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Theoretical basis behind orbital correlation diagram for pericyclic reactions
When discussing pericyclic reactions, the Woodward-Hoffman rules allow one to predict whether a reaction will be thermally/photochemically allowed. This has been discussed in many questions on chem.SE, but essentially boils down to the following statement:

A ground-state pericyclic change is symmetry allowed when the total number of $(4q+2)_\mathrm{s}$ and $(4r)_\mathrm{a}$ components is odd.

What the Woodward-Hoffman rules fail to explain however is why the reaction is allowed. Several explanations exist, such as the Huckel aromaticity of the transition states, however by far the most 'powerful' explanation is supposedly given by a well drawn orbital correlation diagram. Fleming devotes an entire section in his book on the topic of how to draw these diagrams:
"Correlation diagrams provide a compelling explanation, at least for those reactions that have well-defined elements of symmetry preserved throughout the reaction. The idea is to identify the symmetry elements maintained throughout the reaction, classify the orbitals undergoing change with respect to those symmetry elements, and then see how the orbitals of the starting materials connect with those of the product. The assumption is that an orbital in the starting material must feed into an orbital of the same symmetry in the product, preserving the symmetry throughout the reaction. Substituents, whether they technically break the symmetry or not, are treated as insignificant perturbations on the orbitals actually undergoing change."

Fig 1: Fleming, Molecular Orbitals and Organic Chemical Reactions- Reference Edition, 2010, Wiley
As far as I can find, and from recollection, Longuet-Higgins at Cambridge was the first person to use correlation diagrams to explain pericyclic processes (as per figure 2 below), but again, there's no explanation as to why this works.

Fig 2: Early (beautiful) example of a correlation diagram as applied to an electrocyclic reaction. J. Am. Chem. Soc. 1965, 87, 2045. (DOI)
To make an analogy to molecular orbital theory, clearly, correlation diagrams are simply 'mixing' things of the same symmetry (in this case A or S relative to some axis rather than the more expanded set of symmetry labels from group theory), but there seems to be little / no care in correlation diagrams as for the relative energies of the things being mixed. Is anyone able to explain the why of a correlation diagram?

I think it's more than a mixing. It's a continuous mapping. If I'm not mistaken, the argument is that you should be able to continuously distort from reactant to product, and the symmetry at each point of the distortion should match the symmetries of the orbitals. Not sure if this is helping your issue though...

I've always been disappointed by these diagrams for that exact same reason. I've tried to make sense of these energy levels and I ended up  giving up.

Chanced upon this, and I feel like there is also some additional relevant information in here: Barriers to thermally allowed reactions and the elusiveness of neutral homoaromaticity. J. Am. Chem. Soc. 1979, 101 (23), 6797–6802.

Introduction
Instead of the usual cycloadditions (which has some extra complications in terms of the reacting orbitals), let's consider the concerted reaction $\ce{H2 + D2 -> 2HD}$. We'll come back to the cycloadditions at the end.

occurring via a square planar transition state. This can be analysed with the W–H rules and you'll find that it's a thermally forbidden process $(_\sigma 2_\mathrm{s} + {}_\sigma 2_\mathrm{s})$. So, hopefully that's what we find at the end of this discussion.
Ignoring the isotope labels,1 there is $D_\mathrm{2h}$ symmetry throughout the reaction, so we will categorise the orbitals by their symmetry in this $D_\mathrm{2h}$ point group.2 I've defined the coordinate axes such that the reaction takes place in the xy-plane, with the z-axis perpendicular to the plane of the paper.
There is one simplification which can be made: the symmetry of an orbital under the three mirror planes uniquely specifies the irreducible representation under which an orbital transforms. Hence, in order to identify the point group, we don't need to bother with the symmetry under the five other operations (identity, three rotations, and inversion); we simply need to find the symmetry under the three reflections.3

Reactant and product orbitals
The relevant MOs will be constructed from linear combinations of the 1s orbitals on each atom. First, we look at the orbitals of the reactants, labelled $\psi_1$ through $\psi_4$. We will overlap the 1s orbitals on adjacent atoms to produce $\sigma$ and $\sigma^*$ orbitals for both $\ce{H2}$ and $\ce{D2}$. However, since these MOs do not have definite symmetry under all the mirror planes, we need to take linear combinations of these MOs.

Hopefully the energy ordering of these MOs is clear (I actually listed them in increasing order of energy). It can be deduced from the extent of antibonding/bonding character between each pair of 1s orbitals.
The product MOs, labelled with $\phi_1$ through $\phi_4$, have a very similar form. It should not really be surprising, as the system is exactly the same, just rotated 90 degrees.

Orbital correlation diagrams
We can define a reaction coordinate $x$, such that $x = 0$ before the reaction (i.e. reactants) and $x = 1$ after the reaction (i.e. products). Over the course of the reaction, there will always be four MOs; and these four MOs will always transform as $\mathrm{A_g + B_{1g} + B_{2u} + B_{3u}}$, regardless of what the value of $x$ is. This has to be the case because the basis set (four 1s orbitals) transforms as a representation which can be reduced into that sum.
As the reaction proceeds, the positions of the nuclei vary continuously. Therefore, the Hamiltonian, and also the energies of these MOs $\varepsilon_i$ $(i = 1,2,3,4)$, will vary as the reaction proceeds, so $\varepsilon_i$ is a continuous function of $x$. The orbital correlation diagram simply plots how $\varepsilon_i$ varies with $x$. There will be four curves on the diagram: one for the MO with symmetry $\mathrm{A_g}$, one for the MO with symmetry $\mathrm{B_{1g}}$, and so on.
In theory, one could find the MOs and their energies at every value of $x$, and then you would find four curves, one for each $i$. However, for a qualitative understanding, this isn't necessary. We already know the approximate values of $\varepsilon_i(0)$ and $\varepsilon_i(1)$, and we also know the MOs to which these values correspond. So all we need to do is to figure out which reactant MO $\psi_i$ correlates with which product MO $\phi_j$.
From the symmetry labels, this mapping is quite self-evident, as the $\mathrm{A_g}$ reactant MO must become the $\mathrm{A_g}$ product MO, and so on.

In this example, it also makes physical sense that as the reaction proceeds, the reactant MO $\psi_2$ will simply become the product MO $\phi_3$ (pictured above). For other correlation diagrams, there is a subtlety, but it is a point we will return to later on.

State correlation diagrams
Orbital correlation diagrams are all fine and dandy, but they can be put on a much more solid footing by constructing the relevant state correlation diagrams. I have deliberately avoided using the word "state" up until now. Here, "state" refers to an electronic state of the system as a whole.4 The simplest expression for an electronic state is given by a Slater determinant, and we will simply abbreviate it to the usual form that the organic chemists do. For example, in the ground state of the reactants, we can write
$$\Psi_1 = \psi_1^2\psi_2^2$$
where the antisymmetrisation is implicit. The use of capital $\Psi$ instead of small $\psi$ emphasises that this is a total electronic state of a system. The subscript 1 indicates that it is the ground state.
The electronic states of the reactants are well-described by single Slater determinants. For a system with 4 MOs and 4 electrons, there are a total of $8!/4!4! = 70$ states, and so theoretically we need to look at $\Psi_1$ through $\Psi_{70}$. However, it suffices to consider the first three singlet states (the triplet states can be ignored5). These are
$$\Psi_1 = \psi_1^2\psi_2^2 \qquad \Psi_2 = \psi_1^2\psi_2\psi_3 \qquad \Psi_3 = \psi_1^2\psi_3^2$$
which transform as $\mathrm{A_g}$, $\mathrm{B_{1g}}$, and $\mathrm{A_g}$ respectively.6 Likewise we will consider the three product states
$$\Phi_1 = \phi_1^2\phi_2^2 \qquad \Phi_2 = \phi_1^2\phi_2\phi_3 \qquad \Phi_3 = \phi_1^2\phi_3^2$$
which transform as $\mathrm{A_g}$, $\mathrm{B_{1g}}$, and $\mathrm{A_g}$ respectively.
The state correlation diagram plots the energies of these states $E_i$ as they vary with the reaction coordinate $x$, so the question is again: which reactant state $\Psi_i$ correlates with which product state $\Phi_j$.
Let's first make the assumption that a single Slater determinant is sufficient to describe an electronic state, throughout the entire reaction coordinate. In this case, the assignment is very straightforward. Since $\psi_1$ and $\psi_2$ in the reactants smoothly become $\phi_1$ and $\phi_3$, it means that the reactant state $\Psi_1 = \psi_1^2\psi_2^2$ must smoothly become the product state $\Phi_3 = \phi_1^2\phi_3^2$ as the reaction coordinate increases. A similar consideration for the other states leads to the diagram below.

Let's pause for a moment and figure out what this diagram means. It means that this molecular Hamiltonian has three approximate solutions under consideration; we label these solutions as $S_1$, $S_2$, and $S_3$. The first approximate solution, $S_1$, can be described as a Slater determinant:
$$S_1(x) = f(x)^2g(x)^2$$
where $f$ and $g$ are MOs which vary with $x$. At the point $x = 0$, we have $f = \psi_1$, $g = \psi_2$, and hence $S_1 = \psi_1^2\psi_2^2 = \Psi_1$. As $x$ increases, $\psi_1$ continuously distorts towards $\phi_1$, eventually becoming $\phi_1$ when $x = 1$, and likewise $\psi_2$ continuously distorts towards $\phi_3$. Therefore the state $S_1$ continuously distorts from $\Psi_1$ to $\Phi_3$. Similar statements can be made of $S_2$ and $S_3$.
Now, we introduce the key assumption that the reaction must proceed adiabatically. This can be formalised very thoroughly in quantum mechanics, but essentially, it means that the state of the system cannot jump from one curve $S_i$ to another curve $S_{j \neq i}$. I'm not entirely sure about the justification for this, but Patterson has written about it in an article:7

A W–H allowed reaction pathway conserves R (reactant) and P (product) nodal geometries, which in turn guarantees continuous covalent bonding along an adiabatic reaction pathway that takes the reactants through a transition state and on into products. A W–H forbidden pathway can either maintain this continuous bonding (interaction energy between R and P) along an adiabatic pathway and generate products in excited electronic states, or disrupt reactant nodal geometry or bonding to reach diabatically (or nonadiabatically) a lower-energy pathway to products. [...] Continuous bonding is energetically more favorable than either nonbonding or antibonding interactions.

Therefore, if we start the system in the state $\Psi_1$ and allow the reaction to proceed, it has to either end up in the product excited state $\Phi_3$ (as far as this treatment tells us); if it ends up in the product ground state $\Phi_1$, then there must be some disruption of continuous bonding. Both of these possibilities are energetically costly, and therefore the reaction is forbidden.

