text:
Why is methoxy group an electron donating group?
I am studying acidity of carboxylic acids and I found that benzoic Acid is more acidic than p-methoxybenzoic acid. It is written in my book that this happens because methoxy group is an electron donating group. I can understand why electron donating groups can reduce acidity of an carboxylic acid but I can't understand why methoxy group acts as an electron donating group?

As a rule of thumb, if the atom that is attached to the benzene ring is surrounded by more electronegative atoms, the group withdraws electron from the ring. (-NO2, -C=O, -CN, -COOH). And, if the atom directly attached to the ring is surrounded by less electronegative ones, the group donates electron (-NH2, -OR, -OH, -R, etc.)

The lone pair on the oxygen delocalizes into the conjugated aromatic ring.
Edit (based on Jan's suggestion):
Notice that the answer is actually a lot more complicated. A methoxy substituent actually impacts the ring electronics via two competing effects. The oxygen's lone pair is well-placed to delocalize and increase electron density within the ring's conjugated system. This allows delocalization to better stabilize positive charges. So the methoxy is electron-donating from a resonance perspective.
On the other hand, because oxygen is quite electronegative, the methoxy group is electron-withdrawing in an inductive sense via the $\sigma$ bonds.
Take a look at a linear free energy relationship (LFER) plot. For acidity of benzoic acids, the methoxy group has a positive $\sigma_{\mathrm{meta}}$ value but a negative $\sigma_{\mathrm{para}}$ value.
The meta $\sigma$ value measures the effect when there is no resonance structure that delocalizes effectively to the carboxylic acid, and its positive value tells us that the methoxy is electron-withdrawing in that position. The negative para $\sigma$ value tells us that when there is effective conjugation into the carboxylic acid, the methoxy is electron-donating. Notice that it is still electron-withdrawing inductively, but the resonance donation is much stronger.


text:
Does the shielding effect affect atomic bonding?
Electrons within an atom don't experience the full charge of the nucleus due to the shielding effect? Likewise, when creating atomic bonds, does the shielding effect affect the attraction between two bonding atoms?
Thanks for any help offered

Yes, it does. It is due to the poor shielding effect of f-orbital electrons that we see the inert pair effect; that is, $\ce{Pb^2+}$ is more stable than $\ce{Pb^4+}$. Lower oxidation states are more stable as we go down the 13, 14 & 15 group. This affects the number of bonds that an atom can form.

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text:
Bond length comparison between sulfur tetrafluoride and thionyl tetrafluoride

I'm comparing sulfur tetrafluoride and thionyl tetrafluoride. The lone pair has more s character compared to a double bond. Therefore, sulfur in $\ce{SF4}$ should have a greater s character compared to sulfur in $\ce{SOF4}$.
Since $\ce{SF4}$ has a greater s character it's bond length would be comparativly smaller than that of $\ce{SOF4}$.
Therefore, $\ce{X}>\ce{Y}$. But my book says otherwise.
Did I go wrong anywhere?

The simplest way to compare these two compounds is to compare sulfur’s oxidation state. In $\ce{SF4}$, sulfur is in the $(\mathrm{+IV})$ state. In $\ce{SOF4}$, it is $(\mathrm{+VI})$. The higher a compound’s oxidation state, the higher its formal positive charge and thus the more stable (i.e. lower in energy) its orbitals are. Orbitals reduced in energy means contracted orbitals which in turn leads to shorter bonds.
Other than that, the two compounds are pretty identical. Both feature two two-electron-two-centre $\ce{S-F}$ single bonds and one four-electron-three-centre $\ce{F\bond{...}S\bond{...}F}$ bond; the latter can be understood with the following resonance depiction:
$$\ce{F^-\ \ S^+-F <-> F-S^+\ \ F-}$$
Even the structures are comparable although the angles are slightly different.

explain this please:             higher its formal positive charge and thus the more stable (i.e. lower in energy) its orbitals are

@PrakharSankrityayan Orbitals have an orbital energy. A lower energy is called ‘more stable’. Electrons, which populate orbitals, are negatively charged. If the atom is partially positively charged, electrons are better attracted to it. Hence, their orbitals are lower in energy.

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text:
Possible to find expression for the energy of mixing with the Helmholtz free Energy?
I am aware that the energy of mixing can be found using the thermodynamic identity for the Gibbs free Energy and plugging in the expression for the entropy of mixing which finds;-∆G_Mixing≈-TNk (xlnx+(1-x)ln(1-x)) with particles of type A: N_A= (1-x)N and type B: N_B= xN. I was wondering if, when the temperature, number of particles and volume (one type of the particles being in a much greater presence) is held constant, the Helmholtz free energy could be used to find the energy of mixing?


text:
Symmetry forbidden cycloaddition in the Wittig reaction?
My textbook (Klein) describes the mechanism for the Wittig reaction as a [2+2] cycloaddition, followed by a collapse of the ring.

As far as I'm aware, [2+2] cycloadditions are symmetry forbidden, unless the reaction is light-initiated, which the text does not mention. Is the book wrong? If not, what is going on here?

What if the reaction is not concerted? ;)

Then the reaction is not a cycloaddition and a completely different mechanism is needed? (I learned that pericyclic reactions are always concerted, but I may be wrong...)

Very interesting question! I'd like to point out that a thermal $[2_\mathrm{s} + 2_\mathrm{s}]$ cycloaddition is Woodward-Hoffmann forbidden, but a $[2_\mathrm{s} + 2_\mathrm{a}]$ cycloaddition is (e.g. the reaction of a ketene with an alkene proceeds in this manner). See: Why is the Wittig reaction syn stereoselective? Also, Grossman talks about this a bit in his book The Art of Writing Reasonable Reaction Mechanisms. I've attached the relevant pages here.

The other question to ask is: what is the symmetry of the orbitals on the phosphorus ylide?

I think, the simplest explanation here is that the reaction is not concerted. But then remember that the Wittig ylide is not exactly identical to your average alkene either.

There's some evidence that it's not concerted, but far more evidence that it is. At least, that's what I gather from nearly every org chem textbook I've read. March in particular has references to primary literature.

@roymend, as for your earlier comment, all pericyclic reactions are indeed concerted. However not all cycloadditions are concerted, and therefore not all cycloadditions are pericyclic. In this case, the others are hinting that a 2-step cycloaddition can exist, and it would be neither pericyclic nor concerted.

For a somewhat recent review of the mechanism of the Wittig reaction proposed in the past, have a look at Chem. Soc. Rev., 2013, 16, 6670-6696 (DOI)


text:
Is it possible, in principle, for a second row homonuclear diatomic system to show a bond order of 4 in an excited state?
Is it possible, in principle, for a second row homonuclear diatomic system to show a bond order of 4 in an excited state?  If so show the energy level diagram for such a system?
I am confused about this problem, mainly because I don't understand what it would mean for a diatomic system to be in an excited state.  Could I just move an arbitrary number of electrons from the n=2 energy level to the n=3 level?

The normal excited state is one in which a single electron is promoted from a lower orbital into one of the anti-bonding ones. Thus the situation is initially $\cdots g^2$ for 2 electrons in a ground state orbital g (need not be the highest orbital but often is), and, say, after absorbing a photon of the correct energy the situation is $\cdots g^1a^1$ where a is an anti-bonding orbital. This combination of orbitals is called the excited state. (If the electron spin remains paired it is called a singlet excited state, if unpaired ('parallel') it is called a triplet excited state). As one electron is now in an anti-bonding orbital the molecule has overall weaker bonding. (However, the NO excited state may be an exception).
The excited state always decays back to the ground state (it in not a stationary state of the hamiltonian) with a lifetime that depends on the particular molecule or atom and between different molecules the lifetime
can have  a huge range, from less than a picosecond to several seconds.


text:
What's the exact difference between (-)-BTX and (+)-BTX
I quite liked this article, which describes the synthesis of (-)-batrachotoxin as well as a steroisomer, (+)-batrachotoxin. I would like to know what the precise difference is between the two: are they precise mirror images of each other, or is there something more subtle going on? Is there something specific about the (±) notation (as opposed to S and R, for example) with relevance to the difference between the two?
I had a look online but the original paper is thin on structures that I can understand, and just going on google buries me under a slew of (-)-BTX results.

These are descriptors which tell you the direction in which the sample rotates plane polarised light. https://en.wikipedia.org/wiki/Dextrorotation_and_levorotation

From a cursory glance, batrachotoxin has 8 stereocenters. No single Cahn-Ingold-Prelog R or S designation is going to be enough. As a result, we simply identify which enantiomer we're talking about by using the standard optical rotation of the substance. The (+) and (-) forms are enantiomers. They rotate plane polarized light in equal and opposite directions and are non-superimposable mirror images of each other.

Since Pasteur’s famous crystal separation experiment, it has been known that organic compounds can be chiral. Futhermore, it has been known that chiral organic compounds can rotate linearly polarised light beams. Since structure elucidation, especially stereospecific structure elucidation was a very hard task back in the day, it was common to refer to chiral compounds by the direction they rotate polarised light; (+) being dextrorotatory and (–) being levorotatory.
Most early chemical syntheses would be racemic for the simple reason of no chiral substrates being readily available. Racemic compounds were denoted as (±).
The (+) form and the (–) form are always exact enantiomers, i.e. every stereocentre reversed, every chiral axis reversed, every helical chirality reversed. For example, (–)-tartrate is (2S,3S)-tartrate while (+)-tartrate is (2R,3R)-tartrate. (2R,3S)-tartrate or meso-tartrate is a diastereomer of the former two.


text:
Would decomposition take place in an atmosphere without oxygen?
Let's say you placed some sort of biological material (like raw chicken meat, an apple, etc.) in a very large glass container (large enough that moisture could evaporate at a similar rate to the normal atmosphere); filled it with a heavier-than-air, inert gas, like xenon; then, sealed the container and placed it outside.
Question: Would the biological material decompose?
My natural assumption is that it would decompose slightly, but mostly it would dry out and essentially become "mummified."

You can have anaerobic processes. As an example, clostridium botulinum actually finds oxygen toxic.

I think your experiment is along the same lines as one performed in 1680 by Antonie van Leeuwenhoek. His experiment is described in this Wikipedia article:

In his letter of 14 June 1680 to The Royal Society, Antonie van
  Leeuwenhoek described an experiment he carried out by filling two
  identical glass tubes about halfway with crushed pepper powder, to
  which some clean rain water was added. Van Leeuwenhoek sealed one of
  the glass tubes by using a flame and left the other glass tube open.
  Several days later, he discovered in the open glass tube ‘a great many
  very little animalcules, of divers sort having its own particular
  motion.’ Not expecting to see any life in the sealed glass tube, Van
  Leeuwenhoek saw to his surprise ‘a kind of living animalcules that
  were round and bigger than the biggest sort that I have said were in
  the other water.’ The conditions in the sealed tube had become quite
  anaerobic owing to consumption of oxygen by aerobic microorganisms.

