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Difference between ideal & ideal-dilute solution
These two terms has been frequently used in Physical Chemistry by Atkins. He describes Raoult's Law using ideal solution while Henry's law using ideal-dilute solution.
I thought, at first they were same but Atkins in order to contrast the two laws presented graphs in  which one curve was for ideal solution & another was for ideal-dilute solution. Though he wrote two different sections for these solutions, I really couldn't contrast what the difference actually is between ideal solution & ideal-dilute solution. I googled it but to vain! Can anyone please help what he is trying to say by these two terms?

Raoult's Law is for ideal solutions (all of it: solvent and solute):

Raoult's Law only works for ideal solutions. An ideal solution is defined as one which obeys Raoult's Law. "An ideal solution shows thermodynamic mixing characteristics identical to those of ideal gas mixtures [except] ideal solutions have intermolecular interactions equal to those of the pure components." (all quoted from above link)

As stated here concerning ideal-dilute solutions:

"Many liquid mixtures deviate from the ideal behaviour predicted by Raoult's law. Ideal-dilute solutions are those for which the solvent obeys Raoult's law and the solute Henry's law. The difference in behaviour is because the solvent molecules tend to be surrounded by other solvent molecules so that they behave in an almost ideal way. However, the solute molecules tend also to be surrounded by solvent molecules so that their environment and therefore their thermodynamic behaviour is very unlike that of the pure substance." (all quoted from above link)

In summary: Ideal solutions are those which obey Raoult's Law; ideal-dilute solutions are ones where the solvent obey's Raoult's Law and the solute obeys Henry's Law.

Doesn't in ideal solution, solute molecules get surrounded by  solvent molecules? Mean to say what is actually happening when the solute is in ideal solution compared to the solute in ideal-dilute solution? Both the solutions are dilute. So, why does the difference occur?

In an ideal solution, it is assumed that there are no interactions between solvent, solute, and solvent/solute molecules/atoms.  The difference is in the solute: if it's ideal, then the system is ideal; if the solute is not ideal, then it's ideal-dilute. A perhaps silly example is: Ideal = helium solvent, neon solute (very low concentration). Ideal-dilute: helium solvent, oxygen solute (very low concentration).

I have understood , sir. In ideal reaction the interaction between solvent-solvent & solvent-solute is same but not in the case of ideal-dilute solution, right? Can't I use Raoult's Law still for solvent in ideal-dilute solution?

Correct. And yes, you can use Raoult's Law for the solvent in the ideal-solute case.


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What are actually the differences between Sidgwick & Powell theory and Nyholm & Gillespie Theory?
The famous VSEPR theory was proposed by first Sidgwick & Powell. It was modified then by Nyholm & Gillespie.
I was reading Inorganic Chemistry by J.D.Lee where he discusses very briefly the theory of Sidgwick & Powell; I couldn't somewhat conceive the language. So, I googled it but to my misfortune I could nowhere get more info other than this:

The idea of a correlation between molecular geometry and number of valence electrons (both shared and unshared) was originally proposed in 1939 by Ryutaro Tsuchida in Japan, and was _independently presented in a Bakerian Lecture in 1940 by Nevil Sidgwick and Herbert Powell of the University of Oxford. In 1957, Ronald Gillespie and Ronald Sydney Nyholm of University College London refined this concept into a more detailed theory, capable of choosing between various alternative geometries [1]

Everyone would give the historical info that Sidgwick & Powell was the first to introduce the theory which was later modified to the present VSEPR theory. I don't want that!
Can anyone please help me to bring me forward the original Sidgwick-Powell theory? Also the million-euro question: What are the differences between the two theories? In another way to say: What were the modifications that Mr. Nyholm & Gillespie brought forward that "improved the original" theory?
[1]https://en.wikipedia.org/wiki/VSEPR_theory#History

Perhaps you should read the original papers: N.V. Sidgwick and H.M. Powell, "Stereochemical types and valency groups", Proceedings of the Royal Society of London Vol 176, No. 965, pp 153-180 (1940), and R.J. Gillespie and R.S. Nyholm, "Inorganic Stereochemistry", Quarterly Reviews, Chemical Society 11 (4) 339-380 (1957).
To quote out of Gillespie's paper:

Sidgwick and Powell assumed that lone pairs of electrons and bonding pairs were of equal importance and that they arranged themselves symmetrically so as to minimise the repulsions between them. The ligand(or crystal)- field theory is concerned with the influence of the non-bonding electrons upon the arrangement of the ligands which are assumed to be negative ions or dipoles surrounding a central metal atom. Recently the influence of covalent binding has also been considered. These various theories are in fact closely related and an attempt has been made in this Review to integrate them. It appears that the simple theory of electron-pair repulsions receives considerable justification from quantum-mechanics and when refined to take into account differences between the repulsions between lone pairs and bonding pairs of electrons provides a very satisfactory theory of the stereochemistry of inorganic molecules.

That should get you going...

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Why is the resonance concept not required in molecular orbital theory?
In valence bond theory, resonance plays a pivoting role; why isn't such concept needed in MO theory? Why is it told that "MO theory provides a global, delocalized perspective on chemical bonding"?
(This is a follow up to What is actually the difference between valence bond theory and molecular orbital theory?)

Because MO is closer to reality.
Resonance is a convenient fiction used to simplify the visualization of some molecules, and rationalize their actions in the frame work of Lewis structures. But every resonance structure you have ever seen in your life is a lie.
Well, this is not exactly true. It's more precise to say that resonance structures are like blind men trying to describe an elephant: they each give a partial snapshot of truth while being wholly inadequate to describing it all. The "true" structure is a weighted average of all the resonance structures based on how stable they would be if they were the actual molecule.
MO cuts to the chase and attempts to directly describe the molecule in terms of its molecular orbitals: another convenient fiction, mind, but one that is closer to the truth. The reason you don't need "resonance" MO structures is because molecular orbitals are inherently a probabilistic construct: they already take into account the uncertainty in the electrons' positions.

"The reason you don't need "resonance" MO structures is because molecular orbitals are inherently a probabilistic construct: they already take into account the uncertainty in the electrons' positions."- what do you mean by that, sir? Atomic orbitals are themselves  wavefunctions; they are also probabilistic construct & resonance is the superposition of wavefunctions of different electron arrangements for same nuclear frame; it is also a probabilistic construct! What do you actually want to point out among them saying "probabilistic construct"??

Let me make myself plainer. An atomic orbital is probabilistic, but only takes into account the possibilities of an electron being in one atom space. In pure valence bond theory, either the wave function for the electron is over in one atom, or the other. The issue is that the electrons are in fact delocalized. In VBing, the resonance structures are what show this delocalization. However, in MO theory, the delocalization is inherent in the construction of the molecular orbitals: the wave functions are already distributed across all the atoms a given electron interacts with.

In MO theory, the molecule is described by Molecular Orbitals that are approximately calculated from atomic orbitals; I believe the process is referred to as constructing "Linear combinations of Atomic Orbitals" (LCAO).
My understanding is that you use "weighting coefficients" to say how much a particular atomic orbital contributes to a given Molecular Orbital.You also always get out as many MO's as AO's you put in.
Take for example, Benzene. If you start adding up the atomic orbitals it would be understandable to assign each atomic orbital equal weight in the calculation. So in the end you get 6 equivalent MO's.
I would like to hear someone else's perspective on this, but I'm guessing that you sort of "infer" a structure similar to that obtained from resonance because of this equivalency.

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Why is the original Pauling's theory of resonance that uses superposition of wavefunctions not used today?
I have recently studied superposition of states (also the famous Schrodinger's cat), measurement problem, decoherence & so on. I then read the resonance concept from Atkins' Physical Chemistry where it was written as:

It is the superposition of wavefunctions representing different electron distributions in the same nuclear framework. [...] A better description of a molecule of $\ce{HCl}$ is $\psi_\ce{HCl} = \psi_{\ce{H-Cl}} + \lambda\psi_{\ce{H^+} \ce{Cl^-}} $.

Thus this is the general wavefunction which is superposition of all the states. $\lambda$ is a numerical coefficient (can be found out by variation theorem) whose square predicts the probability of being in the ionic state when we inspect the molecule. Just like typical wavefunction-collapse after measurement! This is the original Pauli's theory of resonance. Unfortunately Atkins halts down here; after all it is physical chem-book! Nowhere have I found this approach to resonance; no organic chem book I possess: Morrison, Boyd; Solomons; R.K. Gupta ... They all say it is just the weighted average of all Lewis structures. No, it is not correct! As is said here:

This essay contains information that I have never found in any organic chemistry textbook. This can mean two things: the information is too difficult or you don't really need to know it. Maybe both are true .
[...]  When Linus Pauling developed resonance theory, he defined "resonance" as the superposition of wave functions. Unfortunately, this detail has been lost in the intervening decades, and modern chemists incorrectly treat "resonance" as the superposition of Lewis structures (resonance forms)
[...]Suppose you have a molecular AB that is a resonance hybrid:

  and you want to calculate the electron density on atom A.
The usual (wrong) prediction procedure. The usual procedure says, "combine what you see in each Lewis structure." In other words, combine the electron density (ED) on A in form IV with the electron density on A in form V:
$$ED_\text{wrong} \text{(hybrid)} = 1/2 [ED(IV) + ED(V)]$$
The much-ignored (correct) prediction procedure. You may recall that electron density (ED) is given by the square of the wave function (WF). If we want to know the electron density on A, we must look at the square of the wave function near A:
$$ED\text{(hybrid)} = WF\text{(hybrid)} ^2 \implies ED\text{(hybrid)} = ED_\text{wrong} \text{(hybrid)} + \boxed{WF(IV)WF(V)}$$
The final formula shows the discrepancy between the right answer, ED(hybrid), and the usual (wrong) answer, EDwrong. The correct formula includes WF(IV)WF(V). This term is called the "interaction density" or "overlap density," and is a product of two different wave functions. The interaction density affects the electron density everywhere in the molecule, and its appearance tells us that the electron density obtained by the "usual" procedure is wrong.
Electron density is just one of many properties that interest chemists, but all property predictions -- electron density, bond type, atom charge, dipole moment, geometry, energy, etc. -- have something in common with electron density prediction: the prediction procedure must start with WF(hybrid), and their answers always include interaction terms.

This is one of the great reasons why the modern concept of resonance is wrong; they don't take the "interaction term"!
So, why does only one book (which is not organic) contain the correct interpretation? Why do people follow the wrong version? Resonance is not at all the weighted average of Lewis structures; rather it is just a superposition of wavefunctions? So, why do most of the sources (people & book) advocate the wrong theory? Can anyone tell me which organic chem book still follows Pauling's original theory of resonance?
Added: In my quest to find the resources which still define Pauling's theory of resonance of wavefunctions, Feynman's Lectures, which I have just completed reading, is really worthy to be gold to be noted. Feynman completely describes the benzene molecule as the superposition of base states; it can be found in the CALTECH site.

