text:
Has anyone even taken a picture of a molecule to confirm the geometry predicted by theory?
Have the geometries of molecules been proven, or are we operating strictly on mathematical theory? To put it over-simplistically: has anyone ever taken a picture of a molecule to compare against their math?

Related: Atomic properties

See also X-ray crystallography for another indirect means by which molecular structure has been "seen".

While X-ray crystallography does yield "spots" which can and are mapped to atomic coordinates, it does not resolve electron density to the degree that AFM and STM (and their more advanced counterparts) do - bonds are not exhibited, and until recently, hydrogen atoms have been completely absent from XRD crystal structures. I assert that the degree of inference one uses in interpreting XRD spots is not as pronounced as it is in the images attained from AFM and STM imaging.

Actually, it is the other way around. For anything but the simplest molecules, our pictures are better than our math.  Trying to calculate shapes turns very complicated very quickly.

Yes. Researchers have been using atomic force microscopy (AFM) and scanning tunneling microscopy (STM) for some time for this purpose. Do note that these images are not photographs in the sense that we usually think of "pictures" and are indirect measurements of constituents of the molecule. However, they do yield "pictures" that show the geometry of the molecule(s) in question.
See, for instance, this piece from UC Berkeley News published in 2013 which describes the techniques in a manner that is appropriate for a non-chemist. The same article includes images of molecules before and after a chemical reaction occurs.
I also encourage you (and others) to have a look at images in the IBM STM Image Gallery, which are quite spectacular.
Finally, as others have commented, X-ray diffraction merits mention from both a historical and scientific perspective. As one person notes, the helical secondary structure of DNA was determined by Watson and Crick after examining the fiber X-ray diffraction plates obtained by Franklin's group.
The importance of Franklin's contribution and that of Watson and Crick are monumental achievements. That said, the structure that was elucidated was the secondary structure of a large biomolecule and as such might not fit your criteria for how "complete" the "picture" needs to be to satisfy your question.
Here is the famous Photo 51 result obtained by Rosalind Franklin's doctoral student Raymond Gosling that was subsequently used by Watson and Crick.

Recently, researchers have imaged single atoms using a combination of STM and MRI - see Nature Physics. P. Willke et al. Magnetic resonance imaging of single atoms on a surface. 1 July 2019.

As early examples of "taking pictures of molecules" one might also consider X-ray diffraction images. Such images produced by Rosalind Franklin helped Crick and Watson establish the double-helix structure of DNA.

Rosalind Franklin’s work with DNA definitely merits inclusion in this answer. It would be particularly interesting, I think, since it did not confirm the molecule’s theoretical geometry, since the theorists had not, up to that point, been able to determine what the geometry was. Instead, the images gave hints as to the molecule’s geometry which informed and completed the theory.

Oh, I don’t you should downplay X-ray diffraction that much. It is done for small molecules, too, and high resolutions (lower 1 Å) actually allow seeing atoms, seeing the hole in a benzene ring etc.

@Jan - I mean no disrespect for the technique or the applications, especially for the applications you identify - I have fond memories of the year I spent in a room with a Siemens XRD machine in the early 1990's, gathering data from inorganic carbonates. I am attempting to illustrate (and maybe am doing so poorly) that the quality of XRD images taken in the early 1950's do not yield the highly-resolved images one thinks of as "pictures of molecules" today.

In addition to the answer provided by Todd, there is also the very established technique of single crystal X-ray diffraction.
The basic principle is sending X-rays through a single crystal of a compound. These X-rays interact with the electrons of the molecules or ions in question in a certain way that gives a diffraction pattern (basically, a series of dots in a certain regular order). Using these and a lot of sometimes rather complicated maths you can determine the position of atoms in the compound. This allows you to measure bond lengths and angles.
The caveat of this method is that it requires a crystal, and many compounds are known to display a different structure in solid state than in liquid or gaseous state (e.g. $\ce{SbF5}$ is a monomeric trigonal bipyramid in the gas phase but condenses to $\ce{SbF6}$-octahedra with shared fluorine atoms in solid state). This can be overcome a little bit by resorting to small angle X-ray scattering or SAXS.
SAXS is typically performed on macromolecules in solution. And because of the diffraction maths that I hinted to earlier, the small angle scattering of SAXS primarily gives you the overall shape of a compound, i.e. a ‘potato structure’. Only together with single crystal diffraction can this be translated into an arrangement of atoms — but since SAXS is performed on solutions, the combination of both allows the determination of the structure in solution, which is exceptionally important for proteins.
Anecdotic side note: X-ray diffraction was the technique that lead the way to the concept of ions. $\ce{NaCl}$ was one of the first compounds to be analysed and it was proven that there are no $\ce{Na-Cl}$ molecules but rather a regular three-dimensional cubic crystal structure. Some scientists found this hard to accept and dismissed X-ray diffraction for ‘wanting to disprove $\ce{NaCl}$ molecules.’

In 2005, a field ion microscope captured an image of a very sharp tungsten needle. The small round features are individual atoms.

At even smaller scale than molecules, a quantum microscope has visualized the electron orbital of a hydrogen atom:

last link is dead, though thanks for the great answer


text:
Why is chloroform kept in dark coloured bottles?
Why is chloroform ($\ce{CHCl3}$) kept in dark coloured bottles?

I'm voting to close this question as off-topic because the answer is quite easily found by typing "why is chloroform kept in dark bottles" into the Google search box.

@Todd lack of research on itself isn't a close reason. You can down vote instead.

@DEAD - My bad. VTC retracted. Thanks for the procedural reminder!

good point @DEAD

In the presence of UV light, chloroform can be oxidized by atmospheric oxygen to phosgene ($\ce{COCl2}$), a toxic gas. The dark colored glass blocks UV light so that phosgene is not produced.

source: Stack Exchange

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

sort: votes


text:
Mechanism for the decomposition of HI
I've been trying to find the mechanism for the gas phase decomposition of hydrogen iodide:
$$\ce{ 2 HI (g) <=> H2 (g) + I2 (g)} $$
Anyone know of sources where I can get information on this? I found a 1927 paper in PNAS but wondered if there is anything more recent. This is the 1927 paper:
http://www.ncbi.nlm.nih.gov/pmc/articles/PMC1085202/
Proc Natl Acad Sci U S A. 1927 Oct; 13(10): 720–726.
The Photochemical Decomposition of Hydrogen-Iodide; the Mode of Optical Dissociation

Could you give us a link to that original 1927 paper?

This is a 2006 paper-http://www.cder.dz/A2H2/Medias/Download/Proc%20PDF/posters/%5BGII%5D%20Production%20from%20electrolysis%20&%20thermochemical%20cycles/336.pdf

Related:http://chemistry.stackexchange.com/questions/31225/mechanism-of-photolysis-of-hx

Are we talking about non-catalytic decomposition?

Yes, non-catalytic.

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


text:
Spectrochemical Series - Latest version?
--Studying the Crystal Field Theory [under Coordination Chem.] at school now--
A book my teacher lent me [printed in 2002] has the spectrochemical series in it, and it depicts pyridine [py] as having the same 'strength' as ammonia. But a bunch of other sources [including Wikipedia] display the spectrochemical series albeit it depicts ammonia as having greater 'strength' than pyridine.
Since I have serious trust-issues with Wikipedia, and since that book I mentioned is fairly old, could someone tell me the latest version of the spectrochemical series? Also could you cite a reliable source?
Thanks!

Could you cite both sources you find contradicting?

@Lucademicus Unfortunately, that's going to be a little difficult. That book my teacher lent me was issued by the college he studied at for B.Sc chem students, you won't find it in the open market.....the title doesn't really stand out ['Inorganic Chemistry'] so I don't think that helps....


text:
Why don't covalent compounds dissociate in water?
I understand that ionic compounds dissociate when dissolved in water. Why don't covalent compounds do the same?

Some of them do. For example, $\ce{HCl}$ that is absolutely covalent in gas phase almost completely dissociates in diluted solutions.

Many 'covalent' molecules do dissociate in water, HCl (as pointed out in the comments), phenol, acetic acid, for example, whereas some 'ionic' compounds do not to any appreciable extent, e.g. silver chloride, lead sulfate. The general reason is that the 'system', that is molecule plus solvent, will move into its lowest energy state. This means that if breaking the weakest bond and solvating the ions produced, is of lowest energy then this will occur. In the case of HCL, the polar water molecules can solvate making $\ce{H3O^+_{aq} and Cl^-_{aq}}$ where both ions are surrounded by a few water molecules. Water has a very high dielectric constant (relative permittivity) which means that electric fields are 'damped', i.e they are attenuated to short range so that oppositely charged ions can easily separate and hence exist independently in solution. With carboxylic acids the reasoning is similar, the weakest (O-H) bond dissociates, and the ions are solvated. However, dissociation may not be 100% indicating that the energy gain on dissociating is not that great. In insoluble salts, the lattice binding energy and also bond energy is too great compared to the energy gain expected upon solvation and they remain as molecules in a solid. Organic molecules with strong bonds e.g. cyclohexane, benzene, with only C-C and C-H, hardy dissociate in water. The bonds are too strong compared to energy gain on solvation after dissociation. (If dissociation produced radicals, there will be no effective solvation in a polar solvent so would remain at high energy and hence unlikely to occur)

It all depends on the type of covalent compound used. If the compound used is polar covalent in nature (like HCl, as pointed out) the partially negatively charged oxygen attracts the partially positively charged hydrogen from HCl.
Hence, if a compound is polar covalent in nature, i.e., it shows charge separation, it will dissolve and may dissociate in water, otherwise not.

Then why does Carbon-di-oxide which is non-polar, react with H2O, giving H+ and HCO3- ions

Carbon dioxide is not non-polar. It is non-dipolar. It has a strong quadrupole, and this quadrupole is what reacts (slightly, under ambient conditions) with the water.

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


text:
What does this number indicate in the name of this organic molecule?
When naming organic molecules, what does the 3 in front of the hexene indicate?
(E)-3-methyl-3-hexene

The name "hexene" is ambiguous on its own. (Why?)

So what does the 3 mean?

Location of the double bond.

We should actually write it as hex-3-ene, which allows beginners to understand nomenclature. The 3 shows the location of the double bond.

Also, the current IUPAC recommendations suggest hex-3-ene, too. 3-hexene was an old recommendation.

The number ‘3’ is a locant. It designates an atom in the parent structure, for example to indicate the position of a substituent. Note that the preferred notation has changed in the past; hence, you can find different names in the literature:

trans-3-methyl-3-hexene (1979)
trans-3-methylhex-3-ene or (E)-3-methylhex-3-ene (1993)
(3E)-3-methylhex-3-ene (2013)

In the current version, locants are placed immediately before that part of the name to which they relate. This notation can be useful to avoid ambiguity.
In this case,
‘3-methyl’ indicates the position of the methyl substituent,
‘3-ene’ indicates the position of the double bond, and
‘3E’ indicates the position of the stereogenic unit, which refers to the double bond that gives rise to cis-trans isomerism.

Note that, only the lower locant for the double bond is cited (i.e. ‘3’, not ‘3’ and ‘4’).
Without locants, the name ‘(E)-methylhexene’ would be ambiguous. For example, it could as well refer to (3E)-2-methylhex-3-ene

or to (2E)-3-methylhex-2-ene.


text:
Conductivity of a semiconductor at high temperatures
I understand that increasing the temperature of a semiconductor excites electrons in the valence band into the conduction band. Hence, the resistivity of a semiconductor decreases with temperature.
Is there a high enough temperature where the conduction band would be saturated from electron with the valence band and the resistance of the semiconductor would start to increase like a regular metal?
Assuming the seminconductor couldn't melt, as temperature got aribtrarily high would the seminconductor's resistance get arbitrarily low?