Obviously his experiment was not as tightly controlled as what you propose.  Also, I think the actual size of your "very large glass container" would be the key to microbial activity in your experiment, as anaerobes do require water to thrive. It sounds like van Leeuwenhoek gave his bugs plenty water for the scale of his experiment.
My guess is that there would be some anaerobic decomposition before it pretty much dried out and "mummified".


text:
In a mass spectrum, are there peaks corresponding to a combination of 13C and bromine isotopes?
I just began studying mass spectroscopy and I don't understand why certain peaks are not represented in the mass spectrum graph. Like let's say you have methyl or ethyl bromide. Usually, there are two equal molecular ion peaks at the right-most side, which indicate a $1:1$ ratio of $\ce{^{79}Br}$ and $\ce{^{81}Br}$ isotopes.
But this is only when you assume all carbon atom are $\ce{^{12}C}$. Why doesn't the graph show peaks with combination of $\ce{^{13}C}$ and bromine isotopes?
None of the examples online indicate this.

It does appear. The graphs you're looking at online probably just dumbed it down.
Here's some proof if you like, a compound synthesised by yours truly, with molecular formula $\ce{C7H7BrN2O2}$. I clipped the relevant part of the mass spectrum I got. This was done with ammonia gas CI, so instead of the $\ce{M^{+.}}$ radical cation we expect to see the $\ce{[M + H]+}$ quasi-molecular ion. We expect the following monoisotopic masses (I used this website to calculate these):
$$\begin{array}{ccc} \hline
\text{Molecular ion} & m/z & \text{Predicted relative abundance} \\ \hline
\ce{[C7H7^{79}BrN2O2 + H]+} & 230.9769 & 100.00 \\
\ce{[C6^{13}CH7^{79}BrN2O2 + H]+} & 231.9798 & 8.48 \\
\ce{[C7H7^{81}BrN2O2 + H]+} & 232.9749 & 98.01 \\
\ce{[C6^{13}CH7^{81}BrN2O2 + H]+} & 233.9777 & 8.29 \\ \hline
\end{array}$$

Ahh... That's what I thought. Thank you so much for the clarification!

okay. just saw it


text:
Is the inductive effect always measured relative to hydrogen?
Wikipedia has defined the Inductive effect thus:

"In Chemistry and Physics, the inductive effect is an experimentally observable effect of the transmission of charge through a chain of atoms in a molecule, resulting in a permanent dipole in a bond."

Recently, I learned from a teacher that the inductive effect of any atom/group is always measured with respect to hydrogen.
Under this definition, we would ideally attach the atom/group with an atom of hydrogen and observe the direction of the dipole moment caused by the difference in electronegativity between hydrogen and the atom/group in question. If the electrons-initially positioned at the middle of the bond-are pushed toward hydrogen, the atom/group is termed as a "$+I$", electron-donating group. On the other hand, if the electrons are pulled toward the atom/group, it is termed as a "$-I$", electron-withdrawing group. Thereafter, this atom/group will always exhibit this type of inductive effect. Even if a situation does arise, in which it appears as though an atom/group that exhibits the $-I$ effect with hydrogen is exhibiting the $+I$ effect with another atom/group, this must be the cause of other dominating effects such as the mesomeric effect and in those cases also, it is said that the atom exhibits the $-I$ and not the $+I$ effect.
To illustrate the clear distinction between the definition put forward by the teacher and the one on Wikipedia, let us consider a molecule of $\ce{Cl2}$.
Under the definition on Wikipedia, it is clear that $\ce{Cl2}$ is a non-polar molecule (i.e. No permanent dipole moment has been experimentally observed in this molecule). Hence, no inductive effect is in play and neither of the two $\ce{Cl}$ atoms show either the $+I$ or $-I$ effect.
On the other hand, if we work with the definition put forward by the teacher, we observe that an atom of $\ce{Cl}$ exhibits the $-I$ effect when attached to an atom of hydrogen, being more electronegative than it. Therefore, in the $\ce{Cl2}$ molecule, no permanent dipole exists as a result of the $-I$ effect of both chlorine atoms cancelling away from opposite directions.
It appears that only one of these two definitions is definitely correct and I am confused as to which one that is.

Why should one Cl atom exert inductive effect on the other? They are both equally electronegative, so they will pull the shared electron pair with equal 'force' and so there would be no inductive effect.

I think this is in principle a problem in terminology. You will always have an (a net) inductive effect when you have different electronegativities. The terms +I/-I should only be used in organic chemistry, and there you would indeed compare them to the standard CH bond.

Your definition is falling apart because you are over-applying it. In the definition, you can see that it says:

...transmission of charge through a chain of atoms in a molecule...

It doesn't make sense to apply this in the case of a two-atom molecule, because the inductive effect is used to measure how one bond effects the electron density of other bonds in the molecule. You can assign any bond a dipole moment, but inductive effects are used specifically to explain phenomena observed in organic chemistry, where chains of molecules are (exceedingly) common.
To answer your other question: yes, inductive effects are always measured relative to hydrogen. Linear free-energy relationships are a quantifiable way of measuring a reaction's sensitivity with respect to a certain parameter. The Hammett equation is used to study how sensitive a molecule is in a given reaction to a change in its subsituent's inductive and resonance effects, for which the reference subsituent in all cases is a hydrogen atom. This is perhaps because the $\ce{C-H}$ is the most common bond in all of organic chemistry.
Linear free-energy relationships are particularly useful because similar as to what you said, substituent groups will always have the same relative inductive effects no matter the molecule. A quaternary ammonium group will always have a larger $-I$ effect (and larger $\sigma$ value) than a nitro group than a trifluoromethyl group. A molecule's response to these changes can give great mechanistic insight into organic reactions and the pathways through which they proceed.

A table of the different types of Hammett relations$^{[1]}$:

Their reference reactions:
$\hspace{3.25cm}$

$[1]$ Anslyn, E. V.; Dougherty, D. A. Modern Physical Organic Chemistry; University Science: Sausalito, CA, 2006.

I'm afraid I don't quite understand the last bit completely: "Linear free energy relationships are a quantifiable way of measuring a reaction's sensitivity with respect to a certain parameter." Is it possible to simplify this so that a highschool student can also understand it?

The Hammett equation, for instance, measures how sensitive the ionization of various substances are to changing the substituent para to the carboxyl group. There are other relationships that measure things such as how leaving groups or nucleophiles are affected by changing substituents, how the choice of solvent effects the reaction... the list goes on. If you are interested you can read more in the link I added.

I myself am having a little bit of trouble wrapping my head around the Wikipedia definition which they cite from an organic chemistry textbook. I am specifically troubled by the phrase ‘resulting in a permanent dipole moment.’ According to all definitions of the inductive effect I know, a methyl group is electron-donating ($+I$). Yet, if you consider para-xylene, the molecule suddenly has a centre of inversion (point group $C_\mathrm{2h}$). Therefore, there is no permanent dipole in para-xylene. (Depending on the methyl rotamers, it might also have $C_\mathrm{2v}$, in which case the dipole would be perpendicular to the benzene ring plane — and since there is almost no vertical structure, ignoring the three aliphatic $\ce{C-H}$ bonds, it is predicted to be very small.)
There is no reason to assume that methyl groups suddenly lose their $+I$ effect just because two of them are bonded to two different sides of a benzene ring. The NMR supports the assignment of a double $+I$-substitution with the aromatic protons of para-xylene displaying chemical shifts of $7.05~\mathrm{ppm}$ (compared to $7.36~\mathrm{ppm}$ in benzene; same solvent $\ce{CDCl3}$). Obviously there is additional shielding present in spite of no observable permanent dipole, which can only be attributed to the methyl groups’ $+I$ effect. The same case can be made for other symmetrically para-substituted benzenes such as para-nitrobenzene ($8.41~\mathrm{ppm}$ in $\ce{C6D12 + C3D6O}$), para-dichlorobenzene ($7.26~\mathrm{ppm}$ in $\ce{CDCl3}$) and others. In these cases, especially dichlorobenzene, the point group is actually predicted to strictly be $D_\mathrm{2h}$ — strictly containing inversion.
Thus, when having to choose between your teacher’s and Wikipedia’s definions, I want to go with your teacher’s because it better explains what we discuss in basically every seminar.


text:
Magnetically separating H2 and O2 in a volume of gas
I'm using a method of electrolysis where the anode and cathode are in close proximity (~0.7mm) and therefore the separate collection of H2 from the cathode is very difficult. Since O2 is paramagnetic and H2 is diamagnetic I was thinking I could use permanent magnets to separate the gases in a specially constructed bubbler.
Is it possible to calculate the required magnetic field strength necessary to perform this separation? If so, can you provide a formula or point me in the right direction where I can read about how to calculate this?

You would need a very low temperature and a very strong magnet.

Relevant question

That would be about as practical as making gold from mercury in particle accelerators.

@DHMO or thousands of steps in a cascade (like the processes used to concentrate uranium isotopes). Hardly cheap or practical even compared to fractional distillation of the liquefied gas.

All of your comments have been very helpful, and while DHMO's response is qualitatively correct, the relevant question referenced in #2 provided a formula that seems applicable to my situation. Thank you!


text:
Dependence of reaction rate on concentrations and stoichiometric coefficients of reactants and products
Consider a general equation of a reaction:
$$
\ce{aA + bB -> \text{Products}}
$$
We can say, in chemical kinetics, that the rate of the reaction is:
$$
\text{Rate} = k([\ce{A}]^a [\ce{B}]^b)
$$
where $\ce{[A]}$, $\ce{[B]}$ are the concentrations or active masses of the reactants.
My question is, why is the rate of reaction proportional to the concentrations of the reactants raised to the power of their stoichiometric coefficients ?
(Not really the point of the question, but why to the power and not just a product of the coefficients?)

Related: http://chemistry.stackexchange.com/questions/23642/in-the-elementary-steps-of-a-reaction-why-do-we-assume-that-the-coefficients-of, http://chemistry.stackexchange.com/questions/38167/do-coefficients-matter-in-rate-law


text:
Are the odours added to natural gas completely flammable?
I live in Australia, where both electricity and natural gas are common forms of household energy. An odour is added to the natural gas as a safety precaution since it is otherwise almost undetectable. I haven't been able to find what gas/mixture of gases is used in Australia, with all the sites I've checked just saying "an additive" or "an eggy smell".
The Wikipedia article on natural gas says that both tert-Butylthiol and tetrahydrothiophene can be used as an additive, and the article on tetrahydrothiophene says:

It is also used as an odorant for natural gas, usually in mixtures containing tert-butylthiol.

Whenever I've used natural gas (i.e. burning it for heating or cooking) I've noticed that there is no smell afterwards. Are all the additives completely flammable? Presumably, their combustion products are odourless, but I haven't been able to find any information on that either.

Yes, they burn completely and turn into SO2 and whatever else. While most certainly not odorless, SO2 smells much weaker, so we don't feel it at all.

The odourants are used in very small quantities as their smells are very intense. But, of course, some is emitted before the natural gas catches fire so there may be some residual smell because of that. This doesn't usually last long as the amounts are very very small.

On looking it up on Google and a few more places, it turns out that natural gas is primarily methane, along with a few other hydrocarbons.
Well, I have found out through some searches online that in the case of propane based fuel, like LPG (Liquid Petroleum Gas) in India, ETHYL MERCAPTAN is a popular option. It's smell is described as a

strongly disagreeable odor that humans can detect in minute
  concentrations

And that it's smell resembles that of

leeks, onions, durian or cooked cabbage, but is quite distinct.

