text:
What is the exchange interaction?
As the wikipedia article for the exchange interaction so aptly notes, exchange "has no classical analogue."
How wonderful.
Exchange shows up essentially while enforcing the condition that two electrons should not be distinguishable.
That makes sense as a condition to enforce, but I don't understand how this should influence the energetics of the system.
For instance, when two neon atoms interact, there is a very small amount of energy contribution which is attributed to exchange. I understand that exchange is not a force because there is no force carrier, but there is indeed an interaction taking place which affects the energy of the system. What then am I meant to believe is going on? Am I supposed to believe that two electrons spontaneously switch places so as to maintain their indistinguishability? This seems ridiculous. Especially because such a process ought to happen over a finite period of time and thus be observable, but this doesn't seem to be the case (at least as far as I've heard).
What is exchange then? I understand this may not even be an answerable question beyond the fact that it is an effect which drops out of the math and simply is.
Hopefully someone has some interesting insight for me here.

Also, I have seen this question which I believe is referencing the same effect because exchange is known to play a role in ferromagnetism, but nothing ever really caught hold there probably because the question referenced a specific application and so was very difficult to answer well.

partly related http://chemistry.stackexchange.com/q/58625/16683 If I'm not wrong the interaction between Ne atoms can be understood as an instantaneous correlation of electrons - ie van der Waals interaction

This is really a quantum physics question; there is a difference between the energy of a system with two distinguishable elements and a system with two indistinguishable elements, simply because there are fewer combinations of the latter. This wouldn't mean anything in a classical system, but that's just how the world works - our classical assumptions are simply wrong. There is no classical analogue because in the classical models, there is no true indistinguishability. In reality, two particles can be in a configuration that isn't factorizable - and this has a (predicted!) measurable effect.

In quantum chemistry, probably the easiest way to understand the "exchange interaction" is within the context of the Hartree-Fock model.
$
\newcommand{\op}{\hat}
\newcommand{\el}{_\mathrm{e}}
\newcommand{\elel}{_\mathrm{ee}}
\newcommand{\elnuc}{_{\mathrm{en}}}
\newcommand{\core}{^{\mathrm{core}}}
\newcommand{\bracket}[3]{\langle{#1}\vert{#2}\vert{#3}\rangle}
$

To reduce the complexity of the electronic Schrödinger equation
$$
    \op{H}\el \psi\el(\vec{q}\el) = E\el \psi\el(\vec{q}\el) \, ,
$$
where
$$
    \op{H}\el
    =
    \op{T}\el + \op{V}\elnuc + \op{V}\elel
    =
    - \sum\limits_{i=1}^{n} \frac{1}{2} \nabla_{i}^{2}
    - \sum\limits_{\alpha=1}^{\nu} \sum\limits_{i=1}^{n} \frac{Z_{\alpha}}{r_{\alpha i}}
    + \sum\limits_{i=1}^{n} \sum\limits_{j > i}^{n} \frac{1}{r_{ij}} \, ,
$$
we would like to separate the electronic coordinates from each other by writing down the many-electron wave function as a product of one-electron ones. Unfortunately, such separation won't work due to the presence of $\op{V}\elel$ term in the electronic Hamiltonian. But if instead of $\op{V}\elel$ potential, which prevents the separation of electronic coordinates, a model potential of the form $\sum\nolimits_{i=1}^{n} v_{\mathrm{MF}}(\vec{r}_{i})$ had entered the electronic Schrödinger equation, it would be reduced to a set of $n$ one-electron Schrödinger equations with the many-electron wave function being just a simple product of their solutions, one-electron wave functions $\psi_{i}(\vec{r}_{i})$.

More precisely, since we have to take spin of electrons and the principle of antisymmetry of the electronic wave function into account, the electronic wave function would be an antisymmetric product of one-electron wave functions $\psi_{i}(\vec{q}_{i})$ termed the Slater determinant,
$$
    \Phi
    =
    \frac{1}{\sqrt{n!}}
    \begin{vmatrix}
        \psi_{1}(\vec{q}_{1})  &  \psi_{2}(\vec{q}_{1})  &  \cdots  &  \psi_{n}(\vec{q}_{1})  \\
        \psi_{1}(\vec{q}_{2})  &  \psi_{2}(\vec{q}_{2})  &  \cdots  &  \psi_{n}(\vec{q}_{2})  \\
        \vdots                 &  \vdots                 &  \ddots  &  \vdots                 \\
        \psi_{1}(\vec{q}_{n})  &  \psi_{2}(\vec{q}_{n})  &  \cdots  &  \psi_{n}(\vec{q}_{n})
    \end{vmatrix} \, ,
$$
where the one-electron wave functions $\psi_{i}$ of the joint spin-spatial coordinates of electrons $\vec{q}_{i} = \{ \vec{r}_{i}, m_{si} \}$ are referred to as spin orbitals. Physically the model potential of the above mentioned form represent the case when electrons do not instantaneously interact with each other, but rather each and every electron interacts with the average, or mean, electric field created by all other electrons and given by $v_{\mathrm{MF}}(\vec{r}_{i})$.
Minimization of the electronic energy functional
$$
E\el^{\mathrm{HF}} = \bracket{ \Phi }{ \op{H}\el }{ \Phi } = \bracket{ \Phi }{ \op{T}\el }{ \Phi } + \bracket{ \Phi }{ \op{V}\elnuc }{ \Phi } + \bracket{ \Phi }{ \op{V}\elel }{ \Phi }
$$
with respect to variations in spin orbitals subject to constraint that spin orbitals remain orthonormal eventually results in the set of the so-called canonical Hartree-Fock equations which define canonical spin orbitals $\psi_i$ together with the corresponding orbital energies $\varepsilon_i$
$$
    \op{F} \psi_{i}(\vec{q}_{1})
    =
    ε_{i} \psi_{i}(\vec{q}_{1}) \, ,
    \quad
    i = 1, \dotsc, n \, ,
$$
where $\op{F} = \op{H}\core + \sum\nolimits_{j=1}^{n} \big(\op{J}_{j} - \op{K}_{j} \big)$ is the Fock operator with
$$
    \op{H}\core \psi_{i}(\vec{q}_1)
    =
    \Big(
        - \frac{1}{2} \nabla_{1}^2
        - \sum\limits_{\alpha=1}^{\nu} \frac{Z_{\alpha}}{r_{\alpha 1}}
    \Big)
    \psi_{i}(\vec{q}_1) \, , \\
    \op{J}_{j} \psi_{i}(\vec{q}_1) = \bracket{ \psi_{j}(\vec{q}_2) }{ r_{12}^{-1} }{ \psi_{j}(\vec{q}_2) } \psi_{i}(\vec{q}_1) \, , \\
    \op{K}_{j} \psi_{i}(\vec{q}_1) = \bracket{ \psi_{j}(\vec{q}_2) }{ r_{12}^{-1} }{ \psi_{i}(\vec{q}_2) } \psi_{j}(\vec{q}_1) \, .
$$

Here, the first part of the Fock operator, $\op{H}\core$, is known as the one-electron core Hamiltonian and it is simply the Hamiltonian operator for a system containing the same number of nuclei, but only one electron. The second part of the Fock operator, namely, $\sum\nolimits_{j=1}^{n} \big(\op{J}_{j} - \op{K}_{j} \big)$, plays the role of the mean-field potential $v_{\mathrm{MF}}$ that approximates the true potential $\op{V}\elel$ of interactions between the electrons. The first operator $\op{J}_{j}$ rewritten in the following form
$$
    \op{J}_{j}(\vec{q}_{1})
    =
    \int \psi_{j}^*(\vec{q}_{2}) r_{12}^{-1} \psi_{j}(\vec{q}_{2}) \,\mathrm{d} {\vec{q}_{2}}
    =
    \int \frac{ |\psi_{j}(\vec{q}_2)|^{2} }{ r_{12} } \mathrm{d} \vec{q}_2 \, ,
$$
can be clearly interpreted as the the Coulomb potential for electron-one at a particular point $\vec{r}_{1}$ in an electric field created by electron-two distributed over the space with the probability density $|ψ_{j}(\vec{q}_2)|^{2}$, and for this reason it is called the Coulomb operator.
The second operator $\op{K}_{j}$ has no simple physical interpretation, but it can be shown to arise entirely due to anti-symmetry requirement, i.e. if instead of a Slater determinant one uses a simple product of spin orbitals, termed the Hartree product, there will be no $\op{K}_{j}$ terms in the resulting equations (the so-called Hartree equations). It is for that reason that $\op{K}_{j}$ is called the exchange operator.

To quickly understand why exchange terms appear when using a Slater determinant instead of a simple product of spin orbitals look no further than at the Slater rules. For $\op{V}\elel$ which is a two-electron operator we have
$$
    \bracket{ \Phi }{ \op{V}\elel }{ \Phi }
    =
    \sum\limits_{i=1}^{n} \sum\limits_{j>i}^{n}
    \Big(
        \bracket{ \psi_{i}(1) \psi_{j}(2) }{ r_{12}^{-1} }{ \psi_{i}(1) \psi_{j}(2) }
        -
        \bracket{ \psi_{i}(1) \psi_{j}(2) }{ r_{12}^{-1} }{ \psi_{j}(1) \psi_{i}(2) }
    \Big) \, ,
$$
where the second (exchange) part would be absent if $\Phi$ were a simple product of spin orbitals, rather than a Slater determinant.1)

To the question on why do we interpret exchange terms in the Hartree-Fock as a manifestation of some sort of interaction. On the one hand, it must be clear from exchange terms $\op{K}_{j}$ being part of potential energy terms $v_{\mathrm{MF}}$, that they are indeed related to some type of interaction between electrons. However, it must be said that, unlike for the four fundamental interactions, there exists no force based on exchange interaction. Strictly speaking, I don't even think that the term "interaction" has a definite meaning in physics, unless it simply stands for the "fundamental interaction" in which case it is merely a synonym for "fundamental force".
Exchange interaction is not one of the four fundamental interactions, so the meaning of the word "interaction" here is a bit different than that for fundamental interactions. Oxford Dictionary defines interaction as follows,

1.1 Physics A particular way in which matter, fields, and atomic and subatomic particles affect one another, e.g. through gravitation or electromagnetism.

And exchange interaction indeed is a way by which electrons (of the same spin) "affect one another". This can be readily seen as follows. Consider the simplest many-electron system, a two-electron one, and let us examine the following two important events:

$r_1$ - the event of finding electron-one at a point $\vec{r}_{1}$;
$r_2$ - the event of finding electron-one at a point $\vec{r}_{2}$.

Probability theory reminds us that in the general case the so-called joint probability of two events, say, $\vec{r}_{1}$ and $\vec{r}_{2}$, i.e. the probability of finding electron-one at point $\vec{r}_{1}$ and at the same time electron-two at point $\vec{r}_{2}$, is given by
$$
    \Pr(r_1 \cap r_2)
    =
    \Pr(r_1\,|\,r_2) \Pr(r_2)
    =
    \Pr(r_1) \Pr(r_2\,|\,r_1) \, ,
$$
where

$\Pr(r_1)$ is the probability of finding electron-one at point $\vec{r}_{1}$ irrespective of the position of electron-two;
$\Pr(r_2)$ is the probability of finding electron-two at point $\vec{r}_{2}$ irrespective of the position of electron-one;
$\Pr(r_1\,|\,r_2)$ is the probability of finding electron-one at point $\vec{r}_{1}$, given that electron-two is at $\vec{r}_{2}$;
$\Pr(r_2\,|\,r_1)$ is the probability of finding electron-two at point $\vec{r}_{2}$, given that electron-one is at $\vec{r}_{1}$.