Configuration interaction
That gave us the correct result, but not for the correct reason. The problem is that this assumes that there is no interaction between $S_1$ and other states $S_n$. Mathematically speaking, this means that in the basis $\{S_1, S_2, S_3\}$ the Hamiltonian is assumed to be diagonal,8 such that
$$\langle S_i | \hat{H} | S_j \rangle = \delta_{ij}\langle S_i | \hat{H} | S_i \rangle = \delta_{ij}E_i$$
The question is, is this true? It turns out that, because the Hamiltonian transforms as the totally symmetric irreducible representation (TSIR), the matrix elements $H_{ij} = \langle S_i | \hat{H} | S_j \rangle$ generally only vanish if $S_i$ and $S_j$ are of different symmetry. Therefore, if two states $S_i$ and $S_j$ have the same symmetry, there will often be an off-diagonal element $H_{ij}$. This is the case with $S_1$ and $S_3$ in our example above.
As $x$ increases, the energies of $S_1$ and $S_3$ approach each other, and these two states mix into each other. The new electronic states are therefore linear combinations of our original solutions, $S_1$ and $S_3$. Since $S_1$ and $S_3$ are Slater determinants, the new electronic states are linear combinations of Slater determinants. This process of mixing different electronic states into the wavefunction is commonly known as configuration interaction (CI).
So much for the states; we're not really all that interested in them. The energies are more valuable. It turns out that the presence of this off-diagonal element leads to the non-crossing rule:9 two curves corresponding to states of the same symmetry cannot cross each other. Instead, they are repelled, and there is said to be an avoided crossing. The repulsion is most pronounced when the energies are close to each other, i.e. near the middle of the diagram. The state correlation diagram must be modified to reflect this. The coefficients $a$ and $b$ in the linear combinations must fulfil the criterion $|a|^2 + |b|^2 = 1$.

Within this framework, we can refine our argument for why the reaction is forbidden. In fact, the transition from $\Psi_1$ (reactant ground state) to $\Phi_1$ (product ground state) is adiabatic; however, there is an extremely large electronic activation energy to it, on the order of an electronic excitation energy, which cannot be provided by thermal means.

Real pericyclic reactions
You might notice that the orbital correlation diagram that we drew above is actually similar to that for the disrotatory ring-opening of cyclobutene. The principles are exactly the same, and the state correlation diagram can be constructed exactly analogously (and in fact, the state correlation diagram for this process is found in Fig. 2 of the question), and of course we find that this reaction is thermally forbidden. But, as I alluded to earlier, there is a slight subtlety. Here is the orbital correlation diagram:

Unlike in our first example, there is no obvious one-to-one correspondence between the form of the reactant MOs and the form of the product MOs. For example, it's not obvious why $\pi$ has to transform into $\phi_3$ and not $\phi_1$. It turns out that both $\phi_1$ and $\phi_3$ are actually linear combinations of what $\pi$ and $\sigma$ become in the new geometry.

In the reactant, cyclobutene, the $\pi$ and $\sigma$ MOs cannot mix by symmetry: there is an additional mirror plane (the plane in which the molecule lies), and $\pi$ is antisymmetric and $\sigma$ symmetric with respect to that mirror plane. However, when the reaction proceeds, the symmetry of the system is decreased, and these orbitals are allowed to mix and form linear combinations.
In our earlier example of $\ce{H2 + D2 -> HD + HD}$, the reactant MO $\psi_2$ does physically become the product MO $\phi_3$, as was fairly obvious from the nodal pattern. However, in this situation where there is orbital mixing, it cannot be said that the $\pi$ orbital physically becomes $\phi_3$ as the reaction proceeds. This point was emphasised by Patterson7 and was also described by Woodward and Hoffmann10 in their review of their rules (pp 797 onwards).
So, the correspondence should be determined using symmetry and the non-crossing rule (which also applies to MOs), without recourse to the form of the MOs.
Another example is that of the Diels–Alder reaction. If you look at the $\pi$ MO of ethylene (the second-lowest reactant MO), you find that it becomes the $\pi$ MO of the product cyclohexene. It makes no physical sense to claim that somehow the MO migrated through several carbons over the course of the reaction. Instead, it is more accurate to say that the three symmetric product MOs are linear combinations of what the three symmetric reactant MOs become over the course of the reaction.

Notes and references

If you're really bothered by it, you can change the reaction to $\ce{\color{red}{H}-\color{red}{H} + \color{blue}{H}-\color{blue}{H} -> \color{red}{H}-\color{blue}{H} + \color{red}{H}-\color{blue}{H}}$. But that's just confusing.
The square planar transition state has a higher symmetry of $D_\mathrm{4h}$. However, since $D_\mathrm{2h}$ is a subgroup of $D_\mathrm{4h}$, all the symmetry elements in $D_\mathrm{2h}$ will be preserved in $D_\mathrm{4h}$ symmetry, but not vice versa. For example, the transition state possesses a $C_4$ rotation axis but neither the reactants nor products do. Therefore, we cannot categorise the reactant or product MOs by their symmetry under a $C_4$ rotation; it only makes sense to categorise the orbitals using the more general $D_\mathrm{2h}$ point group. In organic chemistry parlance, this is what is meant by "identify the symmetry elements that are preserved throughout the reaction".
In organic chemistry, it is "conventional to be less formal with the  notation" (Atkins's words; Molecular Quantum Mechanics 4ed, p 400), and so the irrep symbols are replaced with a series of S's (for a character of +1) and A's (for a character of -1). For example, the $\mathrm{B_{1u}}$ irrep would be labelled as ASS since it has characters of $-1, +1, +1$ under the three mirror planes.
The reason why I have never referred to an MO as a "state of an electron" is because of quantum indistinguishability of the electrons, which necessitates that $\Psi$ be properly antisymmetrised. If the Slater determinant (eq. $(1)$) is expanded, it will become clear that the state of electron 1 is entangled with the state of all the other electrons. The wavefunction $\Psi$ cannot be factorised into a product of composite wavefunctions, and so, it is physically incorrect to speak of an electron being in an orbital.
The triplet states can be ignored because spin states are orthogonal. Therefore, for a singlet reactant to end up as a triplet product, there must be a non-adiabatic transition somewhere, and a key assumption is that the process is adiabatic.
The symmetry of a state is obtained by taking direct products of the populated MOs.
Patterson, R. T. An Improved Interpretation of the Woodward–Hoffmann Rules. J. Chem. Educ. 1999, 76 (7), 1002–1007. DOI: 10.1021/ed076p1002.
The diagonal entries of $\hat{H}$ are dependent on $x$, and hence the (approximate) eigenstates $S_i$ and (approximate) eigenvalues $E_i$ also vary with $x$. The basis set $\{S_1, S_2, S_3\}$ is also dependent on $x$, but we don't really need to worry about how it depends on $x$ are we are only looking for a qualitative description of the process.
In the case of two states mixing, the proof is simple: if you use the basis states $\{|a\rangle,|b\rangle\}$ the Hamiltonian is
$$\mathbf{H} = \begin{pmatrix}H_{aa} & H_{ab} \\ H_{ba} & H_{bb}\end{pmatrix}$$
and the eigenvalues are
$$E_\pm = \frac{(H_{aa} + H_{bb}) \pm \sqrt{(H_{aa} - H_{bb} + 4|H_{ab}|^2)}}{2}$$ since $H_{ba} = H_{ab}^*$. Then $E_+$ can only be equal to $E_-$ if $H_{ab} = 0$.
In general I don't want to prove it, because I don't know how to, but a useful reference for this is: Mead, C. A. The "noncrossing" rule for electronic potential energy surfaces: The role of time‐reversal invariance. J. Chem. Phys. 1979, 70 (5), 2276–2283. DOI: 10.1063/1.437733.
Woodward, R. B.; Hoffmann, R. The Conservation of Orbital Symmetry. Angew. Chem. Int. Ed. 1969, 8 (11), 781–853. DOI: 10.1002/anie.196907811

Enough SE for today, I think. [The system actually made me fill in a captcha before it allowed me to submit this...] Disclaimer: this is only based on my own best understanding of the topic, as I couldn't find any good references. I hope there isn't anything wrong, but feel free to point out if there is.


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The nature of a PN junction - Whats preventing full 'diffusion' of electrons?
In this image:

Can the following electron take up 'hole'?
If so, this question is sort of leading onto a bigger question about PN junctions.

What is preventing the electron in the ion from moving to the next hole?
Is it to do with the strength of the covalent bond being stronger than that of the more positive charge in the hole?
Edit:
What I'm trying to understand is why can't the extra electron in the negative boron ion, take up the holes to the left of it in the P (where I've drawn the red arrow in the 2nd diagram). Shouldn't that then allow another conducting electron to take up the hole in the positive ion? And can't this process keep happening till it's fully diffused? I'm still learning all this, this question is more out of curiosity.

Well, that is how holes move.

I don't understand what you are asking. Electrons from covalent bonds are the ones that delocalized holes. Energy level of acceptor impurity is close to valence band energy, which permits electron from a nearby covalent bond to fill it, and remove hole.

@PrittBalagopal Can you explain how the PN junction works then? What's stopping the electron on the ion from moving to the hole like you said?

@Physco111 PN junctions have conduction electrons in n side fill up the holes in p side.

@PrittBalagopal So why then does a PN junction even exist? Shouldn't all the conduction electrons in the N side take up all the holes in the P side?

@Physco111 Please refer a textbook on semiconductors, you should know that a barrier potential exists between the p side and n side, which prevents further movement.

@PrittBalagopal I'm really sorry for bothering you all the time, but you're the closest I've gotten to an answer. What I'm trying to understand is why can't the extra electron in the negative boron ion, take up the holes to the left of it in the P (where I've drawn the red arrow in the 2nd diagram). Shouldn't that then allow another conducting electron to take up the hole in the positive ION? And can't this process keep happening till it's fully diffused? I'm still learning all this, this question is more out of curiosity. If you dont want to keep answering me it's fine. THANK you for your res

When you dope a pure semiconductor like silicon with boron, a few silicon atoms are replaced with boron atoms. Boron, being a trivalent impurity, can only form 3 bonds with the nearby silicon atoms, but what about the fourth bond? A vacancy for an electron is created over there, and is called a hole.

This hole isn't going to stay there. The energy of the acceptor impurity atom (and of the hole) is often so close to the energy level of the valence electrons, that a nearby valence electron can fill up that hole, and move the hole to the initial location of the nearby electron. See this picture that shows holes moving:

As a follow up, a p-n junction consists of two semiconductor wafers pasted on each other, one being of n-type, and the other of p-type. It would look something like this:

The conduction electrons, as you have pointed out, do have a tendency to migrate to the p-side, and will fill up the holes. By doing so, these electrons move from the conduction band on the n-side, to the valence band on the p-side. This movement, however, makes the p-side negatively charged, and the n-side positively charged. An electric field develops in between, which opposes this natural migration of electrons. Ultimately, a field forms strong enough to block all the electrons trying to cross the junction. The potential difference between the junction is called barrier potential, which is about 0.7V for silicon, and 0.3V for germanium.

