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Boltzmann Distribution in Molecular Dynamics Simulation?
The Boltzmann distribution is often used to describe the population of energy states at different temperatures, i.e. something like
$$\dfrac{g_i}{g_j} = \mathrm{exp}\left(-(E_i-E_j)\times \frac{1}{k_B\cdot T}\right).$$
For two given states $i,j$ with energies $E_i,E_j$, we thus have the population given by $g_i$ and $g_j.$
So if I have some observable $\sigma$ calculated for each state $i$, the natural way to calculate the observable of the ensemble would be
$$ \sigma_\mathrm{ensemble} = \left(\sum_i g_i \times \sigma_i\right) / ~\sum_i g_i~.$$
So, if I calculate $n$ ground states, let's say by geometry optimization of several guess structures, I employ the above weighting scheme to calculate the observable for the ensemble.
But here is my problem: If I perform an MD simulation instead, then the states should already be automatically populated according to the Boltzmann population (e.g. higher energy states are less likely to be observed in the MD simulation). So the population of the MD simulation already converges to the Boltzmann population for increasing simulation times $t$
$$ \lim_{t\rightarrow\infty} \frac{g_i^{MD}}{g_j^{MD}} = \frac{g_i}{g_j} $$
Now I am confused, because that would basically mean that there is a continuous transition between e.g. two states that clearly have to be boltzmann averaged up to the "long MD simulation", that is automatically populated accordingly. Where in the latter case, I would accidentally apply the Boltzmann population weighting twice.
Are there any resources on this weird population dichotomy or am I just overlooking something?

What two states do you mean? If you fabricate a system that is already in equillibrium and run an MD on it, (assuming the MD is good,) nothing significant will happen any more.

Within statistical mechanics (SM) a molecular property $X$ is computed by $$\left<{X}\right>_{SM}=\sum^{states}_i X_i p_i$$where $X_i$ is the value of $X$ for energy state $i$ and $p_i$ is the probability of being in energy state $i$ with energy $E_i$:$$p_i=\frac{e^{-E_i/kT}}{\sum_i e^{-E_i/kT}}$$Within molecular dynamics (MD) the corresponding property is computed by$$\left<{X}\right>_{MD}=\frac{1}{M}\sum^M_i X(t_i)$$where $M$ is the number of time-steps and $X(t_i)$ is the value of $X$ at time $t_i$.
You can also compute $\left<{X_{SM}}\right>$ from an MD by finding all the "unique" states (e.g. conformations) and computing their energy.  "Unique" means that even if a state is observed more than once in the MD you only include it once in the sum over states.
In principle $$\left<X\right>_{SM} = \lim_{t \to \infty} \left<{X}\right>_{MD}$$ but this assumes that the method used to calculate the energy and forces are identical in both calculations.
For example, for MD explicit solvent calculations with periodic boundary conditions it not really possible to calculate the energy of a state ($E_i$).  Also, vibrational effects must be included in $E_i$, which is typically done using the harmonic oscillator approximation, while the MD will include anharmonic effects. Similarly, the harmonic oscillator approximation is quantized (QM) while the vibrations in the MD simulation are not (classical).

Thanks for the very detailed and informative answer : ) From that I conclude that I can also Boltzmann average over the states of the MD simulation -- provided that the states are "different enough" (I remembered that there are some difference metrics for 3D geometries). But I guess I will stick to the following approach then: just compute the property  over t and watch until $\lim_{t\rightarrow \infty}(_{MD})$ converges.

This question is actually very broad as the answer can change quite a bit based on what kind of thing you are doing with your MD simulation.
The first thing to note is that the Boltzmann distribution only applies at thermal equilibrium. So if you start an MD simulation from a non-equilibrium state and then try to calculate the average of some property as you suggest, you will notice that the value behaves very erratically and then eventually converges to the expected value. This convergence basically means the system equilibrated so that the Boltzmann distribution actually is obeyed.
Another thing is that it is not true that a system will be immediately obeying a Boltzmann distribution just because you run an MD simulation. Perhaps you are thinking of the fact that many MD codes can begin a simulation by seeding the velocities of the particles according to a Maxwell-Boltzmann distribution. This is by no means required, but is useful because it generally gets you closer to equilibrium than if you just started with velocities from zero. The reason you can't just start from a Boltzmann distribution is because the Boltzmann distribution assumes a knowledge of the states of the system. Generally these states and their properties are what you are trying to find. All the MB distribution asks for, however, is a temperature and the masses of the particles. These are user-defined so no problem there.
A third note is that this answer somewhat depends on if by MD you mean classical molecular dynamics for quantum molecular dynamics. The Boltzmann distribution is somewhat more useful for quantum systems because quantum systems have states of discrete energy which works very naturally with the Boltzmann distribution. That's not to say that you can't use the Boltzmann distribution for classical, but it is somewhat less natural because in order to get meaningful statistics on the dynamics of the higher-energy state you will have to use a constant energy ensemble. And even then, how constant of energy is needed as too tight a constraint might force very non-physical dynamics?
To be extra clear about the main part of your question, you say:

But here is my problem: If I perform an MD simulation instead, then the states should already be automatically populated according to the Boltzmann population (e.g. higher energy states are less likely to be observed in the MD simulation).

The misunderstanding is that an MD simulation does not guarantee higher energy states are less likely to be observed for short simulations times. If you start very far from equilibrium (because how are you going to guess the "equilibrium structure" of a liquid for instance) then you might wander around in high energy states for quite some time before thermal equilibrium is found. This is most definitely not a Boltzmann distribution and the states will not be Boltzmann averaged (and won't follow MB statistics). Everything you say after this is fine, but you implicitly assumed that we started from equilibrium when in fact we do not.
If there is anything more specific I can try to clarify, or if I am missing the heart of your question, please comment and I will edit.

One could of course just monitor the convergence of the observable over the simulation time. I am just wondering if it very wrong to apply a Boltzmann averaging to a short MD simulation.

Well the problem is you will have to choose a size of the bin for the energies because otherwise you will never sample two states of the exact same energy and your statistics will be completely meaningless so you won't be able to calculate an observable.


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Qualitative explanation of how COSY works
I think I understand 1D Proton NMR with the Fourier Transform, qualitatively.  However, I am struggling to understand 2D COSY NMR.
I am stuck on a few things;

In 1D Proton FT-NMR, the "Pulse" I understand to be the application of a transverse magnetic field, applied at a mixture of frequencies simultaneously, like "noise".  This is so that the chemically different protons in the molecule, who each have a different precession frequency, can all resonate from a single pulse.  Then, we turn off the transverse field and measure the FID at a certain sampling rate.

So my first question is - in 2D COSY NMR, are both "pulses" the same?  i.e., do we subject the sample to a "noise" pulse, then a delay, and then a second "noise" pulse, and then measure the FID?
I think we do, because as far as I have read, the only thing we are varying in the second dimension in the experiment is the length of the delay between pulses.

So to my mind, in 2D COSY, during the delay time, the protons are undergoing FID the same as they do in 1D NMR, even if we are not measuring it at this time.

How, then, does varying the delay time lead to the discovery of "correlation" between protons?  What is it about applying a second pulse to partially relaxed protons, and then measuring the FID, that leads us to discover how the protons are correlated, and to the presence or absence of cross-peaks in our 2D spectrum?
If anyone has a link that explains this, that would be fine.  All my searches on the Internet so far just talk about how to interpret the COSY spectrum, with nothing about how/why it actually works.

The wikipedia articles on COSY and Magnetization transfer answer your questions.

In the 1D-proton experiment a hard 90° pulse is applied to the sample before the FID is acquired over some time, t. This set of events is known as the ‘pulse sequence’, and the resulting data is in the time domain, and needs to subjected to a Fourier transform in order to get a 1D spectrum in the frequency domain.
2D experiments are measured much in the same way, except instead of running the pulse sequence once, we run it multiple times, changing some variable and collecting all of the data into one set which will later become the 2D spectrum. [*]
The  standard COSY experiment is one of the simplest 2D experiments to understand. It’s pulse sequence differs from the 1D-proton experiment only by an additional hard 90° pulse. We can explain most (though not all) properties of the COSY experiment using a vector model with no mathematics.
—
Consider a single magnetisation vector, M, which isn't coupled to anything. $\ce{CHCl3}$ would be a good example of this, and leads to a single cross-peak along the diagonal of the  resulting COSY spectrum (a trivial result, but easier to explain).

The experiment consists of 4 steps:

An initial 90° pulse is applied, which transfers M from the z-axis down to the y-axis. This is no different to the pulse used when acquiring a 1D-proton spectrum.
M is allowed to precess for some time, t1. During this time, M ends up somewhere in the xy-plane (having both an x and a y component). The exact position will depend on the length of t1, which is the variable we will differ during the acquisition of our COSY.
A second 90° pulse is applied. This 90° pulse transfers the y-component back onto the z-axis (-z to be precise). The x-component is unaffected (this requires more than a qualitative explanation).
Immediately following the second 90° pulse, we detect. The FID is acquired during t2, which remains constant (just as multiple scans of a 1D proton are acquired for the same length of time). What we are actually acquiring is the residual magnetisation in the x-axis (which wasn't affected by the previous 90° pulse), similar to the standard 1D-proton experiment. If we applied a Fourier transform at this stage, we would indeed just end up with a 1D spectrum. FID in memory to be combined with subsequenc
The above procedure (steps 1-4) is repeated, changing t1 and keeping all other variables constant. For each value of t1, the FID is recorded during t2 - this is the basis of the COSY experiment. The data is combined into a single set, and a Fourier transform applied twice which gives us the 2D-COSY spectrum in the frequency domain. In this simple case, we'd end up with a single cross-peak along the diagonal corresponding to the chemical shift of the proton (in $\ce{CHCl3}$) as expected:

In more complex systems, where multiple protons are present, we observe additional cross peaks in the COSY spectrum off the diagonal corresponding to the coupling between pairs of spins.
The appearance of these additional cross peaks cannot be explained qualitatively using a vector diagram (as per above), but is a consequence of the second 90° pulse that we apply prior to acquisition of the FID.
Where coupling exists, the second 90° pulse doesn't just change the position of the magnetisation vector M, but also causes a population change for the transitions of the spin system through a phenomenon known as coherence transfer.
[*]: There is some potential confusion here, as even in a 1D experiment, we typically run the pulse sequence multiple times in order to improve signal:noise ratio. In this case, all of the variables are kept constant. In a 2D experiment, we run the pulse sequence multiple times, changing some variable in order to build up a 2D data set. We may also run the set of pulse sequences multiple times, to improve signal:noise ratio as per the 1D experiment.

A bit of pedanticism first. Some of your terminology usage is inaccurate, I'd just like to take the opportunity to clear some of it up.