The first two of the above probabilities are referred to as unconditional probabilities, while the last two are referred to as conditional probabilities, and in general $\Pr(A\,|\,B) \neq \Pr(A)$, unless events $A$ and $B$ are independent of each other.
If the above mentioned events $\vec{r}_{1}$ and $\vec{r}_{2}$ were independent, then the conditional probabilities would be equal to their unconditional counterparts
$$
    \Pr(r_1\,|\,r_2) = \Pr(r_1) \, , \quad
    \Pr(r_2\,|\,r_1) = \Pr(r_2) \, ,
$$
and the joint probability of $\vec{r}_{1}$ and $\vec{r}_{2}$ would simply be equal to the product of unconditional probabilities,
$$
    \Pr(r_1 \cap r_2) = \Pr(r_1) \Pr(r_2) \, .
$$
In reality, however, the events $\vec{r}_{1}$ and $\vec{r}_{2}$ are not independent of each other, because electrons "affect one another",
$$
    \Pr(r_1\,|\,r_2) \neq \Pr(r_1) \, , \quad
    \Pr(r_2\,|\,r_1) \neq \Pr(r_2) \, ,
$$
and consequently the joint probability of $\vec{r}_{1}$ and $\vec{r}_{2}$ is not equal to the product of unconditional probabilities,
$$
    \Pr(r_1 \cap r_2) \neq \Pr(r_1) \Pr(r_2) \, .
$$
Inequalities above hold true for two reasons, i.e. there are two ways electron "affect one another".
First, electrons repel each other by Coulomb forces, and, as a consequence, at small distances between the electrons $\Pr(r_1\,|\,r_2) < \Pr(r_1)$ and $\Pr(r_2\,|\,r_1) < \Pr(r_2)$, while at large distances $\Pr(r_1\,|\,r_2) > \Pr(r_1)$ and $\Pr(r_2\,|\,r_1) > \Pr(r_2)$.
In the extreme case when $\vec{r}_{1} = \vec{r}_{2}$, the Coulomb repulsion between the electrons becomes infinite, and thus, $\Pr(\vec{r}_{1}\,|\,\vec{r}_{1}) = 0$ and consequently $\Pr(\vec{r}_{1} \cap \vec{r}_{1}) = 0$, i.e. the probability of finding two electrons at the same point in space is zero.
Secondly, as a consequence of the Pauli exclusion principle, electrons in the same spin state can not be found at the same location in space, so that for electrons in the same spin state there is an additional contribution into the inequalities between conditional and unconditional probabilities above.
This effect is relatively localized as compared to one due to Coulomb repulsion, but bearing in mind the relationship between probabilities and wave functions and taking into account the continuity of the latter, it is still noticeable when electrons are close to each other and not just at the very same location in space.
Now, for a Hartree product wave function
$$
  \psi_{\mathrm{HP}}(\vec{q}_{1}, \vec{q}_{2}) = \psi_{1}(\vec{q}_{1}) \psi_{2}(\vec{q}_{2}) \, ,
$$
one can show that $\Pr(r_1 \cap r_2) = \Pr(r_1) \Pr(r_2)$ if two electrons are in different spin states as well as when they are in the same spin state2). But for a Slater determinant
$$
    \Phi(\vec{q}_{1}, \vec{q}_{2}) = \frac{1}{\sqrt{2}}
    \Big(
    \psi_{1}(\vec{q}_{1}) \psi_{2}(\vec{q}_{2}) -
    \psi_{1}(\vec{q}_{2}) \psi_{2}(\vec{q}_{1})
    \Big) \, ,
$$
$\Pr(r_1 \cap r_2) = \Pr(r_1) \Pr(r_2)$ will hold only for electrons of unlike spin, while when both electrons are in the same spin state $\Pr(r_1 \cap r_2) \neq \Pr(r_1) \Pr(r_2)$2). And this inequality is a clear indication of exchange interaction.

1) I leave it to OP as an exercise. Derivation for a simple product of spin orbitals is pretty trivial, while for the case of a Slater determinant OP can consult, for instance, Szabo & Ostlund, Modern Quantum Chemistry, Section 2.3.3.
2) I leave it as another exercise.

Very nice summary on HF theory : ) But one thing that does not come clear to me yet is how one can understand the exchange 'interaction' between systems as an interaction. Part of the question was if there is some way in which we can kind of "visually" understand the process that is responsible for this energy. Is that possible?

@ketbra, exchange interaction is not an interaction in the same sense as the four fundamental interaction are: there is no exchange force, so that the word "interaction" in "exchange interaction" must not be though as a synonym to "force". Rather, I think, "interaction" here is a synonym to "interdependence" which can be demonstrated by working out the probabilities as I did in the answer.

@ketbra You're trying to understand this from a classical perspective. There is no analogue. You simply have to look at the math and say, yes, there is a contribution from this term, and this term is the exchange piece.

There are three different integrals commonly called the ‘exchange’ integral, which are the resonance integral, the exchange integral itself and the exchange operator which is also an integral. These integrals are related to the Coulomb integral and are conventionally one is  called J the other K or vice versa; textbook authors differ. Almost all authors are coy about giving an explanation as to what the exchange integral represents. Those few that do are confusing so you are not in alone in wanting to understand this.
Resonance integral. This occurs in the description of $\ce{H_2^+}$ and is also sometimes called the exchange integral. In the variational solution to this problem the equations to be solved are
$$\begin{vmatrix} H_{11} - E & H_{12} - ES \\  H_{12} - ES & H_{22}-E \end{vmatrix} $$
where E is the energy and  $H_{ij} $ and S are integrals. The $\ce{H_2^+}$ wavefunctions $\psi$ are found as linear combinations of 1s atomic orbitals
 $\psi = c_1\phi_1 \pm c_2 \phi_2$
where $c_1$ and $c_2$ are coefficients.
The overlap integral is
$S=\int {\phi_1 \phi_2 ~d\tau} $
and
$H_{11} = \int\phi_1~H ~\phi_1 ~d\tau$
and similarly for $H_{22}$. H is the operator
$H=p^2/(2m) +V_{en} + V_{nn}$
where p is momentum and the V are electron-proton (e-p) and proton-proton (p-p) interactions and are of the form $e^2/r$ where r is the appropriate separation, e-p or p-p.
Expanding the $H_{11}$ integral has a term that produces the Coulomb integral
 $$C = \int \phi_1 \frac{e^2}{r_2}\phi_1 ~d\tau$$
which we interpret as the electrostatic potential between proton 2 and the charge distribution of the electron associated with the 1s orbital around proton 1. The integral $H_{22}$ is explained similarly.
Expanding the $H_{12}$ integral produces the exchange or resonance integral,
$$A=\int \phi_1 \frac{e^2}{r_1} \phi_2 ~d\tau$$
which like the Coulomb integral is positive and is where the fun/confusion begins.
If the $H_{12}$ terms were zero, the ground state would be doubly degenerate because by symmetry $H_{11} = H_{22}$. But as the $H_{12}$ are not zero the states based on orbitals $\phi_1$ and $\phi_2$ are not stationary states of the molecule and thus a lower energy is obtained because of the coupling $H_{12}$ between $H_{11}$ and $H_{22}$. A linear superposition of $\phi_1$ and $\phi_2$ has this lower energy.
Some authors (Cohen-Tannoudji, Diu & Laloe , Quantum Mechanics) call this ‘Quantum Resonance’ and describe the fact that $H_{12}$ is not zero as the possibility of the electron ‘jumping’ from the neighbourhood of one proton to that of the other. ‘The electron ‘oscillates’ in time between the two sites under the influence of $H_{12}$ which is responsible for quantum resonance’.
Coulson (Valence) also describes resonance in this way but then warns that it is easy to take the analogy too far and states that there is no possible physical mechanism whereby the electron could oscillate. He then goes on to say that integrals of this form are a consequence of making a linear combination to describe the molecular orbital. So it seems that it has no physical meaning other than it describes that coupling between the two isolated states $\phi_1$ and $\phi_2$ that we started with, and which by forming a molecular orbital lowers the energy compared to these states.
In multi-electron atoms and molecules similar integrals occur but now are mathematically a  more complicated but the interpretation, such as it is, remains the same. In the simplest case of two electrons a and b in i and j spatial orbitals where each orbital can accommodate two electrons, the Coulomb integral is
$$ J=\int \int \phi_i (a) \phi_j (b) \frac{e^2}{r_{ab}} \phi_i (a) \phi_j (b) ~d\tau_a d\tau_b  = \left\langle ij \middle |ij\right\rangle $$
the Coulomb operator is the integral
 $$\int \phi_j(b)\frac{e^2}{r_{ab}}\phi_j(b)~d\tau_b $$
The exchange integral, as its name suggests, swaps electrons a and b between orbitals i and  j and is
$$ K=\int \int \phi_i (a) \phi_j (b) \frac{e^2}{r_{ab}} \phi_j (a) \phi_i (b) ~d\tau_a d\tau_b  = \left\langle ij\middle |ji\right\rangle $$
where the exchange operator is
$$\int \phi_j(b)\frac{e^2}{r_{ab}}\phi_i(b)~d\tau_b $$
and operates on one of the $\phi_i(a)$ functions.
Coulomb and exchange integrals are encountered elsewhere for example in the determining the energy of spectral terms $\mathrm{^1S}$ and $\mathrm{^3S}$ arising from the 1s2s configuration in Helium. The energy of the two levels is  $J\pm K$. The Coulomb term J raises the energy of both energy levels and K splits them. The underlying reason for the energy splitting resides in the respective symmetry of the spatial and spin wavefunctions.

I was just rereading this and I really like the view of exchange as a coupling between states. That has a very nice interpretation for at least something like $\ce{H2^+}$. Thanks for that insight.


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Reasoning retention factor of benzoic acid
In my organic chemistry lab, we ran a TLC plate on silica to determine the identity of an unknown compound. The two possibilities were benzoic acid and acetanilide. I know that the general rule is, the more polar the compound is, the more it absorbs onto the silica and slower and lower it would travel (corresponding to a lower $\mathrm{R_f}$ value). However, one question on my lab report asks:

In practice, benzoic acid tends to have larger $\mathrm{R_f}$ values than acetanilide. Attempt to find an explanation why this is. And note that spots from more concentrated benzoic acid leas to larger $\mathrm{R_f}$ values.

My professor said the answer has something to do with the $\ce{-OH}$ group on benzoic acid.

http://chemistry.stackexchange.com/q/60640/16683

The answer is the dimer formation of the benzoic acid. This is why higher concentrations lead to more extended dimer formation, thus higher $\mathrm{R_f}$ values as the carboxyl group "gets shielded."
One thing of note is that $\mathrm{R_f}$ values are not directly dependent on the polarity. They also depend for example on the size of the molecule (Oswald radius). And the relation between properties of the molecules (as stated for example in the form of molecular descriptors like $\log P$) and the resulting $\mathrm{R_f}$ is very complicated indeed. This is also illustrated by the fact that there are predictors for IR spectra, NMR spectra, solubility and lots of molecular descriptors but I have never heard of a good $\mathrm{R_f}$ value predictor.
This is why I would always be careful to talk about it like "the polarity" of the compound, especially as "the polarity" is not a clearly defined quantity.

