text:
Synthesis Challenge
I have been trying to make a monomer reached a dead end in the middle. A B C are the products I have made so far respective order. After reaching C I wanted  to prepare 2 but could not find a way to do so. haloform had a high chance of brominating Hydroxyl group at other end I speculate. Acid catalysed Haloform on A did not result in 1 but ended up with B again [with suprisingly 2 % more yield than I got with simple base hydrolysis. Now at this point I have no option but to produce 1 and then go for 2. I need references or suggestion to make 2 from C cause that will greatly reduce my work load. I am running short of time for my summer project.

Protect terminal OH, reduce ketone, eliminate to styrene, ozonolysis with oxidative work-up?

@Jan You could probably write this up as an answer

@punarbasu-roy I'm not sure which compounds you need help synthesizing. Correct me if I'm wrong, but you need help with making compound 1 and 2?

The route I would try to prepare 1 would be to purchase 4-Bromo-4'-hydroxybiphenyl which is commercially available (Alfa Aesar sells it - see http://www.alfa.com/en/catalog/A10819)
First Step:
1) Protect phenol moeity with benzyl group via refluxing a solution of 4-Bromo-4'-hydroxybiphenyl (1.0 eq, 1.00 g), milled potassium carbonate (4.0 eq), benzyl bromide (1.1-1.25 eq), and acetone (25-30 mL). Reaction will most likely take 10-20 hours to be complete. Once reaction is complete, cool to ambient temperature and filter reaction mixture. Concentrate the filtrate to dryness.

If you dont have Benzyl bromide, you can use benzyl chloride (1.1-1.25 eq) and potassium iodide (0.1-0.5 eq).

**Alternatively, instead of filtering the reaction mixture, you could add water (50-75 mL) to the mixture to cause the product to crystallize out (this will only work if the product is a solid at room temperature). If you dont see any solids after adding water, you can try adding an ice-water bath under the mixture to encourage crystallization.
Once isolated the material, allow the protected product (benzyl ether) to be dried in a desiccator or vacuum oven to remove traces of water.
Second Step:
2) Perform a Grignard reaction with protected product (above).
   The Grignard reaction can be prepared either in THF or diethyl ether using Mg metal/iodine as catalyst, or using isopropylmagnesium chloride solution in THF).
Once the Grignard reaction has been prepared, pour the reaction onto dry ice. This will form the Protected Compound 1 as the magnesium salt. Add water to solubilize the salt, then wash the mixture with diethyl ether (1-2 times to remove impurities) and then add concentrated $\ce{HCl}$ (37%) until pH 1-2 is acheived which will precipitate out the protected Compound 1.
Third Step:
3) Dissolve protected Compound 1 in ethyl acetate, add small amount of Palladium on carbon, and hydrogenate the mixture under hydrogen pressure (balloon pressure will work just fine too) at room temperature overnight.
Once reaction is complete, degas the mixture with nitrogen or argon, filter, and concentrate to provide the desired Compound 1.

I am not sure how feasable is your reaction scheme. It is just too big. Lately I have found a better way, You can have a look. journal of material chemistry 1991, 1(4), 611-619

If you have the capability,  convert the ketone to an enol ether and ozonolyse / oxidative cleave the alkene portion. You might want to cleave your acetate first, make your ether and then silylate , but you'd be silylating to make the enol ether anyway.


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How to clean glassware containing pesticide solutions
I have pesticide (dimethoate) dissolved in methanol in volumetric flasks and graduated cylinders. How do I clean this glassware properly before moving on to the next dilution of a different pesticide.

I like this question because the cleaning of glassware is an often overlooked science in itself. You want to remove "contaminated glassware" as a possible factor in the interpretation of unexpected results in particular. A lot of brain-power and time can be wasted if you cannot assume that all the glassware (and any other surface coming in contact with your reaction) is clean with respect to the compounds of interest, and of any compounds that could interfere with their reaction.
For the given example, since dimetholate is soluble in methanol, methanol would seem to make a good solvent for rinsing the glassware before it dries.  So long as the solute is still dissolved in the solvent, a thorough solvent rinse should be effective with respect to removing dimetholate.  If the solute has dried, even though equilibrium thermodynamics say methanol should re-solublize  anything from the original methanol solution, the kinetics of this can be limiting/prohibitive. If this is the case (i.e. the solution has dried before rinsing), an aqueous detergent scrub followed by a methanol rinse is probably necessary. As water and methanol are miscible, this should not result in any mixed-solvent induced re-precipitation of anything originally present in the methanol-dimetholate solution.
Keeping to your example, as water an methanol are miscible, even in the case where the solution has not dried you should first rinse with methanol then with DI water. This will ensure any methanol that might precipitate the next standard material will be removed, and any remaining solvent (water in this case) will also not precipitate the water-soluble malathion.
The process is more complicated when you are using the glassware for a solvent in experiment 1 that is not miscible in experiment 2.  Say you a toluene  solution followed by an aqueous solution. As water and toluene are not miscible, any toluene left over from your toluene rinse would form a 2-phase solution in the aqueous solution.  While you always have the option of thoroughly rinsing then thoroughly drying (with the disadvantage of this discussed above), it is generally more desirable to do the following: thoroughly rinse the first solution in toluene, then rinse that with a solvent that is miscible with both toluene and water, like acetone. After the acetone rinse, then you can rinse the acetone with Di water.
If you have a reasonably limited number of solvents, it is good to keep a set of glassware dedicated to each solvent matrix. In my experience, the best procedure, and in fact it was part of an SOP for glassware cleaning, was to keep keep the glassware soaking in the solvent for which it will be used, rather than letting it dry.  Obviously this depends on the use case and there are certainly times that it is necessary to thoroughly dry your glassware. But where that is not necessary, to the degree reasonable, I recommend keeping the glassware separated based on what you use it for (i.e. solvent type), and keeping it soaking in that particular solvent.

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Are all NO3- salts soluble in water? If so, why?
All the examples of $\ce{NO3-}$ salts are soluble in water (all that I know about).
Is it always so or there is some salt which doesn't dissolve in water?
If so what is the reason behind it?

No, the correct way of putting it is
$$\mathrm{Almost~all~of~the~\mathbf{inorganic}~nitrate~salts~are~soluble~in~water. }$$
The families of organic nitrate salts are typically nitrates of azoles and imidazoles. Some bright examples are (R) & (S)-miconazole nitrates, isoconazole nitrate and econazol nitrate. Econazol nitrate (Other names: Spectazole, EN) is the most common organic nitrate salt, and is believed to be very slightly soluble in water, less than $0.1 \mathrm{\frac{g}{100~g}}$ (1, 2, 3)
In inorganic nitrate salts, with a bit of patience, I've noticed that barium nitrate, mercury(I) nitrate, and cobalt(III) nitrate are the least soluble of nitrate salts with a monoatomic cation. They are in the brink of insolubility, but I still wouldn't call them insoluble, since

Their solubility isn't that low at STP; somewhere around $\frac{5~\mathrm{g}}{100~\mathrm{g}}$. ($\ce{Hg2(NO3)2.2H2O}$ decomposes in water, so as RSC says)
Their solubility increases greatly with temperature.

The reason, you say?
Well, let's think of why one compound is soluble in water while the other isn't. When an ionic compound is formed, energy is released. This is called Lattice enthalpy in constant pressure. The process of formation of lattice from separate ions is usually highly exothermic. Therefore, when an ionic compound dissolves in water, and when this lattice "breaks up", the reaction would be highly endothermic.
So, to dissolve in water, the ions have to 'overcome' the lattice energy. How? Of course, the net resulting attractions between water molecules and the ions have to be stronger. This attraction is kind of a formation of a very very weak bond (and its weakness is the reason it's not categorized as such) and thus, is exothermic. This energy release when the dissolution occurs is called enthalpy of hydration, provided that the ions be in the gaseous state.
So, nitrate is a really big anion, with a single charge. The lesser concentration of negative charge results in relatively lesser lattice enthalpies. Also, the possibility of the creation of hydrogen bonds with water enhances solubility by increasing enthalpy of hydration.
That's why almost all of nitrates are soluble.

$\ce{BiO(NO3)}$ could probably be added to the list … to quote a fellow student in the ion lottery practical course. 'Found nitrate, didn’t dissolve; only one insoluble nitrate.' (Note: He meant that we wouldn’t have gotten any others.)

I’ll renew my previous comment. $\ce{BiONO3}$ is sparingly soluble ($3.2 \times 10^{-4}~\mathrm{M}$) at $a(\ce{H3O+}) = 0.025$, with solubility decreasing with higher $\mathrm{pH}$.

Nitron (1,4-Diphenyl-3-(phenylamino)-1H-1,2,4-triazolium inner salt) forms an almost insoluble nitrate.

But the solvation of many nitrates is endothermic, still they are highly soluble why?

The progression of chemical reactions is governed by the change of Gibbs free energy, not enthalpy. They're similar properties, the difference is that the Gibbs energy also includes the change in entropy. Thus a reaction which is endothermic, but increases the entropy sufficiently can move forward.

@MatyasKoszik that is correct. Now, I am far removed from the concepts here, so take this with a grain of salt, but IIRC comparing enthalpies is useful here because (1) dissolutions such as this usually net roughly the same change in entropy, (2) enthalpies are easier to visualize, and measure.

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Solubility and Evaporation - Would the concentration still be 1 ppm?
A certain salt has a solubility of 1 mg/1 L.
I have 1 mg of this certain salt dissolving in 1 L of water, I have a concentration of 1 ppm.
1st Question:
If the solution is left open to evaporation, 0.5 L of water (50 %) evaporated, would I still be getting concentration of 1 ppm while 0.5 mg of this salt became solid again and left at the bottom of the solution?
2nd Question: (if it's a yes to the 1st question)
If I then fill the container back to 1 L mark with water, would the concentration be 1 ppm again after the salts are completely dissolved?

Correct, when the water evaporates you'd have more salt than is soluble in water. The excess would then fall out of solution. Adding the same amount of water that evaporated would leave you with the same kind of solution you started out with, so the concentration of dissolved salt would be the same as when you started (although you might need to stir some to get the salt back in solution).

Thank you Derek. Does this only work with solutes that do not evaporate? What if solute is not a salt but some liquid. Like the pesticide Malathion which is liquid at room temp;

I'm not familiar with Malathion but with liquids the situation is a little different. For two liquids to be miscible, i.e. for them to be able to mix together uniformly, they have to have similar polarities. Since water is a polar solvent, Malathion would also have to be polar for the two liquids to mix. If the two liquids are miscible, then the evaporation of some water wouldn't affect whether Malathion remains mixed with the remaining water.


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How are piezoelectric crystals made?
I want to know how the powdered formed of piezoelectric material is converted to crystal like used in quartz watches. I know that some more chemicals are mixed with the material and then it is pressed in form of tuning fork (for clocks) at particular temperature and pressure. So what are the chemicals, temperature and pressure which are used. How they are made to vibrate at that particular frequency?

