Is it true that an evaporating molecule has the same kinetic energy as a molecule in a pot of boiling water? A molecule on the surface of room-temperature water shoots off the surface of said water, or in other words, it "evaporates." It does so because it gained kinetic energy ${x}$, and ${x}$ was great enough to counteract the pressure of the air surrounding it.

Is it true that this ${x}$ would be similar (the same or greater) as the average kinetic energy level of a molecule in a pot of boiling water?

Reasoning for thinking this: boiling is when the vapor pressure = the pressure of the atmosphere.

Is it true that an evaporating molecule has the same kinetic energy as a molecule in a pot of boiling water?

Well for one the kinetic energy for a given temperature is an average kinetic energy of a distribution of energis. So for the question:

  "Is it true that an evaporating molecule has the same kinetic energy as a molecule in a pot of boiling water?"

The answer is maybe or there probably are molecules that have the same energy in each phase, but I think the question you want to ask is:

  Is it true that all molecules in steam have the same average kinetic energy as all the molecules in a pot of boiling water?

From a thermodynamic perspective, at the boiling temperature the liquid and vapor phases are in equilibrium and have the same temperature. Since temperature is the measure of the average kinetic energy, if temperature is equal average kinetic energy is equal.

So simplistically speaking yes they are the same. You can stop with reading here or veer with me into super technical semantics and minutia—your choice.

  Reasoning for thinking this: boiling is when the vapor pressure = the pressure of the atmosphere.

Now it is commonly taught as you state: "boiling is when the vapor pressure = the pressure of the atmosphere". The thing is though that this is a restatement of "at the boiling temperature the liquid and vapor phases are in equilibrium and have the same temperature" which is similar but not equivalent. You said:

  It does so because it gained kinetic energy ${x}$, and ${x}$ was great enough to counteract the pressure of the air surrounding it.

Boiling occurs at the bottom of the pot which has a higher hydrostatic pressure due to the height of the water $(P = \rho gh)$. So the molecule must not just over come the pressure of the atmosphere but also the pressure of the water height. This increased pressure mean boiling occurs at a higher temperature than water at the top of the pot (see: water saturation pressure). As the steam bubble rises, the pressure decreases ($h$ decreases) and the bubble expands the steam will cool due to the work done by expansion (see: internal energy) lowering the equilibrium vapor pressure and condense at the interface of the bubble. The thermal transport is not instantaneous though so the interface inside the bubble will absorb heat from the latent heat of condensation, and in turn be hotter than the surrounding water thus the bubble will be hotter in the center than at the interface with water. Since the vapor is (very) slightly hotter, the average kinetic energy will be (very) slightly hotter, thus (very) technically it would not be correct that all the molecules in steam have the same average kinetic energy as all the molecules in a pot of boiling water


What is the term used as correction for attractive forces in the Van der Waals Equation? I have seen a question in my textbook which is little bit confusing,

The term that corrects for attractive forces present in real gas in van der Waal equation is
(i) ${nb}$
(ii) $\frac{an^2}{V^2}$
(iii) $\frac{-an^2}{V^2}$
(iv) ${-nb}$

Since the question is about term  "forces" I do not see any of the four options dimensionally correct as (i)&(iv) are related to volume and (ii)&(iii) are related to pressure.
But in my textbook answer the correct answer given is (ii) so how is it dimensionally correct?

Nothing states that the correction is itself a force, only that the correction is for attractive forces. There is thus no need for the correction to have the dimensions of a force.


How is it possible to distinguish antimony from lead? I just bought some (pure) antimony stones from a local store. However, I've heard that some stores deceive their customers by selling lead instead of antimony. I was wondering if there is any simple test that I can perform at home to make sure these stones are antimony and not lead. Since I don't know much about chemistry I thought of asking here...

As mentioned in the comments, you could perform Archimedes' Experiment.

Firstly, take the mass ($m$) (in grams) of your sample.
Then, measure the volume $V$ displaced (according to the first link), in a method summarised in the image below:

You could use any kitchen graduated container. Even though these have graduations for mL (millilitres) and even though the water density changes with temperature (source:USGS), for the temperatures in a home experiment, the following conversion could be used 1 mL = 1 cm3 of water displaced.

Then determine the density by $\rho = \frac{m}{V}$ and compare it to the known values for lead at 11.3 g.cm-3 and antimony at 6.7 g.cm-3.

Another observed feature is that according to the LennTech site for lead is that as opposed to antimony, lead will tarnish on exposure to air.


How can two orbitals constructively and destructively interfere simultaneously? The molecular orbital theory dictates that when two atomic orbitals form molecular orbitals, then two molecular orbitals must form (i.e number of atomic orbitals = number of molecular orbitals). For this to occur there must be bonding molecular orbitals and anti-bonding molecular orbitals. The bonding MO's are caused when the orbitals constructively interfere and the anti-bonding orbitals form as the orbitals destructively interfere. I understand the concept of interference but I do not understand how two atomic orbitals (one from each bonding atom) can simultaneously interfere constructively and destructively. Moreover, I am under the impression that the anti-bonding orbitals exist even if they're not occupied. As a result, is it correct to think that it's the orbitals themselves that are the waves, not the electrons?

In summary:

How can two orbitals interfere constructively and destructively?
Do anti-bonding orbitals exist when not occupied?
Is it the orbitals themselves that are interfering with each other like waves?

They do not simultaneously interfere constructively and destructively.

Let me explain. Since the time-independent Schrödinger equation is not separable for molecules due to inter-electronic interaction terms, we can seek solutions to this problem with the variational method.

Let's examine the case of the $\mathrm{H_2}$ molecule. The most natural/physically meaningful/simple choice is to use a trial function that consists of a linear combination of hydrogen 1s atomic orbitals:
$$E\left(c_{1},c_{2}\right)=\dfrac{\left\langle c_{1}\psi_{1}+c_{2}\psi_{2}\left|\hat{\mathcal{H}}_{\mathrm{e}}\right|c_{1}\psi_{1}+c_{2}\psi_{2}\right\rangle }{\left\langle c_{1}\psi_{1}+c_{2}\psi_{2}|c_{1}\psi_{1}+c_{2}\psi_{2}\right\rangle }=\dfrac{{\displaystyle \sum_{i=1}^{2}}{\displaystyle \sum_{j=1}^{2}}c_{i}^{\star}c_{j}\left\langle \psi_{i}\left|\hat{\mathcal{H}}_{\mathrm{e}}\right|\psi_{j}\right\rangle }{{\displaystyle \sum_{i=1}^{2}}{\displaystyle \sum_{j=1}^{2}}c_{i}^{\star}c_{j}\left\langle \psi_{i}|\psi_{j}\right\rangle }$$

$$H_{ij}\equiv\left\langle \psi_{i}\left|\hat{\mathcal{H}}_{\mathrm{e}}\right|\psi_{j}\right\rangle$$

$$S_{ij}\equiv\left\langle \psi_{i}|\psi_{j}\right\rangle$$

$$E\left(c_{1},c_{2}\right)=\dfrac{{\displaystyle \sum_{i=1}^{2}}{\displaystyle \sum_{j=1}^{2}}c_{i}^{\star}c_{j}H_{ij}}{{\displaystyle \sum_{i=1}^{2}}{\displaystyle \sum_{j=1}^{2}}c_{i}^{\star}c_{j}S_{ij}}$$

After minimizing the energy with respect to $c_1$ and $c_2$, you will end up with two energy levels, each one associated with a wave function:
$$E_{+}=\dfrac{H_{11}-H_{21}}{\left(1-S_{12}\right)}\qquad\psi_{+}=\dfrac{1}{\sqrt{2}}\left(\psi_{1}+\psi_{2}\right)$$

$$E_{-}=\dfrac{H_{11}+H_{21}}{\left(1+S_{12}\right)}\qquad\psi_{-}=\dfrac{1}{\sqrt{2}}\left(\psi_{1}-\psi_{2}\right)$$

In other words, by choosing to model the $\mathrm{H_2}$ molecule with two hydrogen $1s$ atomic orbitals, you find two discrete energy levels, each one associated with its wave function (one anti-bonding and one bonding). So this answers your questions:

The two atomic orbitals do not simultaneously interfere: The MOs are two different wave functions.
Of course the anti-bonding orbital exists when not occupied.

Note that the previous procedure can be generalized for all kind of atomic orbitals.

I want to clarify something about the meaning of the wave function since you're asking: As a result, is it correct to think that it's the orbitals themselves that are the waves, not the electrons?

