text:
Trying to understand the necessity of orbital approximation or the independent electron model
I'm reading the approximations used for solving the Schroedinger's equation for molecules; the first is the the Born-Oppenheimer approximation:

Freeze the molecule with a fixed internuclear separation $R;$ then carry
  out calculations to obtain the total energy, $U,$ and wavefunctions for that $R$ value.$^1$

Then another approximation is used which is known as the orbital approximation to get rid of the inter-electronic repulsion as said by the same source.

Consider each electron to move in some sort of "average potential" which incorporates the
  interactions with the two nuclei and an "averaged interaction" with the other electron.
The electronic Hamiltonian can then be separated into two parts:
  $$\hat{H}_e = \hat{H}_1 + \hat{H}_2 \\ \text{where}:\\ \hat{H}_1 \;\text{is dependent only upon the properties of electron (1) and upon R .}\\ \hat{H}_2 \;\text{is dependent only upon the properties of electron (2) and upon R .}$$

I couldn't understand for what purpose this approximation is meant for. Also, I couldn't able to conceive what average potential actually means.
I then googled the term finally to come before a pdf file where it is written that

$\color{red}{\bullet}$ Electrons repel each other according to Coulomb's law,
  with the repulsion energy $r^{-1} _{ij} .$
$\color{red}{\bullet}$ Hartree-Fock replaces this instantaneous electron-electron
  repulsion with the repulsion of each electron with an
  average electron charge cloud.

Prior to that it is written:

$\color{red}{\bullet}$ It assumes that each electron interacts with an average
  charge distribution due to the other electrons. This is
  wrong!

I've spent a hard time understanding this approximation.
Could anyone please explain why this approximation is taken & how it actually solved that problem? The source cites the reason of inter-electronic-repulsion as the problem; how does this term in the Hamiltonian cause problem & how is it solved by the approximation? Also, what is meant by average potential?
An intuitive explanation would be appreciated.

$^1$ http://www.chem.qmul.ac.uk/software/download/mo/3.pdf

I'm tempted to repeat my advice here to stop switching from one source to another in a hope to find a simple explanation. Pick one (good) book and read it (from cover to cover). It also looks like OP might be trying to jump the gun: it looks just to early for him/her to dive into the quantum chemistry. This question is just another indication of OP lacking a good foundation of quantum theory in his/her head.

The question though is a good one, so I made an attempt to answer is as shortly as possible. :)

You may want to go and check the equations, try to write them down and see what you should put where, instead of fiddling with every single paragraph in your book. Try to understand the actual tools, not the pictures or simple explanations textbooks try to give you. Textbooks often oversimplify, or work ony as a metaphor, math clears things up.

Setting the stage
Once the Born-Oppenheimer approximation is introduced, we have (first) to solve the so-called electronic Schrödinger equation that could be written as follows,
\begin{equation}
    \newcommand{\el}{_{\mathrm{e}}}
    \newcommand{\elel}{_{\mathrm{ee}}}
    \newcommand{\elnuc}{_{\mathrm{en}}}
    \newcommand{\nucnuc}{_{\mathrm{nn}}}
    \newcommand{\op}[1]{\hat{#1}}
    \op{H}\el \psi\el(\vec{r}\el) = E\el \psi\el(\vec{r}\el) \, . \tag{1}
\end{equation}
Here $\op{H}\el = \op{T}\el + \op{V}\elel + \op{V}\elnuc$ is the electronic Hamiltonian, $E\el = U - V\nucnuc$ is the electronic energy, and $\psi\el(\vec{r}\el)$ is the electronic wave function. The electronic Schrödinger equation, which is a partial differential equation, is fairly complicated. One of the most frequently used methods of simplification of such equations is technique known as the separation of variables.
Separation of variables technique
The separation of variables is aimed to broke a partial differential equation of many variables into a set of simpler partial differential equations of fewer variables. Note though, that the separation of variables is by no means a universal approach: for some equations a solution of the form mentioned above make the separation of variables possible, for others it is of no use.
The most common way of separation of variables for a partial differential equation in a function $f(x_1, x_2, \dotsc, x_n)$ of $n$ variables $x_{i}$ is by writing its solution as a product of $n$ functions $f_{i}(x_{i})$ each of which is a function of one variable only
\begin{equation}
    f(x_1, x_2, \dotsc, x_n) = f_1(x_1) f_2(x_2) \dotsm f_n(x_n) = \prod_{i=1}^{n} f_i(x_i) \, , \tag{2}
\end{equation}
and then substituting solution of this form into the equation.
Using algebraic manipulation one could try to break the resulting equation into a set of $n$ independent ordinary differential equations for each function $f_{i}(x_{i})$ and if succeed, one could solve these independent equations and by plugging all $f_{i}(x_{i})$ back into (2) obtain the solution of the starting equation.
Separation of particles coordinates
It is fairly well known that an attempt to separate the coordinates of individual particles in the time-independent Schrödinger equation
\begin{equation}
    \op{H} \psi(\vec{r}) = E \psi(\vec{r}) \, ,
\end{equation}
by making a substitution $\psi(\vec{r}) = \prod\limits_{i=1}^{n} \psi_{i}(\vec{r}_{i})$
will be successful only if the potential energy function can be expressed as the sum of the separate potentials for each individual particle
\begin{equation}
    \op{V}(\vec{r}) = \sum_{i=1}^{n} \op{v}(\vec{r}_{i}) \, ,
\end{equation}
which implies that the particles do not interact with each other. For such a system of non-interacting particles the time-independent Schrödinger equation, partial differential equation of a function of $n$ variables, breaks up into a system of $n$ independent partial differential equations of functions of $1$ variable only,
\begin{equation}
    \op{h}_{i} \psi_{i}(\vec{r}_{i}) = \varepsilon_{i} \psi_{i}(\vec{r}_{i}) \, ,
\end{equation}
where $\op{H} = \sum\limits_{i=1}^{n} \op{h}_{i}$ and $E = \sum\limits_{i=1}^{n} \varepsilon_{i}$. All this is relatively easy to establish, and besides, this derivation is done in many textbooks in great details, so I leave at as an educational exercise to OP.
Separation of electrons coordinates
Electrons, on the contrary, interact with each other, so that a simplification of the electronic Schrödinger equation by means of separation of coordinates of individual electrons won't work. Mathematically, it is the interaction potential $\op{V}\elel$ that will prevent the separation of the electronic coordinates,
\begin{equation}
    \psi\el(\vec{r}\el) \neq \prod\limits_{i=1}^{n} \psi_{i}(\vec{r}_{i}) \, ,
\end{equation}
But what if we can approximate $\op{V}\elel$ with some model potential which will allow the separation of electronic coordinates? Obviously, such an approximate potential should have the form of the sum of one-electron potentials each of which depends only on the coordinates of $i$-th electron as we have already seen. Physically it would mean that electrons do not instantaneously interact with each other, but rather each and every electron interacts with the average, or mean, electric field created by all other electrons. Hence the name of the approximation, the mean field approximation. The corresponding approximate potential can be formally written as follows
\begin{equation}
    \op{V}_{\mathrm{MF}}
    =
    \sum\limits_{i=1}^{n} \op{v}_{\mathrm{MF}}(\vec{r}_{i}) \, .
\end{equation}
The electronic Hamiltonian $\op{H}\el$ is then approximated with the sum of effective one-electron Hamiltonian operators $\op{h}(i)$
\begin{equation}
    \op{H}_{\mathrm{MF}}
    =
    \sum_{i=1}^{n} \op{h}(\vec{r}_{i}) \, ,
\end{equation}
where
\begin{equation}
    \op{h}(\vec{r}_{i})
    =
    - \frac{1}{2} \nabla_{i}^2
    - \sum\limits_{α=1}^{ν} \frac{Z_{α}}{ r_{α i} }
    + \op{v}_{\mathrm{MF}}(\vec{r}_{i}) \, .
\end{equation}
And in contrast to eigenfunctions $\psi\el(\vec{r}\el)$ of the true electronic Hamiltonian $\op{H}\el$, eigenfunctions $\psi_{\mathrm{MF}}(\vec{r}\el)$ of the mean-field Hamiltonian $\op{H}_{\mathrm{MF}}$ defined by the corresponding eigenvalue equation
\begin{equation}
    \op{H}_{\mathrm{MF}} \psi_{\mathrm{MF}}(\vec{r}\el)
    =
    E_{\mathrm{MF}} \psi_{\mathrm{MF}}(\vec{r}\el) \, ,
\end{equation}
can be written as products of one-electron wave functions
\begin{equation}
    \psi_{\mathrm{MF}}(\vec{r}\el) = \prod\limits_{i=1}^{n} \psi_{i}(\vec{r}_{i}) \, ,
\end{equation}
where $\psi_{i}(\vec{r}_{i})$ are solutions of the corresponding one-electron Schrödinger equation
\begin{equation}
    \op{h}_{i} \psi_{i}(\vec{r}_{i}) = \varepsilon_{i} \psi_{i}(\vec{r}_{i}) \, .
\end{equation}
The main problem with the mean field approach described above is that so far we have just assumed that $\op{V}_{\mathrm{MF}}$ exist in one form or another, but we have no idea of how it actually might look like.
There is a well-known workaround, but its a different story.

Note that I ignored spin completely in the discussion above which is, of course, fundamentally wrong, but spin won't make a big difference with respect to the key concept discussed, the mean-field approximation.

First of all, again thanks for answering my question; I've understood that in order to separate the variables, we need to avoid the inter-electronic distance term. But I've not conceived how by not instantaneously acting with each other, we can separate the variable.  Physically it would mean that electrons do not instantaneously interact with each other, but rather each and every electron interacts with the average, or mean, electric field created by all other electrons: why should there be no interaction term  in the average field?

Coulomb interactions between electrons prevents the separation of their coordinates, but if we simply get rid of the corresponding potential, we'll obviously get qualitatively wrong result. So, we proceed more wisely: we not just get rid of Coulomb interactions potential, we also introduce a one-electron additive potential which has to at least approximately recover the energy due to Coulomb interactions between the electrons.

Now, what this one-electron additive potential could mean physically if its aim is to account for energy due to Coulomb interactions? The only sensible meaning that we can give to it is that it is a potential of Coulomb interactions between each and every electron and the average field created by all other electrons in the system. So, the form of the mean-field potential and the role it is supposed to play dictates its physical interpretation.

Sorry, sir, didn't contact as I was in holiday. Okay, I want to ask you that the trial wavefunction $\psi(\vec{r}) = \prod\limits_{i=1}^{n} \psi_{i}(\vec{r}{i})$ is the Hartree product  of $n$ orbitals, but we know that electrons are identical fermions which must be antisymmetrised by using Slater determinant $\psi= |\det(\psi_1\updownarrow_1,\psi_2\updownarrow_2,\ldots,\psi_n\updownarrow_n)\rangle$? So, would the calculation above be different if  the later wavefunction is used instead of $\psi(\vec{r}) = \prod\limits{i=1}^{n} \psi_{i}(\vec{r}_{i})?$

@user36790, antisymmetric product of orbitals is just a linear combination of ordinary products, so the mean-field separation is still possible.

That means antisymmetrisation deosn't effect the separation of variable technique & energy which is be given by $E = \sum\limits_{i=1}^{n} \varepsilon_{i}$? That means I could work with Hartree-product wavefunction & after working, I would then antisymmetrise using Slater dterminant, is it so?

@user36790, no, that's wrong. I've mentioned that an approximate mean-field potential is in general unknown, but what is known is that using the variation method you get a different expression for the mean-field potential when antisymmetry is taken into account comparing to the case when its not taken into account.

