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Nitric acid acting as oxidising agent
In the following mechanism, nitric acid acts as an oxidising agent to remove 2 H atoms from the intermediate (in the blue square)

My first instinct if given only the reagents would be to protonate something, such as the amine group for example, rather than removing hydrogens.
So my question is how can I identify if $\ce{HNO3}$ will act as an acid or an oxidising agent, given only the reagents? Do I just have to try out both and see which one leads to a reaction? Or is there a property of nitric acid that I'm missing?

If compound can be easily oxidised and you add strong oxidating agent...

This is kind of a special case, the dihydropyridine is extremely easily oxidised because the pyridine product is aromatic, nitric acid usually behaves as an acid but it also has oxidising properties so as @Mith said...

@orthocresol so there's no real way of telling, unless I know the full reaction? (eg. knowing that this reaction is thermodynamically favourable due to aromaticity)

The reactions of nitric acid could be due to 1) its acid nature, 2) its oxidizing properties, or both.
Orthocresol was correct that the substrate could predetermine which course was more likely: acid-susceptible substances (like alkalies, or alkaline materials) would respond to the acid nature; oxidizable substances would respond to the oxidizing power (like organic alcohols and sugars).
An interesting case is pure iron: in dilute nitric acid (less than 10%), it dissolves. In concentrated nitric acid, it is oxidized (passivated) and does not dissolve.
Increased temperature would also increase the likelihood of an oxidation reaction compared to an acidic reaction, because the activation energies for oxidations are usually higher.


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How does the Rf value change when changing the eluent composition?

A non-polar molecule X is analyzed by TLC chromatography on silica to determine its $R_f$ value.
The first eluent was hexane:EtOAc = $9:1$ and the $R_f$ value was $0.5$.
In the second run, the eluent was changed to hexane:EtOAc = $7:3$. Will the $R_f$ value be smaller or greater than $0.5$?

I am confused because there are two important interactions: because the 2nd system is more polar, it will form weaker interactions with substance X, and the $R_f$ value will decrease; on the other hand, the molar polar eluent will form stronger interactions with stationary phase, so the $R_f$ value will increase because the substance X is more easily eluted.

The most important interaction is that of the mobile phase with the stationary phase. If you increase the mobile phase’s polarity, that interaction will increase, decreasing the interaction of the solute with the stationary phase. Therefore, a higher ethyl acetate content in the mobile phase will increase $R_f$ values.

What if the molecule  is extremely non-polar and the interactions with mobile phase are dominant? Is this possible? Could I have a reference for your response?

@Uros If the interactions with the mobile phase are dominant, the $R_f$ value in pure hexane will be $> 0.5$. Source: Selling my soul to the chemistry lab for the last five years.

Part two: Yes, in these lab years I had my fair share of compounds whose $R_f$ values in pure isohexane were larger than $0.8$ (which technically makes them impossible to isolate in standard silica chromatography). No matter which other mobile phase mixture I used, their $R_f$ value never sank.

@Jan The capillary action is different for different solvents. How then we can compare the $R_f$ values?

@adosar We basically cannot predict the $R_f$ value of a compound in a different solvent (i.e. if it has $R_f=0.5$ in $3:1$ that only tells me the $R_f$ value will be lower in $4:1$, I cannot give a specific value; both solvents $\text{hexane}:\ce{EtOAc}$). We can, of course, perform multiple TLC experiments and compare the outcomes which leads to the rule of a more polar ratio of the same solvents always giving a higher $R_f$ value for compounds.

As, mentioned by Jan Increasing the polarity of solvent will result in increase in Rf value as the compound in even for polar or nonpolar compound, this is the rule of thumb in most cases, may be except do exist i don't know about it. But for my experience in the laboratory polarity result increase in Rf value.

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Is this equation of state original?
I created this equation of state as a personal undergraduate summer project (here is the link)  seven years ago:
$$\frac{PV}{nRT}-1 \propto \left(\frac{nRT}{PV}\right)^2 \left[\left(\frac{nRT}{PV}\right)^2 - 1\right]$$
I tried to show it to my professors but none of them replied to my emails. I had since forgotten about it until I recently came across it in the cloud. Here is a table comparing its theoretical predictions for the critical temperatures of various gases to the actual experimental values

How do you use it in order to make predictions of $T_c$?

@user1420303 I'm guessing $$\left(\frac{\partial p}{\partial V}\right)_T = \left(\frac{\partial^2 p}{\partial V^2}\right)_T = 0$$ although the link says $T_c = \varepsilon/(k \ln 2)$ (not sure how the derivation works)

@orthocresol Sorry, I was not able to open the link before, it took me to the table image with the SE App . I am going to read it from my PC in minutes.

@user1420303 It was broken just now, I fixed it. ;)

I think it is original, although I'm not sure about some points in the derivation.

@user1420303 Is there any hope of turning this into something publishable?

@KingChem , I hope so, but I honestly have to say that I disagree in some points of the derivation.

@user1420303 Funny that you have said that, for after hearing no reply from my professors I did submit it to this physical chemistry journal but they replied with a rejection saying that "it makes incorrect assumptions". After that I lost interest and discarded it until I rediscovered it. However  it works, therefore either the core of it is correct but it was arrived at in an incorrect way or the existing theory is incorrect. I'm not so big-headed as to believe that I've overthrown decades of chemical knowledge so it must be former. Maybe amended by a professional a journal may accept it?

I can assure that, I hope so. I read your work with enthusiasm but I do not agree with:

Z-1 be proportional to U(r).
Even accepting 1), when you assumed U(r)=0 for ideal gas law behaviour and replaced according to obtain the $NkT/(pV)$, as you imposed ideal gas behaviour that should be 1, and it is valida for every $V$. Also you assumed it in the right hand side but not in the left hand side.

Then after you solve for $V$ ( I don't know how did you do that), you relate the roots to different energy levels, but you first fixed the energy to only one value.
I would don't know how to transform it in something that looks consistent to me. But the way, the results are amazing to me.
I am sorry but I preferred to be sincere.

@user1420303 Thanks for your support. I haven't done chemistry for years so I'll look over the points you made and get back to you tomorrow (it's 1:30 am now). The site is asking me if I want to move the conversation to chat, do you want to do that?

@KingChem , there is no problem, but not today, I have a lot of commitments and don't have extra time until tomorrow

@user1420303 to your first question: U(r) is a measure of the attractive force between molecules, Z is a measure of the average distance between molecules. it follows that if U(r) is increased, Z is reduced making Z-1 increase and if U(r) is decreased, Z is increased making Z-1 decrease. Therefore Z varies in direct proportion to U(r). To your second question: I shouldn't have wrote that, PV=zNkT and that is always valid. To your third question: I'm just saying that it's a quintic so by the fundamental theorem of algebra it will always give five roots.

@user1420303 To your fourth question: I just noticed that the root with the largest value was always P/nRT.

Yes, this equation of state is original. While I cannot absolutely guarantee that it has never been used or proposed before, I can say that I have not encountered it in any “classical” model — and I teach thermodynamics and statistical physics, so I have seen many textbooks and classical exercices :)
Extending beyond the question of originality, I read your demonstration and what bothers me to some extent is the link you establish between $Z$ (the compression factor) and $U(r)$, the intermolecular potential energy. First, let's state that linking $Z$ to $U$ is indeed a common way to derive equations of state. However, the way you are doing it is lacking a sound basis. The sentence “From these explanations it can be said that $Z-1=U(r)$” (page 1) is the problem:

Yes, if $U(r)>0$ then $Z>1$, and the other way around… but that does not mean that the two quantities are proportional.
In fact, $Z$ is a scalar (function of $P$ and $T$), while $U(r)$ is a function of distance (and not thermodynamic variables). So they cannot be linked directly.

You intuited between $Z$ and $U$, which exists, but it needs to be formalized a bit more carefully. Standard approaches to do so rely on perturbation theory or mean-field method. You can check for examples the classical derivations of the van der Waals equation of state to see how this can be done. But it requires some fundamentals of statistical physics…

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How does the relative amount of oxidation of different alcohols affect the amount of energy released?
I am investigating the enthalpy of combustion of different primary alcohols and a website I found briefly mentioned how the relative amount of oxidation of different alcohols affects the amount of energy released. I was wondering if anyone could possibly elaborate on this.
I also found a trend where carbon chain length 1 --> carbon chain length 2 has a great increase in experimental heat of combustion but then there is a smaller increases between carbon chain length 2 and 3. Would there be an explanation for this (I have a feeling it relates to incomplete combustion)?
Thanks!

It doesn't have to do with incomplete combustion, as the enthalpy values you've seen reported are for the enthalpy of complete combustion.
It has to do with the alcohol already being partially oxidized, so there is less energy available for release upon the oxidative process of combustion.  In other words, if a particular carbon atom is already partially oxidized, like in an alcohol, then you can't get as much energy from combustion as compared to a more reduced carbon, like that in a methyl group for example.
This effect is most pronounced for the low-molecular weight alcohols.  Methanol has 1 oxygen per carbon atom, while ethanol has 0.5 oxygens per carbon atom.  That is a much larger relative difference than there is between a $\ce{C10}$ alcohol and a $\ce{C11}$ alcohol.  In other words, the larger a mono-hydroxy alcohol is, the more it resembles a pure alkane with respect to the enthalpy of combustion.


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Why doesn't the bidentate ligand form dative bonding three times?
I was looking at this question where  it asked you to work out the final complex formed when an excess of 1,2-diaminoethane is added to and aqueous solution of copper (II) sulphate.
So the equation I worked out would have been:
$$\ce{[Cu(H2O)6]^2+ + 3(NH2CH2CH2NH2)-> [Cu(NH2CH2CH2NH2)3]^2+ + 6H2O}.$$
Therefore I worked out the complex to be: $$\ce{[Cu(NH2CH2CH2NH2)3]^2+}$$
However the answer was : $$\ce{[Cu(NH2CH2CH2NH2)2(H2O)2]^2+}$$
Why is this?