Let's assume that your system consists of a single particle in a one-dimensional space. The wave function, $\Psi\left(x,t\right)$,
is a complex function that contains all the information about the system. But what is really physical meaningful is its modulus squared, $\left|\Psi\left(x,t\right)\right|^{2}$, which represents the position probability density. What is the position probability density? It's something that multiplied by a certain length (or volume if you're working in three dimensions), let's say dx, gives you the probability for finding the particle between $x$ and $x+dx$
$$P\left(x,x+dx\right)=\left|\Psi\left(x,t\right)\right|^{2}dx$$.
Of course, you can integrate the previous equation to find the probability over a certain region of space, $\Omega$
$$P_{\Omega}={\displaystyle \int_{\Omega}}\left|\Psi\left(x,t\right)\right|^{2}dx$$.


What does the eta notation mean in the naming of a transition metal complex? I am reading a paper (Gutel et al., J. Mater. Chem. 2007, 17, 3290-3292) and I have come across the following reaction:

  In organic solvents, the decomposition under mild conditions of
  ($\eta^4$-1,5-cyclooctadiene)($\eta^6$-1,3,5-cyclooctatriene)ruthenium(0),
  Ru(COD)(COT), under a dihydrogen atmosphere gives ruthenium
  nanoparticles (RuNPs), by releasing cyclooctane (COA).

  $$\text{Ru(COD)(COT)} + \text{H}_2 \rightarrow \text{RuNPs} + \text{COA}$$

What do the $\eta^4$ and $\eta^6$ notations represent?  Do they somehow specify the coordination "connectivity" of the Ru(0) and the COD and COT coordinating ligands?

Below, I have drawn the (two-dimensional) structures of COD and COT.  How would I draw the connectivity of Ru(COD)(COT), or is it not specified in the given name?

What do the $\eta^4$ and $\eta^6$ notations represent? Do they somehow specify the coordination "connectivity" of the Ru(0) and the COD and COT coordinating ligands?

They do.

The $\eta$-notation denotes the hapticity of the ligand, which is the number of atoms in the ligand bonded to the metal center.

In your image, shown above, cyclooctadiene (COD), has two alkene functional groups. A transition metal can coordinate to just one alkene, which would be $\eta^2$, or to both alkenes, which would be $\eta^4$.

Cyclooctatriene (COT) has three alkenes. A metal center can be coordinated at one ($\eta^2$), two ($\eta^4$), or all three ($\eta^6$) alkenes.


How should I construct the orbitals of dilithium using only pen and paper? For a week I have lessons about quantum chemistry and I have a trouble understanding how to arrive at the good answer for the drawing of the orbital $(\ce{III})$ for $\ce{Li2}$ (see the second picture).

In the first picture I did an example of my reasoning with a basic molecule such as $\ce{LiH}$ just to show you how I figure it must work.

And in case someone would like to mention it, yes I haven't drawn the $\mathrm{p}_x$ and $\mathrm{p}_y$ orbitals because it is not the point here.

With LiH

With $\ce{Li}_2$

The question is on this last picture, if you have trouble to read tell me.

When combining orbitals, your first approximation should be to relate similar ones with similar ones. In this case, on each lithium (or nitrogen or fluorine) atom we have:

1 s-orbital
1 p-orbital in the direction of bonding
2 p-orbitals perpendicular to the bonding axis.

We should treat these separately initially, see what the results give us, and then think if we should combine them.

Realise that the point group of the molecule is $D_\mathrm{\infty h}$. However, that is a bad point group to work with. Instead, assume $D_\mathrm{2h}$ and correlate your results back to $D_\mathrm{\infty h}$ according to a table originally presented here:

$$\begin{array}{cc}
\hline
{\text{representation in }D_\mathrm{2h}} & {\text{representation in }D_\mathrm{\infty h}}\\
\hline
A_\mathrm{g} & \Sigma_\mathrm{g}^+\\
B_\mathrm{1g} & \Sigma_\mathrm{g}^-\\
B_\mathrm{2g} + B_\mathrm{3g} & \Pi_\mathrm{g}\\
A_\mathrm{g} + B_\mathrm{1g} & \Delta_\mathrm{g}\\
B_\mathrm{1u} & \Sigma_\mathrm{u}^+\\
A_\mathrm{u} & \Sigma_\mathrm{u}^-\\
B_\mathrm{2u} + B_\mathrm{3u} & \Pi_\mathrm{u}\\
A_\mathrm{u} + B_\mathrm{1u} & \Delta_\mathrm{u}\\
\hline\end{array}$$

Ignore the difference between $\Sigma^+$ and $\Sigma^-$ for now. What is relevant is that both the two s-orbitals and the two p-orbitals along the bond axis transform as $A_\mathrm{g} + B_\mathrm{1u}$ in $D_\mathrm{2h}$, and thus as $\Sigma_\mathrm{g}^+ + \Sigma_\mathrm{u}^+$ in $D_\mathrm{\infty h}$. Which means that we expect a bonding and an antibonding orbital from both — no surprises here. So initially, assume (as it looks you tried to do) a pair of $\unicode[Times]{x3c3}$ and $\unicode[Times]{x3c3}^*$ orbitals for both.

Since the energy difference between s- and p-orbitals is large enough (you can and should always assume that) the antibonding $\unicode[Times]{x3c3}_\mathrm{s}$ orbital should still be lower than the bonding $\unicode[Times]{x3c3}_\mathrm{p}$ orbital. These two now have a different symmetry and thus cannot mix productively.

Fill in your electrons. You have two, which will end up in the lowest $\unicode[Times]{x3c3}_\mathrm{s}$ orbital.

What can still mix is $\unicode[Times]{x3c3}_\mathrm{s}$ with $\unicode[Times]{x3c3}_\mathrm{p}$ and their corresponding antibonding orbitals. This will lead to the fidgety task of determining how much stabilising interaction can still be found. Assume the $\unicode[Times]{x3c3}_\mathrm{s}$ to be stabilised slightly by $\unicode[Times]{x3c3}_\mathrm{p}$, because the former is populated while the latter is empty. That should give the $\unicode[Times]{x3c3}_\mathrm{s}$ orbital a slight look as if it derived from sp hybrid orbitals. (Only weakly though!) Note however, that this secondary mixing can be ignored at first approximation.

Also note that any of this mixing must result in symmetric or fully antisymmetric orbitals with respect to the plane of symmetry perpendicular to the bond axis ($\sigma_\mathrm{h}$).

Taking all this and reviewing your picture:

Your lowest orbital looks good. A fully bonding orbital of the s-type with very slight hints of a p-orbital mixing in.
Orbital 2 should look much more like an antibonding $\ce{H2}$ orbital. Since both antibonding orbitals are unpopulated, mixing is unproductive. If you must, very slightly add a bit of p-contribution on the inner side of the atoms.
Orbital 3 should look almost like orbital 1, but clearly with a much larger p-look; so have two nodal planes, one through each atom. Once you get to and past diboron, this is the primary bonding orbital.
Orbital 4 is technically good. However, as per the initial approximation both lobes on each atom should be equal, and only due to the secondary mixing should the inner lobes be slightly smaller. Draw it, as you may be accustomed to $\mathrm{(sp^3)^*}$ hybrid orbitals.

If you want a shortcut, remember to consider s and p separately. Two s-orbitals mix as you know it from hydrogen, creating a very big spherish blurb and an antibonding double-sphere. Two p-orbitals mix to give something like your orbitals I and II (bonding and antibonding). The two s should be lower than both the p.

In a second step, you can mix the two bonding orbitals (and the two antibonding orbitals; but not bonding with antibonding) to add a little p-contribution to your s-type orbitals. However, the lower should clearly always remain an s-type (with a little p mixed in).


Cleavage of type 1 collagen by pepsin. Where does pepsin cleave a human type 1 collagen molecule? Does it occur at multiple locations or only one?

There are multiple types of collagen variants, so the answer here is somewhat non-specific.
Pepsin is an endo type of serine protease and it has low specificity. Proteolysis depends on residues upstream (P1’) and downstream (P1) of the recognition site. Hydrophobic residues at P1’ and P1 increase the likelihood of successful proteolysis. An aromatic residue such as Phe, Trp and Tyr at P1’ and a hydrophobic residue Phe, Leu and Met at P1 are associated with highest likelihood of successful proteolysis. Positively charged residues such as His, Lys and Arg at position P1 are detrimental to proteolysis.
A collagen fiber contains multiple fibrils that are made up of tropocollagen molecules, strengthened by cross-links. Tropocollagen is in fact comprised of three left-handed procollagen molecules, forming a right-handed triple helical structure.
Treatment of acid-soluble tropocollagen with pepsin releases multiple peptides whose composition mainly includes Glycine, Tyrosine, Leucine and Phenylalanine amino acids among others.


Why the salts in a salt bridge? Take the example of a copper and zinc galvanic cell, connected by a salt bridge of $\ce{KNO3}$. I understand how the reactions will result in positive and negative charges, and that the ions of the salt will migrate to maintain neutrality. But why is the salt needed in the first place?