So, the electron in question doesn't interact with each & every electron but rather it interacts with the smeared out  field created by the other electrons. But could you tell why is it called 'average' field? What is the 'average' there?

Also, is this the shielding effect that decreases the effective nuclear charge that is $(Z- \sigma)e?$

Oh! I think what average meant here is  the smeared spherical charge density around the nucleus. Peter Atkins called this average as spherical mean.

@user36790, you're confusing few unrelated concepts out there.

@user36790, the mean (or average) electrostatic field each and every electron "feels" due to all other electrons has absolutely nothing to do with nuclei.

Yes, I do know that but I've not told anything about nuclei! Just saying that the average field is actually spherically averaged charge density; that's it. You can check this pic; this is what I'm referring to.


text:
Refractive properties of materials?
I've just read this article about meta-materials with 0-refractive index, which potentially may "permit" light with super-luminous rate /under rate is meant the propagation of light's phase, which doesn't transmit information => there is no valuation of the laws of physics/.
Generally speaking, the refractive index of materials depends on the (dominant) molecular interactions within them, on their morphous structure, etc. As you know, these molecular properties govern lots of other chemical features like the viscosity, the density, to some extend the aggregate state, and so on... On this account, it should not come as a surprise that there exists a clear correlation between the refractive index and the properties, mentioned above - e.g. fluids of higher density tend to have higher refractive properties; same goes for viscosity; and so on... Based on those remarks one is able (at least) to guess how they can improve / worsen the refractive index of a given fluid - discussed for instance in this research on aerosols.
My question, though, is how does the refractive index of alloys, polymers, crystals, and solids in general relate to their other chemical properties?
Furthermore, is there a way at least to guess what materials one should potentially try to use in order to create / synthesize a meta-material, like the one from the article, that breaks the limits?

If the above question is too general, I'll rephrase it to make them more concrete...

Refractive index is only connected with so called phase velocity of light - there's no problem with it having as high value as you wish, it could be another question, but off-topic here. Also metamaterials are composites, so it's not that much about synthesis or simple mixing.

As @Mithoron states, the phase velocity can exceed C, just as the imaginary closing point of scissors or the apparent motion  of stellar jets can be superluminal. However, the group velocity (speed of information) is limited to C. BTW, metamaterials can slow light to the speed it has in Pratchett's Diskworld.

@DrMoishePippik Actually group velocity isn't the exactly same as speed of information transfer and in specific situations isn't constrained with c, see https://en.wikipedia.org/wiki/Group_velocity#In_lossy_or_gainful_media

@ Mithoron: I agree with you, but note that my question is more about how the refractive index is related to other physical and chemical properties (e.g. at least for liquids there is a correlation between the carbon residue and the refractivity), not so much how the "super-luminosity" is obtained. The later is a physical question, which is somewhat answered in the article withal.

In your case they write that it's  "made of silicon pillar arrays embedded in a polymer matrix and clad in gold film".

@Mithoron, thanks for comment on group veloicity... I stand corrected.

I can only answer your question about the relation between refractive indices and other properties with two examples.
When looking at dispersion interactions between particles in solution the Lifschitz Theory connects the refractive indices with the Hamaker constants.
The information about refractive indices is even used to choose materials in a way to cancel out dispersion interactions between particles of different kind. (I will attach a paper URL in the next days, just heard it from our professor in the lecture)
Another application of refractive indices is X-Ray NEXAFS spectroscopy and  many more spectroscopic methods involving electron or high energy light radiation. By the Kramers-Kronig relation it is possible to connect absorption constants and refractive indices. Especially one can find for example absorption edges.
Then here I found some probably interesting papers regarding your question. At least their references will be a good starting point. Unfortanetly I have no access and could only read the abstract:
http://www.sciencedirect.com/science/article/pii/0020089165900205
http://www.sciencedirect.com/science/article/pii/0032395083902836

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text:
How many valence electrons do elements in the d block have?
I just noticed that on the periodic table it doesn't say how many valence electrons there are for each column in the d block.
How do I find out how many valence electrons elements in the d block have?

Firstly, it depends on what you count as "valence electrons". If you say that, for all the d-block metals, the ns and (n-1)d electrons count as "valence electrons", then the answer is to just look at the group number. However, that obviously doesn't work for Zn, which effectively only has 2 valence electrons. That's really an extreme case though. It is sometimes said that the d electrons transition from being valence electrons in the early d-block (hence Sc only forms $\ce{Sc^3+}$), to being core electrons in the late d-block (hence Zn only forms $\ce{Zn^2+}$).

Scandium does not only exist in the +3 oxidation state.

The simplest answer is: d-block elements have number of valence electrons equal to their group number, which is equal to the number of electrons in the "valence shell". This works if you are using the definition of valence shell to be the outermost shell.
But it is wrong to apply the concept to determine the valency of the element, because that varies with the compound it is present in. For example, iron can show two valencies, $2$ and $3$ in the compunds iron(II) sulphate ($\ce{FeSO4}$) and iron(III) sulphate ($\ce{Fe2(SO4)3}$) respectively. But from the above definition, iron has $8$ valence electrons.
As a matter of fact, this applies even to many elements outside d-block too, for example chlorine typically shows a valency of ($-1$) in compunds like $\ce{NaCl}$, but it can also show a different valency (more appropriately, oxidation state) in ions like perchlorate anion ($\ce{ClO4^-}$), where its valency is $\mathbf 7$ (while the oxidation state is $\mathbf{+7}$, notice the difference that there is no plus or minus sign in valency).

Iron can actually achieve an oxidation state of +6.

Here is the list of Transition Metals and the generalized information about their valence electrons. The valence electrons (VE) are the electrons in the outer shell of an atom. In the above image valence electrons column imparts about the different orbits of an atom (But again this is a generalized rule there are some exceptions). So in order to find the valence electron you have to check the last digit of the series. at Comparisonofmetals.com you can compare and explore more about these metals with the help of various parameters like their various properties, facts, uses, periodic table details, etc. I came across this website while i was working on my science project. They have some unique and interesting features like comparison of metals, Metals Quiz, plot graphs, compare statistics and much more! I hope it will be helpful.
For example, Scandium metal's electrons per shell are 2, 8, 9, 2 and its outer shell has only 2 electrons. Hence, the valence electron count of scandium is 2.

Hi Pravin, as is, the response here doesn't really answer the question. The table just lists numbers without saying which orbital the electrons are in.

Hi  John Snow, to answer your query; the valence electrons (VE) are the electrons in the outer shell of an atom. In the above image valence electrons column imparts about the different orbits of an atom. So in order to find the valence electron you have to check the last digit of the series. For example, Scandium metal's electrons per shell are 2, 8, 9, 2 and its outer shell has only 2 electrons. Hence, the valence electron count of scandium is 2.

This is not true, because the transition metals can easily lose electrons from their d orbitals as well as the outer s orbitals. This question cannot be answered without considering the orbital electronic configurations of the elements.

Hi Bon, your are right that we can not generalize this rule because there are some exceptions. The image attached above gives generalize idea about the valence electrons of d block element. But if want it the precise information,  please refer this site (http://www.comparisonofmetals.com/)                                                                         Just for an example, the valence elctrons of Gold is 1 according to the table. Gold (Au) is a d-block element and has 1 valence electron (sometimes 3 or even 11 depending on how “valence electron is defined.)

Gold (Au) is a d-block element and has 1 valence electron (sometimes 3 or even 11 depending on how “valence electron is defined.) However Gold most often has chemical reactions with 1 or 3 valence electrons. Reactions with three valence electrons would be marked as gold(III.) If the formula doesn’t connote the usage of gold(III), then gold with one valence electron could be considered.

d- block elements have 1-10 electrons in the d- shell. but the (n+1)th S outer shell has either 2 or 1 electron so as to complete half filled or full filled states of the elements which have one electron less than completion of that state.

This also isn't really true. There are multiple exceptions to your rule; $\ce{Nb, W, Ru, Sg}$; to name but a few. The real rule is the electrons occupy the orbitals that give the lowest energy configuration. Calculating what this configuration is is a complex topic.

Elements in the d block are transition elements, and each posses one or two valence electrons in their respective s orbitals.  Most have two, with several notable exceptions: elements may "steal" an electron from the outermost s block and relocate it to the d block in order to reach a filled or half-filled ($5$ of $10$ electrons) state in the d block.  These elements must be taken on a case-by-case basis.

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text:
What is the industrial yield of paracetamol?
Anyone know how I can find out the typical yield of paracetamol (acetaminophen) from the acetylation of 4-aminophenol?
Please provide the source if you happen to know!

It depends upon the route which of the two are you interested in?

@user1945827 acetylation using anhydride from 4-aminophenol

Details should be in here, http://pubs.acs.org/doi/abs/10.1021/op960007u

For a variety of explanations of industrial techniques as applied to acetylation,  including acetaminophen synthesis,  try here
https://www.google.com/patents/US20120065423
The examples are right at the bottom and are only scaled down lab versions of the industrial technique. Over 90% yield should be easily achievable if acetic anhydride is used (and is achieved in the industrial synthesis of ibuprofen).  The patent examples give yields for the use of acetic anhydride and acetic acid, as well as the technology employed e.g. continuous stirred tank reactor (cstr)  vs plug flow reactor (pfr or tubular reactor)  vs reactive distillation.  Enjoy.


text:
How to calculate the volume of a 0.1 M NaOH(aq) solution necessary to make a 0.5 M EDTA solution with pH = 8?
I was told that EDTA dissolves at a pH of $8$. Therefore I need to use a base to increase the pH to $8$ is what I understand. How exactly would I go about this?
In my head I if I wanted $1~\mathrm{L}$ of $0.5~\mathrm{M}$ EDTA I would take $186.12~\mathrm{g}$ of the EDTA, since its molar mass is $M = 372.24~\mathrm{g\,mol^{-1}}$.
Then I would add $800~\mathrm{mL}$ of distilled $\ce{H2O}$ to dissolve it, but since it wont be at $\ce{pH}=8$, it wont dissolve.
So I assumed I would just add it drop wise until it dissolved while spinning? I lack access to a burrette and only have a crappy pH meter to check pH.
One I observe the EDTA dissolved I simply add up to $1000~\mathrm{mL}$?
I just wish I had a way to figure out how much $\ce{NaOH}$ I needed to add to dissolve the EDTA so I wouldn't have to do it drop wise.

Have you tried to calculate the concentration of a sodium hydroxide solution with a pH of 8?

I saw this on the researchgate.net website:

I followed this procedure for preparing  0.5M EDTA ph8(100ml) . Initially I  warmed around 70ml of water and added an ml of 6MNaoh solution to shoot up the ph to 7.6  with the container on  a hot plate .Then I slowly added the measured quantity of EDTA and Simultaneously kept adding small volumes of 6M Naoh solution while measuring  the ph.Finally around ph 8 the EDTA was completely dissolved and the process was way too quick.."

Bhavani Nagarajan · University of North Texas HSC at Fort Worth


text:
Adding powdered Pb and Fe to a solution
My book states that if we add powdered $\ce{Pb}$ (lead) and $\ce{Fe}$ (iron) to a solution containing $1\ \mathrm{M}$ each of $\ce{Pb^2+}$ and $\ce{Fe^2+}$ it would result in formation of more $\ce{Pb}$ and  $\ce{Fe^2+}$
What I think:
1) Since both $\ce{Pb}$ and $\ce{Fe}$ have negitive standard electrode poential both will have tendency to form $\ce{Pb^2+}$ and $\ce{Fe^2+}$, then as a result $\ce{Pb^2+}$ and $\ce{Fe^2+}$  will increase in solution which is against the result given by book.
2) Another thing which comes to my mind is that since $\ce{Pb}$ has greater standard reduction potential than $\ce{Fe}$ so only  added Fe will convert into $\ce{Fe^2+}$ And the electrons released by it will reduce pre-existing  $\ce{Pb^2+}$  to $\ce{Pb}$ and hence the result as given by book
So my question is what exactly will happen if we conduct the following experiment.