Are you asking about how to prepare the ceramic materials PZT, or asking about the binder added to the the prepared PZT in order to process it?

Also remember that making piezoelectric quartz (which is intrinsically a polar crystal) is a bit different than making piezoelectric PZT (which is intrinsically a symmetric crystal).

Good quality quartz crystals for use in resonators can be found in nature, such as in Minas Gerais, Brazil. They can also be produced by crystallization in a hydrothermal process, simulating in the autoclave the same process that occurred over geologic epochs.
You can easily grow your own Rochelle salt (potassium sodium tartrate, $\ce{KNaC4H4O6·4H2O}$) piezoelectric crystals. Rochelle salt was commonly used in inexpensive crystal headphones and phonograph cartridges. The raw materials are household chemicals and the directions are straightforward. Again, these crystals are grown, not hot pressed.
Tuning forks watches such as the Bulova Accutron did not use crystals, but rather metal tuning forks, made from NiSpan C, with a low coefficient of thermal expansion, similar to invar. These did not rely on the piezoelectric effect, but on magnetic induction.


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how to define proper ppm concentrations for water solvent vs alcohol solvents?

1 ppm is 1 mg / kg
Water density: 999.97 kg/m³
methanol density: 791.80 kg/m³

solution 1 with water as solvent: at 1 $\ce{\mu}$g / mL is considered 1 ppm
solution 2 with methanol as solvent: at 1 $\ce{\mu}$g / mL, should it be considered 1 ppm?
If this is adjusted by the density of methanol, it wouldn't be 1 ppm. I heard it is a common practice that at the ppm / ppb levels, the density of the solvent is often omitted by only focus on the volume. Is this true?

A little bit of punctuation and formatting really helps. It gets hard to follow what you write, specially in the 2nd paragraph.

thank you. When I typed it up late last night it wasn't looking like this. Thanks for the heads up. Going to correct it now.

Could you give a reference on "I heard it is a common practice..." part? The textbooks I usually rely on mention restrict this practice for dilute water solutions for which density can be approximated to 1 kg/l. The IMPCA reference on quantitation in methanol (http://www.methanol.org/Technical-Information/Resources/Technical-Information/Methanol-Specifications-%28IMPCA%29.aspx) actually relies on weighing the methanol sample for trace analysis of chloride.

@PLD This I heard from my friend who is has a master degree in Chemistry and working in the industry as well. Just want to get a second opinion on it here.

I think this is why it's generally not advised to use relative units such as $\%$, $\mathrm{ppm}$ and $\mathrm{ppb}$. If they're going to be used, then it's best to indicate what's the ratio being considered, which is frequently done using $\mathrm{(v/v)}$, $\mathrm{(m/m)}$ or $\mathrm{(m/v)}$, $\mathrm{m}$ being short for mass and $\mathrm{v}$ for volume. For example:
$$\frac{\mathrm{1\ \mu g_{\ solute}}}{\mathrm{1\ mL_{\ methanol}}} = \mathrm{1\ ppm\ (m/v)} = \frac{\mathrm{1\ \mu g_{\ solute}}}{\mathrm{0.7918\ mg_{\ methanol}}} = \mathrm{1.263\ ppm\ (m/m)}$$
What's common practice may be heavily dependent on whom we're talking about. Since we usually prepare solutions in the lab by completing the volume up to a certain amount, one could say it's best to ignore the solvent density and use $\mathrm{(/v)}$ as a rule of thumb. The solute unity will depend mostly whether it's a solid being weighed or a liquid being measured.
As pointed out by @PLD, however, there might be cases where it's important to weigh the solvent, probably when very accurate measurements are necessary and specially if the solvent's volume changes considerably given changes in temperature.
One problem of using the $\mathrm{(m/v)}$ notation is that it doesn't use SI units, and might also be confusing. A more verbose way of indicating this is saying mass to volume. Doing some research for this I even found the "units" $\mathrm{ppm_v}$ and $\mathrm{ppm_m}$, which indicates the solvent's unity, but not the solute's, but definetly aren't as common and are also unclear.
The best solution here is simply to use the full unity, so instead of using $\mathrm{ppm}$ (which actually isn't accepted in many cases, for example most scientific journals), use $\mathrm{mg/mL}$ (or $\mathrm{mg\ mL^{-1}}$) or whatever was really measured. This doesn't leave room for double interpretations and indicates how the solution was prepared.


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Conditions affecting pseudo rate orders
For the reaction:
$$\ce{F
+ Br_2 -> FBr + Br}$$
The gas concentrations as a function of time are given by:
$$kt=\ce{\frac{1}{[Br2]_0-[F]_0}}\ce{\ln(\frac{[Br2][F]_0}{[F][Br_2]_0})}$$
Given that $\ce{[F]_0=4 x 10^{-9}}$ and $\ce{[Br2]=1x 10^{-10}}$ I am meant to show that the rate constant is first order with respect to $\ce{Br2}$. How do I do this?
Fluorine atoms are in excess but is it a large enough excess to assume that the isolation method is valid? I could then cancel terms but would I the have to differentiate to find the rate law?

Maybe it would be best to reorganise that to
$\ce{[Br_2]}=\exp\left[-kt(\ce{[F]_0}-\ce{[Br_2]_0})\right] \frac{\ce{[Br_2]_0}\ce{[F]}}{\ce{[F]_0}}$
and as there is 40 times more $\ce{F}$ than $\ce{Br_2}$, it is safe to assume that $\ce{[F]}\approx\ce{[F]_0}$. I.e. even if all the $\ce{Br_2}$ is gone there is still $\ce{[F]}=3.9\times10^{-9}$. This simplifies the above equation to
$\ce{[Br_2]}=\exp\left[-kt(\ce{[F]_0}-\ce{[Br_2]_0})\right]\ce{[Br_2]_0}$
which is the familiar form of the expression of concentration. It is also easy to see what the pseudo first-order rate constant is.


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Are all molecular structures symmetric?
Are all molecular structure symmetric either in relation to a plane within itself or in relation to other molecules? Are there any completely asymmetrical structures when looking at molecular geometry? Are there any studies into the reason for symmetry?

There is no 'reason' for symmetry as such. Unless you mean symmetric conformations rather than arrangements?

Different molecules have different levels of symmetry - we put in lots of effort into finding out exactly how symmetrical molecules are because symmetry is the basis of lots of other things, most notably molecular orbitals. You can find out more about symmetry elements here

There are a couple of notable types of structurally asymmetric molecules.
Firstly, those that are chiral due to different ligands attached to a central atom such as the improbable $\ce{CClFBrI}$ or 'bromo-chloro-fluoro-iodo-methane'. More generally, many molecules have chiral centers; sometimes multiple ones, like inositol which has 6 in a 6-carbon molecule.
Secondly, there are rarer structures where there are no chiral centers, but the whole thing is chiral. Usually this is because of steric restrictions that force it to be one shape or another. Helicene is a very nice example. Another might be a cage hydrocarbon in the shape of this graph but I don't know if any exist.

are you saying  there are no structures that match Helicane? I am not referring to the position of ligands but the 3d structure specifically.

I'm saying that Helicane comes in two conformations that don't (readily?) interconvert and cannot be superimposed on each other.

Also, I apparently meant http://en.wikipedia.org/wiki/Helicene not "Helicane" which turns out to be a character from Shakespeare!

I guess I am basically asking if there exist any chiral molecules without a matching enantiomer or if some symmetry is always seen when looking at all levels instead of just the individual molecule?

Ah, so theoretically every chiral molecule has a matching enantiomer - but we may not have observed them experimentally, or they might be unstable for some reason. I don't know of any chemically forbidden examples, although there are many reactions that force one enantiomer or another.

Notably, in biochemistry, you (mostly) get one chiral form : L-amino acids, for example. This is covered in http://en.wikipedia.org/wiki/Homochirality but there is a lot more research out there on this topic due to the question of why this happened in the origin of life.

It is not quite clear what do you mean by "in relation to a plane within itself or in relation to other molecules", but I think the answer is no, not all molecules are symmetric. Many of them belong to what is known as $\mathrm{C}_1$ point group, or, in other words they are are totally asymmetric. For more information on molecular symmetry and point groups read this Wiki article.


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What is the role of NaHSO3 in syn diol formation from alkenes with OsO4?
All I have found on this is that NaHSO3 reduces partially the osmate ester, so breaks off Os and leaves a syn diol. Is it possible to show this step with arrows showing the process of electron movement? I suppose this step is not the main part of the reaction, and mayn't be very important, but I would like to know if it's possible - and if it is, how it looks.

After you add the $\ce{OsO_4}$ to make the reaction, you'll need to reduce the remaining $\ce{OsO_4}$ maintaining the $\ce{pH}$ of the solution constant.
In addition:

The hydrolysis of the osmate ester proceeds by simple substitution
  reactions. The $\ce{NaHSO3}$, a mild reducing agent, facilitates ligand
  subsitution by reducing the $\ce{Os}$ further.

The Art of Writing Reasonable Organic Reaction Mechanisms, Robert B. Grossman, p 293
And $\ce{NaHSO_3}$ can do both and so you will use it. You also can add $\ce{Na_2SO_3}$.

I found also something interesting on JACS, it's mentioned at least four times in the document that $\ce{NaHCO3}$ is used as buffer. Also there is a lot of information in the references of this article.

While the reaction is normally run under basic conditions
  ($\ce{K_2CO_3}$, $\ce{pH} = 12.2$, aqueous layer), it is possible to
  buffer the system with $3$ equiv of $\ce{NaHCO3}$ ($\ce{pH} = 10.3$,
  aqueous layer).$^{38, 39}$ Buffering of the reaction mixture does not
  affect the ee, but it can have a beneficial effect on the yield when
  base-sensitive substrates are used or base-sensitive products are
  formed. Thus, the AD of allyl or cinnamyl halides (cf. eqs $1$ and
  $3$) should be performed under buffered conditions to minimize epoxide
  formation (see eq $1$, the yield with allyl bromide as substrate is
  between $61$ and $74%$ under buffered conditions; in the absence of
  $\ce{NaHCO3}$ only $40-50$ % of diol is obtained38). Unfortunately,
  the AD reaction does not turn over if $\ce{K_2CO_3}$ is replaced
  entirely by $\ce{NaHCO3}$.

Hartmuth C. Kolb, Michael S. VanNieuwenhze, K. Barry Sharpless
Chem. Rev., 1994, 94 (8), p 2494


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Different solder composition affects the connection strength?
I was wondering if the tin/lead/other metals bond with the metal parts that are low-temperature soldered. I know that the flux in the solder is used to remove the oxide layer, leaving a clean metal surface ready to form metallic bonds (I think).
So for example, should I use a tin/silver/copper eutectic to bond a silver surface and copper wire? Will that make a difference or can I use tin/lead with the same result? For example, do we get a different connection strength? In every case the solder material is mostly made of tin, which is why I was wondering what's the real impact of the other components.