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text:
Confusion about Ethinyl estradiol IUPAC name
According to Wikipedia, the IUPAC name of ethinyl estradiol is:

17α-Ethynylestra-1,3,5(10)-triene-3,17β-diol

However, in PubChem, its IUPAC name is:

(8R,9S,13S,14S,17R)-17-ethynyl-13-
  methyl-7,8,9,11,12,14,15,16-octahydro-6H-cyclopenta[a]phenanthrene-3,17-diol

My questions:

Which one is correct by latest IUPAC blue book?
How are the locants assigned for this molecule? Specifically, which carbon was assigned the locant "10", since it does not appear in the locants of "octahydro"?

There can be more than one correct IUPAC name.

The first name is from estrane, and the second from phenanthrene.

But how the numbering worked? I struggled to find a sequence.

Have a look

Note that, interestingly, recent (2013) BB does not have a notion of preferred names for natural products: P-100 (…) Preferred IUPAC names (PINs) are not identified for the compounds in this Chapter. The choice between a semisystematic name and a systematic name will be made in cooperation with the IUPAC–IUBMB Joint Commission on Biochemical Nomenclature and will appear in a future publication.

For natural products and derivatives, three levels of nomenclature are recognized in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book): trivial names, systematic names, and semisystematic names. A new compound that can be isolated from a natural source is commonly given a trivial name. These trivial names are usually related to the biological origin of the material.
When the full structure of a natural product or derivative is known, a systematic name may be generated in accordance with the IUPAC recommendations for nomenclature of organic chemistry. However, the systematic name “(8R,9S,13S,14S,17R)-17-ethynyl-13-methyl-7,8,9,11,12,14,15,16-octahydro-6H-cyclopenta[a]phenanthrene-3,17-diol”, which is proposed in the question is not correct. This name implies that the carbon atoms 13 and 17 of the cyclopenta[a]phenanthrene parent structure remain sp2 hybridized; i.e. that a double bond remains between atom 13 and atom 17, which is not possible since both atoms have four bonds (to other skeletal atoms or to substituents). The correct systematic name is (8R,9S,13S,14S,17R)-17-ethynyl-13-methyl-7,8,9,11,12,13,14,15,16,17-decahydro-6H-cyclopenta[a]phenanthrene-3,17-diol.
Such systematic names may be too cumbersome. To overcome this difficulty and show the close similarity to related compounds, a semisystematic name may be formed. The semisytsematic name 17α-ethynylestra-1,3,5(10)-triene-3,17β-diol is based on the retained name estrane for the parent structure (8R,9R,10S,13S,14S)-13-methylhexadecahydro-1H-cyclopenta[a]phenanthrene. (Names of fundamental stereoparent structures are given in Table 10.1 of the 2013 IUPAC recommendations. The corresponding structures are shown in Appendix 3.) Addition of the three required double bonds to the estrane parent structure leads to the name estra-1,3,5(10)-triene, and addition of the missing substituents yields finally yields the name 17-ethynylestra-1,3,5(10)-triene-3,17-diol. To indicate the configuration at chiral centers, the ‘α/β’ system is used as described in IUPAC-IUBMB recommendations for the nomenclature of steroids.

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text:
Deriving the partition function for a harmonic oscillator
This question is in reference to an answer I posted here yesterday.
In it I derived the partition function for a harmonic oscillator as follows
$$q = \sum_{j}e^{-\frac{\epsilon_j}{kT}}$$
For the harmonic, oscillator $\epsilon_j = (\frac{1}{2}+j)\hbar \omega$ for $j \in \{ 0,1,2.. \}$
Note that $\epsilon_0 \neq 0$ there exists a zero point energy.
Let's write out a few terms $$q = e^{-\frac{\hbar \omega/2}{kT}} + e^{-\frac{\hbar \omega3/2}{kT}} + e^{-\frac{\hbar \omega5/2}{kT}} +..... $$
factoring out $e^{-\frac{\hbar \omega/2}{kT}}$
$$q = e^{-\frac{\hbar \omega/2}{kT}} \left( 1+ e^{-\frac{\hbar \omega}{kT}} + e^{-\frac{2\hbar \omega}{kT}} +.....\right) $$
The sum in the bracket takes the form of a geometric series whose sum converges as shown below $$1+x+x^2+... = \frac{1}{(1-x)} $$
 herein, $ x \equiv e^{-\frac{\hbar\omega}{kT}} $
Putting all of this together
$$q = \frac{e^{-\frac{\hbar \omega/2}{kT}}}{(1-e^{-\frac{\hbar\omega}{kT}})}$$
I happened to check Atkins Physical chemistry (10th edition) for the same derivation earlier today.
On page 620, the vibrational partition function using the harmonic oscillator approximation is given as
$$q = \frac{1}{1-e^{-\beta h c \nu'}}$$, $\beta$ is $\frac{1}{kT}$ and $\nu'$ is wave number
This result was derived in brief illustration 15B.1 on page 613 using a uniform ladder. However, in that illustration the uniform ladder starts at 0, but the harmonic oscillator has a zero point energy (which I accounted for in my derivation).
I discussed this with my instructor and he pointed out that as $T \rightarrow 0$ $q \rightarrow 1 $ (since only one state is thermally accessible.
The result derived in Atkins' does that indeed, but mine goes $q \rightarrow 0$.
Now, in the formalism developed in Atkins the do set the ground state energies to zero (on page 605), and basically add non-zero ground state energies to a calculated $\langle \epsilon \rangle$, which is fine.
Anyway, I would love it if someone could weigh in on this and help me resolve this conceptual mess in my head.

By relative ratios, do you mean the relative populations of the excited states? Actually, it shouldn't really be difficult for you to figure it out. The maths works out because the population ratio only depends on the difference in energy: $n_i/n_j \propto \exp[-\beta(\epsilon_i - \epsilon_j)]$ and this is independent of whether the zero point energy is set to 0 or not. Sorry if you are asking about something else. :D Also, if I am not wrong, the interpretation of $q$ as the number of thermally accessible states only makes sense if you set the ZPEs to zero.

Oh that's true. Silly me, the problem with the limits still remains. What you say about the interpretation does make sense. If someone could confirm, then it would be great.

Absolute values of $q$ and energy are not important at all. It is the derived quantities that matter, and they don't depend on your zero energy.

You are correct in your first calculation to obtain $\exp(-h\nu/(2kT))/(1-\exp(-h\nu/(kT))$ because you here choose the zero to be at the bottom of the potential well, if you choose the $j=0 $ level to be zero then you obtain $1/(1-\exp(-h\nu/(kT))$ which is what Atkins assumes.

I'm confused why you're interpreting the partition function as a count of states. It can't be a count; it's continuous.
The zero point energy doesn't actually matter because you can just shift the energy scale so that it starts at zero. The main point of zero point energy is that the ground state of the harmonic oscillator is such that there is energy, and the system is not stationary. I'm going to use it below anyway because you are.
You can get the answer you want, but you'll want to look at the probability $P_i$ of being in state $i$, where the partition function is used to normalize:
$$P_{i} = \frac{\epsilon_{i}}{q} = \frac{\mathrm{e}^{-\beta\hbar\omega (i + 1)/2}}{\sum_{i}\mathrm{e}^{-\beta\hbar\omega (i+1)/2}}$$
Substitution with your convergent sum:
$$P_{i} = \mathrm{e}^{-\beta\hbar\omega (i+1)/2} \frac{1 - \mathrm{e}^{-\beta\hbar\omega}}{\mathrm{e}^{-\beta\hbar\omega/2}} = \mathrm{e}^{-\beta\hbar\omega i}(1- e^{-\beta\hbar\omega})$$
For $T\rightarrow 0$, $P_{i} = \delta_{i0}$, which is exactly what you are looking for.

Thanks great answer! And the interpretation of the partition function as the "number" of thermally accessible states is not mine, it was in the book/given to us by the instructor..but I get what you mean too. This "interpretation" is probably the source of this confusion.

Partition functions are defined as sum over states . If the gap between energy levels is small wrt kT then the sum can be replaced by an integral. This is ok for translational motion and although used for rotational motions in textbooks gives rise to a significant systematic error. The partition number tell us how, in the equilibrium distribution, the systems are partitioned or divided up among the different energy levels. There are $n_1$ systems with energy $\epsilon _1$, $n_2$ with $\epsilon _2$ etc.


text:
Effect of protein hydrophobicity on metal interaction
I'm studying how metals and proteins interact (metal-protein complexes). I've found that almost all proteins could bind with metal at higher $\mathrm{pH}$ values, but also found some proteins showed lower intensity in metal-protein complexes.
I understand $\mathrm{pH}$ has an effect on the charge state of amino acid side chains of proteins, but I really had less information about the effect of $\mathrm{pH}$ on protein hydrophobicity, and hydrophobicity's effect on the protein's ability to bind metals.

Thats a very interesting topic : ) What analytical technique do you use? The pH value will inevitable influence the hydrophobic properties of the protein as amino acid side chains get protonated/deprotonated. If I remember it correctly, at least the log P values are most often specified for different pH values. That would be one way to systematically study the effect of pH on the hydrophobicity of individual amino acids.

So does this question have to do with pH value or hydrophobicity?


text:
Is Lightning Basically a Cathode Ray?
I've heard the term cathode ray pretty much every year in school since they introduced chemistry, and I know that it's a stream(?) of electrons. I'm just having troubles relating it to the real world. I think lightning would be a prime example, but I'm not sure; looking for confirmation.

Except that cathode ray involves vacuum while lightning ionizes the air.

Nope. The lightning flash is more like a plasma, not a beam of electrons like in a cathode ray

As you are already aware, cathode rays are electron beams. You take evacuated glass tube, and  add two electrodes on either end. Once you apply a sufficiently high voltage, a stream of electrons flows towards the "cathode"; this is what you call a cathode ray. Inside the tube there exists near perfect vacuum, so there are barely any other gas molecules, ions etc.
Lightning is similar in the sense that you have a large potential difference that leads to electric discharge. The key difference is that this takes place in air, and not in vacuum. Since air is "insulating", it has to be ionised to for the electric discharge to take place. Indeed this is what happens, when the potential difference is sufficiently high. However, what this creates is a plasma, not a stream of electrons.
A plasma is the "fourth" of matter, and is created by ripping apart some electrons from gas molecules resulting in a collection of electrons and ions.


text:
Identity of unknown metal
I wanted figure out what the identity of this scrap metal was. It seems to be some sheet metal but Wikipedia says sheet metal can be made of a variety of metals. My dad said it might be tin so I thought I'd try and see if it would melt easily since tin melts at a very low temperature. I heated it with a propylene torch but I could only get it red hot. I did notice that it turned yellow when heated and smelled a bit rotten. Any ideas to what this could be?

You could try some simple tests to eliminate some metals.  Is it magnetic? Is it covered with an oxide layer? What is its density (polish then accurately weigh a clean piece; put same into water and accurately measure volume change, calc' density; look up in tables.) File into powder if possible,  try a flame test.  Does it dissolve in acid and alkali?


text:
Why do nanoparticles have a different color than their macro counterparts?
It astonished me when I learned that at nanoscale, gold is no longer "gold",rather it's red. Other elements' nanoparticles also have a different color than their macro counterparts.
Why does this phenomenon occur?