Your second thought is the right one. As for the first, it implies the reaction of both metals with $\ce{H+}$ which releases $\ce{H2}$; for that to happen, the solution must contain some significant amount of $\ce{H+}$, i.e., be acidic, which is not the case.

You should elaborate.

Let us consider everything we have before and after. Before adding the powdered substances, we have a solution containing an iron(II) salt and a lead(II) salt. Both are stable by themselves and they are even stable when mixed — this is because lead(II) is the most stable form of lead in aquaeous solution. Iron(II) might want to oxidise itself to iron(III) but cannot since there is no electron acceptor (hydrogen atoms don’t count, they are too bad an oxidising agent when only present as water).
Consider having pure water and adding powdered lead(± 0) and/or iron(± 0). By a simple check of standard potentials we would assume both to be oxidised to lead(II) and iron(II) respectively. But standard potentials are measured at $\mathrm{pH}\ 0$; at neutral pH the potential of hydrogen equals $-0.41~\mathrm{V}$ — not enough to oxidise lead and only just enough for oxidising iron, but that is easily passivised. Of course, solid lead won’t react with solid iron at room temperature in any way. So no reaction here, either.
(This also means that your first attempt is incorrect: it needs a certain concentration of acid to oxidatively dissolve lead or iron to their respective metal(II) ions. Iron would also further be affected by passivation; not sure about lead.)
Meaning that all we truely need to consider is the reaction between the two lead species and the two iron species. There are six equations we might want to consider:

$\ce{Fe^2+ + 2 e- <=> Fe} ~~~~ E^0 = -0.41~\mathrm{V}$
$\ce{Fe^3+ + 3 e- <=> Fe} ~~~~ E^0 = -0.04~\mathrm{V}$
$\ce{Fe^3+ + e- <=> Fe^2+} ~~~~ E^0 = +0.77~\mathrm{V}$
$\ce{Pb^2+ + 2 e- <=> Pb} ~~~~ E^0 = -0.13~\mathrm{V}$
$\ce{Pb^4+ + 2 e- <=> Pb^2+} ~~~~ E^0 = +1.69~\mathrm{V}$
$\ce{Pb^4+ + 4 e- <=> Pb}$ — value not given.

Of course, reactions will happen if $\Delta G < 0$ and for electrochemical cells we have $\Delta G = - z F \Delta E$ where $z$ is the number of electrons transferred and $F$ is Faraday’s constant. Since $\Delta E = E_\mathrm{red} - E_\mathrm{ox}$ and we want that value to be as large as possible, we need to look for high values for the reducing half-cell and low values for the oxidising one. And finally, since $E = E^0 + \frac{0.059}{z} \lg \frac{[\mathrm{ox}]}{[\mathrm{red}]}$ and the activities of $\ce{Fe^{II}, Pb^{II}, Fe^{\pm 0}}$ and $\ce{Pb^{\pm 0}}$ are all $1$, for half-cells using only those four components, we can simply use the standard potentials.
We immediately see that an oxidation of lead(II) to lead(IV) is not possible: We cannot overcome the $+ 1.69~\mathrm{V}$ required. For the same reason, it is not possible to oxidise iron(II) to iron(III).
When considering something to be reduced, we only have the choice between lead(II) and iron(II) — there is no iron(III) in solution (see above). Out of those two, lead(II) is reduced more easily (higest standard potential), so let’s take that as a starting point. And for the necessary oxidation, we want the lowest possible standard potential which only leaves us with oxidation of iron to iron(II). Thankfully, the overall $\Delta G$ is negative:
$$\ce{Fe + Pb^2+ <=>> Pb + Fe^2+} \\ ~ \\
\Delta E = \Delta E^0 = E^0(\ce{Pb}) - E^0(\ce{Fe}) \\
= -0.13~\mathrm{V} - (-0.41~\mathrm{V}) = +0.28~\mathrm{V} \\
\Delta G = - z F \Delta E = - 2 \cdot 96485~\mathrm{\frac{C}{mol}} \cdot (+0.28~\mathrm{V}) = - 54.0~\mathrm{\frac{kJ}{mol}}$$
It’s nice that this is also the answer your book gives; hence your second attempt is correct. In a nutshell, your reasoning was shorter.


text:
If I want to sample rate constants of chemical reactions, what the distribution will be appropriate?
For example, I have a (chemical) reaction systems, I want to sample the parameter space. What will be the best and validated distribution for sampling? And why?

(Can't comment, not enough rep. ...) Do you mean calculating the rate from a statistical mechanical ensemble (e.g. Molecular Dynamics)?

@InonS Yes, I think this might be one of the possible approaches.

Since you have accepted my proposal of using a free-energy calculation, you can use that in a combination with classical Transition State Theory (think of it as an extension to the Arrhenius equation).
More practically, you can read-up, for example on

GROAMCS
NAMD

For more options, see this overview presentation.


text:
Reaction of one mole of (PO4)3- with one mole of H2CrO4
I am looking for the endproducts of a reaction of one mole of $\ce{PO4^3-}$ with one mole of $\ce{H2CrO4}$ in one litre of water.
First, the following reaction can happen:
$$\ce{PO4^3- + H2CrO4 <=> HPO4^2- + HCrO4-}$$
Then the following reactions may happen:
\begin{align}
\ce{PO4^3- + HCrO4- &<=> HPO4^2- + CrO4^2-} \\
\ce{HPO4^2- + HCrO4- &<=> H2PO4^- + CrO4^2-} \\
\ce{HPO4^2- + H2CrO4 &<=> H2PO4^- + HCrO4^-} \\
\ce{H2PO4^- + H2CrO4 &<=> H3PO4 + HCrO4^-}
\end{align}
Is there a way to determine how many moles there will be of each end product?
Note: this is not homework, rather just a thing I'm interested in. However, I have barely an idea where to start.

Well, you seem to have solid math background, so it should be easier for you to restate this problem in a purely mathematical way. Just look up the pKa values for both acids, and you'll end up with a system of a few equations. Admittedly, it is non-linear, so don't expect to find an analytical solution.

Another thing is, you need to specify the original concentrations, not the amounts. This is because the acid dissociation constants are given in terms of concentrations. (Strictly speaking, it's activities, but whatever)

@IvanNeretin Can you write these equations out, please? I found the pKa values: $\ce{H2CrO4}$: 0.74, $\ce{HCrO4-}$ : 6.49, $\ce{HPO4^2-}$ : 12.32 , $\ce{H2PO4^-}$ : 7.21. Of $\ce{PO4^3-}$ and $\ce{CrO4^2-}$ I can only find the pKb value, 1.68 and 7.51 respectively.

$K_a =10^{-pK_a} = \mathrm{\frac{[A^-][H^+]}{[HA]}}$ for everything that contains H ($\ce{H3PO4,;H2PO4-}$, etc.) As for $pK_b$, you don't need those at all (it is basically the same thing, only the other way around).

To answer this question, you should stop thinking about the acids and bases reacting with each other, but rather they react with water exclusively.
Firstly, the phosphoric acid and its conjugate bases will have the following reactions with water
$$\ce{H3PO4 + H2O <=> H2PO4^- + H3O^+}$$
$$\ce{H2PO4^- + H2O <=> HPO4^2- + H3O^+}$$
$$\ce{HPO4^2- + H2O <=> PO4^3- + H3O^+}$$
Then there is the chromic acid and its conjugate bases
$$\ce{H2CrO4 + H2O <=> HCrO4^- + H3O^+}$$
$$\ce{HCrO4^- + H2O <=> CrO4^2- + H3O^+}$$
So now you have five equlibrium reactions.  You have eight unknowns, $[\ce{H3PO4}]$, $[\ce{H2PO4^-}]$, $[\ce{HPO4^2-}]$, $[\ce{PO4^3-}]$, $[\ce{H2CrO4}]$, $[\ce{HCrO4^-}]$, $[\ce{CrO4^2-}]$, and $[\ce{H3O^+}]$.
The Ka expressions from the above reactions give you five equations.  Then you have the initial concentrations of phosphate and chromic acids, which will give you two more, then the fact that the number of protons is conserved will give you the eighth and final equation.


text:
Why Atomic size decreases left to right in a period in MPT?
In modern periodic table,  atomic size decreases as we move from left to right in a period. As no. of electron increase left  to right but why atomic size decreases?

This is a homework question. We have a policy which states that you should show your thoughts and/or efforts into solving the problem. It'll make us certain that we aren't doing your homework for you. Otherwise, this question may get closed.

Though more electrons are being added, the electrons add to the same orbit. So, as we know nuclear charge is distant dependent, i.e., inversely proportional to atomic radius. And as the no of electrons increases the nuclear charge also increases. With increase in nuclear charge, the atomic size eventually decrease.

The Modern Periodic table is made according to increasing number of their atomic numbers . So if we move from left to right in a period the no of electrons as well as the no of protons goes on increasing . The electrons are added to the same orbit or energylevel but the nuclear charge increases by one in each step due to increase in the number of proton by one in each step. So the inward pull increases and the outer orbital shrinks. As a result the atomic size decreases moving towards right in the modern periodic table.


text:
Which element is undergoing oxidation in the combustion of methane?
Chemical reaction:
$$\ce{CH4(g) + 2O2(g) -> CO2(g) + 2H2O(g)}
$$
The example in our textbook for this type of problem involves the following reaction:
$$\ce{C + 2S -> CS2}
$$
Here, I can clearly see that $\ce{C}$ starts off with an oxidation state of zero and ends with an oxidation state of +4, meaning that it underwent oxidation.
In my given reaction, I see how the $\ce{C}$ in $\ce{CO2}$ has an oxidation state of +4, but I don't quite understand why it doesn't also have an oxidation state of +4 in $\ce{CH4}$. My assumption here is that, in the case of $\ce{CO_2}$, $\ce{C}$ has an oxidation state of +4 and $\ce{O2}$ has an oxidation state of -4; and in the case of $\ce{CH4}$, $\ce{C}$ has an oxidation state of +4 and $\ce{H4}$ has an oxidation state of -4. If that is incorrect, could someone explain why?
Edit: I just realized I overlooked a table that says that hydrogen has an oxidation state of +1, which would mean that in $\ce{CH_4}$, $\ce{C}$ would have an oxidation state of -4. I gather that this means that $\ce{C}$ loses 8 electrons in the reaction, which is why it is considered to undergo oxidation.

How do you know that C is plus and O is minus in $\ce{CO2}$?

A table in my textbook says that O has an oxidation state of -2, and that the sum of oxidation states in a neutral molecule is zero, which suggests that, in $CO_2$, C is +4 and each O is -2.

Does the table say that in respect to a specific compound, or in general?

In general - it is a table showing oxidation states of non-metals (F: -1, H: +1, O: -2, Group 7a: -1, Group 6a: -2, Group 5a: -3

Please use the \ce{...} syntax when referring to chemicals. Plain MathJax, i.e. $ without anything else results in italic type for compounds which is wrong and looks ugly. Also, MathJax does cool things such as automatically turning $\ce{CH4}$ or $\ce{H2O}$ into $\ce{CH4}$ and $\ce{H2O}$, respectively. For more information, check out this meta post.

I believe I have answered my own question - I assumed that H had an oxidation state of -1, when it appears to have an oxidation state of +1.

