text:
What does it mean for meth or any other compound to be 'pure'?
In Breaking Bad, Walter White can cook $99.1\%$ pure meth. In general what does it mean for a substance to be pure? In this case, what exactly does it mean for meth to be pure? Does higher purity mean less unhealthy? More addictive? Giving more euphoria for the user?
For comparison, is the purity of prescribed methamphetamine or methylphenidate to people with ADHD or narcolepsy close to $100\%$? My guess is that purity is how close Walt's product is with legal meth.

One oddity is that very pure drugs can actually be more dangerous than lower-purity ones - if the user doesn't realise that that's what they're getting.  For instance, most heroin that is sold on the black market is actually pretty diluted, in other words it's a small quantity of actual heroin mixed with something a lot cheaper like sugar.  Occasionally a batch of much purer heroin gets on sale, and people have died from not realising the difference and taking the same amount they normally would if it was the dilute stuff - which for pure heroin adds up to a massive overdose.

@A.B. yeah that's what i was indeed part of what i was wondering like if pure is more vs less dangerous or not related, more vs less euphoric or not related, etc

99.1% pure versus, say, 97.4% pure might not make much difference to how strong it was (other than as a boast for advertising purposes!) - the heroin accidents seem to be more a question of some heroin being 50% pure and some 90% pure or more, which means a lot more heroin per gram.  "99.1% pure" sounds more like it would be aimed at customers who were prepared to pay extra to avoid the risk of poisoning themselves... at least, poisoning themselves with anything other than meth.

This article https://www.labmanager.com/business-management/the-seven-most-common-grades-for-chemicals-and-reagents-2655 seems to imply that the American standard for medical-grade chemicals is only 95% (for the raw substances before they're mixed with other things to make pills etc.).  If so, it sounds like 99.1% pure would be well within medical-grade standard.  I'm not sure whether that's what it means, though.  Also, a lot might depend on what the other 0.9% was!  Some chemical processes might lead to nastier contaminants than others.

When a compound is "pure" it means the contents of that compound are exactly what we claim them to be.  If something is $99\%$ pure, that means $99\%$ of it is the expected material (in this case, meth), and 1% is other non-meth compounds.
Purity matters in chemistry and biology because the other compounds can be very nasty.  In theory, if you had a bag of 100% pure meth, and a bag of $50\%$ pure meth, you could simply double the dose with the second bag and end up with the same amount of methamphetamine in your system.  However, you would also have a double dose of $50\%$ "other" stuff.
Sometimes "other" stuff isn't a big deal.  It all depends what that "other" stuff is.  I don't know much about the methamphetamine trade, but the effect of purity is a big deal in the ethanol world.  Moonshiners know to discard the "head" and the "tail" of the distilation, the first and last parts of the distilation, because they contain fusel oils (which taste really bad), and methanol (which can make you go blind in large enough dosages).  They want very "pure" ethanol.
To you talk about the purity of prescription drugs, typically what matters more for the drug companies is that they fully understand what "other" things are in the pills or injections, and they are confident they are safe.  That's why, when you look at many pill vials they will list their "inactive ingredients." They want to make sure you are very confident that the only things you are putting in your body are the "right" things.  Going back to the alcohol industry, "pure" ethanol is actually $95.6\%$ ethanol by volume, and $4.4\%$ water.  That is the limit for how "pure" ethanol can get with fractional distillation.  To have ethanol in any higher percentages than that you have to go to exotic processes to create anhydrous alcohols.  However, that is deemed "pure" because we still know what's in it.  We know its $95.6\%$ ethanol and $4.4\%$ water.
Of course, $100\%$ purity isn't always the desire.  There are myriad alcoholic spirits out there.  All of them are sold at 80 proof ($40\%$ alcohol by volume), so they contain the same amount of ethanol.  Its the "other" stuff that makes each spirit unique.  You pay a great deal of money for the particular impurities that make a great well aged scotch.  You pay very little for the impurities in a cheap tequila that leave your head pounding the next morning.

Except that Absolut (and Finlandia?) are also sold at $50~%$ and absynthe goes all the way up to $90~%$. And I love how we basically crafted core points of our answers around the same general ideas ;)

@BCLC There's certainly a lot to that side of things that I don't fully understand (and will likely choose not to understand).  However, I think the cheap Tequila example may be the closest to something I can relate to.  Cheap Tequila (silver/gold) is mixed with 50% "other" alcohols (it's only 50% agave).  The other alcohols tend to be cheaper, and have impurities like fusel alcohols.  So, while you want the toxicity of ethanol, the hangover is actually mostly caused by the fusel alcohols, even though they make up a small part of the liquid.  I can only assume the solvents used in making...

... meth are particularly nasty, so having some small amount of them left over (impurities) would markedly change the crap you are putting into yourself.  To use the Coke Zero analogy, it's like ordering a Coke Zero, and wanting to minimize how much drain-cleaner is in the liquid you drink.

With pills, also, pretty often the absolute quantity of the active ingredient is tiny, so they make a pill of cornstarch or something else harmless that's physically large enough to handle, and then add just the right dose of the drug.

So in breaking bad, a higher purity for meth is desirable not really because of absolute measures of euphoria or safety but the relative measure of having, say, double euphoria but less than double of the risky other stuff? (Of course the methamphetamine itself is risky)? I like your double dose explanation. thanks!

A pure compound is one that does not have anything else accompanying it — and thus is something entirely unreachable if your detection system is good enough unless you’re going for supercooled $\ce{^3He}$. However, typically lab grade chemicals are sold in purities of ${99+}~\%$ and extra special care is taken so that the remaining impurities do not affect the chemical’s intended use.
After performing a reaction typically a mix of compounds is obtained and some method of purification (e.g. precipitation and filtration, recrystallisation, chromatography) is required to arrive at a pure substance from the reaction mixture. Chemists then routinely record NMR spectra to check for any impurity peaks — and if they did it right, they will only find a water peak and thus deem their compound ‘pure’.
There are two main benefits of using pure compounds, one of which is alluded to above:

More ‘bang for the buck’, i.e. if you have a compound at $95~\%$ and another at $97~\%$, you need to weigh in less of the purer $97~\%$ compound to have the same amount in your flask
Less unintended side reactions from impurities present.

Transferred onto the crystal meth case, number 1 tells us that drug users would need less of the substance to reach the same effect, while 2 tells us that there is less of a potential for side effects. Note however that a higher purity compound can cause worse side-effects if it has impurities present that are responsible for worse side-effects. So $40~\%$ vodka bought from the store ($60~\%$ ‘impurities of the alcohol’; mostly water) won’t do anything aside from getting you drunk but if you run across $45~\%$ ethanol that contains $5~\%$ methanol as an impurity, it is much more harmful although it is purer.
For comparison, most medicines are, by purity standards, very impure — but it of course depends on the method of administration. Most of the time, however, the active ingredient is given in percent on the package; the drug I just checked had only $1~\%$ active ingredient. So technically, going by the above definition $99~\%$ impurities. Of course, in pharmaceuticals all the ‘impurities’ will either help or just be there to dilute the active ingredients. For example, many surface sprays have to contain a base as most active ingredients are added as hydrochloric acid salts. But going into too much detail here would not fit within this margin.

"1." really isn't a concern when you're at 99% vs 97%. The main benefit of using purer compounds is to avoid side reactions. Moreover, medications are typically more impure, but the kinds of impurities present are controlled to be a lot less harmful (ie, free of heavy metal catalysts or toxic side products). Moreover, if a medication has 1% active ingredient it means the rest of it are usually binding compounds that prevent product decay, assist drug delivery, prevent toxicity of overdose, as well as fillers. Those substances are controlled for during production and are not impurities.

@chell You are right about your first point (maybe I should have taken a more extreme example) and the second point should be in my answer somewhere. On the remainder of your comment: of course, drugs are typically manufactured to have the least harmful side-products in them. But from a purely analytical point of view anything that is not ‘the ingredient’ is an impurity whether it serves a purpose or not. Of course, the contents of a pill or a cream are chosen by the pharmaceutical company to have exactly the right dose and smallest size possible, but that’s almost a side point.

So in breaking bad, a higher purity for meth is desirable not really because of absolute measures of euphoria or safety but the relative measure of having, say, double euphoria but less than double of the risky other stuff? (Of course the methamphetamine itself is risky)? I like your bang for buck explanation. thanks!

'But going into too much detail here would not fit within this margin.' --> You can't say this because this chemistry se not maths se. lol.

Organic reactions (reactions involving fairly complex compounds of carbon) like the ones used in drug synthesis, are notoriously complicated and almost always produce various side-products and impurities. The matter is of real concern because almost all such syntheses involve several steps, where A is treated to form B and B is then treated to form C, and then C is treated to form D... (you get the picture.) Any impurities or byproducts produced in any or all steps must be removed to insure product purity, which rare;y happens in illicit syntheses. So important is this purification that Isolation & Purification is a science in its own right in organic chemistry, and very often more difficult that the chemical reactions themselves.
The FDA maintains extremely stringent controls on drugs manufactured for humans and even animals, and every batch manufactured by pharmaceutical makers, whether in the US or abroad for export to the US, whether brand-name or generic, must be exhaustively tested in multiple ways to ensure it passes requirements for purity, content, stability, and even physical state. Not only that, but the purity and identity of all chemicals used in the synthesis must likewise be tested and proven as to purity and identity.
The chemistry in "Breaking Bad" is quite accurate as far as it goes. A common purification technique for drugs of the methamphetamine family is to treat the crude, unpurified product in such a way that the meth forms crystals, which can then be collected and then washed free of the grossest impurities. The result is "crystal meth," which is still too contaminated to ever be okayed by the FDA even if it were legal, but is much cleaner and therefore more potent than the usual slop.

thanks! so this is about safety? do you agree with the bang for buck in another answer and the double dose in another answer?


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What is the process of connecting two molecules under removal of one molecule of water?

Certain molecules can be joined by the removal of water in a process called:

Hydration.

Hydrolysis.

Neutralization.

Condensation.

I know that the process of hydration is defined as, "A compound produced by combining a substance chemically with water."
The process of hydrolysis is defined as, "The chemical breakdown of a compound due to reaction with water."
Neutralization is a chemical reaction in which an acid and a base react quantitatively with each other.
Condensation is the conversion of a vapor or gas to a liquid.
The answer is most likely (1), but I'd like some feedback and explanation of these definitions!

Not that kind of condensation.

Oops, then what is it?

Is this an example of a condensation reaction?  A simple example is the condensation of two amino acids to form a peptide.

Since it is a removal of water, it cannot be hydration which is defined as the addition of water.
Similarly, hydrolysis includes the addition of water to something so that it degrades in some way.
Neutralisation does indeed generate water, but not always one molecule per reaction equivalent (can be two or three) and you do not connect molecules but ions.
Condensation not only has its physical definition you stated but it also means (in a chemical sense) connecting two molecules under formal exclusion of one molecule of water.
Note that in many to most processes, the water exclusion is entirely formal and what actually is excluded is something like $\ce{H2PO4-}$.