Nuclear spins do not "undergo a FID", they undergo free precession. The free induction decay (FID) is the signal that is measured when spins undergo free precession.
"Noise" in spectroscopy refers to unwanted, random signals arising from processes apart from the magnetisation of the spins. It's exactly analogous as the crackling noise you get if you try listening to a radio underground. The music you want to hear is the signal, but the crackling sound is the noise.
From what I understand, you are referring to a pulse that is unselective in terms of the frequencies it can excite. Such a pulse is generally called "strong" or "hard", because it obeys the condition $B_1 \gg \Delta B_0$ (the strength of the pulse is much larger than the reduced external magnetic field).

Many 2D experiments are designed to measure correlations between two nuclear spins, commonly denoted I and S. In the case of COSY both I and S are the same nucleus (typically $\ce{^1H}$, but there are versions with other nuclei). The correlation that is being detected is through-bond coupling.
The pulse sequence of COSY looks easy, but the maths is actually more complicated than it is for something like HSQC. So, if you don't want a mathematical answer, quite a bit of detail has to be omitted. The best non-mathematical way I can explain it is as follows:

The first 90° pulse leads to excitation of spin I.
During the $t_1$ period, spin I undergoes free precession. As $t_1$ is increased, spin I will precess through a larger and larger angle. The rate at which spin I precesses is related to its resonance frequency. So, since we repeat the experiment for many different values of $t_1$, information about the resonance frequency of spin I is encoded in the data set we obtain.
Up till now, it is mostly the same as in a 1D experiment. The difference of course lies in the second 90° pulse. It turns out that the combined effect of the $t_1$ period, as well as the second 90° pulse, also leads to some transfer of the excitation from spin I to spin S. This transfer only occurs if spins I and S have a through-bond coupling.
In the detection period, both spins I and S have been excited and will therefore precess at their respective resonance frequencies. Both can be detected simultaneously (since they are the same nucleus).

Overall, what information do we have? From the $t_1$ incrementation, we know about the resonance frequency of spin I. This means that, after Fourier transformation (which converts from a time domain $t_1$ to a frequency domain $\omega_1$), we will have a peak, centred at $\omega_I$ (where $\omega$ is the resonance frequency).
From the FID obtained in $t_2$, we know about the resonance frequencies of both spins I and S. So, in the $\omega_2$ dimension we will have two peaks, centred at $\omega_I$ and $\omega_S$.
In the 2D spectrum, then, we will see two peaks. One is centred at $(\omega_I, \omega_I)$, and is known as the diagonal peak. The other one is centred at $(\omega_I, \omega_S)$, and is known as the cross peak. Now, the cross peak only appears if there has been transfer of magnetisation from spin I to spin S, which in turn can only occur if spins I and S have a through-bond coupling.
"But wait! Aren't there peaks at $(\omega_S,\omega_I)$ and $(\omega_S,\omega_S)$?" Yes, absolutely. That's because the first 90° pulse also leads to excitation of spin S; after all, spins I and S are the same nucleus, and an unselective pulse necessarily excites both. In an exactly analogous manner to that described above, this gives rise to the other two peaks (you can just switch the labels I and S in my description to see how this happens).

Finally, to answer some of your questions directly:

in 2D COSY NMR, are both "pulses" the same?

Yes, they are exactly the same. (This depends on how advanced an answer you want, though. For the purposes of 2D data processing, e.g. the States method, the phases of the two pulses may be different - i.e. one may be aligned along the x-axis, and the other along the y-axis. But I suppose you can ignore this for now.)

in 2D COSY, during the delay time, the protons are undergoing FID the same as they do in 1D NMR, even if we are not measuring it at this time

As I mentioned above it is free precession and not FID, but yes, this is exactly correct!

How, then, does varying the delay time lead to the discovery of "correlation" between protons?

The thing about varying the delay time, is that we are trying to measure frequencies. These can be either resonance frequencies (i.e. chemical shift), or coupling frequencies (recall that coupling constants are expressed in Hz).
I offer you the analogy of a clock. Let's say you have a clock, and you want to measure the frequency at which the minute hand rotates. If you simply look at the clock for one instant and record the position of the minute hand, you can't tell how fast it is rotating - or whether it is even rotating at all! You need to look at it continuously over a period of time, or you need to look at it at constant intervals and jot down how its position changes with time.
Using only one value of $t_1$, then, is analogous to just looking at the "nuclear clocks" once. Only by using multiple values of $t_1$, can one figure out the frequencies at which the nuclear clocks operate (i.e. the coupling constant) - or whether they are even operating in the first place (i.e. whether they are coupled).

What is it about applying a second pulse to partially relaxed protons, and then measuring the FID, that leads us to discover how the protons are correlated, and to the presence or absence of cross-peaks in our 2D spectrum?

I hope the above discussion was sufficient. And if you want more detail I'm afraid that you need to go into the maths.

When I wrote "noise", I was thinking in terms of "white noise" sound, which is an equal mixture of all audible frequencies.  I thought that the pulse was similarly a mixture of all frequences across our frequency range of interest, so that it resonated with all protons at the same time in modern FT-NMR.  Maybe based on your answer, this is not the case and this aspect is different.

I understand J-coupling in 1D NMR to be based on the possibilities of neighbouring protons being either aligned or anti-aligned with the externally applied magnetic field, and thus either slightly increasing or decreasing the effect of this field on our proton of interest, leading to peak splitting.  It is not clear if this is the same mechanism that leads to "through-bond" coupling in COSY NMR.

Regardless of what the mechanism of coupling is, I am unclear on the reason it gives cross peaks, or rather I am unclear on why cross peaks do NOT occur when the protons are not coupled.  When the transfer of magnetisation occurs from I to S, I assume S continues to precess at its own characteristic frequency.  Is it something to do with I influencing S's phase that leads to the cross-peak?

Firstly regarding the noise bit, if you're interested you should ask a new question, otherwise this comment chain will become too long. I wasn't really correcting your understanding of pulses, it was just the term you used. Ok, as for the cross peaks. You are right, that S will always precess during $t_2$. However it is not simply any S-signal that becomes a cross peak. It has to be magnetisation that precesses at $ω_I$ during $t_1$, and then precesses at $ω_S$ during $t_2$. That's why it is critical that between $t_1$ and $t_2$, there has to be this magnetisation transfer.

Now if I and S are not coupled, all the magnetisation that precesses at $ω_I$ during $t_1$ will simply continue to precess at $ω_I$ during $t_2$. That gives rise to one diagonal peak. Likewise, any magnetisation that precesses at $ω_S$ during $t_1$ will simply continue to precesses at $ω_S$ during $t_2$. That's a second diagonal peak. But since there is no transfer, you cannot get the cross-peaks. / And lastly, through-bond coupling is the same as your J-coupling. It's just expressed this way to emphasise a difference from through-space coupling (important for NoE)

Two pulses are applied to your sample, with a distance $\tau$ between them, which gets incremented by $\Delta\tau$ in, lets say, 128 steps, starting at zero, or, more realistically (because you cant apply two pulses at the same time), at $\Delta\tau$.
There is an FID after each first pulse, which is discarded, and one after the second pulse, which gets recorded. In the end, we have 128 FIDs, each with e.g. 512 points.
Now for COSY, it is elementar that the FID is recorded in quadrature (all modern spectrometers do that), where each peak not only has an intensity, but also a phase, or, mathematically equivalent, a complex signal consisting of a "real" and an "imaginary" part. If you look at 1D NMR spectra, you usually only show the "real" part, and you have to phase the spectrum beforehand.
The absolute phase has no real physical meaning, but from a pulse on a totally relaxed sample, the phase of all peaks is (on an ideal spectrometer) the same, and we wilfully shift it to zero and call the real part "absorptive". Every additive part of the signal is a damped cosine, with slightly different frequencies, and so they all start at intensity $1$ x (number of protons)
Now if you apply the (second) pulse to a sample which is already showing an FID, each signal has a different phase, depending on how far the first FID had evolved yet. If you put all the 128 1D spectra behind each other, you see that the absorptive signal of each peak oscillates between full intensity, zero, negative, and back. (The "magnitude" signal of all 128 would look perfectly the same, just getting a bit weaker.)
They oscillate with the same frequency as in the 1D direction, so if you do the 2D FT, you get the diagonal peaks, and the projection in both directions gives you the ordinary 1D spectra.
Now if two signals (protons) are coupling, they each split into e.g. a doublett in 1D, and in 2D you get those notorious additional crosspeaks, because the magnetisation of each of them feels a coherent modulation with the lamor frequency of the other. The stronger the coupling, the stronger the crosspeaks, and no coupling, no crosspeaks.
If the modulation is close to the actual Lamor frequency, btw., you get those $AB$ spin systems. The modulation can now induce transitions, and so the two protons depopulate certain spin states in each other, depending on their own. In the end (no chemical shift difference), you get no visible coupling at all.

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Relationship between Energy Level and electron position
In Bohr's Model of the atom there are set energy levels that are used to show the energy of the electrons in the atom. In the quantum model there are orbitals that show a probable location of the electrons within the atom.
How does the changing energy level of an electron affect it's position in the atom? For example, Helium has two electrons that both fit in the first "s" orbital. When the Helium electrons absorb the correct quanta of energy they would jump up to the next energy level, but what happens to their position in their orbital? As I understand, they can't move to the second "s" orbital, right, so would their distance from the nucleus just become greater while remaining in the first "s" orbital, or is its change in energy level a difference in the oscillation of the electron within its orbital?

There are a few problems here. One is that it only works for one-electron systems, so it can't be applied to hydrogen. Another is that the position of electrons isn't exactly known, so we need to work with probabilities instead. If you know that an electron is excited from one orbital to another, you can calculate and compare the probabilities for the two orbitals at a given radius from the nucleus.

@pentavalentcarbon guess u meant 'Helium'

The term "position in their orbital" is incorrect. The position of the electron is smeared out over the whole orbital.
I guess, you are still thinking about the electron as a particle, which moves around in the space defined by the orbital. That's wrong!
The orbital IS the electron (at least for one-electron systems; for many-electron systems this is just approximately true). The "probability density" is just about the likelihood of a measurement, if you would do one. Then the electron suddenly behaves like a particle, and needs to decide on a certain position.
One might try to make many of such measurements with very small time steps in between them. Any attempt to connect them to some continuous path or trajectory would fail. The electron would just appear to be randomly jumping around, popping up at different locations, no matter how short the time steps between the measurements.
During a transition to a higher state, the electron does not move. It just changes its wavefunction, instantaneously when absorbing the photon. And as the probability density depends on the orbital (aka wavefunction), the probability density changes as well. There is no need to consider the exact position, as in quantum mechanics we cannot grab it. It is smeared out. It is at every position at the same time (just with varying probabilities).