Jahn-Teller effect

@orthocresol What other transitional metals would show this effect? (is it all elements in row 11 or is this effect unique only to copper (II) )

Look at the Wikipedia page, it has a good explanation... Ag(II) and Au(II) are very rare anyway

Orthocresol already pointed you towards the Jahn-Teller effect. The effect is there in $\ce{[Cu(H2O)6]^2+}$, the hexaaquacopper(II) cation, but more pronounced in the tetraammindiaquacopper(II) cation $\ce{[Cu(NH3)4(H2O)2]^2+}$. Compare an image of the complex as shown below:

Even the naked eye can see that the copper-oxygen bonds are much longer than the copper-nitrogen ones. Measuring the distances in crystal structures containing that cation gives $203~\mathrm{pm}$ for the $\ce{Cu-N}$ bonds and $251~\mathrm{pm}$ for $\ce{Cu-O}$.[1] Why is this the case? To answer that we need to refer to the molecular orbital scheme of a typical octahedral complex:[2]

The three lowest groups of orbitals containing six MOs in total are typically populated by the ligand lone pairs which form the coordinate bond. The next set of two groups — five orbitals in total — are the metal d-orbitals that can (or cannot, in case of $\mathrm{t_{2g}}$) interact with the ligands. For copper(II), we need to fill in nine metal d-electrons meaning that the $\mathrm{t_{2g}}$ level is fully populated while $\mathrm{e_g}$ has three electrons in two orbitals — a denatured state.
This denatured state is not energy-optimal. Energy could be released if we somehow manage to unequalise the $\mathrm{e_g}$ orbitals, give them different energies and have the lower-energetic one be fully populated. Remember that the $\mathrm{e_g}$ orbitals are $\mathrm{d}_{x^2-y^2}$ and $\mathrm{d}_{z^2}$. In a gedankenexperiment, remove the two ligands in $z$-direction symmetrically away from the metal centre. Any orbital that has a $z$ contribution should now be stabilised — most notably, that includes $\mathrm{d}_{z^2}$. The orbital scheme looks slightly different, the complex’ symmetry is no longer $O_\mathrm{h}$ but $D_\mathrm{4h}$ but we have gained energy by being able to differentiate between the $\mathrm{d}_{x^2-y^2}$ and $\mathrm{d}_{z^2}$ orbitals. This is the Jahn-Teller effect.
Consequences of the effect include that ammin ligands preferentially coordinate copper(II) equatorially, and that the tetraammindiaqua complex is the preferred one in high aquaeous ammonia concentrations. Hexaammincopper(II) can be prepared only in liquid ammonia to the best of my knowledge.
Now let’s consider using $\ce{en}$ (ethylenediamine) rather than $\ce{NH3}$. The bridge between the two amino groups is long enough to bridge across one edge of the octahedron but not any further. Replacing all the equatorial ligands with $\ce{en}$ makes sense because we will get another very symmetric complex $\ce{[Cu(en)2(H2O)2]^2+}$, and because these are the ones that are replaced first. If we now wanted to replace the two other water molecules, we would create a rather distorted $\ce{[Cu(en)3]^2+}$ complex, where two $\ce{en}$ molecules have one short, strong and one long, weak bond. This is an unfavourable condition, since the former shorter, stronger bonds have to be weakened to get there. In short, $\ce{en}$ behaves like ammonia and only occupies the equatorial positions, not the axial ones. A consequence again of Jahn-Teller.

[1]: Data taken from Professor Klüfers’ internet scriptum for his general and inorganic chemistry course at the LMU Munich (section 23.4).
[2]: First presented in this answer and originally taken from Professor Klüfers’ coordination chemistry course.


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Supramolecules in Spartan
Maybe someone here is familiar with supramolecules calculations using Spartan? I am new in this area and trying to figure out some things:
First - what's the best way to draw them? I tried to draw one molecule and then using copy-paste append the others but then they overlap and it takes forever to minimize their energy so they make hydrogen bonds in the places I need. Maybe there is any way to 'freeze' hydrogen bonds? Because I now where they shoud be (or at least want to calculate different structures with hb in different places of molecules). If I try to put this molecules near when sketching still when I try to minimize the energy hb forms not in the exact places I need them to be. And I need four molecules to form kinda cyclic-structure.
Thank you in advance!
hb I need between two molecules are:

You can create distance constraints between two atoms (set them to the distance you expect for the hydrogen bonds), and then do a force field minimisation.
It's in the menu called Geometry, and then "Constrain distance". You pick the atoms, and click the little lock in the lower right-hand corner, and then enter the distance you want next to the lock.

Thank you! But if I don't know what should be the distance? How I could calculate it first?

I would just use a typical value (1.6-2.0 Å, Wikipedia) as a first approximation. Then after you get the geometry in the ballpark, you can remove the constraints and do a more accurate minimisation.


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How much pressure does Gallium exert when it freezes?
Assuming the liquid Gallium is in a sealed container, how much pressure will it exert on its container when it freezes?

I suggest to reopen this question. Its intention is quite clear, and it can be answered meaningfully.

related http://chemistry.stackexchange.com/questions/34929/what-happens-if-you-cool-water-in-a-container-too-small-for-it-to-freeze

What's the volume of the container? We'll need to know that, as pressure is force per unit area. The coefficient of thermal expansion for gallium can be used here, but only if we're assuming the container doesn't yield to the expansion forces - is that part of your query, or do you have a material in mind for the container?

Enough to break a glass container.
Gallium is one of the few substances that expands on freezing (like water). The solid is about 3% less dense than the liquid. This is enough to shatter a glass bottle full of gallium on freezing (see this video).
Solids that expand on freezing can exert a lot of pressure (clearly more than enough to break glass) but the exact amount of pressure probably depends on the vessel and the circumstances of freezing.

Suppose we have a substance capable of a phase transition in which it expands. Suppose we put it in a sealed container and change the temperature past the transition point. What would happen? Apparently, some portion of the substance would undergo the transition, which would result in pressure increase, which in turn (according to the Le Chatelier's principle) would attempt to drive the transition back, and so the new equilibrium will be reached.
As to the extent of the effect, let us consider $\Delta G$ of the said transition. When the system is at equilibrium, $\Delta G=0$. If we change our pressure and temperature (not by much, so the linear approximation still applies), $\Delta G=\left({\partial(\Delta G)\over\partial T}\right)_PdT+\left({\partial(\Delta G)\over\partial P}\right)_TdP=\Delta V dP - \Delta S dT$. Now, this new state is also an equilibrium, so this new $\Delta G$ is also $0$. This gives us certain relation between $dT$ and $dP$, namely: $dP = {\Delta S\over\Delta V}dT={\Delta H\over T\Delta V}dT$. By looking up the molar volumes of solid and liquid gallium, and also its enthalpy of melting, you may find out exactly how many hundreds of atmospheres of pressure would we get for each degree of temperature.
Note that this description applies to any phase transition that goes with expansion, be it freezing of gallium (or water, or one of the few more exceptions), or melting of nearly any other substance. The signs of $dT$ and $\Delta H$ in case of melting are opposite to those in case of freezing, but the overall expression is still the same.
Of course, at some point all gallium would freeze, so further cooling will not increase pressure any more. To calculate this point, we need the compressibility figures which are not that easy to come by. Also, at pressures that great we can hardly ignore that the container walls must give in to some extent, no matter what they are made of.


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How to determine mobility when two mechanisms exist simultaneously

Two scattering mechanisms exist in a semiconductor. If only the first
  mechanism is present, the mobility will be $\pu{250 cm2/V s}$. If only the
  second mechanism is present, the mobility will be $\pu{500 cm2/V s}$.
  Determine the mobility when both scattering mechanisms exist at the
  same time.

I'm not sure what the last sentence is trying to tell me. I know mobility is $q\tau_c/m_n$, and I think I need to solve for mean free time here to help, but I'm not sure what I should use for $m_n$ since I don't know what the semiconductor is, and I'm not given a value.
Although, since I'm given two mobilities to start with, should I solve for $\tau_c$ twice and average those? Really not sure what direction I should be going in (if only my professor answered his emails).

I needed to add the inverses of the given mobilities and take the inverse of that to get my final result.
$1/\mu = 1/\mu_L + 1/\mu_I$

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Is there any selectivity between the C-2 and C-4 positions of pyridine in SNAr reactions?
In nucleophilic aromatic substitution (SNAr) reactions on pyridines, displacement of leaving groups at C-2 and C-4 is favoured because the negative charge in the intermediate can be delocalised onto the nitrogen atom.
However, is there any intrinsic difference in the rates of substitution at C-2 and C-4 (assuming the leaving groups at both positions are the same)? If so, why?

True, the para (C4) and ortho (C2) positions give similar resonance structures but consider the fact that:

For an $\ce{S_{N}Ar}$ reaction the electron rich nucleophile must attack. Nitrogen, being highly electronegative, does draw electrons away from the ortho and para positions but because the electron clouds would be larger around nitrogen, it is harder for the nucleophile to approach the ortho than the para. You could think of it as the lone pair on Nitrogen being a steric shield of some sort.
An attack on the para position gives a symmetric intermediate. This would be lower in energy compared to the intermediate produced by an attack on the ortho position.

As such, the para product would be the kinetic product of the reaction whereas the ortho product would be the thermodynamic product. Statistically ortho should be attacked more often as well, as there are two positions whereas there is only one para position.

I really don't think it's that simple. If you look in the literature (pyridine SNAr is very common in med chem for example), there are examples of both selective C-2 and selective C-4 substitution, and I think it really depends on conditions and exact substrate. See e.g. Baran group meeting "Pyridines readily undergo SNAr, usually faster at C-4 than C-2/6, but highly dependent on nucleophile and conditions"

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What is the molecular mechanism behind oncotic pressure?
I can't wrap my head around the idea of oncotic pressure and osmolarity, logically.
I imagine a blood vessel. It is filled with proteins, like a ton of proteins and solutes. So this means, according to oncotic pressure and osmolarity, that water will diffuse into this blood vessel.
I can't understand that. If I have a glass full of water, and I drop a bunch of rocks into the water (my proteins) the water will flow out of the glass. It will exert a force, a very basic force of buoyancy, that will lead water out of the glass.
How exactly is oncotic pressure generated?

blood vessel. It is FILLED with proteins - it's not. It filled with solution. And that solution differs by 0.5% of that osmotic pressure from cells solution. And that allows to move  in- and out- cell, parts of solution from and in blood vessel. And direction of that movement will be directed by small changes(3-4kPa) in pressure. Which allows for blood vessels to do their job. So it's not like rocks and glass, it's like sponge, press harder get water, release and it will suck it. 3-4kPa is pretty close to difference between artery and veins pressures, and that helps direct where waste goes.

The molecular basis of osmotic pressure (oncotic pressure is a special type of osmotic pressure) is similar to filtration but on a very small scale.
Some membranes are partially permeable: they let small molecules pass but block larger ones. This asymmetry means that a flow of small molecules is created. Imagine things at the molecular level where a lot of individual motions are essentially random. The big molecules can't pass through the membrane but small molecules like water can pass through. If one side of the membrane consists of small molecules only and the other side has small and large molecules, then random motion will tend to lead to a net flow from of small molecules into the side containing large ones just because the concentration of small molecules is lower on the large molecule side. There are fewer opportunities for random small molecules to move from that side to the small molecule only side because there are fewer small molecules per unit volume on that side.
Obviously I'm simplifying a little, but this is the essential reason behind osmotic pressure.

I think the OP understands osmotic pressure but wants to know how macromolecules (which actually form a colloidal solution) exert similar kind of effects. Basically they are assuming that a macromolecule suspended in a solvent is like an inert rock in water.

@WYSIWYG You might be right, but my explanation still holds. The explanation at the molecular scale isn't like dropping rocks into a glass: you have to think about the probabilities of water molecules passing through the membrane in different directions.