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Clarification on Oxidation Number and Ionic Charge
I am confused on the differentiation between ionic charge and oxidation number (state).  For example and simplicity, $\ce{NaCl}$ is an ionic compound so it forms $\ce{Na+}$ and $\ce{Cl-}$.  Are these ($1+$ for $\ce{Na}$ and $1-$ for $\ce{Cl}$) both the oxidation states and the ionic charges?  Or is the net of the compound ($\ce{NaCl}$) the ionic charge?
Additionally, when you have something such as $\ce{C2O4^2-}$ in this case a covalent compound , carbon's is $3+$.  Is this carbon's oxidation state?  What then is the net charge of the compound ($2-$) called?

For simple mononuclear ions in ionic compounds, as you have in $\ce{NaCl, Al2O3, ZnS}$ or $\ce{GaN}$, the oxidation state of each element will always equal its ionic charge: $\ce{Na+}$ has $\mathrm{+I}$, $\ce{Al^3+}$ $\mathrm{+III}$, $\ce{Zn^2+}$ $\mathrm{+II}$, $\ce{Ga^3+}$ $\mathrm{+III}$, $\ce{Cl-}$ $\mathrm{-I}$, $\ce{O^2-}$ $\mathrm{-II}$, $\ce{S^2-}$ $\mathrm{-II}$ and $\ce{N^3-}$ $\mathrm{-III}$.
Something like the oxalate anion $\ce{C2O4^2-}$ is a multinuclear anion. Its ionic charge is $2-$, as evident by the superscript. However, you cannot always determine the constituent elements’ oxidation states a priori — the only thing you do know is that the sum of the oxidation states must equal the charge number. In oxalate, carbon has $\mathrm{+III}$ as you correctly mentioned and oxygen has $\mathrm{-II}$, as it should be of no surprise to you. Added up, this leaves us with:
$$2 \times (+3) + 4 \times (-2) = +6 + -8 = -2$$
And $2-$ is the ionic charge as we already know. This rule goes both ways, so you can use the (known) ionic charge of a multinuclear ion to determine an element’s oxidation state if the other oxidation states are known: in $\ce{SO3^2-}$ oxygen has $\mathrm{-II}$, so sulphur can only have $\mathrm{+IV}$.

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How and why are all the substituents in methyl cyclopropane eclipsed?
Why does methyl cyclopropane $\ce{C3H5CH3}$ adopt an eclipsed conformation?

The geometry of the molecule is highly restricted by the presence of the ring.

The presence of the ring caused the $\ce{C-C-C}$ ring to be planar, because every triangle is planar.
Since there is only one carbon atom directly to the right to the circle, it must be eclipsed, so are the hydrogen atoms by VSEPR.

How do you know that "the carbons atoms are eclipsed?"

Because there is only one carbon atom, so it must be eclipsed.

From which perspective are you looking at the molecule? I see 4 carbon atoms, not one carbon atom.

I have just clarified.

why can't cyclopropane pucker and be fluxional to relieve the torisional strain like cyclobutane does?

Why can't cyclopropane be staggered instead of eclipsed?

Because the $\ce{C}$ is already eclipsed. There isn't two $\ce{C}$s, there is just one $\ce{C}$.

If cyclopropane were to arrange itself in a isosceles, right, or equilateral triangle, which one of the three triangular shapes would be the most stable?

bring up my comment please.

also include scalene, obtuse, and acute triangles and all types of  triangles that exist

For example, In cyclopropane, would one angle close to 109.5 degrees and the rest of the angles 70.5 degrees each, be more stable than all angles at 60 degrees?

Drawing a Newman projection shows that the H atoms are eclipsed. Due to its shape, cyclopropane can never exist in a staggered conformation. Note that what looks like a double bond is actually the bond between the carbon on the back to the carbon on top.

why can't cyclopropane pucker and be fluxional to relieve the torisional strain like cyclobutane does?


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Can the following molecules be called meso compounds (cyclic structures which usually exist in non-planar forms)?

In the planar form the following molecules have one plane of symmetry.However in
  the chair-form (the most stable form) for the first molecule and the non-planar form for the second molecule there is no plane of
  symmetry.Will the molecules still be called meso?

BTW I do not think the molecules ever attain completely planar form.

Just for quick reference I'm adding the non-planar forms of the two molecules.

These molecules can never be planar. There are no internal symmetry planes and therefore they are not meso.

I've just tried building 3D models of the two enantiomers of the 1,2,4,5-tetramethylcyclohexane with my molymod kit and they are not superimposable - I tried chair, boat, half-chair and twist-boat conformations. I haven't tried the cyclooctene but I'm pretty sure it's not meso and it can't be planar, neither in cis nor trans.


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Can the following compound having a centre of symmetry be called both optically inactive and meso?

The Centre Of Symmetry (COS) has been shown by a blue dot.

My questions are:

Can this molecule be called optically inactive? I have this doubt because single bonds permit free rotation and the molecule hardly
remains in the given form for long.
Can this molecule be called a meso-compound as it optically inactive?
Slightly unrelated: I'm actually a bit confused with the definition of meso. I learnt that meso compounds must have a plane of symmetry. Am
I correct?

According to Wikipedia a meso compound or meso isomer is a non-optically active member of a set of stereoisomers, at least two of which are optically active.
Update: I realized my mistake in thought procedure after asking the question.(I fell into the "Meso Trap" :-P )
I would like to add this link as it might help future viewers.
http://www.masterorganicchemistry.com/2011/01/12/the-meso-trap/

You are correct that the molecule is not chiral. It is indeed a meso compound due to the internal symmetry plane. Also, don't worry about free-rotation of single bonds.

Ok my doubt is that if it was not meso due to plane of symmetry would it still be optically inactive due to centre of symmetry?

Also can you say why "If you actually draw a Newman projection, the most stable form of the molecule is symmetrical on both sides of the plane." ?

Sorry,  ignore the part about the Newman projection. With the centre of symmetry or also called centre of inversion is a symmetry element,  you should end up with the same molecule after the operation. However,  I am not sure if that's enough on its own to prove thay it is a meso compound.

By the Wiki definition "A meso compound or meso isomer is a non-optically active member of a set of stereoisomers, at least two of which are optically active. This means that despite containing two or more stereogenic centers, the molecule is not chiral." ---- Can we conclude something from this?I think a COS should be enough to prove that it is meso...

I have build a model again and determined the point group which is Cs. Therefore, there is no centre of inversion, there is only internal symmetry plane.

Could you upload a picture of your model? According to the picture a COS exists...

I apologize, it does exist. The point group is Ci. It it obvious that the molecule is a meso compound as it's superimposable on its mirror image and therefore, it seems the centre of inversion should be enough to confirm it as it's the only symmetry element present apart from the identity.

Thank you!Can you answer this question ? http://chemistry.stackexchange.com/questions/61020/stereochemistry-of-cycloalkanes-having-substituents


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Stereochemistry of cycloalkanes having substituents

My teacher said that from $3$ to $6$ member cycloalkanes having any number
  and type of substituents we can always draw the planar structure and
  decide the specifics of stereochemistry like:

Are they optically active or not?
Are they meso compunds or not?
Do they have a plane of symmetry/centre of symmetry/axis of symmetry/alternating axis of symmetry or not?

But I feel that the molecules are hardly in planar form. It would be incorrect to assume they are planar and find the specifics of stereochemistry. Only for cyclopropane the points should be correct as it is always planar. Maybe my teacher said that mistakenly in a hurry.
So regarding points $(1),(2)$ and $(3)$ who is correct? Why? Or are we both partially correct? I need a slightly detailed answer.

For a compound to be optically active it must not be superimposable with its mirror image. It's as simple as that. Also note that stereogenic centres are not always required for chirality. There are other types of chirality such as axial and planar chirality where a compound possesses either an axis or plane of chirality. Sometimes it can be the case that just the sterics of a molecule does not allow for free rotation of bonds and the molecule is locked one configuration. I think in the simplest cases of open chain molecules, you could indeed use the structural formula, but with rings which adopt specific 3D configurations, you must refer to the 3D structure.
Again, as said in 1., to determine if a molecule is chiral, it's better to refer to its 3D structure, especially when with meso, when you are dealing with more than one stereogenic  centre. For a compound to be meso, it must not be chiral, but posses stereogenic centeres (look up the definition).
No, you cannot use the simple structure to determine symmetry elements present if that structure is not really planar. It's not necessary to build a model, but you must use wedged/dashed bonds to show stereochemistry.

If someone thinks differently, please comment and correct me.

Axially and planar-ly chiral molecules do have "stereogenic units", even if they don't have "stereogenic centers".

Yes, I have referred to those as axis and plane of chirality and they are stereogenic indeed as two enantiomers are observed.


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Does bicyclo[3.2.0]heptane have a plane of symmetry?

Most stable form of cyclobutane is the "envelope form" and most stable
  conformation of cyclopentane is either the "envelope form" or the "twist"
  form.

In the above compound do the individual rings exist in their most stable conformation? Or is the actual situation different?
What is the most stable conformation of the above compound? And does it have a plane of symmetry?

I figure that the question "why is the cyclobutane ring planar" is a good question in itself for main site. But you'll probably have to specify that you are looking for an org chem answer and not simply some calculations that say "because it is the most stable conformation".

This question is a prime example of overthinking!

It turns out that bicyclo[3.2.0]heptane does have a plane of symmetry (courtesy PubChem):

According to The Molecular Structures of Bicyclo[3.2.0]heptane and $\Delta^6$-Bicyclo[3.2.0]heptene. An Electron-diffraction Study of Gaseous $\ce{C7H12}$ and Molecular Mechanics Calculations on $\ce{C7H12}$ and $\ce{C7H10}$. Robert Glen, Grete Gundersen, Peter Murray-Rust, and David W. H. Rankin, here are the bond angles in degrees:
The link explains the experimental procedure in deduction of the actual structure.There are two possible stable structures/forms for the molecule namely endo and exo.The endo form is supposed to be more favourable than the exo form.

From here, we can conclude Cs symmetry.

I'm sorry to say.But you have given a very vague answer with no reason at all.Why doesnt the cyclopentane take up an envelope or twist form here?Why doesnt the cyclobutane take up an envelope form?Really vague answer :-(

@ZOZ Satisfied?

No :-(!You gave the bond angles.Ok good.You gave the predicted structure.Good.But you still did not give the reason for the structure's occurence and stability..and you did'nt answer my previous questions..in the comments as well as the original question..

@ZOZ Reason? This is experimentally determined. You could solve the Schroedinger's equation if that counts for a reason. There's no reason. Everything is just rationalization.

@ZOZ Even VSEPR is a rationalization.

It would not be correct to state that there is no reason.Rationalization is a pathway to reasoning.BTW the document you linked was very helpful and tries to explain the experimental procedure along with an attempt to explain why the endo form is more favourable.Thanks for that link.

Just a thought - I suppose more diaxial interactions in exo form could account for its lower stability...

@ZOZ Rationalization is not a reason. The reason is that this conformation has the lowest energy. The reason that this conformation has the lowest energy is because of, you know, too many parameters.