See, it is not like H is always +1 in any compound. Normally people rely on electronegativities to decide which element is in positive oxidation state and which is negative.


text:
How does one solve the following differential equation (mass balance equation)
How does one solve the following differential equation? This is a reduced mass balance equation for a cylinder containing a gas mixture of two gases (here represented by the numbers 1 and 2)):
$$\frac{p_1C_{V_1}^*}{RT_1}\frac{\mathrm dT_1}{\mathrm dt}+\frac{p_2C_{V_2}^*}{RT_2}\frac{\mathrm dT_2}{\mathrm dt}=\frac{\mathrm d}{\mathrm dt}\left(\frac{p_1}{T_1}\right)T_1+\frac{\mathrm d}{\mathrm dt}\left(\frac{p_2}{T_2}\right)T_2$$
For one gas it's easy:
$$\frac{pC_V^*}{RT}\frac{\mathrm dT}{\mathrm dt}=\frac{\mathrm d}{\mathrm dt}\left(\frac pT\right)T$$
$$\frac{C_V^*}{RT}\frac{\mathrm dT}{\mathrm dt}=\frac{\mathrm d}{\mathrm dt}\left(\frac pT\right)\frac Tp$$
$$\frac{C_V^*}{R}\frac{\mathrm d\ln(T)}{\mathrm dt}=\frac{\mathrm d}{\mathrm dt}\left(\ln\left(\frac pT\right)\right)$$
After integrating between the initial and final phase, we get:
$$\left(\frac{T_\text{final}}{T_\text{initial}}\right)^{\frac{C_V^*}R}=\left(\frac{p_\text{final}}{p_\text{initial}}\right)\left(\frac{T_\text{initial}}{T_\text{final}}\right)$$

What does $T_1$ and $T_2$ represent?  In most practical situations, it would be impossible for each gas in a mixture of gases to have separate temperatures.  Also, does $P_1$ represent the partial pressure of gas 1?  Or the total pressure.  (If the latter, what is $P_2$?)

Yes, some gases tend to migrate to warmer regions, others to colder regions. Anyway, for simplicity's sake, you could say that $T_{1}=T_{2}$. $P_{1}$ and $P_{2}$ are the partial pressures of gas 1 and 2 resp (n.b. these aren't equal) and $C_{V_{1}}^{}$ and $C_{V_{2}}^{}$ are the specific heat capacities of each gas at constant volume.

If $T_1=T_2$, then $dT_1 = dT_2$ and you could considerably simplify the equation.  Also if the gases aren't reacting then can't you just calculate an appropriately averaged $C_v$ and then use the single-component gas equation?

yes, but in that case you merge everything and you get the solution I wrote above and I would like to consider the thermal gradient within the cylinder containing the gas in the equation

If you are interested in modeling thermal gradients, why isn't there any spatial variable in your question?  Could you edit your question to expand on exactly what -- chemically -- is confusing you about the equation?  I'm not sure but from the comments it sounds like you are interested in far more than just solving the equation you wrote.  What does it represent?  How did you derive it?  Do you think it has anything to do with spatial gradients?  If you add more info to the question you'll get a better answer.

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text:
Does aniline breakdown in groundwater produce benzene
Can the benzene ring attached to aniline be detached in an aqueous solution (specifically groundwater) under certain conditions? Would the two compounds be detected as separate entities in a laboratory analysis?

You can turn aniline back into benzene but that requires a few steps and usually loses the amino group altogether. Since benzene and aniline are two separate compounds, what makes you think that they would not be detected separately?

In aniline there's no benzene, but phenyl group and amino group, which can't be simply separated.

Simplest way to convert aniline to benzene is $\ce{NaNO2}/\ce{HCl}$ followed by $\ce{H3PO2}$ https://en.wikipedia.org/wiki/Diazonium_compound

Aniline is not easily converted to benzene in aqueous conditions. A google search turns up several papers describing the breakdown products of aniline in groundwater but benzene is not one of them. Methanogenic and denitrifying bacteria seem to play a key part in aniline breakdown, releasing methane and carbon dioxide as byproducts. The other main products seem to be polymerisation products resulting from condensation and oxidation reactions.
In answer to your question about detection; no you would not expect to see benzene as a breakdown product in a sample of groundwater that has been contaminated by aniline.
http://www.ncbi.nlm.nih.gov/pmc/articles/PMC1519315/


text:
What assumptions underlie Hess' law and and its analogues?
Let me first briefly describe the laws as I understand them.
Hess' law states that the enthalpy of reaction can be obtained by taking the sum of the enthalpies of formation for the compounds involved in the reaction, weighted by their stoichiometric coefficient. This is often shortened to $\Delta_\mathrm{r} H = \sum \Delta_\mathrm{fm} H_\mathrm{products} - \sum \Delta_\mathrm{fm} H_\mathrm{reactants}$. A similar law exists for entropy. To get the enthalpy of a compound, one usually looks up the tabulated standard state enthalpy, and then integrates the compound's heat capacity $C_p$ to get the enthalpy at the temperature of interest. A similar procedure is used for computing the entropy. Although I have not seen it done, I guess there is nothing in the way of using such laws in a partial reaction (i.e. equilibrium): by summing the enthalpies of the various compounds at equilibrium and subtracting the enthalpies of the compounds you started with, you should get the "enthalpy of equilibrium".
What assumptions underlie these laws? For Hess' law, I would think "enthalpy of mixing is zero" is one of them, but I've yet to see it stated.
An admittedly wage bonus question: I am here talking about $H$ and $S$, but what about similar laws (if they exist) for other potentials (Gibbs energy $G$, Helmholtz energy $A$ and internal energy $U$)?

The typically unstated assumption in applying Hess's law is that you are going from an initial state of pure reactants (say, in separate containers) to a final  state of pure products (also in separate containers).  In the case of ideal gas reactions, the enthalpy of mixing is zero and enthalpy is independent of pressure.  So, for ideal gas situations, Hess's law can be extended and applied directly to starting and ending states in which the reactants and products are mixed (and even simultaneously present) and the partial pressures are all different from the standard pressure of 1 atm.
Of course, since U, S, and G are functions of state, the strict version of the law, applying only to pure reactants and products, still holds.  However, for ideal gas mixtures, partial molar entropy and partial molar free energy are functions of mole fraction, so this needs to be taken into account for mixtures.

Thanks. Follow up: As far as I understand, measuring heats of reaction is how you determine enthalpies of formation in the first place. But if you're neglecting the contribution from enthalpy of mixing, wouldn't you get inconsistent results by using different reactions to determine a compound's enthalpy of formation? Or are tabulated enthalpies of formation based on the assumption that enthalpy of mixing is negligible compared to the enthalpy of reaction?

In the table of formation enthalpies, no mixing is involved.  The tabulated values are for going from the pure elements to the pure compound.  Any mixing that you may be thinking of is accounted for by including process steps to separate the various materials after reaction  (conceptually using semipermeable membranes, say) to arrive at a pure product compound separate from any reactants that may be remaining.  This is all inherently included in what we call the heat of formation.  For an ideal gas, of course, no separation is necessary.


text:
Metallic bonding and covalent boding
Why do metallic bonds have delocalized electrons, whereas covalent bonds only share electrons? Does it have something to with the atomic masses of the individual non/metals or something else? I'm really interested to know.

It is wrong to think of the covalent bond as a pair of shared electrons between two atoms. The only true difference between metallic and covalent bonds is that the metallic bond’s ‘molecule’ is magnitudes larger.

@Jan That is a really nice way of thinking about it. You should write it up into an answer.

Indeed, the electrons in a covalent bond are also delocalized; it's just that they are delocalized over two atoms. Or maybe three, if we think of a three-centered bond like those in boranes. Or six atoms, if we think of benzene. Or even more, because larger conjugated structures are also known.

Why do metallic bonds have delocalized electrons, whereas covalent bonds only share electrons?
To be somewhat flip this is like asking why are redbirds red and blue birds blue?
In order to pigeon-hole bonding ionic, covalent, aromatic, and metallic bonds are "artificial" constructs used to predict behavior. But there are always exceptions, and no bond is purely in any one of the categories.
So in general a sample which has metallic bonds has delocalized electrons and hence will conduct electricity. Such elements would be metals. If the sample only has covalent bonds then there would not be any delocalized electrons and the sample would be a nonmetal.
For some elements the nature of the bonding yields different results. For example graphite conducts electricity, but diamond won't.
To add another wrinkle in this we can think of gaseous ionic or covalent molecules but in general being metallic is a "bulk" property. So we have to consider not only the bonding in the "molecule", but how "bonding" is distributed through the solid matrix of the sample. For example there are discrete water molecules in ice, but in a diamond there are no discrete carbon "molecules."

The question’s premise is incorrect. While it is nice to think of clearly localised bonds and while it provides an easy access to chemistry at a high school level thinking of just moving single lines (a.k.a. bonding electron pairs) this is not much more than a very simplified model of reality.
The most well-known counterexample is the benzene molecule $\ce{C6H6}$ which is always introduced as not having localised double and single bonds but rather an ‘electron cloud’ in a delocalised π system above and below the molecular plane. But that is only the tip of the iceberg. In the end, all molecules that are not diatomic have all their bonding electrons delocalised across the entire molecule. Check out, for example, the molecular orbitals of water.[1]
It is only through mathematical methods (linear combination) that we can backtrack from molecular orbitals back to orbitals of a specific bond. For example, by taking $a_2 \Psi_\mathrm{mo2} + a_3 \Psi_\mathrm{mo3} + a_4 \Psi_\mathrm{mo4}$ and correctly choosing the coefficients $a_n$ we can construct the wavefunction $\Psi_{\unicode[Times]{x3c3}(\ce{O-H})}$ that corresponds to one of the two $\ce{O-H}$ bonds; by taking $a_3' = - a_3$ we will arrive at the other $\ce{O-H}$ bond. The physical reality is much closer to the entirely delocalised system, though.
That in turn means that the distinction between metals (large, three-dimensional networks of atoms, all somehow bonded to each other), ionic compounds (large, three-dimensional networks of atoms considered ions of opposite charge, somehow also bonded to each other) and molecules (small, three-dimensional discreet entities of atoms, somehow bonded to each other) lies only in the size of the network or in the types of particles involved. Indeed, graphite (the carbon modification) and black phosphorus both exhibit clear metal characteristics even though the elements are clearly non-metallic: absorption of all wavelengths (i.e. black colour), conductivity, high melting points and more.

[1]: Site by professor Zipse, LMU Munich; held in German. Click on one of the mos to access an image of the respective orbital.


text:
Kinetic theory of gases
The intermolecular distance between the molecules are much greater than the diameter of the gas molecules so we neglect their volume compared to the volume of container. "The rapidly moving particles constantly collide among themselves and with the walls of the container. All these collisions are perfectly elastic. This means, the molecules are considered to be perfectly spherical in shape, and elastic in nature". How then are molecules thought to be perfectly spherical in shape although its volume is neglected.

They are thought of as spheres with negligible volume? Which is to say that they do have a volume, but it can be ignored for all practical purposes, because it is small relative to the volume of the container. The average distance separating two such balls is much larger than their combined volume.

The basic explanation that you've recited for the Kinetic Theory of gases is trying to present a simple complete and consistent model which explains the observed behavior. Thus to give a consistent perspective it is easiest to think of gas molecules as elastic spheres. Of course, the assumption about the negligible size of the molecules does indicate that the shape of the molecules is irrelevant to the volume of the gas. But if you consider the molecules as having edges, or different moments of inertia about the three axis then the "simple" model gets too complicated for "simple" mathematics. Think of the all the additional variables that would be needed.
The basic model neglects other factors of course. In the real world gravity plays a part. The atmosphere decreases in density as the altitude increases. So the basic model is trying to illustrate the key factors necessary to understand the kinetic theory of gases, not all the factors.
Edit - The basic gas law PV=nRT has been modified by a number of expressions where correction factors have been used to make the equation more accurate and to give the equation more dynamic range. See some of the equations in the following wikipedia article.