You have a little misunderstanding. An electron bound to an atom can't change position "within an orbital" - in fact it does change position between different orbitals. You are probably confused because people always talk about which orbitals the electrons go into. But these are only where the electrons go at their lowest energy. When electrons get more energy they jump up to higher orbitals. The "position" of an electron and which orbital it is in are the same thing. The only "positions" the electron can have are the orbitals.
You might find it interesting to know that after some time the electrons will jump back down to the "normal" lower energy level (orbital) and release some radiation (light). The gaps in energy levels between the different orbitals correspond to different amounts of released energy and so different wavelengths of light. That's what gives each atom it's characteristic spectral colours.

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What is the nature of the bonding in cyclotriphosphazene?
The bonding in $\ce{(NPCl2)3}$ (and other compounds of the form  $\ce{(NPR2)3}$) has historically been described using the ‘Dewar island model’.
By analogy to benzene, $\ce{(NPCl2)3}$ is π-electron precise, containing 6-electrons in the π-system, and as such was thought to exhibit some degree of aromaticity.
The Dewar model accounted for this benzene-like structure by allowing phosphorous to use an occupied d-orbital to engage in a dπ(P)–pπ(N) interaction, forming the delocalised π-system of the phosphazene ring. This rationalised experimental observations that $\ce{(NPCl2)3}$ is (almost) planar and has equal P-N bond lengths which are somewhere between standard P-N (single) and P=N (double) bond lengths. What it fails to explain however is why other members of the series $\ce{(NPR2)n}$ do not show planarity.
Despite these flaws, and despite it being widely acknowledged that the d-orbitals aren’t capable of being involved in bonding to any great extent, the Dewar island model is still commonly found in inorganic chemistry textbooks and taught to undergraduates (possibly because its an easy to understand explanation and draws analogy to the familiar).

Is there a better model to explain the bonding in $\ce{(NPR2)3}$ that is (1) consistent with the observed properties of the molecule, (2) makes sense in terms of the orbitals/interactions used (i.e. no invoking d-orbitals) and (3) can be backed up by calculation? [*]

[*]: My understanding is that negative hyperconjugation can be invoked, but this doesn't seem to be widely accepted and possibly isn't the most intuitive explanation for bonding by itself.

The bonds between the phosphorus and nitrogen in phosphazenes are best described as having a significant ionic component with each of the N atoms having a negative charge and each of the phosphorus atoms having a positive charge. There is also some negative hyperconjugation resulting from the donation of electrons from the nitrogen atoms' lone pairs to antibonding orbitals  between the P atoms and the R groups. The degree of negative hyperconjugation increases with the electronegativity of the R groups and it imparts a partial double bond character to the bonds between the P and N atoms.
For more information see http://pubs.acs.org/doi/abs/10.1021/ic0511266

It would be nice if you could summarise how the authors got to this conclusion.


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Polarisability of alkynes, alkenes and alkanes
I have noted a little trend in the boiling points of alkynes, alkenes and alkanes. Upon searching for the explanation, I have chanced upon an answer written by M.A.R. (Alkane, Alkene, Alkyne boiling point). In his answer, he explains the boiling point trend which goes in the order alkyne > alkane > alkene by invoking the concept of polarisability of the electron cloud. This makes perfect sense as polarisability is linked to the strength of intermolecular forces and the strength of such forces determines the boiling points of the substances in the liquid state.
However, my point of contention is his explanation of how alkynes are more polarisable due to their more electron-dense triple bond and how alkanes are more polarisable than alkenes because they have two more electrons.
This leads me to wonder what is meant by polarisability. Evidently, I have learnt it in school and I know perfectly well what is meant by it (Refer to https://en.wikipedia.org/wiki/Fajans%27_rules). However, it seems to me that M.A.R.'s "polarisability" is completely different from what I perceive it to be.
M.A.R. says that the more electron-dense triple bond is more polarisable. However, there are so many electrons packed in that small space. That is analogous to a small anion and evidently, small anions are very much not very polarisable.
M.A.R. then seems to contradict himself by not subscribing to the same logic in his explanation of the case of the boiling points of alkanes vs. alkenes. By his previous logic used in the case of alkynes, the double bond being more electron-dense than the single bond would mean that the polarisability of alkenes is more than that of alkanes. Instead, he uses another reasoning that the alkenes have 2 less electrons than the alkanes.
Thus, this leads me to question: Is M.A.R.'s use of "polarisability" correct in a chemistry sense?
On a side note, I would have rationalised this boiling trend using electronegativity and bond dipoles because atoms with sp-hybridised orbitals are more electronegative than sp2 and subsequently sp3. However, this is also flawed because I would not be able to explain why alkane > alkene.

I'm not entirely convinced by the explanation you read, but at the same time I'm not keen on speculating on the reasons why there is such a trend. I think the definition of polarisability is perfectly fine (mathematically it is described with a matrix). The issue is probably more with how the molecular structure links to polarisability.

@orthocresol Indeed, this is just speculation and all we are proposing, be it electronegativity and dipoles or polarisability, are all hypotheses. However, I just want to point out that M.A.R.'s answer seems to be wrong in some areas so that they can be corrected and improved upon. This is so as to prevent users of the site from being misled.

Yeah. You should probably drop him a comment on his answer, since not everybody will see this.


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Radioactivity of materials
Suppose that a raidoactive brick of an unknown element with a half-life of one year (dangerous stuff!) is left on a regular desk for a day. After the day has passed, the brick is removed. Does the desk now exhibit any radioactivity even when artificialy induced radioactivity is considered?(Assume that the brick exhibits alpha, beta, and gamma decay products and ignore the dimensions of the brick and the realism of placing a radioactive brick on a random desk.)

It is possible to irradiate food directly with gamma radiation to sterilize it and delay spoilage for months without refrigeration, all while keeping it completely safe to consume. It's a shame the process isn't used more widely.

As always it seems, it depends... It depends on what kind of radiation the source is emitting, and what the desk is made out of. Also a "brick" is really an undefined size. // A standard size construction brick (9.2 x 5.7 x 20.3 cm) of neptunium-236 is more than the critical mass. So there wouldn't be any desk left...

No. Radioactivity means that the nucleus of a particular atom is unstable, and the nucleus will decay to a more stable state either by decomposing or emitting energy (as a photon). This instability will not make other nuclei unstable unless the decay of one of the radioactive atoms causes a fission process in one of the atoms in the desk. This is a secondary effect though and surely will not happen once the material is removed.
Another way of thinking about this is that there are materials in which people house radioactive substances. These materials do not become radioactive or else there would be no point in having them in the first place. Not just any material will work though because radioactive emission products might be transparent to many materials or are energetic enough to melt/decompose other materials. So, I would bet the desk would get destroyed, but from normal decomposition not from decay.
Finally, the reason that it seems like things become radioactive once there has been radioactive material nearby is because we usually think about the products of nuclear fission. These products will have quite long half-lives and thus the material will stay around for a long time decaying.

What if the desk were made of a material on which artificial radioactivity may be induced?

I don't think there is "a material" where this happens. I think that atoms are capable of capturing a neutron and further decaying, or absorbing a gamma ray and emitting a neutron or some other kind of decay process. This is all going to have a statistical distribution though which will depend on both the desk and the radioactive material, and I believe inducing radioactivity will be quite a small effect compared to the options I have mentioned. It is a good point that this is, in principle, possible though.


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Mechanism of a gold catalyzed addition to triple bond
I needed to draw the reaction mechanism for this reaction:

So, after consulting some literature I drew this mechanism:

But, I am not sure about the last step. Does the $\ce{H+}$ just substitute the $\ce{Au+}$ ion? If so, is my last step correct?
Or, something else happens in the last step?

What happens when you treat grignards with water?

It forms $\ce{R-H}$ and $\ce{Mg(OH)X}$

What does Grignard reagent have do with this?

Replace magnesium with gold and you'll see your final step isn't all that weird after all.

Yeah! I now get it.

You can always self-answer your question.

What about the frontier orbital picture of the last step?

It seems that the addition of gold draws the oxygen to form the final heterocycle and the displaced hydrogen attracts the pi bond electrons to form the saturated terminal sigma bond with the gold. Although I would agree it seems like an unresolved and questionable final product.

In their 2005 The Journal of the American Chemical Society publication, Genêt, et al. have reported a gold-promoted cyclization reaction of bis-homopropagylic diols (see Graphical Abstract below), which afforded bicyclic ketals at room temperature in very good to excellent yields (ref. 1). This cycloisomerization reaction was catalyzed by either Au(I) or Au(II) chloride in methanol:

Abstract: A highly efficient gold-catalyzed cycloisomerization reaction of bis-homopropargylic diols is described. The cyclizations are conducted in the presence of either $\ce{Au^I}$ or $\ce{Au^{II}}$ catalysts in $\ce{MeOH}$ at room temperature in a very short time. The reaction conditions are compatible with functional groups, such as n-butyl, phenyl, allyl, benzyl, and alcohol groups, leading to original strained dioxabicyclo[2.2.1], -[2.2.2], or -[3.2.1] ketals in good to excellent yields.

There was a sequence of compounds has been used as starting material, which consist of different R- groups, one of them been Ph- and $n = n' = 1$ as in this example. A plausible mechanism for the cyclization is given in the paper, which is shown in  following scheme:

In there, 1a and 2a represent the appropriate structures with $n = n' = 1$ and R- = cinnamyl- group. The authors have given the following explanation for the mechanism:

The reaction may be initiated by the formation of the π-alkynyl complex A through the complexation of the unsaturated triple bond to the $\ce{Au}$ catalyst. The coordination of the triple bond, therefore, enhances the electrophilicity of the alkyne as usually advocated for metal-catalyzed reactions of alkynes (ref. 2). The addition of one alcohol may be favored by an intramolecular complexation of the $\ce{OH}$ group to the gold catalyst, as already envisaged by Teles and co-workers (ref. 3) for the addition of methanol to alkynes. The enol vinylgold intermediate B may then be protonolyzed leading to an enol ether, which then undergoes another intramolecular addition of the remaining alcohol leading to the cyclic ketal 2a. This latter step may also be catalyzed by gold, considering the proposed mechanism for the $\ce{Au(II)}$-catalyzed cyclization of α-hydroxyallenes (ref. 4).

In reference 2, it is stated that that the activation of the carbon−carbon triple bond toward nucleophilic attack can be performed by metals with high Lewis acidity (alkyne activation), as illustrated in Equation 1 of the following scheme. It is also easy to realize that an external nucleophile could make an anti-addition as a result and thus, the product should be obtained with trans-stereochemistry (Equation 1). Yet, as envisioned by Teles and co-workers (ref. 3), syn-addition is also possible by an external nucleophile, through complexation with metal, first by alkyne and then by the external nucleophile (see Last Scheme below). However, in many other cases (specifically where a nucleophile would be available intramolecularly as the case in here) the picture is quite different (Equation 2). Here, metal can activate the nucleophile first (nucleophile activation) and the reactions may proceed as a result of the insertion of the alkyne moiety into M−H or M−Nu bonds of “H−M−Nu” complexes, typically with syn-addition as illustrated in Equation 2 of the scheme (This is the case illustrated in A and B).