Good question!
The short answer: if you are soldering bulk silver to copper, 63% Sn/ 37% Pb eutectic would probably be OK, but for silver plating, you might be able to make a successful join using tin/lead, if heat is quickly applied and removed, but might need to use a silver-bearing solder for thin silver plate or if the base metal under the silver is not easily soldered.
BTW, "cold soldered" is perhaps better expressed as low-temperature soldering, though both indium and gold can be joined at room temperature.
A longer explanation: reasons to use a specific solder composition.

To some extent, solder dissolves the surface of the metals to be united. This can be problematic with plated parts, e.g. gold plated over nickel, where the solder can completely remove the plating, making it difficult to adhere to the base metal. For that purpose, solders containing the plating metal are made, but they have other liabilities: expense and brittleness. Similarly, copper-containing solders are used for copper-plated electronic parts.
Not only alloys form between soldered parts, but also intermetallic compounds form, such as $\ce{Ag3Sn}$, $\ce{Cu6Sn5}$ and $\ce{AuSn4}$, some of which are brittle. Changing solder composition may not be practical to prevent some intermetallic compounds from weakening a joint, so barrier coatings and specific temperature cycling and storage may be used to retard their formation.
Pure tin solder, as well as zinc, has the property of forming whiskers, which not only short-circuit electronics, but was rumored to have caught Napoleon's troops with their pants down.

For these and other reasons, soldering can be considered as much a craft as a science. You may have to make some tests of the solder joints if they are critical, as in aerospace or medical applications.


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What is the correct value of the Avogadro constant? And how was it derived?
I found different values of Avogadro constant in different places. So what is the correct value?
$\pu{6.0221367*10^{23}}$ or
$\pu{6.02214129*10^{23}}$ or
$\pu{6.0221415*10^{23}}$ or anything else?

http://physics.nist.gov/cgi-bin/cuu/Value?na

Also, I'm afraid the "how was it derived?" part might be too broad. There various methods and experiments to measure the Avogadro constant, most of which are usually explained well on the Net.

http://en.wikipedia.org/wiki/Avogadro_constant#Measurement

Whenever you're looking for accurate fundamental physical constants, CODATA recommended values are the way to go. As of May 20, 2019, Avogadro's constant is now truly an exact value, with infinite significant digits. Behold!
$$6.022\,140\,76\times10^{23}\ \mathrm{mol^{-1}}$$
It is unlikely this value will change within our lifetimes, which is exciting in its own way!
For historical reasons, I'll leave my previous answer below:

As of 2015, the latest data for the Avogadro constant is from 2014. According to CODATA, the most accurate value is:
$$6.022\,140\,857\times10^{23}\ \mathrm{mol^{-1}}\pm0.000\,000\,074\times10^{23}\ \mathrm{mol^{-1}}\quad\text{(CODATA 2014)}$$
The relative uncertainty in the measurement is thus only 12 parts per billion!
Interestingly, the Avogadro constant may be redefined in the near future to be an exact value, that is, a constant with zero uncertainty by definition, much like the speed of light. This would come as a consequence of redefining the SI kilogram as a function of the number of atoms in an ultrapure monoisotopic $\ce{^{28}Si}$ monocrystalline sphere engineered to extreme precision. A great video on this can be found in the Veritasium YouTube channel.
All that said, I suspect you don't really have to care which constant should be used. All the suggested values differ by one part in a million, which makes essentially no difference for most chemistry.
Edit: As pointed out by Loong in the comments, a few weeks after writing this answer, CODATA released updated values for the physical constants, so I updated this answer for accuracy. The next set of updated values will likely be announced in 2018-2019. For comparison, the previous value was:
$$6.022\,141\,29\times10^{23}\ \mathrm{mol^{-1}}\pm0.000\,000\,27\times10^{23}\ \mathrm{mol^{-1}}\quad\text{(CODATA 2010)}$$
This represents an uncertainty of 44 parts per billion. This means the uncertainty in the measurement has been cut to almost a fourth of its previous value in four years. Go science!

Personally, I like memorizing Avogadro's constant as $6.022\times 10^{23}\ \mathrm{mol^{-1}}$, because it nicely matches the fundamental charge $1.6022\times 10^{-19}\ \mathrm{C}$ and I recall that $22^2=484$, which reminds me of the Faraday constant $96484\ \mathrm{C\ mol^{-1}}$ (actually $96485\ \mathrm{C\ mol^{-1}}$ is a better approximation). Precision to spare for calculations, and pretty mathematical patterns to help my poor memory!

+1 especially for link to Veritasium.  If physicists really wanted to make things easier for everyone, they would get rid of both the Faraday constant and Avogadro's constant, defining a new "Faragadro constant" as being precisely equal to say $10^21$.  That would be both the number of particles in a "new mole" (we need another name for that).  Coloumbs and amperes would be made obsolete and current and charge would be counted in "new moles per second" or "new moles".  Volts would be made obsolete, we could just say "kJ per new mole" instead.

+1, very good answer. For what concerns the redefinition of the kilogram, this is part of the ongoing redefinition of the SI (don't hold your breath, it'll be in 2018 probably). A glimpse of the so-called new SI can be found on this BIPM webpage where there is a draft of the new SI brochure and other information. The new unit definitions are implicit, from the given, exact, values of the constants (the values given in the draft are not definitive).

@MassimoOrtolano Thank you greatly for the incredibly interesting and informative link! It seems the Watt balance is very close to winning the battle to redefine the kilogram, by fixing the value of the Planck constant, but the BIPM nevertheless intends to separately fix the Avogadro constant, too.

So it's 2018 now, any updates (expected or not) to the kilogram or Avogadro's constant that we should look out for? :))

@Equinox If Avogadro's constant is changed to an exact value by definition (which seems likely this year or the next), I'll be sure to perform the last edit ever to this answer with the details!

@NicolauSakerNeto "If Avogadro's constant is changed to an exact value by definition" seems to have happened prior to that comment of yours ;) See Felipe's answer below.

A new definition of Avogadro's constant will replace the current one soon (emphasis mine):

The mole, symbol mol, is the SI unit of amount of substance. One mole contains exactly $\pu{6.02214076 \times 10^{23}}$ elementary entities. This number is the fixed numerical value of the Avogadro constant, N$_\text{A}$, when expressed in mol$^{-1}$, and is called the Avogadro number.

This makes Avogadro number a fixed integer.
The new definition was published on 8 January 2018 as an IUPAC Recommendation in Pure and Applied Chemistry, which is available online.
It has yet to be approved by CGPM (expected November 2018), but "the revised definitions are expected to come into force on World Metrology Day, 20 May 2019" (thanks R.M.!).

It was published as an  IUPAC Recommendation on January 8th 2018. It has yet to be approved by CGPM (expected Nov 2018), and as your article states: "the revised definitions are expected to come into force on World Metrology Day, 20 May 2019".

Thanks for correcting it, I changed the text to reflect that.

Avogadro's number has changed over time because it depends on how we standardize the atomic mass unit.  Formerly, oxygen was used as a standard with a value of 16, and from a comment in another post I recall that physicists used 16 specifically for oxygen-16 (making natural oxygen slightly heavier than 16). When they went to the carbon 12 = 12 standard naturally occurring oxygen dropped below 16, so Avogadro's number represented less mass of oxygen (and everything else) and it dropped accordingly.

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text:
Why does the transverse magnetization in Fourier Transform NMR consist of Ix+iIy ($\hat{I}_{\!x}+\mathrm{i}\hat{I}_{\!y}$)?
More specifically, the expression for the average expectation value of the transverse magnetization has the following form:
$$M^+(t)=N\gamma \hbar\,\text{Tr}(\hat{\sigma}(t)\hat{F}^+)$$
with $\hat{\sigma}(t)$ the density matrix. $\hat{F}^+$ is defined as:
$$\hat{F}^+ = \sum ^{K} _{k=1} \hat{I}^+_{\!k} = \sum^{K}_{k=1} (\hat{I}_{\!kx} + \mathrm{i}\hat{I}_{\!ky})$$
I don't really see how you intuitively obtain such an expression for the transverse magnetisation (in the syllabus it was posited, not derived). On the other hand, if you would take the complex modulus then you would find the magnitude of some 2D vector. It still does not motivate (enough) for me why it should look like this. Doesn't measurement happen in a single direction? Why is it not just
$$\sum^{K} _{k=1} \hat{I}_{\!kx}?$$
It also bears resemblance to the raising operator (which would explain the $+$), perhaps you measure how many quanta are absorbed, which corresponds to measuring the amount of spins that you raise to a higher energy level (but that's not what the raising operator signifies I think)? Or is it a phasor with a $90^\circ$ angle between both components?

You got all your operators, M and F in step operator formalism. You don't have x and y eignestates, that is why you have to do this formalism. Check the whole topic in a physics book, eg. the Cohen-Tannoudji QMech or in an EPR/ESR book. It is something Chemist are generally not thought.

Any modern spectrometer uses quadrature detection, so you don't measure only in one direction. Physically there is only one coil, but you also record the signal phase-shifted by 90°, you get a complex signal in the end. See this chapter from Keeler's book for more details

orthocresol has answered this question beautifully below. I would just add that, you are correct! NMR spectrometers really only measure one transverse component of (the expectation value of) the magnetization. The rest is a mixture of technical convenience and tradition. We can place the local oscillator that drives the down-converting  mixers to one side of the spectrum and avoid complex signals. We can also separate the transmitter frequency and the local oscillator frequency if we really want to.

As Mad Scientist said in the comments, measurement of the transverse signal in NMR occurs in both the $x$- and $y$-directions. This is known as quadrature detection. In practice this is not achieved with two detector coils, but rather with a "trick" which is described more fully in Keeler's Understanding NMR Spectroscopy, 1st ed., section 5.6. For a more mathematical description, see also Appendix A.5 of Levitt's Spin Dynamics, 2nd ed.
Instead of having two separate quantities $M_x$ and $M_y$, it's easier to lump them into one complex quantity $M$:
$$M = M_x + \mathrm{i}M_y$$
For example, if we only have one spin with equilibrium magnetisation $M_0$ pointing along the $z$-axis, then following a $90^\circ (y)$ pulse (which rotates the magnetisation vector to $x$),
$$\begin{align}
M_x &= M_0 & M_y = 0
\end{align}$$
and after a period of free precession $t$
$$\begin{align}
M_x &= M_0\cos(\Omega t)\mathrm{e}^{-t/T_2} & M_y = M_0\sin(\Omega t)\mathrm{e}^{-t/T_2}
\end{align}$$
and $M = M_x + \mathrm{i}M_y = M_0\mathrm{e}^{\mathrm{i}\Omega t}\mathrm{e}^{-t/T_2}$, where $T_2$ is the spin-spin relaxation time. Fourier transformation of this complex signal gives a spectrum $S(\omega) = A(\omega) - \mathrm{i}D(\omega)$. There is a constant of proportionality which doesn't matter and has been dropped. The real and complex components are known as the absorption and dispersion Lorentzians respectively:
$$\begin{align}
A(\omega) &= \frac{1/T_2}{(1/T_2)^2 + (\omega - \Omega)^2} & D(\omega) = \frac{\omega - \Omega}{(1/T_2)^2 + (\omega - \Omega)^2}
\end{align}$$

(Diagram from Keeler's Understanding NMR Spectroscopy, 2nd ed. (p 84).) The absorption component is usually what is shown in the actual NMR spectrum. There are some complexities because the spectrum has to be phase shifted, but the idea holds.
For a single spin it is very easy to use product operators instead of the vector model. After that, to obtain the expressions in your question, you just generalise to multiple spins. I might come back to this some time later, but for now, this will suffice.