I never said rationalization is a reason.But it is a pathway to reason.Atleast it gives a head start and that's all I wanted in an answer.Anyway this is not the place to discuss which definition is correct.That belongs to English SE. Anyway +1.

If the solution of the Schrodinger equation is the only admissible "reason", then chemistry is in a lot of trouble. @ZOZ The cyclopentane ring is already in an envelope conformation, as pH13 notes. Remember that any distortion of geometry in the cyclobutane ring is transferred to the cyclopentane ring via the bridging carbons. Presumably, distortion of the cyclobutane ring towards  a puckered geometry simply creates more strain (bond/angle) in the rest of the molecule than is relieved in the cyclobutane ring.

Most of the discussion about the rationalisation of the most favourable structure is irrelevant to the original question especially since crystals have significant molecule-molecule interactions that might matter (though in this case the molecule has a plane of symmetry within error bars anyway). In solution several conformations most likely equilibrate to an average which has a plane of symmetry unless there are strong barriers to conformational shifts.

The most stable conformation is the endo-form which obeys $C_\mathrm s$ symmetry. The cyclobutane-part is in envelope form and the cyclopentane-part is also in envelope form. The endo version with the methylene group bent towards the cyclobutane moiety is the preferred one.
The next most stable form would then be the exo form, where neither the cyclobutane- nor the cyclopentane-part is in its favorite form, both are slightly distorted. At least after my optimization, this molecule does not have a symmetry, except $C_1$. Though, I did not try a symmetric version.
With a quite high energy gap of ~90 kJ/mol the most unstable form is where both parts are in their twisted forms. Though, this molecule is $C_2$ symmetric.

The structures were optimized with RI-TPSS-D3BJ/def2-TZVP and a following frequency calculation showed that the results are local minima. Those geometries where then used for a single point energy calculation, using the Tight-PNO DLPNO-CCSD(T)/def2-TZVP scheme. All calculations have been made with ORCA 3.0.3.

I see my answer here not really as an answer but more as an additional help for some qualified people to discuss.


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Reaction for the corrosion of iron
Construct the equation for the reaction between iron(II) ions, hydrogen ions ($\ce{H+}$), and oxygen to form iron(III) ions and water.
The answer is $$\ce{4Fe^{2+}(aq) + 4H+(aq) + O2(aq)->4Fe^{3+}(aq) + 2H2O(l)}$$ Can you explain how to get this answer? Thanks.

Rusting of iron is an electrochemical phenomenon.
A certain site on the iron object acts as an anode.
$$\ce{(Fe(s) -> Fe^2+(aq) + 2e ) × 2}$$
Electrons released at this spot go to another site on the object and reduce oxygen in presence of $\ce{H+}$ ions which are either obtained from water out from acidic substances in water:
$$\ce{H2O -> H+ + OH-}$$
$$\ce{CO + H2O -> H2CO3}$$
$$\ce{H2CO3 -> 2H+ + CO3^2- }$$
This site behaves as cathode and reaction is
$$\ce{4H+ + O2 + 4e -> 2H2O}$$
Therefore the overall reaction is :
$$\ce{2Fe(s) + O2 + 4e -> 2Fe^2+(aq) + 2H2O}$$
Now the ferrous ions are further oxidised by atmospheric oxygen to ferric ions which come out as rust in form of hydrated ferric oxides.
$$\ce{Fe^2+(aq) + 2H2O + 1/2O2 -> Fe2O3 + 4H+(aq)}$$

In addition to what is written by user377386, I'd like to add:
$$\ce{Fe^{2+}->Fe^{3+} +e-}$$
$$\ce{4H^{+} + 4e- + O2->2H2O}$$
By multiplying the first equation by four and adding the two equations, you get the equation:  $$\ce{4Fe^{2+}(aq) + 4H+(aq) + O2(aq)->4Fe^{3+}(aq) + 2H2O(l)}$$


text:
How's the IUPAC name of basketane determined?
A book of mine mentions a certain molecule, basketane, and provides its corresponding structure:

Now, I appreciate the book went on to mention this particular compound (I didn't know that organic molecules could be shaped like...well...a basket, so I find this interesting) but the problem is, that it didn't mention basketane's IUPAC nomenclature, much less how to go about naming it.
A quick internet search reveals the following IUPAC name for basketane:

Pentacyclo[$4.4.0$.$\mathrm{0^{2,5}}$.$\mathrm{0^{3,8}}$.$\mathrm{0^{4,7}}$]decane

But I'm still having trouble finding out how people even arrived at this name. Heck, I don't even know if the above mentioned name is even correct. I consulted all my Org. Chem books, and predictably, there was no mention of how to even begin naming basketane. I'm still in high school, so this particular kind of nomenclature is still alien to me (I haven't dealt with any super-scripts or carbons assigned the locant: zero yet).
Would anyone be able to:
1) Validate that name I've come across
2) Explain (citing a decent reference will do as well) how I ought to go about even naming this thing.
EDIT- @DHMO has been kind enough to point out two possible duplicates in the comment section below; but I'd still beg to differ.

Duplicate.

This is cubane but with an additional $\ce{C2H4}$ group that doesn’t really affect nomenclature.

Validation
Let us build this alien step by step:

We will build bicyclo[$4$.$4$.$0$]decane.
We will build tricyclo[$4$.$4$.$0$.$\mathrm{0^{2,5}}$]decane.
We will build tetracyclo[$4$.$4$.$0$.$\mathrm{0^{2,5}}$.$\mathrm{0^{3,8}}$]decane.
We will build pentacyclo[$4$.$4$.$0$.$\mathrm{0^{2,5}}$.$\mathrm{0^{3,8}}$.$\mathrm{0^{4,7}}$]decane.

Bicyclo[$4$.$4$.$0$]decane looks like this:

The part "decane" means that there are $10$ carbons and the molecule is saturated, just as indicated.
The part "bicyclo" means that there are two rings. One ring is $1$-$2$-$3$-$4$-$5$-$6$ and the other is $1$-$6$-$7$-$8$-$9$-$10$.
The part [$4$.$4$.$0$] is based on $\ce{C}1$ and $\ce{C}6$:

The first $4$ refers to the first link between $\ce{C}1$ and $\ce{C}6$, namely $2$-$3$-$4$-$5$ which contains $4$ carbons.
The second $4$ refers to the second link between $\ce{C}1$ and $\ce{C}6$, namely $7$-$8$-$9$-$10$ which contains $4$ carbons
The $0$ refers to the third link between $\ce{C}1$ and $\ce{C}6$ which is just a line with $0$ carbons between.

Tricyclo[$4$.$4$.$0$.$\mathrm{0^{2,5}}$]decane looks like this:

It is now "tricyclo" because there are three rings: $1$-$2$-$5$-$6$, $2$-$3$-$4$-$5$, $1$-$6$-$7$-$8$-$9$-$10$.
The extra part $0^{2,5}$ means that $\ce{C}2$ and $\ce{C}5$ are linked by $0$ extra carbons.

Tetracyclo[$4$.$4$.$0$.$\mathrm{0^{2,5}}$.$\mathrm{0^{3,8}}$]decane looks like this:

(from now on, a strong intuition on topology is required)
The new part $\mathrm{0^{3,8}}$ refers to the new linkage between $\ce{C}3$ and $\ce{C}8$ with $0$ carbons between.

Finally, pentacyclo[$4$.$4$.$0$.$\mathrm{0^{2,5}}$.$\mathrm{0^{3,8}}$.$\mathrm{0^{4,7}}$]decane:

Now you might ask: how on earth is that basketane? Let me show you how:

Not convinced? Here is the proof:

$1$ is connected to $2$, $6$, $10$.
$2$ is connected to $1$, $3$, $5$.
$3$ is connected to $2$, $4$, $8$.
$4$ is connected to $3$, $5$, $7$.
$5$ is connected to $2$, $4$, $6$.
$6$ is connected to $1$, $5$, $7$.
$7$ is connected to $4$, $6$, $8$.
$8$ is connected to $3$, $7$, $9$.
$9$ is connected to $8$, $10$.
$10$ is connected to $1$, $9$.

In the above two pictures, you can verify the connectivities, and they match exactly.

Derivation
Firstly, we try to find the largest ring that can contain the most carbon atoms. In this case, it is already indicated above (trace the ring $1$-$2$-$3$-$4$-$5$-$6$-$7$-$8$-$9$-$10$).
Then, you might ask, "How do we judge which numbering is correct?"
Curiously, the 1993 IUPAC Recommendations do not contain the criteria, but the 1979 IUPAC Recommendations do. Rule A-$32.31$, Nomenclature of Organic Chemistry, Sections A, B, C, D, E, F, and H. Pergamon Press, Oxford, 1979. Copyright 1979 IUPAC.:

$32.31$ - When there is a choice, the following criteria are considered in turn until a decision is made:
(a) The main ring shall contain as many carbon atoms as possible, two of which must serve as bridgeheads for the main bridge.
(b) The main bridge shall be as large as possible.
(c) The main ring shall be divided as symmetrically as possible by the main bridge.
(d) The superscripts locating the other bridges shall be as small as possible (in the sense indicated in Rule A-$2.2$).

Let us look at the first rule:

(a) The main ring shall contain as many carbon atoms as possible, two of which must serve as bridgeheads for the main bridge.

This means that we should maximize the sum of the first two numbers and minimize the sum of the other numbers.
Also, this means that we need to choose two carbons which are each connected to three carbon atoms.
For example, this would be wrong:

Incorrect name: pentacyclo[$\color{red}{4}$.$\color{orange}{2}$.$\color{green}{2}$.$\mathrm{0^{2,5}}$.$\mathrm{0^{3,8}}$.$\mathrm{0^{4,7}}$]
This name is incorrect because the first two numbers, $\color{red}{4}$ and $\color{orange}{2}$, add up to $6$, which is smaller than $8$ as derived from the correct name.

Let us look at the second rule:

(b) The main bridge shall be as large as possible.

The "main bridge" refers to the third number, which is to be maximized.
This is not applicable here, as every number beyond the third is zero.

Let us look at the third rule:

(c) The main ring shall be divided as symmetrically as possible by the main bridge.

This means, that although the first number is to be larger than the second, the difference between the first two numbers is to be minimized.
For example, this would be wrong:

Incorrect name: pentacyclo[$\color{red}{6}$.$\color{orange}{2}$.$\color{green}{0}$.$\mathrm{0^{2,7}}$.$\mathrm{0^{3,10}}$.$\mathrm{0^{6,9}}$]
This name is incorrect because the first two numbers, $\color{red}{6}$ and $\color{orange}{2}$, have a difference of $4$, while the difference between the first two numbers in the correct name is $0$.

Let us look at the fourth rule:

(d) The superscripts locating the other bridges shall be as small as possible (in the sense indicated in Rule A-$2.2$).

This is what A-$2.2$ says:

[...] When series of locants containing the same number of terms are compared term by term, that series is "lowest" which contains the lowest number on the occasion of the first difference. [...]