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Does the reaction between sulphuric acid and ammonium hydroxide give off heat?
I have a mystery solution lab in chemistry, and I was wondering if the addition of $\ce{H2SO4}$ to $\ce{NH4OH}$ gives off heat? Or is there maybe a precipitate or color change?

$$\ce{2 NH4OH(aq) + H2SO4(aq) -> (NH4)2SO4(aq) + 2 H2O(l)}$$
This is an acid base reaction. $\ce{NH4}$ compounds are always water-soluble, so no precipitate, and no color change will be observed. Neutralization reactions like the one above are generally exothermic (they release heat).
This can be proved by observing their heats of formation:
$$\begin{align}
\Delta H_{\mathrm f(\ce{(NH4)2SO4})} &= \pu{-1181 kJ/mol} \times \pu{1 mol} &&= \pu{-1181 kJ} \\
\Delta H_{\mathrm f(\ce{H2O(l))}} &= \pu{-285.8 kJ/mol} \times \pu{2 mol} &&= \pu{-571.6 kJ} \\
\Delta H_{\mathrm f(\ce{H2SO4})} &= \pu{-909 kJ/mol} \times \pu{1 mol} &&= \pu{-909 kJ} \\
\Delta H_{\mathrm f(\ce{NH4OH})} &= \pu{-80.8 kJ/mol} \times \pu{2 mol} &&= \pu{-161.6 kJ}
\end{align}$$
Thus the $\Delta H_{\text{r}}=(\pu{-1181 kJ} + \pu{-571.6 kJ}) - (\pu{-909 kJ} + \pu{-161.6 kJ}) = \pu{-682 kJ}$ is negative, hence the reaction is exothermic.


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Mechanism of polymerization of 4-nitroaniline
Does anyone know the mechanism for/the structure of the product of the explosive polymerization of 4-nitroaniline with concentrated sulfuric acid? My guess is that it is some sort of aromatic substitution reaction, leading to a number of benzene rings connected by nitrogens.

4-nitroaniline?

This article by the scientists of the University of California Santa Barbara should pretty much cover your question.

I was wondering if any more research has been done on the topic.


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Understanding Gibbs free energy and enthalpy
Let's say that a reaction releases -50kJ/mol of Gibbs free energy (delta G), and -75kJ/mol of enthalpy (we know that -25kJ/mol went to make sure that the total entropy of universe increased).
Now let's say we reversed the reaction.
Book says that the minimum energy required to make the reaction occur spontaneously in the reverse direction is the negative of the original change in Gibbs free energy (50kJ/mol).
What bothers me here is that I would think that you would need 75kJ/mol to reverse the reaction - the total energy the original reaction emitted needs to be put back in order to bring the system back to it's original potential energy level. It seems as though enthalpy shows that one amount of energy is needed to reverse the reaction, while Gibbs potential energy indicates a different amount of required energy.
I have no problem solving any of the computational problems relating to these concepts, but I'm curious about how to more fully understand them conceptually. I've noticed that a lot of people are great at solving technical problems related to these concepts, but are not too worried about how to interpret them conceptually. I guess for me, on the other hand, it's more important to understand than to compute...
Thank you for your help!

The equation for Gibbs Free Energy is:
$$\Delta G = \Delta H - T \Delta S$$
The change in Gibbs Free Energy already includes the amount of energy associated with the change in energy. So to reverse the reaction, the total change in energy is given by $-\Delta G$.
Now the question is how much energy do you NEED to get the reaction to go backwards. The minimum amount of energy for it to go backwards spontaneously is $-\Delta G$ because that puts the product on the same energy level as the reactants so in principle they are energetically equivalent. However, what you should really be concerned about was the activation energy of the forward reaction, $E_a$, because this is the mountain that the reaction needs to climb. There is a diagram below, but it is based on the $\Delta H$ values. You can convert this to a graph with $\Delta G$ as well.

Thank you for your answer IT Tsoi.
I wanted to write a comment, but it turned out too long, so I put it as an answer.
It looks like the to have the reaction reach the activation energy you need to give it is more energy than the enthalpy reversal and Gibbs reversal (75kJ and 50kJ in my example). So from the graph you attached, we can see that in practice you may need much more energy than either of these values (especially if you first need to break a bond before the second bond can form).
I was thinking why in principle Gibbs energy input can be less than 75kJ, and the reaction can still, in an idealized circumstance, reverse when given 50kJ.
I think maybe it's like this for reaction in question where the forward is spontaneous (ΔG = -50), and the reverse is not spontaneous (ΔG = +50).
1) You first give it negative ΔG, the entropy change is then bigger than zero, and the reverse reaction can go on.
2) At that point in an idealized circumstance the reaction can suck by itself the remaining  25kJ from the environment to bridge the gap between how much energy you gave it (ΔG) and how much energy is the potential energy difference between the reactants and products of the reverse reaction (ΔH). It can do so because even after taking an extra 25kJ from the environment, the change in entropy of the universe is greater or equal to zero (It's like water evaporating in a dark room by taking energy from the environment because total entropy change is greater than 0).
3) But it can't take more energy from the surroundings than ΔH - ΔG (75 - 50), because by taking more, the entropy of the surroundings will decrease too much making the total change of the entropy of the universe negative. So the (reverse) reaction couldn't take as much energy as is the activation energy, but can by itself get only up to 25kJ from the surroundings (after it's given 50kJ - the negative of ΔG of the forward reaction). To get to the full activation energy it needs an additional energy input, so in practice even after the reaction got 50kJ and then was able to acquire 25kJ more from the surrounding by itself (to a total of ΔH), it still needs more energy to reach the Ea.
So (maybe) in summary: if you give it ΔG, it can suck heat by itself to get to ΔH, but it can't go over ΔH by itself (otherwise ΔS univ will be negative), so to get to Ea it probably needs more energy inputed than ΔG.


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Difference between free energy of formation and free energy of combustion
I was wondering what is the main difference among those two values.

What exactly are you confused about? They are fairly self explanatory.

I think free energy of formation is the change in the Gibbs free energy when a compound is formed from its constituent parts (assuming that the system is in the standard state of 298K, 1atm pressure). What it means is how much work the reaction is capable of doing if it does the maximum possible work it can do (in case it's a negative number) or how much work you need to do on the system to make the formation reaction happen.
I never heard of free energy of combustion yet, but I guess it must be about the change in Gibbs free energy during a combustion reaction. So I think it's the change in Gibbs free energy for a combustion reaction when the reaction goes from the reactants to the products, and should have the same interpretation as the change in free energy for formation in general (probably the delta Gibbs value is going to be negative indicating that the reaction released a lot of energy that can be used for work).
Hope this helps...
(I'm a newb so hope others can verify my answer).


text:
What is the physics/chemistry behind gold dissolution in water/seawater?
Global ocean contains some amount of dissolved gold (around 0.05-0.005 ppb).

Is it ions of gold?
Is it just suspended atomic gold (like water vapor in gases - just suspended molecules)?
Can pure gold (in metal state) be dissolved in pure water at any rate (even it is very slow process)?
If it can be dissolved in pure water what is the physics behind? just diffusion?


text:
Nitrogen bonding and electron affinity
How does nitrogen bond with itself to form N2 if it has no affinity for electrons? If a nitrogen atom can't pull on the electrons of another nitrogen atom how can N2 have covalent bonds.

Why do you say Nitrogen Atoms have no affinity of electrons?

Covalent bonding is not about electron affinity at all. It is about two atoms getting together and creating molecular orbitals, some of which are lower in energy than the starting atomic orbitals, so the electrons may happily go there. It is a win-win situation.

It doesn't matter that the electronegativity difference between nitrogen with itself is 0. It just means the bonds it makes with another nitrogen atom have $100~\%$ covalent character.
A typical chemical bond basically spans the spectrum of bonds with $100~\%$ ionic character, uneven ionic/covalent character, and $100~\%$ covalent character.

$\mathbf{100~\%}$ ionic character: $\ce{A^+B^-}$ or $\ce{A^- B^+}$, with no sharing of electrons at all
Uneven ionic/covalent character (i.e. polar): $\ce{{\stackrel{\delta+}{A}}-{\stackrel{\delta-}{B}}}$ or $\ce{{\stackrel{\delta-}{A}}-{\stackrel{\delta+}{B}}}$, with deviation from perfect sharing of electrons and direction of favored negative/positive end of dipole determined by electronegativity difference
$\mathbf{100~\%}$ covalent character (i.e. $100~\%$ nonpolar): $\ce{A-A}$ or $\ce{B-B}$, with perfect sharing of electrons

It just so happens that $\ce{N2}$ is one example of $100~\%$ covalent character. You can argue the same for $\ce{O2}$, and any similar homonuclear diatomic molecule.
(Granted, there are some exceptions to 100% covalent character, like in iron's metallic bonding, but anyways...)

I’ll respectfully disagree with the first paragraph. I wouldn’t consider metallic bonding as in iron and covalent bonding entirely equivalent, but otherwise nice answer.

Well, I've personally never heard of $Fe_2$. I have heard of iron making an $Fe-Fe$ bond, though, like in diiron octacarbonyl. :)

But I was referring to elemental, solid $\unicode[Times]{x3b1}$-iron ;) Technically, these are very extended bonds.


text:
Composition of desiccant
Most types of desiccation packages use silica gel, but are there any that use calcium oxide instead? Here's my reasoning:

Calcium oxide absorbs water, forming calcium hydroxide.
Adding a (used) package to water produces a slight amount of heat (I may be imagining this).
The solution stays cloudy even after sitting for a while.
The solution almost immediately turns clear upon addition of vinegar.

At this point, I'm pretty sure it is calcium oxide and hydroxide in this package, but I just wanted to make sure first. For point 2, I should note that it is very humid in the place that I live, so it would make sense that most of the calcium oxide is converted into calcium hydroxide, and therefore not much heat is produced.

Yes, calcium oxide can be used as an desiccant.
The following quote is an excerpt from this site:-

Calcium oxide is calcinated or recalcinated lime having a moisture adsorptive   capacity of not less than 28.5% by weight. The
  distinguishing feature of calcium oxide is
  that it will adsorb a much greater amount of water vapor at a very low
  relative humidity than other materials. It is most effective where a
  low critical relative humidity is necessary, and were there is a high
  concentration of water vapor present.
Calcium oxide removes water from a package very slowly, often taking
  days to reach its maximum capacity. As calcium oxide adsorbs moisture,
  it swells. Proper desiccant packaging is required for effective use.
  For these reasons, its use has been limited to primarily the packaging
  of dehydrated foods.

