02\frac{g}{mol}\right)}=2775\ mol}$$
So the total heat we need is:
$$\mathrm{(2775\ mol)\left(40.65\frac{kJ}{mol}\right)=112.8\ MJ}$$
1 watt is one joule per second, so for a 2500 W heat source:
$$\mathrm{\frac{112.8\ MJ}{2500\ J/s}=45\ ks}$$
Or, approximately 12.5 hours. In real life, it may be significantly longer for various reasons, but this is the fastest it could go.

Thank you. I suspected around 12 hours, but never considered the other factors you mention. Your explanation is much appreciated

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


text:
How does a carbocation have a positive charge?
When an alkene bonds with an H+ ion, the electron pair from the pi bond goes towards a new dative covalent bond with the hydrogen ion, leaving, on one side of the old double bond, a carbon bonded to an extra hydrogen, and on the other side, a carbocation with a positive charge. If this gave, for example, a tertiary carbocation, I would like to know how this has a positive charge given that it has 3 bonding pairs of electrons around it, which would give it 6 valence electrons and 8 electrons overall. If carbon has 6 protons, how does this give an ion with a positive charge? I understand that it has lost electrons that were previously in the pi bond, but I don't see how it has a positive charge given that it still has more electrons than protons.

Welcome to chemistry.se! If you have questions about how to beautify your posts, have a look at the [help]. Do you want to know more about this site, please take the [tour].|| As for your question: A $\ce{H-}$ is not a free radical, but an anion with no unpaired electrons. If this adds to an alkene it will also result in an anion. Maybe you want to [edit] your question to change that.

@InquisitiveCheese: try counting up the total number of protons and electrons in your hypothetical ion.

I would like to know how this has a positive charge given that it has 3 bonding pairs of electrons around it, which would give it 6 valence electrons and 8 electrons overall. If carbon has 6 protons, how does this give an ion with a positive charge?

You need to count ALL the protons and electrons in the ion. What about the atoms other than the central carbon?

Well I get what your problem is.
The thing is that earlier C had 4 electrons in its valence shell. Now it is bonded to 3 other atoms giving it a total of 6 electrons in its shell. The octet rule says that it should have 8. Since the other carbon took away 2 electrons this means that our carbon should have a +2 charge.
Well this is where the confusion lies .
The trick is whenever we calculate formal charge we use

Formal Charge= (No.of valence electrons in unbonded state - no of lone
  pair electrons ) - (no. of bond pair electrons/2)

In this case the charge comes out to be (4-0) - (6/2) =+1 We usually associate half of the bond pair electrons to each atom while calculating charge. On the other hand When counting the no of electrons in valence shell we count both the electrons in the bond pair.
Good Luck :)

Even at a lower level, when a positive species is added to a neutral species, math must apply: 0 + 1 = 1 !!! In addition, the atoms must balance.


text:
How can an insoluble compound be a strong electrolyte
Here's a quote from Petrruci General Chemistry (pg. 160):
Silver chloride, $\ce{AgCl}$ is an insoluble ionic compound. When $\ce{AgCl}$ dissolved in water, it is 100% dissociated into $\ce{Ag}^+\text{ and } \ce{Cl}^-$ ions; there are no $\ce{AgCl}$ pairs.
I'm confused:
1) If $\ce{AgCl}$ is insoluble, how is it dissolved in water?
2) If  $\ce{AgCl}$ is a strong electrolyte but insoluble, does it mean a  $\ce{AgCl}$ molecule is ionized, but isn't separated - i.e. the silver and chloride remains close (they're not solvated by water molecules) but each becomes and ion?
I'm confused, can someone explain what's going on?