In one beaker we will have an excess of zinc ions (resulting in positive charge) and the other with an excess of copper ions (resulting in negative charge). If we just have a bridge, perhaps just soaked in water, wouldn't the ions, if not purposefully then randomly, travel across the bridge and maintain neutrality? What is the purpose of the salt then, if it works without a salt?

The purpose of the salt bridge is to prevent the two half-cell solutions from mixing. It is possible to make a really bad galvanic cell by putting both half-cells in a single solution, but they rapidly self discharge as the oxidizing agent ($\ce{Cu^+}$ in your example) can diffuse through the solution and react directly with the other electrode (Zn in the example), with the electrons being transferred directly instead of going through the external circuit.

Having a "bridge" of just water soaked paper is similar to having both half-cells in the same container but poses some issues. First, the diffusion of ions through the bridge is going to be very slow, increasing the cell resistance, second, the self-discharge problem above will occur, and third, because different ions diffuse at different rates, a junction potential develops as the two containers diffuse into one another.

So a salt bridge's job is to not only maintain electroneutrality between the two containers, but to do so without allowing the solutions to mix. An inert salt is used to prevent any reactions from occurring between the salt and the solutions, and things like $\ce{KCl}$ are commonly used for analytical cells as potassium and chloride ions have very similar diffusion coefficients, minimizing junction potential.


Ring closing esterification I've been taught that to make esters, the carboxylic acid has to be activated — for instance as the acid chloride or using DCC.

But, when I read papers on total synthesis, always they use more complex reactions such as a Yamaguchi in order to do a ring closing esterification.

Why are standard esterificaton conditions not suitable for complex molecules?   One though I had was that acid chlorides are so reactive that they're easily quenched by water, but this would just give back a carboxylic acid which could be reacted again (no damage done overall).

The reaction you're describing is known as a macrolactonisation, where a medium/large ring is closed by the formation of an lactone (a cyclic ester).

The real problem with macrolactonisation is that there is a somewhat significant entropic barrier to the reaction, you're taking something with a lot of flexibility and constraining it into a ring, reducing the conformational flexibility of the system.  As the ring size increases, entropy becomes significantly more important than enthalpy.

Because the reaction is so entropically unfavourable, the intermolecular reaction has a tendency of happening (rather than the desired intramolecular ring closure).

To avoid these issues, two things are commonly done:

macrolactonisations are ran with high dilution, high dilution favours the intramolecualar reaction by decreasing the probability of two molecules encountering one another
modern macrolactonisation procedures limit the amount of the activated carboxylic acid present (or indeed the activated alcohol) at any one time, again, preventing the likelihood of two molecules reacting together in an intermolecular sense.

The whole thing is maybe easiest explained with an example. The Yamaguchi macrolaconisation is one of the most common, but many others (i.e. the Shiina) follow the same principle.

In the first step, the seco acid (the molecule with the free OH and the free carboxylic acid) are reacted with the TCBC to form a mixed anhydride.

Although this mixed anhydride is 'activated' relative to the free carboxylic acid, it isn't usually reactive enough, meaning that the mixed anhydride can be isolated (no intra or intermolecular reaction takes place).

The isolated mixed anhydride is then added dropwise to a very dilute solution of DMAP (acting as a nucleophilic catalyst). The DMAP attacks the mixed anhydride, forming a highly reactive intermediate (much more reactive than the mixed anhydride), from which the alcohol may attack the carbonyl, displacing DMAP, and forming the desired macrolactone. By adding the mixed anhydride to the DMAP, it limits the amount of the molecule in the flask which is sufficiently activated to react at any given time, increasing the likelihood of intramolecular reaction further.


What are the currently most accurate sources for lists of the values for elemental properties I have a hopefully simple question. I am a tutor working with a student that wants to know where to find the most recent accurate measurements for elemental properties. He's only in Chemistry I, so he's not looking for every property out there calculated multiple ways. I think he (and I) just like looking at the best sources. So we're looking for the following:

Latent heat of fusion
Latent heat of vaporization
Melting point
Boiling point
Density
Specific heat
Ionization energy
Atomic radius (metallic and covalent)
Atomic mass
Electronegativity
Commonality and uncommonality of charges (like oxygen is commonly -2 but is also -1 in peroxide)

I can point him to wikipedia or https://ptable.com/#Properties/Series for some of the information, but I myself don't know where the most legitimate sources for these values can be found. My CRC handbook (97th edition) lists ionization energies, but for atomic weight it doesn't list the weight of the most common isotope of elements like Ac. Also, it's a few years old.
My question is, what are the current best sources of accurate values for each of these properties?

You should teach your student not to use numerical values with as many significant figures as possible. Knowing how to make sense to decimals is part of the scientific culture. The distance between two cities cannot be given to the least micrometer : it has no meaning. The weight of an object has no use if it is given such as $123,45678910111213 g$. The least figures which can be given for sure is the milligram (here 6). Put on a balance, such an object will display fluctuating next figures. Already the figure $7 (0.1 mg$) will fluctuate at the least draft. To be sure of the next decimals you must put your balance in an insulated and closed room without nobody entering during the whole day, in order to avoid drafts. And even, the least dust grain falling from the air will change one or two of the following figures. So please do learn to use numbers with maybe $2, 3, 4$ or $5$ significant figures. But not with ten or twenty significant figures. It is not the way science works. So why looking for the newest published values which have one or two more significant figures ? The given weight $123.45678910111213 g$ is not better than $123.456 g$. It only gives an irrational feeling of superiority. That's all. It has no meaning for usual applications. Nobody will believe you if you say that you have calculated a weight of $123.4567891011121314 g$, or measured a temperature of $12.3456789 °C$.


How to convert a linear dicarboxylate species to the dimethyl ester? I was wondering about the ozonolysis of cyclohexane and further treatment with $\ce{H2O2}$. I understand that the product to this reaction is hexanedioic acid.

How could we substitute the $\ce{-H}$ on each carboxylic group to a methyl, as in the following structure:

I am pretty sure it would be by adding two moles of methanol (Fischer esterification). Is this procedure correct?

You have a couple of options.

Your acid-catalyzed Fisher esterification is one of them. The downside is that water, which is liberated in the reaction, has a higher boiling point than methanol. You can't just destill it off to drive the reaction to completion.
Zhe and  Hexacoordinate-C mentioned diazomethane and methyl iodide which might work well. You might want to make sure that your DNA does not get methylated.
The Steglich esterification, using dicyclohexylcarbodiimide (DCC) and N,N'-dimethylaminopyridine (DMAP) might be an alternative to the Fisher esterifiction.
You might want to try something different, using dimethyl carbonate as a non-cancerogenic alkylation reagent. The procedure is described in Green Chemistry Letters and Reviews, 2013, 61, 89-93 and uses potassium carbonate as a base, dimethyl carbonate as a reagent (and solvent), and tetrabutylammonium chloride as a catalyst.


Changes in entropy of the universe in a reversible reaction Does entropy of the universe always increase in one of the forward or backward reactions of a reversible reaction and decrease in the other?

If no then how do you explain the entropy of the universe increasing in both directions or decreasing in both directions?!?

If yes, do we get more of the reaction that increases the entropy of the universe? I would presume yes as all real processes increase the entropy of the universe.

First, a defintion of terms:  The meaning of reversible in "reversible reaction" is different from the thermodynamic meaning of reversible.  A reversible reaction is one that is kinetically reversible.  I.e., one where you can adjust the relative concentration of products and reactants back and forth as you change conditions.   By contrast, an example of an irreversible reaction would be cooking an egg, which causes the proteins to denature.  The proteins can't be readily undenatured, even if you change the conditions to thermodynamically favor denaturation because, kinetically, the proteins can't find their way back to their undenatured state.

Anyways, let's suppose we have a "reversible reaction", $$\ce{A <=> B},$$  in which the equilbrium constant is, under certain conditions (e.g., at a certain temperature and pressure):

$$K_{eq} = \frac{[B]}{[A]} = 10,$$ i.e., at equilibrium, we will have 10x as much B as A.

This represents the state at which the entropy of the universe is a maximum under those conditions.  Thus if, under those conditions, $\frac{[B]}{[A]} <10$, the reaction will proceed in the forward direction until it reaches equilibrium, increasing the entropy of the universe in the process.  Likewise if, under those condtions, $\frac{[B]}{[A]} > 10$, the reaction will proceed in the reverse direction until it reaches equilibrium, also increasing the entropy of the universe in the process.

So it is not the case that if, say, $\frac{[B]}{[A]} = 9$, then the entropy of the universe would increase in both directions. Rather, it would increase in the forward direction (until it reached equilibrium), and (theoretically) decrease in the reverse direction.  If we instead had $\frac{[B]}{[A]} = 11$, the opposite would happen: decreased entropy of the universe in the forward direction, increased entropy of the universe in the reverse direction (again, until it reached equilibrium).