(source :Wikipedia)
And also, most important point of all, it is highly explosive.
This is what the New Jersey Department of health and safety has to say:

(Source : This document)
So yeah, You're right. Based on this example, I think it logically follows that these "additives" are flamable and result in products that are ODOURLESS.


text:
Comparison of bond length between sulfur tetrafluoride and selenium tetrafluoride

Compare the bond lengths for selenium tetrafluoride, $x = d(\ce{Se-F})$, and sulfur tetrafluoride, $y = d(\ce{S-F})$:
  (a) $x > y$ (b) $y > x$ (c) $x =y$ (d) none of those.

Since selenium is bigger in size compared to sulfur, therefore the $\ce{Se-F}$ bond in $\ce{SeF4}$ would be shorter in length compared to $\ce{S-F}$ bond in $\ce{SF4}$, thus $y > x$.
But the book says $x > y$. Where did I go wrong?

Bond length is measured from nucleus to nucleus, not from ‘where the core shells end’ or ‘behind the valence electrons’ — these are not defined, you might as well measure to the end of the rainbow. The positions of nuclei can relatively precisely be determined by X-ray crystallography so you have something you can measure.
Naturally, the larger atom must have longer bonds.

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text:
The definition of peptide, peptide bond and protein
According to what I understand the difference between peptide and peptide bond is that peptide is two or more amino acids (up to 19 -inc.) which linked together, while the term "peptide bond" refers to the bond between each two amino acids. Is that true? If it is, the definition of protein is that it contains at least 20 amino acids which are linked together. Isn't it?

Two or more amino acids form a peptide. The peptide has the repeated form $\ce{-CO-NH-C(\alpha)R - }$ where R is an amino acid side-chain. The peptide bond is the amide bond N-C. The $\ce{CONH}$ group is planar. A peptide chain can be artificially constructed in the lab from any of the amino acids, its length and composition is then for you to determine.
A protein is a naturally occurring chain of amino acids folded so as to perform some function. Usually a protein has many amino acids because it has to perform a function in its organism. I'm not sure what the number of amino acids in the smallest protein is, but 20 seems about right. This number is determined by the protein's function, it can be many hundreds. A part of a protein's  backbone is also called a peptide chain.


text:
Which definition of an ore is correct?
In metallurgy, an ore is defined as that mineral from which metal can be obtained with economic viability.
I have also read that ores are rocks from which minerals or metals can be extracted profitably.
 If ores are rocks they cannot be minerals as rocks are aggregates of mineral and vice versa, so which of these is actually correct?

Correct is whatever definition you hold to be correct. "Rock" is of course a stupid category here, because usability as "ore" is given by chemical composition and perhaps microstructure, but not particle size.

@Karl , usability of an ore is given by chemical composition, do you mean to say that usability of an ore is from the mineral since minerals have a definite chemical composition with a crystalline structure? So that would imply that ores be defined in terms of minerals but an ore contains other minerals in addition to the mineral from which metal would be extracted,  then by definition it is an aggregate of minerals which makes it a rock

Minerals (per usual definition)  are unadulterated solids as appearing in nature, while ore is raw material for a plast furnace or similar process. Two completely different categories.

Could you explain that point?  Isnt it that all ores are minerals but not all minerals are ores, is that right?

Processing the ore mineral to the stuff that is finally fed into the blast furnace can involve a number of processing steps. All the way however, it is called "ore".

Frankly you're trying to make a mountain out of a mole hill.
There is no way, in general, to define ore as a mineral or a rock. It depends on what is being mined and the mine itself.

BTW, the definition "mineral from which metal can be obtained" is too restrictive:  it ignores the many non-metallic ores. See http://www.min-eng.com/commodities/nonmetallic/

Some of the terms are pretty vague and it isn't worth getting hung up on a precise definition.
Of the terms you use, "rock" is the vaguest. I means just about any hard thing you find in nature (and even that last qualification isn't precise as some people might call something like concrete an artificial rock). You can't infer anything much from the term "rock". Sometimes geologists might classify particular rocks into useful categories such as sandstone, granite or limestone. But the term is vague enough that rock formations might have more than one type of rock in them.
Ore is a more specific term. It does have a meaning but that meaning can change over time. Generally an ore refers to whether a useful product can be economically extracted from something. So if a rock formation contains a small amount of gold then it would count as an ore if the cost of extracting it was less than the amount the resulting gold could be sold for. But, sometimes, a geologist might refer to a particular mineral as the ore from which the desired substance is extracted. For example, she might call galena an ore from which lead is extracted. But this ignores the fact that galena is found in formations that contain many other minerals and has to be separated from them to be useful. More precise usage would be to define the ore as the rock formation containing enough galena for economic extraction of lead.
A mineral is usually a very precise term referring to a pure substance. Rocks are usually made of mixtures of minerals. Granite, for example, is made from quartz, feldspar and mica. The separate crystals are usually large enough to see with the naked eye.
An ore is simply a rock formation (which might contain more than one type of rock) where one of the minerals in it is economically valuable.

Which definition of an ore is correct?

None. A "definition" is an artificial idea made up by people to describe something. And even then, people have to actually use that definition correctly so it will have any value. Also, the usage of a certain definition differs between fields (geologists, economists, engineers, etc)
I am an ore deposit geologist, so I am familiar with the subject. Usually people describe a rock as a collection of one or more minerals together, where a mineral is a naturally occurring substance with a crystalline structure. Note that this definition does not include obsidian because it's glassy or any amorphous materials because they're not crystalline (where they are obviously rocks!).
Ore is usually considered (by geologists) as a rock with some kind of economic commodity that you can extract at a profit. Engineers will consider ore to be the thing you mine and put into the smelter, and minerals as the stuff that geologists consider ore. These are all applied very loosely. Ore doesn't even have to be extracted at a profit (because this changes with time, technology, and prices). As ore geologists, we name everything that has something even remotely interesting economically as an "ore deposit".


text:
If entropy is decreased when an enzyme-substrate complex is formed, how can the reaction be feasible?
Is it due to the bond formation creating energy that increases the entropy of the surroundings/universe?

There are multiple things going on with this question. First, we don't determine "feasibility" via entropy change of the system (entropy change of the universe, yes, but of the system, no). We determine feasibility via the free energy change of the system (which accounts for the net entropy change of the universe).
Second, you need to recognize that the rate of the reaction is dependent on the free energy difference between any stable intermediate and a subsequent transition state. So, making the enzyme-substrate complex less stable has zero impact on the rate provided it does not change this difference. In fact, it will speed up the reaction if the enzyme-substrate complex converts to the enzyme-product complex via a lower barrier pathway than the substrate to product does without enzyme.
Finally, you should note that enzymes generally don't bind very tightly to the substrate, for two reasons. The first is related to what we just said about barriers. If you decrease the energy of the enzyme-substrate complex relative to the substrate and free enzyme, you're actually slowing down the reaction and making it harder for the enzyme to work. Also, enzyme-substrate binding can be mildly correlated with enzyme-product binding (depending on product). If your product binds too tightly, it will inhibit enzyme turnover and also slow the reaction.


text:
Calculating acid dissociation constant for polyprotic acids
I'm confused with $K_{a1}$ and $K_{a2}$ for a polyprotic acid and how they are found, consider the following:
A solution of $1 M$ of $\ce{H2SO4}$ would yield $1 M$ of $\ce{H3O+}$ and $\ce{HSO4-}$ since $K_{a1}$ is very large.
In the case of the second ionization, $\ce{HSO4-}$ would give $1+x$  M of $\ce{H3O+}$ and $x$ M of $\ce{SO4^2-}$ , if $x$ is known how do we calculate K$_{a2}$
Do we use first concentration + second concentration as following
•   $K_{a2} = (x+1)(x)/(1-x)$
Or only second concentrations
•  $K_{a2} = (x)(x)/(1-x)$
Which one is correct and why?

It should be $1+x$ because the equilibrium accounts for all hydrogen ions in solutions. That's only if you choose to start with $\ce{H2SO4}$ for that. You can also just start with something like $\ce{NaHSO4}$. Since bisulfate is a very weak base, there's no concern with the first dissociation, and there's no extra hydrogen floating around from the first dissociation.

@Zhe so Ka2 for bisulfate would be different for sulfuric acid?

Bisulfate's Ka is sulfuric acid's Ka2.


text:
Why is Diiron nonacarbonyl so exceptional?
My textbook(NCERT) says:

With exception of $\ce{Fe2(CO)9}$, all other metal carbonyls are soluble in hydrocarbon solvents.

Weller, M.; Overton, T.; Rourke, J.; Armstrong, F. Inorganic Chemistry, 6th ed. states:

The most striking exception among the common metal carbonyls is nonacarbonyl diiron(0), which has a very low vapour pressure and is insoluble in solvents with which it does not react.

Other metal carbonyls are well soluble in hydrocarbon solvents but not Diiron nonacarbonyl. Why? What makes Diiron nonacarbonyl so exceptional? Is it because it has low vapor pressure?
There are no explanations regarding the insolubility of diiron nonacarbonyl. Is there any experiment conducted to determine the reason for its insolubility or is it just mere observation?
Update
@Orthocresol has told me to draw comparison of diiron nonacarbonyl with $\ce{[Mn2(CO)10]}$ and $\ce{[Co2(CO)8]}$ because of their similar molecular weight. Based on this fact, I went on to research further on this topic and from various handbooks and research notes I delved upon, I have drawn the following solubility comparison table. [Note: Only binuclear metal carbonyls were considered of form $\ce{[M2(CO)_x]}$ because they have same structure and thus should have same physical properties].
\begin{array}{c|c}
\mathbf{Metal~carbonyl} & \mathbf{Solubility} \\\hline
\ce{Mn2(CO)10} & \mathrm{ether, other~organic~solvents}\\
\ce{Tc2(CO)10} & \mathrm{ether,acetone} \\
\ce{Fe2(CO)9} & \mathrm{Insoluble~in~benzene,ether,petrol.Only~soluble~in~THF}\\
\ce{Rh2(CO)8} & \mathrm{organic~solvents}\\
\ce{Ir2(CO)8} & \mathrm{Ether,CCl_4}\\
\ce{Co2(CO)8} & \mathrm{petrol,benzene,alcohol}\end{array}
Also, a table of comparison for the other two iron carbonyls for a reference.
\begin{array}{c|c}
\mathbf{Iron~carbonyl} & \mathbf{Solubility} \\\hline
\ce{Fe(CO)5} & \mathrm{soluble~in~all~organic~solvents~like~ether,petroleum.Insoluble~in~water}\\
\ce{Fe3(CO)12} & \mathrm{Insoluble~in~water. Soluble~in~non~polar~organic~solvents.
} \end{array}
We can observe that there is a drastic anomaly in case of solubility of diiron nonacarbonyl in both the tables. Any explanation to this?

It's solid not liquid like Fe(CO)5 therefore it has low vapor pressure.

@Mithoron Ok, so why it exist as solid whereas other iron carbonyls are liquids?

Well, almost twice as big molecule make wonders ;)

If it's solely based on molecular weight, then I suppose a good comparison would be $\ce{[Mn2(CO)10]}$ or $\ce{[Co2(CO)8]}$.

@orthocresol. I added a comparison table.