NSF has a nice article about this:  https://sustainable-nano.com/2019/11/12/gold-nanoparticles-color/

There is a fairly simple explanation of why small-enough particles are different from bulk materials. Once the particle becomes similar in size to the wavelengths of light involved then quantum effects start to matter for how the particle behaves.
The actual mechanisms that give specific colour may vary, but the main point is that the size of the particle comes to dominate the bulk properties of the material. In a bulk semiconductor, for example, the electronic properties depend on the band gap between occupied electron levels and unoccupied electron levels. When the particle is small enough, however, the electrons are more constrained and become more like an electron confined to a small box. This can mean that the possible energy levels of the electrons are determined by the physical size of the box not the bulk properties of the material. Hence any emission or absorption of light will be different to the bulk material. The additional physical constraints impose a different set of energy levels on the electrons.
The actual details can be quite complicated but the basic idea is that when particles are small compared to the wavelengths of light, you have to take into account the extra constraints on the system when calculating possible energy levels involved in electron transitions which are what causes colour.


text:
How does one find the probability of tunneling along a reaction coordinate computationally?
The idea for this question comes from Jan's comment over here. I realized I don't know how I would compute the probability of tunneling in the process of a reaction.
When I say "compute" I don't mean something rather trivial like finding the probability of a quantum harmonic oscillator being past the classical turning points which can just be done analytically.
For instance, if I were studying some reaction computationally (to estimate rate constants or something which depends on tunneling) and constructed a potential energy surface for a bond-breaking process, let's say the transfer of a hydrogen, how do I account for the possibility of tunneling through the barrier along this reaction coordinate?
I could speculate, but I imagine I would be wrong. How would one do this using common computational software and techniques?

If the reaction coordinate $\xi$ can be expressed as the change in the relative position between two groups of the molecule (e.g. for instance as a change in a bond length so that $\xi$ is given in units of length) the problem may be treated as one dimensional and one may use computational software to calculate the potential energy $V(\xi)$ along this path. You may then proceed solving the time-independent Schroedinger equation using numerical techniques such as Numerov's integration method. However, in many cases WKB theory provides an accurate estimate.
In terms of WKB, the tunneling probability is given by
$$
T=e^{-2G}
$$
where $G$ is the Gamow factor given by
$$
G=\frac{2}{\hbar}\int_{\xi_a}^{\xi_b}{ \sqrt{2\mu_\text{red}\left [V(\xi) - E \right ]}}d\xi
$$
where $\mu_\text{red}$ is the reduced mass along the reaction coordinate and $E$ is the energy of the incoming particles. The integral runs over the clasically forbidden region from $\xi_a$ to $\xi_b$. The Gamow factor thus represents the square root of the integral over the "reaction barrier" multiplied with the effective mass that tunnels. Of course, you should take zero point energy into account and in the case of light reactants also the vibrational and rotational structure of the products as this may lead to resonances.

There are other approximations, such as Landau-Zener. But according to https://www.sciencedirect.com/science/article/pii/0009261472870684 WKB may take the cake.


text:
Difference between vapour and gas
This is the question i have been asking from months to my friends,teachers,and everyone but i haven't got a satisfactory answer yet(in fact i haven't got any answer).These two terms (vapour and gas) are confusing me from a long time and i can't get the difference.I shall be very thankful if someone of you give me a detailed explaination for the question.I m sorry if this question or a similar question have already been asked on the site and i shall be thankful if you provide me the link of the question so that i can get my answer there.
Finally the question is "What is the difference between vapour and gas".

In physics a vapor (American) or vapour (British) is a substance in the gas phase at a temperature lower than its critical temperature, which means that the vapor can be condensed to a liquid by increasing the pressure on it without reducing the temperature.

is it different in chemistry??

It is a physical phenomenon, not a chemical phenomenon.

Really??But it comes under the chapter states of matter in my chemistry textbook.

Changes in the state of matter does not change the chemical composition of the substance (generally).

Look, i think your and my views are not coinciding,i think change of state have no significance here as i m just asking difference between gas and vapour.

Forget it, there is no difference.

DHMO gave you a comprehensive answer. If you are not interested in the change of state, then there is no difference.

What if i develop my interest in change of state??

from the second comment;   ... and a gas cannot be liquified by pressure alone.

I think vapor have a temperature near to the condensation point so it can become liquid easily but gas have much higher temperature than it's condensation point hence we need a process to make it liquid for example if steam is at around 100°c then it can be termed as water vapor and if you consider nitrogen it liquifies at -198°c or near to that hence it is referred as gas.


text:
Trichloride ion (c.f. triiodide ion)
While triiodide ion is commonly known, why is that trichloride ion (and its salts) almost unheard of? While it does exist according to literature, why is the stability of trichloride ion in aqueous environment much worse than its bromine and iodine counterparts? Aren't elements down the group bulkier and hence forming weaker covalent bonds with longer bond lengths?

http://chemistry.stackexchange.com/questions/27920/why-is-nacl3-possible

As @nilay points out, we can make "polychlorides" with "nonclassical" stoichiometries. But for more typical laboratory conditions it is true that heavier halogens are more favorable for forming polyatomic ions.
This is because polyhalide ions involve two competing forces. On the one hand, the delocalized covalent bonding gives more stability to a polyatomic ion than we would have with the separate monoatomic ions and diatomic molecules. On the other hand, surrounding counterions or dipoles are more strongly attracted to compact, spherical monoatomic anions than to bulky polyatomic ones. With lighter halogens forming the most compact monoatomic anions, the electrostatic effect wins out and we need severe conditions to overcome it. Heavier halogens, with larger atoms and more polarizable electrons, make the polyatomic ions more competitive, as in $\ce{C_5H_6N+Br_3−}$ (http://www.organic-chemistry.org/chemicals/oxidations/pyridiniumhydrobromideperbromide.shtm).

Does it also have to do with the "softness" of iodide ion in a sense that it mingles better with typically "soft" diatomic halogen molecule? Also, have you heard of things like dichloroiodide ion?

With multihalogen species things are more complex; the neutral diatomic species (in this case $ICl$) is polar and has an ion-dipole attraction with more anions (such as $Cl^-$).  This favors association even with smaller "outer" atoms.  Note that you still need a larger halogen ion in the middle to set up the dipole.

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


text:
How to treat skin corrosion by sodium hydroxide?
I'm under training of becoming floorman in the rig. We use a lot of chemicals and one of them is caustic soda or sodium hydroxide. In case it touches my skin can you tell me of ways to treat it? I have thought of neutralizing it by slowly going less basic by using for example household ammonia then hand soap then sea water.

I suggest that you look on the web for the safety data for this any any other compounds you use. Sodium hydroxide will make the skin feel soapy. I would not use ammonia but wash with copious amounts of soap & water as soon as possible. Then wash again when you think you have removed it all.

I don't think you understand how neutralization works. Adding more base (ammonia) will not help you. As with most (but definitely not all) chemicals, the correct response is to flood the affected area with water immediately. You want to get it off you as quickly as possible, not go slow and add more base.

We use a lot of chemicals […]

Before you start to use potentially hazardous compounds and products, get their Material Safety Data Sheets (MSDS) and read them.
Learn about storage, handling, disposal and emergency measures in case of

eye contact
skin contact
inhalation

For a $50\%$ sodium hydroxide solution, a typical MSDS looks like this and states:

Skin Contact:
In case of contact, immediately flush skin with plenty of water for at least $15$ minutes while removing contaminated clothing and shoes. Cover the irritated skin with an emollient. Cold water may be used. Wash clothing before reuse. Thoroughly clean shoes before reuse. Get medical attention immediately.

Forget about the ammonia! Use lots of water!


text:
Remove redness in copper-nickel coin
I had a silver-looking coin (not valuable, don't worry) that was heavily covered in black. I thought it could be silver, so I cleaned it with the electrolysis method (salt + sodium bicarbonate solution; copper on the positive pole and the coin on the negative). This didn't really work, though it seemed to make the coin less black, so I realized the coin was not silver.
I found out it is a copper-nickel coin. Since then I tried immersion in vinegar + salt solution, which made it a bit cleaner, and scrubbing with sodium bicarbonate, which made it cleaner still. But I noticed that, though some parts are silvery, most of the coin has a slight redness as of copper. Is this the copper from the coin, or did I somehow manage to deposit copper on it? How can I make the coin silvery again?

Hydrogen Peroxide? No idea


text:
What is the correct definition of hydration enthalpy and why is it always negative?
The first definition of hydration enthalpy I found on the net is:

The hydration enthalpy is the enthalpy change when $1$ mole of gaseous
  ions dissolve in sufficient water to give an infinitely dilute
  solution. Hydration enthalpies are always negative.

Another definition I found in a book is:

Hydration enthalpy is the energy released when $1$ mole of a compound is
  dissolved in sufficient water to give an infinitely dilute solution.

Now I have three questions:

Which of the two definitions is the correct one? Why?
Why exactly are hydration enthalpies always negative? Suppose I have hydrated copper(II) sulphate which is mostly stable as $\ce{CuSO4.5H2O}$. I convert it to $\ce{CuSO4.6H2O}$ by adding more water forcefully. Shouldn't the hydration enthalpy be positive in this case as it will make the final product less stable than the initial one?
Which definition of hydration enthalpy applies for $\ce{CuSO4}$? I do not think $\ce{CuSO4.5H2O}$ is "infinitely" dilute by any extension of logic. Please explain this if possible.

P.S: I am a novice in thermochemistry. It would be helpful if you could keep the explanations as simple as possible.

I'd just like to note that both definitions are commonly called "enthalpy of hydration", so if you see it in a book, make sure to check the context and figure out which one they're referring to. It's very common to talk about the "heat of hydration of so-and-so ion", which refers to the first of your two definitions.

Could you please answer "Suppose I have hydrated copper(II) sulphate which is mostly stable as $\ce{CuSO4.5H2O}$. I convert it to $\ce{CuSO4.6H2O}$ by adding more water forcefully. Shouldn't the hydration enthalpy be positive in this case as it will make the final product less stable than the initial one?" @orthocresol

Hydration is defined as the following process (with any salt but using copper(II) sulphate as an example):
$$\ce{CuSO4 + 5 H2O -> CuSO4 . 5 H2O (s)}\tag{1}$$
This reaction will only happen if the associated enthalpy $\Delta_\mathrm{hyd} H^0$ — or more precisely, the associated Gibbs free energy $\Delta_\mathrm{hyd} G^0$ is negative. For some compounds, that is not the case; for example sodium chloride does not form hydrates: reaction (2) does not occur spontaneously.
$$\ce{NaCl + n H2O} \nrightarrow \ce{NaCl . n H2O}\tag{2}$$
Thus, we cannot measure a hydration enthalpy. If we can measure it, the process must be spontaneous and thus the enthalpy negative. Thus, all measurable hydration enthalpies are negative.
Sometimes, salts can form multiple hydrates. However, not every hydrate is always possible. Hydration enthalpies only exist for those hydrates which are possible.

Your definitions are basically identical only that they do not measure hydration enthalpies but solvation enthalpies. That is the process as shown in equation (3).
$$\ce{CuSO4 (+ H2O) -> Cu^2+ (aq) + SO4^2- (aq) (+ H2O)}\tag{3}$$
For that process, the enthalpy does not have to be negative. In fact, quite a few ionic compounds dissolve endothermicly, e.g. $\ce{NH4Cl}$.