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Conformations of cyclohexane and flipping
Conformations of Cyclohexane doesn't answers my question
This is what Organic Chemistry by David R. Klein states:

Notice that the chlorine atom is on a wedge, which means that it is coming out of the page: it is
  UP. The methyl group is on a dash, which means that it is below the ring, or DOWN. The two
  chair conformations for this compound are as follows:

Notice that the chlorine atom is above the ring (UP) in both chair conformations, and the
  methyl group is below the ring (DOWN) in both chair conformations. The confguration (i.e.,
  UP or DOWN) does not change during a ring ﬂip. It is true that the chlorine is axial in one
  conformation and equatorial in the other conformation, but a ring ﬂip does not change confguration. The chlorine atom must be UP in both chair conformations. Similarly, the methyl
  group must be DOWN in both chair conformations.

I am having a confusion in the last paragraph. Why should the Chlorine be up even after the ring flip? Moreover, why are we even discussing if it is above the ring or not, as it has already come in the equatorial position?

Essentially, it is stating that a ring flip retains the absolute stereochemical configuration. Ring flipping is a physical phenomenon of 6-membered rings, not a chemical reaction that interconverts stereoisomers. No bonds are broken, though many do get strained in the process.
I'm not sure what you mean by your question about the importance of being in the axial/equatorial positions. Both the methyl and chloro groups are vying to reduce 1,3-diaxial groups by placing themselves in the equitorial position, but doing so for one places the other in the axial position. I think a natural question that arises "is which one wins?"

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NMR spectrum of plasma treated glucopyranoside
This is a 1H NMR spectrum of plasma treated methyl alpha D-glucopyranoside.
Now I'm having problems trying to read the spectra and interpreting if there are any changes to it. What I suspected would happen during plasma treatment is that the methoxy group would break off, leaving it or replacing it with any other possible ROS or RNS. I don't know how to identify the methoxy groups present or to read those peaks to see if there are other possible structural changes.

The usual approach to determining change by NMR following a reaction or procedure is to compare your final spectrum to your starting sample's spectrum, run with identical parameters. That is, look at the spectrum you ran before your plasma treatment and look for changes in peak position and intensity. Within Topspin (which your spectrum here has been processed with) this is very easy to do; you can run a dual display and overlay your spectra on the screen, and even compute a difference spectrum.
Of course, if you are having problems assigning the final spectrum, you should again go back to the spectrum of your starting material and compare to the assignments you made for that. That will make it easy to identify which parts of the molecule have undergone some change. Of course, you might expect that you haven't had complete clean conversion, and there may be a number of products constituting only partial conversion. Otherwise, you are potentially having to characterise a completely new molecule.
You do have the spectrum for your starting material, don't you?

I didn't run a control sample NMR, but seeing as the spectrum should be the same, I obtained one from SBDS. When I'm comparing them they don't really look alike. I can see some differences between the two, my problem is that I am unable to identify and label those differences. I understand that due to different parameters (different solvent, different nmr machine) would make the peaks ppm shift, but I don't think it is that significant. Do you know at which ppm hydroxy and OMe groups would appear on the spectrum?

Yes I do, but you will need to show some evidence of your own efforts to assign and rationalise the spectrum. What can you deduce from the spectrum? Where would you expect the -OH/-Me peaks to come? Your sample looks like it may have been run in D2O; the SBDS is in dmso. Why don't you expect the differences to be significant? I think they probably would be, especially if you are having trouble with assignments to begin with. As it stands, you are essentially asking for someone to assign your spectrum for you. This is not a forum for others to do your work for you.

I understand that. Yeah this sample was done D2O  solvent, as it can be seen from the peak at 4.693 ppm. I have very basic knowledge when it comes to NMR and can only work out structures of simple alkanes or alkenes. This is a good bit over my head. Peak at 3.2741 ppm indicates OMe group as far as I understand. I don't know about the OH group peaks showing in the spectra. Yeah, the SBDS is in dmso, but as far as I understand the only difference is in peak shift. Fact that different solvent was used shouldn't interfere with the spectrum, if the molecule is the same. cont.

cont. Most of the spectrum between my sample and the spectrum on SBDS looks the same except for the multiplet at ~3.5ppm. Hence why I was asking about OH groups, so that I can rule it out as one of the options for it. I can't think of a reason for that multiplet there. My experiment was basically to try and cleave off OMe group, by using plasma generated species. I could identify that, but getting second oppinion on it doesn't hurt. As I mentioned before, I really have close to none experience in reading NMR spectras, hence why I am asking for  help, tp figure out if there are other changes.


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What is the outcome of the following reaction?

The first step is clear to me, making an immine.
However how does the immine react again?
Perhaps it can be reacting with the carbonyl or alkene (to carbonyl over conjugation) I can't get to the target formula.

Iminium --> enamine (basic conditions, alpha proton to lose). The enamine acts as a nucleophile to add into the electrophile in 2.

Once you do the step mentioned above, start looking for intramolecular reactions...

Look up Stork Enamine alkylation.


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How to neutralise path weeder contamination
I strongly suspect my aquarium has been contaminated by a rock sprayed with a path weeder containing Simazine, Amitrole & Ammonium Thiocyanate.
I removed all rock and made several large water changes but fish continued to die one or two a day. When the surviving fish were removed to separate tanks with identical water conditions fish are fine and not one has died for about a week. So it seems to me the chemicals have permeated the crushed coral substrate and filter media and is still leaching into the water.
I have emptied the tank completely & easily can replace the filter media but the substrate, resin decorative pieces, and plastic plants are expensive to replace and I would like to be sure I remove any residue in the tanks and piping etc. Is there anything I can wash the tank, substrate, and decor with to neutralize any remaining toxicity that would not leave a toxic residue in itself?
Would washing everything with vinegar be of any help?
Any advice from those understanding the chemical structure of these toxins and how I might rescue my prized aquarium would be greatly appreciated!

Amitrole is slightly toxic to various species of freshwater fish$^1$. However, it's quite water soluble and should not be difficult to remove from most surfaces with simple warm-water washing.
Ammonium Thiocyanate is particularly water soluble and should also readily be removed via warm-water washes. Thiocyanate ion has lower solubility with some other cations that would be present, but in general, thiocyanates are quite water soluble. Thiocyanate ion has been shown to be toxic to fish at the tens of mg per liter level, however, $^2$.
Simazine has a very low water solubility, on the order of 5 mg/L (4-5 orders of magnitude less soluble than the other compounds) but has been shown to be lethal to fish at levels on the order of its limit of solubility$^3$. I strongly suspect that residual simazine is thus to blame for the persistent toxicity to your fish.
Simazine is about 100 times more soluble in isopropyl alcohol (rubbing alcohol) than it is in water. I suggest obtaining some 99% isopropyl alcohol and scrubbing/soaking all of the decorative material that you value, as well a the tank itself, thoroughly rinsing the alcohol off with warm water, then drying completely. There is no reason to believe that vinegar would be useful in this capacity.
Anything relatively cheap should probably simply be replaced, particularly difficult to clean items like driftwood, high-surface area gravel or sand, and of course all of the filter media. My biggest concern of the items you mentioned is the crushed coral. If it were me, after the cleaning I would return everything to the tank except the crushed coral and once the water is ready, test it out with a couple fish. If things seem OK for a while, then I would set up the tank with everything including the crushed coral.
As a former aquarist, I understand how much work is involved with my suggestions. I strongly believe, however, that you have identified the culprit in the fish die-offs and that the procedure above has a good chance of resolving the issue. Please don't hesitate to ask for any clarifications in the comments below.

US EPA. 1984 (March). Amitrole: Pesticide Registration Standard and Guidance Document. Office of Pesticides and Toxic Substances, US EPA, Washington, DC.
Speyer, M. R. and Raymond, P. (1988), The acute toxicity of thiocyanate and cyanate to rainbow trout as modified by water temperature and pH. Environmental Toxicology and Chemistry, 7: 565–571. doi:10.1002/etc.5620070705
WSSA Herbicide Handbook Committee. Herbicide Handbook of the Weed Science Society of America, 6th Ed. WSSA, Champaign, IL. 1989


text:
Should residual sodium be considered in measuring sodium content of sweat?
I read an article, "Fully integrated wearable sensor arrays for multiplexed in situ perspiration analysis", which describes the design of a wearable band that detects various components in sweat in order to monitor the wearer's health. One of the components detected is sodium.
My thought is that as the wearer sweats and the sweat evaporates, there will be a small amount of sodium left on the skin. When he sweats again, there should be an increase in the concentration of sodium due to the addition of more sodium from the new sweat to the residual sodium left on the skin.
I didn't see this addressed in the paper and am unsure if my thinking is correct. If I'm correct, wouldn't the sensor then need to have a component that does a calculation of how much sodium is due to the new sweat and how much is residual? Would it need to "know" how much sweat was produced to achieve this? What if the sweat under the band doesn't evaporate at the same rate as the surrounding skin due to being trapped under the band?

This is an abstract for an article that probably addresses exactly what you are asking about, but it's behind a pay-wall. I don't know if you have access to this journal; unfortunately the abstract just states the problems, and doesn't itself give any answers.

@airhuff The article does address this, thanks!

Well, you've got me curious as I don't know how they'd get around the "residual" issue. If you've figured it out I hope you will self-answer!

@airhuff I'm trying to get a good grasp of the concept that fixes this problem, then I will definitely self-answer.

I am also waiting for your self-answer. I would hazard a guess and say that you can measure skin conductivity as a function of time and correlate the peaks and dips of conductivity to what was happening during the data collection. Small but persistent increases and decreases in conductivity could be attributed to residual sodium being incorporated into sweat.


text:
How to reduce propyne selectively for the trans product
In my textbook it is given that when alkynes react with sodium in liquid ammonia, hydrogenation takes place and a trans isomer is formed.
It is also given (in another part of the chapter) that when terminal alkynes react with sodium in liquid ammonia, they form a salt with sodium.
I assume the first mentioned reaction doesn't take place in the case of terminal alkynes.
Then, how do we, for example, add deuterium to propyne so as to form trans 1,2-dideuteropropene?

How about $\ce{Na}$ and $\ce{ND3}$?

I am confused if that would result in alkene or sodium alkynide

As you point out, a dissolving metal reduction using $ \ce{Na}$ and $ \ce{NH3}$ in the presence of a proton source such as $ \ce{tBuOH}$ is able to reduce a disubstituted alkyne to the corresponding trans alkene.
The mechanism is understood to go via the following stepwise mechanism:

Both protons on the product  come from the $ \ce{tBuOH}$, rather than the $ \ce{NH3}$ as suggested in the comments, so in theory, using $ \ce{tBuOD}$ would give you the desired product.
One issue with this however, is the ease of exchange between $ \ce{H}$ and $ \ce{D}$– any residual water present would be able to convert $ \ce{tBuOD}$ to $ \ce{tBuOH}$, which would then give you the protonated rather than deuterated product.  Experimentally, you'd probably end up getting a mixture of products with varying degrees of deuteration (mono and di).