Thanks MaxW, i have got another question, sorry to bother you again,is it assumed that collision are elastic(no energy is lost) because it is a closed system.

Elastic collisions and a closed system are mostly independent assumptions. A closed system is mainly implying that neither the gas nor the container is extracting energy from the other. Elastic collisions are mostly about how gas molecules interact with each other. Think of a helium atom (molecule) and an diatomic oxygen molecule. You could really "better" model oxygen as two weights connected with a spring. With such a model calculating how a helium molecule and  oxygen molecule would collide would be vastly more complicated mathematically with scant improvement in the model's accuracy.

If we model molecules connected together by springs. Does that mean that energy would be conserved?.If there is an elastic collision between one particle and another can one particle stop moving after the collision or not?

(1) The spring model isn't used for most models. It just makes the mathematics too complicated. But is not wrong. (2)at a given temperature the energy of the molecules is x-y-z movement, vibrational energy within the molecule, and rotation of the molecule. Thus helium would be expected to conform to the ideal gas laws better than oxygen.  (3) It is possible, not probable, that an individual molecule could stop dead still before quickly having a collision with another molecule.

I edited the answer and added a link to a Wikipedia article that has gas equations with various  correction factors. You'd need to tabulate all the experimental gas data and calculate the coefficients and see which equation gives the best results for the gas in which you have an interest.

@MaxW Is ideal gas a concept only in kinetic theory? Why we call it ideal at all instead of saying the model of gas that kinetic theory assumes?

@adosar - Look how many words are in ideal gas and how many words are in the model of gas that kinetic theory assumes. Which would you rather write 500 times in a general chemistry book? // The ideal gas law isn't just an theoretical abstraction. A lot of gases follow the law well under normal laboratory conditions. At high pressures or low temperatures deviations occur while the compound is still in the gas phase.

I am not a master at this, but from what I know, the volume of the molecule necessarily need not depend on the shape of the molecule. Hence the volume of the molecule does remain negligible when compared to the volume of the gas. Plus, if you must have read about the Vanderwaal's Real Gas Equation although not flawless, accounts for the correction in the Volume of the molecules which we assumed to be negligible in case of Ideal Gases. And as such, in real life we won't encounter a perfectly ideal Gas. So the Real Gas Equation shall help you considerably with your doubt.
Hope you understood:-)

Well wikipedia has a good article (https://en.wikipedia.org/wiki/Real_gas) on various "real gas" equations.  Examination will show that the models just have various "fudge factors" for the basic ideal gas equation PV=nRT. Tables of fudge factors for various models, for a given range of conditions, and for a particular gas would allow more accurate calculations. But the "fudge factors" are determined experimentally, not theoretically. There are models and conditions (e.g. very high pressure) where the volume of the gas molecules is not negligible.

I totally agree with what you have to say. And basically what I would say is that chemistry is a subject where you can't predict a las and then check whether it's true or not. You first experiment and based on observations we make laws. Hence though sometimes a law maybe flawless for a particular case it maybe totally useless for another few.


text:
Does adsorption occur between liquids?
While reading a paper in a field other than chemistry, the author mentions a phenomena that he labelled "adsorption" and describes it as follows:

...I am told (though I do not know which liquids they are), there are
  distinct liquids such that pouring 1 litre of one into a container
  already containing 1 litre of the other does not result in there being
  2 litres of liquid in the container (adsorption).

Now, I know almost nothing about adsorption beyond what I just read on Wikipedia, but I didn't see an  example there where two liquids interacted in the way described. Instead, the examples seemed to be ones where a porous solid (e.g., activated charcoal) adsorbs some liquid.
Does this occur between two liquids? If so, what's an example and how does it work in the absence of pores?

Do you mean aBsorption, as opposed to aDsoprtion? They are not to be confused.

@LiamH Yep, I know. Not sure what the author meant to write, but he definitely wrote aDSorption -- no doubt a large part of what confused me.

The author of the article has named the phenomenon incorrectly. Adsorption can indeed result in a lower volume than might be expected, for example when a gas interacts with a high surface area solid. In the case of hydrogen storage, for example, the adsorbed phase is dense enough at low temperatures that much more hydrogen can be stored in a container filled with a high surface area material, and at lower pressures, than an empty container of the same volume.
However, the situation that is described actually is a result of solvation, not adsorption, and the volume-reducing effect has been discussed in some. other. questions, have a look around!
In essence, adsorption is only discussed in the context of the interface between two phases, such as gas-liquid, gas-solid, or two immiscible liquids.
If someone were feeling really pedantic, they could make an argument that the solvated molecules in the liquid constitute a really small "adsorbent" particle, and the solvent molecules in the first solvation shell are "adsorbates", and more power to them. Really, the forces that are at work in both cases (solvation and adsorption) are very similar.
It might be an interesting study to investigate where the crossover between adsorption and solvation happens as a function of molecule size. After all, graphene is just a really big molecule and its adsorption properties are pretty great. A brief search on adsorption solvation crossover reveals that there is some investigation that has been done on the process of going from a 3D gas $\ce{->}$ 2D adsorbate $\ce{->}$ aqueous solvate here.
tl;dr adsorption does occur at the interface between immiscible liquids, but the OP's article is talking about solvation


text:
Is the 'sweat' on the side of bottles distilled water?
Are the water droplets that form on the side of water bottles (or other drink containers) distilled water? Assume that the sides of the container are completely clean.

It will depend on how clean your bottle is...

Yes, here is the definition for distilled.

purify (a liquid) by vaporizing it, then condensing it by cooling
the vapor, and collecting the resulting liquid.

The sweat on the water bottle is the process of distillation, but only the condensation part of the process is seen on the water bottle.


text:
Electronegativity comparison of group 13 elements
Why is electronegativity of indium slightly  greater than gallium?

Because measurement says so.

Electronegativity is a human construct; it isn't a fundamental physical property of an element. We often like to use electronegativity to explain stuff, which is great when there is a clear-cut difference. So, for example, in LiH you have negative H and positive Li. Here it is not quite so clear-cut: some scales have In more electronegative than Ga and some scales have Ga more electronegative. (This aspect of your question really needs some clarification, btw.) What's the answer? Probably "we don't know", since it depends on what formula you use for this "electronegativity" thingum.


text:
How to do complexometry titration of calcium oxalate?
My titration of  calcium oxalate with NaEDTA doesn't work. I need to calculate the ion calcium within my sample. My essay is to test anticalculi activity.
Why do I say I have problem? Because the colour keeps turning back to its initial colour (pink) more than twelve times. It's never ending and the titration end point is so premature.
In my preparation, I added HCl 0.5 M to my calcium oxalate so it dissolves. Calcium oxalate has very low solubility.

First, I used EBT indicator and ammonia buffer. When i add the ammonia the solution became foggy and there's a precipitation within the sample.
So I used substitution titration with Mg-EDTA (EBT indicator) = Still happens
Then change the indicator to hydroxynaftol blue and calcon = Still happens (I don't use buffer with this indicator, only NaOH, but when I add NaOH it's became foggy again)

Does it have to be complexometric?
You could acidify the sample and do a redox titration of the oxalate with permanganate or dichromate. And since the amount of oxalate is the same as calcium, bob's your uncle.


text:
Are the Phosphate and Sulfate ion diagrams with double bonds an invalid picture?
Whenever you see an image of phosphate, it's always shown having 5 bonds. 3 single bonds to anionic oxygen atoms and a double bond to an extra neutral oxygen.
Not only that, but they're shown being arranged in a tetrahedral manner. All of this additionally applies to sulfate, only that it has 2 double bonds instead of  just one.
Now, from what I've been able to gather from the net, including this website is that this can be explained by invoking sp3 hybridization and a d orbital. The 3 anionic oxygens who already have a charge of -1 pair up with the 3 partially filled hybrid orbitals in phosphorous, while the remaining filled sp3 orbital on phosphorous donates it's electron pair to the empty orbital on an extra oxygen.
Once again, this is exactly the same for sulfate, except that it can only bond to 2 anionic oxygens while also being able to donate 2 electron pairs to other oxygen atoms.
This leaves us with 4 single bonds to oxygen atoms. This doesn't seem unusual at all. Nitrogen can easily take on 4 bonds just like phosphorous, and seeing how row 3 elements have a looser hold on their outer electrons, it makes sense that they can donate their complete electron pairs much easier than row two elements, seeing how their electron pairs would "stick out" more, making them much more available for bonding.
However, in Phosphate when you get to the 5th bond you have to explain it by being a little more creative. You could say that a p orbital on one of the oxygen atoms donates it's electron pair to an empty d orbital on phosphorous, creating and extra bond between phosphorous and one of the oxygens. The same thing happens in sulfate only that for some reason it can form two double bonds. (which I really wonder if that would be pushing the limits of stability at all. I mean wouldn't the two d orbitals kind of overlap at an awkward angle causing some kind of repulsion?)
Now, apparently from what I've heard, experimental evidence has shown that there is very little d orbital involvement in molecules like phosphate or sulfate, and they're practically all single bonds. So my question is, is this true? And if so, then why is the double bond picture still in practically every textbook, diagram, and piece of scientific literature that you come across? How did this picture of these ions emerge? I know it had something to do with linus pauling, but was this just accepted based on theory alone and no experimental evidence at all? Was there some experimental evidence for it in the beginning?
Sorry if it sounds like I'm asking too much, but I just find this whole situation so confusing. I'd appreciate your answers.

You can always go back to your question and [edit] it. Add more context, like linking the question or tags. Keep in mind that once you have obtained an answer it is unwise (and not very welcome here) to go back and change the question. But until then you can try to fine tune your question.

Recent experimental evidence shows that d orbitals are involved in the extra double bond of sulfate or phosphate ions. Therefore a stable sulfate or phosphate ion has 2 double bonds instead of only one to make the formal charge on sulfur or phosphorus zero. Any Lewis structure with expanded octet (periods 3 and higher) should have zero formal charge on the central atom.

It will be very helpful if you could quote those "recent experimental evidences".

Ok, I'll bite.  How do you experimentally identify a $d$ orbital electron versus an $s$ or $p$ one?  Oh boy this is going to be good!

This is completely untrue! There should be half a dozen downvotes here already.

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text:
How do polymer properties change when moving from bulk to nanosized fibers?
Polymer are used in many applications, but how do the properties change when they are fabricated on the nanolevel? Im especially interested in polymeric and polymer derived nanofibers, and why they excel at their respective applications. Thanks!

Nanofiber polymers in contrast with bulk polymers have high ratio of surface area to volume.
So, nanofiber polymers could be useful for any application that needs to maximize the ratio of surface area to volume.
Let's consider their application as scaffolds for tissue engineering: increasing the surface area of artificial-organ scaffolds gives cells more points at which to adhere.
Another application of nanofiber polymers is their use as electrodes  in fuel cells:  the greater the electrodes surface area, the more efficiently they catalyze the reactions that drive the cell.
Another interersting application of polymer nanofibers is in meshes that allow only nanoscale particles to pass through: ultrafine air, water filters, membranes for fuel cells.
Finally, as the capacitance is proportional to the surface area,  capacitors that store electricity, are one example of the possible applications of  polymer nanofibers.
For more details, please see this site and this article.


text:
How to estimate the equilibrium constant of the reaction between NaF and HCl
Since $\ce{HCl}$ is more acidic than $\ce{HF}$, the left hand side of a reaction has a stronger acid than the right hand side.
Since $\ce{NaF}$ is more basic than $\ce{NaCl}$, the left hand side has a stronger base relative to the right hand side.
$$\begin{align}\text{Stronger Acid/Base} & \ce{->} \text{Weaker Acid/Base}\\ \text{(Less Stable)} & \ce{->} \text{(More Stable)}\end{align}$$
So shouldn't the equilibrium constant be larger than $1$?
My textbook says the equilibrium constant is less than $1$.