For example, this would be wrong:

Incorrect name: pentacyclo[$\color{red}{4}$.$\color{orange}{4}$.$\color{green}{0}$.$\mathrm{0^{2,9}}$.$\mathrm{0^{5,8}}$.$\mathrm{0^{7,10}}$]
Let us compare the locants in the incorrect version and the locants in the correct version:

Incorrect: $2,9,5,8,7,10$
Correct: $2,5,3,8,4,7$

The first difference is $9$ vs $5$, in which the correct version is smaller.

Conclusion
Therefore, we have arrived at the correct systemic IUPAC name for basketane:

Pentacyclo[$\color{red}{4}$.$\color{orange}{4}$.$\color{green}{0}$.$\color{purple}{\mathrm{0^{2,5}}}$.$\color{blue}{\mathrm{0^{3,8}}}$.$\color{pink}{\mathrm{0^{4,7}}}$]decane.


text:
Does s stand for sigma bond and p for pi bond?
When we are talking about hybridization, in $\mathrm{sp/sp^2/sp^3}$ hybridization, does s stand for sigma bond and p for pi bond ?

You got it backwards.
The $\mathrm{s}$, $\mathrm{p}$, $\mathrm{d}$, $\mathrm{f}$ orbitals stand for sharp, principal, diffuse, and fundamental. Wikipedia: Electron Configuration § Notation:

The choice of letters originates from a now-obsolete system of categorizing spectral lines as "sharp", "principal", "diffuse" and "fundamental" (or "fine"), based on their observed fine structure: their modern usage indicates orbitals with an azimuthal quantum number, l, of $0$, $1$, $2$ or $3$ respectively.

Then, $\sigma$, $\pi$, $\delta$, $\phi$, being the Greek equivalents of $\mathrm{s}$, $\mathrm{p}$, $\mathrm{d}$, $\mathrm{f}$ respectively, came to designate bonds or molecular orbital symmetry, because $\mathrm{s}$ is the first orbital to form $\sigma$ bonds, $\mathrm{p}$ the first orbital to form $\pi$ bonds, $\mathrm{d}$ the first orbital to form $\delta$ bonds, and $\mathrm{f}$ the first orbital to form $\phi$ bonds.

The $\mathrm{s}$ and $\mathrm{p}$ in $\mathrm{sp^3}$ are precisely the $\mathrm{s}$ orbital and the $\mathrm{p}$ orbital. $\mathrm{sp^3}$ means that $1$ $\mathrm{s}$ orbital mixes with $3$ $\mathrm{p}$ orbitals to create $4$ hybrid orbitals known as $\mathrm{sp^3}$ orbitals.

The last part is a bit incorrect and only applies to ideal tetrahedral molecules like methane strictly. The notation for hybrid orbitals is more accurately $$\ce{sp^{n} {=} s^{\frac{1}{n+1}}p^{\frac{n}{n+1}}}.$$ Therefore one part $\ce{s}$ mixes with one part $\ce{p}$ to form one hybrid orbital $\ce{sp^3}$. There can be different types of hybrid orbitals at the same atom.

It is important to note that while s orbitals cannot form π bonds, p orbitals are very capable at forming σ bonds. Similarly, neither of those two can form δ bonds but d-orbitals can and will form σ and π bonds.
An s orbital does not have a nodal plane. A p orbital has exactly one and a d orbital exactly two. (Note: I am not counting the ‘inner nodes’, e.g. inside a 2s orbital.) This definition was interpreted for bonds to determine whether a bond has zero, one or two nodal planes that fully contain the bond axis. A σ bond is rotation symmetric and does not have a nodal plane. A π bond has exactly one and a δ bond exactly two.
The $\ce{Cl-Cl}$ bond in chlorine, for example, is a σ bond formed by two p orbitals. And similarly, in transition metal complexes $\mathrm{d}^\unicode[Times]{x3c0}\mathrm{p}^\unicode[Times]{x3c0}$ bonds are common.
The $\mathrm{sp}^n$ nomenclature stands for hybrid orbitals; others have already described them.


text:
What is boron classified as a electronegative element and a electropositive element?
What is boron classified as  a electronegative element  or a electropositive element?
Is electronegativity a relative concept ? Why is it that sometimes we treat Boron as an electronegative element and sometimes as a electropositive element?

Electronegativity is relative.

Boron is more electronegative than the metals but less electronegative than the non-metals, generally.

hi, you again! is it why we treat it electronegative and electropositive @DHMO

Yes.${}{}{}{}{}$

Everything is electronegative and electropositive (except caesium and fluorine).

caesium and fluorine are electronegative, i guess

Caesium is the least electronegative and fluorine the most.

are low electronegative elements, electropositive or neutral? @DHMO

Do you know what electronegativity means?

:( i Know that electronegative element attract electrons.

ok got it, cannot be electropositive, but will it be neutral?

Then before it attracts electrons it is neutral and after it attracts electrons it is negatively charged.

"electropositive" and "electronegative" are not charges.

How is it unclear, what Vedant is asking? It may previously have been so, but it seems clear now, and you've only just put it on hold...

When you think about electronegativity, think of it as a relative value, meaning in comparison to atoms of other elements. Therefore, Boron is more electronegative than the metals but less electronegative than the non-metals, in general.

Fluorine is used as the standard with which to compare the relative
  electronegativities of other elements.

Everything is electronegative and electropositive. Caesium is the least electronegative and fluorine the most.

Could you rephrase it instead of just copying my words.

if you want to add your answer go ahead, i will remove mine and also upvote yours @DHMO

No, I just asked you to elaborate instead of copying my words because my words are meant to be short notes to make you think.

ok @DHMO doing jst that


text:
As we go down a fluoride-alkali metal series, why doesn't the boiling point decrease?
As you can see from the graph below, as we go down the blue fluoride-alkali metal series (alkali metal ion is varied from Lithium to Rubidium, which is represented by an increase in ionic mass on the graph), the boiling point increases then decreases.  This is contrary to my expectation that the boiling point should continuously decrease due to an increase in ionic radius.  The charge is obviously not a factor because it is consistent throughout.  What could be the factor that causes this deviation?
Thanks!

You also have to account for the geometry, which is highly influenced by charge (which is not a factor here) and radius (which is a big factor here).

Polarizability of the fluoride ion by the alkali metal ion decreases going down the group, this decreases the covalent character in each ionic bond in question.
See the related questions and answers here:
Does covalent character increase or decrease bond strength?
Melting points of alkali metal halides

Does this mean that since covalent character decreases down the group, sodium fluoride would have a higher b.p. than lithium fluoride, and potassium flouride has even less covalent character regains its ionic properties, thus resuming the 'normal' trend?


text:
Why are many chemical relationships logarithmic?
I've noticed that in many relationships in chemistry, one variable is linearly related to the logarithm of another. Why is this the case?
For example, I carried out an experiment where the dependent variable, the amount of solute adsorbed to a surface, was almost perfectly linearly correlated (Pearson correlation coefficient = 0.93) with the pOH, which is a logarithmic quantity.
Edit: Would this perhaps be better in stats.stackexchange.com?

I don't think it's better suited in stats. However, what I would say is that this is perhaps a very... general question and I am not sure if there is one single answer. Some logarithmic relationships (e.g. in thermodynamics) arise from first-order ODEs of the form $\mathrm{d}x/x = \mathrm{d}y$. Some arise from the fact that the reaction quotient $Q$ is a product of activities, so $\log Q$ is a sum of logarithms of activities. And so on and so forth...

I see your point - this is too general. This specific case involves the adsorption of fluoride to activated alumina, where there was competitive adsorption between $\ce{F^-}$ and $\ce{OH^-}$; in other words, hydroxide ions occupied the surface sites, and the addition of hydroxide meant that fewer sites were available for fluoride to adsorb to.

Regarding your point about reaction quotients, will that mean that I could get a prediction for the relationship by creating some kind of combined equilibrium expression for the two competitive equilibria (fluoride/alumina and hydroxide/alumina)? If that is the case, does it matter that these are heterogeneous equilibria?

Sorry, I'm not sure about absorption, don't really know much about the topic.

Some logarithmic behaviour is built into the way the world works and some is a consequence of how we perceive things which span very large ranges.
The simplest chemical and nuclear reactions (first order ones) are, for example, inherently logarithmic. Take any first order process (like radioactive decay). The process has a fixed probability of happening (a radioactive atom or an unstable molecule might have a 10% probability of decaying in a minute). What you observe in the bulk with processes like this is that the rate of the reaction decays exponentially. You start with a rate of say 12 units per minute. After a certain time half the initial material has gone and the rate is now 6 after the same time again the rate is now 3 and so on. This is exponential decay emerging from a very simple process with a fixed probability. A logarithm emerging from a probability, if you prefer. This sort of thing is very common in chemistry and physics.
The other cause of logarithms is human perception. Our perception of light and sound are both inherently logarithmic. This is partially because there is no sensible way to map the range of magnitudes of many natural phenomena to a linear scale. The intensity of sound and light vary over huge orders of magnitude. The Sun is about 15 trillion time brighter than the faintest visible star and it is fairly inconvenient to divide this scale into equal bands. So perception tends to divide the scale logarithmically with each perceived difference representing a fixed ratio not a linear difference (each star magnitude, for example is about 2.5 times brighter than the next lower one, but represents a just-about discernible difference).

The logarithmic patterns are more a function of math than physical properties.  The logarithmic identity: $\log(x^5) = 5\log(x)$ is responsible for most of your observations.  If I have a property $y$ that is dependent on $x^a$ where $a$ is a constant, I can log both sides to get a relation of: $\log(y) = \log(x^a) = a\log(x)$. This means that for every $1$ increment increase in $\log(x)$, $\log(y)$ will increase by $a$ increments (a linear relation).

On reason is that probability plays an important role in physics, especially at the microscopic scale.
When you combine two systems to a new system some physical quantities simply add up while probabilities multiply.
The arising functional equations
$$f(x \cdot y) = f(x) + f(y) $$
and
$$f(x + y) = f(x) \cdot f(y)$$
are satisfied by logarithmic or exponential functions.


text:
Is it possible to use glycerol DNA/RNA drying during extraction?
Is it possible to use glycerol with ethanol in order to avoid over-drying of DNA/RNA at drying step of extraction protocol?

Glycerol is likely to prevent the evaporation of ethanol.

if glycerol residues stay in DNA, it might interact with later PCR. Overdrying is an underestimated issue and often DNA resuspension is not only long (>3h) but also very partial even after one night resuspension. it is not possible to evaporate ethanol first but reaching the point where sample are still wet but ethanol mostly removed is the best option


text:
What are oxidation states used for?
When I took high school chemistry many years ago, considerable effort was spent on teaching us to compute oxidation states of atoms in various compounds, following a set of rules that looked somewhat arbitrary to me at the time. As far as I remember, we were never told what benefits (other than passing tests) knowledge of oxidation states would give us.
They were used as part of a convoluted procedure for "balancing redox reactions", but I never saw an example of that where it wouldn't give the same results simply to require that the number of nuclei of each element, as well as the total charge, must be the same on both sides of the reaction, and solve the resulting linear Diophantine equations.
Nevertheless, the concept must be useful other than for setting homework exercises -- I see encyclopedias and other sources use much space on classifying compounds based on the oxidation state of this atom or that. Still I don't recall seeing any case where the oxidation states are used for anything (other than computing other oxidation states).
What is this concept actually used for? Just a basic example or two where knowing the oxidation states helps produce a meaningful prediction.
It doesn't seem that "oxidation state" actually encodes any particular configuration change inside the atom that tends to be preserved across reactions. Or does it somehow require effort (energy?) to change an atom from
one oxidation state to another? In which case, what actually changes?
Sorry if this is too basic a question. It may be something "everybody knows". I've tried looking e.g. at the Wikipedia article, but that too seems to be entirely focused on how to determine oxidation states and doesn't explain why one would desire to know them. And all the oxidation-state questions here seems similarly to be about determining the oxidation states, and it is left unsaid why the asker wants to know ...