There is not only one conduction band, but infinite numbers - so you generally do not saturate it/them.

Still, I guess at some point the resistance would start to increase, unless the semiconductor would melt before that.

A number of different things are going on in this scenario.
First, the carrier concentration does increase steeply, although for Si at its melt temperature it is still only about 10$^{20}$/$cm^{3}$, much lower than a metal.
In addition, the band gap is decreasing with temperature (not from carriers, but from changes to the lattice spacing). From a graph in Sze's Physics of Semiconductor Physics, Si goes from 1.12eV at room temperature to a little under 0.9eV at 900K. Extrapolation (probably a bad idea) means that somewhere above 2700K there will be little to no band gap. However, this assumes the physics extrapolates linearly, which it doesn't. (Well, liquid silicon is metallic. However, there is still a large increase in conductivity at the melt point).
Finally, the phonon population increases with temperature, which will increase scattering, decreasing the conductivity (just like a metal). This can be seen from a graph of carrier mobility (also in Sze) with temperature. The mobility drops roughly as T$^{-2.2}$ or so.
Getting the balance of all of these just right under extreme extrapolation, well, that is hard. It is actually easier to point to papers on the electrical characteristics of undercooled liquid silicon (since that has been measured), than to superheated solid silicon (which has not been observed). - for ev

Surely once the band gap has decreased to zero the semiconductor would start to behave more like a normal metal (vibration of metal cations increases resistance more than the increasing overlap of the conduction and valence bands).

@Honwang - that would be my take as well. Just not sure how to extrapolate to where that occurs.


text:
What is the difference between a desiccant and a deliquescent?
I understand that both a desiccating solid and a deliquescent solid are "hygroscopic" in the sense that they bind to water.
But what is the difference? And what are the applications of a desiccant vs a deliquescent?

A deliquescent solid will eventually dissolve in water it absorbed, like, for example, calcium chloride does. Molecular sieves are very good dessicants but they do not dissolve in water, so they are not deliquescent. As you can probably guess, operating with dessicants that are deliquescent may be more troublesome than with non-deliquscent ones, but it doesn't mean that they are inferior.
The choice of a proper drying agent depends on the material you want to dry. For example, non-deliquescent dessicants, like aluminium oxide, silicagel or molecular sieves are best to fill columns for solvent drying, because they won't clog when their drying capacity is exceeded. Deliquescent drying agents are perfect for the use in dessicators, because it is easy to remove them when they are exhausted.


text:
When SiO2 changes state, what forces/bonds are being overcome/broken?
When silicon dioxide changes state, are the covalent bonds being broken (ie. the intramolecular forces) or is it the intermolecular forces?

Intermolecular forces don't break

The very words "intermolecular" and "intramolecular" don't quite apply to $\ce{SiO2}$. They imply it is made of molecules, which is not so.

$\ce{SiO_2}$ is a network covalent compound that has an extremely high melting and boiling point, because many silicon-oxygen bonds have to be broken in order for it to achieve the necessary freedom. To clarify, SiO2, which has a tetrahedra network lattice formation, shows that each silicon is actually bonded to 4 oxygens; each oxygen is bonded to 2 silicon. These excess bonds aside from the ones of SiO2 are broken.


text:
Why aren't neutralisation reactions reversible?
Let's take
$$\ce{3HCl + Fe(OH)3 -> FeCl3 + 3H2O}$$
This would be a normal one-way neutralisation reaction.
How about hydrolysis of ferric chloride (technically iron(III) chloride)?
$$\ce{FeCl3 + 3H2O <=> 3HCl + Fe(OH)3}$$
As this is reverseable, we should be able to right it with the reactants and products flipped.
$$\ce{3HCl + Fe(OH)3 <=> FeCl3 + 3H2O}$$
Now that is nothing but our original reaction …
Why aren't neutralisation reactions reversible?
Also, does anyone have any good links to chemical equilibrium as in comments?

Who says neutralization reaction is one-way? No, there is an equilibrium (admittedly, in many cases shifted pretty far to the right, but still an equilibrium).

@IvanNeretin, so in a way, can we say all reactions are reversible.... just their equilibrium is very far right? The ones we mark as reversable are those with a much closer equilibrium?

Neutralization reactions are reversible. In theory, at least, even if not so much in practice, all reactions are reversible.
If you look at things on a molecular scale, there's a principle called microscopic reversibility. In effect, there's no physical reaction mechanism you can create that's a one-way door. Maxwell's demon doesn't exist, and you can't get around that no matter how many steps you chain in whatever order, or which particular reagents you use.
What you can do, though, is set things up so it's exceedingly unlikely that the reaction proceeds in the other direction. That's what's happening with reactions we normally deem "irreversible". There's something about them which makes it difficult, in practice, for products to go back to reactants.
One common feature is the size of the equilibrium constant. If a particular reaction has an equilibrium constant that's really big, the equilibrium for that reaction is going to be vastly tipped to the product side. So much so, that the reactant side of the reaction will not be present in appreciable amounts. (For example, this is why combustion is viewed as irreversible. The reaction is so favorable toward the products that only parts per million or parts per billion of the reactants will be present at equilibrium.)
Another method for promoting "irreversibility" is for the products to partition to another phase. For example, if the reaction forms a gas or a precipitate, that product is removed from the reaction volume, and the concentration of the products drops, pulling the reaction toward the products. Note that even in this multi-stage setup, each step is reversible. That evolved carbon dioxide molecule could re-dissolve in solution, and then react backwards, but it's terribly unlikely for it to do so. You've basically fixed one of the products at a very low concentration, which draws the equilibrium toward the product side.
Precipitates work much the same way, as the activity of the solid is fixed by surface behavior, and does not increase because you increase the amount of solid you add. (The rate of reaction may increase by adding more solid, but the equilibrium of the reaction does not.)
So in your examples, the reason you get hydrolysis or neutralization is due to how you set up the reaction conditions. Have a large excess of HCl? You drive things toward neutralization. Have a way of getting rid of HCl (for example by driving it off as a gas, or reacting it with a base)? You drive things toward hydrolysis.

In this neutralization reaction, Fe(OH)3 forms a dirty green ppt. which is insoluble in water.This demonstrates one point as to why neutralization reactions are irreversible.
There maybe other points as well, such as specific temperature and pressure conditions,etc.
Please feel free to  my answer, thnx!

I understand your point both don't understand how the hydrolysis will then work.


text:
Selective transfer of least/most stable anion from organometallic reagents
When heterocuprate reagents participate in conjugate addition, the least stable anion is transferred from copper to the electrophile:

On the other hand, palladium-catalysed cross-coupling reactions show a different trend, e.g. in the Stille coupling, the most stable anion is transferred:

How is this dichotomy explained?

It’s not really a dichotomy, it’s just two very different reaction mechanisms.
For the cuprate addition, which also applies to general Grignard-type additions, in principle the reaction is reversible (figure 1).

Figure 1: Reversibility of Michael additions.
However, the reverse direction is only really possible if the anion released is somewhat stable. In your example, thiophenolate may attack but may also be removed again in the course of the reaction. Once an alkyl chain has attacked, there will be no reverse reaction since the eliminiation of $\ce{R-}$ is too unfavoured.
For palladium-catalysed cross-coupling reactions, we need to consider the typical palladium catalysis cycle (figure 2).

Figure 2: typical catalytic cycle of palladium-catalysed ($\ce{M} = \ce{Pd}$) cross couplings. In a Stille coupling as in the question, $\ce{Y} = \ce{SnR3}$. Image taken from Wikipedia where a full list of authors is available.
The important step here is the transmetallation step: One of the ligands of metal $\ce{Y}$ ($\ce{Sn}$) must be transferred to the palladium ($\ce{M}$) centre. This will be the ligand that dissociates most easily — the best stabilised anion. So if the tin centre has three n-butyl ligands and one phenyl one, phenyl will be preferentially transferred since it dissociates more easily.
This is also part of the reason why $\ce{C}_\mathrm{sp^3}\ce{-C}_\mathrm{sp^3}$ cross coupling are so rare.

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


text:
Why are Ka and Kb only used for weak acids and bases, respectively?
The only answer which I found to this is that it's because strong acids and bases virtually completely dissociate, so the dissociation constants would be too large and hence aren't used. This doesn't really seem reasonable to me and I have a feeling there is another answer. Is there really another answer or is this the only answer?

This is absolutely not true. $\mathrm{p}K_\mathrm{a}$ values are frequently quoted amoung organic chemists for very strong bases and sometimes even for very strong acids. For reference, check out the Evans $\mathrm{p}K_\mathrm{a}$ table and the values quoted therein.

The above link has stopped working, I have found this one:  ccc.chem.pitt.edu: evans_pKa_table.pdf, or generally the google.com:  Evans pKa table filetype:pdf

What you are answering yourself is quite correct. To elaborate, the pKa/b of a compound refers to its interaction with water. All strong, and especially magic or superstrong acids and bases react vigourously with water, and the disassociation constant stops being useful. While you can still use the constant in order to... rank two superacids by strength for example, this is hard to use for anything practical.
It also depends on the water. Example; HF is a weaker acid than HCl in water, but stronger than HCl in anhydrous acetic acid. To escape all these conundrums it is necessary to have a way of describing the acid and base potential without including interactions with water.
Instead, for acids and bases outside of this watery scope, you use other methods to calculate - like the Hammett Acidity Function which more directly attacks the parameter in question - the chemical potential of the proton.

I consider this a generally wrong answer because it implies $\mathrm{p}K_\mathrm{a}$ values of strong acids/bases be generally not used (which is wrong). Also, I haven’t seen the Hammett Acidity Function being used outside of superacid researchers and lectures.

Fair enough, but it is clear from the question that Mathematician observed the use only for weak acids and bases. Just because you in your environment, or most people in their environment use pK's does not invalidate his question, and my answer. Truth is that the strength of strong acids and bases are not necessarily reflected by their pKa/b - or else HF would be a benign compound. Which it is not.

Oh, it just means that there is more to the toxicity/aggressivity of $\ce{HF}$ than its ability to act like an acid.

@Jan I rather like that answer (it needs some polish though). You won't use pKa of HCl to calculate pH of its solution, will you? I would be pointless. Also in concentrated solutions pH stops being useful and H0 is needed.

@Mithoron Absolutely correct. I use the $\mathrm{p}K_\mathrm{a}$ of HCl to determine whether something will be fully protonated or not at concentraction $c$. I reverse the right to turn my downvote into an upvote once the required polish has been applied ;)


text:
Is there a commonly agreed threshold at which a compound is considered "insoluble" or "soluble"?
I'm doing a literature review on solubility prediction and this is a bit of information that surprises me in its absence. The only information I've found is through Wikipedia with 0.1 g per 100 ml, but that source throws up a 403 Forbidden and seems to be a university's course notes.

When people are doing predictions, they tend to speak in terms of exact numbers and not to reduce those to a binary “insoluble/soluble” thing.

I don’t think that there is a commonly accepted limit between soluble and insoluble


text:
Acid-Base Conjugate Pairs with multiple Products

Identify the acid-base conjugate pairs in the following reactions:
  $$\ce{H_2SO_4 + HNO_3 <=> HSO_4^- + NO_2^+ + H_2O}$$

The obvious acid base conjugate pair is $\ce{H_2SO4}$ and $\ce{HSO_4^-}$.
If $\ce{NO_2^+}$ and $\ce{H_2O}$ were together, the second acid-base pair would be quite obvious. But the two aren't. What would be the conjugate-acid pair of $\ce{HNO_3}$ in this case?