Last Scheme:

References:

S. Antoniotti, E. Genin, V. Michelet, J.-P. Genêt, “Highly Efficient Access to Strained Bicyclic Ketals via Gold-Catalyzed Cycloisomerization of Bis-homopropargylic Diols,” J. Am. Chem. Soc. 2005, 127(28), 9976–9977 (DOI: 10.1021/ja0530671).
F. Alonso, I. P. Beletskaya, M. Yus, “ransition-Metal-Catalyzed Addition of Heteroatom−Hydrogen Bonds to Alkynes,” Chem. Rev. 2004, 104(6), 3079–3160 (DOI: 10.1021/cr0201068).
J. H. Teles, S. Brode, M. Chabanas, “Cationic Gold(I) Complexes: Highly Efficient Catalysts for the Addition of Alcohols to Alkynes,” Angew. Chem. Int. Ed. 1998, 37(10), 1515–1418 (https://doi.org/10.1002/(SICI)1521-3773(19980605)37:10<1415::AID-ANIE1415>3.0.CO;2-N).
A. Hoffmann-Röder, N. Krause, “Gold(III) Chloride Catalyzed Cyclization of α-Hydroxyallenes to 2,5-Dihydrofurans,” Org. Lett. 2001, 3(16), 2537–2538 (DOI: 10.1021/ol016205+).

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


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What are the products formed by the thermal decomposition of DMPA photoinitiator?
I am using DMPA as a photoinitiator for my experiments to polymerise acryamides  and actylates. I need to know if the DMPA used for the purpose forms any decomposition products if heated at high temperature. I tried to search the literature but could not find any useful information. If anyone knows and could help me I would really appreciate that!

What is "high temperatures"? Heat the pure DMPA, the mixture with monomer, the polymer? What do you expect? What are you afraid of?


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To understand logic and relation of biochemical cycles
I recenty learned the Kreb cycle, in that I saw many chemical are same.
I want to know if there is logic/threrom/technique in which I can write a chemical  cycle to get a product i.e I want basic reactions and how to put it in sequence to make a cycle ?

The Krebs/TCA cycle is not off-topic here per se... but this is something that takes places in a biological system, and if you want to know the "logic" behind it, you'll have to ask this at the Biology.SE [ https://biology.stackexchange.com/ ]. Here, at Chem.SE, we can point at parts of cycle and tell you "Oh, this is oxidation happening here; that's is decarboxylation going on there...ooh, and this one right here is oxidative decarboxylation..." etc, etc. If you're looking for a general idea/pattern as to why biochemical processes like these occur in a given tissue, take it to Bio.SE.

@paracetamol- it seems like he's asking about how to work out what reactions are taking place- much of the elucidation work on these mechanisms was indeed done by chemists, isolating and characterising the intermediates in the cycle!

@Not "...indeed done by chemists..." True, but unless the OP clarifies further (in a future edit, perhaps?) I'm assuming he wants to know why molecules are shuttled around, broken down, rearranged and eventually built up again in biochemical processes (such as the Krebs cycle). Besides, it would be naive to assume that a given process occurs in an identical manner in all relevant tissue, across all life-forms. It's a good question, but is probably better suited on the Bio.SE... unless the OP wants to know the thermodynamics/equilibrium side to the story. ;)

I must say, I really don't understand this question. The Krebs cycle, or any cycle for that matter cycle (eg Suzuki coupling), cannot be a priori worked out just by using logic or a mathematical theorem. Maybe you can use chemical intuition, but that is an entirely different matter.

tl;dr: You can't reasonably 'propose' a biochemical pathway on paper the same way you might be able to for an organic reaction.

Biochemical pathways are highly complex and interwoven. The Krebs cycle doesn't act in isolation, but rather is part of a far larger network which provides it with the components it needs (enzymes, cofactors, oxidatants/reductants etc.) and the regulation required to ensure it works correctly.
The figure below demonstrates a more elaborated Krebs cycle, complete with all of the relevant interactions with various other pathways:[1]

Given their complexity, the discovery of the true nature of a pathway takes a long time, and uncertainty often remains due to the difficulty in directly observing the reactions taking place (though with modern genomics/sequencing its becoming easier to see what might be taking place).
In the case of the Krebs cycle, the key discovery was made by Gyorgyi who won the 1937 Nobel Prize in Physiology for 'his discoveries in connection with the biological combustion processes, with special reference to vitamin C and the catalysis of fumaric acid'.
Using isolated cells from pigeon breast tissue, Gyorgyi was able to identify the importance of succinic acid and fumaric acid in the pathway, noting that the rate of respiration was increased significantly if the tissue was incubated in their presence (as measured by gas evolution, which was already known at the time to be a consequence of respiration).
Building on this discovery, Krebs published the first full account of cellular respiration in his 1937 paper Metabolism of Ketonic acids in animal tissues.[2] In this paper he outlines all of the key steps needed, but even at this stage many of the important details were missing (many of the other necessary pathways weren't known, so Krebs had no idea where the succinate was coming from in the body, for example).
[1]: Gerhard, M.; Schomburg, D. Biochemical Pathways- An Atlas of Biochemistry and Molecular Biology; Wiley: New Jersey, 2012
[2]: Krebs, H. A.; Johnson, W. A. Biochem. J. 1937, 31, 645


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In TOF Mass Spectrometry how does the detector 'know' the difference between ions of different elements?
After the positive ions have been accelerated, I know that they travel through the drift zone at different speeds due to their masses.
But surely, a heavier ion/element that enters the drift zone first may still reach the detector before a lighter ion if that lighter ion is one of the last ions to enter the drift zone. If so, when these ions reach the detector, how does it know that the heavier ion IS heavier than the lighter ion that does not reach the detector until afterwards.
*I'm aware my question is a tad confusing, let me know if I should try and reword it!

TOF needs a pulsed generation of ions, you all the ions into the drift zone at basically the same time.

And one makes the initial pulse width small compared with the transit time difference of the different ion masses desired.

As @DSVA states, the various species must all start simultaneously (e.g. from an atosecond laser pulse). The time-of-flight is from that initial launch until detected. There can even be multiple bounces using electronic mirrors to make the path longer. It's a race, and the leanest and most charged-up species finish in the lead.

I suppose that you are referring to a Wiley–McLaren time‐of‐flight mass spectrometer [Rev. Sci. Instr. 65, 3344 (1994)] which is basically a tube containing an electrode stack for accelerating the ions toward a detector. As mentioned in the comments, one can discriminate different masses if one uses a pulsed experiment, for instance using pulsed lasers or pulsed electric fields to extract the ions.
Another option is to add a magnetic field to the setup in order to bend the trajectory of the ions. The radius of the trajectory depends on the mass-to-charge ratio of the ion and one can record spectra as a functoin of the magnetic field to get a certain mass selected or use a position-sensitive detector to measure several masses simultaneously.
Yet another solution is to use quadrupole–time-of-flight mass
spectrometry, where radio frequencies are applied to four electrode rods placed on the circumference of a circle. Depending on the frequency that is applied to the rods, only ions of a certain mass-to-charge ratio make it out of the quadrupole and the qaudrupole thus serves as a mass filter.
Note that all these techniques are sensitive to the mass-to-charge ratio only and one cannot distinguish between, e.g., O$^+$ and S$^{2+}$.


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How would an increase in the concentration of a sucrose solution affect the rate of its acidic hydrolysis?
Regarding the acidic hydrolysis of sucrose
$$\ce{\underset{sucrose}{C12H22O11} + H2O + \underset{catalyst}{H+} -> \underset{glucose}{C6H12O6} + \underset{fructose}{C6H12O6} + H+},$$
I think that its mechanism can be described as a 2-step reaction consisting of a step which involves the H+, and then the step which involves the $\ce{H2O}$, and I believe the latter step to be a rate limiting step.
My question in detail is:

Would an increase in the concentration of a sucrose solution, with a constant $\ce{H+}$ concentration, have a 1-to-1 influence on reaction rate (for example, if doubling the concentration of the sucrose solution would result in a hydrolysis reaction twice as fast), or would the addition of $\ce{H2O}$ in the second step significantly limit the change in reaction rate?

You mention that you believe the second step of the hydrolysis to be rate limiting already. So that would mean that the concentration of sucrose is not. The second step cannot be varied much, assuming water is the solvent (its concentration cannot be changed then)

The solvent is water, I should have specified so.

I guess my problem has to do more with the concentration dependency of 2-step reactions in general. I an unclear as to how a change in the concentration or quantity of a reactant affects the rate of a 2-step reaction with a rate limiting step.

Looking at the reaction mechanism (of which I understand you are aware)
is good to keep in mind that the reaction rate really depends on the amount of intermediate here, provided the second step is rate limiting.
In the first step, $H^+$ is consumed. Since it is not rate limiting, I can assume the intermediate $[sucrose.H^+]$ to be in pseudo-equilibrium with $[H^+]$&$[sucrose]$, on the timescale of the second reaction step.
$$r_2 = k_2[sucrose.H^+]$$
This means adding more sucrose (provided you have enough acid to protonate it) will indead result in the formation of more [sucrose.H^+] intermediate, and thus, higher kinetics.
Btw, I'm simpifying stuff here by not taking into account the reaction backwards for the second step.

This response does a good job addressing the question, but then unfortunately provides a reaction scheme for subsequent steps that is surely wrong.  The reason is that it shows sucrose (beta-D-fructofuranosyl-a-D-glucopyranoside) being hydrolyzed directly to beta-D-fructofuranose and a-D-glucopyranose, implying either that these two species are the initial reaction products or that they are the major reaction products. Neither is correct. As the discussion in  J. Am. Chem. Soc. (1988) 110, 6372–6376 explains, this is one of the common misleading reaction schemes shown for this classic reaction


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Atomic weights of non-molecular and non-stoichiometric compounds
Consider a composition such as $\ce{FeCo4Sb12}$.
The atomic weight for a mole of substance is usually simply calculated by looking up the atomic weights (g/mole) for each involved element and adding them together proportionally:
$$M_{\ce{FeCo4Sb12}}=1\cdot M_{\ce{Fe}}+4\cdot M_{\ce{Co}}+12\cdot M_{\ce{Sb}}$$
But this compound is not a molecule. It is a representation of the amounts: there are 12 Sb and 4 Co for every Fe atom. I could also have written it as $\ce{Fe2Co8Sb24}$.
Same compound, but the calculation will give a different $M$.
The same issue comes for non-stoichiometric compounds, such as: $\ce{FeNb_{0.88}Hf_{0.12}}$. This just says that there are 88 Nb and 12 Hf for each Fe.
How should I understand the chemical formula when calculation atomic weights of crystalline or other non-molecular compounds as well as non-stoichiometric compounds?