I think I've found a solution - The silver chloride dissolved very slightly, but whatever is dissolved - is 100% ionized. source:
Yahoo Answers

It sounds contradictory, but soluble and insoluble are relative terms. Silver chloride's $K_\mathrm{sp}$ is $1.77\times 10^{-10}$, so one can generally think of it as insoluble, but actually about a milligram will dissolve in a litre of water.
Much like the terms strong acid and weak acid, strong electrolyte and weak electrolyte refer to the dissociation of a substance in a solvent, though they include all electrolytes, not just acids. In the case of silver chloride, though little of it dissolves, what does is present only as $\ce{Ag+}$ and $\ce{Cl-}$, not solvated  $\ce{AgCl}$. In water, all salts are strong electrolytes, but in other solvents, things can be different. For example, while perchloric acid is a strong electrolyte in water, it doesn't dissociate completely in acetic so it's a weak electrolyte.
Source: (1) Electrochemical Dictionary; Bard, A. J.; Inzelt, G.; Scholz, F., Eds.; Springer Berlin Heidelberg: Berlin, Heidelberg, 2008.

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


text:
How is the dissolution of acetic acid that makes its aqueous solution a poor electrolyte?
I'm having a difficulty understanding the following quote from Wikipedia - Dissociation:

Acetic acid is extremely soluble in water, but most of the compound dissolves into molecules, rendering it a weak electrolyte.

To what (non-ionic) molecules can acetic acid be dissolved? If the answer is $\ce{CH3COOH}$ then in what way is it extremely soluble, if it dissolved to itself?

the ethanoate anion, when it deprotonates.

There are two questions being asked here:
Why is aqueous acetic acid a weak electrolyte?
When acetic acid is dissolved in water there is an equilibrium reaction:
$$\ce{CH3COOH + H2O <=> CH3COO- + H3O+}$$
Since acetic acid is a weak acid, the equilibrium position lies well to the left, with only a small fraction of the acetic acid molecules reacting to form ethanoate and hydronium ions. The presence of this small amount of ions results in aqueous acetic acid being a weak electrolyte.
Why is acetic acid highly soluble in water?
Since the vast majority of acetic acid molecules do not dissociate when a sample is dissolved in water, the solubility has to do with the interactions between acetic acid molecules and water molecules. Water has a network of hydrogen bonds between molecules in its liquid phase and so when a substance dissolves in water this bonding is disrupted. However, acetic acid is able to form many new hydrogen bonds to water molecules and so this results in a highly favourable interaction, leading to the high solubility of acetic acid in water.

Many organic molecules such as ethanol and acetone dissolve into water with little or no dissociation, for the reasons bon describes.

The cautionary tale here is to not confuse "dissolution" and "dissociation".

I think the Wikipedia page you quote is poorly worded. What on earth does it mean to ‘dissociate into molecules’? Nothing, in my opinion. A better wording is discussed below.
But first, let’s discuss what actually happens when acetic acid is dissolved in water. As you may know, glacial acetic acid consists mainly of $\ce{H3CCOOH}$ molecules that associate to form hydrogen bonding networks. There is practically no ionisation in glacial acetic acid, i.e. the autoprotonation equilibrium $(1)$ is leaning very strongly to the reactants’ side.
$$\ce{2 H3CCOOH <<=> H3CCOOH2+ + H3CCOO-}\tag{1}$$
So before dissolution, we are dealing with molecules of acetic acid. If we add these into water, most of them just stay being molecules; only a small percentage ionises in water according to reaction $(2)$. The acidity constant shown in the equation is a measure of how many molecules are dissociated; it depends on the concentration.
$$\begin{gathered}\ce{H3CCOOH <<=> H3CCOO- + H3O+}\\
K_\mathrm{a} = 10^{-4.76} = 1.74 \times 10^{-5}\end{gathered}\tag{2}$$
So upon dissolution we have:

most ($> 99~\%$) of the acetic acid molecules remaining acetic acid molecules
a very small subset deprotonating to form acetate anions

Only the latter are charged compounds and thus only they contribute to the solution’s conductivity. Since they are few in number, conductivity is low.
Getting back to the original quote. It seems to imply that dissolving acetic acid in water ‘turns it’ into molecules; which is wrong: the acetic acid molecules remain the same all through. A better wording would emphasise that they do not change. For example:

Acetic acid is extremely soluble in water, but most of the dissolved compound remains as molecules, rendering it a weak electrolyte.
Acetic acid is extremely soluble in water, but only a small fraction is dissociated into ions, rendering it a weak electrolyte.