Will selenium react or bond, ionic or covalent, with vanadium? I was recently asked if $\ce{Se}$ will bond with $\ce{V}$, creating a compound $\ce{SeV}$ . I am not sure if this is even possible but if so I'd like to find out.

The ASM Alloy Phase Diagram Database (at asminternational.org if you have access) shows that $V_{2}Se_{9}$, $V_{5}Se_{8}$, $V_{3}Se_{4}$ $V_{5}Se_{4}$, and $VSe$ are all accepted stable phases in the Se-V binary system.


Linear Combination of Atomic Orbitals In Atkins Physical Chemistry 10th edition in chapter number 10B it is given that within the Born-Oppenheimer approximation, the Schrödinger equation can be solved for the molecular species $\ce{H2+}$ but to simplify things and to obtain a method more readily generalizable to more complex species, we use the linear combination of atomic orbitals concept.

But my question is: how do we know that the wave function obtained from the linear combination obtained (say of two 1s orbitals of each atom) satisfies the Hamiltonian of this new system involving two nuclei? (The modified Hamiltonian is also given on the same page in the book).

Or do we need to make some iterative calculation as in the Hartree-Fock self consistent field method.

PS. I am a beginner so please try to keep the answer technically simple if possible.

Using a Linear Combination of Atomic Orbitals (LCAO) will not exactly satisfy the Schrödinger Equation / be an eigenfunction of the Hamiltonian
We can write down the Hamiltonian for the $\ce{H2+}$ cation with bond length $2R$, nuclei at $\vec{R_1}$ and $\vec{R_2}$, using the Born-Oppenheimer approximation:
$$
H\Psi(\vec{r}) = \left[-\frac{\nabla^2 }{2} - \frac{1}{\vert\vec{r}-\vec{R_1}\vert} - \frac{1}{\vert\vec{r}-\vec{R_2}\vert} \right]\Psi(\vec{r}) = E \Psi(\vec{r})
$$
If you feed in a linear combination of 2 $1s$ orbitals into $H\Psi$ you get a mess that is not a multiple of the linear combination you put in - i.e. a linear combination of 1s orbitals is not the exact solution.
In this case, because we only have one electron, the Schrödinger Equation can be solved to arbitrary accuracy using a variety of numerical techniques for second order partial differential equations in 3D; such as solution in an basis of functions or on a grid in space.
Solving in Practice
A suitable infinite basis of functions can describe all possible functions in 3D.  Two different examples are spherical harmonics with suitable radial components (Slater type orbitals) such as the $1s$ orbital, or Gaussian functions with suitable angular components (Gaussian type orbitals more commonly used in computational chemistry today).
With a infinite number of functions we would have an exact solution, but with only the first few terms in a suitable series we can be remarkably accurate.
In practical electronic structure theory, this means 3-30 functions per atom with suitable chosen parameters, centering the functions on atomic nuclei, as that's where the electrons will be mostly due to attraction to nuclei.
Variational Principle
Seeing as using a finite number of orbitals will not give an eigenfunction of the Hamiltonian, and certainly not the lowest energy ground state, we need another way to confirm a solution.
Instead we solve the Schrödinger equation "projectively".  Hamiltonians are hermitian operators.  One of the properties of hermitian operators is that their eigenfunctions are  orthogonal.  What this means is that for a Hamiltonian $H$ and two exact solutions $\Psi_i$ and $\Psi_j$ the integral:
$$
<\Psi_i\vert H\vert\Psi_j> = \iiint_{\text{all space}} \Psi_i^* H \Psi_j \;\text{d}V = \cases{E_i \text{ if } \Psi_i = \Psi_j\\0 \text{ otherwise}}
$$
Taking the integral of two wavefunctions, one multiplied by the complex conjugate of the other, is often referred to as projecting one wavefunction onto another.
Another property of the Hamiltonian operator is that the value of the integral with some generalised wavefunction $\Psi$:
$<\Psi\vert H\vert\Psi>$, the "expectation value" of the energy of $\Psi$, is minimised by the exact ground state wavefunction.  This property is know as the variational principle.  Hence in self consistent field/Hartree Fock calculations, we solve equations to minimise that integral of our LCAO wavefunctions by varying the coefficents to get the best LCAO approximation to the true wavefunction.


EtG formation outside the liver Ethyl glucuronide (EtG) is formed by conjugation with glucuronic acid catalyzed by the enzyme UDP-glucuronosyltransferase.

My question: Is it possible for EtG to form spontaneously in a urine sample that contains a small amount of ethanol?

No, because the equilibrium for the simple reaction lies on the side of the reactants:

$$\ce{Ethanol + glucoronic acid <<=> Ethyl glucuronide + H2O}\tag{1}$$

The reaction works enzymatically inside the cell because glucuronic acid first reacts with UTP, turning it into a mixed anhydride (phosphate of UDP bound to carboxylic acid of glucuronic acid) plus inorganic phosphate.

$$\ce{UTP + glucoronic acid <=>> UDP-glucuronide + phosphate} \tag{2}$$

Then, the enzyme swaps UDP for ethanol or other molecules.

$$\ce{Ethanol +  UDP-glucuronide <=>> Ethyl glucuronide + UDP}\tag{3}$$

The net difference between hypothetical reaction (1) and the two step process (2) and (3) is the hydrolysis of UTP, a reaction with a negative Gibbs energy of reaction (i.e. equilibrium constant larger than one):

$$\ce{UTP + H2O -> UDP + phosphate} \tag{4}$$

So the formation of UPD and phosphate from UTP drives the two step reaction (it is a coupled reaction).

Attaching glucuronic acid makes molecules more water-soluble, helping to excrete them in the urine 1.


Naproxene syntheses: electrophilic aromatic substitution on activated naphthalene I am studying Syntex's naproxene syntheses and I came across a couple of electrophilic substitutions whose mechanism I can't understand.

The first one is a Friedel-Crafts acylation:
the products are the ones indicated in the picture and they are separated before proceeding. No other byproducts are shown.

Why did the acetyl go in pos. 6? As far as I know, When an electrophilic substitution is made on an activated substituted naphthalene, the electrophile will go in the substituted ring, in the pos. 1 (kinetic control) or in pos. 2 (thermodynamic control), which is in ortho or para to the activating substituent. So I would expect these products, in theory:

I assume the contributing structures that represent the electron-donating effect are the following:

Therefore, although I would expect a 6-acetyl derivative as a byproduct (at least in theory, since (4) might allow it), I don't understand why it is the main product. No clues at all about the 5-acetyl byproduct, since the 5-acetyl intermediate is not stabilized by the electron-donating effect: could it possibly be an error?

The second reaction is a bromination on 2-naphthol:

Likewise, I can think that bromine would first go in pos. 1, and then, since it is only slightly deactivating, another bromine can be added. But again, why in pos. 6? Why not in pos. 3, in ortho to the activating substituent?

And in the end, are these reactions done under kinetic or thermodynamic control?

No other byproducts are shown

I haven't checked the original Syntex patent (if you've ever written a patent you know that sometimes important details are omitted), but it seems odd to me that those two isomers are claimed to be the most significant isomers formed. As you point out, the 2-methoxy-5-acetoxy isomer is not expected to be formed in significant amounts based on electrophilic substitution patterns in substituted naphthalenes (Note: here is a nice concise reference that shows how to apply the same concepts used to predict which isomers will  predominate in electrophilic substitution with benzene to substituted naphthalenes).

In this abstract the Friedel-Crafts acetylation of 2-methoxynaphthalene is studied.  By changing solvents they can shift the major product from 2-methoxy-3-acetoxy to 2-methoxy-6-acetoxynaphthalene.  These two isomers are among those expected and the experiment shows that these two isomers are close enough in energy that small factors can significantly impact the product distribution.  Two other factors to keep in mind

the 6-acetoxy isomer corresponds to "para" substitution
When running a Friedel-Crafts with oxygenated (hydroxyl, methoxy) aromatics, the aluminum chloride will (depending to some degree upon solvent) complex with the oxygen.  In the case at hand this could make the methoxy a bulky substituent disfavoring reaction at the ortho positions.

  But again, why in pos. 6? Why not in pos. 3, in ortho to the
  activating substituent?

Careful low temperature mono-bromination of phenol shows that para-substitution is strongly favored (>95%, reference).  If additional bromine is available a second bromine will be substituted at the ortho position. It seems that the same thing is happening here in your 2-naphthol example.

Like you, I might not have predicted these outcomes, but that's why we run experiments.

  are these reactions done under kinetic or thermodynamic control?

Sorry, without more information (like product composition as a function of time, temperature, etc) I can't answer that.