Atomic weight is a property of atom, not compound. Surely you meant "molecular weight", but that's only good for compounds which have molecules. Go with "formula weight" and don't bother.


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Oxidation States of Actinides
It is observed that actinides do not exhibit +2 oxidation state, contrary to that of some of the lanthanide elements. Why?

I vote to reopen the question as even though it contains traces of homework-alike style, it's quite fundamental and touches the chemistry of $\ce{f}$-elements which is not often discussed here.

@andselisk, I agree it could be a good question if edited, but as it stands there is no effort shown (it clearly falls within the scope of homework)- being interesting / niche doesn't really change this fact

@NotEvans. There are numerous similar questions falling into the same "Here is fact X. Why so?" pattern, though they and many others were accepted. I personally see some inconsistency in decision making.

See also: Why is WF6 stable whereas CrF6 is unknown?
Whenever one wants to compare oxidation states, there are a couple main factors to take into account. To reach a higher oxidation state, one obviously has to pay for it in the form of ionisation energy/energies. However, there is a compensatory effect in that elements in higher oxidation states generally get more out of bonding. For example, the lattice energy of $\ce{FeCl3}$ is larger than that of $\ce{FeCl2}$, or the four covalent bonds in $\ce{XeF4}$ are collectively stronger than the two covalent bonds in $\ce{XeF2}$.
The higher ionisation energies of the actinides tend to be smaller than those of the lanthanides, for the same reasons as explained in the linked question (5f orbitals have one radial node; 4f orbitals do not). This generally favours higher oxidation states and disfavours low oxidation states in actinides.


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Why does acetone have a greater dipole moment than dimethyl ether?
Below are the lewis structures for dimethyl ether and acetone. My chemistry textbook indicates that acetone has a larger dipole moment than dimethyl ether. I am familiar with dipole moments and the fact they are created by an unequal sharing of electron density. However I do not understand why acetone is more polar than dimethyl ether. My best guess is that having an extra carbon provides more electrons which allows the molecule to establish a larger difference in electron density between the oxygen and the rest of the molecule. I am unsure if this is correct and would greatly appreciate a more in depth explanation.
Dimethyl ether

Well the electron density is farther away from the core in acetone?

For acetone, hyperconjugation also comes into effect and the methyl groups donate electron density towards the carbonyl group as well, resulting in an even stronger dipole moment

However, the electron-donating effect of the methyl groups may actually reduce the strength of the dipole moment of the carbonyl group as it neutralises the partial positive charge on the central carbon. So I don't know which perspective is correct...

In a very simplistic view, the dipole moment can be considered the vector addition of all polarised bonds. In an initial step, we can even completely ignore the $\ce{C-H}$ bonds since they are almost unpolar.
In dimethyl ether, you have two polarised bonds that point more or less towards each other at an angle of about $111^\circ$. This is not quite opposite but a large part of the dipole moment will cancel itself out, leaving only a rather small component along the principal $C_2$ axis.
In acetone, the main component of the dipole moment is a bond on this principal $C_2$ axis, which here coincides with the $\ce{C=O}$ bond. No cancelling out can happen.
Additionally, there is only a single carbon atom bound to that oxygen of acetone so this carbon has to deal with the entirety of oxygen’s electronegativity. In dimethyl ether, two carbon atoms can share their electron density with oxygen making each individual dipole less strong.
Therefore, acetone has a higher dipole moment.

Can we completely ignore C-H bonds? Wouldn't there be some hyperconjugation going on?

@TanYongBoon Hyperconjugation would put more electron density on or close to oxygen in this case so yes.

The ether is fairly symmetrical though there is a bond angle of ~111°, "exposing" the oxygen on the side (and its electron cloud).  Acetone, however, has an oxygen off to the side, with its electron cloud offset ~120 picometers (pm), making it more polar.
Remember that oxygen is a strong electron acceptor, so that there will be a "lump" of electron probability surrounding it.

[Sorry about the hodgepodge illustration!]


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Salt Water Metal etching 50 (Euro-)Cent Coin
I want to etch the 50 Cent Euro Coin using this following technique that I came across on some DIY site: Immerse the coin and some disposable piece of metal in salt water such that they do not touch and attach current. Cover the non-to-be-etched parts of the coin with nail polish.
The article proposed to use a 9V battery block and used brass in their example. I understand the 50 Cent coin is made out of Nordic Gold: Is this etching method still viable, and does the 9V block deliver enough energy to at least remove the imprinting of the coin? Is there something else to consider (safety isseus)?
Unfortunately, I have no noteworthy background in chemistry. Thanks a lot in advance for any help.

What stops you from trying it? Chemistry is an experimental science, they say.

I certainly will try.  Unfortunately, my situation (abroad + time pressure) does not allow for extensive experimenting. Are there some obvious Do's and Don'ts or comments on my endeavour?

Don't get electrocuted. That's it

Thank you. What about the energy needed? Will the 9V block (600mAh) suffice or would I be better off with e.g. a 12V 7000mAH motorcycle battery? How many amps do I need to make reasonable progress?

I'm voting to close this question as off-topic because from the nature/tone of your question, it sounds awfully like you have absolutely no idea about the chemistry/electronics of what you're considering doing and are likely to cause harm to yourself/others. The question is also not specifically asking about the chemistry involved.

That is ok. While I wouldn't mind an explanation of the chemistry involved, a suitable (for me) explanation will perhaps be of no benefit to the community. If yes, please close.

ECM (electrochemical machining) is a safe (besides risk of getting electrocuted: keep your hands dry) way to carve metal. However, only electrically conductive materials can be machined. Both aluminum and copper are conductive, therefore you can carve it safely.

Thanks, it works just fine. I‘ve noticed some gases/steam(/smoke?) developing in the process, what is this? Heated water?


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why is the pressure exerted by ideal gas on the walls of the container more than that of real gas
I have learnt that ideal gases have no intermolecular forces. So even at high pressure the volume gets completely compressed.
But in case of real gases,  we have both attractive and repulsive forces. At far distance,  attractive forces dominate.
But at high pressure , when the particles are closer to each other,  the repulsive forces dominate over the attractive forces.
So my question is,  if the repulsive forces dominate over the attractive forces at close range,  then real gas should exert more pressure on the walls of the container.  With this one more doubt arises,  how do gases liquify if repulsive forces are dominant?

At a fixed pressure real gas occupy more volume, that is because of repulsion..

For neutral gases, the intermolecular forces that we observe tend to be attractive. Significant repulsive forces don't really show up until your particles have charge.
The other difference between ideal gases and real gases are that real gases take up space, but the volume deviation from the ideal gas law is usually swamped out by the attractive intermolecular forces. This causes a real gas to exert less pressure than an ideal gas, and also is the reason matter undergoes phase changes.

However,  real gas molecules have some volume. Thus when they come too close,  they tend to repel each other and oppose the compression. This is also the chief reason why volume of the real gas is more than that of ideal gas at a particular pressure. Because as we compress the,  the gas particles prevent each other from occupying the same volume. So if volume increases,  it should be exerting more pressure on the walls than the ideal gas.  This is something that does not  happen in Ideal gases. So how do the real gases compensate this repulsion during change of state?

The repulsion you're talking about is only significant at very very short distances (smaller than even the distance between atoms in a chemical bond). Any larger than this distance and you see attractive forces pull the particles together. This is why a real gas almost always occupies a smaller volume than an ideal gas.

Sir,  kindly have a look at the picture I have added to the question. Here you can see that the at high pressure,  real gas tends to occupy more volume than ideal gas.If repulsion really exists at such less distances,  how can you explain the increase in volume of the real gases at high pressure as given in the graph?

Yes, your graph is accurate, but the positive deviation from IGL is not because the dominant interaction between particles is repulsive. It is only because there is less available volume than postulated in the IGL (due to the nonzero volume of the particles). If you calculate the average distance between gas particles at increasing pressures, you will find that they are still very far apart (until you get to astronomical pressures), and that attractive forces are dominant, even in the region where the pressure of a real gas is higher than an ideal gas.

Sir,  if the volume is less available, then volume of real gas should be less than that of ideal gas.  So excatly why is the volume of the real gas shown more than ideal gas at high pressure. Also,  why at low pressure again volume of ideal gas is shown to be more than real gas.  Can you explain what this graph is exactly trying to Convey that I am unable to get.?

Sir could you answer my question a bit fast,  it urgently need it.

The particles have less distance to travel before they strike another particle, causing the real gas to occupy a larger volume than an ideal gas (at high pressures). At low pressures, the distance between particles is so large, the difference between particles with no volume and particles with volume is negligible. Attractive forces cause the volume to be smaller than ideal gases at low pressures because the particles are "sticky" and pull towards each other. Finally, you cannot apply kinetic molecular theory (how we can think about gases) to phase changes. It's a whole different set of rules.

So exactly can we say that at high pressure, real gas tends to occupy more volume than ideal gas,  causing it to exert more pressure on the walls of the container ( at high pressure only when the repulsive forces dominate?)

That's almost correct. There are no significant repulsive forces that play a part in kinetic molecular theory. The gas takes up more room because each particle takes up space.


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How does the electron configuration correlate with the emission spectrum of an ionic compound?
trying to find a way to correlate the electron configuration of an ionic compound (ie. Cupric sulfate, potassium chloride, Cupric chloride, sodium carbonate, strontium chloride) with its emission spectra.


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Sodium Peroxoborate reacting with Sulfuric Acid and Potassium Permanganate
This is one of the reactions I had to undertake in a lab. I think the reason for the test in the first place is to test for the peroxoborate anion in the solution.
From my understanding what's happening is that sodium peroxoborate will liberate $\ce{H2O2}$ when reacted with sulfuric acid. Adding the $\ce{KMnO4}$, the hydrogen will react to reduce $\ce{Mn(VII) ->Mn(II)} $ and form $\ce{O}$
I'm having difficulty interpreting this into balanced chemical equations. This is what I've tried but not sure if it's correct
$$\ce{Na2[B2(O2)2(OH)4] + H2SO4 ->Na2SO4 +B2O3 +2H2O2 +H2O}$$
This equation shows the liberated $\ce{H2O2}$ but not sure how to write a reaction now where $\ce{KMnO4}$ is added.
Any help would be much appreciated. I've been strugging with this for a while and there isn't much information online.

typo I edited it there

I'd rather write that product as H3BO3.

What about this : $$\ce{Na2[B2(O2)2(OH)4] + H2SO4 +2H2O ->Na2SO4 +H3BO3 +2H2O2 }$$

Good. Now write the reaction between H2O2 and KMnO4, then add both reactions together.