text:
Calculate the root-mean-squared speed of free electrons in the Sun's corona given pressure, temperature and mass of an electron
$m_\text{e}=9.1\times10^{-31}\ \mathrm{kg}$, $T=2\times10^{6}\ \mathrm{K}$ and $p=0.030\ \mathrm{Pa}$ the solution is given as: $5.4\times10^6\ \mathrm{ms^{-1}}$ but I got $9.5\times10^6\ \mathrm{ms^{-1}}$ using the following method (which does not use the value given for pressure. I simply plugged the numbers into the equation below:
$$v_\text{rms}=\sqrt{\frac{3K_\text{B}T}{m}}$$
Where have I gone wrong?

I'll go out on a limb here and say your answer is right and theirs is wrong.
Note that the answers differ by a factor of the square root of 3, so I'm guessing they forgot to include the factor of 3.
There would be a small correction for relativistic effects, but not enough to explain the differences in the two answers.

Thank you very much - well noticed regarding the factor of root 3. Would you possibly be able to take a look at my other questions that I have asked today - I am struggling through a problem sheet and you always give good answers.


text:
How to get the first step of the derivation of the Maxwell-Boltzmann distribution
To derive the Maxwell-Boltzmann distribution the first step requires the knowledge that the equation will take the form $$f(v)=Ke^{-\epsilon/ kT}$$ where $\epsilon$ is the sum of the kinetic energies in the directions of the x,y and z axes. I understand that there needs to be exponential decay and why there will need to a constant. Also, I understand that the eponent needs to be dimensionless but why is KT used? This seems to pop up everywhere without much of an explaination of why (as in this case).  Can you help me understand how to get this first step? Thanks.

You could easily refer this in any text of statistical mechanics. Try helping yourself before asking. Hence I'm down voting this question.

KT comes in from requiring agreement with the ideal gas law.
See "section 1.5.3 $\langle v^2\rangle$ should agree with the ideal gas law" and section 1.3 "The functional form of the dependence of T on $\langle\varepsilon\rangle$ cannot be determined solely from kinetic theory, since the temperature scale may be chosen in many possible ways. In fact, one way to define the temperature is through the ideal gas law: T = pV/(nR)" in Kinetic Theory of Gasses

The Maxwell-Boltzmann distribution of speeds is derived from the Boltzmann distribution, which says that
$$\frac{N_i}{N_j}=e^{-(\epsilon_i-\epsilon_j)/kT}$$
This formula gives the ratio of the number of particles ($N_i$, $N_j$) in states with energy $\epsilon_i$ and $\epsilon_j$. From this it is easy to the that $f(v)=Ke^{-\epsilon/ kT}$ as you noted.

how does the numerator of the exponent change to kinetic energy? Also why is KT in the denominator? Is this derived or found empirically?


text:
Does concentration of salt increase or decrease rate of rusting?
I want to know whether increasing the concentration of salt, specifically $\ce{NaCl}$, increases or decreases the rate of rusting. There are conflicting theories that explain opposite outcomes:

Increasing concentration of salt decreases the amount of dissolved oxygen and thus    decreases the rate of rusting
Increasing concentration increases electron mobility and thus increases rate of rusting

I haven't been able to how rusting works in full detail nor have I been able to understand why there are two contradicting theories present. It seems that people write it more out of opinion than with factual basis.
Can someone please clarify?