The benefit of quadrature recording (which seems to have been cut from Keeler's 2nd ed., but remains in the 1st ed.) is that it allows us to find out the sign of the offset $\Omega$. Let's say that instead of having quadrature recording we only measure along the $x$-axis. Our signal is
$$\begin{align}
M &= M_x \\
&= M_0\cos(\Omega t)\mathrm{e}^{-t/T_2} \\
&= \frac{1}{2}(M_0\mathrm{e}^{\mathrm{i}\Omega t}\mathrm{e}^{-t/T_2} + M_0\mathrm{e}^{-\mathrm{i}\Omega t}\mathrm{e}^{-t/T_2})
\end{align}$$
It's relatively straightforward to show that for a function $M = M(t)$, if $\mathcal{F}[M(t)] = f(\omega)$:
$$\mathcal{F}[M^*(t)] = [f(-\omega)]^*$$
and using this, we can show that if we take the Fourier transform of the one-dimensional magnetisation, we get
$$\begin{align}
S(\omega) = \frac{1}{2}[A(\omega) - \mathrm{i}D(\omega) + A(-\omega) + \mathrm{i}D(-\omega)]
\end{align}$$
and if you take the full expression for this and replace $\Omega$ by $-\Omega$, you will find that $S(\omega)$ is the same regardless of the sign of the offset $\Omega$. This means that our spin of interest could have either a positive or a negative offset, relative to the spectrometer reference frequency $\omega_\mathrm{rf}$, but we would not be able to determine it.
In theory, if you could control the phase of the signal you are detecting, single detection would allow you to extract the sign of the offset. For example, in our example above, if you measure $M_y$ instead of $M_x$, you'll find that there is a difference when you replace $\Omega$ by $-\Omega$. Here's a schematic from Gunther's NMR Spectroscopy, 3rd ed. (p 265):

Here, the author Fourier transformed $M_x$ and $M_y$ separately before adding them up. Since the Fourier transform is linear, this has the same effect as adding $M_x$ and $M_y$ and Fourier transforming the sum. In any case, it shows that if the magnetisation vector starts off along the $y$-axis, you could, in theory, measure only the $x$-magnetisation to determine whether the offset is negative (case a) or positive (case b).
Such a spectrum would be very difficult to interpret, and would probably be impossible to analyse if positive and negative peaks overlapped. On top of that, it would also require you to know the exact phase difference between the initial signal and your single detection coil. I believe this is (due to several reasons) not experimentally possible, so unfortunately, there's no way to extract this information using single detection. (This is also part of the reason why NMR spectra have to be phase corrected – for the average NMR user the computer does this automatically.)
Now, we've established that you need quadrature detection to determine the sign of the offset. However, in the final NMR spectrum, we don't really look at offsets; we look at resonance frequencies. (To be precise, chemical shifts, but those are linearly related to resonance frequencies.) So, if you really wanted to, you could use single detection while placing the reference frequency at the far end of the spectrum. For example, if you want to measure peaks from $400.000$ to $400.001~\mathrm{MHz}$, you can place the reference frequency at $400.002~\mathrm{MHz}$. Then, by virtue of choosing an overly high value for $\omega_\mathrm{rf}$, you know that your offset $\Omega$ has to be negative.
However, a rf pulse centred at $400.002~\mathrm{MHz}$ would be less effective at exciting the spins (as it is further away from being on-resonance), and essentially you are throwing away half of the rf pulse (any components of the rf pulse with frequency greater than $400.002~\mathrm{MHz}$ are wasted). So, it's preferable to put $\omega_\mathrm{rf}$ right in the middle of the spectrum; but as said before, that would necessitate quadrature detection so that you can obtain information about the sign of the offset.
As a final comment, the fact you note about how $\hat{I}_{\!x} + \mathrm{i}\hat{I}_{\!y}$ is the raising operator $\hat{I}_{\!+}$ is reflected in many books when they discuss coherence transfer pathways: often it is said that the detected signal has coherence order $-1$. This corresponds precisely to detecting the $\hat{I}_+$ operator, or specifically, one off-diagonal entry of the density matrix.


text:
Comparing heat of hydrogenation per mole of butenes and 1,3-butadiene
The question asks us to select the compound with the smallest heat of hydrogenation per mole out of the following:
a) 1-butene
b) trans-2-butene
c) cis-2-butene
d) 1,3-butadiene
Since all four compounds form the same product (butane), a comparison without actual data should be possible. 1,3-butadiene is conjugated so it is the most stable compound and therefore its heat of hydrogenation should be the smallest.
But it requires 2 moles of hydrogen for complete conversion to butane. And the answer is actually trans-2-butene.
So would we consider each heat of hydrogenation "per mole of hydrogen" or "per mole of the given compound"? The question isn't clear about this and also, if we do consider it as "per mole of hydrogen" then is it correct to simply half the value heat of hydrogenation for 2 moles?

Related: Why is 2-methylpropene less in energy than its alkene counterparts?

Heat of hydration or heat of hydrogenation?  Only hydrogenation leads to butane as the product.  Hydration would potentially lead to different products for each molecule: (a) 1- or 2-butanol; (b) 2-butanol, (c) 2-butanol,  (d) 1,4- or 1,3- or 2,3- butanediol.

@CurtF. I just realized I used both words in my question. Thanks for pointing it out, I've corrected it.

a comparison without actual data should be possible. 1,3-butadiene is
  conjugated so it is the most stable compound and therefore its heat of
  hydrogenation should be the smallest

No, 1,3-butadiene is not the most stable of these compounds, either for the whole molecule or on a per double bond basis (see below).  The actual numbers (see table below) are too close, we need to look at some data in order to answer the question correctly.

The question asks us to select the compound with the smallest heat of
  hydrogenation per mole out of the following:
So would we consider each heat of hydrogenation "per mole of hydrogen"
  or "per mole of the given compound"?

Here is some data that will help answer the question.

(image source)
On a "per mole of given compound", trans-2-butene has the lowest heat of hydrogenation, 27.6 kcal/mole.
On a "per mole of hydrogen", 1,3-butadiene has a heat of hydrogenation between 28.2 (56.5/2) and 28.6 (57.1/2) kcal/mole, so trans-2-butene still has the lowest heat of hydrogenation at 27.6 kcal/mole.
So in this particular list of compounds, it doesn't matter whether the question is asking per mole of hydrogen or per mole of given compound, the answer is the same, trans-2-butene.
At first I thought "trick question", but then later thought, not really.  The questioner did his homework and wanted the student to work out the answer both ways and see that since the you get the same answer whether you answer on a per mole compound or per mole hydrogen, you don't need to know whether the question is being asked on a per mole compound or per mole hydrogen basis.

is it fair to generalise that almost all conjugated hydrocarbons would have their electrons delocalised over the whole molecule and be able to share and make room for bonding? So making them less stable than a single  (plain) double bond compound? (e.g. the above, 1-butene vs 1,3-butadiene)

Yes and no.  Yes, it is fair to generalize that most conjugated hydrocarbons would have their electrons delocalised over the whole molecule, but this makes them more stable.  Fore example, as shown above, 1,3-pentadiene is more stable than 1,4-pentadiene due to this delocalization.

To saturate one mole of 1,3-butadiene, there is a need of two mole hydrogen gas. In comparison with 1-butene, trans-2-butene, and cis-2-butene, these molecules only need 1 mole of hydrogen gas to be saturated to butane. Thus, 1,3-butadiene has the highest heat of hydrogenation. Stability of the remaining molecules can be used to measure qualitatively the trend of the heat of hydrogenation. Since trans is more stable among them, then it has the least heat of hydrogenation.

So by "heat of hydrogenation per mole", do we consider one mole of the compound to be hydrogenated?


text:
How would I go about making an ethyl ester from vegetable oil?
I'm teaching myself how to make lotion and a lot of lotions contain esters like isopropyl myristate or isopropyl palmitate. I'd like to make an ethyl ester from a popular skin-care oil (olive, apricot kernel, or castor).
Since there's so much controversy about using isopropyl alcohol in skin-care, I'd like to avoid that if possible. It's very difficult to find any info on the process, and the closest I could get was how to make biodiesel. All of those use methanol though.
How would I go about making an ethyl ester from these oils?

The biodiesel process should work well, you will need just slightly more aggressive conditions (i.e. slightly more catalyst, slightly higher EtOH excess, and/or slightly longer time) to make ethyl esters instead of methyl esters, as ethanol is slightly less reactive than methanol.
For skin care, you will need higher purity than biodiesel processes usually provides.  I would imagine that the cosmetics industry uses vacuum distillation to purify their esters.  This would separate the ethyl esters from unreacted triacylglycerols, any base (or acid) catalyst left over from the reaction and soluble in the organic ester phase, organic-soluble glycerol, and organic-soluble impurities.


text:
Determine balloon diameter needed to lift a 7 pound weight with various gases
My question is regarding setting up this equation. The givens are as follows:

Gases are helium, methane and hydrogen. (Test all three)
Assume balloon is spherical.
Assume temperature is $25~\mathrm{^\circ C}$

So I need to find the diameter of the balloon needed to lift a 7lb weight. I setup my equation as follows. $$m=\frac{PVM}{RT}$$
Next, I subtract my first gas ($\ce{He}$) from ambient air and convert my $m$ to equal grams in the weight. so I set it up like so.
$$3730.15~\mathrm{g}=\frac{PVM}{RT}-\frac{PVM}{RT}$$
Next I factor out the pressure and volume and fill in my numbers.
$$3730.15~\mathrm{g}=PV\left(\frac{(28.9~\mathrm{g~mol^{-1}})}{(0.0821~\mathrm{L~atm^{-1}mol^{-1}K^{-1}})(298~\mathrm{K})}-\frac{8~\mathrm{g~mol^{-1}}}{(0.0821~\mathrm{L~atm^{-1}mol^{-1}K^{-1}})(298~\mathrm{K})}\right)$$
This is where I am stuck at as I don't know what makes sense to use for pressure so I can solve for volume? Or maybe this is all setup wrong and makes no sense?

You don't really have to go to this extent  to solve the problem, but I think it will work either way. You can assume that the balloon is a thin flexible membrane with negligible mass, separating the gasses from the rest of the atmosphere. As such, the gas inside the balloon will be equal to atmospheric pressure. How about answering your own question?

That's what I was stuck at. I finished the problem but was wondering if that was appropriate to use 1atm  as the pressure for this equation. You are saying it is appropriate? If there is an easier way to do it I am more than happy to hear it! I spent a long time working on this two different ways and I just am having a hard time wrapping my head around buoyancy.