Well $\ce{KMnO4} $ and $\ce{H2O2}$ react as follows:
$$\ce{2KMnO4 +3H2O2 -> 2MnO2  + 2KOH + 2H2O + 3O2}$$
Adding the two reactions together I get:
$$\ce{Na2[B2(O2)2(OH)4] + H2SO4 + 2KMnO4 + H2O2 ->Na2SO4 + 2H3BO3 + 3O2 + 2MnO2 + 2KOH}$$

You can't have acid and alkali together.

do I need to have a salt and water in the reactants?

It is H3BO3 and K2SO4 in the products, I'd say. Also, it is probably MnSO4 instead of MnO2. Other than that, everything's mostly right.

I appreciate the patients, this is what I have so far and tbh I've spent over 3 hours trying to find resources for this,

$$\ce{Na2[B2(O2)2(OH)4] + H2SO4 + 2KMnO4 + H2O2 -> H3BO3 + MnSO4 + 2KO + H2O + 2NaOH + 3O2}$$


text:
Thermochemical data of methanol
I am trying to find the standard entropy and enthalpy of methanol as functions of temperature, but they are not available on the NIST-JANAF tables: http://kinetics.nist.gov/janaf/
Does anyone know where I could possibly find the data?
Thank you in advance!

This might be what you are looking for: http://wtt-lite.nist.gov/wtt-lite/index.html?cmp=methanol#methanol/A;0,0,508,415;help,about/aa2;50,50,508,415/aa1;491,97,508,415/

Thanks for responding! Do I need to log in to view the data?

as far as I know you don't. I thought when I looked that I could view things without a login.

I took a look again and it doesn't seem to show me any data besides the molecule, property availability, and help tabs. In no place does it say how to access the properties. Am I missing something?

NIST Webbook Chemistry is a reference which I would recommend as a first (as in initial) resource to consult for enthalpies of small organic molecular and inorganic materials.  The entry about methanol lists for one temperature the value in question, the experimental method of acquisition, and the primary literature reference.  The later may contain more data than the one tabulated here, too.
The references might be old, but in a scientific library you will find assistance to figure out who cited these publications which in turn may lead you to more recent determinations.


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Do endothermic reaction have higher energy products than reactants?
If a reaction absorbs energy does  that mean it has a net gain of energy?

There are quite a few different things that may be called "energy".

In a reaction, you have to think about the entropy as well as the net energy content of the reactants and products. An endothermic reaction take energy from the environment but it doesn't necessarily go into the bonds in the product.

The simple answer is yes - as energy is required to be absorbed for the reaction, the products have more energy than the reactants.

We use several quantities called 'Theormodynamic Potentials' to measure the 'energy' or, more accurately, the state of the sustem. Examples would be Gibbs free energy, Helmholtz potential, Internal energy etc. In a general chemical reaction performed in lab (at constant pressure and constant temperature), Gibbs free energy provides an intuitive way of determining the direction of a spontaneous reaction. A reaction always proceed towards minimizing Gibbs Free Energy.
You are probably asking about enthalpy of a reaction being negative or positive, which 'in special cases' corresponds to the heat exchanged during the reaction. In those special cases, exothermic reaction does imply a positive enthalpy change i.e. products have higher 'potential' (referring to enthalpy here) relative to reactants.
Notice how I carefully use 'potential' instead of 'energy' since the latter is quiet vague and may refer to any of the several potentials defined in thermodynamics.

Please refer to Material Thermodynamics by Robert Dehoff for more on the thermodynamics of chemical reactions.

This is not the usual convention for definition of exothermic, afaik.

Exothermic (heat released) means negative enthalpy change by the usual definition of enthalpy.

Once you give enough energy to reactants to cross the activation energy barrier it would react to Form the compounds, if the net energy is released, then the products will have lower potential energy or else (as in your case) the potential energy of products will be more than that of reactants


text:
what is the final concentration of the mixture
In a bizarre twist of chemical fate, an accident results in you results in you mixing 2L of a 3.00M solution of NaCl with 4L of a 1.50M solution of NaCl. what is the final concentration of the mixture?

I'm not going to answer it for you but think of it this way: find out the amount (not to be confused with concentration) of salt in each mixture, and then work out the volume of the final mixture. Paying attention to the units of your quantities can help with conversions, too.

Since I can't post the answer all you have to do is calculate number of moles by the formula $$ M_1*V_1+ M_2*V_2 $$ and divide this by total volume which in this case would be $$ V_1 + V_2 $$

Welcome to Chemistry.SE! Take the [tour] to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax. While we appreciate your contribution to the site, it would be preferable to not answer questions, that do not comply with our homework policy.


text:
Effective Nuclear Charge of Noble Gases
As stated in my textbook, the effective nuclear charge can be calculated by taking the non-valence electrons away from the atomic number, as also stated in the first calculation on Wikipedia: https://en.wikipedia.org/wiki/Effective_nuclear_charge
The textbook also states:
If we assume that the noble gas core is completely shielding, then the 10 inner electrons
of the neon core make the effective nuclear charge 10 less than the nuclear charge.
My understanding is it implies noble gases have 0 effective nuclear charge.
However, as many sources state noble gases have 8 valence electrons, shouldn't their effective charges be +8?

I think that this is basically an accounting problem. The full electron configuration of $\ce{Ne}$ is: $1s^22s^22p_x^22p_y^22p_z^2$. So, what does one now define as core electrons? It could be: "all electrons not in the outermost shell", yielding the $1s$ electrons. An alternative is "all electrons in completely filled shells", yielding all 10 electrons as core electrons. (Note that these two definitions yield identical results for almost all non-noble gas elements, with some difficulties surrounding the $f$-block.) Both definitions have some merit, I would prefer the first one, giving an effective charge of $+8$, although I do not know what one does with that number next.
A word on effective nuclear charge
As noted on the wikipedia page linked in the question, to arrive at more meaningful effective nuclear charges (as might be helpful for the systematic creation of basis sets for computational/quantum chemistry), one has to take the effective shielding strength of the different orbital shapes into account. The $d$-block-contraction, as can be seen in the somewhat different chemistry of $\ce{As}$ compared to $\ce{P}$ and $\ce{Sb}$, is due to this. Thus, in the end, a pen-and-paper calculation as alluded to above can be useful to think about some concepts in a qualitative fashion, but more extensive calculation is needed for quantitative reasoning.

The textbook also states 'effective nuclear charge increases across a period', so to keep that trend, I'd agree with you on the +8.


text:
How do salts of a weak acid + weak base behave in solution?
Say you have Ammonium Formate in solution, the salt formed by the reaction of Formic Acid and Ammonia.  Would both the Formate ion and Ammonium ion protonate and deprotonate, respectively, to form their conjugates until equilibrium is reached?  I would also assume that one of the ions would form more of their conjugate than the other depending on which ion in the pair is a stronger acid/base.  My question arose when researching the Leuckart reaction, which wikipedia states that if ammonium formate is used, it dissociates to formic acid and ammonia before participating in the reaction.  I also found it peculiar that ammonium formate is listed as soluble in diethyl ether, which typically doesnt dissolve salts.  This behavior would make more sense if ammonium formate dissociated into formic acid and ammonia.

Long story short, yes, they hydrolyze, sometimes completely.


text:
Why is the hydrolysis of the conjugate base of a weak acid neglected in buffer solutions?
Consider a solution which contains a weak acid and the salt of its conjugate base with a strong base.
e.g. $$\ce{CH3COOH + H2O <=>[$K_\rm{a}$] CH3COO- + H3O+}\tag1$$
$$\ce{CH3COO- + H2O <=>[$K_\rm{h}$] CH3COOH + OH-}\tag2$$
Now, $\displaystyle{K_\rm{a} = \frac{[\ce{H3O+}] [\ce{CH3COO-}]}{[\ce{CH3COOH}]}}$ and $\displaystyle{K_\mathrm{h} = \frac{K_\rm{w}}{K_\mathrm{a}}}$.
Now, my notes say that for the equilibrium concentration of undissociated acid, the acid coming due to hydrolysis of the salt is neglected and the equilibrium concentration is approximately equal to the initial concentration.
It is also mentioned in a book that

Assume that the extent of protonation of acetate ions and
  the deprotonation of acetic acid molecules is so small that the concentrations of both
  species are nearly the same as their initial values.

Why is this so? If $K_\rm{a}$ is low, then $K_\rm{h}$ should be high. I understand that the deprotonation of acid will be low as it is a weak acid so that we can neglect the $\ce{CH3COO-}$ coming through the acid dissociation but how can we neglect the hydrolysis of the $\ce{CH3COO-}$ coming from the salt?
Edit $-$
My calculation for equilibrium concentrations.
$\ce{[CH3COOH]_\mathrm{eq}} =$ initially added acid $-$ deprotonated acid $+$ hydrolysed salt
$\ce{[CH3COO-]_\mathrm{eq}} =$ initially added salt $-$ hydrolysed salt $+$ deprotonated acid
$\ce{[H3O+]}_{\mathrm{eq}} =$ deprotonated acid $-$ hydrolysed salt
Let the concentration of
initially added acid $= c_1$
initially added salt $= c_2$
deprotonated acid at eq. $= x$
hydrolysed salt at eq. $= y$
Thus, $$K_{\mathrm{a}}  = \frac{[\ce{H3O+}] [\ce{CH3COO-}]}{[\ce{CH3COOH}]} = \frac{(x - y)(c_2 - y + x)}{(c_1 - x + y)}$$
Now, assuming $x << c_1, \, c_2$,
$$K_{\mathrm{a}}  = \frac{(- y)(c_2 - y)}{(c_1 + y)}$$

You may take it into account and compare the results.

@IvanNeretin Please see my edited question.

You said before that you neglect $y$, not $x$.

@IvanNeretin I said that I understand why deprotonation is neglected, i.e. why $x$ is neglected. I don't understand why the hydrolysis, i.e. $y$ is neglected.

Hydrolysis = cleavage of chem. bonds by addition of water (https://en.wikipedia.org/wiki/Hydrolysis). Where is there a hydrolysis here? How can acetic acid be hydrolyzed?

@deusexmachina From the link you've provided, see Salts subsection on Wikipedia

@deusexmachina It states that "A common kind of hydrolysis occurs when a salt of a weak acid or weak base (or both) is dissolved in water. Water spontaneously ionizes into hydroxide anions and hydronium cations. The salt also dissociates into its constituent anions and cations. For example, sodium acetate dissociates in water into sodium and acetate ions. Sodium ions react very little with the hydroxide ions whereas the acetate ions combine with hydronium ions to produce acetic acid. In this case the net result is a relative excess of hydroxide ions, yielding a basic solution."

@deusexmachina The salt $\ce{CH3COONa}$ does hydrolyse as shown in the reaction numbered $(2)$ in my question.