Why would a solution of FeCl2 appear brown/yellow sort of like FeCl3? If we prepare a solution of $\ce{FeCl2}$ by dissolving solid $\ce{FeCl2}$ powder in water, supposing the bottle of $\ce{FeCl2}$ was left open for sufficient time, is it possible for $\ce{FeCl2}$ to have oxidised to $\ce{FeCl3}$? I can't find a suitable reaction for this excecpt one which includes $\ce{HCl}$ which was definitely not present when preparing the mixture.

The yellow/brown color you are seeing in the solution of $\ce{FeCl2}$ is due to formation of various types of hydrated iron oxide and not iron(III) chloride. They may include both the anhydrous($\ce{FeOOH}$)  or the monohydrate($\ce{FeOOH.H2O}$) often referred to as ferric hydroxide($\ce{Fe(OH)3}$). They are generally yellow to brown in color. The reaction is given here:
$$\ce{4FeCl2 + 6H2O + O2 → 4FeO(OH) + 8HCl}$$

Reaction between iron chloride (II), water and oxygen(moisture) to form
iron metahydroxide(Iron(III) oxide-hydroxide) and hydrogen chloride. The reaction takes place at
reflux.

Moreover, if you further heat iron chloride in oxygen at 450-480℃, iron oxide will form.
References(all pdf links)

http://www.jieas.com/fvolumes/vol081-5/3-5-11.pdf
http://www.gfredlee.com/SurfaceWQ/StummOxygenFerrous.pdf
https://repository.kulib.kyoto-u.ac.jp/dspace/bitstream/2433/77056/1/chd061_5-6_335.pdf


Can there be two precipitates in a double displacement reaction? On one of my chemistry homework's, I am doing a double displacement reaction and the question is having me predict the products and make a net ionic equation for copper(|I) sulfate and barium acetate reacting.

$$\ce{CuSO4(aq) + Ba(OH)2(aq) -> Cu(OH)2 + BaSO4}$$

I was able to predict the products of it, but when I was starting to do the net ionic equation, I came across the problem that there were no spectator ions and I am not sure if two precipitates can form out of a double displacement reaction.

Yes, you were right in concluding that there are no spectator ions. How is that a problem? There is no commandment that says "Thou shalt not have two precipitates in one reaction". Indeed, such double precipitations are very much possible, though rare. The most famous example is probably the formation of lithopone.


How do I solve questions pertaining to rate of disappearance and appearance? In a reversible reaction $\ce{2NO2 <=>[$k_1$][$k_2$] N2O4}$, the rate of disappearance of $\ce{NO2}$ is equal to:

  $2\frac{k_1}{k_2}\ce{[NO2]}^2$
  $2k_1\ce{[NO2]}^2-2k_2\ce{[N2O4]}$
  $2k_1\ce{[NO2]}^2-k_2\ce{[N2O4]}$
  $(2k_1-k_2)\ce{[NO2]}$

The answer, they say, is (2). I just don't understand how they got it. I find it difficult to solve these questions. Is the rate of disappearance the derivative of the concentration of the reactant divided by its coefficient in the reaction, or is it simply the derivative?

If someone could help me with the solution, it would be great. And please, don't assume I'm just picking up a random question from a book and asking it for fun without actually trying to do it. I have worked at it and I don't understand what to do.

In general, if you have a system of elementary reactions, the rate of appearance of a species $\ce{A}$ will be

$$\cfrac{\mathrm{d}\ce{[A]}}{\mathrm{d}t} = \sum\limits_i \nu_{\ce{A},i} r_i$$

where

$i$ is each reaction in the system
$\nu_{\ce{A},i}$ is the stoichiometric coefficient of species $\ce{A}$ in reaction $i$ (positive for products, negative for reagents)
$r_i$ is the rate for reaction $i$, which in turn will be calculated as a product of concentrations for all reagents $j$ times the kinetic coefficient $k_i$:

$$r_i = k_i \prod\limits_{j} [j]^{\nu_{j,i}}$$

In other words, there's a positive contribution to the rate of appearance for each reaction in which $\ce{A}$ is produced, and a negative contribution to the rate of appearance for each reaction in which $\ce{A}$ is consumed, and these contributions are equal to the rate of that reaction times the stoichiometric coefficient.

The rate of disappearance will simply be minus the rate of appearance, so the signs of the contributions will be the opposite.

In your example, we have two elementary reactions:

$$\ce{2NO ->[$k_1$] N2O4} \tag{1}$$

$$\ce{N2O4 ->[$k_2$] 2NO} \tag{2}$$

So, the rate of appearance of $\ce{N2O4}$ would be

$$\cfrac{\mathrm{d}\ce{[N2O4]}}{\mathrm{d}t} = r_1 - r_2 $$

Similarly, the rate of appearance of $\ce{NO}$ would be

$$\cfrac{\mathrm{d}\ce{[NO]}}{\mathrm{d}t} = - 2 r_1 + 2 r_2$$

and the rate of disappearance of $\ce{NO}$ would be minus its rate of appearance:

$$-\cfrac{\mathrm{d}\ce{[NO]}}{\mathrm{d}t} = 2 r_1 - 2 r_2$$

Since the rates for both reactions would be

$$r_1 = k_1 \ce{[NO]}^2$$

$$r_2 = k_2 \ce{[N2O4]}$$

the rate of disappearance for $\ce{NO}$ will be

$$-\cfrac{\mathrm{d}\ce{[NO]}}{\mathrm{d}t} = 2 k_1 \ce{[NO]}^2 - 2 k_2 \ce{[N2O4]}$$


Phosphoglucose isomerase mechanism The mechanism for phosphoglucose isomerase changing glucose  6 phosphate to fructose 6 phosphate shows histidine protonating the c5 oxygen and lysine deprotonation the c1 oxygen to form an open chain aldolase. Glutamic acid then deprotonates c2 to create an enediol intermediate and donates the proton to c1. For the proton to be transfered from c1 to c2 the c1 oxygen needs to be protonated and the c2 oxygen deprotonated. Does anyone know of a source that specifies the amino acid residue or residues that facilitate the deprotonation and protonation of the c2 and c1 oxygens?

tl;dr: The active site catalytic residues (BH+ and B’) are thought to be Lys and a His–Glu dyad, respectively. These facilitate the abstraction of proton from C2 and subsequent protonation on C-1.

Some important points to note:

  "shows histidine protonating c5 oxygen and lysine deprotonating c1
  oxygen to form an open chain aldolase
  ”.

The opening of the ring, is facilitated by Lysine and not by Histidine.

  “Glutamic acid then deprotonates c3 to create an enediol intermediate
  and donates the proton to c1”

It is actually the His–Glu  (B’) dyad (see image below) that abstract this proton from C2. See how this process occurs at the same time as the exchange of H+ (protons) with the solvent or medium. In this step C2 loses a proton while C1 gains a proton (Step 3-4)
Finally a proton is added  from the medium and the ring closes

A proposed reaction mechanism for the Phosphoglucose Isomerase reaction involves general acid–base catalysis by the enzyme:

Step 1

Substrate binding.

Step 2

An acid, presumably the Lys ε-amino group, catalyzes the opening of the pyranose ring.

Step 3

A base, presumably the imidazole portion of the His–Glu dyad, abstracts the acidic proton from C2 to form a cis-enediolate intermediate (this proton is acidic because it is α to a carbonyl group).

Step 4

The proton is replaced on C1 in an overall proton transfer. Protons abstracted by bases are labile and exchange rapidly with solvent protons. This step was confirmed by Irwin Rose by demonstrating that $\ce{[2-^3H]G6P}$ is occasionally converted to $\ce{[1-^3H]F6P}$ by intramolecular proton transfer before the $\ce{^3H}$ has had a chance to exchange with the medium.

Step 5

Ring closure to form the product, which is subsequently released to yield free enzyme, thereby completing the catalytic cycle.

Reaction mechanism of phosphoglucose isomerase


How do repulsive solute interactions affect the van't Hoff factor? Context. Consider an ideal solution, one in which the intermolecular solvent-solvent, solute-solute, and solute-solvent interactions are all equal. The presence of a solute therefore produces (solely) an increase in entropy due to the increase in configurational microstates. This increase in entropy leads to the well-known colligative properties, characterized by the van't Hoff factor, the number of effective solute particles per molecule or salt of solute.

Now consider a solution in which the solute-solute intermolecular interaction is changed by $+\epsilon$, so that solutes are more attracted to each other. This decreases the van't Hoff factor due to clumping of the solutes, which we can rationalize from two different perspectives:

probabilistic: solute particles are more likely to be found close to each other (due to the favorable attraction between them), so we effectively have less solute particles as a whole, resulting in a lower van't Hoff factor
statistical-mechanical: microstates in which solute particles are close to each other are weighted more heavily; this perturbation from the uniform distribution of microstates decreases the entropy of the system, effectively decreasing the "strength" of the colligative effect and hence resulting in a lower van't Hoff factor

Both factors agree, so that's good.