For a real balloon, the balloon itself will have mass, and the balloon may have higher pressure inside than atmospheric pressure.
Absent additional information, for a homework problem, you would assume mass of the balloon is zero and pressure is atmospheric pressure.
Note: in the OP 8g/mol is wrong.

Oh my god, I am such a rookie. Helium isn't diatomic... Thank you!


text:
Can all substances be supercooled?
With optimal conditions can we supercool any compound or is it that just water and a certain set of others can be supercooled? If not, why?

Supercooling is a general phenomenon and it is not limited to any particular substance, although the degree you can supercool something will vary. The idea in thermodynamics is that at the melting point the free energy of the solid and the liquid is equal and a lower temperature will favour the formation of the solid.
However, when there is only liquid present, you will first have to form a small crystal nucleus which is not as stable as the bulk solid i.e. has a higher free energy. This is basically a kinetic barrier which prevents forming the solid.
The formation energy of such a nucleus will depend on the interfacial energy of the crystal-liquid system. In a substance where this interfacial free energy is very little there won't be too much supercooling.

So called 'drop tubes' have been used for decades to study the nucleation rate vs temperature of supercooled molten droplets of (mainly) metals in vacuum or inert environments.  An (admittedly old) example is 'A review of long drop tubes as a supplement/alternative to space experiments', R. J. Bayuzick et al., Advances in Space Research 4(5) 85-90 (1984). Nucleation studies of supercooled semiconductors (Si and Ge) was also studied back in the 1980's. Googlingh 'drop tube recalescence' will find more.  Overall, I'd say the literature is quite full of supercooling on any number of different materials.
It is interesting to note that supercooling of a liquid is much much easier than superheating of a solid - the problem is decidedly not symmetric in the general case.  The difference between nucleating a critical volume of solid in a liquid droplet is not the same as initiating melt at the surface of a solid, where there are unsatisfied bonds because of the interface.


text:
Gallium-induced structural failure of aluminium
I have seen youtube videos of people treating some aluminium object with liquid gallium and after letting it sit for a while, they are able to break up the aluminium into tiny little pieces with their bare hands.
Can someone tell me what is exactly going? I haven't been able to find a satisfactory explanation as of yet.
Does the gallium somehow incorporate itself into the metal lattice, disrupting the bonds? or is it forming an amalgam of sorts and actually dissolving the aluminium?
Is it a combination of these two, or are my ideas off completely? Also, what would be the exact mechanics of a process like this?
EDIT: How exactly does gallium "unzip" the grain boundaries?
 Also, besides Ga or Hg, what other metals would produce a similar effect? What mechanism/property allows them to behave in such a manner?

It moves down the grain boundaries and 'unzips' them efficiently.

As you state, it forms an amalgam, just as mercury does. @Jon Custer's explanation is correct, and furthermore, the alloy does not form a protective oxide, so the aluminum oxidizes to $\ce{Al2O3}$ and quickly crumbles. This happens even faster with an Hg-Ga alloy!
BTW, gallium should be safe enough, though expensive, but I would not like to work with mercury without a hood, and without careful containment.
Thanks for the question; this encouraged me to view that amazing demo on Vimeo!


text:
Galvanic reaction between aluminum and carbon fiber composite
I understand that there is a galvanic reaction between aluminum and carbon fiber composite. Structural aluminum parts in contact with carbon fiber composites should be insulated by something like fiberglass.
But what exactly is it that happens, chemically? I know that aluminum has a tendency to be oxidized; aluminum oxidizes readily by itself in air. I also know that carbon fiber itself is a conductor due to delocalized electrons.
So assuming that aluminum is oxidized then the carbon fiber composite must be reduced.
What exactly is reduced? The carbon fiber, or the epoxy? What is the role of the epoxy in all of this? Wouldn't the epoxy be electrically insulating? How exactly does this redox reaction take place across the epoxy?

The reason is that aluminium carbide, $\ce{Al4C3}$ is formed. The reaction occurs at around 500°C on the interface. The problem with this substance is that it is brittle, and it also reacts with water:
$\ce{Al4C3 + 12H2O -> 4Al(OH)3 + 3CH4}$.
More info at this and this wikipedia page.


text:
If given a half reaction, how do you determine if an element can exist in a acidic conditions?
I was given the formulas,
$$\begin{alignat}{2}
\ce{MnO4^2- \;&<=> MnO4- + e-}\quad &&E^\circ = -0.56\ \mathrm{V}\\
\ce{MnO2 + 2H2O \;&<=> MnO4^2- + 4H+ + 2e-}\quad &&E^\circ = -2.27\ \mathrm{V}
\end{alignat}$$
I gathered you have to see if it’s non-spontaneous or spontaneous and compare it with the original reaction, but, when I asked my teacher on how to start, he said “multiply the first reaction by 2 and the second by −1.”
I understand the ×2 to balance electrons, but why do you multiple it by −1? I don’t understand this question, any help would be appreciated. Thanks.

Multiplying a reaction by -1 means that you reverse it; in other words, your reactants become your products and your products become your reactants. By reversing the second equation you then have 2 electrons in your reactants, so that when you add the reactions you start and end with the same number of electrons.

Doing so will result in the two half equations
$\ce{2MnO_4^2- -> 2MnO_4^- + 2e^-}$
and
$\ce{MnO_4^2- + 4H^+ +2e^- -> MnO_2 + 2H2O}$
which you then combine into the total equation
$\ce{3MnO_4^2- + 4H+ <=> 2MnO4- + MnO2 + 2 H2O}$.
The question is which way this goes. You can see that you have loads of $\ce{H+}$ on the left-hand side so you can expect that in acidic conditions the equilibrium will be shifted to the right hand side. You can quantify this by calculating the electrochemical potential from the Nernst equation for the two half cells.


text:
Limitations of Lewis structures
Using Lewis structures to describe chemical bonds inside a molecule is still fairly common start in undergraduate education. Lewis structure however can be misleading, or just simply wrong in many cases. Mostly I see this with aromatic systems, Diels-Alder reactions etc.
My question is if there is a rule of thumb when one can safely use Lewis structure and when we should just  forget them? Or should I just check it every single time and compare them with eg. MOs?

I am not sure what you mean when you say Lewis structures are misleading for aromatic systems and Diels-Alder reactions.  Can you elaborate?

My understanding is that the bond-lengths is poly-cyclic aromatic systems are often unlike what would have predicted from resonant structures. Also, that the explanation of Diels-Alder reaction mechanism needs MO theory.  Please correct me if I wrong.

The usual explanation for the Diels-Alder reaction mechanism is standard electron pushing (e.g. http://en.wikipedia.org/wiki/Diels%E2%80%93Alder_reaction) based on Lewis structures. Wrt bond-lengths in poly-cyclic aromatic systems - can you provide an example?

This is an interesting question for sure, but it seems quite hard. Given the immense complexity and richness of chemical space, I can only imagine that Lewis structures will almost always be inadequate. However, an important factor to consider is whether almost all of these inadequacies will be restricted to exotic compounds we'll never care about.

Dioxygen $\ce{O2}$ is well known as a paramagnetic material. While this is predicted by the MO of dioxygen,  Lewis structure of  $\ce{O2}$ fails to predict this property. So, as Lewis structure fails for a simple molecule like $\ce{O2}$ , I think we should consider it with vigilance.

@YomenAtassi Thank you for pointing out, very good example. +1 My question was more about if we can foresee, based on simple structural features, where we should be careful in the chemistry of first two rows.

@JanJensen Thanks for trying to improve the question. Sorry, I cannot give reference to my unpublished calculations and also the question intend to be general. I am calculating carbon nano-structures and simple theories like Clar's / Lewis resonant structures can give insightful help sometimes (e.g. radical characters), but fail in many cases (e.g bond lengths). If the answer is that e.g. the theory is good, only people use it wrong, or wrong only in very few, special cases, or only wrong when we want to discuss 3D properties (D-A) then so be it. I welcome this as an answer, too.

http://chemistry.stackexchange.com/questions/10215/the-bond-in-coordination-complexes

OK, here goes. Chemists use a variety of representations of molecules to try to predict their behavior and reactivity, but none of them come close to preparing you for what happens in the laboratory. In that sense, none of them are "safe" to use. Lewis structures, and the curly arrow formalisms, are good for organic chemists to quickly explore possible avenues of reactivity and synthetic pathways. But they are not very good at dealing with chemistry south of the second row of the periodic table, nor at phenomena such as aromaticity, electrocyclic reactions, bulk properties, solvation, steric effects, polarity, polarizability, photochemistry, free radicals, etc. So we add in to the mix some "rules", try to think of molecules as being best described as "multinuclear atoms", and try to improve and promulgate our best tools for computational chemistry. For all that, we still need to approach the laboratory with a degree of humility--things that look good on paper usually do not look so good when you try them out in the real world.


text:
What significance and applications does nanotechnology have in chemistry?
Specifically concerning organic chemistry, but an explanation on its significance and applications in inorganic chemistry would be great as well.

The question title as written seems too broad. There are lots of examples of applications of nanotechnology throughout chemistry. There are so many that it would be hard to imagine that there is one definitive answer to this question. Now, for the more localized question of organic chemistry, I think we have a better chance. Perhaps if this question was specific to organic and you asked another question about inorganic chemistry.

There is branch named Chemical Nanoscience and Nanotechnology Which deals with chemistry asspects.
Talking about organic chemistry, nanotechnology is used for Organic synthesis.
With help of Modern Chemistry today we can synthesis almost all complex molecules, thanks to nanotechnology.
Check out Wikipedia-1,
Wikipedia-2 and nano.gov

For organic chemistry, Polymer nanotechnology is developed notably.  Polymer matrix and nanocomposites have become a prominent area of current research and development
Example:
exfoliation

This answer has been flagged as VLQ. While I do not agree with this flag, I and the OP probably too would appreciate and example for the use of nanotechnology in organic synthesis.

@Martin-マーチン I have edited the answer and added example


text:
Why does LiAlH(O-tert-Bu)3 not reduce acyl chlorides to alcohols?
Why do $\ce{LiAlH(O-^{t}Bu)3}$ and similar bulky reducing agents not reduce acidic chlorine further?
$$\ce{CH3-COCl + LiAlH4 + H2O -> CH3-CH=O -> CH3-CH2-OH},$$
 but
$$\ce{CH3-COCl + LiAlH(O-^{t}Bu)3 + H2O -> CH3-CH=O -> no~reaction}.$$

Without being very specific it looks that $\ce{LiAlH(O-^{t}Bu)_3}$ is a lot less reactive than $\ce{LiAlH4}$, mostly because the $\ce{^{-}O-^{t}Bu}$ is a much weaker base than $\ce{H-}$. Also its much bulkier, further reducing reactivity.
The aldehyde being less reactive than the acyl chloride the reducing agent first eats up all the acyl chloride, and then starts reducing the aldehyde. If you were careful to add stoichiometric amounts, you can stop right when you run out of the acyl chloride.
See also this link.


text:
Why can't small trans-cycloalkenes exist?
Why when a ring has 8 carbon's and contains a double bond the trans version is very unstable to exist, for instance a molecule like cholesterol?