@mithoron how does not liquid guarantee low vapor pressure?  I still remember smelling the vapor pressure of moth balls.

@NilayGhosh You mentioned NCERT as your source. In which chapter is this statement mentioned? I checked the two chapters concerning in class 12 Volume 1 i.e. Coordination Compounds & D & F Block but couldn't find this statement

@StackUpPhysics I do not have access to the textbook as of now but I remember the statement being written in the "coordination chemistry" chapter of the NCERT textbook. Maybe the statement got removed in a reprint.

@NilayGhosh I went through the one present at the website of NCERT currently and the hard copy which I possess. Maybe they removed the statement due to lack of conclusive experimental literature to support the claim or inability to explain it to the students at this stage of their studies

Maybe using this as a source would be better- https://books.google.co.in/books?id=Q1bwAwAAQBAJ&pg=PA601&lpg=PA601&dq=With+exception+of+Fe2(CO)9,+all+other+metal+carbonyls+are+soluble+in+hydrocarbon+solvents.&source=bl&ots=etJ2pp4-LJ&sig=ACfU3U0JvKgDdrRU-MzODnu0f4ZrucWuUw&hl=en&sa=X&ved=2ahUKEwjRtOfZ5PvlAhWexzgGHbBQCLwQ6AEwA3oECAgQAQ#v=onepage&q=With%20exception%20of%20Fe2(CO)9%2C%20all%20other%20metal%20carbonyls%20are%20soluble%20in%20hydrocarbon%20solvents.&f=false

This is also an authentic source- https://books.google.co.in/books?id=tUmcAQAAQBAJ&pg=PA555&lpg=PA555&dq=With+exception+of+Fe2(CO)9,+all+other+metal+carbonyls+are+soluble+in+hydrocarbon+solvents.&source=bl&ots=i2dVDwEXdi&sig=ACfU3U2hO44onfYf2EdNklnFVVgmoXgKnw&hl=en&sa=X&ved=2ahUKEwjRtOfZ5PvlAhWexzgGHbBQCLwQ6AEwAnoECAkQAQ#v=onepage&q=With%20exception%20of%20Fe2(CO)9%2C%20all%20other%20metal%20carbonyls%20are%20soluble%20in%20hydrocarbon%20solvents.&f=false

Which chapter, edition of NCERT has the quoted line? I could not find it.

Wikipedia (primary reference) suggests a possible reason for $\ce{Fe2(CO)9}$ dissolving preferentially in THF versus nonpolar solvents:  it reacts according to the scheme
$\ce{Fe2(CO)9 + THF <=> Fe(CO)5 + Fe(CO)4 \cdot THF}$
Such a reaction is invoked to account for the dinuclear complex giving mononuclear products with various ligands in THF.  A similar reaction with the corresponding cobalt and manganese species would have to produce radical or ionic products, due to the odd atomic number of the metal, and so would be less favored.
This would not, of course, explain the lack of solubility in nonpolar solvents or nonvolatility of $\ce{Fe2(CO)9}$.  It is possible, however, that the same $\ce{Fe2(CO)9}$ molecules that bind with the oxygen in THF could also bind intermolecularly through the oxygen atoms, in effect delocalizing the covalent bonding and creating more intermolecular cohesion.
We can represent this type of interaction using the SMILES structural notation.  Fr a single $\ce{Fe2(CO)9}$ molecule it would look like this, taken from Wikipedia:
$\ce{O=C1[Fe]2(=C=O)(=C=O)(=C=O)C(=O)[Fe]1(=C=O)(=C=O)(=C=O)C2=O}$
For two molecules we would have
$\ce{[O=C1[Fe]2(=C=O)(=C=O)(=C=O)C(=O)[Fe]1(=C=O)(=C=O)(=C=O)C2=O][O=C1[Fe]2(=C=O)(=C=O)(=C=O)C(=O)[Fe]1(=C=O)(=C=O)(=C=O)C2=O]}$
Then we have a contributing structure where an oxygen atom from the left molecule could combine with the right one, displacing an iron pentacarbonyl molecule:
$\ce{[O=C1[Fe]2(=C=O)(=C=O)(=C=O)C(=O)[Fe]1(=C=O)(=C=O)(=C=O)C2=\color{blue}{O[Fe]}(=C=O)(=C=O)(=C=O)C=O][[Fe](=C=O)(=C=O)(=C=O)(=C=O)C=O]}$
where the blue atoms form an intermolecular bond in this contributing structure.  The displaced iron pentacarbonyl molecule is in the last set of brackets.
The Case of Hydrogen Bonding
The delocalization of covalent bonds, hypothesized above, may be seen in a more familiar context:  hydrogen bonding.  Usually thus is rendered as an electrostatic attraction between oppositely charged atoms.  But it may also be considered as a molecular-orbital interaction:  a nonbonding electron pair overlaps an adjacent molecule's antibonding orbital in exchange for creating an "intermolecular" covalent bond.  With water it would look like this
$\ce{[HOH][HOH]}$
with the hydrogen-bonded, "intermolecular bonded" contribution
$\ce{[H\color{blue}{O}(H)\color{blue}{H}^+][OH^-]}$
where the intermolecular bond is between the blue atoms.  This is also the autoionized structure for water.

Seems like a possibility however can you share any place else where such a phenomenon is displayed? (The one mentioned in 2nd paragraph)

It's actually analogous to hydrogen bonding where molecules of a substance such as water interact through their hydrogen atoms.

This seems to be a convincing argument, would love to see if you can any provide any data to substantiate the same

This is all my hypothesis, and we do not know enough about diiron nonacarbonyl to truly say.  But xenon difluoride has some interesting characteristics pointing in the same direction.  When I get more time I will discuss this case, hopefully this weekend.

ok will look forward to your results

Yes that is a good start. Looking forward for a complete answer. Also, thanks to @StackUpPhysics for holding a bounty on this question. I had been meaning to hold a bounty
on this question about an year ago but just forgot about it :(.

@NilayGhosh No worries, I found a few questions very interesting hence I added a bounty. I am also the one gaining here :)

Added my elaboration.  Mwikipedia gives an alternate explanation for the odd crystal structure if XeF2, so I discuss hydrogen bonding instead.


text:
What does it mean to say "the 222 space group"?
I know there are supposed to be a few rotation axes, that it is non-centrosymmetric, it belongs to the orthorhombic crystal system, and it has four equivalent positions.
What do the 2s mean?

This is a crystallographer’s shorthand for a certain type of space group.
You may have heard of the 14 Bravais lattices, e.g. cubic, monoclinic, orthorhombic. These lattices all assume perfect sphere’s at the corners of the elementary cell and potentially also in certain central positions (if the lattice is not a primitive one). In actual X-ray structure determination, these perfect Bravais lattice spheres are typically not spheres. Instead, we should consider each sphere a ‘hook’ and a molecule hung to that hook. Since not all molecules are symmetric with respect to all symmetry operations a certain Bravais lattice has, the symmetry of the crystal is markedly reduced. For example, an enantiopure chiral compound cannot form crystals that contain planes of symmetry, so even if that compound were to crystallise in the orthorhombic lattice, there would be no planes in spite of them being required in simple orthorhombic crystals.
This reduction in symmetry leads to a large number of lower-symmetry sub-divisions of the Bravais lattices called space groups. Strictly speaking, the Bravais lattices are merely highly symmetrical space groups. There is a total of 219 or 230 — depending on whether you count enantiomeric space groups as two or one.
The shorthand to unambiguously label these space groups basically describes which symmetry operations are present in a certain direction. Orthorhombic is the easiest to understand: the primitive Bravais lattice has:

in $x$ direction:

a two-fold axis
a plane of symmetry perpendicular to that axis.

in $y$ direction:

a two-fold axis
a plane of symmetry perpendicular to that axis.

in $z$ direction:

a two-fold axis
a plane of symmetry perpendicular to that axis.

You could label this is $\mathrm{P\ \frac2m \frac2m \frac2m}$, but it is tiresome to always include the ‘2 over m’ part. Hence, it is shortened to $\mathrm{P\ mmm}$, the shorthand being ‘m replaces $\frac2m$’. If there is no plane of symmetry, then only the two-fold axis remains and that dimension is given a 2 label instead. Thus, $\mathrm{P\ 222}$ is the same lattice but there are no planes of symmetry in the actual crystal structure. The crystal could also include a screw axis, in which case the symbol $2_1$ would be used, or a glide plane in which a different letter, depending on the orientation of the glide plane, is used.


text:
Naming Ibuprofen
Can someone explain why the ibuprofen medication has the chemical name as (RS)-2-[4-(2-methylpropyl)phenyl]propanoic acid?

What does 2,4,2 refer to in the chemical name?

By the way: the name “ibuprofen” is related to the obsolete name “2-(4-isobutylphenyl)propanoic acid” for 2-[4-(2-methylpropyl)phenyl]propanoic acid.

If we want to name chemicals we need a systematic approach with good rules so whoever names the compound come up with the same name and it also has to be unambiguous. There are two major naming systems, one by IUPAC and one by Chemical abstracts. Both work very similarly.
So how do we come up with that name and what are those numbers? First we find a starting point, we call this the group with the highest priority, which is the acid on the right in this case. Now I won't get into detail here why we do things a certain way, this would take a lot of time and there are resources explaining all the rules.
In this case we start off with the acid and here we got, as our "base system", an acid with a chain out of 3 carbons (including the one of the acid group) so that's a propionic acid. (the purple structure in the picture below) Now attached to this is a phenyl- ring (blue) but we need to write down where it is attached. We start counting at the acid so it's position two, that's the first 2 in the name. On that phenyl- ring we got another group, and this time we count from the position where it's attached to our previous group. So we end up with position 4. And on that 4 position there's the red group, a propyl group and again here on two position there's another methyl.

The outer two refers to the fact that the (4-(2-methylpropyl)phenyl) group is on the second carbon of the propanoic acid. The four tells us that the methylpropyl group is across from the first carbon in the benzene ring (which is the one attached to the propanoic acid group). Finally, the second two refers to the fact that the methyl group on the methylpropyl substituent on the benzene ring is two away from the benzene ring. Alternatively, this compound could be called (RS)-2-(4-(isobutyl)phenyl)propanoic acid, or even (RS)-2-(para-(isobutyl)phenyl)propanoic acid.


text:
Direction of arrows in electrocyclic reactions
We recently started learning about pericyclic reactions and this is a step from a mechanism I'm working on for an assignment. I believe that the arrows on the right are better than the arrows on the left because if I throw in some charges instead of arrows, the outcome I get on the right is somewhat more "stable?" for the reasons stated on the picture. Am I correct in my assumption?
 On the overall reaction,  it's the 6 to 7 step that I'm interested in.

Any reason you didn't draw out the radical version? Frankly, it doesn't make a difference because what experiment would you design to determine this? If you can't, then it doesn't matter.

Also, I don't believe the arrows you drew for the concerted mechanism lead to the partial charges you've drawn.

I meant that they lead to the same bonds being formed in either one direction or the other.

The answer is rather bland and probably not what you want to hear. Electrocyclic reactions are called electrocyclic because they are cyclic. (Please don’t see cyclic circular reasoning in there.) They proceed in a cyclic, i.e. fully symmetric fashion. You simply cannot draw one arrow direction and claim it is the most logical or the only one possible. You could draw these arrows as you drew them, but you could also formulate a radical-based mechanism (each double bond splits into two radicals which combine with the two of the other double bond) at no loss.
Thus, there is no way to say whether left or right. The reaction happens; in a cyclic fashion.