Edit: I'm looking for applications of oxidation numbers, by which I mean rules, laws, tendencies, rules-of-thumb or the like of the general form:

When you have (condition that involves oxidation numbers) then (prediction that can be verified without knowing what oxidation numbers are).

I'm an organic chemist, and we rarely have to balance redox equations, thank God! We rarely have to balance ANY kind of equations, really, other than in a qualitative sense, because most organic reactions are quite messy and produce a host of byproducts and impurities which are only quantified when it's vital to know everything about the reaction and its mechanism and conditions. We do perform numerous oxidations and reductions, but we'll usually just use a slight stoichiometric excess of the redox agent and let the devil see to the details.

I will offer you this bit of mnemonic, though, of the sort we pot-boilers are so fond of, and which my students all seem to find useful: "On the chemical jungle's floor, LEO the lion GERs his awful roar." It's a way of remembering that the loss of electrons is an oxidation (LEO), and the gain of electrons is a reduction (GER). Yes. Sometimes it just comes down to that.

When I took Chemistry at UW-Milwaukee, they skipped the redox part entirely. Years later, when I was a teaching assistant at Princeton U, I was expected to teach redox numbers and states, and I was totally flummoxed. I had to learn it from scratch basically. What I am saying is that the emphasis on redox depends strongly on who is teaching the class and at what university. My opinion now is that redox is primarily useful in analytical and electrochemistry. In a general chemistry class, it should certainly be covered, but no more than one or two lectures. If you are a student and have a hard ti

Of course they are useful. Perhaps you have not come across them yet, but being able to determine the oxidation state of an atom allows us to understand the properties of chemicals and how redox reactions work.
Let's just give two very simple examples. There are tons more. (I don't even want to go into organometallic chemistry, where being able to determine the oxidation state is incredibly important.)

Potassium permanganate, $\ce{KMnO4}$, has manganese in a +7 oxidation state. This means that it would have an electronic configuration identical to the noble gas argon: $\mathrm{1s^2 2s^2 2p^6 3s^2 3p^6}$, and therefore its $\mathrm{3d}$ orbitals are empty.
Now, most transition metal compounds are said to be coloured because of so-called "d-d* transitions". However, this can only occur if the d orbitals are partially filled - if there are no electrons in the d orbitals, like in $\ce{KMnO4}$, then there are no d-d* transitions available.
As such, the intense purple colour of $\ce{KMnO4}$ has to be explained via a different mechanism. In this case, it is explained by ligand-metal charge transfer.

Moving to organic chemistry. Let's say you have an acyl chloride, $\ce{RCOCl}$, and you want to convert it to an aldehyde, $\ce{RCHO}$. If you calculate the respective oxidation states of the carbonyl carbons, you get +3 and +1 respectively.
So, that tells you what kind of reagent you need to effect this transformation: you need a reducing agent. Hydrogen gas, $\ce{H2}$, is one such example of a reducing agent. Why? Well, that's because of oxidation states again. $\ce{H2}$ has hydrogen in an oxidation state of 0, and when hydrogen forms a bond to carbon, it has an oxidation state of +1.
Oxidation states are key to understanding why redox reactions work and why certain species are reducing agents (e.g. $\ce{LiAlH4}$: H(-1)) and why others are oxidising agents (e.g. DMSO, $\ce{(CH3)2SO}$, in the Swern oxidation).

Your first example seems to be one where you say that you would have been able to use oxidation states for something, except that it doesn't work in that case?

For the second case: Of course if you want to take something that doesn't contain any H and make something that does, you will need to provide a H from somewhere. Of course, since the numbers add up to zero, if you have a reaction where one of the numbers increase, there must be other numbers that decrease. But so what? This would be equally true no matter which arbitrary rules you use to assign the numbers. You could assign them pseudo-randomly for each compound and this would still be true.

First example: I am using oxidation states to show that something does not occur. Second example: All I can say is - if you say so. Also, please stop using all caps in your comments, it's not nice.

No, I don't "say so" -- I'm asking what distinguishes the particular numbers you get from the oxidation-number rules in utility from other numbers that you could get from different rules of the same general kind. In order to be able to conclude something from oxidation numbers, there must be a rule in play somewhere that says "if the oxidation numbers are such-and-such, then you can expect this-and-that" -- where "this-and-that" is not a statement about the oxidation numbers themselves. I haven't seen such a rule formulated and would like to see some.

For example, if I want to turn RCOCl into RCHO, I can either get the hydrogen by supplying H2, or by having CH4 turn into CH3Cl in the same reaction. I guess wildly that the latter reaction doesn't happen, but is there a general rule using oxidation states that will predict for me that one of these reactions will happen and the other doesn't? What does that rule look like?

There isn't such a rule, using oxidation states, that says RCOCl + CH4 -> RCHO + CH3Cl is not possible.

Okay, unlucky guess on my part then. Is there any rule, using oxidation states, that says such-and-such reaction will or won't happen?

@HenningMakholm No.There is no rule as such.

I would say the main relevance of oxidation states is qualitative, not quantitative. Sure enough, it is possible to give a detailed quantitative picture of redox reactions in terms of quantum chemistry without ever mentioning oxidation states, and oxidation states ultimately are an extremely simplified reflection of that picture. However, most chemists most of the time aren't at all concerned with the detailed quantum chemistry picture, because they are working at a significantly higher level of abstraction, and so the question about how an oxidation state translates to a binding energy (cf. the comments to Ivan Neretin's answer) just doesn't come up. (I see you have a Stack Overflow profile. Exaggerating the difference somewhat, I might say that such a question is akin to asking of someone writing a Java program what the program is doing with the x86 registries.)
Two accounts, already mentioned in various ways in other answers, of how oxidation states are qualitatively helpful:

Half-reactions (cf. Marcel's answer). They provide a clear picture of the key changes going on in a redox reaction, and they provide a basis for assigning oxidation and reduction potentials. Half-reactions, however, are an abstraction -- an abstraction which wouldn't be nowhere near as meaningful if they weren't interpreted in terms of oxidation states. For instance, in $\ce{3 e− + 2 H2O + MnO4- -> MnO2 + 4OH−}$ the locus of interest for someone thinking in terms of redox reactions is what is going on with the manganese atoms, and the fact that it is their oxidation state that changes, and not that of the hydrogen or oxygen atoms, reflects that. By the way, even though you don't have to use half-reactions to balance redox equations, it always was my preferred method, as it makes enough sense chemically that it makes it much harder to make silly arithmetical mistakes.
Classification of chemical species (cf. Ivan Neretin's answer). Oxidation states can give an approximate, but very useful, idea of some properties of chemical species, such as redox reactivity. For instance, -1 chlorine compounds can be nice and stable (think table salt, $\ce{Cl-}$), while +1 chlorine compounds tend to be quite strong oxidisers (think bleach, $\ce{ClO-}$). No wonder that occasionally species are named after their key oxidation state (cf. Marcel's answer).

If you want a more direct interpretation in terms of a physical property: the oxidation state loosely corresponds to electron density around an atom, relative to what it would be like if that atom was neutrally charged and isolated. You should not expect an actual formula for getting electron densities out of oxidation states (given how loose the correspondence is), nor to compare absolute electron densities around different atoms using oxidation states (given that the values are relative).
To get a better idea of what kind of correspondence I am talking about, picture a neutral chlorine atom. If an electron is added to its electron shell (thus resulting in a chloride anion, $\ce{Cl-}$, which has -1 charge), an increase in electron density happens. Now, suppose that, rather than simply getting an extra electron, the chlorine atom forms a bond with a neutral hydrogen atom, forming an $\ce{HCl}$ molecule. The situation is quite different from the first one, as the pair of electrons forming the covalent bond is shared by both atoms, in a way that can be accurately characterised in quantum chemical terms (molecular orbitals, etc.). Still, the bond is not symmetrical, but skewed towards the chlorine atom, so that the electron density around it is higher than it was in the neutral, isolated state, though lower than in the chloride anion case (and vice-versa for the hydrogen atom). The major simplification involved in assigning oxidation states is ignoring this subtlety: we say chlorine is in -1 oxidation state in both $\ce{Cl-}$ and $\ce{HCl}$. Finally, it is worth noting that adding an electron to the neutral chlorine atom promotes an electronic reconfiguration (cf. the part about permanganate in orthocresol's answer) which happens to lead to a very stable state (a noble gas electronic configuration). That explains why -1 chlorine compounds are relatively stable next to those with higher oxidation states ($\ce{Cl2}$, $\ce{ClO-}$, etc.), which tend to be powerful oxidisers.
The example illustrates how oxidation states are, from the perspective I am adopting here, an approximation (pretending all bonds are either fully symmetrical or fully skewed towards the most electronegative atom) of an approximation (using charges as proxies for electron density). Still, they are useful to the extent they:

Offer a simple model of electron transfer in redox reactions (and note that this is quite independent from all I said about electron densities: Faraday's work on electrochemistry, for instance, predates quantum mechanics and the discovery of the electron by several decades).
Can capture general trends among chemical species of an element, thus serving as basis for heuristics. Given the many layers of approximation involved, these are heuristics, and not hard rules (which explains why no one has given you hard rules here). To go beyond the heuristics, you have to look at the electronic configuration in each particular case, which, other than in very simple cases (such as $\ce{Cl-}$) is merely hinted at by the oxidation state.

Even qualitatively, it is frustratingly difficult to get anyone to explicitly state one of the rules you all keep telling me you use. What is the "what is going on with the manganese atoms" that you learn from oxidation states? Not the oxidation states themselves, but what does those numbers tell you about the reaction? Or about the manganese atoms? Anything that doesn't boil down to "knowing the oxidation states will tell you how the oxidation states change".

@HenningMakholm I hadn't emphasised that because, at first, I didn't see it as the focus of your question. I have added a second part to the answer, which hopefully will make things clearer.

Back when we were small and the computers were huge, math books for kids were full of tricks for quick mental calculation, like "to multiply x by 9, just multiply by 10 and subtract x" or "to find a square of a number that ends with 5, for example, 65, go like $6\cdot7=42$ and append 25". Well, the oxidation states are much like that. Sure, you may balance pretty much anything without them just as well, but we are used to do it our way.
Besides, knowing your oxidation states gives you a rough idea of what to expect from the compound. Will it oxidize everything, including my skin if it gets a chance? or will it be indifferent? or will it reduce everything, to the point of igniting spontaneously in air? will it react with such-and-such?