Related (but not the same): https://chemistry.stackexchange.com/questions/40655/why-birch-reduction-of-alkyne-is-e-selective


text:
Does changing EDTA concentrations affect the amount of calcium ions in milk?
For my chemistry experiment I am titrating EDTA against milk to find out the concentration of calcium ions in milk. I will be using EriochromeT as the indicator. The lab tech said that if I use temperature as my independent variable the end-point of titration will fluctuate. She suggested that I should change the concentration of EDTA instead. Is it a good suggestion? What is a better independent variable to use?

Since this post was a while ago, I'm assuming an answer has already been decided on, but I'll take a shot at answering anyway.
If I were you I would rely on concentration changes of EDTA as the independent variable since it allows for more precision in your titration. If you were to vary temperature, it would have to be the temperature of the titrant such that volume changes could be accounted for (since analyte is usually in an approximate volume of water). While volume fluctuations of titrant are theoretically possible, they are rather difficult in a laboratory setting and not exactly worth the effort unless you have lots of time and expensive equipment to control the temperatures exactly. Not only that, but volume changes of water are only approximately 5% from 0 Celcius to 100 Celcius.1 Thus, as I said, changes in [EDTA] would be more efficient and easier to manage.
Thermal Expansion of Water


text:
What exactly does the "per mole" unit of heat of atomization mean?
I am facing a confusion in the definition of the standard enthalpy of atomization. The confusion arises because of the "per mole" in its units ($\pu{kJ mol^-1}$).
My question: When I say "per mole", do I mean "per mole" of the product or per mole of the reactant?
When I mean "per mole" of the product, my representative reaction for $\Delta H_\mathrm{a}^\circ$ is:
$$\ce{1/4 P4(s) -> P (g)}$$
And when I mean "per mole" of the reactant, my representative reaction for $\Delta H_\mathrm{a}^\circ$ is:
$$\ce{P4 (s) -> 4P (g)}$$
Using either of them leads to different calculations involving Hess's law or other chemical reactions, hence I feel there is a need to resolve this confusion.

To add to this confusion, NCERT Chemistry (Part 1 page 171) writes:

It is the enthalpy change on breaking one mole of bonds completely to obtain atoms in the gas phase.

but then gives a representative reaction as:
$$\ce{CH4 -> C(g) + 4H(g)}$$
which involves breaking 4 moles of bonds instead.

If someone could write the correct definition of the standard enthalpy of atomization, with more than one reputed reference to back their claim, it'd definitely resolve this confusion.

Here per mole means per mole of product formed (as this is atomization). Standard enthalpy of atomization is the enthalpy change when 1 mol of gaseous atoms is formed from its element in its defined physical state under standard conditions (298.15K, 1 atm).

According to Wikipedia,

Standard enthalpy of atomization is the enthalpy change when $1$ mole of gaseous atoms is formed from its element in its defined physical state under standard conditions $(\pu{298.15K}, \pu{1 bar})$.

$\Delta H_{\text{at}}^{\circ}$ focuses on the production of $1$ mole of atoms from the element. Hence it's defined for the reaction:
$\ce{\dfrac{1}{8}S8 ->S}$
but not for the reaction:
$\ce{S8 -> 8S}$
So whenever you see the term "per mole", it is, by definition, implicitly referring to the product and not to the reactant.


text:
Why does branching decrease density and melting point?
The general explanation given is that the chains are further apart, so the intermolecular forces are less, and as they are further apart, they occupy more volume, and the density is lesser. But the length of each chain (and thus its mass) also increases, so how do we know that the density overall decreases?
As for the melting point, longer chains = more electrons = more Van Der Waals forces, so the melting point should increase.
Edit: I'm referring to branching in the context of polymerisation.

As a side note, whenever we talk about melting points, structural symmetry comes into play. More symmetrical molecules pack into the crystal lattice better thus allowing attractive  tendencies of molecules to dominate (facilitated due to a closer approach between molecules)1. As a rule of thumb, as packing efficiency increases, melting point increases. A similar reasoning is given whenever we're to compare the melting points of trans and cis-2-butene.
For instance, let's say you're planning to go to a nice long vacation (ask me, I'd say to Hawaii). But you've decided to carry just one suitcase for the trip. Being a chemist, your love for chemicals is intense, and you've decided that you'll either carry butane or iso-butane for the trip. But which one? More chemicals means more fun, you say. Thus, a molecule with higher packing efficiency would be a clear winner in this scenario. Butane wins this game because it has a symetrical structure and you would be able to "stack" those molecules one over the other which accounts for its higher packing efficiency.  Butane thus would have a higher melting point. Data supports our speculation; melting point of butane is $-140 ^\circ \pu{C}$, while for iso-butane it's $-159.6 ^\circ \pu{C}$.
Longer chains can also mean larger molecular masses, which in turn lead to an increased london force between the molecules, leading to an increased melting point.
Further, a branched molecule doesn't always mean a decreased melting point. Take for example the case of pentane and 2,2-dimethyl propane. The latter has a more "compact" spherical shape, which allows it to pack better into the crystal lattice, resulting in a significantly higher  melting point.
$
\begin{array}{|c|c|}
\hline
 \pu{Compound} & \pu{Melting point (in ^\circ C)} \\ \hline
 \pu{Pentane} & \pu{-130 ^\circ} \\ \hline
 \pu{2,2-dimethyl propane} & \pu{-71 ^\circ} \\ \hline
\end{array}
$

References:

Organic Chemistry, Joseph M. Hornback

Okay, so for the melting point, we have three different effects :

The molecule becomes more sphere-like, so the surface area decreases, and the strength of the Van Der Waals forces decrease.
The length of each chain becomes longer, so the Van Der Waals forces increase.
3)In general, branching decreases the symmetry of the structure, and thus the packing efficiency (and hence m.p ) decrease, but of course, there are exceptions.
Also, I meant to ask this in the context of polymers. Does your answer still apply?
Could you give a similar answer for the density?

I'm not sure that your statement about density is always correct. Looking at octane and its isomers the density changes very little, for example
octane           $703~ \pu{kgm^{-3}}$
2-methylheptane  $698~ \pu{kgm^{-3}}$
2,5-dimethylhexane  $694~ \pu{kgm^{-3}}$
2,2,4-trimethylpentane  $692~ \pu{kgm^{-3}}$
although the melting points do change considerably. In the same order as above they are $-57, -112,-93,-107$ so octane is the exception here; (data from Wikipedia pages).
This can be explained by the fact that melting can be considered to be governed by the repulsive part of the intermolecular potential this having a greater influence than the attractive part when the molecule has an odd shape. Rod like octane has to have deeper intermolecular potential than the odd shaped isomers as it has a higher melting point.
In the isomers the interaction should be dominated by the repulsive part of the potential as the odd shapes must try to push methyl groups into to one another. The potential itself has the same overall geometry, i.e. similar intermolecular minimum distance, as density is the same in all molecules above, but is shallower in nature due to the extra repulsion. Hence the isomers have significantly lower boiling points than octane.


text:
Which is the most stable out of the following linked cyclohexadienes?

I am unable to find the answer to this question. I was guessing it should be (2) due to symmetry but my book says it is (3) without any explanation.
I know that stability is determined by how low the energy of formation is. Does this have something to do with the different sorts of dienes that are present? Like, one type of diene would have less energy than another?
If nothing else, it would be helpful if someone could indicate the names of these compounds, because I don't even have the faintest idea where to start searching for more information.

Lots of conjugation = resonance = stability. There are gradually increasing amounts of orbital involved in conjugation, so they are in order of increasing stability.

You must be aware about the concept of resonance. Similar ly over here as you can see in option number 3 there is longer conjugation between double bonds then compared with any other option because Three double bonds are in conjugation and this conjugation supplies stability to the structure

This is not completely true, the rings are not planar; therefore there are only two bonds in resonance in 3.

Structure $3$ is stabilized through resonance.
Double bond - Single bond - Double bond, can perform resonance.
More the number of resonating (or canonical) structures, more the stability.
Structure (2) also exhibits resonance, but it has less number of resonating structures than structure $(3)$.
Therefore, structure $(3)$ is the most stable one.
If you are not familiar with the concept of resonance, read it. You will encounter many situations where resonance will have a direct link with the stability of certain structures.

3) option has an extended resonance than 2) and 1) which is major cause for its extra stability.

You make a good point, but should elaborate.


text:
What is the link between electrode surface areas and EMF?
I was wondering if someone would help me with my question.
I was completing a question which asked about the effect of increased surface area of a platinum electrode on the EMF of the overall cell.
This is the cell: Pt | H2O2(aq), O2(g) || IO (aq), I2(aq) |Pt
When I answered, I thought the EMF would increase due to the surface area of the electrode being able to now accept more electrons but I found out the answer is that the EMF is unaffected by any change in surface area, why is this?
Thanks for any help anyone can provide!

This is the same difference as between voltage and current. A bigger surface should increase the current.

If by EMF you mean the ideal, thermodynamically determined, voltage then the answer is that increasing the surface area does not affect voltage/EMF. If by EMF/Voltage, the questions means operational voltage then it will increase because increasing the surface area of electrodes decreases the overpotential losses (Tafel equation) by essentially decreasing the resistance due to the kinetics of the reaction thus increasing operational/real voltage of a cell.

When electrode surface area increases,internal resistance decreases.But emf does not change as it is a characteristic of a cell.Emf only depend upon nature of electrodes and electrolyte.

Why is this statement downvoted?


text:
Relative reacting tendency of H2/Pd towards double bond and triple bond in straight carbon chain

If both double bond and triple bond is present in a straight carbon
  chain the which one (double bond or triple bond) will be reduced by
  $1$ equivalent of $\ce{H2(Pd)}$ ?

My guess is that the triple bond should react as it is more reactive but I'm not sure. Also does the answer depend on whether the double bond and triple bond are conjugated with each other or not?
Let us consider two examples (A) and (B):
(A)

(B)

Which bond will be reduced in case (A) and (B) ?

March Advanced Organic Chemistry 7th ed, p 1499 lists alkynes as being easier to reduce than alkenes (in general), although I guess it will probably depend on the exact substrate.

@orthocresol Thanks. Does it depend on whether the double bond and triple bond are conjugated or not? Any ideas?

Triple bond will be more reactive because it can be adsorbed easily on the surface of catalyst through its pi bonds. If double and triple bonds are in congugation then also triple bond is more reactive as in this situation product formed will be conjugated diene.


text:
Magnetic and acidic properties of nitric oxide
I was faced with the following question:

PART C – CHEMISTRY
Which one of the following properties is not shown by $\ce{NO}$?