If you look for the term "valence bond theory" you'll find more books that deal with the full/correct approach.  I have a colleague who's specialty is applying V.B. theory in computational organic chemistry, so there are certainly people who understand and use this approach.  However, there was a historical shift with the introduction of molecular orbitals theory where many people abandoned V.B. theory (M.O. calculations scale better for larger molecules).  Unfortunately, this means many chemists only have a vague/qualitative understanding of V.B. theory.

My two cents: when the concept of resonance is first introduced in general chem, or organic chemistry courses, describing it as a weighted average of resonance structures, is easier both for the instructors as well as for the students to gain a working intuition. Most curricula introduce quantum theory a little bit later.

If I recall correctly VB calculations scale extremely poorly, and not really used with anything bigger then a benzene molecule. In other words, VB may sound like a good idea but full VB treatment makes no practical sense.

why does only one book (which is not organic) contain the correct interpretation?

I bet that there are many books with correct descriptions of  it. Almost any computational chemistry book that makes mention to the subject.

Why do people follow the wrong version?

It depend of the individual. Many because ignorance. Others because they are lazy and prefer do not immerse in modern (¿1920-1930?) ideas of quantum mechanics. Others both of them.

Resonance is not at all the weighted average of Lewis structures; rather it is just a superposition of wavefunctions? So, why do most of the sources (people & book) advocate the wrong theory?

I am unsure about the exact meaning of sentence

It is the superposition of wavefunctions representing different electron distributions in the same nuclear framework.

In principle, any state (approximation to a solution of Schrödinger equation restricted to an antisymmetric subspace) is different to other state because the electron distributions are not the same.
Even more, eigenfunctions of the Hamiltonian operator are vectors that form a complete set. That implies that any state can be represented by a linear combination ("superposition") of eigenvectors (wavefunctions for example).
So "resonance" defined in such way is too vague. Arises so many question with this definition:  When a superposition implies resonance? The wavefunctions used for the combination, correspond to a stationary state? Is Atkins speaking about a particular kind of difference in electron distribution?
There is an awesome amount of literature about general chemistry, inorganic and organic chemistry with very serious mistakes, conceptual mistakes. Almost every book I've read in the subject. I do not believe that they are wrong because of simplification in post of clarity, because I can realize in which sense is good clarity about something that is completely wrong and not useful. Speaking about in-existent thing that contradicts quantum mechanics principles and at the same time do not allow make any prediction is not useful imho.

Can anyone tell me which organic chem book still follows Pauling's original theory of resonance?

Not me, sorry, I am not familiar with organic textbooks. I've seen some titles like "Theoretical and Computational Models for Organic Chemistry" that suggest a good theoretical background, but I can not assure it. I recommend you do not pay too much attention to resonance.

Ha! A nice answer; actually I was quite ignorant then; I had to learn that resonance is the weighted average of all the canonical structures. But in reality, as we all know (which, to me, was kind of revolutionary new idea then), resonance is the quantum superposition of wavefunctions describing different base states- this is greatly summed up in Feynman's Lectures & of course, in 'almost any computational chemistry book'. But really my main concern was the undergrad Organic texts which lack that description. Now, I'm pretty much more literate than the then me & I know quite many books ...

now which bear this more accurate definition of resonance. Though till now, I've not found any undergrad Organic text that deals with this definition- may be due to keeping the facts straight rather than rigorously analysing. I was a bit lazy not to delete this question but, well, it was destined for your answer ;)

Thanks. To my understanding those non rigorous ideas are useful to predict qualitatively with good probability of success. They allow to skip a rigorous treatment that can be very time consuming and sometimes technically  impossible to afford up to date. It is not bad to use them, but it is convenient to be aware of the inaccuracies of this ideas/models.

In my experience it is hard to find people that likes both approach, I suppose that this is the reason why there are not many good books in this sense (I do not know anyone). For the fast progress of chemistry the usage of both approach is a bless. :-) . But for "interdisciplinary" works, sometimes can be very difficult understand each other views, I certify that ;-).

But what are the implications of seeing it as a "superposition of resonance structures" rather than a "superposition of wavefunctions"? Other than the fact that it is inconsistent with the original theory and that it provides a wrong estimate of electron density, as illustrated in the question post above?

@TanYongBoon In practice this is not a problem if they are not taken too seriously. They are not taking into account when computing properties of molecules at equilibrium. However, they correspond to a model which is known to lack of good prediction power, and may lead to misunderstanding.

TL;DR:
It is a useful tool to use experimental values to skip math (and math that is tedious even without considering inter-molecular/environmental interactions like VDW and adsorption)

resonance concept from Atkins' Physical Chemistry where it was written as: It is the superposition of superposition of wavefunctions
$\psi_\ce{HCl} = \psi_{\ce{H-Cl}} + \lambda\psi_{\ce{H^+} \ce{Cl^-}} $.

That's a good interpretation, but a slightly incomplete one again. In here, if $\ce{HCl}$ is the hybrid form we want to study, it exists in a combination of:

a "perfect" covalent bond: $\ce{H-Cl}$ : where the electron cloud is uniformly distributed.
a "perfect" ionised form:   $\ce{H^+}$ separated from $\ce{Cl^-}$ sufficiently for their local electron wave functions to not interact (yes, yes, WFs are infinite - and this separation distance should be infinite too - hence the quotes around "perfect")
and a set of states where the electron cloud around the bond is shifted from ever-so-slightly to heavily polarised.

In effect, you're dealing with an integration problem over $N$ states where each state has a probability of $\lambda i\,\, (1 \Leftarrow i \Leftarrow N)$

incorrectly treat "resonance" as the superposition of Lewis structures (resonance forms)

Again, true, but this gives us three important advantages:

It allows us to represent most bonds/moieties as discrete extreme states. Think of the simplistic representation of the benzene ring and the electron jumping around the circle alternating between a polar $\ce{C^+ - C^-}$ bond and a $\ce{C=C}$ double bond.

Experimentally observed bond energies and polarisation constants give us a good relative approximation to allow us to represent the whole cohort of possible intermediate waveforms as a single:

$$\psi\textrm{(bond)} = \psi\textrm{(extreme-state 1)} + \lambda \psi\textrm{(extreme-state 2)} + \,\,\textrm{( more if the bond requires it)}$$

And given the number of molecules being observed, we can assume that some of molecules must be in these extreme states - and describe chemical reactions in familiar terms. (Bonus: With the tendency of the molecule to revert to a stable superimposition state, we can further explain why reactions that rely on said marginal states can fully use up the reactant)

A better, more concise explanation:
http://chemistry.stackexchange.com/a/38987/28144

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A compound that only absorbs UV-A radiation
As a similar idea to the question A compound that absorbs all visible light, I am curious if there is a compound (or compounds) that absorb UV-A radiation (shown below)?
However, preferable the compound would have relatively good transmission in the UV-B and Visible wavelengths.

Image source

Why would you want to avoid a tan but get burnt? ;)

Well, this question is related to observation of specific wavebands

Yes, there are many such compounds.  "The evolution of UVA protection" is a nice article detailing the progress on UVA blockers.  I'm sure you're aware that  sun care is a big business ($8.7 billion USD worldwide in 2013), so there is no shortage of chemists working in the area. In fact, there is enough (medical and clinical) focus that UVA (315-400 nm) has been subdivided into UVA1 (340-400 nm) and UVA2 (315-340 nm)
Here is a figure from the above link showing the UV spectra of some compounds.  The top part shows UVA absorbers and the bottom half shows broad-spectrum UV absorbers.

and a table with the names of the compounds

BMBM was an early UVA absorber (solid blue line), but was photochemically unstable (dashed blue line), in other words, its blocking power deteriorated with UV exposure time.  More recent UVA absorbers, like DHHB are photostable.
UVA absorbers belong to many different structural classes and contain many different types of functional groups.  Here are few examples.
BMBM

DHHB

TDSA


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What are some good references on conductometry as an analytical technique?
Non of the standard undergraduate textbooks in analytical chemistry (such as Skoog, Harris) discusses conductometry as an analytical technique. I wonder whether this technique is obsolete. Or is it taught under some other area such as physical chemistry ?
Can someone please point me to some good references on this matter ? I am particularly interested in text books.
Thanks.

I am certainly no expert in the field (in fact, I know nothing about it) but I imagine this 225+ page text will prove to be useful for you:
Oscillometry and Conductometry: International Series of Monographs in Analytical Chemistry, Volume 21 by E. Pungor is available via Google Books. You might be able to find a used copy of this book somewhere (it was published in 1965).


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Correct Electron Configuration of Ni-28
I have seen two electron configurations for $\ce{_{28}Ni}$ in my chemistry textbook:
$$[\ce{Ar}]~3d^8~4s^2$$
$$[\ce{Ar}]~3d^9~4s^1$$
Which of those is actually the correct one (in reality)? Why exactly?

NIST suggests here and here (search for Ni) that $\mathrm{[Ar] (3d)^8 (4s)^2}$, specifically in the state $\mathrm{^3 F_4}$, is the fundamental electronic state for neutral nickel atoms in empty space. However, as Wikipedia describes, if you don't specify the term symbol for the configuration, then the average energy of all terms for the $\mathrm{[Ar] (3d)^9 (4s)^1}$ configuration is slightly lower than the average energy of all terms in the $\mathrm{[Ar] (3d)^8 (4s)^2}$ configuration.
Now, why one configuration has a lower energy than another is a much harder question to answer in close cases such as this, as it involves a precise quantitative analysis of electronic energies for every possible electronic configuration of a nickel atom (a multielectronic atom). The general idea is that even though, for $Z=28$, the configuration $\mathrm{[Ar] (3d)^8 (4s)^2}$ is higher in energy than $\mathrm{[Ar] (3d)^9 (4s)^1}$ if you look at the sum of occupied orbital energies, the latter case has more interelectronic repulsions, which causes its total energy to be slightly higher. A lot of information on this can be found in this previous Chem.SE question and the several nested questions within.
Interestingly, the gas-phase fundamental configuration of the monocation $\ce{Ni^+}$ is $\mathrm{[Ar] (3d)^9}$, which further shows how nickel's electronic configuration is teetering on the boundary between the options.

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How to convert ethanal to 2-hydroxy-3 butenoic acid?
How to convert ethanal to 2-hydroxy-3 butenoic acid?
I tried crossed aldol condensation with formaldehyde followed by acid addition.After that how should I proceed ?