This following is a table which compares the properties of various dessicants taken from this site:-

And the following graph depicts the adsorption capacity of various dessicants:-

For more information click on the following sites:-

http://www.ohe-chem.co.jp/e_qa.html
http://www.agmcontainer.com/support/selecting-desiccant
https://www.usaemergencysupply.com/information-center/self-reliance/food-storage-frequently-asked-questions/types-of-desiccants


text:
What do curly brackets {} mean?
I learned that square brackets ex: $[\ce{H2O}]$ can be used to denote the concentration of water molecules in a fluid. The notation  used much in equilibrium problems and such.
However, sometimes I have seen curly brackets {} being used in similar scenarios, but I never really understood the difference. When are curly brackets used and what do they mean?

I found this example from an exam where the problem is to see if a solution of $\ce{Ca(OH)_2}$ in water is concentrated enough to result in precipitation.
Following is what how they solve it in the example:

Solubility constant for $\ce{Ca(OH)_2}$ is given as $K_\mathrm{s} = \{\ce{Ca^{2+}}\}\{\ce{OH^-}\}^2$ (100% saturation point I think)
$Q = \{\ce{Ca^{2+}}\}\{\ce{OH^-}\}^2 \approx [\ce{Ca^{2+}}][\ce{OH^-}]^2$ (actual saturation I think)

Where $Q$ is the actual saturation of the solution. And the answer is that no precipitation occurs since $Q < K_\mathrm{s}$ in this specific example.
So they seem to suggest that {} is something different than the concentration [] and that it can be approximated to the concentration.

Hmm, the place I've seen curly brackets as you describe is for inorganic crystalline substances - see this link for an explanation. That said, they are used to denote the constitution of structural units and not used to enclose chemical species, so the use case I present might not be relevant to yours.

It doesn't seem to have anything with crystalline substances to do. I found an example where they use the denotation which I edited in.

In that context I would guess it is referring to the activity, although I don't think that is a standard notation for it, usually one would write $a_{\ce{Ca^2+}}a_{\ce{OH-}}^2$

Curly braces are often also used to denote encounter pairs, or molecules held together in solvent cages, but that's obviously not what you're talking about

The curly brackets denote "activity of" the species therein.
See the Wikipedia section: Basic definitions and properties of Equilibrium constant


text:
Hess's Law/Enthalpy of reaction
The problem:

Chloroform, $\ce{CHCl3}$, is formed from methane and chlorine in the
  following reaction:
$$\ce{CH4(g) + 3Cl2(g) ->3HCl(g) + CHCl3(g)}$$
Calculate $\Delta rH^{\circ}$, the enthalpy of formation for this
  reaction, using the enthalpies of formation of $\ce{CO2(g)}$
  (-393.509kJ/mol-rxn), $\ce{H2O(l)}$ (-285.83kJ/mol-rxn) and
  $\ce{CHCl3(g)}$ (-103.1kJ/mol-rxn) and the enthalpy changes for the
  following reactions:
$$\ce{(1) \ \ CH4(g) + 2O2(g) -> 2H2O(l) + CO2(g) \ \ \ \ \Delta
 rH^{\circ}=-890.4kJ/mol-rxn \\ (2) \ \ 2HCl(g) -> H2(g) + Cl2(g) \ \ \ \ \Delta
 rH^{\circ}=+184.6kJ/mol-rxn }$$

My attempt at the problem:
Given the hint in the problem statement and by looking at the two given reaction equations i assumed that i needed to create another equation. I created an equation for $\ce{CHCl3(g)}$ from $\ce{CO2(g)}$ and $\ce{H2O(l)}$ and calculated it's $\Delta rH^{\circ}$ from the enthalpies of formation as shown below (Let's call this equation 3):
$$\ce{2CO2(g) + 5H2(g) + 3Cl2(g) -> 2CHCl3(g) + 4H2O(l)}$$
$\Delta rH^{\circ} = -562.502kJ/mol$
Here's where i run into difficulties: I can see that in the 'target reaction' we have HCl on the right and CHCl3 on the right, so it stands to reason that equation 3 will stand as it is presented but equation 2 will switch direction (and the sign of the enthalpy of reaction swop too). I don't want to switch the direction of equation 1 around because it has CH4 on the left and in the target reaction CH4 is on the left. We could multiply equation 1 by 2 and equation 2 by 3 in order to have equal amounts of chlorines and Co2's so we can cancel them, however we are now left with hydrogens on the left and H2O's on the right with no way to cancel them. Hydrogen and water obviously don't appear in the target reaction so i have gone wrong somewhere along the line.
Am i missing a reaction equation? Is there something i can do to cancel the unwanted reactants in the reaction equations? Any help would be greatly appreciated.

If you add up the enthalpies of the following reactions with the given factor and divide the result by two you will get the desired result.
$$\ce{CH4 + 2 O2 -> 2 H2O + CO2} \tag{· 2}$$
$$\ce{2 C + H2 + 3 Cl2 -> 2 CHCl3} \tag{· 1}$$
$$\ce{C + O2 -> CO2} \tag{· -2}$$
$$\ce{2H2 + O2 -> 2 H2O} \tag{· -2}$$
$$\ce{2 HCl -> H2 + Cl2} \tag{· -3}$$

$$\ce{2 CH4 + 6 Cl2 -> 6 HCl + 2 CHCl3}$$

I'm interested in knowing how you knew to create those reactions so i can apply it to similar problems. Was it just trial and error or is there some method to use for problems like this?

It was trial and error. But the problem could also be solved by doing linear algebra.


text:
Why is the water dissociation constant valid for acid/base solutions?
The water dissociation constant was first defined with respect to dissociation of water into hydroxyl and hydronium ions. How then can one say that it is valid for an acidic solution, that is, even in an acidic solution the product of the concentrations of the hydronium and hydroxyl ions is $10^{-14}\ \mathrm{M}$?

That's the nature of chemical equilibrium. There is a reason why constant is defined as $\ce{[H+]\cdot[OH-]}$ and not some other way.

If Kw is written in terms of concentration (as opposed to activity), as [H+][OH-], it is only $10^{-14} M^2$ in the limit of zero ionic strength, but it is approximately correct in general.

image source

Acidity and alkalinity in aqueous solutions is based upon the autoionization of water:
$$\ce{HOH + HOH -> H3O+ + OH-}$$
For pure water, the following equation is valid:
$$[\ce{H3O+}] = [\ce{OH-}] = 1 \times 10^{-7}~\mathrm{M}$$
Thus, to calculate $K_\mathrm{w}$:
$$K_\mathrm{w} = [\ce{H3O+}][\ce{OH-}] = (10^{-7})(10^{-7}) = 10^{-14}$$
If a solution is acidic, then $[\ce{H3O+}]>[\ce{OH-}]$ and if it is alkaline $[\ce{H3O+}]<[\ce{OH-}]$. However, even if the concentrations are not equal the product of molar concentrations $[\ce{H3O+}][\ce{OH-}]$ will always be $10^{-14}$ at $25~\mathrm{^\circ C}$.


text:
How to run a B2PLYP-D3 calculation in Gaussian 09?
Can anyone help me to run a geometry optimization and frequency calculation using B2PLYP-D3 functional including the DFT-D3 dispersion correction? I went through Gaussian manual, however, it was not very clear to me.
%NProcshared=4
%Mem=4gb
%chk=fohsdft3
# B3PLYP/aug-cc-pVDZ opt freq scf=tight

Input file taken from fohsdft

0 1
 O                  -3.46278   2.54045   0.
 H                  -2.50278   2.54045   0.
 H                  -3.78324   3.44539   0.

We have the Gaussian 09 Revision C.01 version.

When I did B3LYP-D3 calculations on protein-ligand interactions in a course project about 5 months ago, I used the following keyword to enforce Grimme's D3 correction: EmpiricalDispersion=GD3. I was also confused about the manual, but here's how I interpreted the manual: "GD3 parameters are defined for the list of functionals in the table, and to invoke these use the keyword EmpiricalDispersion=GD3". This worked for me, but perhaps others can confirm. The manual I refer to is this page: http://www.gaussian.com/g_tech/g_ur/k_dft.htm

Just a reminder that f you've found any of the answers as acceptable, please mark/accept it so that future users can benefit.

Search for B2PLYP on the Moller-Plesset page:

B2PLYPD3 requests the same but with Grimme's D3BJ dispersion [Grimme11, Goerigk11].

The correct D3 parameters will be set for you automatically.
%nprocshared=4
%mem=4gb
%chk=fohsdft3
# opt freq b2plypd3/aug-cc-pvdz scf=tight

Input file taken from fohsdft

0 1
O                  -3.46278   2.54045   0.
H                  -2.50278   2.54045   0.
H                  -3.78324   3.44539   0.

Since B2PLYP-D3 is a method, it goes where the density functional is normally specified.

As Yoda suggested, you have to include empircal dispersion as a keyword on the route section. That specifies you want to include dispersion effects in the calculation. You can just specify the B2PLYPD3 functional in your route section since it is parameterized with dispersion correction.


text:
Is it possible for one specific atom in a molecule to have a non-integer oxidation state?
I am wondering if fractional oxidation states of an atom are possible. I'm not referring to cases such as $\ce{Fe3O4}$ or $\ce{Mn3O4}$ where the average oxidation state is fractional, since these actually comprise a mixture of atoms which are individually in the +2 and +3 oxidation states. What I mean is, is it possible for an individual atom in some compound to have an oxidation state of (for example) 2.5?
To me it doesn't seem possible just because of the way oxidation states are defined. However I have seen some sources which state that fractional oxidation states are possible. I would be interested in knowing if there is some weird compound that has fractional oxidation states?

Note: This is not a duplicate of Are fractional oxidation states possible? I want to know if it is possible for an individual atom in some compound to have a fractional oxidation state, not its average fractional oxidation states.

https://en.wikipedia.org/wiki/Oxidation_state#Fractional_oxidation_states An example with true fractional oxidation states for equivalent atoms is potassium superoxide, KO2. The diatomic superoxide ion has an overall charge of −1, so each of its two oxygen atoms is assigned an oxidation state of −1⁄2, This ion can be described as a resonance hybrid of two Lewis structures, where each oxygen has oxidation state 0 in one structure and −1 in the other.

It depends. Consider various radicals such as the superoxide anion $\ce{O2^{.-}}$ or $\ce{NO2^{.}}$. For both of these, we can draw simple Lewis representations:

In these structures, the oxygen atoms would have different oxidation states ($\mathrm{-I}$ and $\pm 0$ for superoxide, $\mathrm{-II}$ and $\mathrm{-I}$ for $\ce{NO2}$). That is the strict, theoretical IUPAC answer to the question.
However, we also see that the oxygens are symmetry-equivalent (homotopic) and should thus be identical. Different oxidation states violate the identity rule. For each compound, we can imagine an additional resonance structure that puts the radical on the other oxygen. (For $\ce{NO2}$, we can also draw resonance structures that locate a radical on both oxygens and another one that expands nitrogen’s octet and localises the radical there.) To better explain this physical reality theoretically, we can calculate a ‘resonance-derived average oxidation state’ which would be $-\frac{1}{2}$ for superoxide and $-\frac{3}{2}$ for $\ce{NO2}$. This is not in agreement with IUPAC’s formal definition but closer to the physical reality.