@downvoter Why the downvote? Am I unclear in my question?

So your question is why the acidity of a weak acid cannot be neglected? Why don't you just give some numbers, including all concentrations. Then you can just calculate it out.

@deusexmachina I'm somewhat confused about the meaning of the word 'acidity'. My question is "Why do we neglect the protonation of acetate ions(of the acetate ions coming through the salt)?"

Of course you cannot neglect the concentration of $\mathsf{CH3COO^-}$. And assuming $\mathsf{CH3COONa}$ to be completely dissociated should be a good approximation. So obviously, you may not neglect the concentration of the sodium acetate. Basically if you double the concentration of the sodium acetate, the concentration of the hydronium ions will go down to keep equilibrium.

@deusexmachina I'm talking about the neglection of the $\ce{CH3COOH}$ which is coming through the protonation of the acetate ion.

$\mathrm{p}K_\mathrm{a} = 4.75$, therefore $K_\mathrm{a} = 10^{-4.75}$. Thus,
$$K_\mathrm{a} = \frac{[\ce{AcO^-}][\ce{H+}]}{[\ce{AcOH}]}$$
So, let's assume $[\ce{AcOH}] = \pu{1.0 M}$. Just to get a feel, how far the acid is dissociated:
$$K_\mathrm{a} = \frac{x \cdot x}{\pu{1.0 M}} \to x = \sqrt{10^{-4.75}}~\pu{M} = 0.00422 \approx 0.4\%$$
where we have assumed that the contribution from the dissociation of water is negligible, and therefore $[\ce{H+}] = [\ce{AcO-}] = x$.
Now, what happens if I add $\pu{1.0 M}$ $\ce{AcO-}$? Then, effectively, we have:
$$K_\mathrm{a} = \frac{[\ce{AcO-}][\ce{H+}]}{[\ce{AcOH}]} = \frac{\pu{1.0 M}
 \cdot [\ce{H+}]}{\pu{1.0 M}}$$
Therefore, we have:
$$[\ce{H+}] = K_\mathrm{a} \frac{\pu{1.0 M}}{\pu{1.0 M}} = \pu{10^{-4.75} M}$$
If we compare the two cases, before addition of acetate, we had a $\mathrm{pH}$ of
$$[\ce{H+}] = \sqrt{10^{-4.75}} \to \mathrm{pH} = 2.375$$
and after addition of acetate, we have
$$[\ce{H+}] = 10^{-4.75} \to \mathrm{pH} = 4.75$$
Meaning that the solution got more basic after the addition of acetate, as excpected.

Hopefully that provides a good starting point for understanding.

@deuxexmachina After you added 1.0 M acetate ions, how did you take equilibrium concentration to be 1.0 M? How is initial concentration equal to equilibrium concentration?

@ApoorvPotnis Yeah, I figured that this point wasn't really clear yet. So first of all, you can neglect that the acid CH3COOH will dissociate to CH3COO-, as by the above argument, it only dissociates to a very little extent. However, your concern is the other way around, and you argue that CH3COO-Na+ could be protonated to form CH3COOH + NaOH. But: look at the pKa of NaOH and CH3OO-Na: NaOH is a much stronger base! Therefore, you can assume that $\mathrm{[NaCH3COO] = [NaCH3COO]_0}$.

@ApoorvPotnis You can also find a similar description here:  http://study.com/academy/lesson/acid-base-buffers-calculating-the-ph-of-a-buffered-solution.html for a similar explanation

Can you add your comment about $\ce{NaOH}$ being a stronger base into the answer so that I could accept it?

Also, can you show in your answer that without taking $y$ into consideration, $y$ comes out equal to $K_\rm{a} = 1.8 \times 10^{-5}$. Taking $y$ into conideration will not change its value very much. So, we can say that $y << c_1, c_2.$


text:
Coordination Numbers and Hybridization schemes of anions

Draw the molecular structures of
(a) the borax anion, $\ce{[B4O5(OH)4]^2-}$
(b) the peroxoborate anion, $\ce{[B2(O2)2(OH)4]^2-}$

          borax                    peroxoborate

(i) What are the co-ordination numbers of the $\ce{B}$ and $\ce{O}$ atoms in the anions (a) and (b)
(ii) What are the hybridisation schemes used by the $\ce{B}$ and $\ce{O}$ atoms in the anions (a) and (b)

I: For (a) $\ce{B}$ has 2 four-coordinate ($\ce{BO4}$) and 2 three-           coordinate atoms ($\ce{BO3}$).I'm assuming $\ce{O}$ is 0 since its not a central atom.  , (b) $\ce{B} = 4$ and again I'm not sure but I thought coordination numbers only applied to one atom in a molecule. In this case the boron atom.
II: For (a)  $\ce{B}$ has 2 $sp^3$ hybrid orbitals and 2 $sp^2$ hybrid orbitals, $\ce{O}$ has just $sp$  orbitals. , (b) $\ce{B}=$ has just $sp^3$ orbitals and $\ce{O}$ has $sp$ orbitals.
I would just like some clarification on the coordination numbers of the $\ce{O}$ atoms in both anions and to see if my hybridisation schemes are correct.

Related: https://chemistry.stackexchange.com/questions/59273/structure-of-borax

https://chemistry.stackexchange.com/questions/81836/what-is-the-hybridisation-of-o-in-borax


text:
Dissolution of protein powder in water
I am using this Mocha Coffe.
When I try to dissolve it in a cupful of cold water, it forms a globule of powder and does not dissolve.
But if I add a small amount of water to the powder and mix it, it makes a paste which will dissolve in more water. What is happening here ?

A lot of powders show this behaviour. Try it with flour or 100% cocoa powder, and you get the same result.
What all these powders have in common, is that the grains contain some hydrophobic material.
When you put a bunch of powder in water, the outside layer of powder gets wet and sticks together, but doesn't let the water through, trapping an air bubble inside. Then when you try to stir, the waterproof bubbles just float around, easily avoiding your spoon.
But if you only put a small amount of liquid, the stirring is much better, and you can wet all the grains.
Source: this entry in the Hitchhiker's Guide to the Galaxy (of all places).


text:
Effect of pressure on the freezing temperature of water
A few lines are written in a book by B.H. Mahan regarding the change in freezing point of water on changing the pressure.

The vapor pressure of water at triple point of water is 4.579 mm. What would happen if the external pressure applied to the system by the piston were to be increased above 4.579 mm ? First of all the va pour present would be completely converted into liquid and solid, to restore the equilibrium pressure.

I'm in total agreement of this. But the next lines appear to be odd.

It is found experimentally that as the pressure on the system is further increased the temperature of the system must be decreased, in order for both  ice and water to remain in equilibrium.

I don't understand  the second statement. If the equilibrium is further disturbed why the system should not  behave the same as before? Why do we have to change the equilibrium constant  to restore the equilibrium?
Thanks in advance.

From the triple point (273.16K, 611.657 Pa), if pressure is increased and temperature maintained at 273.16K, all the water being liquid is the equilibrium condition.
To maintain the solid and liquid phases in equilibrium, as pressure is increased above the triple point pressure, temperature must decrease slightly (to 273.15K if the pressure has increased to atmospheric pressure, for example).

Can you please explain this from the perspective of equilibrium and le Charlie's principle?

@gowreeshmago Liquid is the most dense phase of water, so increasing pressure favors the liquid phase with respect to the vapor or solid phase.


text:
effect of different groups on Amine inversion
I know different groups attached to the nitrogen atom affect the amine inversion barrier, but is it due to steric factors or orbital interaction factors, eg. $\ce{(i-Pr)2NMe}$ vs $\ce{NCl3}$.

Both of those matter.


text:
Definition of an intensive variable
The IUPAC definition of an intensive variable is:

Physical quantity whose magnitude is independent of the extent of the system.

Why is it defined for "a system", and not for a homogenous system, as for a heterogenous system the quantity (e.g. density, temperature) might be different for the different extents/parts of the system?

There's no truly "homogenous" system.

I don't see why being heterogeneous would prevent a system from having an (average) density, temperature, etc.

@S.Chevalier.  But if there is a clear boundary between the substances/phases, even the average temperature, density will not remain same for different parts of the system.

For a homogeneous material, an intensive variable is independent of the amount of material.  Even non-homogeneous materials approach homogeneity in each  small local region.  And in these regions, intensive variables have the same values as if you had if you had a large homogeneous region under the same local conditions.

The intensivity of a variable doesn't depend on the homogeneity of the system.
Consider system A, a metal rod of length 2 with a gradient of temperature :

This metal rod has an average temperature, which you could calculate.
Now consider system B, a metal rod of length 4 with the same gradient of temperature :

This metal rod also has an average temperature. Furthermore, it is the same temperature as system A. (if you are unsure about that try to pick a gradient (I arbitrarily chose T = exp(-x) and T = exp(-x/2) for the graphs) and do the calculations yourself)
System B is bigger (it would be twice as massive) but its temperature is the same. Ergo the temperature doesn't depend on the size of the system and is an intensive variable.


text:
Why does alkyl rearrangement happen in the Hofmann rearrangement?
In the mechanism of Hofmann rearrangement, one step involves migration of an R−group from carbon to nitrogen.

This reaction proceeds even if the migratory aptitude of the R-group is poor (like in the case of ethanamide). What motivates the migration of the R-group and makes the reaction proceed?

Why do you say that migratory aptitude is poor? Poor relative to what?

The ‘motivation’ is that migration is the only way in which the reaction proceeds in a productive manner towards the product. From seeing the mechanism alone, you have no idea if the reaction of a particular substrate is rapid or slow, if it requires higher or lower temperatures etc. Only if you start taking these into account, you can tell whether the compound in question reacts rapidly or not.
I would not be surprised to see vastly different reaction times between compounds, depending on whether the $\ce{R}$ group easily migrates or not.


text:
Gel in water machine
I have a hot and cold water dispensing machine from my water company. I use 5 gallon water jugs delivered ever 3-4 weeks. The machine has a tank in it where water is kept cold until dispensed by gravity into whatever container I choose to use.
In the bottom of the tank I found a gelatin looking liquid in the bottom.  There was also ice formed where that tank is cooled. My company says its from the hot part of the machine heating water then the cool part cooling it causing the minerals to separate and not dissolve back into the water. I have never seen gelatin minerals before but wondered it this is something I need to look into further.


text:
Do gases behave ideally near the critical temperature?
In class, my TA said the closer the temperature of a gas is to the critical temperature, the more ideal it is. I'm struggling to understand this, as I thought more ideal behavior occurs at high temperature and low pressure. I am probably missing something really obvious :(

You're right. Pressure and temperature interact in a compressible gas. So any sweeping statement about either pressure or temperature alone is spurious.