Note that 2-hydroxybut-3-enoic acid has four carbon atoms and ethanal has two. What is the product of the aldol self-condensation of ethanal? Does that give you any hints?

Once you have got your acrolein from the aldol condensation of ethanal and formaldehyde you can react it with $\ce{NaCN, HCN}$ under kinetic control to get direct addition of the cyanide to the enone, forming a cyanohydrin[1]. The nitrile can then be hydrolysed in aqueous acid to yield the carboxylic acid.


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Difference in acid strength of oxalic acid and malonic acid
Despite very similar skeletal structures, the difference in acid strength between oxalic acid (ethanedioic acid) and malonic acid (propanedioic acid) is quite significant. What is the reason for that?
It seems that the inductive effect plays a role, but what is the explanation for the inductive effect in oxalic acid as there is same group on opposite sides?
How does the inductive effect plays different roles in both acids?

Inductive effect decreases significantly with no of atoms in between the inductive group and the group on which we are checking the effect on. In case of malonic acid there is one carbon atom between the two carboxyl group. As the -I effect decreases so the stability of carbanion and so the acidic strength. I think this may some what explain

Linear saturated dicarboxylic acids
$n$ = number of $\ce{-CH_2}-$ groups in chain
 n                        pKa1     pKa2   Ka1/Ka2
 0  Oxalic acid           1.25     4.14   776
 1  Malonic acid          2.83     5.69   724
 2  Succinic acid         4.21     5.41    15.8
 3  Glutaric acid         4.34     5.41    11.7
 4  Adipic acid           4.41     5.41    10.0
 5  Pimelic acid          4.50     5.43     8.51
 6  Suberic acid          4.526    5.498    9.4
12  Dodecanedioic acid    4.45     5.05     4.0

Dodecanedioic acid
Aromatic carboxylic acids
                                   pka1     pka2    pka3
 benzoic acid                      4.202    ----    ----
 ortho-phthalic acid               2.89     5.51    ----
 meta-phthalic acid                3.46     4.46    ----
 para-phthalic acid                3.51     4.82    ----
 o,o'-bibenzoic acid
 p,p'-bibenzoic acid
 benzene-1,3,5-tricarboxylic acid  3.12     3.89    4.70

Linear unsaturated dicarboxylic acids
$n$ = number of $\ce{-CH=CH}-$ trans groups in chain
n                            pka1      pka2
0  Oxalic acid               1.25      4.14
1  Fumaric acid              3.02      4.38
2  trans,trans-Muconic acid  3.87
3

Straight-chain, saturated carboxylic acids
$n =$ total number of carbon atoms in molecule ($n \geqslant 2$ has terminal methyl group).
n                     pKa
1  Formic acid         3.77
2  Acetic acid         4.76
3  Propionic acid      4.88
4  Butyric acid        4.82
5  Valeric acid        4.82
6  Caproic acid        4.88
15 Pentadecanoic acid  4.8

most $\mathrm{p}K_\mathrm{a}$ data from Wikipedia.

How does the inductive effect plays different roles in both acids?

First think of what happens when the carboxylic acid group of a straight-chain, saturated carboxylic acids ionizes to form the carboxylate ion. The $\ce{C=O}$ and $\ce{-C-O^{-}}$ bonds of course hybridize. This hybridization occurs faster than hydrogen bonds to the solvent form. The net result is effectively two oxygen atoms with 1/2 a charge each. This increases the ability of the  carboxylate ion to solvate, which increases its acidity.
In malonic acid the $\ce{C-C}$ bond "shares" some of that hybridization. This is the "inductive effect." The additional delocalization of the electrons from the carbon atoms makes the carboxylate ions more electronegative than they would be in a long linear saturated dicarboxylic acid. So the first ionization ,  the second ionization

Is this still under development? Maybe since it was reactivated now, it would be a good time to remove that note...

@Martin-マーチン - Thanks for the heads up. I took off the under development text. I'm not doing anything else with this. If some else wants to find some additional values then they are free to add them to the tables.

First, notice that a carboxylic carbon ($\ce{COOH}$) has an oxidation state of $\mathrm{+III}$ (except in formic acid where it is $\mathrm{+II}$). This means, that the carbon is very electron deficient and wishes to draw electrons towards it from sources that are not the two oxygen atoms (because they are stronger). We term this an inductive effect, as we cannot easily draw mesomeric structures to explain it.
Since the inductive effect is mainly based on electrostatic interactions, it gets exponentially weaker with increasing distance. In oxalic acid, there is one one bond separating the two carboxylic groups, so they can exert a strong inductive effect. In malonic acid the effect is weaker due to the separating $\ce{CH2}$ group. And in succinic and subsequent diacids it is almost unmeasurable due to the distance between the groups.
But how do we rationalise each group exerting a $+I$ effect on the corresponding other group? Well — we cannot deprotonate twice immediately (I am introducing this as an axiom, but it does have reasoning that I feel too lazy to explain right now). So we have to pretend one side is going to be inert while the other is deprotonated. And if we do that, we see that the inductive effect of one side is strong in drawing away electron density from the $\ce{COOH}$ group we are deprotonating. Since this means there is a lower effective negative charge, deprotonation happens more easily, i.e. at lower $\ce{pH}$.
We can use this simple picture since the groups are isotopic, i.e. they can be transformed into each other by a $C_2$ rotation of the molecule’s inherent symmetry.

Acidic nature is directly proportional to strength of conjugate base.
In oxalic acid, the hydrogen oxalate ion and the oxalate ion (Which are conjugate bases in first and second dissociations) are resonance stabilized. The negative charge on oxygen atom is in conjugation throughout the molecule (with two pi bonds). Hence it is more stable.
In malonic acid, the conjugate base is relatively less stable, as the conjugation is restricted to only one pi bond.
Resonance effect is more stronger than inductive effect and moreover, resonance effect in the conjugate base of oxalic acid takes place throughout the molecule.(there is a negative charge, pi bond, pi bond, lone pair conjugation ).
Hence oxalic acid is relatively more acidic as compared to malonic acid.

Please show me the resonance structure which transfers the charge from one carboxylic unit to the other.

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text:
Preservation and bleaching of sliced mushrooms with hydrogen peroxide
According to the article Extending the Shelf Life of Fresh Sliced Mushrooms, it states that 5% (v/v) hydrogen peroxide is an effective means for the preservation of sliced mushrooms.
In regards to the preservation:

In the USA hydrogen peroxide
  is an approved bactericide for some dairy products and is used for disinfecting fruit
  and vegetables. Its activity is due to its oxidising effects on
  bacteria and it also bleaches mushrooms during the soaking period.

How does oxidising effects cause the bleaching, hence preservation of sliced mushrooms?

Hydrogen peroxide is quite a strong oxidizer. First, let's address the bleaching effect: It bleaches by oxidizing colored compounds, making them colorless. From Wikipedia:

An oxidizing bleach works by breaking the chemical bonds that make up the chromophore. This changes the molecule into a different substance that either does not contain a chromophore, or contains a chromophore that does not absorb visible light. This is the mechanism of bleaches based on chlorine.

This is also the mechanism of hydrogen peroxide as a bleach, since both chlorine-based bleach and hydrogen peroxide are oxidizers.
As for the antibacterial effect: hydrogen peroxide in this particular concentration (5%) may or may not be effective against bacteria (and viruses, fungi, etc.) due to the presence of the enzyme catalase in some bacteria and fungi. In those organisms which it is effective against, hydrogen peroxide would probably work by causing oxidative stress in the organism.


text:
Safety/health hazards: working with 2-aminobenzimidazole, iminoesters, and 1,3,5-triazine
How dangerous is it working with 2-aminobenzimidazole, iminoesters and 1,3,5-triazine? Has there been any cases of health problems recorded concerning those three compounds?

You should edit your question immediately - eliminate personal info and state problem more precisely.

I wanted to do it myself but wasn't sure how to state it. IMO, if promoter gives you such work, you shouldn't worry too much - you need to be cautious in lab on daily basis and probably worked with more dangerous stuff not even knowing about it.

Your previous revision stated that you studied some stuff about their hazard $\ldots$ could you provide a link to what you studied?

Uhm. This is really really weird. You want to quit your chemistry study because you worry about lab safety? Laboratories are generally safer than freeways and should pose no hazard when being properly guided, which is in principle what a university is for. I'm not sure whether I believe the story of yours. It's also none of my business. Just saying that we are not here to help you synthesize your own pharmaceuticals...

@user19185 - If you are very concerned about how to proceed if you believe your research has become hazardous to your health, you could ask a question at Academia. I would avoid personal details and the specifics of your research so that the question could be general enough to be useful. Also, be clear that you are not looking for legal or medical advise. The internet is a terrible place to find either.

Looking up the chemical you're working with is always a good idea, but you need to look closer and evaluate the specific danger those chemicals present to you. What you want to look at are the danger symbols and the hazard and precautionary statements. Those will tell you the specific dangers a chemical presents.
If you look up your first substance, 2-aminobenzimidazole, it looks pretty tame for organic chemistry standards. It has an exclamation mark as danger symbol, which doesn't tell you much, but the absence of a skull and crossbones and a health hazard symbol tell you that this probably won't kill you immediately. This doesn't mean the stuff is harmless, but it is also not extremely toxic. If you look at the hazard statements of this chemical you'll find that you shouldn't get it into your eyes, on your skin or breathe it. All of that is a good idea when working with any chemical, and this is exactly what you should learn when studying chemistry. This is a really tame chemical compared to a lot of stuff organic chemists work regularly with, but you still should avoid exposing yourself to it.
Your second item is an entire category of chemicals, I'll skip it for this reason. The third one is a bit more dangerous than the first one, it has a corrosive and a health hazard danger symbol. You want to avoid contact with it, but that's what all your protective equipment like safety goggles, your lab coat and your fume hood are for.
I'm not saying those chemicals are harmless, they're not. But they are less dangerous than many other common chemicals in an organic chemistry lab.

I've answered this in part based on an older revision of the question, as the newer one doesn't have enought context.


text:
Why do we not see silicon in a structure like graphite?
Silicon exists in a structure similar to diamond, with 4 silicon atoms bonded to each other in a tetrahedron.
Why do we not see it exist in a structure like graphite?
Is it even possible to get silicon to exist like this? If so could we make graphene but with silicon atoms?
Not sure why you would, since I understand carbon graphene will still be cheaper. But I would just like to know.

An answer is this: http://www.madsci.org/posts/archives/1997-08/866552381.Ch.r.html.

It's possible, and it's already been done! The compound is called Silicene.