In other words, oxidation states are just a useful model that works quite well with a few tweaks, but they are not "reality". For the superoxide "bound" electrons, the question "which of the two possible electrons is participating in the binding" doesn't make any sense, and in fact, they are in a superposition that observably changes the binding energy according to QM predictions. The simplified oxidation states work reasonably well when you pretend that the superoxide anion is one "unit" with a definite "oxidation state" of -1, considering each atom separately doesn't work quite as well.

IUPAC recognizes that there are fractional oxidation states, but asks that you avoid writing them.
IR 4.6.1 says not to write an oxidation state "where it is not feasible or reasonable to define" because:

This avoids the use of fractional oxidation states.
Examples:

$\ce{O2-}$

$\ce{Fe4S4^3+}$

See also IR-5.4.2.2 which says "oxidation numbers are no longer recommended when naming homopolyatomic ions" because "ions such as pentabismuth(4+) (see Section IR-5.3.2.3) and dioxide(1—) (see Section IR-5.3.3.3), with fractional formal oxidation numbers, could not be named at all"

According to the IUPAC Gold Book[1], oxidation state is defined as:

A measure of the degree of oxidation of an atom in a substance. It is defined as the charge an atom might be imagined to have when electrons are counted according to an agreed-upon set of rules ...

Since the definition explicitly involves counting electrons, it is not possible for an individual atom to have a fractional oxidation state since it cannot have a non-integer number of electrons.

Thanks for the answer. From Rodriquez comment on the superoxide ion, am I correct in saying that atoms involved in resonance structures will technically have a fractional oxidation state?

@Nanoputian I hadn't considered resonance here. I think the IUPAC definition still doesn't allow fractional oxidation states but Jan's point about them being more realistic is very true.


text:
Car Parinello advantages and disadvantages
A look at wikipedia and some online sources state that Car Parinello is less computationally demanding than self consistent methods. The original journal "Unified approach for Molecular Dynamics and Density Functional Theory" by Car and Parinello state that their method is appropriate for

(i)
  compute
  ground-state
  electronic
  properties
  of
  large
  and/or
  disordered
  systems
  at
  the
  level
  of
  state-of-the-art
  electronic
  structure
  calculations;
  (ii)
  perform
  ab
  initio
  MD
  simulations
  where
  the
  only
  assumptions
  are
  the
  validity
  of
  classical
  mechanics
  to
  describe
  ionic
  motion
  and
  the
  Born-
  Oppenheimer
  (BO)
  approximation
  to
  separate
  nuclear
  and
  electronic
  coordinates.

Both of these advantages arise from the fact that the Car Parinello method is computationally less demanding However, I would like to know about its disadvantages. There must be a reason why self consistent methods are still widely used. I would also like to know if there are more advantages than it being computationally less demanding.

In the 1980s when it was introduced, being computationally less demanding was key. Even then you could only simulate a small number of atoms. Now you don't need to sacrifice much to do a full DFT on bigger systems.

In CPMD, the electronic degrees of freedom propagate together with the nuclei, this means small time steps, controlling the electron temperature, etc. The BOMD is conceptually much simpler and with good wavefunction projector the SCF converges in few iterations and the timestep of simulation depends only on nuclear degrees of freedom. Compare CPMD with CP2K.

Car Parinello is an Molecular Dynamics (MD) model. The MD models of chemicals I've worked with use a projection of the expected wave-function using classical mechanics to approximate what the atoms and molecules in the model are doing. Essentially assigning a force constant based on distance between the atoms in the system to approximate the energies involved in the simulation. This kind of simulation can give semi-qualitative numbers to energies of association or dynamics of a reaction in some cases where the motions of all the atoms are essentially classical in nature. However, if there is proton motion involved in the reaction coordinate or if you would like chemically meaningful energies of association MD models will not be workable for your goals. MD models can, typically, only get a "feel" for whether or not an association is possible or if a reaction could take place. And if there is a quantum mechanically allowed and classically forbidden step, say proton tunneling, the model will fail.
So MD models are great for very large or distributed systems. For example modelling a solute in solvent or large solid systems, both of which are beyond most ab initio Quantum Mechanics (QM) calculations due to the number of individual calculation steps needed for convergence. There are some "high level" (nit-picky) calculations that require precision beyond what a classically built model can give which is why DFT and ab initio calculations are still run today.
Now I have not ever used Car Parinello in my work, (I'm part of one of those nit-picky groups) so I cannot give any real numbers for the errors in this method as compared to other MD simulations. I would, however, say that the writer of the Wiki page is not un-biased (neither am I, see above regarding nit-picky group) and is trying to show how great this method is compared to other MD simulations.  This particular MD simulation seems to be using some SCF calculations and long range basis set functions for outer shell electrons and so I think it is mainly for use in solid ionic systems rather than the small isolated molecules my group typically models.
For DFT (or essentially all modern QM calculations) the electronic wave function is approximated using basis functions. These are pre-programmed functions that can be mixed together with little computational effort to approximate the true wave function. The basis sets used by Car Parinello are plane wave basis function which means the wave function repeats at long range and allows (forces) assignment of electrons far removed from one another. This means that a self consistent state is quickly reached for solid crystalline systems.
My personal bias (use a grain of salt for this section) is against such basis sets due to the fact that it forces a pseudo-crystalline pattern on the system (at least at long range). For most chemically interesting systems it is the oddities on a surface or the internal structure which give rise to chemical change. And so, this model may fail to represent these systems well due to the choice of basis set. Once again, not having used this model I cannot say it would fail to represent these interesting features, I just doubt it would give reasonable energies or wave functions after completing the calculations.

This compares Car–Parrinello MD with classical MD and static DFT calculations. I think the original question was more along the line of comparing Car–Parrinello MD with Born–Oppenheimer MD…


text:
Why are electrochemical cell potentials the same for reactions with same overal equation?
Why are cell potentials the same for reactions with same overal equation?
e.g. hydrogen-oxygen fuel cell in acidic and alkaline conditions
It is from the chemistry past paper

Actually, cell potentials will vary according to changed conditions. This is outlined by something called the Nernst equation, which states that
$$E_\text{cell}=E^\circ_\text{cell} - \frac{RT}{zF}\ln Q_\mathrm r$$
Because of this, a change in $\mathrm{pH}$ could affect the $\ln Q_\mathrm r$ term, causing a change in cell potential. In addition, the values for reduction potentials that you look up online or in a textbook are for a reaction occurring at $25\ \mathrm{^\circ C}$, meaning that your reduction potential will likely change with a change in temperature.

please see my added pictures of the question thanks for replying! @Benzene

Given the above fuel cell, a change in pH would have no effect on the concentrations of the reducing or oxidizing agents, as neither of them are protic. As such, the reaction remains the same and the value of Q is unchanged by the change in pH.


text:
How can we accurately carbon-date samples from La Brea?
TL;DR - given a sample of bone which has been encased in asphalt for tens of thousands of years and has become saturated with the asphalt, how can you get an accurate age for the specimen through carbon dating?
This past weekend I took my kids to the Page Museum at the La Brea Tar Pits in Los Angeles. While there we saw an astonishing number of fossilized Pleistocene animals. The interesting thing is that all of these fossils have a lovely (to my eyes, at least) brown color which is staining from the asphalt in which these bones have lain for thousands of years.
While there, we saw a presentation which explained that the asphalt seeps into the porous bones and it was explained that we know how old the specimens are through carbon dating. Which got me thinking - if the specimens are all highly contaminated with asphalt, how can they know that the sample they're testing contains carbon from the bone and not from the asphalt (which is presumably much older than the bone)? I did try asking one of the docents, but they're really not equipped to handle such a technical question.

Various articles in the scientific literature confirm your assumption that the removal of asphalt from bone samples poses a significant challenge for accurate radiocarbon dating.
Several methods and results are summarized in O’Keefe, F. R.; Fet, E. V.; Harris, J. M.; Compilation, Calibration, and Synthesis of Faunal and Floral Radiocarbon Dates, Rancho La Brea, California Contrib Sci 2009, 518, 1–16. All techniques involve a combination of solvent washes (e.g. Soxhlet treatment) followed by the isolation of the bone collagen (since bone collagen does not exchange carbon with the environment) under different hydrolysis conditions (time, temperature, acid strength, etc.)

According to Radiocarbon dating of petroleum-impregnated bone from tar pits at Rancho La Brea, California. Science  30 May 1969:
Vol. 164, Issue 3883, pp. 1051-1052 :

Extensive bone dating was not possible before a method was devised for the
  removal of the impregnating petroleum...Bone is now routinely dated from its collagen


text:
Computational Chemistry Negative Lattice Vectors
This is a computational chemistry question.
I've recently been using the Materials Project API to get VASP input data, that I've been processing. The POSCAR format is as so, for those who do not know: POSCAR
Title
Scaling Factor
Lattice Vector A
Lattice Vector B
Lattice Vector C
etc (Not relevant to question)

On a few rare cases, i get Lattice vectors with negative values in them:
0.872808 2.468674 1.511748
-2.623232 -0.003400 4.543571
1.745616 -2.468674 0.000000

which displays fine using VESTA. Using pymatgen I converted it to a CIF file, which says:
_cell_length_a   3.02349597
_cell_length_b   5.24646459
_cell_length_c   3.02349568

and looks identical to the POSCAR file. However, I do not understand where these values come. For example, Lattice vector A:  0.872808+abs(-2.623232)+1.745616 != 3.02349597.
Seeing as VESTA displays it correctly, and pymatgen converts it correctly, I feel like it is just a concept that I do not understand. I tried googling but found nothing helpful. How do I convert from the given lattice vectors with negative values to absolute values as shown in the CIF format? What is the math behind this?
SOLVED:
Use Vector Magnitude/Pythagoras Theorem: sqrt(LatvA[0]^2 + LatvA1^2 + LatvA[2]^2) = 3.02349597

Ever heard about Pythagorean theorem? Also, vector $\vec a$ is the first row, not first column. Also, welcome to Chem.SE.

@IvanNeretin Of course..... but that doesn't answer the question.

The data in your $3\times3$ table are the coordinates of certain vectors. To get the length of a vector from its coordinates, use the said theorem.


text:
Is this reaction of 4-chloro-1,1-dimethylcyclohexane SN1, SN2, E1, or E2?
In the following question, which I tried to solve, I became confused by which way I should proceed.
Here is my thought process:

SN1: If Cl detaches and forms a secondary carbocation, then there's no nearby groups to stabilize it or make a shift to turn it into a tertiary cation.
SN2: I can't make out how exactly SN2 would occur
E1: I can't make out how exactly E1 would occur
E2: There's no stereospecific data given in the question whether the methyl group and the hydrogen lie in one plane or not, so I couldn't make out a way to do anti-elimination.

The Answer is given : E1

so @benzene , This is the E1 product as per the mechanism , Am i right ?