In general, salt (particularly NaCl) will increase the rate of corrosion (rusting).
To understand why, consider metallic iron $\ce{Fe}$ which rusts (oxidises) to iron(II) oxide $\ce{Fe2O3}$ in the presence of oxygen $\ce{O2}$ and water $\ce{H2O}$.
$$\ce{4Fe +3O2 +6H2O->4Fe(OH)3}$$
Corrosion (rust) is a 'redox' reaction, which means it involves reduction of oxygen into hydroxide ions and oxidation of metallic iron to iron cations.
For any redox reaction to take place, electrons are transferred. In this case, from the metallic iron which loses electrons to oxygen. The presence of salt (or any electrolyte) in the water accelerates the reaction because it increases the conductivity of water, effectively increasing the concentration of ions in the water and so increasing the rate of oxidation (corrosion) of the metal.
The situation is complicated by the fact that salt dissolved in water actually reduced the amount of dissolved oxygen present in the water. This is because the water molecules are attracted to the dissolved ions from the salt (solvation), which has the tendency to decrease the weak affinity of non-polar oxygen molecules to water, thereby deriving dissolved oxygen out. So if the metal is totally submerged in water, higher salt concentrations can actually reduce the rate of corrosion, if the water is not aerated.
So it depends upon the amount of oxygen getting into the water and onto the metal. For example, underwater plants can release oxygen in salty water which will accelerate the rate of corrosion of the submerged metal. Of course, if the metal is not completely submerged in salt water but only partially submerged, or is affected by salt water 'spray', it will be still be exposed to oxygen and will rust faster.


text:
How do derive the equation for relative velocity of gas molecules
The relative velocity is $$v_{rel}=\sqrt2v_{mean}$$ where $v_{mean}=(\frac{8KT}{\pi m})^{\frac12}$$$v_{rel}=(\frac{16KT}{\pi \frac m2})^{\frac12}$$ because the reduced mass of two of the same particles is half the mass of one. In general: $$v_{rel}=(\frac{8KT}{\pi \mu})^{\frac12}$$
However, my question is: what is the need for reduced mass? If $v_{rel}=\sqrt2v_{mean}$ where $v_{mean}=(\frac{8KT}{\pi m})^{\frac12}$. Wouldn't that just leave you with: $v_{mean}=(\frac{16KT}{\pi m})^{\frac12}$ Where does reduced mass come into it?

Your equation $v_{rel}=(\frac{16KT}{\pi \frac m2})^{\frac12}$ is incorrect.
$v_{mean}=(\frac{16KT}{\pi m})^{\frac12}$ (your last equation) is also incorrect.
Instead $v_{rel}=(\frac{16KT}{\pi m})^{\frac12}$ and $v_{rel}=(\frac{8KT}{\pi \frac m2})^{\frac12}$ would both be correct.
The need for reduced mass is for when the particles are not the same mass, usually in the context of considering collisions between two different molecules or atoms.
$\mu = m_1m_2/(m_1 + m_2)$ by definition.  If $m_1 = m_2$ then $\mu = m/2$
When there are two different masses, there are two different mean velocities.
To see where the equation with the reduced mass comes from, look at appendix 1.4 of Kinetic Theory of Gases, along with pages 1-30 to 1-34.

Thank you. The reduced mass must also be used for when the particles are the same (i.e use $\frac m2$) to get the required equation rather than the one I wrote at the end of my question with the 16.

Is reduced mass just always used for two-body problems?

@RobChem I added to my answer, see if it is clear now.

Perfect, thanks. Sorry to be a pain but would you please take a look at http://chemistry.stackexchange.com/questions/22037/calculate-the-frequency-so-that-the-wavelength-of-sound-in-air-would-be-equal-to


text:
Martensitic transformation in ceramics
What is the specification of Martensite (displacive) transformation in ceramics comparing to the metals? I just know about Martensite transformation in metals. For example, these transformations are thermal and diffusionless. Also, the kinetic and morphology are dominant by strain energy.

Take a look at Kelly, P. M., Francis Rose, L.R., The martensitic transformation in ceramics — its role in transformation toughening, Progress in Materials Science, 2002, Vol.47(5):463–557, doi:10.1016/S0079-6425(00)00005-0.
Basically, a transformation is martensitic if it occurs from a collective shear motion of the atoms where each atomic displacement is less than the distance between atoms.  There is no diffusion (atoms moving to other lattice sites). The definition is the same as in a metal, but in a ceramic you have multiple atom types in the crystal, that's all.


text:
Which of the following is the strongest base?

Which of the following is the strongest base?

$\ce{LiOH}$
$\ce{CH3Li}$
$\ce{LiNH2}$
$\ce{LiF}$

According to me the answer should be $\ce{LiF}$ as it has max number of lone pairs but the answer says its 2). I am confused.