Arrow-pushing mechanisms are crude oversimplifications at best.

@Jan Can the same be said about sigmatropic rearrangements e.g. [1,5] hydride shift?

@Anton Quite obviously so. Which also neatly explains why they can be written both as hydride transfers and as proton transfers.


text:
Ammonium is a weak acid but isn't it considered a strong conjugate acid as a salt's ion?
I'm learning about salt hydrolysis in school and this just jumped out at me. How can $\ce{NH_4^+}$ be both a weak acid and a strong conjugate acid? Since $\ce{NH_4^+}$ is a weak acid it has a strong conjugate base $\ce{NH_3}$. However, as part of a dissolved salt, since $\ce{NH_4OH}$ is a weak base, then $\ce{NH_4^+}$ is considered a "strong" conjugate acid. How can we reconcile these facts? Are the definitions slightly different for a salt ion and a normal acid?

Just because something is a weak base does not mean its conjugate acid has to be a strong acid. Likewise, just because something is a weak acid does not mean that its conjugate base is a strong base.
The only connection between acidity of an acid and basicity of a conjugate base, henceforth labelled $\ce{HA}$ and $\ce{A-}$, is that they are related by the ion product of water, $K_\mathrm{w}$ by the following equations:
$$\begin{align}\ce{HA &<=> H+ + A-}&\quad K_\mathrm{a} &= \frac{[\ce{H+}][\ce{A-}]}{[\ce{HA}]}\tag{1}\\[0.8em]
\ce{A- + H2O &<=> HA + OH-}&\quad K_\mathrm{b} &= \frac{[\ce{OH-}][\ce{HA}]}{[\ce{A-}]}\tag{2}\\[0.8em]
K_\mathrm{a} \times K_\mathrm{b} &= \frac{[\ce{H+}][\ce{A-}]}{[\ce{HA}]} \times \frac{[\ce{OH-}][\ce{HA}]}{[\ce{A-}]}\\[0.6em]
&= \frac{[\ce{H+}][\ce{A-}][\ce{OH-}][\ce{HA}]}{[\ce{HA}][\ce{A-}]}\\[0.6em]
&= [\ce{H+}][\ce{OH-}] = K_\mathrm{w}\tag{3}\\[1.2em]
\Longrightarrow \mathrm{p}K_\mathrm{a} + \mathrm{p}K_\mathrm{b} &= 14\tag{4}\end{align}$$
Therefore, the sum of the acid and base constants must equal $14$. But if $\mathrm{p}K_\mathrm{a} = 6$, then $\mathrm{p}K_\mathrm{b} = 8$ — meaning the conjugate base of a weak acid is a weak base. This basically mirrors ammonia’s case.


text:
Which one is the healthiest to operate a lantern with: kerosene, diesel or sunflower oil?
Given a lantern (originally made for kerosene) however, I'm looking for the fuel which is the safest for indoor use.
I've already tried different kinds, and found three of them that burn just as nicely: kerosene, diesel and sunflower oil, which means, all of these seem to produce nearly the same light, the same heat, and none of them becomes smoky on a medium flame.
The only differences I experience:

Kerosene stinks quiet badly, and costs 5-6 times as much as the other two.
Diesel stinks just a little bit and is pretty cheap.
Sunflower oil doesn't stink at all and is the cheapest.

So from my experiences, I'd go with sunflower oil.
But the most important question is, based on which I will decide: Which one produces the least poisoning stuff when burnt? So what might be in the air I can't see and should be afraid of.
(once again: The lantern is always set to a flame that doesn't produce any visible smoke at all.)

While I don't doubt your intentions in posting this, we do shy away on this site from providing advice that is (broadly) "medical" in nature, and that includes queries such as yours. You might want to rephrase your query a bit. That said: (a) we cannot comment on anything in the air you need to be afraid of short of an air sample from you; and (b) I would avoid using diesel fuel in a lantern (at all), especially in a place where I'm breathing.

https://en.wikipedia.org/wiki/Mineral_oil is what you want. It is the undyed, unodorised version of what i think you mean by "kerosene". You have to find out what it is called in your country. It should be rather cheap.

Karl, thanks for the tip, I didn't know about this kind of oil and it seems very promising based on what I read about it. However, it turns out that for some reason it  is not commercially available in my country in its normal industrial form as a first distillate, only its refined version for medical uses, which is quite expensive unfortunately. But I'm still investigating as it seems the best fuel so far for indoor uses. It's a pity I cannot just buy some to try it out..

If you forget to ventilate your room, all three will kill you after approximately the same time due to lack of oxygen.

@Jan Wikipedia cites a source suggesting that a 37-lumen kerosene lantern burns about 25 ml/hour or about 20g/hour. My answer here suggests that this would consume about 68 g of oxygen per hour, which is pretty small -- more than you'd use at rest, but definitely less than while exercising. So you should be fine, unless your house is literally sealed so tightly you can't have company over without asphyxiation. (I still recommend good ventilation, with or without a lantern.)


text:
Electron has volume or not?
Sometimes we say that electron has volume, as in orbital we can define it's volume (at least for some fraction, as orbital itself has no limit, it is spread all over the space expect nodes).
On other hand in general chemistry books often spin quantum no. is referred as electron spinning on its axis. But this is absurd, because this magnetic angular momentum is an intrinsic property of the electron. Also that an electron is an elementary particle, it is 1-D (no volume), so it's not possible for an electron to spin on it's own axis.
So what's the matter, do electron has volume or not?
What I think that it's a particle, but we can't describe it's position what we can do is probabilistic approach, we can only describe some region where we can find it more likely or less likely. We defined it's volume, otherwise electron themselves has no volume. Even it is more weird that electron is everywhere in the orbital ;D
Can somebody explain it?

So far, in the standard quantum physics, electron is treated as a point-like particle.

Spin has  properties of regular angular momentum, but it is not. That is usually the first thing you are told when it is introduced in class.

Related question on Physics.SE - (http://physics.stackexchange.com/questions/24001/what-is-the-mass-density-distribution-of-an-electron)

@Karl but the electron can have regular angular momentum, too...

@Greg The orbital angular momentum is also a quantum mechanical entity. And if the electron is flying in a synchrotron, then it has of course regular, non-quantum mechanical angular momentum, but so what? It doesn't need a volume for that.

related http://chemistry.stackexchange.com/questions/62009/do-electrons-in-an-atom-revolve-around-the-the-nucleus-clockwise-or-counterclock/62054#62054

@Karl No one was talking about synchrotrons. L has non-zero expectation value for many electronic states.

Does this answer your question? Do electrons have some volume, area or shape?

If volume is understood classically, i.e. as the quantity of three-dimensional space enclosed by a closed surface, then quantum systems in general do not have a certain volume. One could, in principle, redefine the notion of volume for a quantum system to be, say, an area chosen so that there is a certain probability (for example, 90%) of finding the system within it, but that would be a completely different quantity, of course. It should also be noted that in quantum theory statements about a system having or possessing certain values of physical quantities has a lot of issues, even when the quantities are well defined. For instance, does a quantum system have (possess) certain values for physical quantities before the corresponding measurements are made?

On other hand In general Chemistry books often spin quantum no. is
  referred as electron spinning on its axis.

You already know that such picture is an absurd, and you are perfectly right here: electrons are not spinning. So there is no any contradiction there: they don't have a classical volume and they are not spinning.

What I know, any quantity has multiple defined values of physical quantities, can be shown as of if it has its own space. Eg. Momentum space. It can attain any of the value from its quantized space.

An Orbital is not the volume of an electron. An orbital is the volume in space where the probability to find a certain electron is 99%. The electron itself is treated as this probability density in QC, and as a point charge when considered a definite particle.
In reality, an electron probably has a volume (since it has a mass, sometimes). In reality, there are no "orbitals" either...
On the other hand, particle-wave-duality...
And the concept of volume for electrons doesn't fit with Einsteins theory of relativity.
Well the world of quantum mechanics is complicated...

"In reality, there are no "orbitals" either..." What do you mean by that?

@Mockingbird an orbital is a mathematical concept describing a volume where the probability to find an electron is 99.99%. The fact that there are different localizations of orbitals which "look" extremely different is also a hint that orbitals might not be a "real" object.


text:
What is the reaction between plain soda and sugar?
Today I was having lunch and I ordered plain soda. And some amount of sugar was added, suddenly all the liquid fell of the glass. I thought there should have been a reaction between plain soda and sugar. What is it?

Soda is carbonated water - it has carbon dioxide dissolved under high pressure. Carbon dioxide is a gas that would like to escape. Shaking the bottle of soda, for example, provides a suitable mechanism to allow release of the gas.
Adding dry sugar also provides a mechanism for this to occur through nucleation. The sugar crystals give the carbon dioxide a surface via which to escape the solution. This is not a chemical reaction per se, but a physical reaction. It is essentially the same process as the coke-mentos reaction. Sugar crystals are great for this, as they have a rough surface and therefor a very high surface area available.

There is no longer any pressure on the soda once it’s in the glass.


text:
Lanthanide bond lengths and dissociation
Due to lanthanide contraction, covalent character increases and ionic character decreases in lanthanides. Since a covalent bond is weaker than an ionic bond, $\ce{Lu(OH)3}$ should dissociate more easily than $\ce{La(OH)3}$. But that is not the case. Why?


text:
Reaction of sodium hydroxide with ethanol
I am carrying out a titration to determine the solubility of a carboxylic acid in mixtures of water and ethanol. I will be titrating a diluted (previously) saturated solution of water, ethanol and the acid against sodium hydroxide.
Can the sodium hydroxide interact with the ethanol so as to give an incorrect result?

The hydroxyl group in ethanol has a pKa of about 16 (see wiki), which means you will not be able to deprotonate it using sodium hydroxide. Thus, the result of your titration should not be altered by the amount of alcohol in the solution.
Depending on the amount of ethanol I could envision that you are not determing solubility in pure water but in the sepcific mixture with ethanol, but that seems to be assumed in this kind of experiment, right ?

Thanks for your answer - I came to the same conclusion, but forgot to add it as an answer. Nevertheless, it's nice to have confirmation! The experiment consists of determining the solubility of oxalic acid in mixtures of water and ethanol in different proportions. One follow-up question: I have used volume units to make up the solutions (i.e. 50mlH2O and 100mlC2H5OH) - is this appropriate?

Sure why not. In case of water organic mixtures i usually use weight concentrations for calculations because volumes are not additive in this case and one has to be careful not to forget that.


text:
Is it true that allylic free radical is more stable than allylic carbocation?
I have been studying reaction mechanisms involving allyl groups and it seems that the chances of forming of an allylic free radical are higher than forming a carbocation, I know that these things might be specific to reaction occurring, so I just wanted to find if there is any such relation between stabilities of radicals and cations.

This question as stated can't be answered because you can't compare the stability of two different species. You can give their energies relative to something, but since one of these species has an extra electron, you're comparing two different reactions. In fact, you're ionizing the radical to get to the carbocation. Sure, the ionization energy is positive, but you still can't compare these two species head-on.