E1 is the correct answer. There are a couple of factors that lead me to this conclusion. First of all, heat favors elimination over substitution, meaning that this reaction will either be E1 or E2. Now that we know that this is an elimination reaction, it is just a matter of deciding if it is E1 or E2. E2 requires a strong base to occur and since $\ce{H2O}$ is a week base, we know that this reaction must follow an E1 mechanism. While E1 is favored by a tertiary carbocation intermediate, both E1 and E2 can occur with a secondary carbocation.

Funny that your account name is Benzene but your account picture is a tesseract.


text:
Why does KCl come in amber coloured bottles?
I remember from school that photosensitive chemicals are kept in amber coloured bottles. Is KCl photosensitive? I read up the impurities list on the bottle and it says that it may have bromides too? So is it because of the KCl or the bromides? If it is due to the bromide, what sense does it make since it would react just to form another impurity?

I suspect that when manufacturers order bottles, sometimes they have a few extra amber ones and sometimes they have a few extra clear ones.  In cases where it doesn't matter chemically which bottle is used, silly logistical factors like this might be decisive.

On a sample of one selling site, amber bottles are cheaper than similar clear bottles, for some moderate to large quantities, at least today.  KCl is fairly cheap so when selling the stuff in small quantity every penny saved counts towards profit.  So: Economics could be the cause.

I would totally believe this. Also if amber bottles are cheaper, then just stocking one color bottle simplifies things. You have 1/2 the bottles in inventory which is huge. // I'd guess that clear bottles must have much purer ingredients.

Recycled glass makes amber bottles - mix up the green glass, brown glass, clear glass, whatever.  It's the colour of play-doh when you mix it all together. Clear glass needs clear feedstock - clear means pure and pure means refined and refined means expensive-er.  Some things legitimately benefit from being stored in bottles that block short wavelength light, preventing photodegredation, but KCl is certainly not one of them.

I had to look at the price charts a few times to see that the images were making the same point that the text was.  The prices are kind of interesting:  the amber is cheaper when you buy 128-999, 5000-9999, or 10000-25000, but the clear is cheaper when buying 1-127, or 1000-4999.  That's kind of interesting.

@JoshuaTaylor - yeah, I should have done some SO-style "markup" with freehand circles drawn with a pen tool in Paint ... but ... got lazy.

I can't imagine what I would do with 25000 bottles.

@AliCaglayan - here are some ideas!

The potassium chloride solutions in my lab are in transparent glass bottles, as was the potassium chloride solution in my inorganic students’ lab at the LMU in Munich. If I recall it correctly, the solid potassium chloride we had there was in a transparent bottle, too, but don’t quote me on that.
There should be nothing in potassium chloride that will react under irradiation. And these types of impurities are generally not a reason to use such a bottle. But maybe the supplier of your choice just happens to have too many amber bottles?

Haha! Thanks. Maybe I was wrong to generalize based on just the bottle in my lab :)


text:
How does diluting an acid change pH?
By the Henderson–Hasselbalch equation, log (salt)/(acid) term is a ratio of the concentrations. So, how can dilution change the pH of the acid as the concentrations of both salt and acid would change with dilution and the term would remain the same?

I don’t see how you arrived at that maths …

Consider a strong acid such as $\ce{HCl}$ which — within experimental error — can be considered fully dissociated in aquaeous solution. This can be expressed as:
$$c(\ce{H3O+}) = c_0(\ce{HCl})$$
Assume you have a concentration $c = 1~\mathrm{\frac{mol}{l}}$. pH is defined as:
$$\mathrm{pH} = - \lg [\ce{H3O+}]$$
Thus our solution has a pH of:
$$\mathrm{pH} = - \lg [\ce{H3O+}] = - \lg 1 = - 0 = 0$$
(Remember that to calculate pH we need dimensionless concentration. In this post, $c$ shall always contain a dimension while square brackets shall not.)
Take $1~\mathrm{l}$ of this solution and add $1~\mathrm{l}$ of water. Our new solution has a concentration of $c = \frac{1~\mathrm{mol}}{2~\mathrm{l}} = 0.5~\mathrm{\frac{mol}{l}}$. Our pH is:
$$\mathrm{pH} = - \lg [\ce{H3O+}] = - \lg 0.5 = - (-0.301) = 0.301$$
The solution’s pH value obviously changed with concentration change.

Now, let’s do the same thing with a weak acid such as acetic acid ($\mathrm{p}K_\mathrm{a} = 4.76$). But first, I need to do a bit of maths. Remember that we cannot use the Henderson–Hasselbalch equation, because it assumes a buffered system. A simple acid is unbuffered. Instead, I will start at the definition of the acid constant:
$$K_\mathrm{a} = \frac{[\ce{H3O+}][\ce{OAc-}]}{[\ce{HOAc}]}$$
And since chemistry dictates that $[\ce{H3O+}] = [\ce{A-}]$.
$$K_\mathrm{a} = \frac{[\ce{H3O+}]^2}{[\ce{HOAc}]}$$
Because $c_0(\ce{HOAc}) = c(\ce{HOAc}) + c(\ce{H3O+})$, we can transform to:
$$K_\mathrm{a} = \frac{[\ce{H3O+}]^2}{[\ce{HOAc}]_0 - [\ce{H3O+}]}$$
Which finally gives us an equation with only one unknown. Solving this:
$$K_\mathrm{a} = \frac{[\ce{H3O+}]^2}{[\ce{HOAc}]_0 - [\ce{H3O+}]}\\
K_\mathrm{a} \left ( [\ce{HOAc}]_0 - [\ce{H3O+}] \right ) = [\ce{H3O+}]^2\\
[\ce{H3O+}]^2 + K_\mathrm{a} [\ce{H3O+}] - K_\mathrm{a} [\ce{HOAc}]_0 = 0$$
Since the determinant is positive, we can immediately write:
$$[\ce{H3O+}]_{1/2} = \frac{- K_\mathrm{a} \pm \sqrt{K_\mathrm{a}^2 + 4 K_\mathrm{a} [\ce{HOAc}]_0}}{2}$$
Now plug the values $c_{0,1}(\ce{HOAc}) = 1~\mathrm{\frac{mol}{l}}$ and $c_{0,2}(\ce{HOAc}) = 0.5~\mathrm{\frac{mol}{l}}$ as above. You will need to add the value of the square root or you will get a negative concentration. This is what you should be getting (assuming OpenOffice isn’t buggy):
$$\begin{array}{ccc}
\hline
\text{expression} & c_0 = 1~\mathrm{\frac{mol}{l}} & c_0 = 0.5~\mathrm{\frac{mol}{l}}\\
\hline
K_\mathrm{a}^2 + 4 K_\mathrm{a} [\ce{HOAc}]_0 & 6.95 \cdot 10^{-5} & 3.48 \cdot 10^{-5}\\
\sqrt{K_\mathrm{a}^2 + 4 K_\mathrm{a} [\ce{HOAc}]_0} & 8.34 \cdot 10^{-3} & 5.90 \cdot 10^{-3}\\
- K_\mathrm{a} \pm \sqrt{K_\mathrm{a}^2 + 4 K_\mathrm{a} [\ce{HOAc}]_0} & 8.32 \cdot 10^{-3} & 5.88 \cdot 10^{-3}\\
[\ce{H3O+}] & 4.16 \cdot 10^{-3} & 2.94 \cdot 10^{-3}\\
-\lg [\ce{H3O+}] & 2.38 & 2.53\\
\hline
\end{array}$$
Obviously, once again, the pH went up when diluting the solution.

The Henderson–Hasselbalch equation, as I skimmed past earlier, is meant for buffer systems. The concentrations $[\ce{A-}]$ and $[\ce{HA}]$ are those concentrations you added to the buffer system. Since we did not add any acetate ions (other than acetic acid), the logarithmic argument would become zero causing an undefined error. We cannot use Henderson–Hasselbalch here.
However, if you have a solution containing $1~\mathrm{\frac{mol}{l}}$ acetic acid and $1~\mathrm{\frac{mol}{l}}$ sodium acetate, the logarithm’s argument will be $1$, therefore the logarithm will disappear and the pH value will be $4.76$ — exactly the $\mathrm{p}K_\mathrm{a}$ value of acetic acid. This buffer system is concentration independent. Diluted acids to whom the conjugate base has not been added as a salt are not!

Just what I was looking for! Diluting a buffer wont change its pH but diluting an acid would. Thanks a lot :)

@A.K. A logarithm needs a base. $\log$ by itself is meaningless. The answer often uses $\log_{10}$ for which the shorthand notation $\lg$ exists.


text:
How can I permanently write on HDPE plastic?
So I just spent the better part of an hour graduating a food grade HDPE 5 gallon bucket with permanent marker so I can easily fill it with water of a certain amount for homebrewing.
I'm all done. The only problem is that permanent marker wipes off of HDPE quite easily. I guess I should have tested that first...
Anyways... what CAN I use to write on HDPE plastic? Or will I be forced to etch the markings in with a nail or something?
I tried a Google search for "How to write on HDPE plastic?" and came up empty.

Could you just label the bucket with a writable/printable sticker?

I could, but the that would come off when I power wash the bucket.

From experience, one good way to permanently mark on HDPE is a paint marker. The following is just the first one I found online; no endorsement intended:

   Promotional image from uline.com
They tend to be stinky and a tad messy, but they usually leave clean, visible lines and are basically indelible, requiring a lot of scrubbing with alcohol, acetone, etc. to remove. No idea how well it'll survive power washing, though.

Flags answer for spam :D

@paracetamol I knew some joker'd go there. :-P

There's 2 or 3 "permanent" ways. Paint ain't one of them -but- I'd agree that if you're careful, that paint is thick and adhesive enough to tolerate a lot of abuse before it chips, cracks, rubs or just falls off. It may be good enough, ymmv.
Way #1. Hot melt. Few consumer hot melt guns and sticks get hot enough, there's commercial guns which actually use colored HDPE (or PP) to weld to HDPE.
Way #2. Dremmel, or nail or razor or even a soldering gun tip. You're removing some of the material that's already there with obvious consequences for strength.
Way #3. Plasma treatment. I have this for completeness, it's not realistic for consumers
Way #4. Ok, I know I said 3, and even then we need to be loose about what "writing on" means. The 4th way is to use sandpaper to make the surface rough enough so that some of the pigments from a permanent marker gets caught in the nooks and crannies - this could obviously be combined with #2...but sanding isn't writing per se.
I typically use adhesive labels if I want to mark my HDPE bottles, the large surface area gives plenty of contact and the labels I use have a coating (some of the fancier ones come with an additional clear layer, but I typically use a couple of wrap-arounds of 2" wide transparent packaging tape)


text:
Nature of urea and it's relation with uric acid
My question is what is the nature of urea. I checked it on the Internet and according to my results, the nature of urea is basic. Then why is there uric acid? For example when we add magnesium oxide to water, magnesium hydroxide (a base) is formed. Similarly when we add sulfur trioxide to water, sulfuric acid is formed. Thus, if the nature of the substance decides whether it will form an acid or a base then why is this case not followed by urea as urea is basic but it forms uric acid?