In order of decreasing basicity they are ranked $\ce{LiCH3}$ > $\ce{LiNH2}$ > $\ce{LiOH}$ > $\ce{LiF}$. We are dealing with the ions $\ce{F-}$, $\ce{OH-}$, $\ce{NH2-}$ and $\ce{CH3-}$ and so the factor which influences their basicity is the stability of the ion compared to its conjugate acid.
The stability of the ions is determined by the energy of the lone pairs. Carbon has the least effective nuclear charge ($Z_{\text{eff}}$) and so the lone pair on the $\ce{CH3-}$ ion has the highest energy, making it the least stable ion and so the strongest base. The $Z_{\text{eff}}$ increases as we move across the period until we get to fluorine which has the greatest $Z_{\text{eff}}$ and so its lone pairs have the lowest energy, making $\ce{F-}$ the least basic ion.

Look at the conjugate acid of each of the anions (the base). If the conjugate acid is a strong acid (low pKa), then the anion is a weak base. If the conjugate acid is a weak acid (high pKa), then the anion is a strong base.
Let's list the anion or base, the conjugate acid and the pKa
\begin{array}{|c|c|c|} \hline
\text{Base }& \text{Conjugate Acid}& \mathrm pK_\mathrm a \\ \hline
\ce{OH-} & \ce{H2O} & 15.7\\ \hline
\ce{CH3-} & \ce{CH4}  & 48 \\ \hline
\ce{NH2-} & \ce{NH3} & 38 \\ \hline
\ce{F-} & \ce{HF} & 3 \\ \hline
\end{array}
The conjugate acid with the highest pKa is methane ($\ce{CH4}$) it is the weakest acid of the group. Therefore the $\ce{CH3-}$ in $\ce{MeLi}$ is the strongest base in the group.

And indeed, working with alkyl lithium reagents is dangerous as they are pyrophoric!


text:
Why six C atoms are usually seen in cyclic compounds?
When it gets to Carbon-based molecules, one very possible structure when there are more than six C atoms is the hexagon; though not mostly perfect, it emphasizes that six Carbon atoms tend to bond with each other.* Why?
I assume this is because of the "stability" that is attributed to the hexagon. Explainers have discussed mainly about this issue.
Neither me nor my teacher were convinced by the description; (Actually, all of this is because of my teacher!) since he asked that for example, why not octagon? No response.
I wondered that if I asked this in here what ideas I am going to face.... Sorry, but I'm accustomed to doing lots of "whys".
*For example; look at the pic below:
In the image above, the number of hexagonal rings is twice the number of pentagonal rings and eight times more than octagonal rings. This macro-molecule is the micro of what's happening in the nature. I don't have citations; but it takes a minute or two to see how much the number of hexagonal rings is surpassing the number of other rings. Perhaps it was so obvious, nobody mentioned it.
Update: According to @ron, and with a little payment of thought, I reached the fact that these rings aren't necessarily geometric; the important thing is that six C atoms, instead of five or eight or whatever else number of C atoms are the majority.
Image credit: Pseudopotential model for Dirac electrons in Graphene...

Could you be more clear about what you are asking, perhaps by giving some examples. The shape of a ring system depends on how many atoms are in it but I am not sure if this is what you are asking.

You see how most of organic compounds have a hexagonal ring if there are more than 6 Carbon atoms; (if the structure is not linear) For example, an organic compound with 8 C atoms is more likely to have a hexagonal ring with two other C atoms attached rather than an octagonal structure. There are compounds with other rings, but they are the minority. My Q is about the reason C atoms mostly preferred to be hexagonal. I am adding a pic for demonstrating what I mean.

@MARamezani  Those 6-membered rings in your drawing are not geometric (e.g. perfect) hexagons.

@ron, do you think it's better to modify the Q like this: Why is it usual that six C atoms are mainly bonded to each other to form organic compounds?

@MARamezani  If you have 6 carbon atoms, you could make 6 methanes, 3 ethanes, 1 butane and 1 ethane, 1 benzene, etc. 6 carbon atoms can be arranged in a variety of ways, one way is not more or less "usual" than another.