Now consider a solution in which the solute-solute intermolecular interaction is changed by $-\epsilon$, so that solutes are less attracted to each other. Consider the two perspectives again:

probabilistic: solute particles are less likely to be found close to each other (due to the unfavorable attraction between them), so we effectively have more solute particles as a whole, resulting in a higher van't Hoff factor
statistical-mechanical: microstates in which solute particles are close to each other are weighted less heavily; this perturbation from the uniform distribution of microstates decreases the entropy of the system, effectively decreasing the "strength" of the colligative effect and hence resulting in a lower van't Hoff factor

Questions.

Why don't the probabilistic and statistical-mechanical viewpoints agree in the latter case?
How can the incorrect viewpoint(s) be modified to make them correct?

I am inclined to believe the statistical-mechanical viewpoint is correct, since it's more fundamental. I'm also somewhat leery of the implication that solute particles being less likely to be close to each other leads to effectively more solute particles.

If this implication is indeed wrong, why is it wrong? How about the converse, that because solute particles are more likely to be found close to each other, we effectively have less solute particles as a whole?
If this implication is actually right, why is it right?

I think that your first probabalistic argument is correct, i.e that as the solute interaction increases the effective number of them is reduced. In the second case, as the interaction become repulsive, the effective number of them cannot increase over the maximum number there. They are restricted in their configurational entropy because the repulsion  means that the chance of finding one next to another is greatly reduced. Thus the probabilistic aligns with your stat-mech one.

You can calculate the entropy of mixing in an ideal solution then in a 'regular' one in which the components interact with one another.

In the ideal solution a statistical argument based on number of ways of placing particles on a grid,  leads to the entropy being given by

$$\Delta S_{mix}= -R[n_1\ln(x_1)+n_2\ln(x_2)]$$

for $n_1,\,n_2$ moles and $x_1,\,x_2$ mole fractions of species 1 and 2.

In a solution in which there is interaction between (equal sized) molecules the free energy is

$$\Delta G_{mix} = R[n_1\ln(x_1)+n_2\ln(x_2)] +(n_1+n_2)x_1x_2E$$

where E is is an energy of mixing. This can be calculated by considering that one solute molecule interacts with energy $\epsilon _{aa}$, the solvent with energy $\epsilon _{bs}$ and between solute and solvent with $\epsilon _{ab}$. Again a statistical argument based on molecules being on lattice sites means that nearest neighbour interactions of the type a-a  and b-b occur with probability $x_a^2$, and $x_b^2$ respectively and between a and b as $2x_ax_b$ and the 2 arise because a_b and b-a interactions are counted.

The total mixed energy is then $E_{tot} \approx (\epsilon_{aa}N_a^2+ 2\epsilon_{aa}N_aN_b+\epsilon_{bb}N_b^2 )$$

but the change on mixing is this energy less that due to the two components unmixed which is $\approx (N_a\epsilon_{aa}+N_b\epsilon_{bb})$

which means that the extra mixing energy is $E = zN_0(2\epsilon_{ab}-\epsilon_{aa}-\epsilon_{bb})$

where $z$ is related to the coordination number on the lattice.

The entropy on mixing is then

$$\Delta S_{mix}= -R[n_1\ln(x_1)+n_2\ln(x_2)] -(n_1+n_2)x_1x_2\frac{\partial w}{\partial T}$$

where the second term $\displaystyle \Delta S^E =-(n_1+n_2)x_1x_2\frac{\partial w}{\partial T}$ is a positive number about $0.2R$ J /mol/K for benzene/cyclohexane and $0.01R$ for benzene /$\ce{CCl4}$

Changing $\epsilon_{bb}$ as you suggest will still increase or decrease $\Delta S^E$ acting through the derivative.


Uncertainty Equations Confusion I am confused about the uncertainty equations calculations.

If I have the equation X1 * Y1 = X2 * Y2 and I want to find the uncertainty of X1 , I convert the equation to X1 = (X2 * Y2)/(Y1) .  Now, I would know how to calculate the uncertainty if it was a straight division or multiplication uncertainty but this equation is both.  How do I go about calculating the uncertainty if the equation contains both multiplication and division? Do I just do the multiplication uncertainty first and then the division uncertainty?

Thanks any advice is appreciated.

You have three variables, for simplicity let's call them, x, y, z each with a (standard deviation) uncertainty $\sigma_x, \sigma_y, \sigma_z$. The formula you want to find the error on is then $f=xy/z$ which will have a standard deviation $\sigma_f$ The method that always works is the  the formula that gives the square of the uncertainty
$$ \sigma_f ^2 = \left( \frac{\partial f}{\partial x}\right)^2_{y,z}.\sigma_x^2+\left( \frac{\partial f}{\partial y}\right)^2_{x,z}.\sigma_y^2+\left( \frac{\partial f}{\partial z}\right)^2_{x,y}.\sigma_z^2$$

which looks horrid and intimidating  but is very easy to use as it only involves taking the derivative of f with x, y and z in turn, while keeping the other variables constant.

In your example $\partial f/\partial x = y/z$, $\partial f/\partial y =x/z$ and $\partial f/\partial z =-xy/z^2$. Putting the parts together gives
$$\sigma_f^2= \frac{y^2}{z^2}\sigma_x^2 + \frac{x^2}{z^2}\sigma_y^2 +\frac{x^2y^2}{z^4}\sigma_z^2 $$

You can simplify this a bit but you should have all the values to plug into this. Don't forget to take the square root at the end.


Is poison still poisonous after its 'expiration date'? As we all know that,

  Any poison is nothing but a chemical compound.

And as discussed in the question: Chemicals-do-have-an-expiry-date!

So, my question is: Is poison still poisonous after its 'expiration date'? and Is every poison always as poisonous as just after it was synthesised  ?

please feel free to update with the suitable tag(s).

It depends on what the poison is.
If we take the colloquial use of the word and include toxins and venoms, many are things like proteins that will certainly denature or otherwise degrade, eventually becoming harmless. e.g. tetrodotoxin, ricin, botulinum, etc.
I would expect that type of poison to have the shortest shelf-life as they are relatively fragile.

Many other poisons are small organic molecules.
These can often be degraded by oxidation in air, exposure to UV, hydrolysis etc. and would include things like nicotine and nerve agents like sarin and VX.
Many nerve agents, have shelf lives of a few years and research has actually been done to extend them for use in munitions.

Several metals are known to be poisonous (like lead, mercury, and cadmium) and are problematic because they are toxic in not only their elemental forms, but also in inorganic and organic compounds.
There may be a great difference in toxicity of the different forms, (see elemental mercury vs methylmercury), but most forms remain at least somewhat toxic.
These may last a very long time because reactions likely to occur under normal conditions may not render them safe, e.g. a chunk of cinnabar ($\ce{HgS}$ mineral) sitting on your desk will not undergo any significant change to render it safe, even on a geological timescale.


A little help understanding some basic algebra(I think) in conversion pathways Attached is a picture from my textbook.

Here is my question:

Must the 3.18cm(1.25in X 2.54cm) have to be cubed(^3) in order to cancel the cm in the 1cm^3 that is beneath the 22.59g Osmium?

Is that a standard rule in algebra? And if it is, under what section might I find it.

I've looked in Blitzer's 6th edition of Introductory Algebra for a clue but didn't find anything there. I've checked out Principles of General Chemistry by Martin S. Silberberg for a similar conversion but didn't find anything similar; all there was in that chemistry text book were much easier conversions.

I've even gone onto youtube and searched "multiplying fractions with unit and exponents" but couldn't find anything.

I find it strange that the entire cm^3 doesn't cancel out and just the cm does.But cubing the 3.18cm is necessary to arriving at the given answer. I could assume that the 1cm^3 should be cancelled, but we cannot cancel the cm^3 on one side and not cancel the cm^3 on the other side, but if we did that, we would be removing the ^3 which is necessary to arriving at the given answer.

Thank you all

Yes, 3.18cm has to be raised to power 3.
$$Density=Mass/Volume$$ or upon rearrangement
$$Mass=Density*Volume$$
The information is provided about edge of the cube made of Osmium and its density. So in order to obtain mass of the cube, it is necessary to find the volume of the cube(since the information about density is supplied). The volume of the cube is edge raised to the power 3,i.e.,
$$Volume of cube=edge^3$$
Since the information regarding edge is provided in inches it has to be converted into cm because the units of density is provided in $g/cm^{3}$.
Once the the volume is also obtained in $cm^{3}$, the mass can be obtained in g.