Uric acid is not what you get when you add water to urea. Go and look at the structures of the two compounds.

The names of uric acid and urea are both derived from urine in some way which is whence the chemicals were first discovered. This is even more evident in e.g. their German names Harnstoff (urea) and Harnsäure (uric acid), literally meaning urine-compound and urine-acid. They have no further relationship.
Apart from uric acid containing the word acid, there is no clue as to the acidic or basic property given in the name.


text:
What is the name of the crinkly plastic used in clear parts of envelopes?
I've always wondered what it is, as it makes a lot of noise on being deformed. No other common plastic that I've come across has quite the same properties.

In the old days it was cellophane, and may still be (or just generically called cellophane whether it is or is not).

It's more commonly referred to as 'envelope window film'. I've read somewhere that about 90% of all envelope window films are made from polystyrene, but sadly I cannot find the source at the moment. The remainder is made from polyester, polypropylene or glassine.
For examples see this manufacturer or this one. I'm not affiliated with either company, these are just the only ones I could find that provide specifications and proof my point.

It usually is BOPS: Biaxially oriented polystyrene.

Yes I just took a look at an envelope... with IR spec. It's atactic polystyrene.


text:
boric acid + isopropyl alcohol
I have found recommendations for making a soldering flux using boric acid mixed in with ethyl alcohol in 50/50 ration or 60/40 depending on the consistency desired.
Some have recommended that methyl alcohol is better to use as more of the boric acid can be dissolved in it? Is this correct?
If so, how do I calculate how much boric acid I can dissolve in isopropyl alcohol? What would be the best solvent to create a paste of boric acid to use as a soldering flux?

For safety reasons, I would stick with ethanol, as it is less flammable than methanol (your using hot soldering equipment) and much less toxic.

For safety reasons, I would stick with ethanol or isopropanol, as they are less flammable than methanol (your using hot soldering equipment) and much less toxic, even though methanol is a better solvent for $\ce{H3BO3}$. If it weren't for the expense, due to heavy taxation of ethanol, it would be preferable to isopropanol. From the table below, you'll need almost twice as much isopropanol (though 1-propanol is the solvent listed, it is an isomer of the more common 2-propanol, AKA isopropanol).

Solubility of $\ce{H3BO3}$ in various alcohols at 25 C, percent by weight:
Methanol:     22.66%
Ethanol:     11.96%
1-Propanol:    7.34%

And methanol flames are hard to see, so it might be on fire and you might not notice until things get exciting...

In general, boric acid dissolves well in alcohols.  BUT the usual recommended solvent is ethylene glycol, which dissolves boric acid very well and has a low vapor pressure (i.e. low flammability).  I'm guessing it would also dissolve well in propylene glycol.


text:
Optical Isomerism
I have been told that structurally symmetric molecules are optically inactive irrespective of the atoms that are being used.
However I see optical activity as the force that turns plain polarised light by some angle and I think factors like mass, electrons, etc. should influence that.
Any explanation would be great

Possible duplicate of Molecular chirality and optical rotation

Mass, electrons, and whatever other factors are already considered by symmetry; if the molecule has certain symmetry, then all these factors try to turn the plane clockwise and counterclockwise with exactly the same force, and so end up not turning it at all. Also, welcome to Chem.SE.

Thanks but that is the result and I know it, I want to know how we come to that result

http://physics.stackexchange.com/questions/15503/how-do-optically-active-compounds-rotate-plane-polarized-light

The article is about how light rotates and I understand I want to know how symmetry cancels the factors like mass of atom, charge etc.

So are you basically asking- how do optically active compounds polarise light?

Symmetry is strict. We only consider a molecule symmetric if the plane of symmetry transforms identical atoms onto identical atoms. The atoms transformed onto each other must have the same bonding partners, the same geometry, the same oxidation state etc.
There are cases where it is not immediately obvious. $\ce{NO2}$, for example, has a plane of symmetry traversing the nitrogen atom which transforms the oxygen onto each other. In simple figures, one would draw the molecule as $\ce{O=N-O.}$. However realise, that this is only one possible mesomeric depiction, another one being $\ce{.O-N=O}$ with the oxygens reversed. Even though they don’t look it, they are symmetrically identical. Therefore, we need no longer be concerned with charge or electronics.

How this plane of symmetry works: Remember that we have a macroscopic solution, usually in the range of $50~\mathrm{mmol/l}$ to measure optical activity. That is some $10^{21}$ molecules, all randomly oriented. The polarised light beam will interact with every molecule it meets, and either be polarised to the right or to the left. However, since the molecules are symmetric and numerous, for almost every molecule the beam interacts with there will be another one oriented in exactly the opposite way (as in mirror image), reversing the former’s effect for an overall zero-effect.
Thereby, things such as conformational variants are cancelled out by sheer numbers being averaged.

One final aspect remaining is mass. Many elements come in different isotopes that differ by mass and of course, if one of two symmetry-equivalent atoms happens to be a different isotope than the other that will make that molecule ‘chiral’. But remember again that we have sheer numbers we can average and that the isotope distribution is generally random. Thus, even different isotopes being present in different molecules do not change the overall macroscopic picture of an achiral molecule.
If you somehow manage to stereoselectively destroy a molecule’s symmetry by, e.g. having the re-methyl group $\ce{^{12}C}$ and the si one $\ce{^{13}C}$ then that molecule should be (weakly) chiral.

Different atoms would have different electron clouds(with diff atoms), so a tri-molecular symmetric structure should be active and that is exactly my poin

@PrabhdeepSingh You really have to clarify what you mean there. Electron clouds (i.e. orbitals) are always symmetric or antisymmetric with respect to every single $\sigma$ a molecule has. (And light cannot distinguish between symmetric and antisymmetric.)

Molecules interact with plane polarized light through their electron clouds, not by the mass of their molecules or their charge.
So if you have a bulk solution of symmetric molecules, when the plane polarized light passes through there is no 'bias' for the light and it continues along its way.
This is similar in the case of a racemic mixture, where we have equal amounts of enantiomers because the little nudging to the left and right of the individual molecules are cancelled out in the bulk solution.
However, when you have an optically pure compound in bulk solution, when the light strikes it although the optically pure compound in bulk solution is randomly arranged, there is a very clear bias which causes a net rotation of the light which is observable.
You can imagine this by taking the example of arranging multiple fan blades fixed in a water pipe randomly. The mass of the fan blades themselves do not cause the rotation of the water - at most it causes turbulence. However, the orientation of the fans (which have a certain asymmetry) causes the flow of water to spiral in a certain direction. Regardless of whether the fan blade is facing forward, backward, perpendicular, to the flow of the water, the blades will cause the water to rotate in a certain way.
It's a bit of a challenge to visualize, but I hope this helps.

Different atoms would have different electron clouds, so a tri-molecular symmetric structure should be active and that is exactly my point

But that makes no sense because the (1) electron cloud of the molecule is a property of the molecule, no longer of the individual atoms, and (2) if it is symmetric, there is no way for the light to distinguish left or right-handedness so there is no 'bias' for the light to rotate towards.

@PrabhdeepSingh this is a long shot but you can imagine that if light strikes a symmetric molecule, then it doesn't matter which way it rotates (left or right) because it takes the same amount of energy either way. However, if it strikes an asymmetric molecule, it is easier for it to rotate in one direction rather than the other, so more of the light rotates left than right, and then you have observable rotation to the left.


text:
How to calculate probability of finding an electron at a point?
In school i learned how to calculate probability of finding electrons in some volume but how can we calculate the probability of finding a electron at a particular point.
Point here has negligible volume.

The probability you want is $0$.

An electron doesn't have any meaningful presentation of being at a point. At that level the world is inherently fuzzy, so it does not make sense to talk about where an electron is. With the quantum mechanics used in chemistry, the best interpretation we can have is: in which region is the electron likely to be present?

Consider the 1D case for simplicity.
The modulus square of the wavefunction is interpreted as a probability density in the sense that
$$
|\psi(x)|^2\mathrm{d}x
$$
gives you the probability of finding the electron in within $x$ and $x+\mathrm{d}x$. Now if you want to consider the probability to find the electron exactly in $x$ you have to take the limit $\mathrm{d}x\to 0$ and this gives you zero probability. Formally, this corresponds to compute
$$
P(x=x_0) = \int_{x_0}^{x_0} |\psi(x)|^2\mathrm{d}x = 0,
$$
which is zero since the limits of the integral are the same. Therefore there is $0$ probability to find an electron at exactly one point.


text:
Stability of ferrous ions and iron(II) hydroxide in acidic conditions
I was reading a chemisty textbook when I came across the following statement.

$\ce{Fe(OH)2}$ is not stable, even more so in basic conditions. [---] $\ce{Fe^2+}$ ions are also less likely to oxidise in the low pH range.

First attempts
No explanation was given. At first, I was hoping for a simple case of Le Châtelier – Braun principle. However, the equilibrium
$$\ce{4[Fe(OH2)6]^2+ (aq) + 4H3O^+ (aq) + O2 (g) <=> 4[Fe(OH2)6]^3+ (aq) + 6H2O (l)}$$
would suggest the opposite. Next, I considered a simple approach to complex stability. As a general guideline, metals in lower oxidation states favour cationic complexes. The opposite is also true. Therefore, one might conclude then that if anionic complexes are not preferred, $\ce{Fe^2+}$ ions ought to be stable.
One example would be that the hydrolysis is more hampered in acidic media.
$$\ce{[Fe(OH2)6]^2+ (aq) + H2O (l) <=> [Fe(OH)(OH2)6]^+ (aq) + H3O^+ (aq)}$$ Furthermore, in very basic conditions the hexahydroxidoferrate(II) would form. $$\ce{[Fe(OH)2(OH2)4] (s) + 4OH^{-} (aq) <=> [Fe(OH)6]^4- (aq) + 4H2O (l)}$$
However, this rule breaks down for $\ce{[Fe(CN)6]^4-}$ which is rather stable with $K_\mathrm{instability} \approx 10^{-37}$.
Half-reactions
As you might imagine, I was trying to avoid using half-reactions and the Nerst equation. Iron chemistry is rich and there are a myriad of possibilities. Before we move forward, let us note that the corresponding ion electronic structures are
$$\ce{Fe^2+:[Ar]{3d^6}};\\
  \ce{Fe^3+:[Ar]{3d^5}}.$$
So it seems that ferric ions are indeed more stable. Yet the reduction half-reaction
$$\begin{align}
  \ce{Fe^3+ + e- &-> Fe^2+} &
  E^\circ &= +0.77~\mathrm{V}
\end{align}$$
 implies the opposite. (Is this due to solvation?)
I did find a useful Pourbaix diagram (Andel Früh: Pourbaix diagram of Iron; $c(\ce{Fe}) = 10^{-6}~\mathrm{mol/l}$, $T = 25~^\circ\mathrm{C}$; wikimedia.org):

What we can tell is that ferric ions themselves are not favoured at a higher pH. In fact, the opposite is true. It just requires a strong oxidizer. This does not mean ferrous ions are not oxidized in basic conditions, it is just that the stable species are iron(III) oxide hydrates (probably iron(III)oxidehydroxide hydrates as well).
Problem
I have not really thought about this particular issue before, so I would definitely love to hear your ideas.