However, silicene generally isn't perfectly flat. (Then again, as  Karl Ratzsch points out in the comments, graphene isn't perfectly flat either). The Wikipedia article notes that it usually takes a puckered shape.
I don't know the reasons for the puckered shape, but I recall being quite surprised that silicene could be synthesized at all. The reason why is that I was taught in high school that silicon very rarely forms double bonds due to its large size. For a visualization, please refer to the absolutely awful image below (if anyone could provide an actual graphic I'd be grateful).

The black atoms on the left are carbon. The $p$ orbitals overlap to a significant degree, which means that pi-bonds can easily form. The green atoms to the right are silicon. You can see that, because of the larger atomic radius, the p-orbitals don't overlap as much, meaning that double bonds are generally not favorable in silicon. This explains why compounds like silenes are relatively unstable.
Pi systems are crucial to the chemistry of carbon--they allow carbon to form multiple bonds, flat rings (like in graphene), and generally be stable in a lot of configurations. The lack of pi-bonding in silicon is sometimes given as a rationalization for why silicon-based lifeforms can't exist, and explains why it's so hard to get silicon into the flat sheets that form silicene.

Graphene is not flat, either. Cannot be, because only 3D crystallites have perfect long-range order. Landau-Peierls instability this is called.

@Karl I've just asked Are all stable 2D Xenes and binaries buckled?


text:
What is Enthalpy?
At school, they teach the following two formulas: $$\Delta H = -mc\Delta T$$ and $$q= mc\Delta T$$ What i am not sure is that my teacher says that q and $\Delta H$ are the same thing but to think of q as a magnitude and hence it is always positive. However what I am confused about is how can q always be positive? What if the change in temperature is negative? Also, what is the difference between q and $\Delta H$. My current understanding is that they are the same thing and they measure the change in heat energy. Is this right?

Welcome to chemistry.SE! If you had any questions about the policies of our community, please visit the help center.

Enthalpy is thought as the total potential energy of a system. The actual thermodynamic equation for enthalpy is: $$\Delta H = \Delta U + \Delta {(pV)}$$ where $\Delta U$ is the change in internal energy. Internal energy can be thought as the total amount of heat energy there is in a system. Hence $\Delta U$ can be thought as the total energy that is required to make a system, while $\Delta {(pV)}$ can be thought as how much energy is required to make space for the system.
Q on the other hand is actually a different thing to enthalpy. Q refers to the heat transfer of a system. Note that this means that q can be positive or negative (if heat energy moves from the system to the surroundings) and hence it is not always positive. The correct formula for heat transfer is: $$q = \Delta U -w $$ So actually, q doesn't always have to equal to enthalpy change. However at a school level it is taught that they equal the same thing. This is because of we assume that pressure is constant (which is a common assumption made as usually most experiments have a constant pressure), enthalpy becomes: $$\Delta H = \Delta U + p\Delta V$$ Also since $w = p\Delta V$, enthalpy becomes: $$\Delta H = \Delta U - w = q$$ Hence, enthalpy and q equal the same values only when pressure is constant.
The equation that your teacher has given you is actually an equation for heat transfer, not for enthalpy change. However because enthalpy equals heat transfer in most cases, it is commonly said that enthalpy change also equals to that equation. However the one mistake is that there shouldn't be a minus in the equation for enthalpy change. The equation for heat transfer and enthalpy should both be the same.


text:
Physical Interpretation of Gibbs Free Energy
I know that $\Delta G = \Delta H - T\Delta S$ and that if $\Delta G$ less than $0$ a reaction will be spontaneous. What I am wondering is what is the actually physical interpretation of Gibbs free energy. I mean, if the Gibbs free energy of a particular reaction $20\ \mathrm{J\ mol^{-1}}$, what does that actually mean?

I'd be careful with the word 'spontaneous' - as an example, graphite has the lowest free energy of the carbon solid phases at room temperature and pressure.  Graphite is about 2.9 kJ/mol more stable than diamond.  Yet, diamonds don't mysteriously turn in to chunks of graphite - you need to get over a nucleation barrier and that barrier is pretty high.

Gibbs free energy is a  measure of the total non-expansionary work that a system can do. So in the case of your example, that system has the potential to do 20 joules of non-expansionary work per a mole of substance. Non-expansionary work refers to work other than pressure-volume work, such as electrical work. Hence not only is Gibbs free energy is important to tell whether a reaction will be spontaneous or not, it is also important to engineers as they tell how much non-expansionary work a system, such as motor engine, can do.
Gibbs free energy is really derived is from the Clausius inequality which is: $$dS\geq \frac{dq}{T}$$ Basically is states that in a reaction, the change in entropy must always be equal (in the case of an ideal gas) or greater than the transfer of heat energy divided by temperature. Therefore, for a reaction to be spontaneous, the following relation must be satisfied: $$dS-\frac{dq}{T}\geq0$$ Now, lets assume that the pressure is constant so we can simplify this relation.
Constant Pressure
Enthalpy is given by: $$\Delta H_p = \Delta U + p\Delta V$$
and internal energy by: $$\Delta U = q + w$$
Now by substituting $\Delta U$ in the expression for enthalpy we get: $$\Delta H _p= q + w + p\Delta V$$ Now, since w = $-p\Delta V$, the above equation simplified so that: $$\Delta H_p = q$$ So, now in the Clausius inequality, we can substitute dq with dH and multiply by $T$ to get: $$TdS \geq dH$$ Now by integrating both sides and putting all the variables on one side, we get the question for gibbs free energy: $$\Delta G = \Delta H - T\Delta S$$ and hence this is why $\Delta G$ has to be negative for a reaction to be spontaneous

Thanks for the answer. Also, I am also interested in the derivation of the formula. Could you please show me the derivation for Gibbs free energy also.

@benzene The Gibbs free energy can be derived by doing two subsequent Legendre transforms on the formula for U... Is that enough to get you started?

What are Legendre transforms?

@benzene Eh I think it's easier to say that Gibbs energy is defined to be $G = H - TS$. That's what most texts would write. At const $T$, $\Delta G=\Delta H-\Delta (TS)=\Delta H-T\Delta S$. The reason why we choose this definition of $G$ is because it's convenient to have the intensive variables $p$ and $T$ as the differentials: for a reversible process $dG=V,dp-S,dT$, at const. $p$ and $T$, $dp=dT=0$ so $dG=0$, which is the condition for equilibrium. Legendre transformations are just another way of saying the same thing. https://en.wikipedia.org/wiki/Legendre_transformation#Thermodynamics

@benzene Consider if we stuck with the internal energy $dU = T,dS - p,dV$. The condition for equilibrium would then be $dU = 0$ at const volume (which is still okay) and const entropy - but there's no way you can make sure a system stays at const entropy. To convert this to something usable, you need to do the Legendre transformation - since $T$ and $S$ are a pair, you just add $TS$ to $U$. You could stop here to get the Helmholtz free energy $A$, but for us, doing one more step and subtracting $pV$ to get $G$ makes things even easier since we like dealing with const $p$, not const $V$.

I'm going to concentrate on the $T\,S$ term (which you likely find the most mysterious) and hopefully give a little more physical intuition.
The $T S$ term arises roughly from the energy that is needed to "fill up" the rotational, vibrational, translational and otherwise distractional thermal energies of the constituents of a system. Simplistically, you can kind of think of its being related to the idea that you must use some of the energy released to make sure that the reaction products are "filled up" with heat so that they are at the same temperature as the reactants. So the $T S$ term is related to, but not the same as, the notion of heat capacity: let's look at this a bit further.
Why can't we get at all the energy $\Delta H$? Well, actually we can in certain contrived circumstances: let's look at these circumstances to find out why in a meaningful definition we must leave behind an amount $T\,\Delta S$. Let's think of the burning of hydrogen:
$$\ce{H2 + 1/2 O2 -> H2O};\quad\Delta H \approx 143\:{\rm kJ\,mol^{-1}}\tag{1}$$
This is a highly exothermic reaction, and also one of the reactions of choice if you want to throw three astronauts, fifty tons of kit and about a twentieth of the most advanced-economy-in-the-world’s-1960s GDP at the Moon.
The thing about one mole of $\ce{H2O}$ is that it can soak up less heat than the mole of $\ce{H2}$ and half a mole of $\ce{O2}$; naively this would seem to say that we can get more heat than the enthalpy change $\Delta H$, but we'll see this is not so. We imagine a thought experiment, where we have  a gigantic array of enormous heat pads (all individually like an equilibrium “outside world") representing all temperatures between absolute zero and  $T_0$ with a very fine temperature spacing $\Delta T$ between them. On my darker days I find myself imagining an experimental kit that looks eerily like a huge pallet on wheels of mortuary shelves, sliding in and out as needed by the experimenter! We bring the reactants into contact with the first heat pad, which is at a temperature $T_1 = T_0 - \Delta T$ a teeny-tiny bit cooler than $T_0$  thus reversibly drawing some heat $\Delta Q(T_1)$  off into the heat pad. Next, we bring the reactants into contact with the second heat pad at temperature $T_2 = T_0 - 2\,\Delta T$, thus reversibly drawing heat $\Delta Q(T_2)$ off into that heat pad. We keep cooling then shifting to the next lowest heat pad until we have visited all the heat pads and thus sucked all the heat off into our heat pads: see my sketch below:

Now the reactants are at absolute zero. There is no heat needed to "fill them up" to their temperature, so we can extract all the enthalpy $\Delta H$ from the reaction as useful work. Let's imagine we can put this work aside in some ultrafuturistic perfect capacitor, or some such lossless storage for the time being.
Now we must heat our reaction products back up to standard temperature, so that we know what we can get out of our reaction if the conditions do not change. So, we simply do the reverse, as sketched below:

Notice that I said that $\ce{H2O}$ soaks up less heat than the reactants. This means that, as we heat the products back up to standard temperature, we take from the heat pads less heat in warming up the water than we put into them in cooling the reactants down.
So far, so good. We have gotten all the work $\Delta H$ out into our ultracapacitor without losing any! And we're back to our beginning conditions, or so it seems! What's the catch?
The experimental apparatus that let us pull this trick off is NOT back at its beginning state. We have left heat in the heat pads. We have thus degraded them: they have warmed up ever so slightly and so cannot be used indefinitely to repeatedly do this trick. If we tried to do the trick too many times, eventually the heat pads would be at ambient temperature and would not work any more.
So we haven’t reckoned the free energy at the standard conditions, rather we have simply calculated the free energy $\Delta H$ available in the presence of our unrealistic heat sink array. To restore the system to its beginning state and calculate what work we could get if there were no heat sink array here, we must take away the nett heat flows we added to all the heat pads and send them into the outside World at temperature $T_0$. This is the only "fair" measure, because it represents something that we could do with arbitrarily large quantities of reactants without the presence of artificial, finite resources like heat pad arrays.
But the outside World at $T_0$ is warmer than any of the heat pads, so of course this heat transfer can’t happen spontaneously, simply by dent of Carnot’s statement of the second law!
We must therefore bring in a reversible heat pump and use some of our work $\Delta H$ to pump this heat into the outside world to restore standard conditions: we would connect an ideal reversible heat pump to each of the heat pads in turn and restore them to their beginning conditions, as sketched below:

This part of the work that we use to run the heat pump and restore all the heat pads, if you do all the maths, is exactly the $T\,\Delta S$ term.
The above can also cast some light on an endothermic reaction. In such a reaction, we imagine having an energy bank that we can borrow from temporarily. After we have brought the products back up to temperature, we find we have both borrowed $-\Delta H$ from the energy bank and put less heat back into the heat pads than we took from them. So heat can now flow spontaneously from the environment to the heat pads to restore their beginning state, because the heat pads are all at a lower temperature than the environment. As this heat flows, we can use a reversible heat engine to extract work from the heat flowing down the gradient. This work is, again, $-T\,\Delta S$, which is a positive work gotten from the heat flowing down the temperature gradient. The  $-T\,\Delta S$ can be so positive that we can pay back the $\Delta H$ we borrowed and have some left over. If so, we have an endothermic reaction, and a nett free energy: this energy coming from the heat flowing spontaneously inwards from the environment to fill the higher entropy products (higher than the entropy of the reactants). A good thought experiment here is to imagine towing an iceberg into warmer latitudes, and then putting it into contact with the outside world through heat engines which convert a fraction $1-T_\mathrm i/T_\mathrm o$ of the heat flowing into work.
Take heed that, in the above, I have implicitly assumed the Nernst Heat Postulate -the not quite correct third law of thermodynamics - see my answer here for more details. For the present discussion, this approximate law is well good enough.


text:
How to convert ethanol to 1,1-dichloroethane?
How to convert ethanol to 1,1-dichloroethane?
My approach: Convert ethanol to ethene followed by $\ce{HCl}$ addition. But that's no use. Well if it is possible to create gem-diol from here, then maybe it can be done. But not getting a suitable reaction. Can't seem to figure out anything. Please help!

Well you have to at least tell us what your thoughts on this is.

Addition of chlorine gas with UV light to methane forms polychlorinated methane, so something alone these lines is nearing your goal.

convert ethanol to 1,1-dichloroethane?

A standard route for producing gem-dichloro compounds involves treatment of a ketone or aldehyde with $\ce{PCl5}$ (see the second to last paragraph in the "Chlorination of organic compounds" section here and page 18 here).

(image source)
Therefore, you could carefully oxidize your ethanol to acetaldehyde and treat it with $\ce{PCl5}$.


text:
Methods to prepare lactic acid from acetaldehyde
I got one method [here](using acetaldehyde and carbon monoxide)1.
But is there any alternative method (using specific catalyst)?

What is the downvote for?

I'm not the downvoter or the upvoter, but it would be really nice if you add a summary of what's written there here. I could assume the downvote is because someone wasn't inclined to click on an external link to find out what your question really is.

A fairly simple two step synthesis involves nucleophilic addition of $\ce{HCN}$ to the carbonyl group and then aqueous hydrolysis of the nitrile to yield the carboxylic acid product.

Acetylene + $\ce{HCN-Ba(CN)2}$ → vinyl cyanide
Vinyl cyanide - hydrolysis → acetaldehyde cyanohydrin
Acetaldehyde cyanohydrin - dil HCl → lactic acid


text:
Mechanism for oxidative cleavage of tertiary alcohols by NaOCl
I came across this today

An ester $\ce{C4H8O2}$ on reaction with $\ce{CH3MgCl} $ and hydrolysis gives an alcohol. The alcohol on reaction with $\ce{NaOCl}$ and $\ce{H3O+}$ gives acetic acid. Identify the structure of the ester.

But what I'm facing a problem with is the role of $\ce{NaOCl}$. My teacher said it can cleave and oxidize a tertiary alcohol, say tert-butyl alcohol, to acetic acid.
But I couldn't understand the mechanism. Any help will be appreciated.

There are not so many possibilities for your ester that yield an alcohol which could be oxidized to acetic acid. Which esters could you imagine from $\ce{C4H8O2}$?

My problem is not with the ester...but how will NaOCl oxidize the the alcohol obtained...the mechanism..

I don't see any evidence in SciFinder that this is an actual transformation. Can you confirm with your teacher that a simple tertiary alcohol can be oxidized to an acid?

I actually think you are misinterpreting the problem, but I'll wait for confirmation about the tertiary alcohol oxidation.

@jerepierre Maybe i misinterpreted what my teacher said.But it will take atleast a week before i can confirm with him.But what why did you say i'm "misinterpreting" the problem?What do you think will be the solution?What is the particular ester C4H8O2 according to you?What will be NaOCl mechanism according to you?Maybe i'm having a conceptual problem...but its always good to discuss :-)

Reaction of an ester with a Grignard reagent gives two products: (1) a tertiary alcohol coming from the carbonyl carbon and two equivalents of the Grignard reagent and (2) an alcohol liberated from the ester. I believe it's this liberated alcohol that is reacting with bleach, since a tertiary alcohol is typically resistant to oxidation. However, the liberated alcohol might be primary, secondary, or tertiary depending on the identity of R2.

The information given says that after treatment with bleach, acetic acid is formed. This tells us two things: (1) that liberated alcohol is a primary alcohol, since the oxidation goes all the way to an acid and (2) that liberated alcohol has two carbons, since acetic acid has two carbons. That implies that the liberated alcohol is ethanol. The starting ester has four carbons total, which leaves two carbons for the carboxy fragment. The original ester was ethyl acetate.


text:
Does alcohol hydrogen in the enol tautomer of a β-dicarbonyl have sp² character?
My reasoning for the somewhat implausible "$\mathrm{sp^2}$" character of the hydrogen between these two oxygens on the enol structure of the right comes from recognizing that if it were, then we would have 6 $\mathrm{sp^2}$ atoms with 6 π-electrons, so maybe that extra bit of instability from having a higher orbital on hydrogen could be sort of taken care of by being stable like an aromatic compound and forming a hydrogen bond.
So how could I go about figuring out if this is true, either with MO theory or something else?

(source: ucdavis.edu)

So are you asking whether the hydrogen $2p$ orbitals are involved in bonding?

Yes, I think so, or if they are occupied ever, for instance would I expect this compound to experience a chemical shift when put into a magnetic field similar to an aromatic compound.

Not a chance. 2p are just too far away. Look at the energy levels diagram.

As @IvanNeretin points out, the hydrogen is s-hybridized like normal.  This $\ce{O..H..O}$ structure is just another example of a 3-center-2-electron bond involving hydrogen.  Diborane ($\ce{B2H6}$) is another example.  See here, and links therein, for more information on such bonds involving hydrogen.

In reference to the possibility of downfield NMR shifts, the C-13 and H-1 spectra from SDBS show no signs of aromaticity.

Wait a minute. Diborane is electron-deficient, but this thing is not. See, we had a normal 2-electron O-H bond, then another O went in with its lone pair (2 more electrons). If anything, this should be a 3-center-4-electron construction.

@IvanNeretin Yes, I agree, I should have written "3-center-4-electron" in my earlier comment.  Thanks for catching that.

Yes, it has 3c-4e character, but it is limited, as here's asymmetric h-bond - one of contributors much more dominating


text:
Titration of H3PO4 with NaOH
I came across an ionic equilibrium problem stating:

Find the pH when 150 ml 1 M $\ce{NaOH}$ has been added to 100 ml 1 M $\ce{H3PO4}$.

I'm stuck with this question. What I know:
$$\ce{H3PO4 + NaOH -> NaH2PO4 + H2O}$$
$$\ce{NaH2PO4 + NaOH -> Na2HPO4 + H2O}$$
Finally 50 millimoles $\ce{Na2HPO4}$ and 50 millimoles $\ce{NaH2PO4}$ remains.
Should I apply acid buffer equation?

The answer to your question does not need a single calculation.
You have got your 1 molar phosphoric acid that will get deprotonized by the 1 molar sodium hydroxide solution that is added. As both solutions have got the same concentration, you will reach the first inflection point after 100 mL and the second inflection point after 200 mL.
Now what is between those two volumes, at your desired volume of 150 mL?
A simple buffer system that can be described using the Henderson–Hasselbalch equation:
$$\mathrm{pH}=\mathrm pK_\mathrm a-\log_{10}\frac{\ce{c(H2A- )}}{\ce{c(HA^2- )}}$$
As both concentrations $\ce{c(H2A- )}$ and $\ce{c(HA^2- )}$ are equivalent, the logarithmic expression gets zero and your final pH equals the second $\mathrm pK_\mathrm a$ value of phosphoric acid, which is 7.21 (according to the German Wikipedia).
$\hskip3.5cm$

This is a simple question  involving a mixture of strong acid (not that strong) and a strong base. Since $k_a$ values are not given the book might have wanted you to solve it like this. They want you to assume complete dissociation of both compounds. That is, 5 moles of $\ce{NaOH}$ gives 5 moles of $\ce{OH^-}$ ions after complete dissociation.
Find the equivalents of $\ce{H^+}$ from phosphoric acid (a) and the equivalents of $\ce{OH^-}$ from $\ce{NaOH}$ (b). Also calculate the total volume. Number of equivalents=$NV$. But be careful since concentration is given in molarity.
If $b>a$, then the resulting solution is basic. Calculate the equivalents of $b$ that remain. From this you will get pOH.
$$[\ce{OH^-}]=\frac{b-a}{V_1+V_2}$$
If $a>b$, then resulting solution is acidic. Calculate the amount of $a$ that remains. From this, you will get pH.
$$[\ce{H^+}]=\frac{{a-b}}{V_1+V_2}$$
Where N is normality and equivalents of acid is greater than equivalents of base.


text:
Urban air pollution and the smell of freshly cut lawn grass?
Recently, all the lawns in the apartment complex that I live in have had their lawns cut - giving that pleasant smell of freshly cut lawn.
Now, I am aware that the smell arises from a chemical distress signal from within the plant itself - are the chemicals from freshly cut lawn contributors to urban air pollution?
Related: How to determine source of air pollution

The information page from the Compound Interest page of the same name describes it well (below).
A summary of the main points are that grass already emits small amounts of VOCs (volatile organic compounds) - mostly comprising of 6-carbon atom based aldehydes and alcohols.
When the lawn is cut, the main smell is from (Z)-3-hexenal, at which only 0.25 parts per billion can be sensed (smelled) by humans - this compound is short lived, hence why the smell is a freshly cut smell.
As a little twist, in research presented in the paper Air Pollution and the Smell of Cut Grass, the authors suggest that a significant proportion of aerosol VOC's in urban air pollution actually originate from the release of these chemicals when lawns are cut - and given the sheer number of manicured lawns in an urban centre, this starts to make sense.
Some of these released chemicals are highly photoreactive. The flux of VOCs is seasonal as well in many places, contributing to seasonal variations of VOC-based photochemical smog.
Within the article, VOC flux measurements taken in the urban area of Melbourne Australia indicate that 6% of the flux arises from uncut grass; however, this flux doubles when the lawn is cut. The combined flux is comparable to industrial sources.
It is suggested in the article, that

within an
  urban airshed, during the growing season, about one third of the emissions of
  photochemically reactive VOCs come from grass and cut grass.


text:
What happens to the complex part of the wave function when doing LCAO?
I understand that in general the wave function can take complex numbers, however when talking about combining atomic orbitals to form molecular orbitals we talk about the phase of the orbital being positive or negative to create constructive or destructive interference. It seems as if we have lost the aspect of the wave function taking on complex phase, what happened to it?