It combines with oxygen to form nitrogen dioxide

Its bond order is $2.5$

It is diamagnetic in the gaseous state

It is a neutral oxide

I am confused about (3) & (4). I know $\ce{NO}$ reacts with water & oxygen to produce $\ce{HNO2}$, and then $\ce{NaOH}$ to produce $\ce{NaNO2, N2O, H2O}$. So, obviously it's acidic.
But there is one unpaired electron in the $\ce{N}$ atom. So, I see no reason for $\ce{NO}$ to be diamagnetic in gaseous state.
But, these questions do not have two correct answers. So, one of them has to be wrong. I am not sure which one.

Related: https://chemistry.stackexchange.com/questions/71080/why-dont-no-react-with-h2o

That's true.

The bond order in $\ce{NO}$ is 2.5

The oxidation state of $\ce N$ is +2, and this is a colourless, paramagnetic gas (i.e. it is a radical, or contains an unpaired electron). (source)

Neutral oxides are those oxides which show neither basic nor acidic properties when they react with water. Examples include carbon monoxide $(\ce{CO})$ and nitrous oxide $(\ce{N2O})$ which are only slightly soluble in water, and nitric oxide $\mathbf{(NO)}$ which is appreciably soluble in cold water. (source)

So, option 3 is the correct answer as it is an inaccurate statement.

but NO reacts with NaOH. So, why isn't acidic?

NO itself doesn't react with NaOH, so it is not acidic. It is only the downstream reaction products of NO with O2 that are acidic.

Also, I take issue with that first citation. The oxidation of NO is second order in the NO concentration, so it is only a "quick" reaction when the concentration of NO is high.

http://chemiday.com/en/reaction/3-1-0-8020 So is this wrong?

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text:
Why does steric hindrance cause a d8 complex to have a tetrahedral geometry rather than a square planar geometry?
I'm currently working on a question based on two isomers of a nickel complex, where nickel exists in the oxidation state Ni(II) (i.e. has d-electron configuration d8). It has been deduced that the "normal" isomer is square planar and the "iso" isomer is tetrahedral.

The first of the two isomers (top) is the "normal" isomer and the second (bottom) is the "iso" isomer.
Apparently steric hindrance causes the "iso" isomer to be tetrahedral rather than square planar, but I am unsure how to explain this. Is it due to the bond angles causing less strain?

Welcome to Chemistry StackExchange. Presumably the change in geometry is driven  by steric hindrance between the O atom and the $\ce{C(CH3)_2}$ groups, even though there is potentially free rotation about the NC single bonds meaning that the methyl groups will be able to rotate to minimise  any steric clashes.

@polyphyrin What's the IUPAC name of this compound?

Thank you! I suppose what I really want to know why it is more sterically favourable to take up a tetrahedral geometry then?

@porphyrin those are not $\ce{C(CH3)2}$ groups.  They are $\ce{CH(CH3)2}$ groups.  With the bond lengths and angles in the $\ce{O-Ni-N-C}$ loops that additional hydrogen atom can mess you up if you try to tuck both methyl groups away with bond rotation.

@Oscar Lanzi I think we are saying the same thing, the methyl groups will get in the way in a square planar geometry, but can rotate to minimise interaction, but even so they are still bulky enough to force a change to tetrahedral.

For a given number of ligands surrounding a central atom there is always an arrangement in which the distances between the ligands is maximised and which typically also results in the maximum possible angles between then. You can deduce them logically if you start at a single ligand (which has no steric clashes whatsoever).
A second ligand should approach the same centre from exactly the opposite side leading to a linear arrangement and a bond angle of $180^\circ$.
A third ligand should approach the same centre perpendicular to the existing bond axis. Upon its approach, the two ligands already present will ease away from the approaching one. The final arrangement is trigonal planar with bond angles of $120^\circ$.
A fourth ligand should approach the same centre in a perpendicular manner again; this time relative to the plane of the three existing ligands. Again, they will ease away to make more room for the incoming. The final arrangement is a tetrahedron with bond angles of $109.5^\circ$.
Going from a tetrahedron to a square-planar arrangement means squashing the ligands together i.e. reducing the space between them. Thus, if there are no electronic arguments favouring a different arrangement any tetracoordinated entity should assume a tetrahedral environment to minimise steric strain.

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


text:
Synthesis of sodium methoxide
For synthesis of sodium methoxide, can we use sodium hydroxide in methanol and slow evaporation after dissolving sodium hydroxide?

No, all that does is give you back your NaOH. Its not a strong enough base to completely deprotonate methanol.
I use sodium metal in methanol, cooled in ice/water bath and stirred under nitrogen for safety.


text:
Mechanism - Dialkyne boron complex + Iodine?
Exam tomorrow and I was flicking through one of our handouts to find this reaction (D). I'm certain we don't need to know the mechanism but I was quite inquisitive as to how the iodination works.
The hydroboration part is fine, I just don't understand how it fuses the two alkynes.
I've tried to work it through as in the reaction above (C)

I am not that sure but I suppose first iodine acts upon one of the triple bond . Now the carbon attached to iodine as well as boron gets partial positive charge due to which the the bond between the other carbon and the boron probably rearranges to bond with the carbon connected to Iodine and boron. Later I suppose another partial positive iodine attracts the partial negative iodine and the bond goes back to form triple bond thereby breaking the C-B bond too and left with the product....


text:
Rearrangement of double bonds
Consider:

I found these rearrangements, but I could not understand what the mechanism is behind this.
When I searched on the Internet I found the Wikipedia page Di-pi-methane rearrangement.

This could be due to hydride shift.

In the first image, the pi-bond shifts on the carbon from which $\ce{H}$ is eliminated. As the pi-bond leaves the first carbon, it becomes positively charged due to lose of electron, and the $\ce{H}$ gets converted to $\ce{H^-}$(hydride ion) as it accepts electron. Therefore, it is called hydride shift. It also stabilises the molecule as the resulting molecule is conjugated. The same can be applied for second and third images.
This is for the second image. The third one you can do it yourself. All the resulting molecules are stabilised due to conjugation.

This makes sense as far as the stability of the final product (except maybe the third example), but there still has to be something to overcome the initial activation energy. Does it need to happen at reflux? Is a catalyst needed?

@NH. I think it mostly happens in presence of light or when heated.

No, no, no... Hydride shift does happen here but you need strong acid to reversibly protonate alkene - it's simple carbocation rearrangement.

@Mithoron As no reagent was given, so this can also be a mechanism for it. But I don't understand your talk about carbocation.

@Mesentery has already proposed a possible mechanism, and @CupC_56 has tried to explain why it might happen. I found @Cup_C56's points to be correct, but the explanation to be quite less than ideal, so to speak, so I'll try to provide an alternative and I hope somewhat more sensible explanation as to why this happens.
It is indeed due to the resonance-stabilization of the conjugated products formed due to the rearrangements you quote. Resonance-stabilized compounds have lower energy than non-resonance stabilized ones, but why is that? One of the possible explanations is that resonance "averages out" energy minima and maxima between the resonating parts of the compound. This "smoothed-out" compound usually has more symmetry - and in most cases, symmetrical configurations are more stable than asymmetrical ones if both are under comparable amounts of steric stress.
The rearrangement is to a conjugated diene because conjugation facilitates resonance - if the $\pi$-bonds are too far apart, the $\pi$-clouds cannot overlap, which is resonance (or electron delocalization over $\pi$-bonds). "Delocalization" quite literally means that the $\pi$-electrons, instead of being associated with a particular bond between 2 atoms of the molecule, are part of an extended $\pi$-cloud over the molecule (an extended molecular orbital formed by the overlap of partially- and fully-occupied molecular orbitals and unoccupied atomic orbitals (near the $\sigma$-bonds) - the premise of Molecular Orbital Theory - please do find the time to read about it!). The overlap probability and extent, which determines the overall stabilization, falls off very quickly with the separation between the overlapping orbitals - usually beyond one bond it is very weak, and beyond 3 it is negligible.
So now you realize why it is favorable for an organic diene system to undergo rearrangement to a conjugated system - the latter is stabilized by $\pi$-cloud overlap, or "resonance".

Are you familiar with delocalization of pi-bonds? Well, if not, i suggest you read about that as it will help you in understanding such mechanisms, and also various other reactions related to Benzene and its homologues.

All the structures which you have depicted in your diagram are conjugated compounds (alternate double and single bonds), and hence the delocalization of pi-bonds becomes fairly easy, as compared to normal unsaturated compounds. Delocalization is basically the breakage of pi-bond at one site, and shift of the electrons to another site (forming a double bond in the process at the same site) in the same molecule.

WHEN conjugated compounds exist, the delocalization is easier because the intermediate complexes formed are quite stable, and hence lower energy is required for the breaking of such pi-bonds. (I strongly suggest you read more about stability of conjugated compounds as it is too broad a topic to be mentioned here).
NOTE: Not all compounds in your diagram are conjugated ones. To be more specific, the product of each reaction is.
This is one more clue as to why conjugated compounds are more stable. These reactions obviously proceed to form such products which are more stable than the reactants.

Delocalization is basically the breakage of pi-bond at one site, and shift of the electrons to another site  - no it isn't. See here for details on this common misconception.

Well, im pretty sure that my explanation holds good for the said question. What i have defined for delocalization is accurate, and i strongly disagree with the answer provided in your link.
Please do consult Morrison Boyd's book for Organic Chemistry to clear your doubts.

I don't think that anyone can claim that one is more wrong than the other, but I agree with @bon, as MOT provides a more sensible (and more in line with experimental observations) description of what we call "resonance". Also, I take strong offence to the implication that bon had any doubts - from his answer, it does not seem so. On the contrary, your answer is somewhat suspect - note intermediate complexes formed are quite stable, and hence lower energy is required for the breaking of such pi-bonds - should it not be that more energy is required for breaking stronger bonds, not less?

And I very much disagree with your definition of delocalization - it seems quite like the "lies-for-children" definitions that teachers are likely to give to younger students. See Wikipedia: In chemistry, delocalized electrons are electrons in a molecule, ion or solid metal that are not associated with a single atom or a covalent bond. Most definitely not what you say it is.

@TamoghnaChowdhury "intermediate complexes formed are quite stable, and hence lower energy is required for the breaking of such pi-bonds - should it not be that more energy is required for breaking stronger bonds, not less? –"
Obviously, if it is going to a lower energy state, then the energy required for the breakage of pi-bonds  is compensated because of the greater energy released during product formation. Is that that hard to comprehend?

Furthermore, i believe the  Molecular Orbital concept is not one which a person should use in initial explanations. The definition which i gave is fair enough for answering the question on hand (which is what is the goal here, not arguing over which concept is superior to the other). Surely, the MO theory is more refined, but is not necessary to explain every question regarding resonance. It is used for much higher reasoning, not quite pertaining to the scope of this question. If anyone here wants to explain the concept of MO to the person who asked this question, be my guest.