Hint: if you see an acid on the left, probably there should be an acid on the right.

I'm not sure I understand. Conjugate acid pairs differ chemically by one hydrogen ion - $\ce{HNO_3}$ does not differ by one hydrogen ion when compared to either of $\ce{NO_2^+}$ or $\ce{H_2O}$. This is where I am stuck.

Ok. What is the conjugated acid for $\ce{NaOH}$ (if any) in $\ce{H2SO4 + NaOH = NaHSO4 + H2O}$  ?  The situation is pretty similar. Alternatively: as you said, the problem is that $\ce{NO2+}$ and $\ce{H2O}$ are not together, but does they dissociate right away, or is there some state in between, not present in the equation, but still existing?


text:
Isomerisation of alkanes under Lewis acidic conditions
My textbook says:

n-Alkanes on heating in the presence of anhydrous aluminium chloride and hydrogen chloride gas isomerise to branched chain alkanes.

But no mechanism is given. After a little research, I found this: https://goenkanscienceblog.wordpress.com/2015/01/29/suggested-mechanism-of-isomerisation-of-alkanes/
However, it doesn't explain the role of the HCl gas. Is the mechanism correct and if so, what is the role of HCl?

Interesting question, reminded me of the synthesis of adamantane, so I googled it. The answer along with lots of discussion can be found here - http://pubs.rsc.org/en/content/articlepdf/1974/CS/CS9740300479 basically trace alkene is present in the alkane sample; AlCl3 abstracts Cl- from HCl, H+ protonates alkene to form carbocation which undergoes 1,2-alkyl shifts.

@orthocresol    Wouldn't the carbocation increases the number of carbons in the original alkane?

Don't get what you mean. The website you linked also goes via a carbocation intermediate.

@orthocresol     There methyl shift is within the alkane whereas here trace alkene is acting as carbocation.

e.g. in n-butane there is butene impurity - protonated to form butyl carbocation

@orthocresol    Thanks got it. Only last query : how does reaction further propagates. Am i right if I say that carbocation will take hydride from alkane thereby producing another carbocation and so on?

Yeah, probably.

@orthocresol So if we somehow prepare pure n-Butane, it definitely wouldn't show this reaction?

The mechanism for butane is shown here, instead of a general n-alkane.
Initiation

As the comments have mentioned,  it probably begins by the protonation of butene impurities.
The Lewis acid, $\ce{AlCl3}$, stabilizes the chloride ion from $\ce{HCl}$ when the $\ce{H}$ is used to protonate the butene.
A secondary carbocation is formed, which undergoes methyl shift to form a primary carbocation. I have no idea why a primary carbocation can be formed.
Propagation

An alkane attacks the carbocation, and supplies a "hydride" to the carbocation to complete the carbocation.
Meanwhile, another carbocation is formed which also undergoes methyl shift to regenerate the carbocation at the beginning of this step.
Termination

I am going out of my way here to suggest a isobutyl chloride (1-chloro-2-methylpropane) termination product. I can't find other Lewis bases to complete the carbocation for the termination step.

So is this a first order reaction?

Maybe the initiation step involves a methyl and a hydride shift? That would explain the stability

Possibly the AlCl3 extracts a methyl anion from the main chain? That would explain the formation of a more branched alkane due to carbocation stabilisation by a hydride shift.


text:
Why does cobalt have a higher atomic mass than nickel?
I understand that this deviation from the general trend of increasing atomic number and atomic mass is because cobalt's isotopes tend to have more neutrons than nickel's. Why is that the case? How would one go about predicting similar cases like this?
I've read a bit about the shell model. I understand that 28 is a "magic number" of nucleons, i.e. a filled shell. In that case, wouldn't it be the most stable for both atoms to mainly be found with 28 neutrons; and so, cobalt-55 and nickel-56 would be the most stable isotopes of each element. Why is that not the case? Am I missing or misunderstanding something?

Heavier nuclei need more neutrons than protons to be stable, (proportion rises to 126/82 for Pb) but I'm no nuclear physicist and even they have problems with prediction of stability of heavy nuclei.

https://en.wikipedia.org/wiki/Isotopes_of_nickel https://en.wikipedia.org/wiki/Isotopes_of_cobalt

Protons in each nucleus repel each other electrostatically.
As atomic number increases, there is more repulsion and an increasing advantage of spreading out the protons with neutrons.
Calcium-40 is the last (greatest atomic number) observationally-stable isotope with an equal or greater number of protons than neutrons.
Cobalt-55 and Nickel-56 are both unstable to beta decay, because it is favorable to emit a positron converting a proton into a neutron.
The natural abundance of isotopes are not necessarily in proportion to the stability, but instead stellar and cosmic processes.  As explained in the 1963 article Computer Guides Professor Hoyle to New Hypotheses:

Working with Professor W. A. Fowler of the California Institute of Technology, he had begun with an analysis of the distribution of isotopes of the common metals, and found that nickel presented an unusual case; in the natural metal on Earth, nickel-58 is some three times more abundant than nickel-60, which is contrary to what one would expect from considerations of nuclear stability. He found the solution to this puzzle in the effect of neutrino emission from highly evolved stars, leading to extremely short life times, in the range of 20 minutes to 10 hours, and also in the phenomenon of electron-positron pair production. Nickel-58 could be made in an explosion in a large star which occurred so rapidly that the process of synthesis was cut short.

So, while the details of that article may be out of date, the reason cobalt has a higher natural abundance atomic mass than nickel is a combination of nuclear stability and stellar processes.
See On the e-process - Its components and their neutron excesses for further discussion of natural abundance ratios of Co and Ni isotopes.


text:
How to find pH of an aqueous ammonia solution at high pressures?
Here's the simplified question: I start with $15~\%$ of $\ce{NH_3}$ and I want to find the $\mathrm{pH}$ of the following reaction at $25~\mathrm{^\circ C}$ and $5\ \mathrm{kbar}$:
$$\ce{NH_3 + H_2O -> NH_4^+ + OH^-}$$
Now I believe I understand the basic procedure. I was able to find the pH at STP, but there are a few things that confuse me about finding the pH at high pressures. Here's the procedure I followed for STP:
$$[\ce{NH3}]_\mathrm{eq} = 0.15-x\tag{1}$$
$$[\ce{NH4+}]_\mathrm{eq} = x\tag{2}$$
$$[\ce{OH-}]_\mathrm{eq} = x\tag{3}$$
$$K_\mathrm{b} = \frac{x^2}{0.15-x}\tag{4}$$
where $x = [\ce{OH-}], \mathrm{pOH} = -\log[\ce{OH-}],$ and $\mathrm{pH} = 14 - \mathrm{pOH} = 11.2$
Now for my questions about applying this to $5~\mathrm{kbar}$:

Unfortunately I'm not entirely clear about what they mean by $15~\%$—whether it's a mass fraction or in g/ml or what.
But I did manage to get the right $\mathrm{pH}$ using $0.15$ as my initial $[\ce{NH3}]$ … though maybe that was just a coincidence. Actually if I assume that it's a mol fraction and then convert it to molarity, I do get something close to the right answer but not quite … if anyone has any ideas about this please let me know.
Also if I am missing some conversion from $15~\%$, is there any reason $[\ce{NH_3}]_\mathrm{initial}$ would be different at STP vs $5\ \mathrm{kbar}$? If both start with $15~\%\ \ce{NH3}$?
From what I've read, the value for $K_\mathrm{b}$ shouldn't have any pressure dependence at all. And for this specific reaction, $K_\mathrm{b}$ should be equal to $K_\mathrm{p}$ since there are two molecules on the left and two molecules on the right sides of the equation.
However when I calculate $K$ using $\Delta H$ values at $5~\mathrm{kbar}$, I do get a different value (at STP, $K = 10^{-4.754}$ and at $5~\mathrm{kbar}$ $K = 10^{-2.81}$). So does $K_\mathrm{b}$ really not change with pressure or is this just an approximation at low pressures? Or am I dealing with two different $K$ values? Does the $K$ that I calculated from $\Delta H$ only equal $K_\mathrm{b}$ at STP?
Actually now that I think about it, the value for $K_\mathrm{w}$ definitely changes at high pressures. At STP $K_\mathrm{w} = 10^{-14}$ and at $5~\mathrm{kbar}$ $K_\mathrm{w} = 10^{-12.559}$
So if $K_\mathrm{w}$ changes at high pressures, shouldn't that mean that $K_\mathrm{b}$ does as well?
Is there some sort of step I'm missing at the end? Because I tried solving for $\mathrm{pH}$ using both K values (the value at STP and the value I found at $5~\mathrm{kbar}$) but I still don't get the right value for $\mathrm{pH}$.
If I had to guess, I'd say that $[\ce{OH-}] = a_\ce{OH-}$ at STP which is why $-\log a_\ce{OH-}$ gives me the right $\mathrm{pOH}$ and $\mathrm{pH}$ at STP. But at $5~\mathrm{kbar}$ $[\ce{OH-}] \neq a_\ce{OH-}$ so I'd need the activity coefficient? But if activity coefficients were needed, their source should have been cited which it wasn't … Unless there's a straightforward way to solve for the activity coefficient?

Let me know if any clarifications are needed.
Edit: I've just found that if I take the ratio of the vapor pressure of Ammonia (~10.1 bars) to the total pressure (5000 bars) and multiply that by .15, I do get very close to reproducing the 5 kbar results. Any ideas why that might be?

Hi, I’ve cleaned up the MathJax in your post. You should really use \ce{...} (i.e. mhchem) for chemical formulae. It’s much easier to type and it also produces upright (non-italic) letters. Find out more on [meta].

A possibility may be that you're expected to use $P_{\text{NH}_3}$ rather than $[\ce{NH3}]$.

I thought about that but aren't partial pressures only used for gases? At these pressures ammonia would be a liquid. Though I suppose if the pressure is high enough, it might be compressible... Thanks for the suggestion!

Although I suppose ammonia is a gas at 1 bar pressures so I'd probably have to take that into consideration for that calculation... Hmmmm


text:
Crystal Field Splitting of d-Orbitals in Octahedral and Tetrahedral Ligand Fields
We're learning about the Crystal Field Theory (CFT, under coordination chemistry) at school now, and my teacher mentioned that the 5 d-orbitals split into orbital sets, namely the $\mathrm{t_{2g}}$ (triply degenerate-gerade) and the $\mathrm{e_{g}}$ (evenly degenerate-gerade) for an octahedral ligand field; whereas for the tetrahedral ligand field, the notation is the same except that the subscript 'g' is removed, indicating it is ungerade (That bit was my interpretation … I think I'm wrong though)
Now a lot of the books and sites I've looked up mention the same gerade notation for both octahedral and tetrahedral ligand fields.
So naturally my question would be "Which one's right and which one's wrong?" and a reason if possible.

A lack of a subscripted g doesn’t mean ungerade, it means that gerade/ungerade cannot be established due to the lack of the corresponding symmetry element (centre of symmetry, if I’m not mistaken) in the compound as a whole.

@Jan - yes, the lack of the 'g' in tetrahedral point groups is exactly because there is no inversion center in the compound as a whole.