"knowing your oxidation states gives you a rough idea of what to expect from the compound" -- how? You seem to claim that these values can be used to make predictions of something that is not ITSELF about oxidation states, but the examples you give ("will it oxidize everyting") are themselves in terms of oxidation states and do not seem to lead out of a closed world where the only "facts" you get are things about "oxidation states" themselves.

You either have the reaction, or you don't. Say, $\ce{Mn^7+}$ is a strong oxidant, hence it belongs to the first category. It will oxidize your skin. This is a real, very palpable fact. It is not about oxidation states. It is about those nice brown spots on you palms which would stay for about a week.

What is the rule you use to predict something from knowledge of oxidation states here? You seem to be claiming that you can conclude from oxidation states that a certain reaction will happen -- what is the rule for getting from "the oxidation states are such-and-such" to "such-and-such reaction is likely to happen"?

Well, it's not much of a rule, but... Long story short, I know my elements and what oxidation states they "prefer" to be in; if they are in unusual states, then something is likely to happen.

Are you getting at something like "here's a way to use oxidation numbers to estimate binding energies of a compound"?

No, far from that. Everything I said is just qualitative. Moreover, it is about elements, not compounds. "In these conditions Fe is going to be +3" - that's a likely conclusion, for example.

Well, then that's not the direction I'm looking for. I'm asking for the step where you go from "Fe is going to be +3" to a conclusion that's not about oxidation states.

If you want a definitive conclusion, then go for quantitative things, like oxidation potentials and Gibbs energies.

I just want some conclusion that's not about the oxidation states themselves.

@HenningMakholm How about "Cu is going to be +2 therefore this compound should be black instead of red"?

(Actually, scratch that, because you could also look at tell by whether you have Cu2O or CuO. But consider a solution instead where you have etiher Cu+ or Cu2+ ions)

Oxidation states are used for "electron bookkeeping". This is useful in redox reactions, where we need to know what species are oxidised and reduced. By extension, oxidation states are an important tool for determining reacting ratios.
For example:
$\ce{Zn(s) + Cu^2+(aq) -> Zn^2+(aq) +Cu(s)}$
$\ce{Cu: +2->0}$
$\ce{Zn: 0-> +2}$
Hence we know that zinc is oxidised, and copper is reduced. While this is obvious from the reaction equation, this will not be the case in "messier" redox reactions, and oxidation states help us understand what is going on.
They also allow us to distinguish between compounds by using oxidation numbers - for example it is the oxidation number that allows us to tell the difference between iron (II) oxide and iron (III) oxide.
We should keep in mind, though, that oxidation states are a purely theoretical concept, and do not actually exist as physical quantities. When we use oxidation states, we are essentially assuminig that all bonds have a purely ionic character - this is rarely the case.

This is another "how do I find them?" explanation that says nothing at all about "what do I use them for once I have found them?"

I gave two examples of uses for oxidation states: differentiating between different compounds of the same two elements, and keeping track of electron transfer in redox reactions.

"differentiating between different compounds of the same two elements" apparently mean that you use the oxidation states to know what the oxidation states are! That's not an application as far as I can see. Your "keeping track of electron transfer in redox reactions" d0esn't look like an application either. You write down those two extra lines under the reaction, but WHAT DO YOU GET OUT OF WRITING THOSE LINES? Again, the claimed benefit of knowing the oxidation states simply seems to be that now you can write them down -- what then?

We distinguish between the compounds by using the knowledge that they have different oxidation states.

A change in oxidation states is the very definition of a redox reaction - hence we need oxidation states to keep track of what is happening in the reaction.

(2), yes, so saying "redox reaction" is not an answer to how one uses oxidation states to predict something that is not itself about oxidation states. Saying that "we need oxidation states to keep track of the oxidation states" is just tautological. I'm asking for an application where oxidation states makes a prediction where determining whether the prediction came true or not does not involve knowing anything about oxidation states.

And (1) is again just a way to say "we need to know the oxidation states because otherwise we wouldn't know what the oxidation states are".

(1) is a way of differentiating between compounds.
I suggest you take a look at Orthocresol's answer, this is going nowhere. You seem to be under the impression that oxidation states are an intrinsic property of nature - this is not the case. They are simply a way we have invented for describing  and understanding interactions that involve electron transfer.

"a way of differentiating between compounds" -- again if the only thing the oxidation states tell you is that some compounds have some arbitary numbers attached to them and other compounds have other arbitrary numbers attached to them, then it's just a parlor game. Knowing what the oxidation states in different compounds are is not an application. An application would be something that allowed me to conclude a difference between the compounds that is not about oxidation states.

Even if oxidation numbers had been arbitrary, which they are not (see Orthocresol's answer below), they are still a way of distinguishing between compounds of the same element. Hence they have an undeniable practical use - for example, farmers need to know whether the fertilizer they are using contains iron (II) oxide or iron (III) oxide.

Wait, when you say "distinguishing", do you merely mean that oxidation numbers are used for naming compounds?

In that specific case, yes.

For simple compounds of elements with vastly different electronegativity oxidation states are useful to distinguish compound by oxidative/reductive power. Said powers may, to some extent, be used to predict how likely particular reactions to proceed. For example, halogens in higher oxidation states are all good oxidizers and carbon is a reducer. Thus, oxohalogenates are likely to react with carbon, sometimes violently.

Oxidative power manifests as electrode potential and thus objectively exists. Sometimes, however, electrode potential is really hard to obtain or can be measured only in exotic conditions.
Similarly, each element has 'preferred' oxidation states, corresponding to most stable compounds.
This use, however, is relevant only in inorganic chemistry of simple compounds of elements with highly different electronegativity. When catenation is involved or elements connected have close electronegativity, formal electronegativity loses predictive power. The situation can be somewhat salvaged by ad-hoc 'corrections', like certain atom neighbors stabilizing higher/lower/nonstandard oxidation states, but ...

Oxidation states are useful for balancing redox equations, though there the order of electronegativity used to calculate them does not matter.

Admittedly, it IS possible to balance equation without them, either ad-hock, or using half-reaction method.
Oxidation states are of marginal use in organic and metaloorganic chemistry and in chemistry of polyhedral clusters. On the other hand, definition 'compounds of elements with strongly different electronegativities' covers most common chemicals and everyday substances. Thus, oxidation states ARE useful.

For (2), can you give an example where the oxidation states gives us something that is not already a consequences of counting nuclei of different elements?

For (1), "distinguish compound by oxidative/reductive power" is itself a statement about oxidation states. I agree that if you know the oxidation states assigned to atoms in different compounds, you can figure out whether that number increases or decreases during a reaction, but so what? Does nature care about that number increasing or decreasing? I'm looking for an application that is not about oxidation states themselves, one that predicts whether this or that will happen, where "this or that" is not just a rephrasing of a claim about oxidation states themselves.

1.use of oxidation number in nomenclature.
You must have heard of Iron(III) sulphate Fe2(SO4)3,the reason of writing it as Iron(III) sulphate is because each iron atom have a oxidation state of +3 here.
Further,we indicate Magnetite (Fe3O4) as iron(II,III)oxide this is because two iron atom shows an oxidation state of +3 and one iron atom shows an oxidation state of  +2 in Magnetite.
2.Oxidation states of elements are helpful in balancing redox reaction.The elementry reactions such as Mg+O2 ----->MgO can be balanced by simply hit and trial but when it comes to balance a redox reaction,the oxidation states of elements involved should be known.It simply helps in determining which copound will behave like oxidising agent and which will behave like reducing agent.
I hope you got my message.

Oxidation states are based on oxidation numbers. An oxidation number is the charge, or apparent charge, of an atom or group of atoms when combining in a compound.
The oxidation number of an atom is determined by the octet rule. This states that atoms are particularly stable when they have eight electrons in their outer shell. Outer shell is somewhat of a simplification compared to the quantum mechanical model of the atom but it works very well for particular purposes. What's important is that these outer electrons are the way the atom presents itself to the world and interacts with other atoms. As you progress up the periodic table, each atom has an additional electron to electrically balance the additional proton. These electrons are added in such a way that there are never more than eight in the outer shell. One notable exception to the octet rule is the very first shell which can only hold a maximum of two electrons.
Let's begin with oxygen which has eight electrons. Two of these are in the innermost shell leaving six in the 2nd shell which in oxygen's outer shell. Byt the octet rule, oxygen needs 2 more electrons to give it an octet. Its oxidation number is -2. Why negative? If it gains these electrons to complete its octet, it will then be negatively charged.
Let's now look at sodium. 2 electrons in the 1st shell, 8 in the 2nd and 1 in the 3rd shell. So Na needs to gain 7 electrons to form an octet in the 3rd shell. That's asking for too much, its nucleus isn't up to the electrical task. But what if it loses that lone outer electron? Then the 2nd shell becomes the outer shell and, since it has an octet, the Na will be stable. It will also be charged +1 which is sodium's typical oxidation number.
So where does oxygen get the two electrons it need to form an octet? To where does sodium dump its electron to form an octet? One way is for two sodium atoms to each give up an electron to one oxygen atom. This means sodium and oxygen would be expected to combine in a 2 to 1 ratio. Indeed, Na2O is formed since the now negative oxygen is attracted to the now positive sodiums. In this compound the oxidation state of oxygen is -2, that of the sodium is +1.
It's simplistic to imagine the electrons are completely transfered except in the case where ions are formed. In water, electrons are shared between two hydrogen atoms and one oxygen atom. Each hydrogen needs one more to fill its outer shell (the 1st shell needing only two to fill). The oxygen, as usual, needs 2 nore to make an octet. So, H2O results. But the electrons are not shared equally, the oxygen has a tighter pull on them and gains a partial charge of -2, its oxidation number in water. The hydrogens get to share one of those electrons of oxygen to complete its outer shell but both that electron and the one the hydrogen had of its own are pulled away from the hydrogen toward the more powerful attraction of the oxygen. This results in a partial positive charge on the hydrogen compared to its atomic state. Oxidation number, and oxidation state, -1.
This is just the beginning of a complete discussion of this. The bottom line is that oxidation numbers help us understand, and write chemical formulas for, chemical compounds. Proper formulas help us write and balance chemical equations. Proper chemical equations help us pass Chemistry class and graduate.

It's been years since I got my BS in chemistry (which I regrettably haven't used since), but there's two contexts where oxidation states come up --
(1) as emphasized in many of the answers, this can provide a shorthand for balancing chemical equations in general. In this usage, it's not that great or that necessary at this point in time.
(2) More specifically, it's useful for characterizing the states of metal atoms in primarily ionic bonds. This is in a sense a variation on the more general answer, but here it can genuinely be useful. From what I remember, there are many occasions where you're looking to trade out the metallic part of an ionic bond (of course "ionic" vs. "covalent" is a bit of an abstraction), but metal ions behave differently depending on oxidation state. One specific different is their electromagnetic properties -- the difference between oxidation states reflects a difference in the arrangement of their electrons (to an abstraction). Thus, ionization states provide a way to eyeball that you're going to balance things out by just looking at that number. Want to precipitate out gold? Knowing The oxidation state will tell you the easiest reaction.
Particularly when dealing with salts, we often don't care (too much) about what's on the other side of the Cl- we are going to use for our reaction, but we do care what the oxidation state is to know how much X+Cl- to add.