Are you asking why a trans configuration of a double bond in an eight membered cyclic system is less stable than the cis configuration?

Yes, i was asked a question whether a molecule of cholesterol displays cis-trans isomerism at the C=C bond.

Because you're stuck in the ring structure.
The trans isomer has branches that, in their natural state, would want to rest antiparallel to one another. But in double bond 6 membered ring structure (like in cholesterol), these bonds would need to be twisted a lot so that they could sit in the octagonal ring structure. Given how a double bond is rigid and will already resist rotation, this theoretical trans isomer will have such a high energy requirement for it to exist that it will simply refuse to do so.

trans-cyclooctene is stable at room temperature.


text:
What causes the iridescent colour in laboratory grown bismuth?
Artificially grown bismuth crystals are arguably magnificent, as seen in the image below:

Image source: Minerals.net
The website states that:

They have hopper-like growths in pseudocubic crystals and are usually coated with chemicals to prevent tarnish, thus maintaining the silver-white color. Sometimes the coating gives a colorful effect on the bismuth.

What coating gives the colourful iridescence? and what chemical processes cause the colourful iridescence to occur?

The colour is from a thin film of bismuth(III) oxide that forms on the surface if the crystals are formed in air. At the elevated temperatures used to melt bismuth, the oxide forms quite quickly. The iridescence is a result of thin film interference—light waves constructively or destructively interfere as they bounce off the bismuth-oxide and oxide-air interfaces. The interference is dependant on the wavelength and incidence angle of light as well as the film thickness, which results in different wavelengths of light being attenuated by different amounts, making the surface appear coloured and the colour changing with viewing angle.

Though not nearly as stunning, regular aluminium bars can sometimes also show the same effect with the correct illumination and viewing angles. However, thin film interference can be greatly increased in many metals by electrochemically forced oxidation, forming a layer of oxide with precisely controlled thickness. This process is called anodizing or anodization, and can be used to colour metals without using dyes. A great video explaining this process, by Bill Hammack the Engineer Guy, can be found here.

Yes, there are a few metals that make very nice colours from interference effects with oxide coatings, especially titanium and niobium.


text:
Why are Group 11 elements unreactive?
Group 11 metals namely $\ce{Cu, Ag, Au}$ are known for their low reactivities. They are thus rightly called coinage metals and have been historically used to make currency owing to this very property of theirs. My question is, what is the reason behind this behaviour?
Is it because of the stability of fully filled and half filled orbitals? But then the counterexample of $\ce{Cu-Zn}$  presents itself. These are neighbours in the periodic table, and the differentiating electron in each goes completes a full or half-filled orbital, yet $\ce{Zn}$ is quite reactive.
This behaviour can be seen (sort of) between $\ce{Hg-Au}$ as well which differ in one electron and yet are completely different in terms of their properties.
EDIT: The other two questions referred to as duplicates of this one cite different properties of the elements that make them unreactive. Gold has strong $\ce{Au-Au}$ bonds (due to relativistic effects) and $\ce{Zn}$ is more reducing than copper due to factors like hydration energy and the enthalpy of atomization. But this does not answer my question because:

I was looking for the root cause of this behaviour. If gold is stabilized by relativistic effects, why don't these effects operate on other transition metals? If zinc has high hydration enthalpy, what is the reason?
Since all the coinage metals occur in a single group, I really wanted an explanation that could apply to all of them, such as a peculiarity of their electronic configuration that renders them inert to most available reagents.

http://chemistry.stackexchange.com/questions/25192/why-is-zinc-more-reactive-than-copper?rq=1, http://chemistry.stackexchange.com/questions/21810/why-is-gold-unreactive-when-only-one-electron-in-outer-shell

Yeah...$\ce{Zn^{2+}}$ is a pseudo-inert ion...due to filled orbital

The first step toward answering this question is understanding the extent to which the statement "Group 11 metals are unreactive" is or isn't true.
As stated in How do plants respond to copper deficiency? Plant Signaling and Behavior vol. 3, pages 231-232:

The transition metal copper is essential for all organisms.

It is not just that copper is present in every living thing.  Copper is reacting in every living thing. Copper ions cycle between the +1 and +2 oxidation states in Plastocyanin,  Copper/zinc superoxide dismutase and cytochrome oxidase. So as you are saying copper is unreactive, it as reacting in cytochrome oxidase in every mitochondrion of your body, and in Cu/Zn superoxide dismutase in the cytosol of cells through your body.
One way to quantify the reactivity of metal ions in through reduction potential, here listed for reduction of the ion to the zero oxidation state metal.  (limiting to d-block metals at least as unreactive as Cu)
\begin{array}{c @{:} c} \\
\color{blue}{\text{Ion}} & \color{blue}{\text{Reduction potential}} \\\hline
\mathbf{\color{red}{Cu^{2+}}} & 0.3419  \\\hline
\ce{Tc^2+} & 0.400   \\\hline
\ce{Ru^2+} & 0.455  \\\hline
\ce{Cu+} & 0.521 \\ \hline
\ce{Rh+} & 0.600  \\
\ce{Rh^2+} & 0.600  \\
\ce{Rh^3+} & 0.758  \\ \hline
\mathbf{\color{red}{Ag^+}} & 0.7996   \\\hline
\ce{Os^2+} & 0.85  \\\hline
\ce{Hg^2+} & 0.851  \\\hline
\ce{Pd^2+} & 0.951   \\\hline
\ce{Ir^3+} & 1.156  \\\hline
\ce{Pt^2+} & 1.18   \\\hline
\mathbf{\color{red}{Au^{3+}}} & 1.498  \\
\ce{Au+} & 1.692  \\
\end{array}
Looking at this list, we see that there are a substantial number of d-block metals that are less reactive than copper.  A few are less reactive than silver, and gold is the least reactive on the basis standard reduction potential.
The overall trend in d-block metals, is down and to the right in the block is less reactive. The exception is that, for a given row, the group 12 metal is more reactive than the group 11 metal, but due to the formation of a 2+ ion, not a 1+ ion.
Next we should look at ionization energies of d-block metals to see how much of the reduction potential corresponds to properties of an individual atom, versus bulk or solvent effects.
Looking at first-ionization energies, for each row of the d-block, the highest ionization energy is always the group 12 element (Zn, Cd, Hg). This corresponds well with the fact the Zn, Cd and Hg do not form +1 ions.  The group 11 elements have significantly lower first-ionization energies, and do form +1 ions.
Then looking at second-ionization energies, group 11 has the highest second-ionization energies for every row of the d-block!
Much of the reactive/unreactivity of d-block is explained by the increase of nuclear charge to the right and down the d-block.  The complete d-shell of neutral group 11 atoms shields the valence s electron well enough that it may be removed to react.  However, to lose 2 electrons, the group 11 elements would need to lose a d-electron from a completed d-shell, which is difficult.  For group 12, two s valance electrons can be lost, and the complete d-shell maintained.

I see your point. Even if their unreactivity isn't anything exceptional, there must be something special about the group 11 elements. These are widely used for many applications, whereas their neighbours are relatively unknown by comparison.

@Gerard Ag, Cu and Au are the top 3 best electrical conductors at room temperature.  This is probably the main reason for the high production rate of copper today.

@Gerard Also, Cu, Ag, Hg, and Au were the least reactive metal known to ancient people (before 1500).  The other metals listed in my answer: Ru, Rh, Os, Pd, Ir, Pt, Tc, were first understood to exist  generally in the 1740-1940 time period.

The classic paper of Dr. Norskov explained this phenomenon very well. (Off topic: Dr. Norskov is like the godfather of catalysis and I had a chance to meet him at a summer school and that was a great experience for me.)
The inertness of metal surface to adsorb depends on two factors:

Degree of filling of $d$ orbital
Degree of overlapping of orbital

Gold and Copper both has filled $d$ orbital ($d^{10})$. That means the main contributor to the reactivity of transition metal is totally absent here.
Now the second factor is very crucial to explain the inertness. In case of hydrogen adsorption on $\ce{Au}$ and $\ce{Cu}$ the anti-bonding orbital is below Fermi level, that means anti-bonding is filled for the adsorption of $\ce{H}$ on $\ce{Au}$ and $\ce{Cu}$.

From the first picture, we can see that for noble metals Fermi level is too high and in the second (actual SCF DOS) picture we can see that $\ce{Au}$, $\ce{Cu}$ had anti-bonding DOS which is below the Fermi level. Now, How it can explain the inertness? Fermi level is almost analogous to HOMO.
Fermi level is the energy level where the orbital has equal probability of being filled or unfilled (Fermi-Dirac statistics). From the left picture of the 1st picture you can see the energy profile. To better explain I tried to use some abstract numerical number to define the importance of anti-bonding orbital.
Suppose you want orbitals $A$ and $B$ to form the MO $AB$. Now $A$ has energy -10 and $B$ has energy -15, so they have total -25 unit energy. Now for the stability, the system will always try to minimize energy. The bonding orbital has -30 unit energy and anti-bonding orbital has 7 unit energy. So they have total -23 unit energy.
So the question is why the bonding + the anti-bonding energy > $A + B$. The reason is when an atomic orbital forms a molecular orbital, there is an energy penalty associated due to the orthogonalization of the orbitals. So, in case of $\ce{H2}$ formation, there is no electron in antibonding orbital, so total energy is reduced and that's why $\ce{H_2}$ is more stable than $\ce{H}$ atom; while in the case of $\ce{He}$, both bonding and anti-bonding orbitals are filled, so it's energetically more favorable for it to be in an isolated state.
Now, In $\ce{Cu}$ and $\ce{Au}$ the antibonding state is below the Fermi level, that means anti-bonding states are filled and that's why it is energetically unfavorable. But in case of $\ce{Ni}$ and $\ce{Pt}$ antibonding orbital is above Fermi level and hence the adsorption process is more favorable. There are some other terms to explain it from the quantum mechanics viewpoint (like the overlap matrix $V_{sd}^2$ and some other fundamental terms) but the basic idea is the DOS of bonding and anti-bonding orbitals are the main reason for this inertness.
Reference: http://www.nature.com/nature/journal/v376/n6537/pdf/376238a0.pdf

I'm taking a stab at this.
Ok, so the issue with the group 11 metals, as far as I can see, is that while, yes, their s orbitals are not filled, they do not exist in isolation: they have neighbors who all have only 1 electron in them. So, they form very strong allotropes. In other words, gold would have been reactive, but there are other golds right there it can bond with and so share electrons with it. So the group 11 elements form large networks of allotropes, where electrons freely flow and are shared with other atoms of their kind. These bonds are very stable, and ergo, very hard to break. In fact, Au-Au bonds have been found to stabilize some very strange molecules: see here.


text:
Comparison of atomic radii
My teacher told me that Neon has a larger atomic radius than Fluorine.I am of the understanding that it is merely a consequence of the way we define the atomic radius and that we use Van der walls radius for neon which is larger that the covalent radius of fluorine.
He also said that the radius of a fluorine 1- anion is more than a neon atom by using the fact that they are iso-electronic species and that since neon has a higher nuclear charge, it will pull on the 10 electrons more than fluorine.
But what definition of atomic radius am i using for neon now? Aren't we always supposed to use the Van der walls definition for noble gases? In case we do use the Van der walls radius of the neon atom and the ionic radius of the fluorine anion, how am i supposed to compare these two and come to the conclusion that the fluorine anion is larger?