The labels you are talking about refer to the orbitals’ irreducible representations. Which irreducible representations can occur is dictated by the point group of the entire compound. For octahedral compounds, the point group is the very symmetric $O_\mathrm{h}$, which basically includes every element of symmetry you can have in a cube. Tetrahedral complexes belong to the $T_\mathrm{d}$ point group with a considerably lower symmetry (i.e. a smaller number of symmetry operations).
Octahedral complexes are probably more present in the average non-coordination chemist’s brain. Since $O_\mathrm{h}$ includes a centre of symmetry, all irreducible representations have a subscripted g or u to indicate whether equal phases of an orbital are mapped onto each other upon inversion (g for gerade) or whether phases are mapped onto their opposites upon inversion (u for ungerade). Due to the d-orbitals having two nodal planes, they are automatically g, which is why they as a whole transform as $\mathrm{e_g + t_{2g}}$. Since, as I said, this is the more prevalent case in general, this is what a lot of chemists have in mind.
The tetrahedral point group $T_\mathrm{d}$ does not contain a centre of symmetry. This is easily seen when looking at a tetrahedron, which is not inversion-symmetric. Thus, the irreducible representations do not gain a subscripted g and u. The d-orbitals thus transform as $\mathrm{e + t_2}$. Adding a g to these is wrong, but common (see above).
Thus, you were correct in assuming: ‘just because everyone does it does not mean it is correct.’ The same thing can be said for $\ce{S=O}$ double bonds in sulfate.


text:
Formation of complexes by ferric ions is more frequent than ferrous ions. Why?
Looking at the electronic configurations of ferrous and ferric ions they are d6 and d5 systems respectively. Hence, ferric ions are more stable. But in octahedral complexes like hexathiocyanates, the electron count of ferric ions reaches only 17 while for ferrous it reaches 18. In the case of thiocyanate , ferric complexes with the ligands while ferrous does not react at all.

And how do you measure "frequency"?

@Mithoron Can you be more clear?

That's exactly what I'm asking from you... Why do you say that one type of complexes would be more common/frequent or whatever?

@Mithoron my mistake.. I have edited the last part.

Why should 18e rule apply to typical coordination complexes... It does not. $\ce{[Ti(H2O)6]^3+}$ has 13e and $\ce{[Zn(H2O)6]^2+}$ 22e and everything in between exists... Is $\ce{[CuCl4]^2-}$ unstable because it has 17e only? Perhaps better to have 18e $\ce{[CuCl4]^3-}$?

@orthocresol Arguably, $\ce{[Zn(H2O)_n]^2+}$ is not really a defined complex and most definitely not a defined octahedron, but I’ll agree that some zinc cations will have that structure for an observeable amount of time ;)


text:
Classes of Organometallic Compounds - Where do I put Grignard Reagents?
[Note- I've already seen this question Why do Magnesium and Lithium form *covalent* organometallic compounds? as user Mithoron was kind enough to point out, but the problem that I face is different though. MY problem here, is that if I'm asked whether Grignard reagents are a)Ionic compounds of electropositive metals or are they b) Sigma bonded Complexes what should my answer be?]
My school textbook defines Organometallics (A section under the chapter 'Coordination Chemistry') as "Compounds where the metal atom is directly linked to a carbon atom(s) " , and then it proceeded to give the following classes of Organometallics:
1) Ionic Compounds of Electropositive Metals

Includes compounds like Alkyl Sodium, Phenyl Lithium and Alkyl Magnesium Halides (Grignard Reagents)
2) Sigma Bonded Complexes

Includes the likes of Trimethyl Aluminium
3) Pi Bonded Complexes

Includes Zeise's salt and Ferrocene
4) Sigma and Pi Bonded Complexes

Includes transition metal carbonyls like Ni[(CO)_4]
Now the thing is, some other (relatively unheard of) reference material places Grignard reagents under Sigma bonded complexes. Now I'm not sure which one's right, or if both are right, or if both of them are wrong, since in my brain the R-Mg bond in the grignard reagents are polar covalent, i.e- smack in the middle, between covalent and ionic bonding.
So going by convention (if there even is a convention for something like this), I reiterate my question, which one's right?
[Additionally if there are more classes of Organometallic compounds that I haven't mentioned, I'd really appreciate it if someone would add on to it...]
Thanks!

http://chemistry.stackexchange.com/questions/429/why-do-magnesium-and-lithium-form-covalent-organometallic-compounds

@Mithoron I've already seen that question, but MY problem here, is that if I'm asked whether Grignard reagents are a)Ionic compounds of electropositive metals or are they b) Sigma bonded Complexes what should my answer be?

related http://chemistry.stackexchange.com/a/31646/9961

Sigh Exact bonding depends on organic group, also you can have ether complexes. I would rather but them in a bag with organoaluminum then organosodium, but why do you really care?


text:
How does one optimize the molar concentration of electrolyte in terms of the rate of electrolysis?
There are many textbooks/recipes for various electrolyte solutions in the electrolysis of water. I know there is a minimum voltage to kick off electrolysis, and I'm supposing a minimum concentration of electrolyte like $Na_2SO_4$.
But isn't there an optimum concentration of electrolyte that maximizes the rate of electrolysis? Are these recipes based on that optimum, and if so how can one calculate what that would be for any given salt or liquid electrolyte?


text:
What is the compound I made from KNO3 and table sugar which subsequently exploded when I exposed it to a flame?
I mixed 75% $\ce{KNO3}$ and 25% $\ce{C12H22O11}$ (by volume) in boiling water to make slow burn fuses. Some of the product became a thick brown paste.
After drying, a hard cake of an approximate 1 tablespoon amount exploded when lit. What compound did I make?

Rocket candy.

It's not a new compound - it's just an intimate mix of fuel (sugar) plus oxidiser (nitrate). It's not the most stable of situations to have...


text:
Are the following two compounds isomers or not?
Are the following two compounds isomers or not?
I feel they are enantiomers(non-superimposable mirror images) however my book states that they are not isomers?
I cannot understand why?

Probably a typo. You are correct that they are enantiomers.

This is an example of a compound with no chiral center, but it is chiral.  Compare 1,1-diethyl 3-methyl propa-1,2-diene (did I get the substituents in the right order?).

http://chemistry.stackexchange.com/q/42999/16683 #3

You are right, these two molecules are enantiomers. If you rotate the bottom molecule 180 degrees, you can see that it is the mirror image of the top molecule. Enantiomers are stereoisomers, so the only reason your book would say they are not isomers is that it (unfortunately) contains a typo.

Or possibly they missed the concept that a molecule can have not a (single atom) chiral center and yet be chiral.


text:
What makes formaldehyde 'safe' or 'dangerous'?
I know that formaldehyde naturally occurs as a metabolite and can be found in our own and other animals' bodies, plants, and unprocessed food.
But formaldehyde is also registered as a carcinogenic VOC and is considered bad when it's in plywood and cosmetics.
Is this judgement based on dosage, form (e.g. gas/liquid), or amount released?

Well, dosage, yeah. But it’s also not seen well as a metabolite and one of the causes of a hangover after having drunk too much.

@Jan I think you thought about acetaldehyde. If you drank CH3OH hangover would be your last problem ;)

@Mithoron I thought there was some small-scale side-metabolism that created a small amount of formaldehyde, too, but it was on TV and it was a long time ago ;)

As for why it is dangerous - formaldehyde is widely in biology used to fix cells as it causes crosslinks between proteins to other proteins and to DNA.

The toxicity of formaldehyde is dose-dependent. The famous saying "the dose makes the poison" is one of the rules of thumb about toxicology. Your body can effectively detoxify small amounts of formaldehyde in a given amount of time, however if you are exposed to concentrations that result in you ingesting/inhaling more than what your body can metabolize, then toxicity will begin to occur.
Most consumer products with detectable levels of formaldehyde (I'm thinking of shampoos and body washes particularly) do not directly have formaldehyde added to them. As formaldehyde is a fairly reactive gas that will polymerize & evaporate easily, formulation chemists will use compounds that decompose slowly to formaldehyde plus other mostly inert material. What this does is provide a constant, low level of formaldehyde present to halt microbial growth. (Formaldehyde is a wonderful disinfectant - few things if any will survive treatment with concentrated formaldehyde - and it turns out you need a LOT of formaldehyde to kill a human as compared to a microbe!)
Formaldehyde is also a natural trace compound formed in fermentation/decomposition of other organics, so it's detectable in the biosphere if you want to go looking for it - I'm sure you can find detectable amounts in fruits, wood products, even the air around a garden.

Why would formaldehyde be more poisonous to microbes than to the average mammal, after scaling? Concentrated H2CO kills everything, why would a level low enough to be totally harmless to humans effectively stop microbial growth?

@Karl that's an insightful comment. There's 2 important reasons why microbes are more susceptible to formaldehyde poisoning by shampoo than humans. One is that microbes absorb everything through their cell walls  while mammals have thick skin. The other is that microbes are literally swimming in the toxins. Try drinking a bottle of shampoo and it'll do you no good at all. Similarly, nicotine is used as insecticide. An insect on a plant covered in nicotine spray will easily ingest the mg/kg dose that would be lethal to a human The lethal per kilo doses do obviously vary between organisms though

Formaldehyde will be absorbed by the skin pretty well, as are other, similar substances, eg. alcohol. I take the point that the microbe is immersed in the shampoo or whatever for it's lifetime, while a human has it on his skin for a minute or two per day. Different timescale.


text:
Concentrations appearing in the Ka of buffer solutions
I've just finished studying buffer solutions and I think I understood more or less how they work, but there's something I can't make sense of:
In the $K_a$ of equilibrium of a buffer solution made from Sodium acetate and acetic acid, we just have the concentrations of acid, $\ce{H+}$ and acetate ion. So, why isn't the concentration of $\ce{Na+}$ taken into consideration in the $K_a$? Why just the others?
Moreover as $[\ce{Na+}] = [\ce{CH3COO-}]$ (the acetate ion in solution is almost only the one dissociated from the salt) shouldn't in the Henderson-Hasselbach solution occur a $a^2$ factor? So in the numerator having $[\ce{CH3COO-}]^2$ and no just a single one?
Thank you

Because Na+ is a spectator ion in this situation.
$$ \ce{Na+ + CH3COOH <=> Na+ + CH3OO- + H+}$$
In other words, Na+ is on both sides of the chemical equation, so [Na+] would appear in both the numerator and denominator of the equilibrium constant if Na+ is included.
$[\ce{Na+}][\ce{CH3COO-}][\ce{H+}]/[\ce{Na+}][\ce{CH3COOH}] = [\ce{CH3COO-}][\ce{H+}]/[\ce{CH3COOH}]$

I don't have any objections to make here, but I'd step away if people deem this answer a bit low quality. I mean, short answers are generally just not acceptable.

@Rubisco my female teacher would say "your answers should be like a skirt: long enough to cover the subject, but short enough to be interesting".

I agree with the sentiment, but something gets triggered when people see that they don't need to scroll down to see the whole answer. And I don't have the rep to vote to undelete if anything happens.

I might go a little deeper and explain why sodium is a spectator ion - e.g. what's the critical reaction under discussion and why sodium is not involved. (At the very least, the questioner would be helped by a short discussion of how to identify spectator ions in similar chemical scenarios.)

So the answer below about "not changing much" is wrong? Also I can't understand why is Na a spectator ion while others  aren't in this case as there are no major reactions going on...I see things Na does also done by other ions? Thank you

@NurBedeir No, it's not wrong.  The equilibrium constant, when expressed in terms of concentration rather than activity, is not truly constant, but instead is a function of ionic strength.  To calculate ionic strength, all the ions need to be considered, including Na+.  But this is a small effect compare to what you are suggesting in the question.