Question: How many potassium and how many calcium do I need to neutralize sulphuric acid?

Answer without oxidatin numbers: How the hell I can know?
Answer #2 without oxidation numbers: Just mix it and see what happens.
Answer with oxidation numbers:

Sulphuric acid contains sulphur in oxidation state $\ce{S^{VI}}$ and acids, except for halogen ones, are formed with appropriate ammount of oxygen and hydrogen. Rule of thumb taught to kids is "divide the oxidation number of the [suplhur] by two. Round it up and you have number of oxygens. then add one hydrogen. If you don not need to round, add one oxygen and two hydrogens."
Therefore sulphuric acid is
$$\ce{H_2SO_4 = H_2^IS^{VI}O_4^{-II}}$$
and in water it dissolves
$$\ce{H_2SO_4 + 2H_2O -> (SO_4)^{-II} + 2(H_3O)^+}$$
Now move to our question, how to neutralize this acidic solution? Write reactions:
$$\ce{(SO_4)^{-II} + 2(H_3O)^+ + Ca^0 -> Ca^{II}(SO_4)^{-II} + H_2^0 + 2(H_2O)^0}$$
$$\ce{(SO_4)^{-II} + 2(H_3O)^+ + 2K^0 -> K_2^{I}(SO_4)^{-II} + H_2^0 + 2(H_2O)^0}$$
The answer is that for neutralisation of one mole of sulphuric acid one need one mole of calcium or two moles of potassium.
tl;dr
The oxidation numbers and other rules tells you thats $\ce{C_2O_5}$ doesn't exist and why. Same for $\ce{CaSO_4, KS, HO_4}$, etc.

Without oxidation numbers? If I have H2SO4 and deprotonate it fully, I'll end up with a sulfate ion SO4(2-). In order to make that into a neutral salt together with Ca(2-) ions there needs to be one calcium per sulfate. In order to make it into a neutral salt with potassium I'll need two K- per sulfate. No oxidation states there, just plain old ion charges.

Oxidation states are where we pretend that sulfate is made from one S(6+) ion and four O(2-) ions. This pretense seems to be to be completely useless in your example; the sulfate never breaks apart during the reaction and can conveniently be treated like an unit when doing the balancing.

@HenningMakholm Got you! You have used oxidation numbers. $SO_4(2-)$ addresses that.

x @Crowley: No I haven't. -2 is the electric charge of the sulfate ion, and I don't dispute that charges are useful to know.

@HenningMakholm Try to make the molecule neutral using available electronic configurations of the atoms.

x @Crowley: Why? Simply balancing the charges works perfectly in this case.

@HenningMakholm Then what is the difference between ionic charge and oxidation number?

Let us continue this discussion in chat.

x @Crowley: An oxidation state is a number assigned to a single atom, which is what its charge would be if all of the covalent bonds it participates in were ionic instead. I'm asking what this pseudo-charge is useful for. I agree that actual charges are useful.

@HenningMakholm: It useful to operate with covalent bonds without calculating the electron density functions in the bond. It is like using $S=\pi r^2$ for circle instead of calculating the integral everytime you need the area of circle.


text:
Henry's law and O2 concentration in alveoli blood
Can Henry's law be applied to find $\ce{[O2]}$ in alveoli blood? This kind of question appeared in an another forum and I've been struggling to find a definitive answer. My initial thought was 'no', because Zumdahl's Chemical Principles book says that

Henry's law is obeyed most accurately for dilute solutions of gases that do not  dissociate or react with the solvent. $\ce{O2}$ will bind to the haemoglobin in the blood. Given this reaction in the solvent, $\ce{O2_{(g)}}$ in blood does not follow Henry's law.

However, a quick google search turned up multiple sources that use Henry's law directly to find the dissolved oxygen concentration (which is about $2\%$ of total oxygen):
http://www.umich.edu/~projbnb/cvr/O2transport.pdf
https://www.ncbi.nlm.nih.gov/books/NBK54103/
What I failed to understand is why the dissolved oxygen is linearly proportional to the partial pressure while the oxygen dissociation curve is not. Is there something I'm missing?

Possibly helps: https://chemistry.stackexchange.com/questions/98048/measure-of-partial-pressure-on-blood

Henry's law states that the amount of a gas that dissolves in a liquid
is directly proportional to the partial pressure of that gas.
Oxygen has a larger partial pressure gradient to diffuse into the
bloodstream, so it's lower solubility in blood doesn't hinder it
during gas exchange.  Therefore, based on the properties of Henry's
law, both the partial pressure and solubility of the oxygen and carbon
dioxide determine how they will behave during gas exchange.

I encourage you to read the article below.
Source: Boundless. “Henry's Law.” Boundless Anatomy and Physiology. Boundless, 21 Sep. 2016. Retrieved 16 Oct. 2016

Thanks, the article was very insightful but did not properly address my question. But it got me thinking that air pressure in the lungs might stay fairly constant and thus the oxygen dissociating into hemoglobin will not affect the dissolved oxygen predicted by Henry's law. Might this be a correct line of thought?

Old, but I'll take a stab at explaining my take on your question about linear proportionality for any future readers:
Henry's law gives us the direct proportionality for dissolved oxygen in the plasma. The key here is understanding that we are talking about dissolved oxygen in the plasma. This is distinct from oxygen bound to hemoglobin. As oxygen partial pressure increases, more of it is "pushed" into the plasma. If you just had plasma with no RBCs, that would be it. There would be a specific coefficient of Henry's law for the plasma, but it would otherwise be no different than dissolved oxygen in water.
The nonlinear part comes in with the Hemoglobin. Hemoglobin is not directly effected by the partial pressure of oxygen, but by the amount of oxygen available to it in the plasma (which we know is driven by partial pressure of oxygen). Hemoglobin essentially removes takes some of the dissolved oxygen from the plasma. At low pressures, there isn't much oxygen around, so hemoglobin can only slowly (and nearly linearly) increase its uptake. The more available to it, the more it will take (to a point). s partial pressure of oxygen increases, more oxygen dissolves in the plasma, so hemoglobin has more to take, and it does The cool thing about hemoglobin is the "positive feedback" that occurs. Once one molecule of oxygen is bound the next is more likely to bind. This gives the curve its exponential-like region. But remember, it will only continue to bind to a point. It maxes out at 4 molecules oxygen per molecule of hemoglobin. This is why the sigmoid curve has an upper asymptote. As more and more molecules of hemoglobin "max out", the curve flattens out.
Figure 1 in this article is a nice graph of the two distinct mechanisms of oxygen carrying by the blood.

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text:
How to calculate the ratio of rate constants of competing reactions?

There are two compounds $\ce{A}$ and $\ce{B}$ that react with one common reagent $\ce{C}$ following first order kinetics.
  Suppose 99.9% of $\ce{A}$ must react before 0.1% of $\ce{B}$ has reacted then what is the minimum ratio of their respective rate constants?

My Attempt at the question:
We know that the reactions follow first order kinetics, hence I tried to find the time at which the given percentage of $\ce{A}$ and $\ce{B}$ is left using the formula
$$ t = \frac{2.303}{k} \cdot \log\left(\frac{C_0}{C_1}\right)$$
where $t$ is the time, $k$ is the rate constant for each reaction and $C_0$ and $C_1$ are initial and final concentration respectively.
After this I tried to find the ratio of rate constants directly but couldn't arrive at the answer.
I derived the above formula using the integrated rate law.

Your approach seems right. You know the values of $A_{0}$ and $A_{t}$; and $B_{0}$ and $B_{t}$. You can equate the two $t$'s to compute the ratio between $k_{A}$ and $k_{B}$.

I tried that. But not arriving on any answer. Try out the calculations and steps. It's tricky and confusing after a few steps. Try it out for yourself and tell me your approach

For the minimum ratio of rate constants, we take the first instant of time that can be measured. Wherein, 99.9% of A has reacted and 0.1% of B has reacted.
Therefore for A
$$\dfrac{C_0}{C_1} = \dfrac{1}{1 - 0.999}$$
Similarly for B, we have
$$\dfrac{C_0}{C_1} = \dfrac{1}{1 - 0.001}$$
Therefore,
$$k_At = 2.303 \log\left(\frac{1}{(1-0.999)}\right)$$
$$k_Bt = 2.303 \log\left(\frac{1}{(1-0.001)}\right)$$

This article may be helpful too "Determination of the Rate Constant Ratio in Competitive Consecutive Second-order Reactions" by
W. G. McMillan

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text:
Is it possible for the forward and reverse reactions to have different reaction orders?
For example, in the case of non-dissociative adsorption, the rate of the forward process is given by
$$\frac{\mathrm{d}\theta}{\mathrm{d}t} = kCN(1−\theta)$$
where $C$ is the concentration, $N$ is the number of surface sites, and $\theta$ is the fraction of filled sites.
The reverse reaction is given by
$$\frac{\mathrm{d}\theta}{\mathrm{d}t} = −kN\theta.$$
Hence, it seems like the first reaction is second-order, while the second is first order. This seems strange to me.

But $\mathrm N$ is constant?

N is constant, but $\theta$ is not

The usage of \ce should be limited to typesetting chemicals and chemical equations - please don't use it to enclose mathematical expressions - these ought to be in italics.

@orthocresol Thanks! What does the \mathrm you used mean?

Math, roman type; i.e. no italics. In this case, it is only suitable for the $\mathrm{d}$ in differentials, which should be upright.

Why is this strange?
Consider:
$$\ce{2A <=>[$k_1$][$k_{-1}$] A_2}$$
Forward rate is second order: $k_1[\ce{A}]^{2}$.
Reverse rate is first order $k_{-1}[\ce{A2}]$.

IMHO it is very problematic to speak about 'the reaction order.' If you just say 'the order' of the reaction without specification against what the order is specified, this will lead to confusion very quickly. After all, there can be several reactants/products involved in the reaction. And at equilibrium the order of the reaction and reverse reaction should be the same against every component, respectively.

It's not that bad when there's only one reactant, but yes, in general, we should specify the reaction order w.r.t. some reactant.


text:
EDTA Titrations
I was doing some questions on EDTA titrations and was wondering how you actually derive the second equation here:
$$\begin{align}
α_{\ce{Y^4-}} &= \frac{[\ce{Y^4-}]}{[\ce{H6Y^2+}] + [\ce{H5Y+}] + [\ce{H4Y}] + [\ce{H3Y-}] + [\ce{H2Y^2-}] + [\ce{HY^3-}] + [\ce{Y^4-}]}\\[10pt]
&= \scriptsize \frac{K_1K_2K_3K_4K_5K_6}{[\ce{H+}]^6 + [\ce{H+}]^5 K_1 + [\ce{H+}]^4 K_1K_2 + [\ce{H+}]^3 K_1K_2K_3 + [\ce{H+}]^2 K_1K_2K_3K_4 + [\ce{H+}] K_1K_2K_3K_4K_5 + K_1K_2K_3K_4K_5K_6}
\end{align}$$
where $\ce{Y^4-}$ is the fully deprotonated form of EDTA and $K_i$ represents the $i$-th dissociation constant of $\ce{H2Y^2+}$.