I don't think this post should be closed. It shows initiative, and a fleshed out version of @Zhe's comment would be a valuable answer for future readers.

@ringo Good idea. I'm on it.

This is trickier than it sounds because you have to compare things that are similar or maybe you won't get the comparison you want.
First, you can't compare the energies of two things that don't at least have the same components. In this case, the allylic radical has an extra electron over the allylic cation. Let's say you generated the radical by fragmenting propene via removing a hydrogen atom or a hydride anion, respectively, as the other product. Notice that you there is energy in the other product, and you can't really just compare the radical and cation without taking into account the other product.
Second, you're in a situation where you make one species from the other. In fact, if you ionize the allylic radical, you will make an allylic cation. Notice that since you are ionizing a neutral species, the ionization energy is positive: you're going to need to add energy into the system to remove the electron. This makes sense because that electron is stabilized by the 8 nuclei that are present.
So, while you can't really compare the two species in your question head-to-head, it's clear than you can create the cation from the radical by adding energy to remove an electron. But this doesn't mean the cation is higher in energy; it just means the cation and the electron separated at infinite distance are higher in energy than the radical.
Hope this helps.

Thank you so much, this does seem like a quite posible explanation for the question.


text:
Validating Phase Diagram of a mixture
In lab I did several different runs to create a phase diagram for the mixture between biphenyl and naphthalene. My phase diagram is attached, just for visual purposes. My question is I'm supposed to validate my phase diagram by using the following formula to calculate the enthalpies of fusion for both compounds.
$$\left(\frac{\mathrm dT}{\mathrm dX}\right)_{(T=T_\mathrm m)} = \frac{RT^2}{\Delta H_\text{fusion}}$$
I understand how to do the math, however I'm trying to figure out where that formula comes from to better understand the relationship between the enthalpy of fusion and the phase diagram. Thanks!
A side question is where could I find an accepted value of the eutectic temperature for this mixture to compare my results with?

What $X$ is ? $\text{}$

T refers to temperature, and X refers to mole fraction.

From the Gibbs-Helmholtz equation we usually write
 $$\left (\frac{\partial \ln(X)}{\partial T}\right )_p = \frac {\Delta H^0_{fus}}{RT^2}$$
where X is the molec fraction of solute. Integrating this equation gives
$$ \ln(X)=\frac{\Delta H^0_{fus}}{R}\left(\frac{1}{T_{fus}} - \frac{1}{T}  \right)    $$
thus a plot of log mole fraction vs reciprocal temperature should produce the enthalpy. The eutectic point is the point where the melting curves meet with the solid phase, its about 0.4 to 0.5 on your figure. On the left should be the melting curve of liquid with solid biphenyl as you plot T vs naphthalene mole fraction, and on the right, liquid plus solid naphthalene. Solid biphenyl + naphthalene should be observed below $\approx 30 $ C at all mole fraction and liquid above the blue dots.


text:
How does alcohol act as a dehydrant?
I was doing histological slide preparation with rat's liver. There as a part of the microtechnique the fixed tissue had to be dehydrated using alcohol in the order 30%, 50%, 70%, 90% and 100% for 20-30 min each.
This had to be done to get rid of the water (possibly extracellular) before it was embedded in paraffin for microtomy because paraffin is immiscible with water.
My question is how does the alcohol extract water out of the tissue?

It's not about extraction. Alcohol just gradually replaces water.

@IvanNeretin What is the need to treat the tissue with increasing concentration of alcohol solution? Why is it not treated directly with 100%  alcohol?

The alcohol displaces water gradually through the process of osmosis. To prevent rupture of cells from the osmotic pressure, samples are immersed in a series of baths of increasing alcohol concentration.
After immersion in 95% or stronger alcohol, it is sometimes the practice to displace the alcohol with toluene or other non-polar solvent so that the specimen can be permanently mounted with paraffin or balsam.

I have a series of questions do you mind answering them? Like won't the cells shrink instead of getting ruptured?  You know water moves from higher conc to lower conc i.e from inside of the cell to the water-less and acgohol rich extracellular region.

The alcohol infiltrates the cell more rapidly than water leaves, in most cases. http://www.madsci.org/posts/archives/2006-03/1143666101.Cb.r.html If the movement of water were more rapid than ethanol, the cell would shrivel, as you describe.


text:
Does a technical umbrella term exist that covers the different kinds of biomolecules?
I want to write a sentence along the lines of:

There are four ____ in biochemistry: proteins, lipids, carbohydrates and nucleic acids.

Does a specific technical term exist that I can use to fill the void? Or am I relatively free in my choice?

Refer to them like what? Group or type of chemical compounds? Why not? You're asking about basic semantics...

Yes, I want to write a sum about the topic but I don't know the terminology.  Refer to them like groups. I mean to ask if I can divide the main biochemical molecules to 4 families.

Well, they are already divided - these are 4 groups of compounds. Group or type is better word then family here IMO, but what's here to ask about...

Are they the only groups of compound in biochemistry?

"Classes of biomolecules".

I don’t know what your background is. Maybe it is organic chemistry, where the term functional group is used in a very specific way and using what would seem to be a synonym instead is wrong. Maybe it is biology (taxonomy) where it makes a big difference if a ‘group’ is an order, a family, a genus or something else.
The typical primary metabolites in biochemistry — carbohydrates, peptides/proteins, nucleic acids and lipids — do not have an overarching, widely accepted technical umbrella term. You can call them groups, classes, kinds or even types and be well understood.
If I had to suggest something, I would go with classes of primary metabolites. But I wouldn’t call that a strict technical term like functional group (organic chemistry) or species (taxonomy).

Adding on to what Jan mentioned I looked it up in a few of the well-known textbooks. (Emphasis mine.)
Alberts et al. Molecular Biology of the Cell 6ed, p47 writes

"four major families of small organic molecules"

Berg & Stryer Biochemistry 7ed, p2 is a bit more careful

[...] two different classes of molecules: large molecules such as proteins and nucleic acids, referred to as biological macromolecules, and low-molecular-weight molecules such as glucose and glyccerol, referred to as metabolites [...]

Lehninger Principles of Biochemistry 6ed, p15 writes

Many biological molecules are macromolecules... shorter polymers are called oligomers... Table 1-1 shows the major classes of biomolecules in an E. coli cell. [Entries in Table 1-1: Water, proteins, nucleic acids, polysaccharides, lipids, monomeric subunits and intermediates, inorganic ions.]

Voet & Voet Biochemistry 4ed, p15 writes

[...] E. coli, and living things in general, contain only a few different types of macromolecules: proteins, [...] nucleic acids, and polysaccharides. [...] Lipids, the fourth major class of biological molecules, are too small to be classified as macromolecules but also have a modular construction.

Two weeks ago I suggested "classes of biomolecules" in my comment. In lieu of what I have quoted above, I think that might be a little bit imprecise since there are more types of biomolecules than simply those four classes. So, "major classes of biomolecules" would be a good idea.


text:
Is there a chemical compound can transform heat into light at 70 °C?
Is there is compound that can transform heat into light as phosphorescence / thermally generated photons at around 70 °C, in a reversible reaction?
For example, is there a phosphor which can be visible at 70'C and not at 20 degrees?
I'd very much appreciate some information. Thanks!

It might be possible in a few ways:

In the process of thermoluminescence, a previously-activated material gives off light when warmed. This is used to date materials such as ceramics, which accumulate energy from terrestrial radiation and cosmic rays. Though ceramics have to be heated quite a bit to release this stored energy, some substances exhibit thermoluminescence at room temperature or lower. $\ce{BeO:Zn}$ has peak emission at 307K, ~34oC.

N.B thermoluminescence is the release of energy previously stored in a material, not the direct transformation of heat energy into electromagnetic.

In an anti-Stokes process, a substance absorbs two or more photons of lower energy to release one of higher energy. This is used in IR beam visualization cards. However, it is unlikely that thermal energy would be sufficient to produce visible light.
Non-blackbody radiators can emit more energy at short wavelengths than would be expected according to Planck's law. "Limelight" owes its intense white light to candoluminescence, or non-blackbody radiation. Using a "heat mirror", the efficiency of incandescent lamps can be greatly improved. That said, though theoretically a material might exist that could prevent emission of almost all wavelengths longer than visible light, it is doubtful that a detectable amount of visible light would be emitted at 70oC.

The energy that can be transferred by collisions between particles at temperature $T = 343~\pu{K}$ is about
$$ E_{therm.} = \frac{1}{2} k T = 0.015~\pu{eV} \tag{1}$$
The energy that corresponds to the lowest energy of visible light with a wavelength of $\lambda = 800~\pu{nm}$ is about
$$ E_{h\nu} = h \nu = \frac{h \nu}{\lambda} = 1.55~\pu{eV} \tag{2}$$
This means that thermal energy at $70~\pu{°C}$ is at least two orders of magnitude too low for producing visible light. So the answer to your question is no.

Could those who have downvoted my answer please explain why?


text:
How can I tell whether or not a molecule is planar?
I am doing several questions involving judgment on the planarity of a compound.

Which of the following is not a planar molecule?

$\ce{H_2C=C=CH_2}$
$\ce{H_2C=C=C=CH_2}$
$\ce{H_2C=C=O}$
$\ce{NC-HC=CH-CN}$

I had the idea that the compound in with central atom has $\ce{sp}$ hybridisation is planar or the compound in which all the atoms has the same hybridization. But it is not working in this case. I am aware of finding out the hybridization of a atom in a compound but I feel trapped to decide the planarity of certain compounds. The question does not address this purpose.  Anyone has any idea to solve this and many related questions?

For an sp hybridized atom, the p orbitals that are used to form pi bonds will be perpendicular. That makes the pi bonds formed either side of an sp atom perpendicular to each other.

The answer to your question is here: Why are the two CH2 groups in allene perpendicular to each other?

Unfortunately, I think the answer to the general question is: learn chemistry. Because there are all sorts of factors that influence geometry...

@orthocresol can you give reference  to some other questions whee this problem has been discussed

I think these general rules work:

If there is an $\ce{sp^3}$ hybridized carbon (or nitrogen), the molecular is NOT planar.

2) If there are no $\ce{sp^3}$ hybridized carbons (or nitrogens), and there is only one $\ce{sp^2}$ hybridized atom (carbon or nitrogen), it will be planar.
3) If there are no $\ce{sp^3}$ hybridized atoms, and there are 2 $\ce{sp^2}$ hybridized atoms (carbon or nitrogen) that are separated by an even number of double bonds and no single bonds, then the molecule will not be planar.
So a general simple rule is that:

the molecule will not be planar if there is an $\ce{sp^3}$ hybridized carbon (or nitrogen) atom or two $\ce{sp^2}$ hybridized atoms of carbon/nitrogen which are separated by an even number of double bonds and no single bonds. Otherwise, its structure allows it to be planar.

Even though the molecule will have a structure that allows for it to exist in a planar conformation, there may be some/many that do not persist in a planar conformation due to steric effects, or complex three dimensional geometries.