What i mean to convey is that resonance explained by MO theory or the simple electron-shifting theory is analogous to the explanation of molecular structure based on VBT or MO. The former may not be the exact explanation for the structure, but more often than not is used in place of MO for explanation of lower-reasoning questions

@CupC_56, why yes, it is - at least in the way you have stated it in your answer. Your comment clarifies your point, which I do agree with - but I'd recommend that for the ease of understanding of future readers that you edit the phrasing used in your answer currently to the one used in your comment.


text:
Can Aquatabs be used to clean swimming pool water?
Can Aquatabs be used to prepare/chlorinate water for swimming?  They are used  to clean water for drinking so is it safe to assume that if used in proper quantity they can also clean water for swimming?
We are currently at a place where there is no pool and the temperatures are soaring, our 3 year old son has a small pool which we take along on picnics.  I was not able to find pool chlorine somewhere over here but we do have adequate quantity of aquatabs with us.
I tested this and the water smelled/tasted just like it does in a properly chlorinated pool but want to be sure if they are adequate.
Just in case, the label says

Each tablet contains 500mg Sodium Dichloroisocyanurate (300mg Free available
  chlorine)

I am not really worried about maintaining a pH level over the matter of days, its more of a one time thing then we can change water.

What is the source water? Is it turbid (cloudy, silty)?And FYI, taste/smell should not be an indicator of a properly chlorinated pool, because if done properly a pool should not have that "chlorine smell" (it's actually chloramines you're smelling in a bad pool).

Its very clean and theres no visible trace of dirt or contamination in it. People use it for drinkng as well after boiling it but we use Aquatabs for precaution

Aquatabs can certainly be used as a disinfectant in a small swimming pool like you've described. The active ingredient has many uses, including drinking water treatment as you've pointed out, but it is also used as a disinfectant in swimming pools.
According to this Wikipedia article regarding the active ingredient of Aquatabs, sodium dichloroisocyanurate:

As a disinfectant, it is used to sterilize drinking water, swimming pools, tableware and air, fight against infectious diseases as routine disinfection.

A study was done[1] comparing the effectiveness of sodium hypochlorite (common household bleach, also used for pool chlorination) with sodium dichloroisocyanurate. From the abstract:

A comparison of commercial sodium hypochlorite (NaOCl) and sodium dichloroisocyanurate (NaDCC) products was made. Solutions of NaOCl and NaDCC containing the same levels of available chlorine (av.Cl) exhibited very similar bactericidal activities, despite significant differences in pH.

The ideal level of active chlorine for a swimming pool at neutral pH is about 0.5 to 1.5 mg of active chlorine per liter of pool water. So, if you aim for the middle of this range, you would want to add one Aquatab tablet for every 300 liters of water in your pool. Also, note that chlorine is less than half as effective at a pH of 8 as it is at a pH of 7, so I would pay some attention to the pH anyway.

D. Coates, J. Hosp. Infect. 1985, 6 (1),31-40. A comparison of sodium hypochlorite and sodium dichloroisocyanurate products. DOI: 10.1016/S0195-6701(85)80015-3

Great! Also, I have one comment on a comment made to your question; you definitely should smell some chlorine. It is true that the smell of chloramines can fool one into thinking their pool is properly or even over chlorinated when in fact it's just been neglected and bodily fluids and other stuff have used up the chlorine, leaving the stinky chloramine byproduct. As you've described your use case (changing water after a few days) I wouldn't worry about this. Again, the smell of chlorine is perfectly normal when you've done everything correctly.

Yep, that is why i didnt argue on that comment. I swim frequently enough to be able to distinguish properly chlorinated pool's smell vs properly contaminated pool :)


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Anion-exchanged compound - smaller 1H NMR integrations
I anion exchanged one of my compounds (Br- to PF6-) and the NMR integrations changed as shown on the two NMRs below. The aliphatic region peaks integrate to smaller values in the second one. Are there any known cases where PF6- somehow influences the relaxation time?
The other possibility is that this product decomposed to a similar compound, which would slightly throw off the integrations...

"Are there any known cases where PF6- somehow influences the relaxation time?" Well, you could just use longer dwell time or some kind of experiment that get's rid of such effects (qNMR).

As above - if you think it's a relaxation effect, acquire another spectrum with a longer relaxation delay.

On face value, there is nothing here that necessarily suggests a change in relaxation times. Before thinking too hard about what chemistry is happening, you should assure us that your data is reliable and consistent. You should provide details of your experimental conditions; are you using a calibrated π/2 pulse for excitation or a shorter pulse approximation? What relaxation delay are you using? Have you measured T1 relaxations for these nuclei? You could consider repeating your experiment using a much greater relaxation rate between scans to see whether the integrations change.
These are more interesting phenomena in the spectra that need rationalising; loss of fine structure for downfield peaks for instance.
As for your question though, there are lots of examples of counterions changing relaxation times. This is a major tool used for probing outer sphere complexation, and can be used to determine how anions and cations interact around the ligand. It is used widely in protein structure (complexation with paramagnetic cations) and investigations with ionic liquids. If this is a counterion-induced relaxation change, you should be able to account for this by repeating your NMR experiment using appropriate parameters.


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Can activated charcoal retain adsorbtion properties even when submerged in liquids?
This is following a question on Skeptics Stackexchange, where apparently medical science claims that activated charcoal can adsorb poison when ingested.
One of the members posted a related question on this forum asking about burnt toast having properties of activated charcoal.
Although the question on the toast isn't answered properly, at least could anyone here validate if activated charcoal can actually retain its adsorptive properties even when it is soaked in human saliva, digestive juices and tea and/or water?

Given that activated charcoal is used in emergency rooms on a regular basis, I'd say it likely has some effect.

I searched for papers that prove how exactly it works, but couldn't find anything, which is why I asked here. I just hope the doctors aren't just following hearsay just because they didn't have the time to perform lab tests to check the validity of the claim. Perhaps the activated charcoal / burnt toast only has a placebo effect? btw, even homeopathy is said to only have a placebo effect on people: https://en.wikipedia.org/wiki/Homeopathy

I think this is very similar to using activated carbon in swimming pool and aquarium filters. In this case I can attest to the clarification of aquarium water with the overdue change of an activated carbon filter. It's just a matter of the partition coefficient between water and the carbon for the compound of interest. Some compounds preferentially adsorb to the carbon, some preferentially stay dissolved in the aqueous solution. These adsorption coefficients have been published for hundreds if not thousands of organic compounds (for water/activated carbon at least, maybe not carbon/bile).

Also, activated carbon is used as a stationary phase in liquid chromatography. This wouldn't work if different organic compounds didn't adsorb to the activated carbon in the liquid (usually aqueous) phase.

Could you share a link to the adsorption coefficients? I saw a lot of articles about the adsorption properties of activated charcoal on Google Scholar. So it seems like the claims are true after all.


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Treatment of 2-vinylbicyclo[2.2.2]oct-5-en-2-ol with potassium hydride
What happens when this bicyclic compound is treated with KH in THF?

Does $\ce{KH}$ simply act as a base here? $\ce{H-}$ from $\ce{KH}$ could potentially act as a nucleophile, but I don't find any electrophilic centre in the compound. What happens?

The reaction in question is one of the early examples of an oxy-Cope rearrangement [1] in which a 3-hydroxy-hexadiene undergoes a [3,3]-sigmatropic rearrangement to afford a ketone.

The general mechanism is shown below:

Source: Organic Chemistry, Oxford University Press, 2 ed. Clayden, Warren and Greeves
In a general sense, the reaction is a [3,3]-sigmatropic rearrangement (if you look at the curly arrows, you'll see that we're essentially transposing a sigma bond from one end of the pi system to another).
The presence of the oxygen provides a driving force for the reaction, since the enol(ate) formed during the rearrangement is able to tautomerise, yielding the ketone product.
In the specific case you've given, the presence of a base means that during the sigmatropic rearrangement, the oxygen is deprotonated and therefore anionic. This specific case, using an alkoxide, is a specific kind of reaction known as an anionic oxy-Cope. The anion is thought to help speed up the reaction, to the extent that minimal heat is usually required.

References:
[1] J. Am. Chem. Soc., 1964, 86,5019.  DOI: 10.1021/ja01076a067

If you want to challenge yourself further with this reaction, consider what the product would be if we substituted one of the hydrogen atoms on the vinyl group for a deuterium atom.


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Why is sodium carbonate less soluble in water than sodium bicarbonate?
Why is sodium carbonate less soluble in water than sodium bicarbonate? If you think about their structures, the only thing that is different is that sodium carbonate has two sodium atoms, while sodium bicarbonate has one sodium atom and one hydrogen atom instead of the second sodium atom. Sodium and hydrogen have different electronegativity values, giving sodium bicarbonate polar character.
Does the difference in solubility have to do with the dipole moment of the molecules? If so, how can it be explained in more detail?

related https://chemistry.stackexchange.com/questions/5128/making-a-saturated-solution-of-na%e2%82%82co%e2%82%83

I suspect that it's a matter of the carbonate ion having a greater formal charge, causing it to be better solvated than the singly charged bicarbonate ion. Keep in mind that if I were certain of this, I would have written a formal answer ;)

It isnt, as another answer has stated. To my knowledge this is only true for Mg and Ca (bi)carbonates. The trend in the alkali metals are oppostie: carbonates will be always more soluble than bicarbonates.

Oops! The question contains an incorrect assumption. But chemists are amazing - they can still answer such a question:

Sodium carbonate is way more soluble than sodium bicarbonate at all temperatures above about freezing. Na2CO3 has several crystalline hydrates that are temperature sensitive and make the curve less smooth than most other salts. (I suppose the caffeine is in there to wake us up.)
So, I woke up to the fact that when we calculate the weight of "sodium carbonate" which dissolves at 0 C, it's the decahydrate, whereas sodium bicarbonate is anhydrous, so at 0 C, the solubility of sodium bicarbonate is 0.833 moles per liter, while the same weight of sodium carbonate decahydrate dissolved in a liter is only 0.245 moles.
Now I'm more puzzled by why Na2CO3 has such a complex hydrate system, and NaHCO3 doesn't!

Likely, this is due to hydrogen bonding. $\ce{NaHCO3}$ has one hydrogen bond donor site, but $\ce{Na2CO3}$ has none.
For an organic compound, one hydrogen bond for each four-five carbons may be enough to make it water soluble.

NaHCO3 has AFAIK a polymer anionic structure, bound by hydrogen bonds.

According to the Born–Landé equation, the lattice energy is proportional to Z+Z-. Therefore in $\ce{Na2CO3}$ the Z+Z- value is 2 where as in $\ce{NaHCO3}$ the value of Z+Z- value is one.  So the lattice energy is higher in $\ce{Na2CO3}$ than $\ce{NaHCO3}$. In addition to that the size of bicarbonate ion is higher than carbonate. Therefore packing efficiency will be higher in sodium carbonate than sodium bicarbonate. Thus more energy will require to disrupt $\ce{Na2CO3}$ than $\ce{NaHCO3}$. Therefore $\ce{Na2CO3}$ is less soluble in water than $\ce{NaHCO3}$.