Even the process $\ce{Na^+(g) \rightarrow Na^+(l)}$ is called hydration. Is the enthalpy for it always negative?

FYI "Hydration energy (also hydration enthalpy) is the amount of energy released when one mole of ions undergo hydration which is a special case of solvation. It is a special case of dissolution energy, with the solvent being water."  So the last example you gave falls under hydration only. See  https://en.wikipedia.org/wiki/Hydration_energy

I don't agree with "If we can measure it, the process must be spontaneous and thus the enthalpy negative." Spontaneous is about Gibbs energy, not enthalpy. I guess the entropy is expected to be negative (constraining water molecules to the crystal lattice) but unless you say that explicitly, the enthalpy could be positive or negative for a spontaneous reaction or process.

@KarstenTheis Of course, it's not easy to think oneself back into a five year old answer ... However, I think I wrote that because starting material (ionic compound) and product (hydrate of the ionic compound) are both solids and somehow I recall that in those reactions the entropy term would become zero and only the enthalpy ends up determining $\Delta G$ and thus the spontaneity? It's been a long time and I could have drawn wrong conclusions from prior knowledge.

Definition 2 sounds like the definition for enthalpy of solution.
In that case, both are correct: they measure different things.
Hydration enthalpies are always negative because hydrating an ion would always stabilize it. For hydration enthalpies, note that the waters of hydration are absent: you are hydrating the gas phase ions.
You can also measure both for $\ce{CuSO4}$. There is no hydration enthalpy for the hydrate.

No, $\ce{CuSO4}$ does not have a hydration enthalpy. Look at the definition again. You need to dissolve the component ions to infinite dilution. That preludes the ability to "hydrate" the hydrate.

To me both definitions are correct. The complex cation hexaaquacopper(II) cannot undergo further hydration and therefore it is infinitely dilute given that all 6 ligands have been provided. From the definition, since hydration enthalpy is a measure of energy released from a mole of compound then the final compound possesses less energy than the initial one. enthalpy final minus enthalpy initial in this case will always be negative. in terms of stability the pentaaquacopper(II) complex should be unstable because 5 is not a common coordination number for the copper metal ion so the complex should be less stable and more energetic than the hydrated one.


text:
What's the van 't Hoff factor of dilute aqueous ferrous sulphate solution?
Is it more than one or less than one?
Here is what I think: since it is a dilute and aqueous solution, so $\ce{FeSO4}$ will be hydrated and new formula will be $\ce{FeSO4*7H2O}$.
Thus, the abnormal mass would be greater than the normal molar mass.
Therefore, van 't Hoff factor would be less than $1$. Is that correct?

Policy on AMIRITE questions

Since it is dilute, the ions will dissociate: $\ce{FeSO4 -> Fe^2+ + SO4^2-}$, and whence hydration?

@DHMO Firstly, it's not policy yet; secondly, even though it's tempting, please don't make obnoxious edits like that. All of us are very frequently tempted to leave snarky comments. Whether you rise above that to be a better person is up to you.


text:
What is the best way to remove/break carbon carbon bonds?
I am working on synthesis homework and I can not figure out how to remove a carbon substituent from a cyclohexane.

What is the best way to remove the three carbons and exchange them for a ketone?

A possible pathway is as follows:

At high temperatures, formation of the more substituted alkene will be favored. Catalytic amounts of base provide a kinetic pathway for this rearrangement. $\ce{OsO4}$ creates a vicinal diol, while $\ce{NaIO4}$ cleaves the $\ce{C-C}$ bond and oxidizes the alcohols to carbonyls. Additionally, $\ce{NaIO4}$ oxidizes $\ce{(HO)2OsO2}$ back to $\ce{OsO4}$, allowing the reaction cycle to continue.

Ok thanks, I didnt know you could use base+water to rearragne a double bond like that. Very helpful

Quick question, what program are you using to draw out the compounds?

@unrnick ChemDraw. If you are at an uni, you should be able to get it for free

Worth noting that OsO4 is a nasty (and expensive) chemical, which leads to the main benefit of using NaIO4 as a co-oxidant: you get to use OsO4 in catalytic quantities.

@orthocresol I wouldn’t be sure periodate is a co-catalyst here. From what I learnt, $\ce{OsO4}$ generates the diol which is periodate-cleaved to two carbonyl groups.

@Jan that's what's said in the answer, at least. I looked it up: JOC 1956, 21, 478 "Catalytic amounts of osmium tetroxide are sufficient because periodate oxidises osmium in its lower valence forms to the tetroxide, thus regenerating the hydroxylating agent."

@orthocresol Oh yeah, I should have read better =D. Of couse, you can use pretty much any co-catalyst, but using periodate has the benefit of co-performing the cleavage.

Here's a variation of ringo's method:

Treat the alkene with $\ce{OsO4}$/$\ce{NaIO4}$ to obtain a cyclohexyl methyl ketone
Convert the ketone to cyclohexyl acetate by using a peracid (MCPBA).
Cleave the ester
Oxidize the cyclohexanol to cyclohexanone with a mild oxidant

I can offer another two-step synthesis:

isomerise the double bond as ringo mentioned.
Criegee reaction ($\ce{O3}$) followed by workup with $\ce{PPh3}$.

Ozonolysis followed by reduction, elimination of water next ozonolysis (repeat till you have cyclohexanone) looks like the best solution for me.


text:
Maximum force that hydrogen-hydrogen bond can withstand
For $\ce{H-H}$:

bond length is $74~\mathrm{pm}$
bond energy is $436~\mathrm{kJ~mol^{-1}}$

(Source: article chemical bond on Wikipedia.)
The University of Waterloo page that the article cites says:

The amount of energy required to break a bond is called bond
  dissociation energy or simply bond energy.

and

The bond energy is essentially the average enthalpy change for a gas
  reaction to break all the similar bonds.

So I'm assuming that as the average enthalpy change increases, so does the force that stresses the bond (and thus the forces on each hydrogen atom).
For a single molecule of $\ce{H-H}$, what is the maximum force that the bond can withstand before breaking?

You do understand that the bond dissociation energy is an energy. The magnitude of the force does not matter. Only the work done matters. You can (theoretically) use the tidal force from the black-hole itself on a hydrogen molecule for a tiny fraction of a second and it would not break.

@DHMO - I would disagree. Unless you're somehow applying a black hole symmetrically to the $\ce{H2}$ molecule, the high force is almost certainly applied differently to each atom, and will greatly exceed the force constant of the potential energy curve. There's a whole (new) area of mechanochemistry and we can clearly break bonds with applied forces.

DHMO seems to be saying that if you apply a huge force for a short enough time, the molecular bond can survive because not enough energy will have been placed into the atoms to break the bond. That is correct.

The force and the bond enthalpy are not necessarily directly related.
The bond dissociation energy is the amount of energy released when the bond is formed, or the energy required to break the bond.
The force constant of the bond derives from the curvature of the potential energy curve. For example, if we assume a harmonic oscillator (spring) for a bond, then:
$$ V(x) \approx \frac{1}{2}k_f \; x^2$$
Real bonds aren't purely parabolic, but clearly the second derivative of the potential energy curve near the equilibrium bond length will give the needed force constant.
Here I've plotted two fictional bonds with the same length. The blue curve gives a bond with a higher bond enthalpy and force constant (i.e., the curvature is higher than the red curve).

It should be clear that this isn't necessarily true. We can easily imagine some bond that has a shallow curvature but a deep enthalpy.
I should indicate that in many cases, these are related - stronger bonds typically have higher force constants and are harder to distort, but the point here is that the bond dissociation energy does not directly tell us anything about the mechanical strength of a bond.

One comment on the question - as the enthalpy increases, the molecule simply goes to a vibrationally excited state - increasing the bond length. I'm not sure I'd want to imply an intuition of forces occurring on the atoms. The enthalpy simply ensures the vibrational amplitude increases - it stretches more.

Hmm. So, to directly answer the question, would that mean that the force required is simply $k_\mathrm{f}\cdot x$ where $x$ is the displacement (away from the equilibrium bond length) that one wants to stretch the H-H bond length to? (Assuming harmonicity, of course.)

@orthocresol - Well yes, by unit analysis, force times displacement = energy. So yes, if you have a way to pull on a molecule with a set force over a displacement, you're putting energy into the bond and can break it. (For example, there are AFM "pulling" experiments.)


text:
Why would breathing pure oxygen be a bad idea?
My textbook mentions that SCUBA tanks often contain a mixture of oxygen and nitrogen along with a little helium which serves as a diluent.
Now as I remember it, divers take care not to surface too quickly because it results in 'the Bends', which involves the formation of nitrogen bubbles in the blood and is potentially fatal.
If that's the case, why not use pure oxygen gas in SCUBA tanks? It seems like a good idea since it would

a) Enable divers to stay underwater for longer periods of time (I keep hearing that ordinary SCUBA tanks only give divers a pathetic hour or so of time underwater.
b) Possibly eliminate the chances of developing 'the Bends' upon surfacing. Well, it seems plausible, that is if the diver were to take a 10 minute deep-breathing session with pure oxygen to flush out whatever nitrogen's there in his lungs before hooking up a cylinder of pure oxygen and going for a dive. So if there's no gaseous nitrogen in his lungs and blood, then he wouldn't have to worry about nitrogen bubbles developing in his system.

Now those two possible advantages aren't hard to overlook, but since no one fills SCUBA tanks with pure oxygen, there must be some reason that I've overlooked, that discourages divers from filling the tanks with pure oxygen. So what is it?
Also, I hear that the oxygen cylinders used in hospitals have very high concentrations of oxygen; heck, there's one method of treatment called the Hyperbaric Oxygen Therapy (HBOT) where they give patients 100% pure oxygen at elevated pressures.
Hence I doubt whether the increase in pressure associated with diving is the problem here. So I reiterate:
Why is it a bad idea for divers to breathe pure oxygen underwater?

I guess most of the recent answers have kinda missed a main point, so I'll rephrase the question:
Why is it a bad idea for divers to breathe pure oxygen underwater? If it is indeed due to pressure considerations as most sources claim, then why doesn't it seem to be a problem when patients are given 100% pure oxygen in cases like the HBOT (which is performed at elevated pressures) ?

You will be breathing the same air pressure, but it will all be oxygen. This will increase the amount of dissolved oxygen in your blood. It's a little bit analogous to putting a burning tissue into an atmosphere of pure oxygen, I would think. I don't know exactly what causes the toxicity, but I'd guess that it's related to that increased rate of oxidation.

You can breathe pure oxygen in low pressure without problem. The Apollo 1 used this, but after the Apollo 1 accident many things have changed

The amount of air-time which SCUBA tanks provide underwater depends on the depth at which you are breathing.  This can be much less than 1 hour for deeper dives.  The volume that is contained in your lungs does not change, but the pressure at which that gas is supplied depends on depth.  Thus, the static volume of gas contained in your tanks will provide a lower number of breaths at deeper depths (higher pressure). The time which you can spend at a particular depth without the need to decompress is also dependent on the depth.

There is a vast difference between what is considered safe when directly under the care and observation of a doctor in a hospital (Hyperbaric Oxygen Therapy) using medical grade equipment and what is reasonable for unsupervised recreational diving.

Not mentioned so far, but for long operations underwater you can use rebreathers. They work by recycling the nitrogen, scrubbing the CO2 and replacing it with fresh oxygen. Complex and risky - it introduces a whole new set of failure modes.