It is all about minimizing the energy of a molecule.
In the case of carbon, the only molecule that adopts a perfect hexagonal geometry in its ground state is benzene (and its derivatives that possess a 6-fold rotational axis).  In this case the hexagonal geometry is adopted because all of the carbons are $\ce{sp^2}$ hybridized.  The ideal geometry (lowest energy) for an $\ce{sp^2}$ hybridized carbon involves three 120 degree bond angles around the carbon atom.  This can be accomplished in the case of benzene by placing 6 carbon atoms in a plane and connecting them as shown to produce the benzene structure.

Other carbon-based molecules may project as hexagons, but they are not hexagons when viewed in 3 dimensions.  For example, cyclohexane is not flat like benzene, but rather typically exists in what is called a "chair conformation" in its ground state where the C-C-C angle is around 109.5 degrees.

why not octagon?

Some other regular polygon shapes can be found in molecules.  For example, the cyclopentadiene anion exists as a regular pentagon, and the cyclooctatetraene dianion exists as a regular octagon.  Both of these molecules contain only $\ce{sp^2}$ hybridized carbons.

As mentioned above, these atoms prefer to exist in a geometry with 3 equivalent 120 degree bond angles.  However, the internal angle in a regular pentagon and a regular octagon are 108 and 135 degrees respectively.  Since these angles deviate from the 120 degree ideal, these molecules will be strained, unlike benzene.

What other ring shapes are preferred, low-energy minima?

We've already discussed the $\ce{sp^2}$ case where the preferred 120 degree angle is found in benzene which exists as a hexagonal structure.  $\ce{sp}$ hybridized carbon is most stable when the X-C($\ce{sp}$)-X bond angle is 180 degrees.  Clearly small or medium size rings comprised solely of $\ce{sp}$ hybridized carbon cannot exist as stable entities.  In the case of $\ce{sp^3}$ hybridized carbon, the closer we get to the ideal tetrahedral angle (an angle whose cosine = -1/3, ca. 109.47 degrees) the lower the energy of the molecule.  We've already mentioned cyclohexane and while it's C-C-C angle is close to the ideal tetrahedral angle, it is not exactly the tetrahedral angle (because each of its carbon atoms does not exist in a perfect tetrahedral environment).  If we fuse a few more cyclohexane rings onto cyclohexane, we can produce the molecule adamantane with 4 cyclohexane faces all in the chair geometry.

There is less strain energy in adamantane (per carbon) than in cyclohexane.  The geometry at each carbon atom is closer to the ideal tetrahedral geometry, but still not quite there.  How can we further elaborate adamantane into a ring system where all carbons have the ideal tetrahedral angle? Does adamantane look vaguely familiar?  By continuing to add more and more cyclohexane rings we build the following ring structure...

... Diamond, where all of the $\ce{sp^3}$ carbon atoms are in the ideal tetrahedral environment, an extremely stable arrangement.
EDIT: response to OPs comment

So, the lesser energy state must be the reason that after six number
of atoms, five is the common one, right?

Kinda.  Here's a table showing estimated ring strains for saturated hydrocarbon rings of various sizes.

[Image source  Also, see this link for a discussion about why the ring strain for cycloheptadecane (-3.4 Kcal/mol) is suspect and probably should be higher.]
5- and 7-membered saturated hydrocarbon rings have similar strain energies, yet 5-membered rings are more common.  We must consider the free energy of the process, enthalpy and entropy changes. In addition to strain (an enthalpic factor) we must also consider entropy.  In ring forming reactions it is entropically more difficult for a 7-membered chain to bring both ends together for closure to a ring, then it would be for a smaller 5-membered chain.


text:
If something contains large amounts of cyclopentane and cyclohexane, what is it called?
If something contains large amounts of cyclopentane and cyclohexane, what is it called? I'd like to know what is the something as well. It's a chemistry olympiad question for this year. See here, Question 3-g. Just let google translate it for you. I find some questions quite weird as well, but what can you do. Otherwise its pretty good.

Flammable. Is this seriously an Olympiad question? I highly doubt it has a specific name that is well accepted.