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How does a chemical "know" which other species are present and where equilibrium is?
Please forgive me if this is a vague question. I have always wondered how a chemical reaction "knows" where its equilibrium should be. For example, using a basic example of Le Chatelier's principle, for some theoretical reaction
$$\ce{A + B <-> C + D}$$
 If I were to add extra A and B to the tube, the reaction would proceed to the right until a new equilibrium was established. How do the chemicals "know" that I have added extra A and B?
Am I thinking about this incorrectly? Is it not actually a matter of the chemicals "sensing/knowing" which other species are present, but rather their own individual properties and how likely they are to exist in one form or another?
Related to this quandary, I learned in my metabolism course that NADH and NADPH are used to regulate different reactions because the NAD+/NADH and NADP+/NADPH ratios need to be different, and that merely adding the phosphate allows these separate species to have different equilibria. Can NAD+ really distinguish NADH from NADPH?

There is no sensing or distinguishing at work here, it is just a matter of statistics and probability. If your equilibrium is $\ce{A + B <=> C + D}$, equilibrium means that both the forward and backward reaction have the same likelihood of occurring in a given time. Now if you chuck in more A and B, it simply makes it more likely that A and B collide to form C and D, than for C and D to collide to form A and B. Ergo, you end up producing more C and D.

This is a good question.  We spend a lot of time teaching equilibrium chemistry, where we assume equilibrium is reached before we proceed to the next step.  This makes sense because it's  lot easier to teach, but it does lead to some odd assumptions about how we must reach the equilibrium.

It doesn't. Think of how a chemical reaction actually proceeds deep down - one molecule hits another molecule, and it maybe sticks (A + B -> C). At the same time, random collisions and vibrations may break apart one molecule into two molecules (C -> A + B). The more A and B you have, the likelier it is they will collide in a way that creates C (simply because the favourable collisions are more frequent) and vice versa. When the rate of the two reactions is the same, you get an equilibrium - the reactions still proceed, but in both directions equally.

What we got here is a dynamic equilibrium, not a static one. For a static one the molecules would need to know when to stop reacting or when to react in a certain way, in a dynamic equilibrium you don't need to know.
Here's a ELI5 explanation of what's happening:

Let's say you've got a garden with an appletree and hundred of apples
  are on the ground. You don't want them there so you throw them to your
  neighbor. he's an old man and doesn't want them either, so he throws
  them back. You are much faster than him and there are hundreds of
  apples around you, so you don't even need to move to throw them and
  you can throw them very fast. He on the other hand has only a few
  apples laying around and has to move quite far, so he's throwing them
  back much slower. But what happens after time?
Well, there will be a point where there are more apples on the old
  mans side than on yours, so you have to run a lot to get to those
  apples while the old man has to move only some steps to reach a new
  apple. Eventually it will level out and both of you will throw apples
  at the same rate. That's a dynamic equilibrium. Because there are more
  apples on the old mans side he can throw them back as fast as you can
  throw them to him, even if you are faster.

Now here the chemical explanation:
Reaction rates are dependent on the concentration. High concentration equals fast reaction, low concentration equals slow reaction. At the beginning you got only starting material, it will react fast to the product. The product will react back but there is extremly low concentration so it's slow. But after some time you will reach a point where the concentration of the starting material is so low that the reaction is so slow, that it matches the speed of the back reaction which got faster because you got more and more product. At that time there will equal amounts react in both directions, making it look like the reaction is standing still.
For example: If we got the reaction $\ce{A + B \rightleftharpoons C + D}$ and the reaction rate of the forward reaction is $r_1=k_1[A][B]$ and the backwards rate is $r_2=k_2[C][D]$ then we will reach an equilibrium $r_1=r_2$ and that's the case if $k_1[A][B]=k_2[C][D]$.

Okay, great analogy. So to extend my question into the dissociation of water, it's not that neutral water somehow knows "oh there isn't enough OH- and H3O+" its just that (at a given temperature) Water has a 1/10^7 probability to be dissociated?

@Arcadium:  water dissociation is not a good example because we don't have a clear intuitive idea how it happens.  In the example $\ce{A + B \rightleftharpoons C + D}$, the forward rate is proportional to the product of the concentrations of $A$ and $B$ because that is how often they meet.  If you add $A$ and $B$ the rate will increase because they meet more often.  That will generate more $C$ and $D$.  The forward will dominate until the concentration of $C$ and $D$ rises to balance.

Yeah, yeah, but what's the chemical analogue of the old guy calling the cops on you, huh? :-)

@DavidRicherby I believe that would be when the old guy's reaction causes your glass reaction vessel to burst all over the floor.

@Arcadium Each molecule of $\ce{H2O}$ has some (small, constant) probability at each unit of time to dissociate. However, the rate at which $\ce{HO-}$ and $\ce{H+}$ re-associate are dependent on their concentration (that is, how frequently they bump into each other). For pure water, it turns out the rate of re-association of $\ce{HO-}$ and $\ce{H+}$ matches the (relatively constant) rate of disassociation of $\ce{H2O}$ when $\ce{[HO-][H+]} = 10^{-14}$. (It's a little more complicated than that, as you actually need multiple water molecules to get dissociation, but that's the gist of it.)


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Spin interaction term for a two-particle system of fermions with a potential that contains spin-spin interaction
Consider a two-particle system consisting of two identical fermions in a potential $$V(\vec{r})\vec{\sigma_{1}}\cdot\vec{\sigma_{2}}$$ where $V(\vec{r})$ is the spatial part of the potential and the indices $1$ and $2$ correspond to each particle respectively and $\vec{\sigma}=\sigma_1i+\sigma_2j+\sigma_3k$ with $\sigma_{i}$ the $i$-th Pauli matrix.
I want to find the contribution of the spin interaction term to the energy of the system. Now, working with singlet and triplet states, I know that the energy will be analogous to the eigenvalue of each state respectively.
Working things out, I have found that the eigenvalue of the singlet state for $\vec{\sigma_{1}}\cdot\vec{\sigma_{2}}$ is $-3$ times the eigenvalue of the triplet state.
By also trying out as $V(\vec{r})$ a potential used in physics(Yukawa potential), I also found that the triplet state has the same energy as the case in which we have two bosons in that potential.
So, denoting the ground state energy of that potential for two identical bosons as $E$, the triplet state gives $E_{triplet}=E$ and the singlet state gives $E_{singlet}=-3E$.
So, the question is:
What is the physical explanation of this?.
And, why is the spin-spin interaction more "intense" for the singlet state than for the triplet state(by taking the absolute value of each state's eigenvalue)?
EDIT: Since I don't consider spin-orbit interaction, I think that the answer might have to do with spin-spin coupling because of the dependence on $\vec{\sigma_{1}}\cdot\vec{\sigma_{2}}$.

Hello and welcome to Chemistry.SE. I think your questions lies in that fuzzy region where chemistry and physics overlap. Did you try asking this on Physics.SE? (I see you are rather active on that site, so just curious). I'm certainly not saying you won't get a good answer here, but that's what I'd try next if it doesn't get enough good attention here. It definitely lies outside of my expertise, but hopefully someone here will have a good answer for you...best of luck!

Also, we try to avoid using MathJax formatting in the titles of questions, I believe because the search engines don't like it. I'll leave it to you to [edit] that as you see fit.

@airhuff I have just added it to physics.SE too. Is this allowed? I think that chemists who do quantum chemistry should have a good intuition that is needed for answering this question, so I double posted it. I think both communities could offer something on this, but if it's not allowed, I will delete one of the two.

Cross posting is strongly discouraged on the network - better to delete one, and if you don't get a good response on Chemistry (for example), then we can always migrate it to Physics. I think it would be on topic on both sites, so you can take your pick. If you need moderator help, feel free to flag your post for mod attention (or drop by in chat).

This post gives a good discussion of your options for posting, including "tailoring" the question to best fit each site. It seems that this is one acceptable option, where, as orthocresol said, identical cross-posting is not. You could also just leave the question on Chemistry, then if you're not getting good responses you can ask a moderator to migrate it to Physics if you want (or delete and re-post yourself). I hope this was helpful...

@airhuff  I only left one here. So, I'll be waiting for answers here! Thanks

What type of interaction will couple the spin and spatial extent of the wavefunction via the potential V?  Normally spin & spacial parts are treated separately as we have no idea what the spin wavefunction is is spatial terms so I don't see how they could couple together other than by SO coupling or similar.  Have a look at the answers here https://chemistry.stackexchange.com/questions/60093/identical-particles-why-is-triplet-oxygen-more-stable-than-singlet-oxygen/60096#60096

and to this question https://chemistry.stackexchange.com/questions/58625/what-is-the-physical-basis-for-hunds-first-rule

Could also be connected to another recent question I asked: https://physics.stackexchange.com/questions/332021/intuitively-why-is-the-energy-decrease-of-the-singlet-state-three-times-larger

The answer to ' why is the spin-spin interaction more intense for a singlet than triplet'  is due to the fact that any internal interaction cannot change the overall energy  of the system. Put informally its a sort of 'centre of gravity' thing and applies in all cases. The total energy change is $-E_S + gE_T$, where the $g=3$ is the multiplicity of the triplet. The energy change is zero as $E_S = 3E_T$.
The spin-spin interaction has a standard calculation for example, for the hyperfine structure splitting of hydrogen atom ground $1$s $^2\mathrm S_{1/2}$ ground state, which produces the 21 cm line of radio astronomy and the hydrogen maser.  The potential in which the spins exist (a proton and electron in this case) only comes into the calculation in determining the overall magnitude of the splitting and then as $|\psi(0)|^2$ sometimes called the Fermi contact interaction.
(By standard methods I mean the interaction is of the form $H=A\vec I\cdot \vec J$ where the magnetic moment is generated by nuclear spin angular momentum $\vec I$ and $\vec J$ that due to electron spin. The total angular momentum of the atom is $\vec F= \vec I + \vec J$ is constant and the expectation value is the energy $E=A<\vec I\cdot \vec J>= \frac{A}{2}(F(F+1)-J(J+1)-I(I+1))$. In the case of two spin $1/2$ particles $J=1/2$ and $I=1/2$ and the energies are $A/4$ for $F=1$ and $-3A/4$ for $F=0$. The scaling term A contains constants and  $|\psi(0)|^2$ )


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Calcium detection using XRF
I have a mixture of metal oxides in powder form. Mainly Iron oxide, silicon dioxide, and calcium oxide. I know for certain calcium oxide is in the mixture, but my handheld XRF is not detecting it.
I am analyzing my sample through a very thin plastic wrap, polypropylene.
I have even taken a sample of pure lime, and my handheld XRF is not detecting any calcium. Just a bunch of others such as $\ce{Si, Al, P}$ and so on.
How might this be?
Is the plastic having an adverse effect?
My XRF is in 'general metals' mode; does it need calibrating to detect the $\ce{Ca}$?