Chemistry is based on the good-enough as opposed to theoretical perfection. Although $\ce{Na+}$ doesn't affect the pH directly, it does change the ionic strength of the solution. The pKa of acetic acid would be affected slightly by the ionic strength of the solution as a secondary correction. Textbook problems ignore such factors because it leads to calculation complications that require extra equations that in turn require iterative computer programs to solve.


text:
Water solubility of hydroxbenzoic acids
why is the log P of the esters lower than the alcohols?
Is it because of the formation of dimers?

Guess OH group is more water-friendly than ester, due to formation of hydrogen bonds. Indeed, look at a simpler pair: ethanol is infinitely soluble in water (in other words, miscible), while ethyl acetate is not.

But the logP values of the esters are lower than the alcohols!

Oops... Well, let's wait for someone to undo my explanation and provide a better one.

What is your source for the logP values?

Source is in brackets below.


text:
Stability of iron(III) complexes

Amongst the following the most stable complex is:
A) $\ce{[Fe(H2O)6]^3+}$
B) $\ce{[Fe(NH3)6]^3+}$
C) $\ce{[Fe(C2O4)3]^3-}$
D) $\ce{[FeCl6]^3-}$

I know that the answer should be $\ce{[Fe(C2O4)3]^3-},$ but I could not find a satisfactory explanation.

It is better to hold something with two hands than with one. Therefore bidentate ligands generally have an advantage over monodentate. Also, welcome to Chem.SE.

This is known as the chelate effect. Complexes with bi- or polydentate ligands — chelating ligands to be exact — are observed to be more stable than their nonchelating counterparts.
To understand this, we need to compare apples with apples, so your example is not the best one. (However, based on the trend we can still give a prediction; more on that later.) Instead, I will base my argument on comparing $\ce{[Fe(NH3)6]^3+}$ and $\ce{[Fe(en)3]^3+}$, where $\ce{en}$ is ethylenediamine $\ce{H2N-CH2-CH2-NH2}$.
Consider the following equilibrium:
$$\ce{[Fe(NH3)6]^3+ + 3 en <=> [Fe(en)3]^3+ + 6 NH3}\tag{1}$$
To determine which side is favoured (‘more stable’), we need to take a look at the equilibrium constant $K$. This is defined as
$$K = \frac{[\ce{[Fe(en)3]^3+}][\ce{NH3}]^6}{[\ce{[Fe(NH3)6]^3+}][\ce{en}]^3}\tag{2}$$
But to get a hint on its numeric value without measuring it, we need to apply the Gibbs-Helmholtz equation:
$$\Delta_\mathrm{r}G^⦵ = - RT \ln K\tag{3}$$
Equation (3) tells us that $\Delta_\mathrm{r}G^⦵$ will be negative (exergonic) if $\ln K$ is positive ($R$ and $T$ are positive by definition) — so if $K >1$, i.e. the product side is more favoured than the reactant side. So far no new information. We need to apply another equation, the definition of the Gibbs free energy:
$$\Delta_\mathrm{r}G^⦵ = \Delta_\mathrm{r}H^⦵ - T\Delta_\mathrm{r}S^⦵ \tag{4}$$
For the enthalpic term $\Delta_\mathrm{r}H^⦵$, we need to ask ourselves which bonds are broken and which are reformed. We break six $\ce{Fe\bond{<-}N}$ bonds and form six $\ce{Fe\bond{<-}N}$ bonds over the course of the reaction. Thus we can assume:
$$\Delta_\mathrm{r}H^⦵ \approx 0\tag{5}$$
Therefore:
$$\Delta_\mathrm{r}G^⦵ = - T\Delta_\mathrm{r}S^⦵ \tag{4'}$$
The entropic term $\Delta_\mathrm{r}S^⦵$ is most influenced by the number of particles in solution. The reactant side has four free particles in solution while the product side has seven. Since $7>4$, we can assume the product side’s entropy to be larger. Thus giving:
$$\begin{align}\Delta_\mathrm{r}S^⦵ > 0\tag{6}\\
- T \Delta_\mathrm{r}S^⦵ < 0\tag{7}\\
\Delta_\mathrm{r}G^⦵ < 0\tag{4''}\end{align}$$
Therefore, the reaction as written in equation (1) is exergonic, the products are more stable then the reactants.
Applying this to the problem, we see that most of the ligands are similar enough to not warrant any other stronger effects. Thus, the complex with three bidentate, chelating ligands is most stable.


text:
Acid/alkali + salt reaction?
Why some acid/alkali can react with some salt, forming another acid/alkali? For example:
$\ce{2NaCl +H2SO4->Na2SO4 + 2HCl}$
$\ce{Ca(OH)2 +Na2CO3->CaCO3 +2NaOH}$
but not the reverse.
What determines this? I heard someone said the weaker acid/alkali produced is favourable, but HCl is stronger than $\ce{H2SO4}$. I also heard that only if the products are insoluble or lost from the solution, the reaction happens. Someone suggested that the problem is concentration, as sulfuric acid is 98% while HCl is 38% Is that true?

Yes, that's the point. $\ce{CaCO3}$ is insoluble, and $\ce{HCl}$ is a gas.

This question is difficult to understand.   Also is this question similar?  http://chemistry.stackexchange.com/questions/57814/why-arent-neutralisation-reactions-reversible?rq=1

@ Agriculturist, I don't understand why the sulphuric acid 'displaces' hydrochloric acid from chloride salt, but hydrochloric acid can't 'displace' sulphuric acid from sulphate salt.

Concentrated HCl solutions can only reach about 40%.  So say you add NaCl to 98% sulfuric acid.  You quickly end up supersaturated with respect to dissolved HCl and it goes into the gas phase which thus very much limits the ability for that reaction to go in the reverse direction.

The sulfuric acid reaction gives sodium bi-sulfate, maybe?

Hydrochloric acid solutions can only readily reach about $\pu{40\%}$ concentration. So say you add $\ce{NaCl}$ to $\pu{98\%}$ sulfuric acid. You quickly end up supersaturated with respect to dissolved $\ce{HCl}$ and it goes into the gas phase which thus limits the ability for that reaction to go in the reverse direction.
The situation is similar for the second reaction, except in this case rather than removing carbonate from the system via volatilization, you are forming the insoluble $\ce{CaCO3}$, preventing the reaction from happening in the reverse direction.

It has been observed that:
A STRONGER ACID or BASE can produce a WEAKER (usually more volatile) from a reaction with given salt.
While H2SO4 is stronger (though less strong than HCl) and less volatile it can give HCl from NaCl
and HNO3 from NaNO3 etc.
[kind of we have seen in nitrating mixture NaNO2 + HCl to generate HNO2]
Ca(OH)2 though weaker than NaOH and being less volatile can give the latter from NaCl.
[Think of NaOH giving off NH3 from all ammonium salts!]
The cases governing may be:
1. STRONG acids as well as bases give way to weaker ones (when reacted with suitable salts
2. More volatile may be given off by less volatile ones
3. Solubility

I disagree with your explanation. $\ce{HCl}$ is actually a stronger acid than $\ce{H2SO4}$. $\ce{HCl}$ has a $pK_a$ of -6.3 while $\ce{H2SO4}$ has a $pK_a$ of around -3, indicating that $\ce{HCl}$ is a stronger acid than $\ce{H2SO4}$.  The key point is that hydrogen chloride formed is insoluble in conc. $\ce{H2SO4}$ and escapes as a vapor from the mixture, driving the reaction forward as the reverse reaction cannot occur. In the second reaction, the $\ce{CaCO3}$ is insoluble and does not dissolve into ions again, so the reverse reaction cannot occur and the reaction proceeds forward.

I don't think the strength of the acid is a factor. A weak acid like acetic / ethanoic acid (vinegar) can displace a strong acid (eg HCl) from its salt - if you mix salt and vinegar you end up with sodium acetate and hydrochloric acid. It has to do with the volatility. A less volatile acid displaces a more volatile one from its salt.


text:
Is polymerized oil soluble in oil?
"Burned food is bad for you." This basic tennet appears to be firmly held by specialists.
Hence burnt food should be carefully cleaned from pots before subsequent use, and likewise on grilling equipment.
Yet the standard care for cast-iron cookware involves coating with a thin layer of oil and heating. The desirable temperature is variously indicated to be between 250F and 400F, possibly depending on the type of oil. The temperature for this reaction, polymerization, is just short of carbonization.
What is the stability of polymerized oil? Is it established whether it is soluble during subsequent cooking with new oil?
Previous relevant question(s):

Molecular structure

Once the food turns completely black, it's usually no longer harmful. It's highly crosslinked and carbonised, and is not taken up by the body during digestion. "polymerisation" is not the correct term for that, btw. And the black stuff on your cast iron pan is not just carbonised oil, but also a varying mixture of iron oxides, phosphates, etc. And yes, it sticks very well.


text:
Asymmetric Suzuki coupling - saturated alkyl chain
9-BBN is applied for Suzuki coupling of a saturated alkyl chain with some alkenyl,alkynyl or even secondary alkyl halide like the modification attached:

(Saito, B.; Fu, G. C. Alkyl−Alkyl Suzuki Cross-Couplings of Unactivated Secondary Alkyl Halides at Room Temperature. J. Am. Chem. Soc. 2007, 129, 9602–9603. DOI:10.1021/ja074008l.)
For asymmetric hydroboration another bulky group is used - Ipc:

Is there an asymmetric version of Suzuki coupling employing saturated groups where instead of 9-BBN, Icp or a similar chiral group would be used or would the stereochemical information be lost during transmetallation?
EDIT: I've found this reaction sequence in Carey, Sundberg Advanced Org. Chem:

The product could be used for a Suzuki couping, but the question whether the stereochemical information is lost during coupling remains open.

Is this the sort of thing http://www.ncbi.nlm.nih.gov/pmc/articles/PMC2929007/ your asking about?

That article explains the enantioselectivity in the case of chiral halide, I am asking about chiral organoboranes.

As of 2010, when I wrote my thesis and papers on the alkyl-alkyl couplings, I do not believe there were any asymmetric variants using a chiral borane.

I found this reference, but I don't have access to read it: http://pubs.acs.org/doi/abs/10.1021/jo201263r Amusingly, I worked with Liz also when we were both in the Jacobsen lab.

Oh, so it is possible.

Maybe that's not surprising. The transmetallation and reductive elimination steps are most likely concerted and not reversible.


text:
Deriving the volumetric thermal expansion coefficient from other properties?
I have a lot of gas properties in a model but I discovered that he volumetric thermal expansion coefficient is not among them. I've found little about how the constant relates to other properties - I only know of how it's used, and there's abundant tables of them.
Can the constant be derived from any combination of the following (some parameters vary with temperature, etc)?

Temperature
Pressure
Density
Average molecular weight
Specific heat
Entropy
Enthalpy
Internal energy
Sutherlands constant (viscosity)
Lambda constant (viscosity)
Static and kinematic viscosity
Speed of sound
Bulk modulus
For each molecular component:

Molar/volumetric percentage

Mass percentage

Molecular weight

Partial pressure

Density

I'd rather derive the thermal expansion coefficient from other existing values if possible.  If not, I guess a lookup table will be in order.

It is derived from the equation of state.  That, is, the relationship between P, V, and T.

Hmm. I actually don't have that.  Both my temperatures and pressures (from which density is derived) are empirical and don't represent the result of instantaneous fluctuations, but rather different environments equalized over long periods of time.  The partial pressures and such come from the gas percentages (also empirical) in the differing environments. Is there no other way to get it?  I don't have any data on instantaneous pressure-volume-temperature relations - anything on the subject would be nothing more than an added lookup table.