In the problems you listed above, using this rule:

Not planar because there are no $\ce{sp^3}$ and the two $\ce{sp^2}$s are separated by an even number of double bonds.
Planar because there are two $\ce{sp^2}$s but they are separated by an odd number of double bonds (3) (and no single bonds)
Planar because there are no $\ce{sp^3}$s and only 1 $\ce{sp^2}$s that make 3 or more bonds (C or N).  The orbital geometry is NOT planar because the $\ce{sp^2}$ oxygen is separated from the $\ce{sp^2}$ carbon by an even number of double bonds.
Planar because 2 $\ce{sp^2}$s are separated by an odd number (1) of double bonds (and no single bonds)

I think these rules would make $\ce{C60}$ planar...

True.  Interesting point.  C60 might be described as having approximately planar geometry-it is basically a planar sheet cut and contorted into a spheroid.

Butatriene (which is listed above in the problem) is planar but that doesn't mean that the hydrogens will not be contorted out of plane.

Basically C60 is contorted graphite

I find that claim dubious, but instead of dwelling on that, here's another one. How about biphenyl? Planar or non-planar?

Biphenyl is only conformationally aplanar because of steric  interaction.

Also, in both C60 and other aromatics, there are not really any carbon-carbon single bonds and not really any carbon-carbon double bonds, so they are allowed to be a-planar by the rules above because no two SP2 hybridized carbons are separated by an odd number of double bonds.

Exactly, and steric interactions do a lot in terms of conformations that you haven't accounted for here. I'm just saying this is super-complicated.

Since Benzene is planar we can see that the bonds between two carbon atoms can not be defined as double bonds, but your example of biphenyl gives me an idea for an exception that does violate the letter of the rules.

Let us continue this discussion in chat.

I am a little bit confused.According to this linka) is not planar but according to your rules,it is the other way.Also you mentioned in your answer that butatriene is non planar but in your comments,you mentioned it as planar.Please resolve the issue.Thanks.

Sorry, I got my evens and odds reversed in the prior edit.  I have fixed them as well as the answers.  So basically if two SP2s are separated by an odd number of double bonds (1, 3 etc) then their single bonds will be in the same plane, but if they are separated by an even number of double bonds (with no single bonds in between) they will be rotated $90^o$

Note that b) butatriene is "textbook" planar, (it is a line of carbons with hydrogen fishtails), the hydrogens will have a tendancy to get twisted out of the same plane.

Likewise d) 1, 2 dicyanoethylene (both forms) is textbook planar but has a trans and a cis form and the cis form will probably twist mildly out of plane.

http://pubs.acs.org/doi/abs/10.1021/ed082p1329

Shouldn't then [10]-Annulene be planar(all sp2, no sp3)?


text:
Will glycerol and citric acid react? If yes, under what conditions?
I would like to know whether the reaction will take place if glycerol and citric acid are merged? In particular, what will happen if both substances are heated up to high temperature (+100°C)?

Well, they would create esters, but very slowly in r.t.

As @Mithoron said, it will form esters. Actual name is Citric acid esters of tri-glycerides or citroglycerides.

(source)
Where at least one of R1, R2 or R3 represents a citric acid moiety,
  one represents a fatty acid moiety and the remainder may represent
  citric acid, fatty acid or hydrogen.
Obtained by esterification of glycerol with citric acid [...] acid;
  consists of mixed esters of citric acid and edible fatty acids with
  glycerol; may contain minor parts of free fatty acids, free glycerol,
  free citric acid and mono- and diglycerides; may be wholly or
  partially neutralized with sodium hydroxide or potassium hydroxide.
They are white to ivory coloured, oily to waxy material. They are used
  as stabilizer, emulsifier, dough conditioner, antioxidant synergist.

According to this site:

The influence of various heating methods without catalysis to prepare
  polyesters from citric acid : glycerol blends were studied. In the
  presence of short-term microwave treatments, i.e., 60 s at 1200 W,
  blends of glycerol and citric acid invariably formed solid amorphous polyesters. Fourier transform infrared spectroscopy showed that
  citric acid and glycerol blends can form highly stable polymers
  composed of ester bonds. The glycerol citrate polyester polymers
  exhibited the least degradation in water, more in acid solutions
  (0.1–1.0M HCl), and the most deterioration in strong alkaline
  solutions (0.1–1.0M NaOH) after 72 h soakings. Polyesters of glycerol
  and citric acid were studied with differential scanning calorimetry
  and thermal gravimetric analysis. The polyesters were found to be
  thermally stable (up to 313°C).

A noble application of glycerol + citric acid esters or citroglycerides is in the making of a biodegradable thermoset polymers.
1. Source 1

A new biodegradable thermoset polymer, was made from simple,
  economical, biocompatable monomers without the need for a catalyst.
  Glycerol and citric acid, non-toxic and renewable reagents, were
  crosslinked by a melt polymerization reaction at temperatures from
  90-150°C. Consistent with a condensation reaction, water was
  determined to be the primary byproduct. The amount of crosslinking was
  controlled by the reaction conditions, including temperature, reaction
  time, and ratio between glycerol and citric acid. Also, the amount of
  crosslinking was inversely proportional to the rate of degradation. [...]

2. Source 2

The notable increase in global consumption of plastics and their long
  residence time in the atmosphere show the great need for biodegradable
  products. In this project, we developed biodegradable polymers based
  on citric acid and glycerol. Their synthesis was carried out at
  different conditions of constant temperature and concentration and
  they were synthesized using an economically viable method.
  Characterization was undertaken using: Acid number, FTIR spectroscopy,
  refractive index, viscosity, impact analysis, tensile test, hardness,
  calorimetry, the % humidity (oven method with recirculating air), and
  density determination, along with qualitative evidence to corroborate
  their biodegradability. These elaborated polymers were mixed with
  medical grade PVC formulation, obtaining a hybrid polymer, showing
  their mechanical properties were changed.


text:
Why can't the strength of superacids be measured in water?
I learned about acid strength, that the strength of an acid increases with it's degree of ionization when solvated. So, in water, a strong acid is one where $\ce{[H_3O^+]}$ is large, which is equal to a low pH: $\mathrm{pH=-log[H_3O^+]}$.
Considering extreme cases, such as superacids, I have found out that other methods are used to measure their acidity (methods I don't really understand). My question is why is it impossible to simply get super high concentrations of $\ce{[H_3O^+]}$ in aqueous solutions of superacids, and use this to determine the acid strength. Also, is pH used as a measure of acidity outside of aqueous solutions?
I have come over the leveling effect, but I don't think I fully understand it. The way I understand it (for the case with water as solvent) is that basically any acid in water will protolyze $\ce{H2O}$ to $\ce{H3O+}$, making this the effective acid. I don't understand why this would affect the measured pH, as it is $\ce{[H_3O^+]}$ you are measuring.

related http://chemistry.stackexchange.com/questions/15421/why-is-magic-acid-a-superacid

Any acid-base reaction is always an equilibrium:
$$\ce{HA^1 + (A^2)- <=> (A^1)- + HA2}\tag{1}$$
and for each pair of acids $\ce{HA^1}$ and $\ce{HA^2}$ you could calculate a $K_\mathrm{a}$ value to determine one acid’s strength with respect to the other. This $K_\mathrm{a}$ value is typically calculated according to equation $(2)$ if $\ce{(A^2)-}$ (which does not have to feature a negative charge; I just wanted to avoid different descriptions for the two acids) is the solvent.
$$K_\mathrm{a} = \frac{[\ce{HA^2}][\ce{(A^1)-}]}{[\ce{HA^1}]}\tag{2}$$
If $\ce{(A^2)-}$ is $\ce{H2O}$ (i.e. $\ce{HA^2}$ is $\ce{H3O+}$) and the acid $\ce{HA^1}$ is a strong acid or a superacid, then this equation is of little use; the equilibrium $(1)$ will be strongly shifted towards the product side, the concentration $[\ce{HA^1}]$ will be very small and thus $K_\mathrm{a}$ very large. With ‘very small’, I mean that $[\ce{HA^1}]\approx10^{-10}$ — acid $\ce{HA^1}$ is said to be fully dissociated.
Exact measurements of $K_\mathrm{a}$ become very hard if not impossible because you would need to determine extremely minute concentrations. The concentrations $[\ce{HA^2}]$ and $[\ce{(A^1)-}]$ are of little help since:
$$[\ce{HA^2}] = [\ce{(A^1)-}] \approx c_0(\ce{HA^1})\tag{3}$$
This is the levelling effect of water: no matter how strong the acid, there is no acidic species stronger than $\ce{H3O+}$ that can survive in aqueous solution for extended time. Most importantly, consider two very strong acids with different $K_\mathrm{a}$ — e.g. one with $\mathrm{p}K_\mathrm{a} = -15$, the other with $-10$. They have a very different acidity, yet the concentrations of all relevant species in aqueous solutions will be nigh identical.
Therefore, a new, more expandable way to measure and compare the strengths of very strong acids was required.

The reason we can differentiate weak acids in aqueous media is because the acids themselves are able to exist in the solution. Some dissociate more than others, so the same concentrations of different acids have different $\ce{[H3O+]}$.
Any acid that has a $\mathrm{p}K_\mathrm{a}$ lower than that of $\ce{H3O+}$ ($-1.74$), however, is said to be a strong acid, and cannot exist in aqueous solution. They dissociate completely, so for a solution of any strong acid ($\ce{HA}$), $\mathrm{pH}=-\log\ce{[HA]}$. This means that a $1\ \mathrm{M}$ solution of $\ce{HCl}$ ($\mathrm{p}K_\mathrm{a}=-6.3$, a strong acid) will have the same $\mathrm{pH}$ as a $1\ \mathrm{M}$ solution of trifluoromethanesulfonic acid ($\mathrm{p}K_\mathrm{a}=-14.7$, a superacid).
And to answer your other question, yes, $\mathrm{pH}$ is a quantity that is only used to measure aqueous solutions. A solution that doesn't contain water will always have $\ce{[H3O+]}=0$.

This gets a little fuzzy.  Most people identify nitric acid as a strong acid and yet Wikipedia gives pKa=-1.4>-1.74.  The "cutoff" for strong acids should reflect typical dilute concentrations instead of the total concentration of water, so should be $0$ (1M) or maybe even $+1$ (0.1M).

Acidity is solvent dependent, and water is but one of numerous solvents within which one may design an acidity scale.
More than 400 acidity scales have been developed over time, so you need to realise acidity in water is but one example! Aqueous acidity takes centre stage because of the importance of water in human life, and on planet earth, but it is not the only one!
Auto-protolysis of H2O being on the order of $10^{-14}$ gives rise to the aqueous pH scale of 14.
By definition, a superacid has what is called a "Hammett" acidity, $H_\mathrm o$, of $-12$ or lower. There is no equivalent of $H_\mathrm o = -12$ on the aqueous pH scale.
In water, the strongest acid you can generate, is $\ce{H3O+}$ (because $\ce{H2O}$ levels either to the acid, $\ce{H3O+}$, or the base, $\ce{HO-}$) . Any acid which is in fact stronger than the strongest acid in water, will simply level water to produce $\ce{H3O+}$, with an acid strength occupying the lower limit of pH in water (relative to the natural extent of $\ce{H3O+}$ or $\ce{HO-}$ as dictated by its auto-protolysis reaction) .
A superacid will simply be out of (off of) the scale of aqueous acidity, sort of in a class of its own, relatively speaking, with acidity strength which lays at lower pH levels than is accessible via aqueous pH,as a result of being a much stronger acid than $\ce{H3O+}$.
There is Brønsted-Lowry acidity in whatever solvent you want, aqueous and otherwise. Even liquid HCl (as its own solvent) has an auto-protolysis similar to that of $\ce{H2O}$.