This is assumption, so not making an official answer for it, BUT: Phosphorous and Ca share peaks so if the amount of Ca and P is within the same order of magnitude it could be you need a calibration or better equipment to handle that mixture. I know because at my place of work we are managing the same issue, only we are interested in the P content. There is a special set of calibrations and control samples to reduce the problem.

Ca cannot be detected in General Metal Mode. You have to change to "Mining Mode"
In this mode, you can measure Ca properly.


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Thermodynamic Derivation of Ideal Gas Law
The ideal gas law is perhaps the best-known equation of state, and admits both a derivation via the kinetic theory of gases and via statistical mechanics. But these are both microscopic theories, and the ideal gas law is a macroscopic equation.

Can the ideal gas law be derived from macroscopic thermodynamics alone, without recourse to microscopic theories? If so, how? If not, why not?

Assume now that $$U = \frac{3}{2}Nk_\text{B}T,$$ as is the case for a monatomic ideal gas.

Can we now derive the ideal gas law?

Perhaps something to consider regarding the second question. From thermodynamics, $$P = -\left(\frac{\partial U}{\partial V}\right)_{S,N} = -\frac{3}{2}Nk_\text{B}\left(\frac{\partial T}{\partial V}\right)_{S,N} \stackrel{?}{=} \frac{Nk_\text{B}T}{V} \quad \Longleftrightarrow \quad \left(\frac{\partial T}{\partial V}\right)_{S,N} \stackrel{?}{=} -\frac{2}{3}\frac{T}{V},$$ where $\stackrel{?}{=}$ is read "is to be shown equal to." This is an interesting identity, I suppose, but I'm not sure if or where I can proceed from here. As usual, any help is appreciated.

for the second part, you must interchange differentials $ \frac{dU}{dT}=\frac{3}{2}Nk\frac{dV}{dV}=3nR$ ,  then $dU=\frac{3}{2}nRdT $--> $$U=\frac{3}{2}nRT$$ (within a constant, i suppose). If N=Navogadro, U=$\frac{3}{2}RT$

Classical thermodynamics does not admit that molecules exist thus if the gas law is derived it must be without recourse to any molecular ideas regarding the nature of matter.
The chemical potential of a perfect pure gas at temperature T and pressures p and $p^{\mathrm{o}}$ is defined as
$$ \mu = \mu^{\mathrm{o}} + RT\ln(p/p^{\mathrm{o}})$$
We can choose $p^{\mathrm{o}}$ to be unity then
$$ \mu = \mu^{\mathrm{o}} + RT\ln(p)$$
This is still dimensionally correct as $p^{\mathrm{o}} = 1  \pu{~bar}$ and as we have chosen the temperature, $\mu^{\mathrm{o}}$ is the standard chemical potential which is a function of temperature only.
Differentiating the chemical potential at constant T gives
$$ \left(\frac{\partial \mu}{\partial p}  \right)_T = \frac{RT}{p}$$
To find another expression for the derivative we can use
$$dG=-SdT+ Vdp +\Sigma\mu_idn_i$$
 for $dn_i$ moles of species i gives. For a pure species we can drop the subscript i and replace the summation with $\mu dn$ then
$$ \left(\frac{\partial \mu}{\partial p}  \right)_T=\frac{\partial}{\partial n}\left(\frac{\partial G}{\partial p}  \right)_T= V_m$$
where $V_m$ is the volume per mole.
Comparing these two derivatives gives $V_m=RT/p$ or
$$pV=nRT$$
Thus the definition of the chemical potential implicitly contains the perfect gas law.

thank you for the response. In this framework, is it possible to motivate the definition of the chemical potential?

Actually I find this derivation unsatisfactory as it is somewhat circular in nature, but at present I can't see a way round it.

The answer is almost: starting from the $1^{st}$ law $dU=TdS-pdV$ partial differentiation gives you $$\left(\frac{\partial U}{\partial V} \right)_T = T\left(\frac{\partial S}{\partial V} \right)_T - p \tag{1}\label{1}$$ and now apply the Maxwell relation to the right side and get
$$\left(\frac{\partial U}{\partial V} \right)_T = T\left(\frac{\partial p}{\partial T} \right)_T - p \tag{2}\label{2}$$
Note that for an ideal gas (Joule's law) $U=f(T)$ and thus $$\left(\frac{\partial U}{\partial V} \right)_T = 0 \tag{3}\label{3}$$ so that $T\left(\frac{\partial p}{\partial T} \right)_T - p = 0$ and also  $\left(\frac{\partial p}{\partial T} \right)_T = \frac{p}{T}$. This we can  integrate to be $$ \text{ln} p = \text{ln}T + g(V) $$ or $$ ph(V) = T \tag{4}\label{4} $$ for some function $h(V)$ of the volume$V$.
Now to finish we need Boyle's law (this is why the answer is "almost") according to which for the ideal gas the product $pV$ depends only on the temperature. When applied to $(4)$ this gives us $V \propto h(V)$, and that is just the ideal gas equation of state $$ pV \propto T \tag{5}\label{5} $$


text:
Physical Chemistry: Calculating ΔG, ΔH, ΔS on a given equation
This question was brought in class, but I have no clue if my answer is correct or not.
I wonder if you guys can double check my answer.

Between $\pu{0 ^\circ C}$ and $90 ^\circ C$, the potential of a cell with $n=2$, is given by:
  $$\frac{E}{v} = 0.35510 - 0.3422\times 10^{-4} T - 3.2347\times 10^{-6}T^2$$
where $T$ is the Celsius temperature. Calculate $\Delta G, \Delta H,
 \Delta S$ for the cell at $50 ° C$.

My answer:
\begin{align}
  E &= 0.35510 - 0.3422\times 10^{-4}\cdot 50 - 3.2347\times 10^{-6} (50)^2  \\
    &= 0.35510 - 1.711\times 10^{-3} - 8.08675\times 10^{-3} \\
    &= \pu{0.3453 V}\\[2ex]
\left(\frac{\partial\Delta G}{\partial T} \right)_P
   &= -\Delta S\\
   &= -nF\left( \frac{\partial E}{\partial T}\right)_P \\
   &= -nF \cdot (-0.3422\cdot 10^{-4} -6.4694\cdot10^{-6}\cdot T) \\
   \text{at $50 °C$,} \implies &= nF\cdot(3.5769\cdot10^{-4}) \\
   &= 69.02 = -\Delta S\\
   \therefore \Delta S &= \pu{-69.02 J//mol} \\[2ex]
 \Delta G &= nFE \\
   &= -2\cdot \pu{96485 C//mol}\cdot \pu{0.3453 V}\\
   &= \pu{-66633 kJ//mol}\\[2ex]
\Delta H &= \Delta G + T\Delta S\\
   &= \pu{-66633 kJ//mol}
      + (\pu{50 ^\circ C} + \pu{273.15 K})\cdot(\pu{-69.02 J//mol})\\
   &= \pu{88.926 kJ}
\end{align}

Method/numbers  seems to be ok except for little typo; it should be  $\Delta G =-nFE$.


text:
Effect of noble gas configuration on polarising power
I don't understand this statement written in my textbook about the polarising power of cations:

If two cations have the same size and charge, then the one with pseudo noble gas configuration ( with 18 electrons in the outermost shell) has greater polarising power than the other with noble gas configuration (with 8 electrons in the outermost shell). This explains why $\ce{CuCl}$ is more covalent than $\ce{NaCl}$.

Why does this happen?

Recall the fact that the basic difference between an ionic bond and a covalent bond, is basically just charge separation. In ionic bonds, charges are well separated, and the bond arises due to opposite charge attractions. Covalent bonds, on the other hand, have the electrons shared in between them, and it is these electrons that hold the nuclei of the two bonded atoms together.
But sometimes, a covalent bond isn't completely covalent, nor is an ionic bond completely ionic. Even an ionic bond may have some degree of "covalentness" in them. You can compare the covalentness of two ionic bonds with the help of a few rules formulated by Fajans. A few of these rules would be:
Rule 1: Smaller the cation and greater the charge, more the covalent character.
Why is this true? If you consider a cation to be similar to a uniformly charged sphere, the electric field generated at it's surface would be:
$$E=\frac{1}{4\pi \epsilon}\frac{Q}{R^2}$$
Where $Q$ is the charge on the cation, and $R$ is it's radius. It's quite clear now that highly positive cations and smaller cations can generate higher electric fields, and can strongly pull the electrons of the anion into the space between the ions, which gives rise to a partial covalent character.
In your example, it was given that "cation with pseudo-inert gas configuration can polarize better than a one with inert gas configuration". This is very much true, and can be explained using screening effects of orbitals. In the case of $\ce{Na+}$, it's valence electrons are in the $p$ orbitals, while in the case of $\ce{Cu+}$, they are in the $d$ orbitals. It's a known fact that $d$ orbitals shield the electrons from nuclear attractions poorly compared to the $p$ orbitals (see Why do screening effect decreases due to d-orbital? for more details). As a consequence, the nucleus attracts the valence electrons much more strongly in case of $\ce{Cu+}$, leading it to have a smaller size, and more polarizing power.
Rule 2: Greater the size of the anion, more is it's polarizability and greater is the covalent character.
Now the first question that would arise is, what is polarizability and why does it affect covalent character? Polarizability refers to the ability of the anion to get polarized, which is charge separation within the ion. Larger anions allow more room for the electron clouds within them to move around, which causes the cation to polarize it better. See the picture below for better clarity:

The reason why cations with pseudo inert gas configuration have more polarising power is because of the poor shielding effect of the 3d orbitals which increases the ability of cations to distort the electron cloud of anion, thus more polarising power.

Considering the example of NaCl and CuCl:
In NaCl, the electrons effectively shield (or screen the outermost electron from the attraction of nucleus) or one can say the shielding is $100\%$.
Whereas in case of CuCl, there are 10 3d electrons present which ineffectively shields the nucleus, so the value of screening decreases.
The increasing order of shielding: s>p>d>f
Thus, CuCl has configuration : $\ce{[Ar] 3d^10}$ (pseudo noble gas configuration)
NaCl has: $\ce{1s^2 2s^2 2p^6}$ or [Ne] (noble gas configuration)
Hence, $\ce{Cu+}$ will distort the cloud of $\ce{Cl-}$ to a greater extent than the distortion caused by $\ce{Na+}$.


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Why does the stability of ascorbic acid decrease (increased rate of oxidation to dehydroascorbic acid) when pH increases?
I am doing a Chemistry IA on the stability of ascorbic acid in different pH, and found that the stability decreased (there was a lower final concentration of ascorbic acid) as pH increases. Why is this so?

The decrease of ascorbic acid is probably mostly due to oxidation.
Oxidation is loss of electrons.
Increased pH means more ascorbate, thus negative charge and higher electron density.
Higher electron density means easier loss off electrons, thus more oxidation and less ascorbic acid at the end.