Hyperbaric oxygen therapy isn't safe for a healthy person! Oxygen is very toxic, and a brutal oxidising agent - it's dangerous enough breathing it at standard pressure (oxygen radicals are one of the main causes of genetic damage), and it only gets worse at higher partial pressures. HBOT is used on people who have some trouble with normal oxygen absorption - for example, due to embolism or carbon monoxide poisoning. Those problems make it harder for oxygen to attach to red blood cells, which means you need a higher partial pressure of oxygen to keep the blood oxygen content proper.

Im sorry, Why this question doesn't have an accepted answer?

The other answers here, describing oxygen toxicity are telling what can go wrong if you have too much oxygen, but they are not describing two important concepts that should appear with their descriptions. Also, there is a basic safety issue with handling pressure tanks of high oxygen fraction.
An important property of breathed oxygen is its partial pressure. At normal conditions at sea level, the partial pressure of oxygen is about 0.21 atm. This is compatible with the widely known estimate that the atmosphere is about 78% nitrogen, 21% oxygen, and 1% "other". Partial pressures are added to give total pressure; this is Dalton's Law. As long as you don't use toxic gasses, you can replace the nitrogen and "other" with other gasses, like Helium, as long as you keep the partial pressure of oxygen near 0.21, and breathe the resulting mixtures without adverse effects.
There are two hazards that can be understood by considering the partial pressure of oxygen. If the partial pressure drops below about 0.16 atm, a normal person experiences hypoxia. This can happen by entering a room where oxygen has been removed. For instance, entering a room which has a constant source of nitrogen constantly displacing the room air, lowering the concentration -- and partial pressure -- of oxygen. Another way is to go to the tops of tall mountains. The total atmospheric pressure is lowered and the partial pressure of oxygen can be as low as 0.07 atm (summit of Mt. Everest) which is why very high altitude climbing requires carrying additional oxygen. Yet a third way is "horsing around" with Helium tanks -- repeatedly inhaling helium to produce very high pitched voices deprives the body of oxygen and the partial pressure of dissolved oxygen in the body falls, perhaps leading to loss of consciousness.
Alternatively, if the partial pressure rises above about 1.4 atm, a normal person experiences hyperoxia which can lead to oxygen toxicity (described in the other answers). At 1.6 atm the risk of central nervous system oxygen toxicity is very high. So, don't regulate the pressure that high? There's a problem. If you were to make a 10-foot long snorkel and dive to the bottom of a swimming pool to use it, you would fail to inhale. The pressure of air at your mouth would be about 1 atm, because the 10-foot column of air in the snorkel doesn't weigh very much. The pressure of water trying to squeeze the air out of you (like a tube of toothpaste) is about 1.3 atm. Your diaphragm is not strong enough to overcome the squeezing and fill your lungs with air. Divers overcome this problem by using a regulator (specifically, a demand valve), which allows the gas pressure at the outlet to be very near that of the ambient pressure. The principle job of the regulator is to reduce the very high pressure inside the tank to a much lower pressure at the outlet. The demand valve tries to only supply gas when the diver inhales and tries to supply it at very nearly ambient pressure. Notice that at depth the ambient pressure can be much greater than 1 atm, increasing by about 1 atm per 10 m (or 33 feet). If the regulator were to supply normal air at 2 atm pressure, the partial pressure of oxygen would be 0.42 atm. If at 3 atm, 0.63 atm. So as a diver descends, the partial pressure of oxygen automatically increases as a consequence of having to increase the gas pressure to allow the diver to inflate their lungs. Around 65 m (220 ft), the partial pressure of oxygen in an "air mix" would be high enough to risk hyperoxia and other dangerous consequences.
Now imagine a gas cylinder containing 100% oxygen. If we breathe from it at the surface, the partial pressure of oxygen is 1 atm -- high, but not dangerous. At a depth of 10 m, the partial pressure of supplied oxygen is 2 atm -- exceeding acceptable exposure limits. This is a general pattern -- raising the oxygen fraction of diving gasses decreases the maximum diving depth.
And you can't lower the partial pressure much because the lower limit, 0.16 atm, isn't that much lower than the 0.21 atm of sea level atmosphere.
One general category of solutions is to change gas mixes at various depths. This is complicated, requires a great deal of planning, and is outside the scope of your question. But it is certainly not as straightforward as just simplifying the gas mixtures or just raising the partial pressure of oxygen.
Additionally, compressed oxygen is a relatively annoying gas to work with. It is not itself flammable, but it makes every nearby organic thing flammable. For instance using grease or oil on or near an oxygen fitting risks spontaneously igniting the grease or oil. Merely having grease on your hand while handling oxygen refilling gear (with a small leak) can burn your hand.

Whoa, nice answer! But would you also happen to know why patients subjected to HBOT (Which is at pressures exceeding atmospheric pressure) are given 100% pure oxygen? Yet again, thanks for the answer, it was pretty good! ^_^  Though it didn't answer everything in the question :/

@AaronAbraham: Some of the use-cases for HBOT are things that reduce blood oxygen (e.g. CO poisoning or lung injuries), but  I'm curious about other use-cases.  Maybe they monitor for signs of oxygen toxicity and stop the treatment if necessary?  Otherwise, maybe it's worth the risk or small amount of damage?  Maybe it matters that you're at rest during HBOT, not exercising by swimming around?  (breathing more shallowly?)

" It is not itself flammable, but it makes every nearby organic thing flammable." - define "flammable".

Grease on your hand does not ignite when it gets in contact with pure oxygen at ambient pressure. An oily cloth is something different, there the heat can build up.

So if air mixtures become toxic around 220 ft, do technical dives below that depth have to use lower oxygen mixtures? I know that there are dives that go that deep, but wouldn't they be risking hypoxia during their descent/ascent/safety stop?

@AaronAbraham : I am not a doctor, but my understanding from various readings is that tissue damage, especially to the lungs, can impede the uptake of oxygen leading to sub-normal oxygen uptake.  (The equilibrium concentration of oxygen in the blood falls because the source term, oxygen in the air, is situationally reduced.  This is vaguely like effusion.)  One solution is to push more oxygen at higher pressure to increase the source term and establish a higher concentration in the blood.  Note that CNS oxygen toxicity is a known hazard of HBOT.

@Karl : The oxygen escaping from a leak at a tank refilling station will not be at ambient pressure.  Typical SCUBA tanks carry 175-300 atm and are filled by compressors working with gasses up to around 350 atm, although this maximal pressure is usually only present in a small part of the compressor.  Jab in a leak at a random place, though, and you can expect to receive gas initially at 200 atm.  I haven't directly observed the effect of placing organic valve grease in such an environment, but I have seen the charred bits of a compressor that had been part of such an "experiment".

@JanDvorak : Don't really want to get pulled into that quagmire.  However, the NFPA placard for  oxygen lists Flammability: 0.  The USCG CHRIS code lists oxygen as "not flammable".  The DOT places compressed oxygen in class 2, "Non-flammable gas".  So, by appeal to various authorities, I'm comfortable calling (even compressed) oxygen "not flammable".  (Several of these note that oxygen can enhance combustion, a sentiment with which I heartily agree.)

@nworb99 : Deeper dives involve switching gasses to mixtures with lower oxygen partial pressure.  You may find the term "bottom gas" to describe the mixture to be used at maximal depth.  (Beware "nitrogen" narcosis, a different hyperbaric inert gas hazard not discussed in this answer.)  Trimix 12/52 (12% oxygen, 52% helium, 36% nitrogen) has a partial pressure of ~1.3 atm at 100 m (equivalent to air at 43 m).  (Although helium is not magic.  See high pressure nervous syndrome.)

@EricTowers Once it is out of the leak, it IS at ambient pressure. A running compressor has hot parts, that's a bit different from your 35°C fingers.

@Karl : Once it is out of the leak, it expands and cools while decreasing its pressure from 200 atm to 1 atm.  This is not instantaneous.  The rate is thermodynamically limited.  See Joule-Thomson expansion.  Additionally, I'm not interested in any argument that directly contradicts provided photographic evidence.

Oh, I am also not interested in this argument, im my case because it involves hearsay physics and anecdotal evidence. Good we agree on something. ;-)  Oh, you are also right that oxygen is not flammable.

I can't get what do you mean here: "If you were to make a 10-foot long snorkel and dive to the bottom of a swimming pool to use it, you would fail to inhale. The pressure of air at your mouth would be about 1 atm, because the 10-foot column of air in the snorkel doesn't weigh very much. The pressure of water trying to squeeze the air out of you (like a tube of toothpaste) is about 1.3 atm. Your diaphragm is not strong enough to overcome the squeezing and fill your lungs with air. " With an oxygen pressure of 1.4 atm, it is bigger than the external pressure, so inhalation will be possible...

@santimirandarp : There is no oxygen pressure of 1.4 atm.  There is you, trying to inhale.  Surrounding you is water at the bottom of a column of 10 ft. (about  3 m) of water, squeezing you (and especially your torso) from all sides with 1.3 atm of pressure.  At your mouth is a snorkel.  It contains air at the bottom of a column that is (depending on your definitions) 100 km + 3 m, so is at a pressure of 1 atm.  Your diaphragm has to expand against 1.3-1 = 0.3 atm compression.  Your diaphragm has to inflate your chest against 0.3 ton/sq. ft., which won't happen.

@EricTowers But you are describing a person with a pure oxygen supply at 1.4 atm, isnt it? that's what I get from the text; otherwise, why do you say 'So, don't regulate the pressure that high? There's a problem'

@santimirandarp : A person at the bottom of a swimming pool with a snorkel poking up in to the atmosphere is breathing air, not pure oxygen.  The problem this person experiences is that they cannot expand their lungs because of the difference in water pressure and air pressure.  It does not matter what the gas is, they cannot inhale.  This is just at 10 feet, at greater depths, the gas must be supplied at greater total pressure.

Yes I think I have problems with your English, I got that perfectly, thanks...

Our body is used to the environment around us. Once you change part of the environment, you have to be ready for the consequences.
Inhaling pure oxygen is the cause for what is known as oxygen toxicity.

Oxygen toxicity is a condition resulting from the harmful effects of breathing molecular oxygen $\ce{(O2)}$ at increased partial pressures.

High concentrations of oxygen result in a considerable concentration of various radicals, one of which is hydroxyl and can engage in a chain reaction that deteriorates lipids in cell membranes.

By Tim Vickers, after Young IS, McEneny J (2001). "Lipoprotein oxidation and atherosclerosis". Biochem Soc Trans 29 (Pt 2): 358–62. PMID 11356183.Vectorized by Fvasconcellos. - w:Image:Lipid peroxidation v2.png, Public Domain, https://commons.wikimedia.org/w/index.php?curid=1728531 Click for a larger version
Different settings have been proposed to minimize the damage of oxygen toxicity, one of which is hyperbaric oxygen therapy.

As a certified SCUBA diver, I learned that breathing pressurized pure oxygen leads to oxygen toxicity, which can be fatal. However, I'm not anywhere near an expert on the mechanism of oxygen toxicity, but I believe it has to do with resulting in a lot more reactive oxygen species which can cause oxidative stress and lipid peroxidation. I'm not really offering this as a good answer (since I don't know much about it) but more as an invitation for someone who does know more about it to expand on it.
Additionally, since the air we normally breathe is about 78% nitrogen and only 21% oxygen, it doesn't really seem healthy in my opinion to up that concentration of oxygen from 21% to 100%.
Here's a picture from Wikipedia that shows oxygen reacting with a lipid radical to create/propagate lipid peroxides, which cause cell damage.