If you have a multicomponent mixture, then there are ways of deriving the equation of state, but it isn't simple (unless the mixture behaves as an ideal solution).  Also, there are empirical mixing rules that can be applied.  See Moran et al Fundamentals of Engineering Thermodynamics


text:
Mass transfert in process chemistry engineering, two film theory
I'm having troubles to understand a part of my lesson. I have no idea where to ask my question, but it is consider as chemistry so here we go.

Let me introduce some notations :
$x$ and $y$ are the mole fractions
of the same compound but respectively in liquid phase and in gas
phase.
$k_{c_l}$ and $k_{c_g}$ are the coefficient of exchange of matter in
$\ce{m.s^{-1}}$
$N_{1n}$ is the molar flux density of the compound $1$ in the normal position

Let now me introduce (translating from french to english) you the part of the lessons in which I have troubles :
No phenomenas of production or consommation of the compound $1$ occur in the interface $I$, and we are in stationnary state then the molar flux density is the same in both parts and we have $$N_{1n}=c_lk_{c_l}\left(x_1^{I}-x_1\right)=c_gk_{c_g}\left(y_1-y_1^{I}\right)$$
The fluids at the interface are supposed to be in thermodynamical equilibrum then we have $$y_1^I=K_1^{eq}x_1^I$$

We can for example define a fictive gas phase which will be in equilibrium with the liquid phase (or the contrary). The mole fraction of the compound $1$ is by definition $$y_1^*=K_1^{eq}x_1$$
The total driving force is then $$\begin{align} (y_1-y_1^*)&=(y_1-y_1^I)+(y_1^I-y_1^*)\\ &=(y_1-y_1^I)+ K_1^{eq}\left(x_1^I-x_1\right)\end{align}$$ if we suppose to have a unique average equilibrum constant in the liquid film.
Now I'm having troubles their :
We can then define a global transfer constant in the ralation to the driving force in gas phase : $$\frac{N_{1n}}{K_{c_g}c_g}=\frac{N_{1n}}{k_{c_g}c_g}+\frac{N_{1n}K_1^{eq}}{k_{c_l}c_l}$$
I really don't understand this last relation, if someone can help me, I would be glad. I apologize in advance for the troubles you can have too, to understand what I tried to translate here.
Thank you
PS : I have really no idea for the tags, feel free to edit them

You already derived this relation (eq 1),
\begin{align} (y_1-y_1^*)&=(y_1-y_1^I)+(y_1^I-y_1^*)\\ &=(y_1-y_1^I)+ K_1^{eq}\left(x_1^I-x_1\right)\end{align}
Now, to keep the molar flux density between the gas phase concentration $y_1$ and fictitious conc. on the other side $y_1^*$ we assumed a fictitious overall transfer coefficient $K_{cg}$. So, now we have another relation for molar flux density
$$N_{1n}=K_{cg}c_g\times(y_1-y_1^*)$$
that follows,
$$y_1-y_1^*=\frac{N_{1n}}{K_{cg}c_g}$$
Similarly,
$$y_1-y_1^I=\frac{N_{1n}}{k_{cI}c_I}$$
and
$$x_1^I-x_1=\frac{N_{1n}}{k_{cg}c_g}$$
Plug these above three relations into eq 1 and you will get your last equation.

Thank you a lot. I was on this for more than 4 hours... studies ... bam


text:
Piece of equipment similar to syringe for gas collection
I'm looking for the name of a piece of equipment similar to a syringe. The only difference is that this item is made entirely from glass. The purpose of this equipment is to collect gases.
Here is a photo of two of these pieces of equipment connected to a gas-washing bottle:

This piece of laboratory glassware is a gas syringe. Apparently, gas syringes are not standardized; however, typical models have a capacity of 100 ml. Some models include a GL 14 screwthread tube, a stopcock, or a three-way stopcock.


text:
can protons or H+ appear as stable gas?
Can H+ appear as stable gas or they react to form H2? Can H+ appear as stable gas or they react to form H2? Can H+ appear as stable gas or they react to form H2? Can H+ appear as stable gas or they react to form H2? thanks!

And can you tell me why you wrote 3 times the same sentence?

There's a reason that the minimum character requirement is in place.  It usually takes at least that much text to set up the background and explain what you have tried.

http://chemistry.stackexchange.com/questions/46882/do-bare-protons-exist-even-transiently-in-aqueous-solution http://chemistry.stackexchange.com/questions/32558/why-are-hydrogen-ions-always-associated-with-another-molecule

@Mithoron it says 'Adding a bare proton to a molecule is a real event in the gas phase' so is it possible to produce a proton beam as we do easily with electron beams? even in vacuum, I don't care

Off course you can make proton beam, it's even used for therapy https://en.wikipedia.org/wiki/Proton_therapy but it's by no means a gas, not even plasma, but separated particles. In the future you need to word your questions properly and say exactly what you're asking. Also this is rather physics than chemistry.

A bare naked proton is one of the most reactive species known.  Finding them in the gas phase is unlikely in terrestrial conditions.  Additionally, there are some charge balance issues
However, hydrogen plasma is known to be abundant in the universe, forming the primary composition of most stars.  This would be H+ ions swimming in a sea of electrons.  It requires intense energy to keep this state, and without an input of heat the plasma will condense to hydrogen atoms, which will then form hydrogen gas.

though H+ are very common in liquids and cells, yet you say it's impossible to ionize H+ gas?

it's easy to produce a beam of electrons, isn't it that easy to produce a beam of protons?

Free H+ are not common.  Electrostaticly interacting H+ (such as with a solvent) are very common.


text:
Why, in an applied magnetic field, is electron with spin parallel to the magnetic field higher?
My lecturer today said that when an applied magnetic field, $B_0$, is applied, the electron with spin parallel to $B_0$ is higher in energy than the electron with spin antiparallel to $B_0$. Similarly, in protons, it is the opposite; parallel spin is lower in energy and antiparallel spin is higher in energy. Why is that?

Do you want to know why there is an energy difference to begin with, or why electrons and protons behave differently?

I understand that there exists a energy difference, but I just don't understand why when the spin is parallel to the direction of the external magnetic field it is higher in energy

In addition to the other answer, the energy  of a magnetic moment in a magnetic field is $E=-\bar{\bf{\mu}}.\bar{\bf{B}}$ where the dot indicates the dot product between the two vectors $\bar{\bf{\mu}}$ and $\bar{\bf{B}}$, thus the energy is lowest when the magnetic dipole is parallel to the field $B_0$; $E=-\mu B_0$. The sign of $\mu$ is opposite for nuclear and  electron spin angular momentum.The magnetic moment of a nuclear spin is defined as $\mu=+\gamma I$, where $I$ is the nuclear angular momentum and if the field is along only the $z$ direction  $\mu_z=\gamma I_z=m_z\gamma \hbar$ (Joule/Tesla) where $\gamma$ is the magnetogyric ratio, (unit radians/Tesla/sec) the components are $m_z$, the projection, magnetic or azimuthal spin quantum number on the $z$ axis, the axis parallel to the magnetic field $B_0$. $I_z$ is the $z$ component of the nuclear spin angular momentum. (The magnetogyric ratio is $\gamma = g\mu _0/\hbar$ where $\mu_0$ is the nuclear magnetron and $g$ the nuclear spin factor, different for each type of nucleus). The energy is then $E=-m_z\gamma\hbar B_0 $. For a proton $m_z=\pm 1/2$.
The spin magnetic moment of the electron is defined in a similar way to that for nuclear spin but with a negative sign; $\mu_e=-\gamma _e I_e$ where $I_e$ is the electron spin angular momentum and $\gamma _e$ the electron magnetogyric ratio. (  $\gamma _e=g_e\mu_B/\hbar$,  $g_e$ is the electron $g$ factor, $g_e =2.0023..$, and $\mu _B$ the Bohr magnetron. Some authors give a negative $g_e$, in this case the negative sign is removed from $\mu _e$).
(The nuclear magnetogyric ratio can be positive or negative, e.g. $\ce{26.75.10^7 rad T^{-1} s^{-1}}$ for protons. For a few types of nuclei e.g. $\ce{^{15}N, ^{29}Si}$ $\gamma$ is negative. The magnetogyric ratio for the electron is much $\ce{larger 1.76 10^11 rad T^{-1} s^{-1}}$.)

In short: electrons and protons have magnetic moments. Electrons are negatively charged, and the direction of their magnetic moments is reversed from that of protons.
Intuitively, the potential energy is lowest when the magnetic moment is aligned with the magnetic field, as is the case with the proton. Because electrons are negatively charged, however, their magnetic moment points in the opposite direction, and the potential energy is then lowest when the magnetic moment is anti-aligned with the magnetic field.


text:
How do I find the formula of the compound given the ratio of its constituents by weight?
Problem : An organic compound containing carbon, nitrogen and oxygen will have a weight ratio $9:1:3$. The molecular weight of the compound is $108$. What is the formula of the compound?
I thought of calculating the empirical formula, but percentages are given for that, and not weight.

It's an algebra problem.
108 = 9x + 1x + 3x
Solve for x.
Weight of carbon per molecule = 9x
Weight of nitrogen per molecule = 1x
Weight of oxygen per molecule = 3x
Then convert from weights to numbers (I assume you know how to do that  :)  ).
Note that the result is rather nonsensical, as it gives you fractional atoms.

The answer is given as $C_6H_8N_2$.


text:
Effectiveness of esterification catalysis's [Cu, Sulphuric acid, Acid Zeolite], in Whiskey distillation
If you go on a tour of a distillery, you will normally be told at some point how the (beautifully polished) copper pot stills catalyse the esterification of ethanol and higher (fusel) alcohols, with acids from the fermentation process improves the flavor of the whiskey. (And how their still geometry creates a superior product)
Obviously the distillation process, removing water and most of the higher (fusel) alcohols contributes to this as well.
How effective is Copper in that scenario? A still would not have more than say 100 m2 of copper surface area and a batch load would be on the order of 5-50 thousand liters
How much more effective would using Sulphuric acid as a catalysis? (Obviously this would not be useful/food grade, but nice to know as a baseline)
Now (the much more interesting part)
How effective would Acid Zeolite be as a catalysis? Either as a distillation column packing medium, or adding a 'handful' to a barrel before leaving it to sit in a ware house for a couple of years/decades?

Sulfuric acid would obviously be the best catalyst (when storing in barrels), mainly because it would be dissolved in the alcohol, while solid catalyst would require constant stirring/agitation. And a copper pot is a bad catalyst because of its low surface area.

@vapid Maybe, but it would not be very palatable :P & I would worry about what it might do to the barrel. Over the time scale of years I suspect (hope) normal Brownian mixing would be sufficient with Acid Zeolite. Good points though :D


text:
Van der Waals real gas equation
Van der Waals corrected the ideal gas equation for real gases and gave the following equation:
$$\left(p+\frac{an²}{V^2}\right)\left(V-nb\right)=nRT$$
I’m very confused about one aspect of this equation, which is this: why add the pressure term but subtract the volume term?
Edit: If somebody could please just explain the physical significance of the corrected equation and also, exactly what Van der Waals was trying to do by correcting this equation, it would be tremendously helpful for me.

Where do you see that you substract the volume?

Ideal gas molecules are mere dots with no volume, and don't interact with each other. Real gas molecules have certain own volume which eats up some of the space, hence $-nb$. Also, they repel each other and thus push outwards harder than they would otherwise, hence $+an^2/V$.