The atomic radius of neon is smaller than fluorine. http://www.periodictable.com/Properties/A/AtomicRadius.v.wt.html

Atomic radii cannot really be uniquely nor accurately defined.  However, there is generally a very big difference between covalent, van der Waals, and ionic radii.
So, by most any definition of the respective terms the covalent radius of fluorine will be smaller than the van der Waals radius of neon.  Similarly, the ionic radius of fluoride ion will be larger than the van der Waals radius of neon. The reason for the latter is the higher nuclear charge on neon.
However, most sources list a slightly smaller van der Waals radius for fluorine compared to neon, but a slightly larger covalent radius for fluorine compared neon. So when comparing radii you have to carefully specify what kind of radii you are talking about.

So we can generalize it to say that the atomic radius of fluoride is greater than neon (as long as we aren't comparing  van der Waals radius.) ?


text:
How the heat a large amount of water that is safe to use?
I would like to know how heat up a large amount of water ( assumption : 100 liter ) that people could use to take their bath and also it would not be dangerous even if they accidentally ingested it ? .
My idea is that to design a metal container and have the chemical in it . I'll react the chemical that is inside the container and then I'll put the container into the water , which will eventually heat up the water through thermal equilibrium .

possible duplicate of Can you heat water with additives?

Why don't you heat it normally, using cooker, boiler, or stove - 100 liter isn't that much.

An appliance exists for this. According to the Lowes website, the smallest hot water heater they sell is 30 gallons, which will be about 115 liters. You can buy heaters up to 80 gallons (~300 litres) or probably larger.


text:
Electron transfer in reaction between ammonia and hydrochloric acid
$$\ce{NH_3 + HCl -> NH_4Cl}$$
The electronic structure of ammonia is such so that the nitrogen atom has an outer shell with a lone pair of electrons.
When ammonia reacts with hydrochloric acid (above), the hydrogen from the $\ce{HCl}$ moves to the ammonia. However, it leaves its electrons behind because there is already a lone pair of 2 electrons on the $\ce{N}$ atom, allowing it to retain a $\mathrm{1s^2}$ shell.
Why doesn't the hydrogen take its electron over and nitrogen doesn't then "donate" an electron to the chlorine?

This is an acid base reaction : strong HCl acid with weak base NH3 and it's a neutralization. Here the Cl- is behaving as a spectator ion.

Chlorine is the more electronegative species. It would thus retain the bonding electrons when the hydrogen is evicted. This hydrogen then forms a coordinate bond with the lone pair on nitrogen.

authors: Tħëø Śėläłėdi


text:
Highest osmotic pressure
I’d like to ask a question and clear my misunderstanding.

Which of the following 0.01 M solutions have the highest osmotic pressure?
  -$\ce{NaCl}$
  -$\ce{C12H22O11}$
  -$\ce{BaCl2}$
  -$\ce{CO(NH2)2}$
  -$\ce{[Cr(NH3)4Cl2]Cl}$

I know that the equation for osmotic pressure is
$$\Pi = cRTi$$
I have eliminated sucrose, sodium chloride, and urea from my answer, because sucrose and urea are non-electrolyte. And sodium chloride has smaller van ’t Hoff factor than barium chloride. I ended up in barium chloride and the complex compound. I’m not really sure with the complex compound since from what I learnt, a complex compound is easier to dissolved in water (my teacher in solubility chapter taught me that $\ce{AgI + NH3 -> [Ag(NH3)2]I}$ have a high solubility because it is a complex compound). However I’m not sure if it has a high van ’t Hoff factor as well.

You are correct in thinking that urea and sucrose would have van't hoff factor (i) = 1, since they are non-electrolytes and don't undergo disassociation(or ion pairing).
What you are left with are the following: $\ce{NaCl}$ , $\ce{BaCl_2}$ and $\ce{[Cr(NH_3)_4Cl_2]Cl}$.
Let us begin with the the two chloride salts: $\ce{NaCl}$ , $\ce{BaCl_2}$. we assume 100% disassociation for the two (strong electrolytes) and we have i = 2 and i = 3 respectively for both.
Now, the complex would indeed disassociate but not completely (the extent would be determined $K_f$ for the complex). Thus 1 < i < 2.
Based on this, I believe you can order them on the basis of increasing osmotic pressure.


text:
Do extreme homeopathic dilutions still retain quantities of the starting materials?
This Indian Institute of Technology Bombay  press release quotes some research papers which claims to prove that homeopathic remedies do retain original substances even when diluted.

The paper reports the fascinating observation that certain high potency (highly diluted)
  homeopathic remedies made from metals still contain measurable amounts of the starting
  material, even at extreme dilutions of 1 part in 10 raised to 400 parts (200C).  It is well known that a series of 1:99 dilutions done sequentially will produce a significant dilution of the starting material in very short order. Specifically, if the starting material is at one molar concentration (6.023x10e23 molecules per liter), then at about the 12th dilution (12C) there should be no or very nearly no molecules left of the starting material. At 200th dilution (200C) the likelihood of there being even one atom of the starting material approaches zero. However, dilution does not work so simply, according to this paper. Using electron microscopy (TEM), electron
  diffraction, and atomic spectroscopy, Chikramane et al. found that, contrary to the arithmetic, there are nanogram quantities of the starting material still present in these ‘high potency’ remedies in the form of nanoparticles.

This was published in a homeopathy journal and not a chemical engineering journal, although the department was chemical engineering.
A later paper, Why Extreme Dilutions Reach Non-zero Asymptotes: A Nanoparticulate Hypothesis Based on Froth Flotation suggests that nano-particles are distributed unevenly, floating to the top of the container, which may interfere with dilution predictions,  public in a journal from the American Chemical Society.

However, physicochemical studies of these solutions have unequivocally established the presence of the starting raw materials in nanoparticulate form even in these extreme (super-Avogadro, >1023) dilutions. In this article, we propose and validate a hypothesis to explain how nanoparticles are retained even at such enormous dilution levels.

Another researcher in a Letter to the Editor that the original element was found may be be because those original elements were stuck to the inner walls of lab glass:

Further-more, soluble compounds can be captured by nanosized structure, often nanobubbles, lining the boron-silicate glasses with which dilutions, subjected to mechanical stresses, are made. Therefore, an imprecise fraction of compound may be dispersed into the solvent from the glass wall whenever a mechanical stress is applied to the dilution, possibly giving gelsemine concentration far from the theoretical evaluation into pure water.

(The "mechanical stress" presumably refers to the succussion during homeopathic preparation.)
In practice, are some of the original materials still found in 200C dilutions?

I see this was previously asked on Skeptics.SE. If a solution is not mixed perfectly in each step of successive dilution and sampling, then there's no way to be sure what the last solution's concentration is. Poorly mixed solutions could presumably carry solute far beyond the Avogadro limit of dilution. Whether this is actually the general case in homeopathy, I don't know. Either way, it is far from enough to prove its soundness.

In short, the paper in Langmuir shows that, if one tries to treat a suspension as a real solution, then dilutions won't work as arithmetic predicts. Which is quite obvious. The paper's contribution is in the fact that they show how these are not solutions, but really non-homogeneous suspensions. This explains how homeopaths may mistakenly make solutions that still contain some active principle, through poor technique when carrying out the dilution process. It in no way supports the extravagant claim that extreme dilution would retain any of that material, if it were done properly.

Think about it … a single belladonna dropped into a river out of which a town like Leipzig takes its drinking water … would create a higher atropine concentration than is sometimes ‘present’ in highly diluted homeopathetica (typo on purpose).


text:
Are all complexes with a polydentate ligand examples of chelation?
I'm getting a little confused about the definition of chelation and its implications (I should probably point out I'm only an A2 Chemistry student). The IUPAC defines chelation as the following:

The formation or presence of two or more separate coordinate bonds between a polydentate (multiple bonded) ligand and a single central atom

Does this mean that any ligand that is not monodentate (e.g. even ethanedioate) is an example of chelation? I thought chelation was only for complexes with a large coordination number (e.g. $\ce{EDTA^4-}$), it would seem odd to call a single metal ion with 1 ethanedioate ligand chelated?

You  dare doubt in IUPAC definitions?  ;)  It's just like it says.

I think you should not focus on the ligand to define chelation. You should instead focus on the complex that is formed. In fact, the very same multidentate ligand may participate in binding in more than one way:

by forming more than one bond with the same central atom, therefore forming a chelated complex;
by forming only one bond with that central ion, even though it could form two; this may be the case when a vast excess of this ligand is present with respect to the metal ion;
by bridging between two adjacent central ions.

Only the first of these possibility is referred to as chelation. In other words, forming a chelated complex doesn't depend solely on the fact that the ligand is multidentate. Being multidentate is a necessary, but not sufficient condition for a ligand to form chelated complexes.

Since this is the accepted answer, I'll comment here. There are some single molecules such dimethylglyoxime which ligate fairly specifically with one cation ($\ce{Ni^{2+}}$ in this case) because the size of the "hole" in the molecule into which the metallic ion must fit has a very specific size.

This is what my textbook says:

All polydentate ligands are the example of chelating ligands.
However, it must be noted that $\ce{NH2NH2}$ and $\ce{N(CH2CH2)N}$ cannot act as chelating ligands due to the formation of a three membered ring and locked structure respectively.

(Also, if you're looking for a little extra information - it isn't essential that a polydentate ligand will always give a chelating compound. If it is a flexidentate ligand (i.e. it can show variable denticities) and can hence be monodentate as well a polydentate, then it will obviously not display chelation in its monodentate state.)

Could you also mention which textbook you're talking about?

This answer seems contradictory. You state that All polydentate ligands are the example of chelating ligands and then go on to say that $\ce{H2NCH2CH2NH2}$, a bidentate ligand, cannot act as a chelating ligand. Please can you clarify your answer.

@bon i think the bidentate ligands are exceptions..

So the statement All polydentate ligands are the example of chelating ligands is wrong.

@M.A.R.: It's from the book Concise Inorganic Chemistry by J.D. Lee (Adapted by Sudarsan Guha). In my version, this is mentioned in page 177.

Chelation is simply foratiom of a complex by a ligand which has two or more then two sites within a molecule,the word is derived from crabs claw as it grips the central metal atom and all polydentate ligands can act as chelating agents...


text:
Cleaning an overhead water tank using Alum or other chemicals?
Alum is generally used in water tanks to settle down solid impurities. Since it contains aluminium I suspect it to be toxic to humans. So are there any safe chemicals to clean large overhead tanks without removing the water or  doing much effort. I am asking about solid impurities and not disinfection .