I'd say critical behavior is the farthest possible from ideal.


text:
Dangers of expired bleach?
I have some unopened Clorox bleach bought over a year ago (at $8.25~\%\ \ce{NaOCl}$) and read on Wikipedia that bleach naturally decomposes according to
$$\ce{3 NaOCl (aq) -> NaClO3 (aq) + 2 NaCl(aq)}$$
Now, while there potentially might be some further decomposition as
$$\ce{2 NaClO3 (aq) -> 2 NaCl (aq) + 3 O2 (g)}$$
Would I be at risk of filling the air with toxic chlorate intermediates after using this bleach to clean up an ultrasonic ("cool mist") humidifier?

And why do you think chlorate would be any more dangerous then NaClO ?

Because its detection threshold may be above safe exposure limits

Here's another interesting link to consider http://blogs.edf.org/health/2017/02/03/perchlorate-risks-from-bleach/

That's perchlorate( $\ce{ClO4-}$ ), the article is talking about. You are interested in chlorate($\ce{ClO3-}$).

Yup I'm aware, just possibly yet another intermediate to consider though. Not sure how reliably either decomposes to safe exposure limits (especially for airborne/inhalation routes!)

I doubt that perchlorate forms spontaneously from chlorate, IIRC, perchlorate is made by electrolysis.

To put it simply: old bleach does not contain NaClO, hence it is not useful for cleaning. Do not use it.

You have outlined the principal decomposition path of hypochlorite in the question. Chlorate in itself is rather stable under ambient conditions and would probably not decompose.
The greatest danger is if the bleach solution is no longer alkaline. If acidified, both hypochlorite and chloride as well as chlorate and chloride can comproportionate to give chlorine gas according to equations $(1)$ and $(2)$.
$$\begin{align}\ce{Cl- + OCl- + 2 H+ &-> Cl2 + H2O}\tag{1}\\
\ce{ClO3- + 5 Cl- + 6 H+ &-> 3 Cl2 + 3 H2O}\tag{2}\end{align}$$
This can happen if carbon dioxide is absorbed from the air by hydroxide according to $(3)$.
$$\ce{CO2 + OH- -> HCO3-}\tag{3}$$
However, you should be able to smell chlorine early enough if it starts to get dangerous. If upon opening the chlorine smell is stronger than usual, open windows immediately and leave the room.
All other decomposition products should not cause any great harm. However, chlorate’s oxidative ability is less than hypochlorite’s to the best of my knowledge, so you might find that the partially decomposed bleach does not clean as well as the fresh one.

Bleach decomposition can proceed with two pathways i.e. one is forming chlorate and other one is forming chloride.
$$
\begin{align}
\ce{3NaOCl -> 2NaCl + NaClO3} & \tag{R1}\\
\ce{2NaOCl -> 2NaCl + O2} & \tag{R2}\\
\end{align}
$$
R1 is dominant although both reaction requires heat and pH control.

These two decomposition reactions of $\ce{NaClO}$ solutions are maximized at pH around 6. The chlorate-producing reaction predominates at pH above 6, while the oxygen one becomes significant below that. For example, at 80 °C, with $\ce{NaOCl}$ and $\ce{NaCl}$ concentrations of 80 mM, and pH 6–6.5, the chlorate is produced with ∼95% efficiency. The oxygen pathway predominates at pH 10. This decomposition is affected by light and metal ion catalysts such as copper, nickel, cobalt, and iridium. Catalysts like sodium dichromate $\ce{Na2Cr2O7}$ and sodium molybdate $\ce{Na2MoO4}$ may be added industrially to reduce the oxygen pathway, but a report claims that only the latter is effective.

Point to note that bleach can react with carbon dioxide and moisture if it left untouched for a long time. Hypochlorous acid that is created is somewhat unstable and it might decompose to give chlorine fumes or further react with hypochlorite to form chlorate.
$$\ce{OCl- + 2HOCl -> ClO3- + 2HCl}$$


text:
Thermodynamics Vapor Tables Problem

3-29 A piston-cylinder device contains $\pu{0.85 kg}$ of refrigerant-134a at $\pu{-10^\circ C}$. The piston that is free to move has a mass of $\pu{12 kg}$ and a diameter of $\pu{25 cm}$. The local atmospheric pressure is $\pu{88 kPa}$. Now, heat is transferred to refrigerant-134a until the temperature is $\pu{15^\circ C}$. Determine (a) the final pressure, (b) the change in the volume of the cylinder, and (c) the change in the enthalpy of the refrigerant-134a.

This is my work so far...

These are the superheated R-134a vapor tables that are given:

For some reason this question is all over the internet and there are numerous solutions readily available from a brief Google search (here and there). Assuming there is no way you haven't seen them yet, probably you should explicitly state what is unclear to you and what is hard to understand. I'd suggest to handle units accurately, to start with, and don't mix apple and oranges.

Those are the same answers I got, so I suppose the solutions manual I'm using is either wrong or using EES yields more precise answers. Thanks.

Yep, I honestly looked only very briefly, but it may just be the matter of rounding off numbers. The only thing that draws attention is inconsistency in typography (e.g. pa, Pa, Pa), but hopefully you'll sort it out before submitting the papers.

Did you see my answer (below)?  Any questions?  Another (equivalent) way to do it accurately is to interpolate v with respect to 1/p rather than p.

Sorry, I just read your replies. You're right, I tried the second method (interpolating with respect to the inverse of p) and it worked as well. The only thing I'm having a problem understanding now is when to apply it. Is it done when the gas is behaving ideally? I tried repeating the same problem using the Nelson-Obert generalized compressibility chart and got Z1 = 0.975, which would yield v1 = 0.2313 when using v=Z1R'T/p

I got the same answer that the book got.  Linear interpolation on v like you did is not accurate enough.  You expect the behavior to be close to that of an ideal gas.  So pv should be nearly constant.  Try interpolating on pv rather than on v and see what you get.
At p=0.06, pv=0.021029
At p=0.1, pv=0.020743
So, at p = 0.0904, pv=0.020812
v=0.020815/.0904 = 0.2302


text:
Does the amine group participate in resonance in aniline?
When we draw resonance structures of aniline, 8 electrons participate in resonance, making it antiaromatic. But in our school textbook, it is said to be aromatic. Does the lone pair of $\ce{NH2}$ participate or not? If it doesn't, then it would make sense.

Who said that 8 means antiaromatic? More importantly, when does this apply?

The lone pair of nitrogen is also part of the extended π-system; the molecule itself is still aromatic. As I already explained in the question about fulvene, aromaticity is nothing you can determine based on simple rules, especially not counting electrons.
Structural indicators like bond lengths and angles give a good idea how well established resonance is. Additionally you can draw a qualitative molecular orbitals diagram and look if it is similar to benzene. Eventually the determination of aromaticity is an experimental one, for example with NMR spectroscopy.
From the crystal structure (M. Fukuyo, K. Hirotsu, and T. Higuchi, Acta Cryst. 1982, B38, 640-643. DOI: 10.1107/S056774088200363X) we know that the carbon-carbon bond lengths are all similar and between 136 and 141 pm. The NMR spectrum clearly shows the protons have aromatic shift, see chemicalbook.com.
The molecular orbitals of aniline (calculated at DF-BP86-D3BJ/def2-SVP; Gaussian 09 D.01) show that the nitrogen is part of the π-system. They are also very similar to benzene. Here are the orbitals of the π-system in decreasing energy. Virtual (unoccupied) orbitals are red/yellow, and occupied orbitals are blue/orange.

For aromaticity, you need several rules to be satisfied:
1) The compound must be cyclic;
2) It must be ring-conjugated (no cross-conjugation!);
3) The number of electrons on π-orbitals in ring conjugation must obey Huckel's rule and be equal to a number of (4n+2) electrons, where n is a whole non-negative number.
Al that means that we need to count only those electrons, which are in aromatic cycle, but we do not care for "exo" electrons.

You can read more about quinone non-aromaticity here, here, and some general information about aromaticity here.
Back to your question:

Resonance forms 2-4 do not have any kind of ring conjugation, as when the aminogroup is involved into resonance, it gives its non-bonding pair for bond formation, and makes the compound cross-conjugated. Since there is no ring conjugation the forms 2-4 are non-aromatic.
Aminogroup does participate in resonance, but it has little influence on aromaticity.

1,4-Benzoquinone certainly is aromatic as the NMR shows.

You cannot view resonance forms as independent structures, therefore "Since there is no ring conjugation the forms 2-4 are non-aromatic." is just plain wrong. Either the molecule is aromatic (as a whole) or it is not.

I won't agree about benzoquinone aromaticity. You can predict aromaticity of compounds by chemical shifts of hydrogens, but the NMR is not always the only sign, as many papers say. You also need to consider ring currents, right? There is certainly a lot of debates in this topic, but I would think that quinone is much closer to unsaturated ketones with all their properties, than to aromatic compounds.

Regarding to your resonance remark: aniline is aromatic, and its stabilization energy is even larger than the one for benzene because of delocalization. It never loses its conjugation, but following the formal rules, I can't call these structures fully aromatic.

Then you are using an old set of rules, or you did not understand them. You can always ever only discuss the resonance hybrid as a whole. And Hückel's rules simply do not apply to this compound.

In fact, there are exceptions to each of the points 1), 2) and 3). What sense do these rules make, if for every rule, there are exceptions?

@Martin-マーチン when you say it is aromatic because of the NMR, what do you mean exactly? Are the 1H shifts sufficiently different from that of a typical α,β-unsaturated ketone such that it may be attributed to a ring current?


text:
Can carbocation rearrangements have kinetic and thermodynamic products?
As the figure in this webpage says, at low temperatures, the major product of hydrohalogenation of 1,3-butadiene is the kinetic product, the 1,2-addition product. At high temperatures, the major product is the thermodynamic product, the 1,4-addition product, in which the pi electrons have transferred to the inner carbon skeleton to form a more substituted alkene product. The carbocation rearranged due to resonance.
However, as the figure below from David Klein's Organic Chemistry 2e book tells us--the major hydrohalogenation product of 3-methyl-1-butene is the rearrangement product, which is the more stable product. Here, the carbocation rearranged due to its inherent stability at highly substituted carbons. The temperature of the reaction was not specified.

My questions are:

Is it safe to assume that the hydrochlorination of 3-methyl-1-butene give a product ratio as shown in the figure at all temperatures (except at temperatures where decomposition may occur, if it can)?
Do alkene addition reactions also have thermodynamic and kinetic control, like the butadiene reaction exhibited, or is the thermo and kinetic product production only due to resonance?
Can carbocation rearrangements have kinetic and thermodynamic products, like the one for the hydrohalogenation of butadiene?

These are fundamentally different. Resonance isn't a "rearrangement".

Confused by title. Being under thermodynamic control implies not being under kinetic control.