@GM: Clearly, we are subtracting $nb$ from $V$.

@Ivan Neretin: Are you sure about this? In most texts, the situation is described quite differently; the molecules close to the walls of the container experience a force of intermolecular attraction from the neighboring molecules and hence, the net force with which they collide with the walls decreases, hence decreasing the pressure.

The equation does not have any wall-related parameters in it. It describes an idealized situation without walls at all. Otherwise things might have been different.

Without a boundary surface(walls), how is the pressure of the gas defined?!

Think of an infinite expanse of gas at certain pressure and temperature, where such-and-such volume contains the prescribed amount of gas... No, that's too far-fetched. OK, think of a gas in a container with idealized walls which do not interact with anything.

@IvanNeretin I respectfully disagree with your second comment, the repulsions act over relatively short distances, and the potential on average is attractive at intermediate distances.

@getafix OK, then why $+an^2/V$?

@IvanNeretin I have edited my answer to include a discussion on this very issue. Correct me if I'm wrong :)

Got it. You were right, and I was not. Of course it is attraction that matters. It is attraction that is responsible for the critical point and the emergence of liquid state after that, even though the liquid itself is not described.

Let me begin my discussion by first introducing the notion, using some intuitive model building, and will try to derive, or perhaps guess the van der Waals equation of state.
The perfect gas equation, $ PV = nRT $ doesn’t allow for any interactions between molecules. The van der Waals equation is a refinement of this model, in the sense that it introduces the possibility of both repulsive and attractive interactions.
Imagine the gas molecules are small, hard, impenetrable spheres. This accounts for repulsive interactions, as now they can’t get arbitrarily close to each other. Consequently, they are no longer free to move about in a volume, $V$, and only have access to a much smaller volume $V-nb$ where $nb$ is roughly the volume taken up by the molecules.
The modified equation now becomes:
$$P = \frac{nRT}{V-nb}$$
Now to calculate the excluded volume:
$$V_\text{molecule} = \frac{4}{3}\pi r^3$$
where $r$ is obviously the radius.
Now note, a particle is surrounded by a sphere of radius $2r$ (two times the original radius) that is forbidden for the centres of the other particles. If the distance between two particle centres were to be smaller than $2r$, it would mean that the two particles penetrate each other, which, by definition, hard spheres are unable to do.
The excluded volume for the two particles (of average diameter $d$ or radius $r$) is: $\frac{4}{3}\pi (2r)^3$ or, $8V_\text{molecule}$. Divide this by two to get the volume excluded per molecule and we get $b \approx 4V_\text{molecule}$
Note: Of course, molecules are not infinitely hard, as van der Waals thought; there exists some “softness”. Thus, the factor four yields an upper bound; empirical values for b maybe lower.
Now turn on an attractive force between the particles.
Assume, notwithstanding the existence of this force, the density of the fluid is homogeneous.
And assume that a large majority of the particles are always surrounded by other particles, which attract each other (unless they get too close, when they repel; thus, the bulk of the particles do not experience a net force pulling them to the right or to the left. They feel that the container is infinitely large without any walls.
Now, the surface particles feel a net force from the bulk particles pulling them into the container, because there are no particles on the side where the wall is (and we assume here that there is no interaction between walls and particles, which is not true in reality).
This net force decreases the force exerted onto the wall by the particles in the surface layer. The net force on a surface particle, pulling it into the container, is proportional to the number density:
$$F = \frac{N_\mathrm a}{V_\mathrm m}$$
Assuming homogeneity, the number of surface particles is also proportional to the number density. So the total force on the walls is decreased by a factor proportional to the square of density.
Since pressure is force per unit area, it's decrease proportional to:
$$a'C^2 = a'\left(\frac{N_\mathrm a}{V_\mathrm m}\right)^2 = \frac{a}{V_\mathrm m^2}$$
Now, the modified equation becomes:
$$ P = \frac{RT}{V_\mathrm m-b}- \frac{a}{V_\mathrm m^2}$$
Derivation using Statistical Mechanics
For a system of $N$ particles, in 3 dimensions
The partition function:
$$ Z = \frac{1}{N!} \int \mathrm dp^{3N}\mathrm dx^{3N}\mathrm e^{-\frac{\beta}{2M} \Sigma p^2_n}\mathrm e^{-\beta u(x)} $$
This integral factorises into two, an integral over $p$, and an integral over $x$, i.e. a kinetic energy contribution and a potential energy contribution.
I write the result as
$$Z = \frac{z^N}{N!}$$
for an ideal gas, this result is
$$\frac{V}{\Lambda^{3N}}$$
where
$$ \Lambda = \frac{h}{(2\pi mk_\mathrm bT)^{1/2}}$$
(thermal wavelength)
For a van der Waals gas,
$$
U(r) =
\begin{cases}
\infty, & \text{when }r<d \\
-\epsilon\left(\frac{d}{r}\right)^6, & \text{when }r \ge d
\end{cases}
$$
basically attractive up to a certain distance, and repulsive when they come too close. r is the distance between the spheres and d is the distance where they touch. Also, each particle moves independently in an average potential field offered by the other particles
Also, for an ideal gas we integrate over all space i.e. the entire volume of the container, however, we cannot do that anymore. Instead our volume of integration is $V - Nb'$ (or $V – nb$, same as what we did before, $n$ is number of moles instead of number of particles).
$$b'= \frac{2\pi d^3}{3}$$
Now, assume that in a shell of thickness $\mathrm dr$ there are $ \frac{N}{V} 4 \pi r^2\mathrm dr$ particles, i.e. the position of the particles is averaged (reality, like always is different)
So for a single particle, the attraction felt is
$$\phi = \int_{d}^{\infty} u(r)\frac{N}{V} 4 \pi r^2\mathrm dr$$
and
$$z = \frac{(V-Nb')\mathrm e^{-\phi/2k_\mathrm bT}}{\Lambda^3}$$
I will omit the computations and quote the results,
$$\phi = -2a' \frac{N}{V}$$
where
$$a' = \epsilon (2/3) \pi d^3$$
going back and putting things together,
$$\ln Z = N\ln(V-Nb') + \frac{N^2 a'}{Vk_\mathrm bT} - N\ln(\Lambda^3) - \ln(N!)$$
Now,
$$ P = k_\mathrm bT \frac{\partial \ln Z}{\partial V}$$
which turns out to be,
$$ P = \frac{Nk_\mathrm bT}{V-Nb'}- \frac{N^2 a'}{V^2}$$
UPDATE: Physical Interpretation
I am writing this section because I get the impression (from the discussion the in the comments) that my answer hasn't adequately addressed this.
In the limit $P \to 0$ (or $V_m \to \infty $) we can expect perfect gas behaviour, i.e $PV = \text{constant}$
However, say at high pressures, we observe that real gases would have larger molar volumes than expected, so we remove a term from the volume.
Similarly, the pressure is lower than ideal gas pressure, so we add a term to it to compensate.
So, instead of saying $PV_m = \text{constant}$, we say $ (P +\text{"something"})(V_m - \text{"something"}) = \text{constant}$
It also accounts for critical behaviour, simply solve the following system to obtain critical constants (since the critical point, is a point of inflection on the PV diagram)
$$ \frac{\partial P}{\partial V_m} = 0 $$
$$ \frac{\partial^2 P}{\partial V_m^2} = 0 $$
Of course, it would be too optimistic to expect a simple, elegant model such as this to truly determine all properties of gases with complete precision; the van der waals equation of state does have its failings, but if nothing else, it improves at least the qualitative discussion of properties/behaviours of real gases

Comment have been moved to chat: http://chat.stackexchange.com/rooms/44503/discussion-on-answer-by-getafix-van-der-waals-real-gas-equation.  Please do not hold extended discussions in comments.  Thanks!

When you see the ideal gas equation, derieved from the kinetic theorey of gases, there are many assumptions taken for the ideal gases, a few like their Potential energy is 0, the molecules are spherical, and so on. What different real gas equations attempt to do is to reduce the number of assumptions taken by providing "corrections". Van der Waal attempted to correct the pressure and volume terms.
According to Van der Waal, his equation of real gas should be of this format:
{ Pressure on walls of container without any intermolecular attraction }. { Volume available for motion of molecules }  = nRT
In ideal gases, we assume that there are no intermolecular attractions between any of the molecules, thus the pressure of the gas is used in the equation. Also, we assume that the volume available for motion of the molecules is the volume of the container itself.
What exactly is the volume available for motion of molecules?
It is the space in which the molecule can move in. If two molecules are near each other, then neither can travel into the space occupied by the other molecule. Thus the volume available for motion is actually less than the volume of container.
Thus, Van der Waal introduced two correction terms, $\frac{n^2a}{V^2}$ and $nb$, which are the pressure correction term and the volume correction term respectively.
If we account for intermolecular attractions, the pressure of real gases increases. If we assume molecules to occupy space in real gases, the volume available for motion decreases.

But are you absolutely sure about the fact that pressure increases? Most texts describe the situation quite differently. Please refer to my response(comment)to Ivan Neretin's comment.

@Kaumudi, I'm certain that pressure increases.

But most of the texts seem to point to the contrary. Even Khan Academy!

@Kaumudi, the corrected pressure we plug in the real gas equation increases. Real gases' molecules have intermolecular attractions, which decrease the overall pressure of the real gas. If we were to treat it as an ideal gas, we would only put the overall pressure. But since it is a real gas and we have to exclude the intermolecular attractions, pressure increases.

Since it is a real gas, we are required to  take the intermolecular forces into consideration! You seem to have got it all wrong, I'm afraid. Read getafix's answer.

Suit yourself sir. And please avoid the insulting language.

Please refer to the answers given above. You must be prepared to accept that you may be wrong at all times and this time, you were wrong. I am not attempting to win this argument; I am simply begging you to remove your prejudices and accept my viewpoint with an open mind.

It is difficult to express my viewpoint over these restrictions between us. I am a 11th standard student studying for IITJEE with not a lot of time to answer completely. I believe it is understandable one will get frustated by this incomplete communication.

No, it is not understandable that one gets frustrated and calls somebody else an "idiot" merely because you didn't agree with what they had to say.

Im sorry I'm now in my senses please leave me alone I'm sorry for trying to answer.


text:
Role of hydrochloric acid and sodium hydroxide in isolation of citric acid from lemon juice
I'm trying to extract citric acid from squeezed lemons.
After studying the subject, I found a lab manual [1, p. 68]:

ISOLATION OF CITRIC ACID

Reagents:
Lemon juice (ca. $\pu{100 mL}$ – 3 lemons)
$\ce{CaCl2}~\pu{5 g}$
$10\,\%$ $\ce{NaOH}$
$\pu{2 M}~\ce{H2SO4}$
$\pu{2 M}~\ce{HCl}$
$\pu{2 M}~\ce{NaOH}$
Glassware:
Magnetic stirrer
beakers $\pu{250 mL}$ (3)
Measuring cylinders (2)
filtering system with Büchner funnel
flask $\pu{100 mL}$
Pipettes (2)
Pasteur’s pipettes
beaker $\pu{50 mL}$
glass rod
Place a beaker (v. $\pu{250 mL})$ on a stirrer and pour into it $\pu{100 mL}$ of fresh squeezed lemon juice (weight it!).
Add dropwise slowly and carefully $10\,\%$ aqueous $\ce{NaOH}$ basifying
the mixture up to $\mathrm{pH} = 8.$
You will recognize this moment by changing of the colour from light yellow to light orange.
Filter the obtained mixture on a Büchner funnel.
(Caution! Change from time to time the filter papers.
The pores of filter paper become blocked, so it is necessary to replace the filter paper with a new one.
Repeat it as often as necessary.
A clog paper can cause an explosion of the filter flask!).
Move the obtained transparent layer to the beaker, place it on a stirrer and add $\pu{50 mL}$ of $10\,\%$ aqueous $\ce{CaCl2}.$
Stir for 15 minutes, then heat it to boiling and filter calcium citrate
$(\ce{Ca3C12H10O14})$ from the hot mixture on a Büchner funnel.
Wash the obtained product on funnel with hot water.
Dissolve the obtained crude product in small amount of $\pu{2 M}~\ce{HCl}~(\pu{5 mL}).$
Then neutralize the solution with $\pu{2 M}~\ce{NaOH}$ to approximately $\mathrm{pH} = 7.5$ and boil the mixture.
Separate the sediment on Büchner funnel and dry on the air.
Weight the product and calculate the yield relative to the amount of the fresh lemon juice.