Aluminum is remarkable safe for human consumption, also considering that it is present in high concentration essentially everywhere, and it is not known to have any significant specific biological roles. Wikipedia happens to have a nice well researcher section on aluminum's possible effects on human and animal health: Aluminium: health concerns. Alum is also routinely added to water in low concentrations for elimination of solid particulate, as you mentioned.
My main concern for your application is the fact that typical methods of water purification from solid particulate, just like addition of alum, will tend to form flocs in your water tank that will then need to be removed! Depending whether this scum will float on top of the water, or sink to the bottom of the tank, the removal may be more or less easy, also depending on what kind of access you have to this tank. I doubt if any of those solutions would count as not "doing much effort"...

Thanks , but Al is toxic even the article says so, and Alum has been recently implicated in A new study, dated Feb 5 2015, links the brain translocation of alum particles to a Trojan horse mechanism, that obeys to CCL2, signaling the major inflammatory monocyte chemoattractant.[94]

Hmm, the fact that one study hypothesizes that injectable alum may be biopersistent is a far cry from showing that aluminum is toxic. The article explicitly says that "Aluminium salts are remarkably nontoxic, aluminium sulfate having an LD50 of 6207 mg/kg (oral, mouse), which corresponds to 500 grams for an 80 kg (180 lb) person." This is from Ullman's Encyclopedia of Chemical Technology.

Marco B so if a peice of a kilo solid piece of alumif dropped in over head tank would be sufficient to clean. Tge water and non toxic to us and children ?


text:
What is the difference between these equations for the Contact process?
Some webpages describe the contact process as this (I left out $\ce{SO2}$, $\ce{SO3}$ production):
$\ce{SO3 + H2SO4 -> H2S2O7}$
After this, it is controllable to add water to the oleum therefore making liquid sulfuric acid. However, some other sites describe this:
$\ce{SO3 + H2O  -> H2SO4}$
I am confused about what really happens in the absorption tower. I've read about the wet acid, but it states that adding water causes gaseous $\ce{H2SO4}$ to form, which is condensed.
EDIT: In the absorption tower, trioxide sulfur enters as gas, and sulfuric acid comes down as liquid. As a result the liquid comes out as oleum (first eq.). The other contact process states different: the liquid entering is 98% sulfuric acid, and 2% water, this water reacts with the sulfur trioxide (this little water avoids the heat problem), giving strengthened sulfuric acid (let's say 99.5% sulfuric acid, and 0.5% water). Still confused.

Does this website/video help to describe the different stages of the contact process?
http://www.kentchemistry.com/links/Kinetics/Contact.htm

In principle, your sources are correct and not contradictory.
In order to avoid the formation of problematic sulfuric acid mist, sufur trioxide is not directly absorbed into water. Instead, it is absorbed in sulfuric acid of 98 %–99 % concentration, which approximately corresponds to an azeotrope where the partial pressures of sufur trioxide, sulfuric acid, and water are relatively low.
Sulfuric acid reacts with the absorbed sulfur trioxide to form disulfuric acid:
$$\ce{H2SO4 + SO3 <=> H2S2O7}$$
The disulfuric acid can be converted to sulfuric acid by reaction with added water:
$$\ce{H2S2O7 + H2O <=> 2 H2SO4}$$
Thus, the overall net reaction is as follows:
$$\begin{align}
\ce{H2SO4 + SO3 \;&<=> H2S2O7}\\
\ce{H2S2O7 + H2O \;&<=> 2 H2SO4}\\
\hline\ce{SO3 + H2O \;&<=> H2SO4}
\end{align}$$

Thank you, but sources state either one of the equations as the contact process, which is what happens in the absorption tower. And adding the water occurs later, outside the tower.

The answer to your question is thoroughly explained in Chemistry of the Elements (Second Edition), A. Earnshaw and Norman Greenwood, 1997, p. 708:

(...) The  $\ce{SO3}$ gas cannot be absorbed directly in water because it would first come into contact with the water-vapour above the absorbant and so produce a stable mist of fine droplets of $\ce{H2SO4}$ which would then pass right through the absorber and out into the atmosphere. Instead, absorption is effected by 98% $\ce{H2SO4}$ in ceramic packed towers and sufficient water is added to the circulating acid to maintain the required concentration. (...)

Two pages following, the process is represented as:
$$\ce{SO3 + H2O\ ($\text{in 98% }$H2SO4) -> H2SO4}$$
In the Industrial Manufacture of Sulfuric Acid box (p. 708-710), the $\ce{H2S2O7}$ species is not mentioned. However, the discussion about concentrated Sulfuric Acid states:

It is clear that "pure" anhydrous sulfuric acid, far from being a single substance in the bulk liquid phase, comprises a dynamic equilibrium involving at least seven well-defined species. The concentration of the self-dissociation products in $\ce{H2SO4}$ at 25º (expressed in millimoles of solute per kg solvent) are:
$$
\begin{array}{lcr}
\ce{HSO4-} & \ce{H3SO4+} & \ce{H3O+} & \ce{HS2O7-} & \ce{H2S2O7} & \ce{H2O} & \text{Total}\\
\hline
15.0 & 11.3 & 8.0 & 4.4 & 3.6 & 0.1 & 42.4
\end{array}
$$

Considering the above, you can assume that the absorption of $\ce{SO3}$ in 98% $\ce{H2SO4}$ will lead to the formation of $\ce{H2S2O7}$ but also several other species. If you are really interested in industrial production details and properties of sulfuric acid, this reference is really valuable.

Thanks. This is what I read how they add water to the sulfur trioxide in the absorption tower. Which is different to the other  contact process equation, hence my confusion.


text:
Which relatively simple molecules violate the octet rule?
I'm learning how to draw Lewis diagrams. Everything I've read emphasizes the octet rule. However, to the best of my knowledge, that rule only applies to elements in the first three periods.
On our exam, I expect our professor will ask us to draw some Lewis diagrams of molecules whose atoms do not comport with the octet rule. However, I haven't seen any diagrams of these; it'd be useful to get a sense of what they look like so that my gut doesn't disapprove of what will otherwise be strange looking diagrams.
What are some relatively common molecules that comprise atoms that comprise a great number of valence electrons?

I don't think that it's really that any molecules have over 8 valence electrons, but some electrons do get shoved into the d orbitals. One example is $\ce{XeF4}$, a square planar with single bonds on all fluorines and two electron pairs on xenon.

Haha...Whenever anyone talks about violating the octet, one of the first molecules they shove into our face is $\ce{BF3}$. Or if they fail in that mission, we'll get hit with $\ce{BCl3}$.

@M.A.Ramezani I'm afraid that I don't get your point here.

@Macro The title is asking for  a common and easily understandable molecule that violates octet. I gave him two. :)

http://chemistry.stackexchange.com/questions/444/can-an-atom-have-more-than-8-valence-electrons-if-not-why-is-8-the-limit

There are 3 types of octet rule "violations" or exceptions

molecules with an odd number of electrons, such as nitric oxide

(image source)

molecules with less than 8 electrons around an atom, $\ce{BeCl2}$ and $\ce{BH3}$ serve as examples

(image source)

molecules with more than 8 electrons around an atom, such as $\ce{PCl5}$ or $\ce{SF6}$

Take a look at the image source links for other examples as well as some practice problems.
The bonding around electron deficient atoms such as the boron in $\ce{BH3}$ is best explained in terms of 3-center 2-electron bonds (reference; SE Chem example_1, SE Chem example_2)).
Sometimes, arguments based on d-orbital involvement are advanced to explain the bonding in atoms surrounded by more than 8 electrons.  However d-orbital involvement in non-transition metals is now generally considered to be unlikely.  Instead hypercoordinated (or hypervalent) bonding in which 3-center 4-electron bonds are formed has become a more generally accepted explanation.  A number of examples have previously been discussed here on SE Chem (for example, see here).

Nitrous oxide, N2O also violates the octet rule by most counts.

Nice answer. All I have to add is that molecules like $\ce{PCl5}$ can be drawn in accordance with the octet rule.

@Jan  Do you mean using a resonance structure where one of the $\ce{P-Cl}$ bonds is shown as ionic?

In a rather different vein are aromatic carbocations such as the cyclopropenyl cation

Source:  http://www.chem.ucla.edu/~harding/IGOC/C/cyclopropenyl_cation.html
As can be seen, each of the contributing structures has a positively charged carbon atom that violates the octet rule.
What really happens is, the molecular orbitals are so constructed that the $p$ orbitals lying outside the plane of the ring are shared among all three carbon atoms in the ring.  Then one orbital is strongly stabilized and the other two orbitals are destabilized.  You think you need two pairs of p-orbital electrons to "meet the octet rule" with ordinary valence bond structures, but only one pair is actually stable so the octet rule has to give way.


text:
Element with No Proton?

On seeing this graph closely I noticed the there a element with no proton. But with no proton there will be no electron too and don't think that an lone neutron would form an element. So is that given a half-life on neutron OR there exist a element (isotope) with no electron?

No, that is not a chemical element, it is an isolated neutron with a lifetime of few minutes. It undergoes beta decay into a proton, an electron and an antineutrino.

It's but a matter of convention. You may consider there is an element with atomic number 0 (BTW, recently a tetraneutron has been supposedly discovered, which would be another isotope of it). Instability is not a big deal; a good many respectable elements are unstable. But it would be an element without chemistry.

A free neutron is not considered to be an element and there is no element that does not have at least one proton.
Free neutrons could be included on the graph as they are unstable and decay with a half life of around 10 minutes ($611.0\pm1.0~\mathrm{s}$) although most versions I have seen do not usually include them as they aren't really nuclei which is what the graph is about.


text:
What are ω and ν in this expression?
In the picture below a proportionality equation is shown:
$$\omega \propto \nu \exp\left(\frac{-\Delta G_m}{RT}\right)\propto \dots$$
While $\Delta G_m$ is the Gibbs free energy, $R$ is the gas constant, and $T$ is the temperature, what are the parameters $\omega$ and $\nu$ exactly?

You can see the context on the picture: Yttria-stabilized-zircona.

Looks like $\omega$ is ionic conductivity and $\nu$ is a frequency factor, but I'm reluctant to provide this as an answer without more information - can you provide more information from the text/publication photo, above?

$\Delta G_m$ is the free energy barrier for the motion of atom from one site to another. The equation overall looks like a transition state approach to this activated process, and thus $\nu$ is the frequency at which this motion is attempted, i.e. a $\nu$ is a characteristic lattice frequency. In such models, it is often assumed to equal the Debye frequency.
Finally, $\omega$ is the rate of transition from one site to another, or something proportional to it (because the proportionality symbol is used rather than an equality). $\omega$ can be thus be the ionic conductivity, or a diffusion constant.
You can find the same expression and graph in this chapter of undergrad lecture notes, for confirmation.