Have you tried to make the calculus? Which values of $\ce{K_a}$ have you got? And obviously it depends on the values of each concentrations.

I am not allowed to use equilibrium constant tables. I am supposed to come up with an answer based on reasoning but for some reason my reasoning is wrong.

My textbook doesn't mention anything about concentrations for that question.

Thats strange, let me think.

What solvent is the reaction supposed to occur in?  The answer will be different if it is say neat HF vs. if it's water.  Or maybe it is liquid HCl at cryogenic temperatures?  Or acetonitrile?

@Yashas how more stable is concluded from the weak acid/base ?

@ProblemDestroyer Weak acid/base don't readily react in acid-base reactions unless a strong complementary base/acid is used.

By the way is this reaction equlibrium possible ? I mean as HCl is strong acid it will dissociate completely so there is nothing called eqb which can be established isnt ? @Yashas

@ProblemDestroyer In principle, we can have an equilibrium for nearly all reactions. Some reactions might strongly favor the products or reactants but there is still the notion of equilibrium. In this particular case, even though HCl is a strong acid, we can still talk about equilibrium. In fact, that's how we define pH!

Understood @Yashas .

We can entirely ignore the sodium cation; it is only a spectator ion. This means, the reaction we are observing is:
$$\ce{HCl + F- <=> Cl- + HF}\tag{1}$$
The equilibrium constant for this reaction is:
$$K = \frac{[\ce{Cl-}][\ce{HF}]}{[\ce{HCl}][\ce{F-}]}\tag{2}$$
We can expand this equation:
$$K = \frac{[\ce{Cl-}][\ce{H+}][\ce{HF}]}{[\ce{HCl}][\ce{H+}][\ce{F-}]}\tag{2'}$$
And then we realise that that is nothing else than:
$$K = \frac{[\ce{Cl-}][\ce{H+}][\ce{HF}]}{[\ce{HCl}][\ce{H+}][\ce{F-}]} = \frac{[\ce{Cl-}][\ce{H+}]}{[\ce{HCl}]} \cdot \frac{[\ce{HF}]}{[\ce{H+}][\ce{F-}]} = \frac{K_\mathrm{a}(\ce{HCl})}{K_\mathrm{a}(\ce{HF})} \tag{2''}$$
So the equilibrium constant is the fraction of the acidity constants of $\ce{HCl}$ and $\ce{HF}$. We know that $\ce{HCl}$ is a much stronger acid than $\ce{HF}$. This is reflected by the inequation:
$$K_\mathrm{a}(\ce{HCl}) > K_\mathrm{a}(\ce{HF}) \tag{3}$$
Since the numerator is larger than the denominator, the value of the fraction must be larger than $1$. Therefore, the product side is preferred.
If your book arrives at any other conclusion, it is disregarding experimental results and should be replaced.

@CurtF. Fair point. Since the equation balances without $\ce{H2O}$, I removed it.

Sir although i agree eith your calculation , shouldnt in real life it would not have a eqb between HCl and NaF as such HCl is a strong acid so when it is together in water with NaF it will dissociate completely isnt ? So no equlibrium can be established  right ?

@ProblemDestroyer Have you ever taken a bottle of HCl and taken a sniff? There are acidic fumes caused by the equilibria $\ce{H3O+ + Cl- <<=> H2O + HCl(aq) <=> H2O + HCl (g)$ Or in other words: HCl an only leak into the gas phase if there is an ever so minute concentration of undissociated HCl in the aquaeous phase. Therefore, your assumption is wrong. (For practical intents and purposes, however, hte equilibrium is so strongly product-sided that we can ignore undissociated HCl in almost all calculations.)

I guess it's late, but better late than never.
Technically, HF should be the stronger acid, as its parts are already strongly partially charged. So strong in fact, that even after dissociating into water the iones are present rather as ion pairs (H3O+.F-) than as seperate ions. This however inhibits fluoridic acids property as acid.
The proper reasoning for these two acids therefore would be simply per electronegativity difference between their compounds, as both acids have only 2 atoms, of them being hydrogen and the other a member of the halogen class.

This is not true, HCl is a stronger acid than HF.

Electronegativity doesn't matter here http://chemistry.stackexchange.com/a/34829/9961

No, HF is a weak acid with a dissolution equilibrium of $6.8*10^{-4}$


text:
Does aluminium dissolve in soda glass?
Specifically, if a piece of aluminized glass is heated to the melting point of the glass will the coating dissolve into it or otherwise react with it?

I think nothing will happen - you get molten salt polymer and molten metal phases - very different properties.

@Mithoron would a thermite-like reaction happen? http://blog.teachersource.com/2011/03/02/silicon-from-sand/ or would it have to be well above melting point?

At least with quartz it's supposed to be hard to trigger - http://www.researchgate.net/publication/257912633_Thermit-type_SiO2-Al_reaction_in_arc_discharge - obtained by arc discharge (!)

The aluminum would like to oxidize, and may end up grabbing some oxygen from the glass. It will likely end up as a alumina-silicate glass (with the sodium/boron/whatnot as well).

Since soda[-lime] glass is specified, then Al will react with the alkali forming an aluminate, e.g. $\ce{Na2Al2O4}$. This may or may not dissolve in the molten glass, depending on concentration of the aluminate(s), temperature and cooling rate. If it precipitates, the glass would have a "frosted" appearance.

sodium aluminosilicate glasses definitely exist.


text:
Explanation of calculating equilibrium constant

Hydrogen iodide is a colourless gas that will decompose into colourless hydrogen gas and purple iodine gas according to the following endothermic reaction.
$$\ce{2 HI (g) <=> H2 (g) + I2 (g)}$$
A $\pu{1.0 L}$ glass container was filled with $\pu{0.60 mol}$ of hydrogen iodide gas. When equilibrium was established, there were $\pu{0.25 mol}$ of iodine gas present in the container. Calculate the equilibrium constant for this reaction.
  $$
\begin{array}{|c|c|c|c|}\hline
&\ce{HI}&\ce{H2}&\ce{I2}\\\hline
\text{Initial}&0.60&0&0\\\hline
\text{Used/made}&0.50&0.25&0.25\\\hline
\text{Equilibrium}&0.10&0.25&0.25\\\hline
\end{array}
$$
  $$K=\frac{[\ce{H2}][\ce{I2}]}{[\ce{HI}]^2}=\frac{0.25\times0.25}{0.10^2}=6.25$$

This is what I understand:

Product always goes on the top, reactant on bottom
Balancing coefficients eg $\ce{2HI}$ become powers $\ce{-> HI^2}$
Initially, 0.6 moles of hydrogen iodide gas was placed, then at equilibrium 0.25 moles of iodine gas.
If there was 0.25 moles of iodine gas, there has to be 0.25 moles of hydrogen gas aswell thus the top part of the fraction becomes $0.25 \times 0.25$
I have no idea where the 0.1 moles come from …

I don't want to 'understand' these as much as I just want to be able to do them.

Your reasoning is correct. Initially, you saw that there was no iodine in the vessel. After reaching equilibrium, you saw 0.25 mol. Where did this come from?
Answer: Hydrogen iodide broke down to form this iodine and hydrogen. How much hydrogen iodide? Well this is a super simple moles problem.
0.25 moles of iodine was formed, and the mole ratio for $\ce{HI:I_2}$ is $2:1$. For 0.25 moles of iodine to be formed, 0.5 moles of hydrogen iodide must have been used (just look at your balanced equation).
If 0.5 moles of hydrogen iodide was used, how much hydrogen iodide is left? 0.60 moles minus 0.50 moles to leave 0.10 mol.
Then, as you did correctly, you plug it into the $K_c$ formula.

While you have stated that you just want to know how to do it (which you should be able to now), it is useful to know what's going on. Basically, the $K_c$ is a measure of where the equilibrium is. If your $K_c$ value is high, there will be a higher concentration of the products at equilibrium.

You’re pretty good up to step 4. However, your step 5 is somewhat premature:

If I have generated $0.25~\mathrm{mol}$ of $\ce{H2}$ and $\ce{I2}$ (each), how much of the original $\ce{HI}$ did I use up?
If I used that much $\ce{HI}$, how much of the original portion is remaining at equilibrium.
Now that I have amount $x$ of $\ce{HI}$, enter it all into the equation.
Calculate, write the answer, score in the exam.
Profit.

I’ll leave it to you to decide which of my steps your step 5 corresponds to.


text:
Definition of the B3LYP functional in common QC programs
I wanted to use custom functionals with Gaussian and came up with some interesting definitions of the B3LYP functional within Gaussian, Orca and Turbomole, ...
$$\small\begin{array}{lcccccc}
\hline
  &  \text{Gaussian}  &  \text{ORCA}  & \text{Turbomole}  &  \text{NWChem} & \text{Molpro} & \text{GAMESS(US)}\\
\hline
\text{HF}  &  0.2  &  0.2  &  0.2  &  0.2  &  0.2  &  0.2\\
\text{Slater}  &  0.8  &  0.8  &  0.8  &  0.8  &  \color{red}{0.08}  &  0.8\\
\text{Becke}  &  0.72  &  0.72  &    0.72  &  0.72  &  0.72  &  0.72\\
\text{VWN-x}  &  {\color{red}1}  &  {\color{red}{1}}  &  0.19  &  0.19  &  0.19  &  0.19\\
\text{LYP}  &  0.81  &  0.81  &   0.81  &  0.81  &  0.81  &  0.81\\
\hline
\end{array}$$
It seems to be clear, that all programs define it somehow similar, but there are also some points that made me curious about the definitions. Let's start with the overall structure of this functional:
$$\text{XC-Functional}=a E_x^\text{local}+(1-a)E_x^\text{HF}+b E_x^\text{non-local}+c E_c^\text{non-local}+(1-c) E_c^\text{local}$$
The Gaussian documentation for DFT inputs specifies how one could change the amounts of each "sub-functional" by using the IOp-Statements through:
IOp(3/76=mmmmmnnnnn) IOp(3/77=mmmmmnnnnn) IOp(3/78=mmmmmnnnnn)

Where mmmmm and nnnnn will get divided by 10000 and 3/76=$P_1P_2$, 3/77=$P_3P_4$ and 3/88=$P_5P_6$. Those $P_{1,..,6}$ values are part of the following equation, which is a modified version of the equation above.
$$\text{XC-Functional}=P_2 E_X^\text{HF} + P_1 \left(P_4 E_X^\text{Slater} + P_3\Delta E_x^\text{non-local}\right) + P_6 E_C^\text{local} + P_5 \Delta E_C^\text{non-local}$$
Combining both, this yields: $P_1 = \{1,0\}$, $P_2 = 1-a$, $P_3 = b$, $P_4 = a$, $P_5 = c$ and $P_6 = 1-c$.

Now looking at Gaussian and ORCA, they both use 100% VWN and 81% of LYP, which should actually be combined to give 100% (instead of 181%) as is done by all other mentioned programs through using 19% VWN and 81% LYP.
It seems, that I am lacking some important information to understand that definitions in both, Gaussian and ORCA, and I hope seriously, that someone can enlighten me.