A barrel :D Merry christmas!

Which Olympiad was this in? National or international. I don't recall the question

Yes, this years house round of Chemistry Olympiade. http://www.iuventa.sk/files/documents/2_olympiady/cho/51.%20ro%C4%8Dn%C3%ADk/%C3%BAlohy%20a%20rie%C5%A1enia/dom%C3%A1ce%20kolo/ch51dkcteul15.pdf Question 3-g. Just let google translate it for you. I find some questions quite weird as well, but what can you do. Otherwise its pretty good.

The most likely name would be naphthenes. It could also be called cycloalkanes.
Added:
Since the reference to the question is about the coal and oil production, the answer would be naphthenes. When the coal and oil were distilled, a layer (called naphthene) was removed which was found to contain mainly cyclopentane and cyclohexane.
I would complete the statement as:
If a fraction of distillate contains large amounts of cyclopentane and cyclohexane, it is called naphthene.

The answer should be much more concrete than this I assume. I need to replace the word something as well as figure out what it is.

Thanks for the answer man, I have a nice teacher so I can consult it with her without a problem. Fortunately enough, in the next rounds inspecific questions such as this one dont occur, they are more practical.

Also I suppose for better specification, I think it would be better like this: If a fraction of distillate from oil and coil contains large amounts of cyclopentane and cyclohexane, it is called naphthene. Surely not all fractions of distillates are called napthenes, right?


text:
Clean tin with aluminium
I've been using aluminium foil and table salt to clean my silverware for a long time without actually concerning myself with the chemistry behind it. A couple of days ago i started digging into the chemistry and found this page on wikipedia
Reactivity series
and when I looked closer i saw that both silver and tin where less reactive than aluminium. Does that mean that I can safely use the same technique with my tinware as with my silverware?


text:
What is the chemistry behind cleaning silverware with aluminium foil
This question has a accepted answer on this site already:
What happens during cleaning silverware?
but the reason I ask the question again is because I found different "recipes" for the technique and also different reaction formulas in other sources.
The "recipe" given in the linked question is

"Put [the silverware] into a pot with some club soda and a piece of aluminium foil and pour over a hot water."

And the reaction formulas given for "blackening" is the following
$$\ce{3Ag2O(s) + 2Al(s) -> 3Ag(s) + Al2O3(s)}$$
and for the reaction with the aluminium foil
$$\ce{3Ag2O(s) + 2Al(s) +3H2CO3(aq) -> 3Ag(s) + 2Al(OH)3(aq) + 3CO2(g)}$$
But I also found the following "recipe" in a Swedish newspaper
"Put aluminium foil in a bowl. Put some baking powder on the foil. Put in the silverware. Add more baking powder. Pour over hot water and wait."
I assume this recipe is more or less equivalent to the one given earlier since the baking powder essentially turns the water into carbonated water?
The reaction formula for the "blackening" of the silver given in the newspaper is
$$\ce{Ag + sulfurcompunds + O2 -> Ag2S + H2O}$$
and the reaction with the aluminium is described as following:
$$\ce{2 Al(s) + 3 Ag2S(s) + 6 H2O(l) - 6 Ag(s) + Al2O3(s) + 3 H2S(g)}$$
The newspaper also links to a small piece written at a University in Sweden with regards to this question.
where they explain the blackening of the silver in the same way as the newspaper but they give a different reaction formula for the aluminium reaction
$$\ce{2 Al + 3 Ag2S -> 6 Ag(s) + Al2S3}$$
So my question is: Is any one of these reaction more "true" (occurring more often naturally) than the others or is it the case that a little bit of everything is happening?

I've actually done this experiment before on a slightly tarnished silver ring, and I'm pretty sure I could smell a small amount of $\ce{H2S}$ being evolved when I put my nose right up against the container.

Is any one of these reaction more "true" (occurring more often naturally) than the others or is it the case that a little bit of everything is happening?