Follow these steps:

Convert the length into same unit as the the length unit in density(here it is $cm^{3}$). So, 1.25 in=3.18cm
Now to obtain volume use the formula for volume,i.e., to cube the length. So, volume=$$(3.18cm)^{3}\approx 32.16cm^{3}$$
Multiply volume with density to obtain mass. $$Mass=32.16cm^{3}*22.59g/cm^{3}\approx723g$$


Tunneling corrections in reaction rates I've studied Transition State Theory since I asked this a while ago, but so far I have not been able to find a detailed explanation of how tunneling corrections are implemented into reaction rates computations.

I am thinking specifically in reactions involving proton transfer along a hydrogen bond. Electrons don't pose any trouble, because they are treated quantically. But for protons, which are usually beheld as classical particles, semiclassical corrections accounting for tunneling must be made, since a full quantum treatment of protons seems undoable.

First idea: If one performs a molecular dynamics simulation, one could at each step think of the proton "just for a moment" as a quantum particle, find an approximated wavefunction thanks to the WKB method and perform a Monte Carlo simulation to decide "where" will we find the proton at the next step. This seems doable, but it's not what is done, as far as I know (sometimes performing molecular dynamics simulations is not possible, or not useful).

Second idea/ problem: Imagine we have a Free Energy surface (I am concerned with two-dimensional problems, where there are two relevant reaction coordinates to be tracked), which informs us about the precise changes in free energy along infinitely many reaction pathways. For each of those, a classical reaction rate can be computed by means of Eyring equation and then we could take averages over some sets of paths via path integrals or something like that. Even better, one can employ Variational Transition State Theory to derive estimations for the reaction rate, once the Free Energy surface is known. I expect this will not be hard. But I don't have a clue as to how to make the tunneling corrections for the reaction rate in this scenario, although I suspect is in this setting where most tunneling corrections are made.

I've certainly found a lot of bibliography dealing with Quantum Transition State Theory and Tunneling Corrections, but absolutely nowhere do they explain with all the mathematical and physical details the concrete formulas they are employing. Instead, they write a lot (without much equations), give you other references, where you can find still some more references -and so on.

So, in brief, my question is: could anyone provide me with a reference containing a thorough, complete and detailed explanation, not lacking the mathematical details, of how does one compute the tunneling corrections for reaction rate constants?

Have you checked the original papers by C. Eckart, R. P. Bell, H. Shin, J. O. Hirschfelder and E. Wigner? These contain a fair degree of mathematical and physical treatment.

There is also a nice paper from S. G. Christov called "Some aspects of the theory of proton-transfer processes" published in J. Res. Inst. Catalysis 1968, 16 (1), 169-194 (pdf).

Other possible references:

Johnston, H. S.; Heicklen, J. Tunnelling Corrections for Unsymmetrical Eckart Potential Energy Barriers. J. Phys. Chem. 1962, 66 (3), 532–533. DOI: 10.1021/j100809a040.

Some papers published by D. G. Truhlar together with T. N. Truong, B. C. Garret, R. T. Skodje.


DC Cell + Copper Wire + Ground Water DC Cell + Copper Wire + Ground Water produces bubbles from wire and water turns rust red gradually? What is happening?

I can see something is settling down...

Groundwater contain besides other things $\ce{Fe^2+}$ a lot of organic material Lenntech, Natural Variations in the Composition of Groundwater. Oxygen released during electrolysis denatures organic proteins and the $\ce{Fe^2+}$ oxidize to $\ce{Fe^3+}$ that resulting to the reddish colorization.


To what state would copper wiring revert if abandoned for centuries? Would copper wires from power transmission or telephone lines acquire a patina but remain largely intact, or oxidize into dust after a century or more without replacement or maintenance?

Obviously environmental factors would be enormous; ideally I'd like to know what the metal's state would be in ideal (underground?), typical (temperate zone), and worst-case (near salt water?) scenarios. Clearly some circumstances allow copper objects to survive largely intact for several thousand years, since such artifacts have been found by archeologists, but does this require very exceptional circumstances?

This question is inspired by an answer on the Worldbuilding site, which claims that a post-apocalyptic society would have a difficult time developing metallurgy as the rusted remains of human civilization would be more difficult to smelt than the naturally-occurring ores which were relatively abundant during the Stone Age. Hopefully by asking about this copper wires specifically, I have an appropriate question for this site. If so, I may ask another question about iron and steel.

I did not have much luck in finding relevant tags. Any assistance there is appreciated!

Let me talk first about how iron corrodes (as in rusts) and then compare that to copper. The simple answer is that when iron corrodes, the iron oxide (iron which has chemically bonded with oxygen, which it loves to do) that forms on the surface of the pure iron underneath, does not seal the surface against further intrusion of more oxygen. When copper corrodes, the copper oxide does seal the surface protecting underlying copper against further corrosion. I don't know if iron oxide is simply more porous to oxygen than copper oxide is but I do know the following. Iron can chemically combine with oxygen in multiple ratios. When you examine a highly corroded iron surface you'll notice that the first layer of rust is very red. If you brush or maybe lightly sand that red layer of rust away you'll notice a black layer underneath it. That's still rust. If you sand your way through that you'll get to the pure iron underneath which has more of a dark silver color. The red rust has more oxygen in it than the black rust but can pass some of that oxygen to the black rust turning it red. Similarly, the black rust can pass some of its oxygen on to the pure iron underneath it. That process continues until oxygen can leach all the way through the iron. Chemically, iron has more than one oxidation state. Oxidized copper atoms are much more reluctant to release their oxygen to adjacent pure copper and act as a barrier to the intrusion of more oxygen.


Colour due to F-centre is same as that produced in a flame test, why? I have noticed that the colours that the F-centres impart in Alkali metal halides are the same that the corresponding alkali metals impart to their flames.

$\ce{NaCl}$ when heated in an atmosphere of $\ce{Na}$ becomes yellow.

$\ce{LiCl} \rightarrow$ pink

$\ce{KCl} \rightarrow$ violet/lilac

Why ?

Why are the colours the same as the metal. Basically, the anion doesn't affect the colour, but the cation does. And since the cation is just the metal missing an electron or several, of course, they exhibit similar behaviour!

Colours we see are determined by the transitions made by electrons "falling" from a high energy state to a lower energy state, and in the process releasing one or some photons with a frequency in what we call the visible spectrum.

When an F-centre is formed in the described conditions, we have effectively extended the lattice with additional cations the same as those already present. But rather than additional anions that match as well, free electrons take the place in the lattice that would be occupied by one.

Now the anion already has its orbitals filled - there is nowhere for an associated electron to move into when excited, nor somewhere for other electrons to enter when relaxed, and thus there is no way for them to release energy as photons that we sense as light. So the anion doesn't affect the colour.

However the cation has empty orbitals when neutral, let alone when positively charged (i.e. has an electron(s) removed). This means any excitation or relaxation will involve these orbitals.

The result is that the electron movements in the  F-centre lattice are the same as those that would occur in the ionisation of the metal which  the cation is derived from, and hence the colours are also the same.

Why those specific colours for the respective compounds? That would be a completely different question, and probably found in a quantum mechanics discussion rather than a chemistry area.


What explains the stark white colour of my iodometric titration analyte solution? During my Advanced Higher (secondary school) investigation, I reacted pure copper with concentrated nitric acid to acquire a copper nitrate solution. This was neutralised with excess sodium carbonate and brought back to a clear solution with ethanoic acid.

I then reacted the solution with potassium iodide in a conical flask before titrating it with sodium thiosulfate.:

$\ce{2Cu^2+(aq) + 4I^-(aq) → 2CuI(s) + I_2(aq)}$

$\ce{I_2(aq) + I^-(aq) ⇌ I_3^-(aq)}$

$\ce{I_3^-(aq) + 2S_2O_3^2-(aq) → S_4O_6^2-(aq) + 3I^-(aq)}$

This analyte solution was at first brown, but when titrated it turned a pale coffee colour. When starch was added just before the end point, it turned a dark grey-brown colour, as expected - the iodine forming a complex with starch. https://en.wikipedia.org/wiki/Iodine_test

But when all of the iodine had reacted with the sodium thiosulfate, I was left with a stark white solution. I tested to make sure no iodine was left over by adding in another drop of starch - no reaction. So, this seems to be the endpoint.

Copper iodide seems the likely explanation, but the wikipedia page shows that it is mostly insoluble in water. It is soluble in iodine solutions but all the iodine is gone, presumably.

Does it take time for the copper iodide to precipitate? I quickly rinsed out the conical flask. There was some white staining on the sides which I think is copper iodide.

You’re correct that this is just a suspension of copper(I) iodide. If CuI were soluble in water, the solution would be clear; the white colour is a sign of (very small) particles of the solid suspended in solution.


About the use of pascal or atmosphere in chemistry I was wondering if there are specific formulae in chemistry for which it is "compulsory" or convenient to use atmosphere (atm) rather than pascal (or even mmHg) as a unit of measurement.