PS My discussion with Martin came up with the possibility, namely that there might be some missing brackets in the second equation, giving the following:
$$\text{XC-Functional}=P_2 E_X^\text{HF} + P_1 \left(P_4 E_X^\text{Slater} + P_3\Delta E_x^\text{non-local}\right) + P_6 \mathbf{{\color{red}(}}(1-P_5)E_C^\text{local} + P_5 \Delta E_C^\text{non-local}\mathbf{{\color{red})}}$$

Appendix: Information from the program output or the documentary

Gaussian 09 Rev. A.02 output
IExCor=  402 DFT=T Ex=B+HF Corr=LYP ExCW=0 ScaHFX=  0.200000
ScaDFX=  0.800000  0.720000  1.000000  0.810000

ORCA 3.0.3 output
Fraction HF Exchange ScalHFX         ....  0.200000
Scaling of DF-GGA-X  ScalDFX         ....  0.720000
Scaling of DF-GGA-C  ScalDFC         ....  0.810000
Scaling of DF-LDA-C  ScalLDAC        ....  1.000000
Perturbative correction              ....  0.000000
NL short-range parameter             ....  4.800000

Turbomole 7 define
 b3-lyp          | HYB  | 0.8S+0.72B88   | 0.19VWN(V)     | 1-3,5,6,10
                 |      | +0.2HF         | +0.81LYP       |
 b3-lyp_Gaussian | HYB  | 0.8S+0.72B88   | 0.19VWN(III)   | 1-3,5,6,10
                 |      | +0.2HF         | +0.81LYP       |

NWChem
Molpro
B3LYP   EXACT:B88:DIRAC:LYP:VWN5    0.2:0.72:0.08:0.81:0.19

the 0.08 is probably a typo
GAMESS(US)

ORCA can perform both B3LYP and B3LYP/G, the first corresponding to the TurboMole implementation and the second corresponding to the Gaussian implementation.

@Jonathon In ORCA B3LYP vs B3LYP/G only changes the VWN from VWN-5 to VWN-3 and does not change sth about the amount of VWN, like what you can see from my appended Turbomole information.

I just checked the outputs and it looks like you are right, I guess the manual is wrong then. I will file a bug report for them to fix the wording.

Q-Chem has both B3LYP (corresponding to Gaussian) and B3LYP5, as noted in the Q-Chem manual

It's one of the real annoyances of B3LYP as a functional...

Tangentially related to the question: ORCA, Turbomole, Molpro and GAMESS(US) are now all interfaced to Libxc (along with ~30 other programs), which provides common and portable implementations of ~600 density functionals including several variants of B3LYP. If you want to implement a custom density functional, this is quite simple to do in Libxc. Moreover, e.g. Psi4 and PySCF which are both open-source/free-software packages for Gaussian-basis quantum chemistry allow you to define and customize the exchange-correlation functional in the input file, making it easy to use custom functionals with

As a former contributor to Orca who had access to Turbomole sources before that, I can say with confidence that the internal definitions of B3LYP and B3LYP_gaussian are consistent among them and Gaussian (where available). So no calculations should go wrong.
While I cannot speak for Gaussian, for Orca I am certain that the ScalLDAC parameter is badly printed here (and not relevant for many of the newer functionals).
As to what you identified as a typo in Molpro -- It may be that B88 was shoe-horned into the more mainstream definition of GGAs, which can be loosely worded as "a multiplicative correction to the LDA energy" (the energy density, actually). Becke, on the other hand, described the B88 functional originally as an additive correction ($\Delta E$). NWChem makes the distinction explicit in the keyword nonlocal, see the manual section on exchange-correlation potentials. So for Molpro, the exchange accounting should read: $0.2+0.72+0.08=1$, whereas the others use: $0.2+0.8=1$ and keep the $0.72$ implicit.


text:
Buffer capacity of different types of milk
Currently I am tasked with designing an experiment that that I will have to perform in 2-4 weeks, but I don't have to decide on on a topic soon. I was thinking of performing an acidometric titration (most likely with hydrochloric acid) of different types of milk such as skim, organic, fat free and from it make a $\mathrm{pH}$ curve. From here I would see how these different types of milk acted as buffers relative to each other, by analyzing their buffer regions in the $\mathrm{pH}$ curve.
I have heard that milk can act as a buffer, but I cannot find information about what buffers regulate the $\mathrm{pH}$ the most.
I would greatly appreciate if any of you could share your knowledge or some sources about buffers constitute milk.


text:
Why anhydrous acetic acid is differentiating solvent?
Anhydrous acetic acid is differentiating solvent in which strong acids show different strength but water isn't. Why anhydrous acetic acid shows the strength of acids clearly, but water solvent cannot do so?

see: https://en.wikipedia.org/wiki/Leveling_effect

Every basic solvent will "weaken" dissolved acid which is stronger than solvents conjugated acid, but protonated acetic acid is very strong.

@Mithoron thanks your point is great for answer.


text:
Shelf Life of Methylcyclopentadiene
We are making organometallics in the lab, but instead of just regular Cyclopentadiene dimer, we crack methylcyclopentadiene dimer to make methylcyclopentadiene.
I cannot find the shelf life of methylcyclopentadiene, because we are going to be using a lot, and it takes quite a bit of time to make, and we need to make some in advance.
Does anyone happen to know the shelf life of methycyclopentadiene?
Thanks!

If you google "methylcyclopentadiene, rate of dimerization" you will find some useful links.  This one provides the rate constant and activation energy for the dimerization of MeCPD, with this information you can calculate the shelf half-life at your storage temperature.

This link says that MeCPD can be stored at -30 °C for a few days.


text:
Detect the presence of non stereochemically annotated atoms in mol files
I'm facing a problem now and I don't know where to start from. I'm treating databases in which there are mol-files that have stereo-chemical carbons, which, sometime, are not annotated as being R/S or Z/E in the stereochemistry column.
I would like to be capable of finding all asymmetric carbons to check if the stereochemical center is annotated or not. Does anyone knows of a software, a library or elegant piece of code to figure out this problem. I'm working with RDkit (Python) and I can access to chemaxon if needed.

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

With the latest rdkit, version 2022.9.4, it is possible to obtain the information about all the unassigned chiral carbons from a MOL file, using the rdkit.Chem.FindMolChiralCenters function along with the argument includeUnassigned set to True.
A sample MOL that I use for demonstration: 2-methylbutan-1-ol.mol
Actelion Java MolfileCreator 1.0

  6  5  0  0  0  0  0  0  0  0999 V2000
    3.4641   -0.5000   -0.0000 C   0  0  0  0  0  0  0  0  0  0  0  0
    2.5981   -0.0000   -0.0000 C   0  0  0  0  0  0  0  0  0  0  0  0
    1.7321   -0.5000   -0.0000 C   0  0  0  0  0  0  0  0  0  0  0  0
    1.7321   -1.5000   -0.0000 C   0  0  0  0  0  0  0  0  0  0  0  0
    0.8660   -0.0000   -0.0000 C   0  0  0  0  0  0  0  0  0  0  0  0
    0.0000   -0.5000   -0.0000 O   0  0  0  0  0  0  0  0  0  0  0  0
  2  1  1  0  0  0  0
  3  2  1  0  0  0  0
  4  3  1  0  0  0  0
  5  3  1  0  0  0  0
  6  5  1  0  0  0  0
M  END

A sample code is shown below:
from rdkit.Chem import rdmolfiles, FindMolChiralCenters
mol=rdmolfiles.MolFromMolFile('2-methylbutan-1-ol.mol')
chiral_carbon_info = FindMolChiralCenters(mol, includeUnassigned=True)

print(chiral_carbon_info)
# >> [(2, '?')]

The output means there is a chiral carbon whose atom id is 2 and the question mark means that the chirality configuration is not defined.
PS:
The sample MOL file was created using this website, which allows you to create a MOL file either from SMILES string or manually drawing the various atoms in the molecule.


text:
What is more reactive towards a cyanohydrin formation an aldehyde or an Ketone?
Which is more reactive—aldehydes or ketones—towards the nucleophilic addition by cyanide to form cyanohydrins?

Aldehydes will generally form cyanohydrins more readily than ketones, but some ketones such as trifluoromethyl or acyl ketones may behave more like aldehydes due to the electron deficiency at carbonyl carbon.


text:
Nucleophilic attack at a carbonyl - angle of attack
What is the specific angle between the carbonyl carbon and the nucleophile during the nucleophilic addition reaction?

Have a look at the Wikipedia article concerning the Bürgi-Dunitz angle.

Let's take the following approach to answer your interesting question:

Use the bent bond model to describe the carbonyl
Calculate the bond angles around the carbonyl carbon
Treat the reaction as an $\mathrm{SN2}$ reaction
Use trigonometry to deduce the angle of nucleophilic approach

The bent-bond model is widely used in chemistry to describe highly-strained systems such as cyclopropane and double bonds (ref_1, ref_2).  In the case of a double bond, we wind up with a two-membered ring (which nicely explains the high heat of formation of such compounds). Here we will use it to describe the carbonyl double bond.  Here is a picture of what the bent bond description of a carbonyl looks like (it's the top left diagram that's of interest, note too that it is a picture of an amide carbonyl).

(image source)
In a simple carbonyl compound like formaldehyde the $\ce{H-C-H}$ bond angle is around 118°.  Using Coulson's theorem (ref_2, ref_4) we can use this bond angle to estimate the hybridization of the carbonyl $\ce{C-H}$ bonds.
$$\ce{1+\lambda^2cos(118)=0}$$ where $\ce{\lambda^2}$ represents the hybridization index of the $\ce{C-H}$ bond. From this equation we find that $\ce{\lambda^2=}$ $\mathrm{2.13}$, or the carbonyl $\ce{C-H}$ bonds are $\ce{sp^{2.13}}$ hybridized.
We can now sum up the s-character in the 4 bonds around the carbonyl carbon (it must add up to 1) and deduce the hybridization of the 2 $\ce{C-O}$ bent bonds.
$$\ce{\frac{2}{1+\lambda^_{C-H}^2}+\frac{2}{1+\lambda_{C-O}^2}=1}$$
Solving this equation we find that $\ce{C-O}$ bent bonds are $\ce{sp^{4.54}}$ hybridized.
Substituting this information back into Coulson's theorem allows us to estimate the $\ce{O-C-O}$ bond angle as being equal to 102.7°.
In an $\mathrm{SN2}$ reaction the nucleophile approaches 180° from the leaving group ("backside attack").  In this case, one of the carbonyl $\ce{C-O}$ bonds is the leaving group.
Using simple trigonometry we can estimate what the angle is for 180° attack of the $\ce{C-O}$ bond relative to the plane of the carbon and oxygen nuclei. Half of the $\ce{O-C-O}$ bond angle is 51.4°, leaving 38.6° to reach 90°. So the angle for backside attack relative to the plane containing the $\ce{C}$ and $\ce{O}$ atoms is $\mathrm{90°+38.6°=128.6°}$
The actual angle for attack has been found to be around 107° (see @Philipp 's comment on the Bürgi-Dunitz angle).  This angle was first determined by extensive, detailed crystallographic investigations and later supported by computational studies. The approach used here is no more than a quick, "back of the envelope" method. Yet it still provides an answer within about 20° of Bürgi and Dunitz's work. For such a quick, crude method, I would consider this result to be in qualitative agreement. It supports the back side attack $\mathrm{SN2}$ model, illustrates how a $\ce{C-O}$ bond participates as a leaving group and how we wind up with a tetrahedral carbon.


text:
Destabilization of antibonding MO vs stabilization of bonding MO
My textbook writes that

The net energy stabilization due to the occupation of a bonding MO is equal to the net energy destabilization due to the occupation of the corresponding antibonding MO.