I think the best answer is "a little bit of everything".  Silver sulfide forms faster but requires exposure of the silver to sulfur-containing materials (like human skin, food, etc.).  Silver that isn't exposed to sulfur would still be oxidized by the air, but at slower rates.  So depending on the life history of the silver object in question, different amounts of the sulfide and oxide could be present.


text:
Why do esters actually smell?
We all know the very famous esterification reaction, where a carboxylic acid and an alcohol react to give an ester and water. But why do they actually smell? What is the reason behind its aromatic character?
I initially thought that it must be due to the conjugation between the two oxygens present. I do know that aromaticity is decided by the conjugated lone pairs present in the compound where the compound must follow Hückels rule. But it's only valid for a cyclic and a planar compound.
Why do esters have that fruity smell? What is the chemical explanation for a smelly non planar compound?

Don't confuse the term aromaticity when used to describe electron configuration and the term aromatic when used to describe smell. Smell came first as a simple way of characterising chemicals. When structures of the chemicals started to be understood the term became associated with a particular subset of compounds like benzene and used to describe their electronic configurations. There are a lot more compounds with interesting smells than compounds with aromaticity in their electronic configuration.

@matt_black The question is not the difference between the two. The basic question is that why does the $ -COO- $ group make anything smell at all.

Oh I got it. I've asked for its "aromatic" character there. It's not to be confused with aromaticity, of course.

Why things smell at all is a big subject about how the sense of smell works. It has nothing to do with electronic aromaticity as you seemed to assume in the question.

I did not assume that. Like I said in the previous comment I meant "aromatic" in the smelly way.

The term aromaticity originated with the discovery of unusually stable hydrocarbons that also happened to have strong smells. Many hydrocarbons smell, but not all are aromatic.
Nowadays, a compound being classified as "aromatic" has little to nothing to do with its smell and everything to do with its electron configuration. There are also non-aromatic compounds and anti-aromatic compounds. If we try to relate these to smell as well, what would an anti-aromatic compound smell like? Bad? But isn't that a little too subjective for science? Doesn't benzene smell "bad" too? Then why is it aromatic (i.e. unusually resistant to chemical change?) Why not anti-aromatic?
Esters smell partly because they exhibit weak intermolecular forces. This allows ester molecules to enter the gas phase and reach your nose. Esters don't exhibit intermolecular hydrogen bonding, unlike alcohols, for example. These are no strongly positively polarized hydrogens in esters to participate in hydrogen bonding. Consider for example ethyl butyrate, which smells like pineapples.

Most of the molecule resembles a plain aliphatic hydrocarbon! And we know these only exhibit weak van der Waals intermolecular forces. No wonder esters smell (good and bad) - they're volatile and reach our noses easily!

Thanks for the answer. I always thought that compounds were classified because they smell. But it's actually because they were smelly that's why they were classified...nice.

This is not an answer to the question. This is not because molecules are volatile than they smell !

@9-BBN you can always give your own answer by writing one below

@Rohinb97 I won't because I have that kind of knowlegde. I found your answer great but it is more an answer about why can we smell some molecules and not some others. And as I said this is not because molecules are volatiles than they smell. In my town they add a smell in the gas to be able to smell it if there is a problem.

Dissenter is correct about aromatic compounds and that the basic requirement for a compound to be detected by smell is for it to be volatile enough to reach the nose. This is why things like dry salt don't really smell like anything. Once they reach the nose, things become more complex.
Why many esters smell fruity while most thiols smell rather unpleasant is a rather interesting question. Just as most sugars taste sweet, many molecules that have structural similarities also smell similar. Humans have about 350 different olfactory receptors. These are G protein-coupled receptors that change conformation when they bind a compound and activate a signal pathway that is eventually sent to the brain for processing. However, since we only have 350 different receptors and there are far more than 350 odoriferous compounds, there is not a one-to-one mapping of receptors to compounds—there isn't an "acetone receptor". Instead, olfactory receptors have partial and overlapping selectivity. Each receptor type responds to more than one compound, thiols with a certain general shape, for instance, and different receptor types can respond to some of the same compounds. A given compound then has a more or less unique pattern of receptor activation. All this information gets sent to the brain and is processed into the perception of smell. Compounds that have similar structural properties activate many of the same receptors, so they smell similar.