@Joseph.L Was this equation given or did you find it somewhere?

I posted the general solution for the degree of dissociation here:
$$\ce{\alpha(H_{n-m}A^{m-})} = \frac{x^{n-m} \prod_{j=1}^{m}k_{j}}{\sum_{i=0}^{n}x^{n-i} \prod_{j=1}^i k_{j}}$$
with $x = [\ce{H+}]$, $n>0$ as the number of protons and $0 \le m \le n$ as the number of dissociated protons. But from the beginning.

If EDTA would be a simple acid that has four protons, this would look like:
$$\alpha(\mathrm A^{4-}) = \frac{k_1 k_2 k_3 k_4}{x^4+x^3 k_1+x^2k_1 k_2+x k_1 k_2 k_3 + k_1 k_2 k_3 k_4}$$
This derives from solving a linear equation system based on the mass balance:
$$c_0(\ce{H_nA}) = \sum_{i=0}^n c(\ce{H_{n-i}A^{i-}})$$
and all the respective equilibrium/acid constants:
$$k_{a,i} = \frac{c(\ce{H3O+})~c(\ce{H_{n-i}A^{i-}})}{c(\ce{H_{n-i+1}A^{(i-1)-})}}$$
What you will get from solving this, is an equation for the concentration of every deprotonated EDTA subspecies, as well as for EDTA itself:
$$c(\ce{H_{n-m}A^{m-}}) = \frac{c_0(\ce{H_nA}) \prod_{i=1}^{m} k_i}{\sum_{i=0}^{n}x^{n-i} \prod_{j=1}^i k_{j}}$$
As the degree of dissociation is defined as "the concentration of the most deprotonated species" devided by "the sum of concentrations of all species of the respective acid in solution", it would be for EDTA with four protons:
$$\alpha(\ce{A^{4-}}) = \frac{c(\ce{A^{4-}})}{c_0(\ce{H_nA})} = \frac{c_0(\ce{H_nA}) \prod_{i=1}^{m} k_i}{\sum_{i=0}^{n}x^{n-i} \prod_{j=1}^i k_{j}} \left(\sum_{i=0}^n c(\ce{H_{n-i}A^{i-}})\right)^{-1}$$
Which then ends up into the equation as stated above:
$$\alpha(\mathrm A^{4-}) = \frac{k_1 k_2 k_3 k_4}{x^4+x^3 k_1+x^2k_1 k_2+x k_1 k_2 k_3 + k_1 k_2 k_3 k_4}$$

As in my eyes one can easily see now, is that your equation somehow derives as mine, but with the assumption that EDTA has six protons that can get deprotonated. But following my way, it would look a bit different:
$$\alpha(\mathrm A^{6-}) = \frac{k_1 k_2 k_3 k_4 k_5 k_6}{x^6 + x^5 k_1+x^4 k_1k_2+x^3k_1 k_2 k_3+x^2 k_1 k_2 k_3 k_4+x k_1 k_2 k_3 k_4 k_5 + k_1 k_2 k_3 k_4 k_5 k_6}$$

If at all, I think, the proper way of handling EDTA completely would be to take it as an amphoteric compound that can either act as an acid and "lose" four protons or act as an base and "take up" two more protons. The mass balance is nearly the same but the two protonation steps are added:
$$\ce{c_0(H_4A)} = \sum_{i=0}^4 c(\ce{H_{4-i}A^{i-}}) + \sum_{i=0}^2 c(\ce{H_{4+i}A^{i+}})$$
And the system of all four equilibrium/acid constants will be expanded with the two equilibrium/base constants for the protonation steps. So for $i=1\ldots4$:
$$k_{a,i} = \frac{c(\ce{H3O+})~c(\ce{H_{4-i}A^{i-}})}{c(\ce{H_{4-i+1}A^{(i-1)-})}}$$
and for $i=1\ldots2:$
$$k_{b,i} = \frac{c(\ce{OH-})~c(\ce{H_{4+i}A^{i+}})}{c(\ce{H_{4+i-1}A^{(i-1)+})}} = \frac{k_{\mathrm W}/c(\ce{H3O+})~c(\ce{H_{4+i}A^{i+}})}{c(\ce{H_{4+i-1}A^{(i-1)+})}}$$
In that case, the result would be:
$$\alpha(\mathrm A^{4-}) =\\ \frac{x^4 k_\mathrm W^2}{x^6 k_{\mathrm b,1} k_{\mathrm b,2}+x^5 k_{\mathrm b,1} k_{\mathrm W}+k_{\mathrm W}^2 \left(x^4+x^3 k_{\mathrm a,1}+x^2 k_{\mathrm a,1} k_{\mathrm a,2}+x k_{\mathrm a,1} k_{\mathrm a,1} k_{\mathrm a,3}+k_{\mathrm a,1} k_{\mathrm a,1} k_{\mathrm a,3} k_{\mathrm a,4}\right)}$$


text:
Why some hydrates are coloured while other are not
I was studying about different vitriol and noticed that some hydrates are coloured while some are colourless even after having water in their crystals.
For example; $\ce{FeSO4.7H2O}$ (iron sulfate heptahydrate) has a green colour while anhydrous iron sulfate has no colour, similarly $\ce{CuSO4.5H2O}$ (copper sulfate pentahydrate) has a blue colour while anhydrous copper sulate has no colour. When we come to zinc sulfate,it has no colour in all of its three hydrates (the most common one being the heptahydrate).
Why does zinc sulfate not have any colour like $\ce{FeSO4.7H2O}$ or $\ce{CuSO4.5H2O}$ regardless of whether it has water of crystallisation?

i don't know which tags r suitable so please add suitable tag if u know.

For $\ce{Zn^2+}$, the answer is simple: Zn(II) is $\mathrm{d^{10}}$, therefore no d-d* transitions.

it's either d-d transitions or charge transfer transitions and it all depends on the d-orbital occupancy and the geometry. I feel like this question has likely been answered on here already though.

Is this d-d transition related to d block elements only??


text:
Is entropy change of the system for reversible process same as entropy change of the system for irreversible change?
Is entropy change of the system for reversible process same as entropy change of the system for irreversible change? Is Q for reversible change equal to Q for irreversible change? Or the change in entropy of the surrounding is same in any case because surrounding is bigger than system, so the change in surrounding is small?
And finally, does internal energy of the system depend on entropy of the system and how?
I registered just so I could ask this question. I would be very grateful for answer.

There are a lot of questions here. Can you narrow it down a bit?

Supposing the heat is transferred from surroundings to the system irreversibly, if we want to calculate entropy change of the system we have to find reversible path for the change, so that means that change in entropy for the reversible path found and previous irreversible is the same. But shouldn't for spontaneous irreversible process  entropy change be greater than for reversible path?

For the same initial and final states of the system, the entropy change doesn't depend on whether it is a reversible path or an irreverisible path.  However, for an irreversible path, the integral of dq/T (where T is the temperature of the surface through which the heat transfer is occurring) will be less than for the reversible path (because entropy is being generated within the system to make up the difference).

Being a state function, entropy is independent of path, so the entropy change of the system is indeed the same for both a reversible and an irreversible process. The transferred heat, however, is not, and does depend on process.
You've probably heard mentioned that the entropy change through a reversible process is less than that through an irreversible process. At first glance this proposition seems suspicious, because we've just said that as a state function entropy is independent of path$-$the resolution is that the entropy change of the system remains the same regardless of process, but the entropy change of the environment is indeed different, such that the total entropy change is greater for an irreversible process than for a reversible process. [Interpreted differently: the environment is in a different state after the system undergoes a reversible process than after it undergoes an irreversible process.]
The internal energy does depend on the system's entropy$-$by the first law we may write $$\mathrm{d}U = T\,\mathrm{d}S + \mathbf{f\cdot\mathrm{d}X}$$ (where $\mathbf{f\cdot\mathrm{d}X}$ represents a generalized work term; commonly this is $-p\,\mathrm{d}V$), so we see that $U$ is a natural function of $S$ and $\mathbf{X}$; $U = U(S,\mathbf{X})$. To give a functional form for $U$ as a function of $S$ and $\mathbf{X}$, we'll need in general an equation of state, an equation for energy, and a type of process$-$let's restrict ourselves to the case of an ideal gas, for which the two equations are $$pV = nRT \quad \text{and} \quad U = \frac{3}{2}nRT$$ (assuming also a monatomic gas), and work out how $U$ varies over an isobaric process. We want to evaluate $\mathrm{d}U = T\,\mathrm{d}S - p\,\mathrm{d}V$; let's start by working out $-p\,\mathrm{d}V$: $$-p\,\mathrm{d}V = -p\,\mathrm{d}\left(\frac{nRT}{p}\right) = -nR\,\mathrm{d}T = -\frac{2}{3}\,\mathrm{d}U.$$ Now plug that into our starting equation, substitute, and rearrange... $$\frac{5}{3}\,\mathrm{d}U = T\,\mathrm{d}S = \frac{2U}{3nR}\,\mathrm{d}S \quad \Longleftrightarrow \quad \frac{5}{2}nR\,\frac{\mathrm{d}U}{U} = \mathrm{d}S,$$ which integrates to $$\Delta S = \frac{5}{2}nR\ln\left(\frac{U_2}{U_1}\right) = \frac{5}{2}nR\ln\left(\frac{T_2}{T_1}\right).$$ This is no different from working out $q_\text{rev}$ for such a process and then using $\Delta S = q_\text{rev}/T$; the role of energy is simply more explicit in this approach.


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Attack at ketone vs carboxylic acid carbonyls

Why in this the nonbonding electrons of N group does not attack C2 as I think it is more electropositive ?
Then how pyruvic acid is converted to alanine.

Firstly your numbering is the wrong way round. Carboxylic acid should have higher priority. In any case the carboxylic acid carbon is less electronegative (mesomeric donation of OH oxygen lone pair into C=O $\pi^*$) and this is especially so when you consider that with ammonia, the carboxylic acid should be mostly deprotonated.

@orthocresol then how pyruvic acid is converted to alanine

It is a process called reductive amination, look it up.

@orthocresol so here reductive amination takes place or not

Did you consider reduction to lactic acid (A) and subsequent formation of lactide (C) upon heating?

@KlausWarzecha no I mean reduction of original compound that is pyruvic acid .

The keto group is more reactive compared to the ester group due to the mesomeric effect (+M effect) caused by the hydroxyl group in the ester functionality, as already pointed out by orthocresol. See also  https://en.wikipedia.org/wiki/Carbonyl_reduction. Generally:
$$
 \mathrm{acylchloride} > \mathrm{anhydride} > \mathrm{aldehyde} > \mathrm{ketone} > \mathrm{ester} \approx \mathrm{carboxyl} > \mathrm{amide}
$$
A very detailed discussion on the subject is given for example in Warren Organic Synthesis: Strategy and Control or in Francis A. Carey Advanced Organic Chemistry - Part A.

Does here reductive amination takes place or not