Why are ferrous ions and iron(II) hydroxide more stable at lower pH?
Why is the ferrous ion generally more stable, even though the electronic structure suggests otherwise?
This Pourbaix diagram does not take into account possible complex formation. If someone could explain the issue via complexes as well, I would greatly appreciate it.

States of aggregation should not be subscripted, it is not wrong, but the recommendations (Sec. 2.1.) are different.

@Martin-マーチン: Hmm, that's interesting. Never thought it should be written in any other way. Not to argue with IUPAC, but to me it seems not using subscript unnecessarily gives the states too much attention while completely omitting them would have the opposite effect. Subscript is kind of the middle ground :). Nevertheless, thank you for editing, I'll keep it in mind in the future (if possible).

The stability of iron(II) hydroxide J. Chem. Educ., 1957, vol. 34 pages 178-179 shows that the book isn't entirely correct.

...Iron(II)
  hydroxide would thus appear to be unstable in neutral
  solutions and to possess increasing stability as the
  basicity of the medium increases.
The foregoing reasoning led to the attempt to prepare
  iron(II) hydroxide from the chloride under such
  conditions that the hydroxyl ion concentration was not
  allowed to fall below approximately one molal until the chloride ion had been removed. The hydroxide was precipitated from strongly basic solution by the addition of dilute iron(II) chloride to a saturated solution of
  sodium hydroxide in an air-free system with vigorous
  stirring. The precipitate was repeatedly washed with
  molal sodium hydroxide (oxygen and carbonate-free)
  until a test for chloride ion with silver nitrate could
  no longer he obtained. Five additional one-liter washings with normal base completed the preparation of the
  iron(II) hydroxide used for the solubility studies.
  This technique led to the formation of a coarse-grained
  precipitate, white when viewed by reflected light and
  slightly greenish tinged by transmitted light. Dilutions
  of the material were made to one-tenth molal base
  without evidence of darkening when maintained under
  an oxygen-free atmosphere.

Sorry for a delayed response. So, basically, the book wrongly generalised the stability trend of ferrous ions to iron(II) hydrokside? Also, what about the inherent stability of the ions in solution? Would the situation be reversed in vacuum, as predicted by the electronic structure? Thanks for your reply! Sadly I am unable to retrieve the complete study you cited. The quote helped.


text:
Test for diethylether or tertiary alcohol?
What is a simple chemical test to tell the difference between a tertiary alcohol (e.g. 2-methylpropan-2-ol) and an ether (specifically diethylether/ethoxyethane)?
Neither of them can be oxidised by reagents such as potassium dichromate, Tollens reagent or fehlings reagent.
I came up with a test in which a less than stoichiometric amount of ethanoic acid is added and left to react, then some sodium carbonate is added and if there is no effervescence that means an ester has formed hence it must have been the alcohol, otherwise it was probably the ether. However I am unsure about the extent to which this reaction completes and it's simplicity.

See Lucas' reagent. In general, alcohol contains an $\ce{-OH}$ group, hence it would react with many things that ether would not (alkali metals, for example, though this one would hardly make a good practical test).

The problem with testing for alcohols is that ether normally contains a small amount of water which may otherwise give a false positive for a poorly chosen case.  Dry the chemical first with sodium or magnesium sulphate.

Place a sample in a test tube and immerse it in an ice water bath. If it freezes, it is the alcohol. If not, diethyl ether. This nose knows the difference.

As mentioned in the comments, using Lucas' Reagent is the best option in this case because you're dealing with a tertiary alcohol vs an ether, but let's consider your proposal as well.
In principle, it should work because an ether will not react with ethanoic acid whereas an alcohol would. However, the formation of an ester simply by the addition of ethanoic acid to alcohol is not likely to complete (you generally need a dehydrating acid like $\ce{H2SO4}$ to drive the reaction). In other words, when you put in the bicarbonate you will most likely see effervescence.
If you're thinking of looking for effervescence, using thionyl chloride $\ce{SOCl2}$ might be an option since it will react with the alcohol to give a sulfur dioxide gas and HCl:
$$\ce{R-OH + SOCl2 -> R-Cl + SO2(g) + HCl}$$
You have to make sure the solvent is dry though.
In the specific case of t-Butanol vs diethyl ether, though, you could simply heat the solution slightly to about 40 degrees Celsius. If it boils, then you've got diethyl either. If it doesn't, it's t-Butanol.

Diethyl ether is minimally soluble in water (6 grams per 100), but t-butyl alcohol is completely miscible with water, according to Wikipedia.
Selecting between these two compounds can be made by mixing 5 mL of H2O with 5 mL of the unknown. A single phase means t-butanol; two phases means diethyl ether.


text:
Why is nylon formed using carboxylic acids, while aramids are formed with acyl halides?
I was reading about carbon polymers and noticed something interesting about the difference between the reaction used to form nylon and the reaction used to form aramids: in the former, the monomers are an amine and a carboxylic acid, while in the later the monomers are an aromatic amine and an aromatic acyl halide.

Formation of an aramid

Adipic acid, used to create Nylon 66
This struck me as interesting because it seems like the reaction should be roughly the same for the creation of either polymer, with the only difference being the presence of the aromatic ring in the aramid in place of a simple carbon chain. Is the halide required for polymerization in the aramid, but not the nylon, or is there another, perhaps economic, reason for its use? Why are halides used in the creation of aramids, but not in simpler nylons?

Aromatic amines are less nucleophilic than aliphatic amines because the aromatic amines have the non-bonding electron pair "distracted" in the resonance with the aromatic ring. Therefore, they need a more reactive counterpart (acyl halide instead of the acid).

Good thinking - to make nylon, you would first need to convert the carboxyl groups on adipic acid into an acylating agent such as the acid chloride. The standard "nylon rope trick" demo uses adipoyl chloride. See for example:  https://depts.washington.edu/chem/facilserv/lecturedemo/NylonRopeTrick-UWDept.ofChemistry.html

You don't need to convert nylon monomers to acid chlorides. The major manufacturing route works fine with acids and amines.


text:
Understanding three-dimensional structure of common salt
I started reading through The Feynman Lectures and came across this diagram. What do the numbers next to the nodes in the diagram represent? It's not explained in the text. All that is mentioned is: "Figure 1–7 is an illustration of the three-dimensional structure of common salt, sodium chloride."

A direct link to the text is here, just Ctrl-F "crystal of salt".

Note that 1, 3, 5, and 7 are white circles, while 2, 4, 6, and 8 are black.

Note also that they (the numbered nodes, I mean) represent the exact contents of one unit cell. Other than that, I don't think they have any meaning.

The numbering most definitely does not define a unit cell.

I never said it does. I said it represents the contents of one unit cell (in one of the possible ways), and I stand by that statement.

I'm sure he wanted to explain something with the numbers, but later forgot to. What, I cannot fathom. The black and white circles each form a primitive tetrahedron. ?!?

The numbers have three features in common:

Odd numbers are on one type of ion (anions) and even numbers on the other (cations).

Each number is at a distinct position in the periodic structure of the unit cell. For instance, the top center and bottom center anions are periodic translations of each other and thus do not represent distinct positions, so only one gets the number 5.

If we use (2) to put the translated numbers into one of the small cubes, we find that 1-2-3-4-5-6-7-8-1 follows a skew-polygonal line along the selected small cube. This may be what detetmined the ordering. The use of even and odd numbers for the two types of ion (1) is a corollary.

Anyway, if I except the point (1), it is easy to see that there are no points on the Ox and Oy axis, but one point ((2)) on the Oz axis. On the faces, there are only one point ((5)) in the Oxy plane, then 3 points ((2), (3) and (4)) on the Oxz plane, then 3 points ((2), (7), and (8)), on the Oyz plane.  Why is it so ?

There is most likely no significance in the numbering of the nodes. During Feynman's lecture, he did not speak about these numbers.
A voice recording of Feynman's lecture is included on the website posted by OP. This recording is distorted in some areas, which is why a transcription is also provided.
Reading/listening to the lecture should show that this image simply served as a way to introduce solids, crystals, and in particular ions. The first lecture of "The Feynman Lectures on Physics" most likely is not going to explain crystallography or at least not in great detail.
Seeing as Feynman only specifically spoke about sodium chloride and nothing about the other salts in the legend, one can imagine that this image was not necessarily made for this lecture. The existence of the different salts and their spacing was not discussed.


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Is molecular weight of proteins additive?
I have an elementary question regarding calculation of molecular weight of a complex of two proteins. Essentially, I have two proteins binding as
$$\ce{A + B <=> AB}$$
I have the concentration of [A] and [B] in ng/ml, which I can convert to nM using its molecular weight (available in kDa). I have only concentration of [AB] in ng/ml, but I am not able to convert that into nM. Is the assumption that molecular weight of [AB] is equal to sum of molecular weights of [A] and [B] correct assumption?
I am not able to see the intuition behind this logic, and if someone can help me here, I would be very thankful.

The molecular mass in Dalton, amu or whichever unit you prefer is equivalent to the molar mass $M$ in $\mathrm{g/mol}$ that chemists prefer to use for this discussion. Both are strictly additive. The molar mass of a larger complex is the sum of the molar masses of all its components.
This is because of how the unit is defined. Focussing on molecular mass, we could see ot as mass of a molecule. The mass of a molecule is put together by adding the masses of the atoms that this molecule is made up of. If that molecule now forms larger aggregates, the mass of the resulting thing (still a single ‘thing’ albeit larger) is the sum of the masses of its constituents. And so on.
You just need to beware in case you have aggregates that are not strictly $\ce{AB}$. Consider two proteins that form an $\ce{AB2}$ complex: In this case $M(\ce{AB2}) = M(\ce{A}) + 2 \cdot M(\ce{B})$, because we have two $\ce{B}$ in the final complex.

Thank you for your answer and for editing the question. I tried to see if MW was additive using the example of $$\ce{NaCl}$$ and it turned out to be additive. Wasn't so sure about large proteins though, thanks!


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Can fluid catalytic cracking work on a smaller scale in a laboratory?
Fluid Catalytic Cracking usually requires a very large and tall structure. Is there a way of miniaturizing the process to fit inside a chemical lab with 10 ft ceilings? Specifically: can a fluidized bed reactor operate on this scale?

I guess there are probably such pilot reactors. https://en.wikipedia.org/wiki/Pilot_plant