At five meters depth (1.5 bar absolute pressure), pure oxygen becomes toxic after ten-15 minutes or so already.  After an hour or two, the effects can set in at snorkeling depth. That makes it rather not useful for any diving purpose. For continued exposure, the partial pressure of oxygen should be kept below 0.6 bar.
For medical use (hyperbaric medicine), high oxygen partial pressure can be useful, although I belive there is some discussion about how useful it actually is. Very few things in medicine have no drawbacks and dangers.

Maybe for medicine they use pure oxygen but not higher than a critical absolute pressure?

@santimirandarp High absolute pressure is cumbersome, because you need a slow decompression afterwards. Hence the use of pure oxygen at only moderately increased abs. pressure.

Yes, I just was saying because it possibly answer the last part of OP question...

I'm not a chemist but I want to chime in for the SCUBA part.
Normally recreational divers dive with compressed air (21% O2).
There is a special education to learn and use 32% and 36% O2 air mixures. This is called Nitrox (at least at PADI).
The benefit is you can dive longer and make more dives per day (since it displaces N2) but you have to dive shallower. This is popular on multi dive occasions like dive safaris on ships.
This is also not always avaliable on site.
Further, since no one mentioned it yet, the apparatus to provide and refill 100% O2 for recreational use would be quite dangerous.

Industrial/Technical divers who dive very deep use helium mixes and what not, to simply go very deep.
I heard that, military divers sometimes use very high O2 mixtures to go long distances very shallow (4m max) - but I'm not sure thats true.

Oxygen, especially pure oxygen likes to steal electrons.
By stealing electrons it "corrodes" other elements into their oxides. Think of iron rusting.
So, as we are composed of elements, too much oxygen for too long will increase our rate of oxidation, damaging our cells.

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


text:
Chalk and vinegar
So, I wanted to make calcium acetate, through the reaction of calcium carbonate and vinegar.
My proposed reaction:
$$\ce{2CH3COOH(aq) + CaCO3(aq) -> Ca(CH3COO)2(aq) + CO2(g) + H2O(l)}$$
It seems that there is not much reaction between the chalk (source of $\ce{CaCO3}$ and the vinegar (source of acetic acid), except intermittent particles of chalk floating to the top.
The vinegar I am using is white distilled vinegar, with a $95\%$ concentration. I diluted it because the amount of chalks I would be using would not fit the amount of vinegar that would come from the $95\%$ solution.
Quantitatively, I'm using $11.4~\mathrm{ml}$ of vinegar (before dilution), and $11.4~\mathrm{g}$ of chalk. I have left it for over a day, but still there has not been much apparent progress.
What should I do? I tried heating up the solution in hopes of increasing the rate of reaction, but calcium acetate is less soluble in hot water than in cold water.
I do not want to precipitate the calcium acetate out of the solution, and onto the calcium carbonate sediment at the bottom.

Calcium carbonate reacts with acids (including vinegar) quite readily and visibly. This means that at least one of your reactants is not what you think it is.

I've let it sit for a day by now. There is a rather noticeable change: the smell, once vinegar, has evolved to a coke-vinegar mix. I'm assuming the coke-ish smell is coming from the CO$_2$ which has been produced. Which, in that case, is a good sign.
Small bubbles have coated the side of the beaker. There is a reaction, it's just not readily happening.

Are you taking into account that vinegar only contains 4 to 8% of acetic acid?

I just searched online and found no information agreeing with that. Where did you get that from?

https://en.wikipedia.org/wiki/Vinegar#Distilled

I am strongly suspicious that you have 95% acetic acid. That would be very close to glacial, and that's certainly not how vinegar comes at the store.

@Variax Good finding - "distilled" vinegar has 5-8 % and after dilution it may be too weakly acidic. Also only 95 % in shop is alcohol ;)

I am assuming by chalk you meant you were using blackboard chalk. In which case Ivan's comment was close, but all the other comments have completely missed the issue. The reason you saw no reaction is because chalkboard chalk isn't actually chalk, it is typically gypsum which is chemically calcium sulfate ($\ce{CaSO4}$) which is inert to acid. Chalk the mineral is calcium carbonate and chalkboard chalk used to be made from this, but today the sticks for writing are made of gypsum. Calling them chalk is a holdover from the old days.
I too fell for this trap many years ago.

So what other source of calcium carbonate would be better? Maybe some egg shells or sea shells?


text:
What is the chemical structure of ceramic nonstick coating on pans?
There are lots of pans nowadays that have a ceramic nonstick coating instead of a PTFE coating. However I can not find anything online describing what kind of ceramic such nonstick coatings are. Does anyone have an idea what these nonstick ceramics are on a chemical level?

There is a near continuum (and thus a potential infinitude) of relative concentrations of the different components of ceramics and their means of fabrication, but US Patent US7879449 B2 (PDF) gives a thorough description of their own non-stick ceramic invention as well as descriptions and references to others.  Below are several excerpts from the cited patent:

The novel coating includes a $\ce{SiO2}$ ceramic matrix that includes the condensation product of hydrolyzed alkoxysilane and a colloidal silica sol. A series of new abrasion resistant and high hardness non-stick coating materials have been developed with ceramic matrix and organic compounds by sol-gel process.
This non-stick material has two main component of silicon dioxide. One is silica sol which is pure $\ce{SiO2}$, another is an organoalkoxysilane which is an organic-inorganic hybrid material. Both materials can make strong chemical bond from chemical reaction, condensation. $\ce{(OR')_{4-x}}$ groups are hydrolyzed with water and then make condensation with silica sol and each other. The ceramic non-stick hybrid coating is applied by a sol-gel process. Two kinds of non-stick organic compounds make a chemical bond with silica ceramic network or matrix and it can exist in the ceramic structure in a stable state.
The main matrix is composed of a network of silica formed by the chemical reaction or condensation of colloidal silica sol (water base) and an organoalkoxysilane such as MTMS (methyltrimethoxysilane). The colloidal silica sol is a very stable state because of the electrostatic repulsion of silica particle surface.
The organoalkoxysilane preferably has the general formula $\ce{R_xSi(OR')_{4-x}}$. Wherein $\ce{R}$ is one or more organic radicals chosen independently from among methyl, ethyl, and propyl, $\ce{R'}$ is a low molecular weight alkyl radical, and x is at least one and less than four. Commonly x is one, so that the organoalkoxysilane has three hydrolysable sites. This increases cohesion and bonding of the components and also provides a thickened layer and increases the links with the inorganic silica surface.
The surface of colloidal silica is coated by organoalkoxysilanes, in other words, hydrolyzed organoalkoxysilane make a thin layer or coating on the silica surface as a result of the condensation reaction with the hydroxyl groups of the silica surface.

Again, this is just an overview of a single process. The patent is well written and diagrammed and I suggest reading in detail to best answer your question (Images > Full Pages).


text:
What does 10 Vols mean?
I am currently working on a project and the protocol says the following, "Wash resin 3 times 10 vols of the extract buffer." This is very confusing to me and I am not exactly sure what the 10 vols part means. I tried to look it up, however, the few explanations I found were still confusing. I am new to research, and have not been exposed really to the "lingo".

Some experimental procedures are only good enough to be chucked at the heads of the people who wrote them for inclarity — your’s is an example.

I think you mean "wash resin 3 times with 10 vols of the extract buffer".
Vol means volume strength. It tells you how much of a certain element in milliliters will be RELEASED from a certain substance. In this case, your buffer is supposed to release 10 vol of something. That's all it means.
However, I am surprised about the usage of vol in this context. Volume strength is usually a term directly associated with hydrogen peroxide, and dealing with the release of a certain volume of oxygen.


text:
Semi-numeric or analytic Hessian for DFT (GAMESS)
I'm trying to understand what parameter in the $FORCE group is better to use in computing vibrational frequencies (NVIB=2) of compounds after optimization. I'm using GAMESS Aug'16 on Ubuntu 14/16.04.
Initially, three separate methods for both gas and water (PCM) are used to optimize. Later, higher levels of theory are used.

B3LYP/6-31+G*
ωB97X-D/6-31+G*
PBE0 (PBE1PBE)/6-31+G*

Of course, analytic is the default for B3LYP and PBE0, but for ωB97X-D, seminum must be used, from what I've found the hard way. Is it reasonable to use ANALYTIC for the former two and SEMINUM for the latter or should I use SEMINUM across the board to avoid misconstrued results? I'm not seeing a very large difference in only phenol but a difference is noticeable. Considering my target compounds are much larger than a single ring, they may become more noticeable.
Thank you for any input.

I think, yes, you should use SEMINUM for all your calculations.

Well, those are different functionals, result will be different anyways. What would be the reason for using it for every fnctional?
I would first try if the results are actually different between analytical and semi-numeric.

Another good resource for GAMESS (US) is the google group for GAMESS


text:
Who are some writers that also had careers in chemistry?
I love chemistry, but I am interested in so many other things too. I believe it's important to have a complete intellectual life, and it always makes me happy to find examples of people who took their love for science and created great art with it. Many would be surprised at how many great authors were also chemists. Here are a few of my favourites:

Isaac Asimov: Professor of biochemistry, Boston University School of Medicine
Kurt Vonnegut: biochemistry major, not a successful chemist though.
Robert M. Pirsig: biochemistry major

Can anyone help add to my list?

I'm voting to close this question as off-topic because it is not about chemistry.

got it. I wasn't sure whether or not this counted as "about chemistry", but I totally understand. Where is a good place to post this?

Within the chem.SE network, your best option is to come into chat. It's never a guaranteed answer, but neither is it guaranteed on main site. Otherwise, you'd have to look elsewhere because list questions are not on-topic here. I figure that /r/chemistry would quite like this question.

Before this gets closed: Carl Djerassi, Primo Levi ;)

I liked the question! Although, everyone else is right when they say it shouldn't have been posted here :/ [ P.S-  Danke @KlausWarzecha ]

@KlausWarzecha Carl Djerassi seems interesting! Primo Levi also got mentioned on the reddit thread. Also; What about Oliver Sacks? I know he's not a chemist per se. He was a neurologist, but a he was also a very enthusiastic amateur chemist, and he wrote a book about it: Uncle Tungsten. I haven't read my copy yet, but one thing I will say is that the hardcover edition is a beautiful print. I'm looking forward to reading it.