You pretty much answered your own question, but perhaps I can help clarify: (1) Two very (but not equally) strong acids will both be effectively 100% dissociated in water, so the concentration of hydronium ion will be equal to the starting concentration of acid for either acid. Hence, the measured pH will be the same for the two acids. (2) Remember that the concept of pH as a scale of acidity from 0-14 becomes less relevant with increasing concentrations of acid (or other solutes) relative to water.

So, to differentiate between two strong acids (compounds which easily dissociate), you need a less alkaline solvent (which less easily take up protons) => the two acids protolyze to dissociate to different degrees?

There you have it.


text:
Why do cookies prevent toothpaste from foaming?
If I eat cookies and subsequently brush my teeth such that there is still small amounts of cookie material stuck to my teeth, the toothpaste will not foam as well as usually, making the process less enjoyable. This effect is consistently reproducible with most (sweet) cookies but any not other foods I have tried so far. The toothpaste contains typical foaming agents, such as sodium lauryl sulfate. What could be the mechanism behind this effect?

You might want to try with other fat-rich foods (say, buttery seared fish?) to verify @iad22agp 's answer and advance your observations.

My first take on this is that the cookies contain fat, which "consumes" the surfactants in the toothpaste (like sodium lauryl sulfate) so it does not foam up the way it normally would. You  might want to see what Wikipedia says about micelles and how they interact with fats. A similar effect can be seen when washing really greasy dishes with an insufficient amount of dishwashing liquid - in which case no suds will form. In your case I don't understand why the effect would not also be observed with other fatty foods such as chocolate or popcorn.

I tried this with chocolate today and it definitely seemed to reduce the foaming, so your hypothesis seems plausible. Maybe it is the fattiness combined with the tendency to stick to teeth that makes the effect so prominent with cookies.


text:
Should the measurement uncertainty be included in the random uncertainty?
So, the other day I was carrying out a titration and got the titres $33.4cm^3$ and $33.3cm^3$. I calculated the random uncertainty to be $0.05cm^3$. Now, the burette itself had an uncertainty of and I'm wondering whether I should add these two uncertainties together to find the uncertainty for the titres. Also, if I have to use the average of the titres to calculate the percentage uncertainty or whether I need to calculate the percentage uncertainty individually for the titres.

Related: Significant figures when standard deviation is 0

Via interpolation you should be able to read a burette to 0.01 ml. It depends on the class of burette, but I'd think all would be better than 0.05 ml.

Yes - to calculate the total uncertainty when adding or subtracting measurements, it's the room-sum-squared of all the uncertainties.
$$
e_{tot} = \sqrt{(e_{meas})^2 + (e_{rand})^2}
$$
The more generalized form of this will be the root-sum-squared of all sources of error and uncertainty.
$$
e_{tot} = \sqrt{\sum{(e_n)^2}}
$$
To get the quantity "$e_{meas}$" you apply the root-sum-squared formula of all the uncertainties of your measurement devices used to transfer liquids (transfer beaker, graduated cylinder, burette, etc...).
$$
e_{meas} = \sqrt{(e_{beaker})^2 + (e_{burette})^2+(e_{grad.cyl})^2+(e_{burette})^2...etc}
$$
=========
To answer whether you apply the burette uncertainty to your measurement, the answer is also yes. Assuming that the burette has an uncertainty of $±0.01 cm^3$, the actual reading of your measurements will be: $33.3cm^3 ± 0.01 mL$ and $33.4cm^3 ± 0.01 mL$. The two uncertainties in the measurement will be applied to the $e_{tot}$ equation above.

You can't end up with a measurement that has multiple kinds of units (i.e. statement However, in the case that your measurement devices use different units). If you're adding and subtracting then you use rms of error. If you're using multiplication and division then you use the rms of the % error.

Oh yikes, I've cleared that up to not cause confusion and specifying that these error calculations pertain only to addition and subtraction of measurements.


text:
Properties of a free radical scavenger
Antioxidants are free radical scavengers and are able to reduce free radicals to make them stable. What I am interested in is why some molecules make for superb free radical scavengers and why others don't. For example why Vitamin E is a better scavenger then say methane (I know the comparison is silly but for arguments sake).
If its possible could the answer be as simplified as possible as I have not yet reached an advanced enough level in chemistry. I would also like to stay away from the biological aspect of free radicals would like the answer to be in purely chemistry terms.


text:
Multiple solutions in the balancing of redox reactions
While looking at the reaction between potassium permanganate and hydrogen peroxide in the presence of sulfuric acid a friend and I were attempting to find a general equation to yield the coefficients of the balanced reaction depending on varying levels of acidity. I solved using half reactions and came up with equations for each coefficient in terms of the mole amounts for the manganese(II) ions and manganese(IV) oxide produced. My friend however, went about the problem by setting up a system of equations from the initial skeleton equation and yielded answers which depended on 3 variables.
From this, his method gave the answer
$$\ce{H2O2 + 4MnO4- + 6H+ -> 4O2 + Mn^2+ + 3MnO2 + 4H2O}$$
as a possibility, while my method yielded
$$\ce{7H2O2 + 4MnO4- + 6H+ -> 7O2 + Mn^2+ + 3MnO2 + 10H2O}$$
from the same ratio of manganese(IV) oxide to manganese(II). His reaction appears to balance correctly. Is there something he is missing that prevents that reaction from occurring with his coefficients, or does the half reaction method hide some possible answers?

What are the half reactions?

The first thing that you should realise is that under specific conditions (here: acidic ones) only one product is formed in this reaction. Permanganate is not reduced to both manganese(IV) oxide and manganese(II) cations, it will either cleanly give manganese(IV) oxide (if the acid concentration is very low) or cleanly give manganese(II) cations (if the acid concentration is sufficient). The only way to generate both is to have sufficient acid at first which is subsequently used up by the reaction and drops below a threshold. Thus, if you were titrating and notice the dark precipitate of $\ce{MnO2}$, immediately stop, throw away the analysis and start again as it has become worthless.
Therefore, there is only one simplest solution to the problem, depending on which half-reaction you prefer:
$$\begin{align}\ce{MnO4- + 8 H+ + 5 e- &-> Mn^2+ + 4 H2O}\tag{Red1}\\[0.6em]
\ce{H2O2 &-> O2 + 2 H+ + 2e-}\tag{Ox}\\[0.6em]
\ce{2 MnO4- + 6 H+ + 5 H2O2 &-> 2 Mn^2+ + 8 H2O + 5 O2}\tag{Redox1}\\[1.6em]
\ce{MnO4- + 3 e- + 4 H+ &-> MnO2 + 2 H2O}\tag{Red2}\\[0.6em]
\ce{H2O2 &-> O2 + 2 H+ + 2 e-}\tag{Ox}\\[0.6em]
\ce{2 MnO4- + 2 H+ + 3 H2O2 &-> 2 MnO2 + 4 H2O}\tag{Redox2}\end{align}$$


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Why is the strong acid written in a net ionic equation?
Strong acids dissolve and break apart into ions, so in a net ionic equation, why are they written?

If the strong acid starts not in the aqueous phase, then you can't cancel out the ions that it dissociates into. Also, these ions may participate in a reaction, for example, neutralization with a base or precipitation via its conjugate base. In such cases involving double displacement reactions, the component ions cannot be both cancelled out.
Example 1:
$$\ce{HCl_{(aq)} + NaOH_{(aq)} -> H2O_{(l)} + NaCl_{(aq)}}$$
Only $\ce{Na+}$ and $\ce{Cl-}$ are spectator ions that could be cancelled out.
Example 2:
$$\ce{HCl_{(aq)} + AgNO3_{(aq)} -> AgCl_{(s)} + HNO3_{(aq)}}$$
Only $\ce{H+}$ and $\ce{NO3-}$ are spectator ions that could be cancelled out.

So is it just an exception to the rule?

What's this rule you speak of?

The rule is that all aqueous ions are cancelled out on both sides on a net ionic equation

But they have to be aqueous on both sides of the equation. These are not.

So when a acid combines with a base, they produce a solid?

In water, acid base neutralizations produce water, which is a liquid.


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General Alkaloid Extraction Process
I'm getting into alkaloid extraction from various ethnobotanicals. It is a bit frustrating trying to extract alkaloids from ethnobotanicals without good guides out there, and I wanted to make my own techniques and experiments. However, I need to know what the basic outline is for extraction. Are there any guides or basic principles I need to know?
I'm interested in extracting for example Leonurine from Wild Dagga or Mitragyna and 7-OH-Mitragyna from Kratom.

What I was looking for was Acid-Base or Liquid-Liquid extractions.

Your question appears to be a little broad as written. Can you provide a narrower scope (such as the alkaloids you're interested in)? Also, what is "teks"?

Tek is short for techniques in various forums. I'm interested in extracting for example Leonurine from Wild Dagga or Mitragyna and 7-OH-Mitragyna from Kratom. I know there are several steps people use such as modifying the pH, defatting using different solvents, etc.

7-Hydroxymitragynine
The extraction and crystallization of the parent alkaloid, mitragynine, from dried leaves of Mitragyna speciosa (Kratom), has been described in great detail by A. H. Beckett, E. J. Shellard and A. N. Tackie in Planta Med., 1965, 13, 241-246 (DOI).
(Interestingly, the procedure also furnished speciofoline, a spiroindolo indolizidine alkaloid.)
Leonurine
A first extraction was seeming reported by S. Kubota and S. Nakajima in Nippon Yakubutsuqaku Zasshi, 1930, 153. I don't have a DOI and the source is most likely in Japanese.
I don't have personal experience with the extraction of guanidines from biological material and would probably rather try to synthesize leonurine, following the procedures described in Tetrahedron, 1969, 25, 5155-5161 (DOI).

General guidelines
I have given some hints on extraction in an answer to a question on the use of TLC in alkaloid identification.


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Oxidation numbers of atoms in Ti(SeO3)2?
What are the oxidation numbers of atoms in $\ce{Ti(SeO3)2}$? I'm having trouble figuring this out since I don't know the charge for either $\ce{Ti}$ or $\ce{Se}$.

The selenite anion ($\ce{SeO3^{2-}}$) has a net charge of -2. Titanium's oxidation state should be obvious given this information. As for selenium's oxidation state, since we know each oxygen has a charge of -2, and since the entire polyatomic anion has a net charge of -2, the following equation must balance:
$$\ce{Se}+3 \cdot (-2)=-2$$

Is that just one of the polyatomic ions i should know by memory or is there a method of knowing that?

Selenium polyatomic anions are analogous to the sulfur polyatomic anions; e.g. sulfate ($\ce{SO4^{2-}}$) and selenate ($\ce{SeO4^{2-}}$), sulfite ($\ce{SO3^{2-}}$) and selenite ($\ce{SeO3^{2-}}$)

Selenite is maybe not one that you should have memorized, but sulfite $\ce{SO3^2-}$ is. You can use what you know about the similarities in chemical behavior of the elements in the same group to make an inference about $\ce{SO3^2-}$.