It got very scared because sh*t just got real and squared away ;)

Also, generally chemists prefer working with real wavefunctions, which are obtained by taking appropriate linear combinations of the complex wavefunctions. https://en.wikipedia.org/wiki/Table_of_spherical_harmonics#Real_spherical_harmonics This is allowed because any linear combination of degenerate eigenfunctions is also an eigenfunction of the Hamiltonian with the same eigenvalue. Another benefit is that these real wavefunctions have convenient directional properties e.g. the p orbitals are aligned in the x, y, and z directions. Same goes for the d orbitals.

I understand that in general the wave function can take complex numbers [...]

It is not that it can, rather it should: by the very postulates of quantum mechanics wave function is a complex-valued function. So, both the spin orbitals (that are one-electron wave functions) and the Slater determinant built out of them are in principle complex-valued functions.
However, in solving the HF equations by the SCF procedure it is quite typical to impose some constraints on the spin orbitals, for instance, restrict them to be real-valued functions rather then complex-valued ones. As with any other constraint this restriction might lead to what is called SCF instabilities, i.e. situations when relaxing the constraints lead to a different variational solution of lower energy.
To elaborate a bit more let us mention few other well-known (as well as lesser-known) constraints for the SCF procedure:

The infamous restricted Hartree-Fock (RHF) model with the requirement that spin orbitals come in pairs: two spin orbitals corresponding to two different pure spin states are constructed from the same spatial orbital:
$$
\begin{aligned}
  \psi_{2i-1}(1) &= \phi_{i}(1) \alpha(1) \, , \\
  \psi_{2i}(1)   &= \phi_{i}(1) \beta(1)  \, .
\end{aligned}
$$
In the unrestricted Hartree-Fock (UHF) model the requirement above is relaxed and we exclusively use spatial orbitals from one set to construct $\alpha$ spin orbitals and spatial orbitals from another set to construct $\beta$ spin orbitals:
$$
\begin{aligned}
  \psi_{2i-1}(1) &= \phi_{i}^{\alpha}(1) \alpha(1) \, , \\
  \psi_{2i}(1)   &= \phi_{i}^{\beta}(1) \beta(1)   \, .
\end{aligned}
$$
A bit less known is the fact that the unrestricted Hartree-Fock method is not so unrestricted. In fact, there is a constraint here: each and every spin orbital describes an electron in a pure spin state, either $\alpha$ or $\beta$, while in general (in accordance with the postulates of quantum mechanics) electron can be in superposition of this states:
$$
  \psi_{i}(1) = \phi_{i}^{\alpha}(1) \alpha(1) + \phi_{i}^{\beta}(1) \beta(1)
$$
This most general setting is known as the general Hartree-Fock (GHF) method.

And all these three methods (RHF, UHF, GHF) exist in two variants: the real one, in which spin orbitals additionally required to be real-valued functions, and the more general complex one. All this give rise to six variants of the HF method with many possible instabilities between them discussed it some details in the seminal paper by Schlegel & McDouall1:

There is also a very similar earlier paper by Seeger & Pople.2 which suddenly is not freely available.
Its important to realize that any constraint of the mentioned above can only raise the electronic energy: with a great deal of certainty, we may expect that if any constraint is relaxed, the variational procedure will result in a lower energy due to greater variational freedom. Thus, for instance, for the most usual different real variants of the Hartree-Fock method we have
$$
    E_\mathrm{e}(\mathrm{RGHF}) \leq E_\mathrm{e}(\mathrm{RUHF}) \leq E_\mathrm{e}(\mathrm{RRHF}) \, .
$$
The same, of course, is true for the corresponding complex variants of these methods,
$$
    E_\mathrm{e}(\mathrm{CGHF}) \leq E_\mathrm{e}(\mathrm{CUHF}) \leq E_\mathrm{e}(\mathrm{CRHF}) \, .
$$
And for any of the three formalism
$$
    E_\mathrm{e}(\mathrm{C}x\mathrm{HF}) \leq E_\mathrm{e}(\mathrm{R}x\mathrm{HF}) \, , \quad \text{where} \quad x = \mathrm{G}, \mathrm{U}, \mathrm{R} \, .
$$

RHF/UHF/GHF and correct spin symmetry
The possible rise of electronic energy that accompanies introduction of more and more constraints in the GHF -> UHF -> RHF sequence seems to be quite contrary to the goal of the variation method. However, UHF and RHF constraints are nothing but symmetry constraints: they arise by requiring an approximate electronic wave function to have the same spin symmetry as the exact non-relativistic one, i.e. to be an eigenfunction of spin operators, the total spin-squared operator $\hat{S}^2$ and the $z$-component of the total spin operator $\hat{S}_{z}$.

With respect to $\hat{S}_{z}$ operator it is well known that any Slater determinant built out of spin orbitals corresponding to pure spin states is an eigenfunction of $\hat{S}_{z}$. Thus, RHF and UHF wave functions are eigenfunctions of $\hat{S}_{z}$, but GHF wave function is not.
The situation with $\hat{S}^{2}$ is a bit more complicated, but all in all only in RHF (but not in UHF or in GHF) formalism it is possible to construct an approximate electronic wave function which is an eigenfunction of $\hat{S}^{2}$.3

To conclude, as it was pointed out by Löwdin, we always face with a dilemma here: should we seek a solution that is a true variational minimum or should we seek a solution with the correct spin symmetry?

1) H. B. Schlegel and J. J. W. McDouall, Do You Have SCF Stability and Convergence Problems? in Computational Advances in Organic Chemistry: Molecular Structure and Reactivity, Springer Netherlands, 1991, pp. 167-185. DOI: 10.1007/978-94-011-3262-6_2. Free PDF from wayne.edu.
2) Seeger, R., & Pople, J. A., Self‐consistent molecular orbital methods. XVIII. Constraints and stability in Hartree–Fock theory. The Journal of Chemical Physics, 66(7), 1977, 3045-3050. DOI: 10.1063/1.434318.
3) This can be done even for an open-shell electron configuration, although it would require a linear combination of Slater determinants with constant coefficients, none of which alone is an eigenfunction of $\hat{S}^{2}$.


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How do I calculate the cost of a single Aluminum atom?
I am doing a Chemistry assignment in which I was given a piece of aluminum foil which I had to measure and use for my calculations. I am supposed to calculate the cost of a single aluminum atom using the given data from my teacher and the data which I have measured from my piece of aluminum foil. I understand the basic concept of how I am supposed to do this, but I am having some trouble. Here is the data:
Data given by teacher:
  Cost of a roll           $5.53
  Roll size                30.5 cm x 30.5 m
  Density of Al            2.70 g/cm^3
  Molar Mass Al            26.98 g/mol

Data measured from my piece of Aluminum foil:
  Mass     1.35 g
  Size     18.6 cm x 12.5 cm

I have determined a way to do this, but I am having trouble with it. First of all, I have determined the amount in moles which is $\mathrm{0.0500~mol}$. Then I multiplied the amount in moles with Avogadro's constant ($\mathrm{6.02 \cdot 10^{23}~mol^{-1}}$) to determine the number of aluminum atoms in the piece of aluminum foil I received which is $\mathrm{3.01 \cdot 10^{22}}~\text{atoms}$. Now I just have to find the cost of a single aluminum atom, but this is where I am stuck. I cannot use the cost of the aluminum foil roll to divide it by the number of atoms since I was given a piece of aluminum foil so I need to find the cost of the piece of aluminum foil and then divide it by the number of atoms.
So basically my question is: How do I find the cost of the piece of aluminum foil? If I can find the cost then I can divide it by the number of atoms to determine the cost of a single aluminum atom.
However, if there is a way to find the number of atoms in the roll of aluminum foil, I could then use the cost of the aluminum roll to find the cost of a single aluminum atom. Please explain if there is a way to do this using the data given by the teacher.

you could always look up the bulk price of aluminium per tonne (this will be a lot less than the foil but will still be valid). Aluminium is a commonly traded commodity and will be sold by weight. You already know the weight of the sample...

@matt_black I cannot use any other data other than the ones I was given and the ones I measured.

You can compute the area of aluminum foil in the roll as:
$$30.5\;\rm{cm}\cdot 3050\;\rm{cm} = 93025\;\rm{cm}^{2}$$
You have measured the mass per square centimeter as:
$${1.35\;\rm{g}\over 18.6\;\rm{cm}\cdot 12.5\;\rm{cm}} = {1.35\;\rm{g}\over 232.5\;\rm{cm}^{2}}$$
You have been given the cost of the roll (and thus the cost per area):
$${\$5.53\over 93025\;\rm{cm}^{2}}$$
Using those numbers, the molar mass for aluminum, and Avogadro's constant (here, it's just the number of atoms so I designate it as such):
$$\rm{cost\;per\;atom} = \left({\$5.53\over 93025\;\rm{cm}^{2}}\right)
\left({232.5\;\rm{cm}^{2}\over 1.35\;\rm{g}}\right)
\left({26.98\;\rm{g}\over 1\;\rm{mol}}\right)
\left({1\;\rm{mol}\over 6.022\cdot 10^{23}\;\rm{atoms}}\right)
$$
$$\rm{cost\;per\;atom} = \$4.59\cdot 10^{-25}$$

One more thing, could you explain to me why in the equation it is 232.5/1.35 and above the equation it was 1.35/232.5? Why was it flipped? I can't seem to understand that. Thanks!

It's just to make the units come out (dimensional analysis). It can be as I wrote it initially or inverted:  it's the ratio that matters. How it's written depends on what units cancel and make it through to the end. Same with Avogadro's number here.