Does anybody know the reason for adding hydrochloric acid and sodium hydroxide in the last paragraph?
This looks very useless to me, since in this case the alkaline and acid neutralize each other.
Reference

Przybył, A. K.; Kurek, J. Laboratory Of Organic Chemistry: Natural Products And Pharmaceuticals; Adam Mickiewicz University: Poznań, 2013. (PDF)

True, they would neutralize, and the authors actually did bother to emphasize that: "Then neutralize the solution with 2M NaOH..." That's the very point of it. If you could add just the right amount of acid on the first try, you probably wouldn't need this step, but that's hardly possible.

Thanks for your comment. But why should you add hydrochloric acid and sodium hydroxide, after we already obtained calcium citrate? After the following step: "Stir for 15 minutes, then heat it to boiling and filter calcium citrate (Ca3C12H10O14) from the hot mixture on a Büchner funnel..." we could also go to the next steps on page 69, without adding an acid and alkaline?

Theoretically, we could. But see, they are still adding acid there on p.69.

Is this maybe a purification step? But which impurities are removed
and on which chemical principle is this based?

I have also found recipes, without the last step on page 68 (without adding HCl and NaOH). But I would like to know what the reason was.

@ChemStudent: Yes, the addition of HCl is to purify the crude calcium citrate. The crude material is washed with hot water, then dissolved with HCl - presumably leaving some contaminant on the filter paper - and precipitated with NaOH more gently (pH 7.5 instead of the earlier 8). BTW, a clogged filter should pose no hazard to the filter flask, just take more time (and it will decrease your yield and make more mess to put in a new filter). You could pretest your flask by evacuating it wrapped in a towel.

The $\ce{HCl}$ is added to dissolve calcium citrate by protonating the citrate to form (neutral) citric acid:
$$\ce{Ca3(C6H5O7)2(s) + 6HCl(aq) -> 3 Ca^{2+} (aq) + 6 Cl- (aq) + 2 C6H8O7 (aq)}$$
To recrystallize, you have to form citrate again. The $\mathrm{p}K_\mathrm{a}$ values of citrate range from 3.1 to 6.4, so at pH = 7.5 citric acid deprotonates fully to form citrate again:
$$ \ce{ C6H8O7 (aq) <=> C6H5O7^3- + 3H+(aq)}$$
On the other hand, you don't want the pH too basic because then calcium hydroxide will precipitate. In the presence of sufficiently high concentration of calcium ions, calcium citrate will precipitate again:
$$\ce{ 3 Ca^{2+} (aq) + 2 C6H5O7^3- <=> Ca3(C6H5O7)2(s) }$$
Each step in crystallization will have less contaminants, with some loss of the desired product. So it might look like you are going around in circles and losing product. However, the purity of the product will increase with every recrystallization step.

$100$ g lemon juice contains $6.08$ g citric acid, and $0.3$ g malic acid $\ce{(CHOHCOOH)2}$, according to K. Diem, Scientific Tables, CIBA-Geigy SA, Basel, 1972, p. 508
The first precipitation produces an impure precipitate of calcium citrate containing about $5$% calcium malate $\ce{CaC4H4O6}$. When redissolved in $\ce{HCl}$, neutralized and reprecipitated, the malate ions are eliminated. The second precipitate contains pure calcium citrate.


text:
Why doesn't a halide anion react with primary or secondary alcohols using SN2?
In the tenth edition of Organic Chemistry by Solomons and Fryhle, section 6.18A of the "Ionic Reactions" chapter, in the last paragraph, they have mentioned that:

Use of a weakly basic anion such as a chloride ion ($\ce{Cl^-}$ ), or an acetate ion ($\ce{CH3CO2^-}$) or a weakly basic and a highly polarizable one such as $\ce{Br^-}$, $\ce{I^-}$ or $\ce{RS^-}$ increases the likelihood of substitution (SN2). (emphasis mine)

But when we react alcohols with concentrated hydrohaloic acids (is that the right term?) with alcohols, we see that the rate of reaction decreases from a tertiary to a primary alcohol, indicating that the reaction is SN1. For primary alcohols, it is shown in the textbook to proceed through (SN2) pathway. [There is no such reaction mentioned with hydrobromic or hydroiodic acid (atleast I haven't come across such a reaction)].
Elsewhere, I have read about the Lucas test, which involves reacting the alcohol with hydrochloric acid and zinc chloride, where nearly no reaction occurs with primary alcohols.
My question is, why does this reaction not occur using an SN2 pathway for secondary and primary alcohols in the Lucas test? Why is it biased towards an SN1 reaction?

I would have thought that especially the primary alcohols react via an $\mathrm{S_N2}$ mechanism, but you’ve gotten me curious.

The issue is that when carrying out the Lucas test you have a polar protic solution (due to the alcohol which you are testing), in which case Cl$^-$ is well solvated and thus not a very good nucleophile. As you are not using a strong nucleophile, you disfavor the S$_\text{N}$2 mechanism, but since you have a primary alcohol, you are not really favoring S$_\text{N}$1 either. Hence, why no reaction happens and the Lucas test works to identify primary alcohols. I think your question of Why is it biased towards an S$_\text{N}$1 reaction? might be a bit misleading because the reaction is neither biased towards S$_\text{N}$1 nor S$_\text{N}$2 for primary alcohols - it simply does not happen at all (at room temperature). In the case of secondary and tertiary alcohols, however, I would definitely say the reaction is biased towards S$_\text{N}$1 since you have polar protic conditions, a weak nucleophile, and a more stable carbocation intermediate.
Upon raising the temperature, I suspect that the reaction would then proceed via an S$_\text{N}$2 mechanism for a primary alcohol (as, despite not having a strong nucleophile, the barrier would be lower than via S$_\text{N}$1), but remain S$_\text{N}$1 for secondary and tertiary alcohols. Hope this helps!

Thank you for your answer! That makes a lot of sense. I might add that for chlorination using $\ce{PCl3}$, the chloride ion attackes the alkyl phosphite intermediate thru SN2 mechanism, and it is able to do so because the solvent is polar aprotic (I assume so!). This was a part of the reason why I had asked for the difference!

Hydroxyl group is not good leaving group that's why the reaction in the Lucas test with primary alcohol don't occur via SN2 reaction.
In the Lucas test Hydroxyl group first, form hydronium ion in the presence of strong acid and ZnCl2. Hydronium is better leaving group thus form carbonium ion immediately. As, you know tertiary carbonium ion is much more stable than secondary and primary carbonium ion.
Lucas test also occur with primary and secondary alcohol at elevated temperature but with tertiary alcohol the reaction with HCl occur even at room temperature.

Once the hydroxyl group has been protonated, why doesn't the halide ion displace it by SN2? That is my question. The reaction could switch to an SN2 mechanism for primary alcohols, but that doesn't happen!

It is not always the case, there are some other important criteria for SN2 reaction as well that you have to considered. Under these condition that you have mentioned the prefer pathway is the formation of carbonium ion irrespective of primary, secondary and tertiary  carboinium ion.

What important criteria for SN2 are you referring to? Also, AFAIK, carbocations (carbonium ions) don't form in case of primary alcohols unless the conditions are pretty extreme. My question is for room temperature conditions.

SN2 reaction normally occur in polar aprotic solvent like DMSO, DMF, whereas SN1 reaction usually occur polar protic solvent like Alcohol.

Oh, okay! We use water as solvent for the Lucas test (isn't it?). So that promotes SN1. You can add this in your answer, I think. Thanks for hearing me out!

@FreezingFire, exactly, Always welcome


text:
IUPAC name of a salt made of both cationic and anionic complexes

The IUPAC name of the complex compound [Ni(NH3)4][NiCl4] is

In my opinion the correct answer should be Tetrachloronickel(II) tetraamminenickelate(II), because nickel has got more oxidation number when there are chlorine atoms and therefore it should be named first. But the correct answer seems to be Tetraamminenickel(II) tetrachloronickelate(II).

Clearly, the "-ate" ending goes to the anion. As for the oxidation state of nickel, it is the same (+2) in both.

I thought I saw a question of this type before and maybe I even answered one somewhere. Alas, I can’t find it …

Related: How to determine the oxidation states of the metal atoms when both cation and anion are coordination complexes?; @Jan's answer can be found there. It's also useful!

@IvanNeretin: May I know how did you determine that oxidation state of nickel is the same in both? Was that based on your experience or is there any specific algorithm?

What else could it be? Well, I guess it is based on experience, after all.

The coordination compound $\ce{[Ni(NH3)4][NiCl4]}$ contains two coordination entities, $\ce{[Ni(NH3)4]^2+}$ and $\ce{[NiCl4]^2-}$, which can be named separately.
The rules for formulating and naming coordination compounds are presented in Chapter IR-9 of the current version of Nomenclature of Inorganic Chemistry – IUPAC Recommendations 2005 (Red Book). An abridged version is included in the IUPAC Technical Report Brief guide to the nomenclature of inorganic chemistry. Pure Appl. Chem. 2015, 87(9–10), 1039–1049 as well as in the corresponding four-sided lift-out document, which is available as supplementary material.
Generally, ligand names are listed as prefixes to the name of the central atom along with any appropriate multipliers. Names of anionic coordination entities are furthermore given the ending ‘ate’. The oxidation number of the central atom may be indicated by a Roman numeral appended in parentheses to the central atom name; alternatively, the charge on the coordination entity may be indicated.
Therefore, the systematic name for $\ce{[Ni(NH3)4]^2+}$ is tetraamminenickel(II) or tetraamminenickel(2+), and the systematic name for $\ce{[NiCl4]^2-}$ is tetrachloridonickelate(II) or tetrachloridonickelate(2−).
In a salt made of both cationic and anionic coordination entities, the line formula for each ion is separately enclosed in square brackets. The cation is placed before the anion and no individual charges are shown:
$\ce{[Ni(NH3)4][NiCl4]}$
The systematic name of the entire compound is formed by combining the name of the electropositive constituent, cited first, with that of the electronegative constituent, both suitably qualified by any necessary multiplicative prefixes. The two parts of the name are separated by a space.
Therefore, the systematic name for $\ce{[Ni(NH3)4][NiCl4]}$ is tetraamminenickel(II) tetrachloridonickelate(II) or tetraamminenickel(2+) tetrachloridonickelate(2−).
Note that the ligand name ‘chloro’ is obsolete. According to current IUPAC recommendations, the correct ligand name is ‘chlorido’.

Shouldn’t ammine come before chlorido considering “alphabetical order”?