Essentially, it claims that these two quantities are equal.

(Image source: SparkNotes)
Shouldn't the antibonding orbital be destabilized more than the bonding orbital is stabilized, though?

I agree with your notion. If there was no energy lost by creating bonding and antibonding orbitals and populating them both, helium would easily form compounds (and all our thinking towards antibonding electrons being bad would be null and void).

Related: How can antibonding orbitals be more antibonding than bonding orbitals are bonding?

Searching for the truth where there's no one, trying to make scientific sense of what isn't even scientific... All these models are way oversimplified, so there is very little sense in being nitpicking as well as asking about their truth: the general chemistry MOT description is generally miles away from reality. To learn the real things one better studies quantum theory rather than asks all these non-sense like "Isn't the antibonding orbital more antibonding than the bonding orbital is bonding?"

By the way, does the author say a word about $\ce{He2}$ instability? I mean, yeah, at this particular page he says that the energy differences are equal, but maybe he elaborates on that matter a bit later and corrects the initial oversimplified view.

TL;DR

"The destabilisation of an antibonding MO is equal to the stabilisation of a bonding MO, relative to the constituent AOs."

Several key assumptions are inherent in this statement.

Born–Oppenheimer approximation (not discussed further)
"Orbital approximation": wavefunction of $\ce{H2, He2, ...}$ can be expressed as an antisymmetrised product of one-electron wavefunctions (called MOs)
Said MOs take the form of the stationary states of $\ce{H2+}$, without modification
LCAO approximation: stationary states of $\ce{H2+}$ can be expressed as a linear combination of 1s orbitals on both hydrogens
The overlap integral $\langle \mathrm{1s_A} | \mathrm{1s_B} \rangle$ is neglected

If we make all five assumptions, then the statement above is correct.
If we only make assumptions 1 through 4, then MOs will still have the same form as above (i.e. everything else about the MO diagram is correct), but the statement above will be incorrect: the antibonding orbital will be more destabilised than the bonding orbital is stabilised.
The question linked in the comments does give a qualitative explanation, so depending on what you are looking for, it may be worth checking out.

Assumptions
The MOs that you see are one-electron wavefunctions that are obtained by looking at the single-electron species $\ce{H2+}$. Finding the exact stationary states of this system (by analytical solution of the Schrödinger equation) is a bit complicated, so the strategy is to find a set of approximate stationary states for the electron in $\ce{H2+}$, which we can label as $\{|i\rangle\} (i = 1, 2, 3 \cdots)$. These states are called molecular orbitals.
For multi-electron molecules, like $\ce{H2}$, a further approximation is usually invoked, in expressing the electronic wavefunction of $\ce{H2}$ as a properly antisymmetrised product of these states $\{|i\rangle\}$. Therefore, the MOs, which were only "rigorously" derived for $\ce{H2+}$, may be extended to molecules such as $\ce{H2}$. Very loosely speaking, we may assign one stationary state $|i\rangle$ to one electron, which corresponds to placing an electron in an orbital. Because of the Pauli exclusion principle, each stationary state may only be assigned to a maximum of two electrons.
These are obviously crude approximations and it is a surprise that it even works decently well. In particular the issue of electron-electron repulsions in multi-electron systems is glossed over; in this very simple model, the MOs, being derived from a one-electron system, are entirely unaffected by them, which doesn't really make sense. As the comments on the question have alluded to, there are much more sophisticated ways of obtaining MOs nowadays, but since the claim in the question is derived using this model, we have to use it.

MOs of hydrogen molecule ion
Once we have assumed that the stationary states (i.e. MOs) are of the form
$$\psi = c_1\phi_1 + c_2\phi_2$$
where $\psi$ is an MO and $\phi_1, \phi_2$ 1s atomic orbitals on the two hydrogens, then one can use the variational principle to optimise the two coefficients $c_1, c_2$. The idea is that one wants to minimise the energy of the wavefunction $\psi$:
$$\mathcal{E} = \frac{\langle \psi | H | \psi \rangle}{\langle \psi | \psi \rangle}$$
with respect to both $c_1$ and $c_2$. To this end, both the following partial derivatives must be equal to zero:
$$\left(\frac{\partial \mathcal{E}}{\partial c_1}\right)_{\!c_2} = \left(\frac{\partial \mathcal{E}}{\partial c_2}\right)_{\!c_1} = 0$$
The maths will be omitted and if one is interested, one may find it in a variety of textbooks. For example it is covered in Atkins & Friedman's Molecular Quantum Mechanics 5th ed. (search for Rayleigh–Ritz method), or in Coulson's Valence 3rd ed. It is also described thoroughly in this handout from QMUL. The key result is the secular equations, which are a set of generalised eigenvector equations:
$$\mathbf{Hc} - \mathcal{E}\mathbf{Sc} = \mathbf{0}$$
where the $2 \times 1$ vector $\mathbf{c}$ simply contains the coefficients $c_1$ and $c_2$, the elements of the Hamiltonian matrix are $\mathbf{H}_{ab} = \langle a | H | b \rangle$ and the elements of the overlap matrix are $\mathbf{S}_{ab} = \langle a | b \rangle$. The basis set used is $\{|\phi_1\rangle, |\phi_2\rangle\}$. For a non-trivial solution (i.e. $\mathbf{c} \neq \mathbf{0}$) we require the secular determinant to vanish:
$$|\mathbf{H} - \mathcal{E}\mathbf{S}| = 0$$
Now we can use symbols to fill in these matrix elements:

$H_{11} = H_{22} = \alpha$ (this must be so, as both hydrogen atoms are equivalent by symmetry)
$H_{12} = H_{21} = \beta$ (assume that $\beta$ is real, so by the hermiticity of $H$, we must have $H_{12} = H_{21}$)
$S_{11} = S_{22} = 1$ (the atomic orbitals are normalised)
$S_{12} = S_{21}$ (again, assume that this is real. Assumption 5 is that this goes to zero, but for now we will not neglect it; later we can set it to zero to see what happens.)

The exact form of the Hamiltonian, as well as more explicit expressions for these quantities, are given in Atkins MQM. For now they need not concern us. Solving the determinant we obtain two values of $\mathcal{E}$ and two associated vectors $\mathbf{c}$, which can simply be written as the MOs $\psi = c_1\phi_1 + c_2\phi_2$. The MOs (i.e. approximate stationary states) are now quoted without proof, with their eigenvalues (i.e. their approximate energies):
$$\begin{align}
\psi_1 &= \frac{1}{\sqrt{2 + 2S}}(\phi_1 + \phi_2) & & & \left\langle \psi_1 \middle| H \middle| \psi_1 \right\rangle &= \mathcal{E}_1 = \frac{\alpha + \beta}{1 + S_{12}} \tag{1} \\
\psi_2 &= \frac{1}{\sqrt{2 - 2S}}(\phi_1 - \phi_2) & & & \left\langle \psi_2 \middle| H \middle| \psi_2 \right\rangle &= \mathcal{E}_2 = \frac{\alpha - \beta}{1 - S_{12}} \tag{2}
\end{align}$$
$\mathcal{E}_1$ is the ground state energy (i.e. $\mathcal{E}_1 < \mathcal{E}_2$) since $\alpha$ and $\beta$ are both negative quantities (again MQM has more info).

Stabilisation and destabilisation energies
Now we have actually reached the part where the original question can be answered. What is the stabilisation of the bonding MO? The energy of the constituent AOs is $H_{11} = H_{22} = \alpha$. So, the stabilisation must be:
$$\begin{align}
E_{\mathrm{stab}} &= \alpha - \mathcal{E}_1 \\
&= \alpha - \frac{\alpha + \beta}{1 + S_{12}} \\
&= \frac{\alpha S_{12} - \beta}{1 + S_{12}}
\end{align}$$
Similarly, the destabilisation of the antibonding MO is:
$$\begin{align}
E_{\mathrm{destab}} &= \mathcal{E}_2 - \alpha \\
&= \frac{\alpha S_{12} - \beta}{1 - S_{12}}
\end{align}$$
The quantity $S_{12}$ is positive (again it should be in MQM), so $E_{\mathrm{destab}} > E_{\mathrm{stab}}$.
The "incorrect" piece of information comes from a further simplifying assumption - namely, that $S_{12} \ll 1$. That is not the greatest assumption, since the values of overlap integrals can sometimes be quite non-negligible (for a $\ce{H-H}$ bond length of $74~\mathrm{pm}$, $S_{12} = 0.753$!), but it is a useful way to predict approximate energies and the general forms of the MOs.
If $S_{12} \ll 1$, then $1 \pm S_{12} \approx 1$ and equations $(1)$ and $(2)$ become:
$$\begin{align}
\psi_1 &= \frac{1}{\sqrt{2}}(\phi_1 + \phi_2) & & & \left\langle \psi_1 \middle| H \middle| \psi_1 \right\rangle &= \mathcal{E}_1 = \alpha + \beta \tag{3} \\
\psi_2 &= \frac{1}{\sqrt{2}}(\phi_1 - \phi_2) & & & \left\langle \psi_2 \middle| H \middle| \psi_2 \right\rangle &= \mathcal{E}_2 = \alpha - \beta \tag{4}
\end{align}$$
Now it turns out that, with the additional neglect of overlap approximation, $E_{\mathrm{stab}} = E_{\mathrm{destab}} = -\beta$. That's where the "destabilisation of antibonding MO = stabilisation of bonding MO" idea comes from.
The arguments above can be extended to the case where you have constituent AOs which are different. The main difference is that $H_{11} \neq H_{22}$, so you have to write $H_{11} = \alpha_1$ and $H_{22} = \alpha_2$. The result you obtain will be rather similar: the destabilisation of the antibonding MO is greater than the stabilisation of the bonding MO, but in the limit of $S_{12} \to 0$, they become equal.

"The true wavefunction for H2HX2 does not need to have an electron in one "orbital" or another. But that is a story for another day (and for someone else to say...)" Want to know more about it. Will the hyperlinked book work?

@Mockingbird it's been a while since I wrote this answer. Fundamentally what I meant is that we cannot solve the Schrodinger equation for $\ce{H2}$. When we construct a total wavefunction from constituent molecular orbitals, it is merely an approximation to the true wavefunction, which we can't obtain. I'd recommend reading Atkins MQM (I removed the other references).

I cleared the earlier comments as I revised the answer thoroughly; I think it is accurate now, but if there are still mistakes please drop a comment.

Why is there a consensus regarding the Schrodinger equation of $\ce{H2}$ can't be solved?

I liked your previous answer better. It's probably due to my inadequate grasp of QM.

"Orbital approximation": wavefunction of H2,He2,…HX2,HeX2,… can be expressed as an antisymmetrised product of one-electron wavefunctions (called MOs)" I think you should elaborate it further.

@Mockingbird no, I don't want to elaborate further. This answer is long enough and I don't want to spend any more time on it, please go read a QM book if you want more detail, or ask a new question

What is the smallest possible value of the overlap integral for any interaction between two atoms? I would think for ionic bonds with very high ionic character (e.g. $\ce {CsF}$), the overlap integral would be very small in value.

@TanYongBoon, technically, depending on how you choose the phase of the orbitals, the overlap integral may be negative as well, so the most negative value is -1, if you take the overlap between an orbital and the negative of itself. And the smallest value in terms of magnitude is 0, for two orbitals which are orthogonal to one another (any pair of orbitals with different symmetry labels, for example). But if you're asking about real life examples of smallest but nonzero overlap, I don't know - CsF is a good idea, though.