I did this using one estimate for the number of aluminum atoms in 4.55 grams of aluminum, scaled it up to a kilogram, and divided aluminum's current price per kilo (US\$1.87) by my estimate for the number of atoms in a kilogram of aluminum, yielding a price per atom of about $\\\$8.453 \cdot 10^{-26}$, or around eight sextillionths of a penny.
Of course, I'm only in eighth grade, so my calculations may be way off.


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Can linear SOCl2 exist?
Specifically, instead of its normal structure, this Lewis structure seems to be valid as a constitutional isomer:
$\ce{Cl-O-S-Cl}$
So what prevents this from existing, or if it already exists how is it made?

Anything with Cl-O bond is a strong oxidizer, sulfur in the oxidation state +2 is a reducing agent, so the thing will pretty quickly transform into common $\ce{SOCl2}$.

Just for clarification: You do mean ‘linear’ $\ce{Cl-O-S-Cl}$ as opposed to trigonal pyramidal $\ce{O-SCl2}$, correct?

Not for long. Sulfur is in this case a reducing agent (love to give electrons away). And Oxygen is an odixidizing agent. It's just not stable or favorable. But sure you can make it.

Ivan's comment is better than this answer - you need to back up claims about possibility of creation in the answer

Not quite. I failed to add a discrediting note like "...should this thing exist, it will..."

@Ivan Neretin Yes, it looks like it would be better, but still...

I highly doubt that it could exist. The co-linear arrangement, $C_{\infty\mathrm{v}}$, of $\ce{Cl-O-S-Cl}$ is a second order saddle point on the DF-BP86+D3(BJ)/def2-SVP level of theory on the $S=0$ potential energy surface. It has two imaginary bending modes, $\nu_1=371.39\mathrm{i}$, $\nu_2=108.23\mathrm{i}$. (Please note that they are actually perpendicular to each other.)

That basically means, that distortion along these modes will lead to an arrangement of the nuclei which is lower in energy. Since this is a second order saddle point, this arrangement can itself only be a transition state (first order saddle point), which also cannot be considered stable. (Compare also here.)
Therefore distortion among $\nu_1$ would probably lead to $\ce{Cl-O}$ and $\ce{S-Cl}$ fragmentation. While distortion along $\nu_2$ might preserve the $\ce{S-O}$ bond, fragmenting the chlorine atoms/ chloride ions - but that is pure speculation of a hypothetical case.

Let's try a very tempting ad-hoc explanation of why this is the case. In that case let us assume all atoms are $\mathrm{sp}$ hybridised and the bonding axis is $z$. The Lewis structure that seems to be valid has therefore three lone pairs ($\mathrm{sp}$, $\mathrm{p}_x$, $\mathrm{p}_y$) at each chlorine and two lone pairs ($\mathrm{p}_x$, $\mathrm{p}_y$) at the oxygen and sulfur. The three σ bonds are formed by the $\mathrm{sp}$ orbitals.
That would result in four lone pairs that are co-aligned in the same plane. Overlap is inevitable, and in the most simplest terms of VSEPR theory, we would call that lone pair repulsion. (Please note that for strictly VSEPR theory we would have to assume tetrahedral hybridisation.)
In other words, same spin electrons have to avoid occupying the same space and since the orbitals are so close together this will rise the energy a lot. You can probably call it Pauli repulsion.
The MO picture is too messy, so that I won't show it here. Basically most of the occupied orbitals are anti-bonding with respect to one of the bond. The HOMO is of π symmetry and anti-bonding with respect to every bond.

This is a good answer, but unfortunately I wasn't really interested in sp hybridized linear, I meant it in the "straight chain alkane" linear sense where the oxygen and sulfur are likely sp3 hybridized.

@Kainui you mean in a zig-zag fashion? Well then it is not linear, is it?

It's not linear in that symmetry-phys chem sense, but in sense of org chem, i.e. 'linear alkane chain', it is somehow "linear".

Yeah exactly @ph13 , I mean it seems kind of funny to imagine an oxygen in a truly linear state, I can't think of a single case where oxygen is truly 'linear' in that sense lol.


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Difference between 'homogeneous mixture of several chemical compounds' & 'non-homogeneous system'
I just started reading Thermodynamics by Enrico Fermi; he started defining state of a system for homogeneous mixture of several chemical compounds & then non-homogeneous system as:

I thought non-homogeneous system to be a mixture of several homogeneous chemical compounds separated by boundaries. But Fermi seems to have done it otherwise; he has defined the state separately for homogeneous mixture of several chemical compounds & non-homogeneous system that is, mixture of several homogeneous chemical compounds. I am not getting how they are different at all. Can anyone please explain how they are different? What's the difference between  homogeneous mixture of several chemical compounds & non-homogeneous system?

You are right that normally a non-homogeneous system would be one where there is boundary between two internally homogeneous phases (like milk which is basically oil and water mixed at a very fine scale).
But Fermi as also allowing for the possibility that there may be only one phase where the concentration of its components varies continuously in space (if there is more than one phase the variation is discontinuous by definition).
It certainly isn't a common situation if the phase in in equilibrium where we would expect things like diffusion to create a uniform concentration of things over time, at least in liquids and gases). But it is very common in the real world where equilibrium is rarely reached (imagine the state of a poorly stirred cup of tea where the concentrations of tea-stuff will be larger near the tea bag than far away from the tea bag).
It is also very common in places where it is hard to get to equilibrium which is often true in solids (I don't know whether this is what Fermi was thinking of). Pyrex glass, for example, is apparently a single phase but the structure is deliberately made different near the surface than far away from the surface to create tension in the surface layers that protect the glass from thermal shock. But the properties vary continuously.
The examples I can think of may not be relevant to what Fermi was saying as they mostly are metastable and won't be at equilibrium which is often what thermodynamics restricts itself to worrying about.


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Chemical Test for tin
How can I distinguish whether a metal that I have is tin ($\ce{Sn}$)? I am specifically looking for a chemical reaction that will happen only with tin and not with other metals such as $\ce{Al,~Cu,~Fe,~\ldots}$. Do such reactions exist? What are they?

I think this question would be incomplete without mentioning what metals the sample could be made of, i.e. all of the metals.

@inɒzɘmɒЯ.A.M It could be made of: $\ce{Al,~Ti,~V,~Fe,~Co,~Ni,~Cu,~Zn,~Sn}$.

Beside determining the volumic density (tin: 7.31 g/ccm, aluminum: 2.7 g/ccm, etc. pp) and as asked here on chemistry SE, there is a quick pre-test.  In German it is known as "Leuchtprobe", literally "illumination probe".  Be reminded to protect yourself properly (safety googles, fume hood, gloves).
In a first instance, you have to reduce the tin compound.  For this, a little of your solid under investigation is finely mixed together with some grains of zinc (Zn). In a porcelain dish, to this mixuture is added 5 mL of 20 m% aqueous HCl.  Await this reaction ceased.  Zn will produce hydrogen, and because it is freshly created (statu nascendi), it is of superiour reducing power compared to bottled hydrogen gas, and the solution is cooled to room temperature.
Then dip a cool test tube, partially filled with cold water, into this solution.  While still wetted by the solution, rapidly  place the test tube into the upper part of the non-luminous flame of a Bunsen burner.  Blue fluorescence is an almost specific evidence for tin (caused by $\ce{SnCl2}$, and stannanes), already at low concentrations.
Presence of arsenic may hamper this fluorescence.  Niob-compounds yield some similar fluorescence, too.  Beside using tin-free chemicals and equipment to be used -- and prior to the test of your unknown -- do at least once the test in known absence of tin, and once with a sample with tin to learn how to discern the two instances.

One can better detect the fluroescence by adding potassium permanganate into the test tube to create a dark background. (Or so I heard; I let others do that test in the ion lottery practical course.)

I agree with this comment; a test tube partially filled with a cold aqueous potassium permanganate solution increases the contrast compared to a test tube simply filled with blank water. Own practical experience in the lab (not only during the initial (in)organic Marcha analítica separation process

but as TA, too) that a test tube hold too long unshaken eventually leads to a bumping solution, including a stain on cloth and skin.

Is there any other analytical method to recognize tin?

My initial answer assumed you look for a qualitative test with as little as possible techniqual equipment and chemicals.  To go further depends on the sample matrix and often, there are reliable procedures to do this.  The U.S. american NIST, for example, developped digestion standards how to proceed properly, like this more senior reference (http://nvlpubs.nist.gov/nistpubs/jres/21/jresv21n1p95_A1b.pdf).   Atomic absorption spectroscopy (AAS) would represent an alternative (https://en.wikipedia.org/wiki/Atomic_absorption_spectroscopy), tin being listed in the table "GFAA method development".

By coincidence, AAS was a recent answered question, too: http://chemistry.stackexchange.com/questions/30892/atomizers-in-aas

Don't look at the fluorescence too long, because the water in the tube locally overheats within ten seconds: splash!

@Karl And safety glasses/goggles.  And chemically speaking, to perform the test at least once with a blank (no tin) and once with presence of tin as each measurement is a comparison against a reference. The illumination in the lab (fluorescent tube / LED lumiophore / day light) can influence how well it is seen, too.

We always put some KMnO4 into the tube with cold water, so it's practically black. Makes it much easier to see the fluorescense. And gives brown stains everywhere in case of spillage. And yea, safety glasses! ;)


text:
Why does oxidized copper become green?
Why does oxidized copper turn green? I am sorry if this question is too basic, but am really stumped here. It would be helpful if one could provide a research paper or website with your answer.

Copper oxide is not green, but green verdigris, basic copper carbonate (or acetate) forms on copper exposed to air and moisture.
Heating copper in air can form a near-transparent film of copper oxide, which can exhibit all the colors of the spectrum due to interference of light. The same effect cause a "rainbow" in oil slicks on water.

I don't think this answers the question, just says how not why. I am imagining some kind of answer with d-orbitals.

The question is a misstatement. Read it

Huh, I still don't see it, it sounds like someone's asking for an explanation for the color change of copper when it's oxidized. Oxidized does not mean "gains oxygen".   I think Aaron is trying to understand why copper carbonate and/or copper acetate are green.

Copper when is exposed to air, it gets oxidized and a tarnish of green color is formed on it.As Cu is a transition metal, it possesses color according to the "crystal field theory".
So, Cu reflects green color when it comes in contact with oxygen.


text:
Is there a relationship between an electron's energy and its distance from the nucleus?
I have read that in Bohr's model of the atom electrons in the same atomic orbital have the same energy, and that this energy (as a result of the electron existing in a fixed atomic orbital) is related to the distance of the electron from the nucleus. However, using the quantum mechanical model of the atom, in which an orbital is the area of space around an atom where the probability of interacting with an electron is greatest, it is implied that there is no relationship between an electron's energy and its distance from the nucleus. Is there no relationship, or have I misunderstood something else?
Also, it is often mentioned that an atom is 99.99% empty space etc., but since the probability of interacting with an electron is described by its wave function (which suggests it is possible, though unlikely, to interact with an electron very close to the nucleus), this suggests to me that this isn't necessarily true. What am I misunderstanding?