@gannex I don't recall that I ever read anything by Oliver Sacks, but I remember that I enjoyed Cantor's Dilemma by Carl Djerassi a lot. Love and despair, academic misconduct,  the wonderful life of postgraduates, weird aspects of the peer-reviewing system and academic life... Anyway, all of this doesn't belong on Chem.SE, and probably not on academia either. orthocresol is probably right with reddit! Btw, if you can't wait for the novels, have a look at the real life on  http://retractionwatch.com/ ;)


text:
Chemical Reactions at the Cathode of the Leclanche Cell
In the dry cell, the overall reactions that occur are:
$$\textbf{anode:}~~~ \ce{Zn -> Zn^{2+} + 2e^-}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$$
$$\textbf{cathode:} ~~\ce{2NH4+ + 2MnO2 + 2e- -> Mn2O3 + 2NH3 + H2O}$$
I have read that the overall reaction at the cathode actually consists of two separate reactions. However different sources have conflicting chemical equations for these two reactions.
Some sources says that the hydrogen is ammonium gets reduced first and then there is a separate redox reaction between the manganese dioxide and hydrogen gas:
$$\ce{2NH4 + 2e- -> 2NH3 + H2}$$
$$\ce{2MnO2 + H2 -> 2Mn2O3 + H2O}$$
While other sources say that the manganese first gets reduced and then the produced hydroxide ions are neutralised by the ammonium:
$$\ce{2MnO2 + H2O + e- -> Mn2O3 + 2OH-}$$
$$\ce{NH4+ + OH- -> NH3 + H2O}$$
Which of these reactions are correct?
Also, if the first set of equations are correct, does the redox reaction between the manganese dioxide and hydrogen gas also contribute to the voltage produced by the battery, or is it only the the redox reaction between the ammonium and zinc that contribute to the $1.5\ \mathrm{V}$ that is produced.


text:
Pyro- Prefix in Chemistry
I was wondering is someone could please help answer this question for me. I was recently reading the ingredients on my toothpaste container, and one of the compounds in it was 'Sodium Pyrophosphate'; what does the 'Pyro-' prefix mean? I see it elsewhere but I've never come across such a nomenclature such as this. What does it mean in terms of Chemistry?

It means that there are two phosphates connected together under loss of water, i.e.:

If you have any questions about the definitions of chemical nomenclature, have a look at the IUPAC gold book. See here. It basically says, that a pyro compound is one that is formed by heating another compound, usually under the loss of simple molecules like $\mathrm{H_2O}$ or $\mathrm{CO_2}$.

So sodium pyrophosphate would be the sodium salt of the above acid ; )


text:
Significant figures in pH calculations: does pH 0.30 count as 2 or 3 significant figures?
Just say I wanted to give a pH calculation to three significant figures.
Eg: the $\mathrm{pH}$ of a $0.500 M$ solution of $\ce{HCl}$ is $-\log(0.5) = 0.3010299957...$
Would 3 s.f. be $0.301$ or $0.30$?
I know normally when zero comes before a number, it isn't considered a significant number (i.e $0.00034$ and $0.34$ both have two significant figures). But since pH is a scale, can I treat $0.30$ as three significant figures?

Why do you think, it being a pH scale has to use different significant figures?

you need to use normal convention

Short answer, no: it doesn't matter that it's a 'scale', you handle its sigfigs exactly as you would any other numerical quantity. Also, beware: never in my experience have I differentiated numerical things by whether they were "a scale" or "not a scale". You may have picked up a false notion there somewhere.

This is a very tricky question, because pH is a log value. So, only the mantissa means anything in terms sig figs. You really need error bars, but I would go with 2. Importantly, note that 4.30 pH has the same sig figs as 0.30 pH.

I hope my newly edited answer helps. I also come to the conclusion that 2 decimal places are fine assuming that the error is $\pm$ 0.005 M : )

@Zhe thanks, that really helps. What you said about 4.30 pH having the same sigfigs as 0.30 pH makes sense to me, as in this case the zero isn't just a placeholder while in 0.30 mol it would be

Short Answer:

The number of significant figures in the mantissa of a value expressed in scientific notation equals the number of significant figures to the right of the decimal in the logged value. So in your example, assuming the likely error is the last digit, I'd use two.

Now back to your question, you have $[\ce{HCl}]=0.500$ assuming, the error is in the last digit, I'd recommend using 2 significant figures, $\pm$ error in the last.
Additionally, just because pH is a scale doesn't mean you can treat $0.30$ as a number with three significant figures, it still has two.
The general principles are given below.
Concepts like this are best illustrated with an example.
Consider, a random number $0.0000273$ Let us write this in scientific notation for the sake of clarity, $0.0000273 = 2.73×10^{–5}$ We have three significant figures.
Now take the log on both sides
$\log(0.0000273)= \log(2.73×10^{–5}) = \log(2.73) + \log(10^{–5})$
Now let us examine the two terms separately,
First, $\log(2.73) = 0.436$: the answer has three significant figures, reflecting the possible error in the last digit of $2.73$
$\log(10^{–5}) = -5$  This is an exact number and has infinite significant figures, i.e there is no error.
Going back, $\log(0.0000273)= \log(2.73) + \log(10^{–5}) = 0.436 -5.0000...$
In this situation the rules for significant digits for sums, dictate that we can only add to the same decimal place as the value with the least significance. In this case, that is the third decimal place from 0.436, so
$\log(0.0000273)= \log(2.73) + \log(10^{–5}) = 0.436 -5.0000... = -4.564$
$\log(0.0000\overline{273}) = -4.\overline{564}$  (the overline indicates the 3 significant figures)
End of example. Now back to your question, you have $[\ce{HCl}]=0.500$ assuming, the error is in the last digit, I'd recommend using 3 digits after the decimal, $\pm$ error in the last. So, $0.301 \pm 0.00\epsilon$

Answer is saying that OP has "pH=0.500", but instead [HCl]=0.500.

@DavePhD Thanks for your comment, sorry my mistake. Careless, careless me

thank you so much for all the time and effort you must have put into this great answer!

If 0.500 means "between 0.495 and 0.505" then
-log[HCl] is between 0.305 and 0.296
so statistically 3 digits beyond the decimal point should be used.  The uncertainty in the third digit is almost identical in both values ([HCl] and -log[HCl]).
Systematic error from approximating pH as -log[HCl] will be much greater. 0.3 is wrong by more than 0.1 pH units because the activity coefficient of 0.5M (H+, Cl-) is about 0.75.


text:
Dipole Moment of Normal Water vs Heavy Water
The question is in the title itself.
My guess: Higher for Normal water.
My Reasoning: Internet told me Deuterium is more electronegative than Protium. So,  there should be less difference between the electronegativity values of $\ce{D}$ and $\ce{O}$. Consequently, the dipole moment for $\ce{D2O}$ should be lower.
Is this a valid reasoning or should I take into account factors like lesser vibration amplitude of $\ce{D}$ than $\ce{H}$ that forms more stable bonds.

Dipole moment= Delta × distance
So the more the distance the more dipole moment. H2o have more dipole moment due to greater distance, so more vibration.

There is no measurable difference.
HOH 1.8546 ± 0.0006 D
DOD 1.8558 ± 0.0021 D
Dipole moment of water from Stark measurements of H2O, HDO, and D2O J. Chem. Phys. 59, 2254 (1973)

they are indeed very close to each other. But if I am to predict theoretically, how would I predict that? My reasoning seems to not match witg the data

@AmritanshSinghal you could think about bond length differences and bond angle differences, in addition to what you were already thinking about.  But I think it's completely unreasonable to decide which factor predominates when there is no measurable difference.


text:
Why do not metals gain electrons and non-metals donate electrons?
My teacher said that metals lose electrons and non-metals gain electrons to complete the noble gas configuration because of stability as it consumes less energy but why are systems having high energy less stable as compared to systems having less energy.
Why do metals and non-metals lose and gain electrons respectively instead of gaining and losing them respectively?

Why, they do, just not that often. As for the reasons, for example, look at sodium: how many electrons it has to lose to achieve the noble gas configuration? How many electrons it has to gain, if it would go the other way?

Because being metallic and rather giving up electrons than taking in more has the same fundamental reasons, which your teacher has not (yet) bothered to explain. Also it's not a law, but rather a rule of thumb.

First, it is a high-school rubbish that was kept only because of lack of desire to revise basic courses.
Can metal gain electrons? Certainly, YES. Sodium has several compounds with negative oxidation state, many transition metals have extensive chemistry where the metal is in formal negative oxidation state and so forth.
Can nonmetal lose electron? Certainly, as in nonmetal-nonmetal compounds it is unavoidable.
So, what is it really about? What is the difference between metals and non-metals?
Before we continue, we have to revisit definition of what metal is.
Metallic solid (sometimes shorthanded as metal) is a solid with metallic conductivity. It arises from having continuous half-filled orbitals allowing electrons to move freely. Metallic solid perfectly can be a compound (say, $\ce{Ag2F}$). Furthermore, many non-metals under heavy pressure undergo transition into metallic phase.
However, when we are talking from chemical PoV, metal is an element. But what kind of element? The problem is, that the definition of metal in chemistry has a history, and the term was introduced long before many of metals known today were discovered. As such, only common metals were considered and no exotic compounds was known, and certainly before extremely high pressures became available.
Consequently, a metal was characterized by its ability to lose electron (but many elements are capable of doing so) and having metallic conductivity when in form of simple compound. The border cases were either discarded or counted as 'metalloids' - a special kind of non-metal.
For example, tin has metallic and non-metallic allotropes at ambient pressure. Oups, is it a metal or non-metal ? Well, legally it is considered as a metal. However, antimony, with pretty much the same case, is typically considered as metalloid.
There is however a strict difference between tin and antimony in than tin is capable of forming normal salts and having a basic oxide, while antimony has only slightly acidic oxides. The reason for discrimination DOES exist. On the other hand, some heavy transition metals, such as rhenium and tungsten do not have true basic oxides.
So, it would be best to define metals by enumerating them. It is, however, easier to enumerate non-metals: commonly recognized non-metals are boron, silicon, arsenic, tellurium, iodine, everything to right and up of that elements and hydrogen. Optionally, germanium and antimony may be included. (Polonium, astatine and some others are questionable cases, since their chemistry is virtually unexplored thanks to their high radioactivity. But for that reason they can be safely ignored)
Said elements have some things in common: they have relatively high electronegativity, form covalently bound or molecular solids, form acidic oxides (if form at all) and do not form simple cations stable in water. All this comes from them having relatively high number of electrons in valence shell and tight binding of said electrons.
By no mean you should assume that sodium, for example, is eager to lose an electron. Nope, this process results in energy consumption. Only subsequent stabilization of by electron affinity of its partners and ionic packing results in net energy gain. On the other hand, addition of electron to an atom often results in slight energy release.
TL; DR. Non-metals typically has compact electron shells tightly tied to their nucleus and as such are uneager to loose them.

Since you mentioned "metallic solids", are there any examples of solids that exhibit such properties, like free delocalized electrons, but aren't actual elemental metals themselves? Can electrides satisfy this condition?

@PrittBalagopal Yes, plenty. One of the simplest would be $\ce{VO2}$ . It is also common to induce metallic conductivity by impurites, such as in $\ce{(In,Sn)O2}$ (indium-tin oxide) films. Molibdenum (and tungsten) bronses  $\ce{K_x(Mo/W)O3}; 0<x<1 $ also qualify.   AFAIK, solid electrides do not qualify.

Thanks for this info. Is there anything that's organic and shows metallic conductivity at the same time?

It is very rare for organics. To behave as metal, a solid needs a continous network of partially filled orbitals. Most organics is molecular, not having such network. Polyacetilene doped with iodine does show electrical conductivity, though, and I guess some other polymeres may have it too.

You should only use that definition in the context of ionic compounds. To form ionic compounds, you typically want to ‘transfer electrons completely’ (high school terminology) from one element to the other and thereby create charged compounds.
Typically, these charged ions are energetically least unfavourable if they display the so-called noble gas configuration — valence electrons like the closest noble gas. Now if you remember that the periodic table ‘wraps around’ (sodium is the element following neon), it may make sense that it is easiest for some elements termed metals to give off electrons to achieve this state, while others, termed non-metals, have an easier time gaining electrons for the same purpose.
This, however, is as permeakra verbosely wrote, a crude generalisation and only valid in the high-school ionic-bond context.