Erase Bohr's model from your head - it leads down many seedy back alleyways that end in nothingness... The quantum wave function is the electron for all practical purposes.

Also, since the introduction of the wave function, there is no such thing as free space.

I will start off by addressing Jon's comment above. Yes, the Bohr model is flawed. I think it is still worth learning about it just from a historical standpoint, to see how we discovered the quantum mechanical description of the electron, but when you are studying it you absolutely have to remember that all of what he said was effectively rubbish.
Now on to your questions.

Is there a relationship between an electron's energy and its distance from the nucleus?

The first thing that should be said is that in QM, unlike in the Bohr model, there is no well-defined "distance from the nucleus". Electrons exist everywhere (except at nodes) and the probability density is given by $\lvert\psi\rvert^2$. However, it is possible to calculate the "most probable distance from the nucleus" using the radial distribution function, $r^2 R^2$, where $R(r)$ is the radial part of the wavefunction. Essentially, you take the derivative of this with respect to $r$ and set it to 0 - you can read more here. For a 1s electron in hydrogen, you will find that the most probable radius is approximately $52.9 \text{ pm}$, a quantity called the Bohr radius and denoted $a_0$.
(As an aside: oddly enough, this correlates with the flawed Bohr model, which said that the radius of the 1s electron in hydrogen was $52.9 \text{ pm}$. However, what Bohr said was that the electron was always at this distance, which is wrong. QM says that the electron is most likely to be found at this distance.)
Anyway, you will find that there is some degree of correlation between the most probable radius and the energy of the electron. If we consider the 2s orbital of hydrogen, you can perform exactly the same calculation as you did for the 1s orbital, and you will find that the most probable radius is $4a_0$.

Also, it is often mentioned that an atom is 99.99% empty space etc...

That's just a simplified view presented in popular science (or elementary chemistry classes) that ignores the quantum mechanical model. No serious chemistry book would write something like that. I don't know exactly how it's calculated, but I can make a good guess. You take any atom you like, let's say hydrogen. You look up the atomic radius - Google tells you it's $52.9 \text{ pm}$. (No surprises there.) So you can think of the atom as a sphere with that radius, and calculate the volume of the sphere: $V = \frac{4}{3}\pi r^3$. Then you go and look up the volume of an electron and a proton (a physicist would get a headache from just reading the phrase "volume of an electron"), and add it up, and you find that the "atom" is "99.99% empty space".
As Ivan succinctly said - with the QM description of the atom, there is no longer such a thing as free space since the electron "exists everywhere". Apart from that, just like how the distance from the nucleus is no longer well-defined in QM, the "volume" of an atom is also no longer well-defined. We pretty much threw that view out of the window.


text:
Why does Avogadro's Law hold true for ideal gases?
Avogadro's Law states that the volumes of ideal gases are proportional to the number of molecules that are present in the gas (or the number of moles in the gas). I've been trying to find out more about it, but I can't seem to find out why this holds true, at least for ideal gases.

Avogadro's Law actually states that the volumes of ideal gases at a fixed temperature and pressure are proportional to the number of molecules that are present. This is actually a sub-case of the more general ideal gas law, which states $PV = nRT$ (or $PV = nkT$, if you're working on a per-particle basis) .
The ideal gas law actually falls out of a basic statistical mechanics analysis of the behavior of a collection of molecules which interact solely by elastic collision. (Here's a 25 minute video of one such derivation.)
Basically, the more particles you have in a given region with a certain fixed speed (i.e. a fixed temperature), the more frequently they collide - not only with each other, but with the walls of the container. This means that putting more particles in the (same size) container (at the same temperature) increases the outward pressure on the container walls. In fact, the rate of collisions (and thus the pressure) is directly proportional to the number of particles (doubling the number of particles at a fixed temperature doubles the rate of collisions).
But we want to hold the pressure constant! If we can't change the temperature, one way to reduce the pressure is to increase the volume (e.g. let the container expand like a balloon.) As the volume grows, each particle has to travel farther before it can collide with the walls of the container, meaning that the rate of collisions and thus the pressure on the walls gets less and less.
It just so happens that this relationship between pressure and volume is inversely proportional (that is, Boyle's Law). So to halve the pressure, we need to double the volume. So if we double the number of particles, to keep the same pressure we need to also double the volume. Or more generally, the volume is increased in direct proportion to the number of particles - that is, each mole of gas takes up a certain volume (at a fixed temperature and pressure). If we try and put it into a different volume, the pressure (or the temperature) will change.
Okay, but why doesn't a fixed number of different particles take up different volumes? For example, why doesn't doubling the mass of the particles make it take up half the space (or double, or a quarter ...)? - Heavier particles might hit the walls of the container harder, but at any given temperature heavier particles are moving slower then lighter ones, so the collision rate is less. When you run the numbers and calculate the velocity of the particles, the rates of collision, and then how hard they hit the container walls, the mass cancels out. At a fixed temperature, the contribution to pressure from a single particle colliding with the walls of the container is independent of its mass.
You can do the same sort of analysis with other properties. They are either 1) are explicitly discounted by the assumption of "ideal gas" (e.g. no particle interactions) 2) don't factor into the calculation at all (e.g the absorption spectra of the atoms) or 3) precisely balance out - leaving you with $PV = nRT$ being the relation for an ideal gas.

I think you could have made a bit more forceful statement -- By definition, ideal gas molecules interact solely by elastic collision which allows basic statistical mechanics analysis to model their behavior.


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Separation of 2 compounds ( Powder )
Suppose I have a mixture of 2 pure binary compounds, a white powder. What is the best method to separate the 2 compounds?
What I have thought of so far is to dissolve powder in DI water and carry out filtration if one compound is soluble and the other is not. However, I have not been able to think of any other methods to separate the 2 compounds if both are either water-soluble or water-insoluble. I considered fractional distillation but the compounds may become anhydrous. Furthermore, the boiling points of the compounds are not given and they could end up being too close for fractional distillation to work.
Is there any efficient, and preferably simple, method to achieve the above-mentioned desired result?
Edit: Is there any general method of approach for separating such 2 compounds as there is for separating water-soluble and water-insoluble compounds?[Filtration]  Or does it have to be achieved purely by trial and error of various methods?

I think, for the most part, designing a separation procedure depends on you knowing the chemical identity of the two compounds. Like if you have AgCl and AgBr (not exactly a white powder, but I'm just trying to make a general point) you can separate them by addition of ammonia. The key is to find a reagent that reacts differently with both solids. Water is a possibility (e.g. Na2CO3 and AgCl), acid is a possibility (e.g. Ca(OH)2 and BaSO4). But if you don't know what the solids are, you're not going to know what that reagent is.

Could also vary the volume and temperature used for dissolving the solids. Cold di water extractions? Or whatever solvents you are using That's if you don't want other information of your unknowns. Without solubility info, if it's weak acid/base, would be doing lot of trial and error and quantitative analysis just to see if any the solvents you may find are reactive. Maybe try using a spectrometer to get better idea of the compounds?

Your question is about as broad as chemistry itself. Like you said, you may try dissolving them in water, if one is soluble and the other is not. If it's not working with water, chances are it will work with ethanol, or acetone, or... see, there are many solvents on the shelf. No luck? Maybe by now you've found something that dissolves them both, so you may try fractional crystallization. Still no luck? Maybe the compounds differ in density or selective adhesion of something, so flotation might help. Still no luck? Chromatography, by any chance? No? Distillation, like you said (provided they both would survive until boiling point), or maybe sublimation (good for $\ce{NH4Cl + NaCl}$)? No? Well, there is a multitude of chemical processes; some of them might change (that is, destroy) one compound or both, but probably in such a way that they can be regenerated later.
And if everything else fails, you can always look at the mixture under a microscope and separate the crystals manually.

"And if everything else fails, you can always task an undergraduate with looking at the mixture under a microscope and separating the crystals manually," you mean? ;-)

If the powdered compounds are of different density you can separate them based on density in a liquid that does not dissolve them. For example, if you want to separate caffeine (density=1.36 g/cm3) from lidocaine HCl (density=1.20 g/cm3) you can tune the density of you separation medium to be between the two compounds in density. The more dense compound will sediment and the less dense will float achieving separation. A good hydrophobic separation medium in this case could be a mixture of tetrachloroethylene (density=1.62 g/cm3) and hexane (density=0.661 g/cm3)-tune the ratio of the two to tune the density of the separation medium. If the compounds do not dissolve in water you can tune the density of water solutions for used as separation mediums using compounds such as sodium chloride, or zinc bromide or similar to achieve high density solutions if necessary. In this case collect your separated fraction and rinse it with DI water to get rid of the salts.


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Micelles absorb fat, is there an adjective for this?
"Hygroscopic" means the capacity to absorb water. Is there an adjective to describe the same idea except for the absorption of lipids? Like, when you wash your hands with soap and the soap micelles absorb the bacon grease, for example.
I'd like to say that micelles have ____lipid-absorbing_____ properties.

Well... Lipophilic?

Amphiphilic, rather than lipophilic, in the particular case of soaps/surfactants, though amphiphilicity implies lipophilicity. There's no real analogy to hygroscopicity, though, since lipids generally are not found in the atmosphere / gas phase.


text:
Syn-addition or anti-addition to fumaric acid?
In my textbook, the addition of alkaline $\ce{KMnO4}$ to fumaric acid (trans) is stated as a syn-addition even though $\ce{OH}$ is present on opposite sides in the product.
But,the defination of syn-addition is:

Addition of electrophile and nucleophile on the available sites on the same side of molecule.

Isn’t this contradictory?

Ever heard of rotation around single bonds?

@Mithoron,yes.Molecules can freely rotate around single bond.

I don’t know what molecule you ended up with, but the way I draw it, I end up with (2S,3S)-2,3-dihydroxysuccinic acid as shown below.

The syn-addition step would technically be the addition of the electrophile permanganate to the double bond. From there, the oxygens are in place and are not moved. If you reorder the structure to the identical zig-zag representation, you should arrive at a clear syn-orientation.
Note that there are two rotations in this: the first step includes a rotation of fumaric acid by $90^\circ$ so that I can draw the five-membered ring in a plain. After cleaving off $\ce{MnO4^3-}$ (which further reacts to $\ce{MnO4^2-}$ in basic conditions) I arrive at the syn-product.
With maleic acid, everything would be the same, except for the orientation of the carboxylic groups in the second and third structures. That would lead to the hydroxyl groups pointing in different directions even though we are still dealing with a syn-addition.