Of course, we can convert one into the other, but the choice of using one of the two also conditions other units, such as in the case of volume.

Can someone provide me with a good picture of the relationship between these units and the others that we may need to use in the same context? (e.g. we use L  with atm; $\mathrm{m^3}$ with Pa).

Generally, the use of quantity equations is preferred and is strongly recommended. They have the advantage of being independent of the choice of units.

For example, instead of “speed in kilometre per hour is 3.6 times the quotient of distance in metres by time in seconds”, which corresponds to the numerical value equation

$$v/(\mathrm{km/h})=3.6\frac {l/\mathrm m}{t/\mathrm s}$$ or $$\left\{v\right\}_{\mathrm{km/h}}=3.6\frac{\left\{l\right\}_{\mathrm m}}{\left\{t\right\}_{\mathrm s}}$$

you should write “speed $v$ is the quotient of distance $l$ by time $t$”, preferably in form of the quantity equation

$$\displaystyle v=\frac lt$$

Generally, you can work with any units you like; however, this requires the use of conversion factors. The use of conversion factors is never required when only coherent SI units are used; therefore, the use of coherent SI units (the SI base units and the SI coherent derived units, including those with special names) are preferred.

In order to avoid large or small numerical values, decimal multiples and submultiples of units may be formed with the SI prefixes. However, when prefixes are used with SI units, the resulting units are no longer coherent, because a prefix effectively introduces a numerical factor in the expression for the unit in terms of the base units.

Furthermore, various non-SI units are accepted for use.

In the International System of Quantities, pressure $p$ is defined as
$$p=\frac{\mathrm dF}{\mathrm dA}$$
Where $\mathrm dF$ is the force component perpendicular to the surface element of area $\mathrm dA$.

The coherent derived units in the SI for the pressure has the special name pascal and the symbol $\mathrm{Pa}$. According to the definition of pressure, the pascal can be expressed in terms of other SI units as $1\ \mathrm{Pa}=1\ \mathrm{N/m^2}$. The pascal can also be expressed in terms of SI base units as $1\ \mathrm{Pa}=1\ \mathrm{m^{-1}\ kg\ s^{-2}}$. The last form is especially useful to check whether units in equations and calculations are converted correctly.

The SI prefixes may be used with any of the names and symbols, but when this is done the resulting unit will no longer be coherent.

The bar (symbol $\mathrm{bar}$) is included in the list of non-SI units that are used by special interest groups for a variety of different reasons. Although the use of SI units is to be preferred for reasons already emphasized, authors who see a particular advantage in using these non-SI units should have the freedom to use the units that they consider to be best suited to their purpose. In particular, the bar is included because since 1982 one bar has been used as the standard pressure for tabulating all thermodynamic data. (Prior to 1982 the standard pressure used to be the standard atmosphere, equal to $1.01325\ \mathrm{bar}$, or $101\,325\ \mathrm{Pa}$.)

The following other non-SI units can be used, which requires the use of conversion factors. However, the use of these units is deprecated.

$$\begin{array}{lll}
\hline
\text{Unit} & \text{Unit} & \text{Conversion} \\
\text{name} & \text{symbol} & \text{factor} \\
\hline
\text{standard atmosphere} & \mathrm{atm} & 1\ \mathrm{atm} := 101\,325\ \mathrm{Pa}\\
\text{kilogram-force per square metre} & \mathrm{kgf/m^2} & 1\ \mathrm{kgf/m^2} = 9.80665\ \mathrm{Pa}\\
\text{technical atmosphere} & \mathrm{at} & 1\ \mathrm{at}=1\ \mathrm{kgf/cm^2} = 98\,066.5\ \mathrm{Pa}\\
\text{conventional millimetre of water} & \mathrm{mmH_2O} & 1\ \mathrm{mmH_2O}:=10^{-4}\ \mathrm{at} = 9.806 65\ \mathrm{Pa}\\
\text{conventional millimetre of mercury} & \mathrm{mmHg} & 1\ \mathrm{mmHg} \approx 13.5951\ \mathrm{mmH_2O} \approx 133.3224\ \mathrm{Pa} \\
\text{torr} & \text{Torr} & 1\ \mathrm{Torr}:=\frac1{760}\ \mathrm{atm} \approx 1\ \mathrm{mmHg} \approx 133.3224\ \mathrm{Pa} \\
\text{pound-force per square inch} & \text{psi}, \mathrm{lbf/in^2} & 1\ \mathrm{lbf/in^2}\approx6\,894.757\ \mathrm{Pa} \\
\hline
\end{array}$$

Note that expressions for units shall contain nothing else than unit symbols and mathematical symbols. Any attachment to a unit symbol as a means of giving information about the special nature of the quantity or context of measurement under consideration is not permitted.

For example, write “the gauge pressure is $p_\mathrm e=0.5\ \text{bar}$”, not “$p=0.5\ \text{bar(g)}$”.

(The symbol $p_\mathrm e$ is recommended for gauge pressure, defined $p-p_\mathrm{amb}$, where $p_\mathrm{amb}$ is the ambient pressure. Thus the gauge pressure is positive or negative depending on whether $p$ is larger or smaller than $p_\mathrm{amb}$.)


What does "gamma" in gamma-oryzanol mean? The structure of gamma-oryzanol is:

What does the "gamma" in its name mean?

No designations such as alpha-oryzanol, beta-oryzanol, etc. appear to be in use, and according to this WebMD entry, as well as this PubChem document, another name for gamma-oryzanol is simply "oryzanol".

There seems to be some ambiguity as to just what oryzanol, or gamma-oryzanol refers to. According to Chemical Book:

  Gamma Oryzanol is a mixture of 6 substances derived from rice bran oil, including sterols and ferulic acid, called Cycloartanyl Ferulate or Triterpene alcohol ferulate.

But then other sources, including this US National Institutes of Health publication, consider gamma-oryzanol composed of four components:

  There are four main components of γ-oryzanols, namely, cycloartenyl ferulate, 24-methylene cycloartenyl ferulate, campesteryl ferulate, and sitosteryl ferulate [original publication:  Lerma-García M. J., Herrero-Martínez J. M., Simó-Alfonso E. F., Mendonça C. R. B., Ramis-Ramos G. Composition, industrial processing and applications of rice bran γ-oryzanol. Food Chemistry. 2009;115(2):389–404. doi: 10.1016/j.foodchem.2009.01.063

Further confusing the issue is that the PubChem article referenced above gives a single structure for gamma oryzanol (the same as that given in the OP), but does not include any physical properties that one would expect for a single pure compound (i.e. no melting point, solubility, vapor pressure data) although it does give the molecular weight. Again, PubChem lists "oryzanol" as a synonym and does not have any other oryzanol-named compound in it's database.

Summary TL/DR:
The gamma in "gamma-oryzanol" doesn't refer to a specific structure or isomer and there is no corresponding compounds named alpha-oryzanol, beta-oryzanol, etc. Rather, the term gamma-oryzanol is also simply called "oryzanol" and it refers to a combination of several different moieties as described above.


What is the length of a unit cell of CuCl assuming that it is fcc? The density of $\ce{CuCl}$ is given – $x\ \mathrm{g/cm^3}$
The crystal structure is assumed to be fcc.

My teacher is claiming the we can apply the formula

$$\rho=\frac{Z \cdot M}{a^3 \cdot N_\mathrm A}$$
And he took $M = 35.5\ \mathrm{g\ mol^{-1}}+63.5\ \mathrm{g\ mol^{-1}} = 99\ \mathrm{g\ mol^{-1}}$,
$Z = 4$ as it is fcc.
But my question is how can we take $M = 99\ \mathrm{g\ mol^{-1}}$? It means every particle in the fcc has mass $99\ \mathrm u$ doesn't it?

Well, yes, depending on the detail you want to go into. You can find a representation of the crystal structure of CuCl at, for example, http://www.webelements.com/compounds/copper/copper_chloride.html - you will see that the Cu atoms are indeed a simple fcc lattice.  The Cl atoms are on sites that are an interpenetrating fcc lattice.  The end result is the zinc blende (ZnS) lattice prototype, very familiar to semiconductor folks as the crystal structure for GaAs an associated III-V compounds. If all sites were occupied by the same atom, this is known as diamond cubic, commonly seen in diamond, Si, and Ge (amongst others).

But, yes, one way to consider the crystal (particularly for your purposes here) is an fcc crystal with a Cu-Cl basis. Just don't expect the x-ray diffraction to correspond to fcc.


How do I make 3d molecular graphics similar to those shown on Wikipedia? I want to make 3d molecular graphics similar to those found on Wikipedia. In the page for benzene, you can see the structure looks like this:

However, if I go to the interactive image as shown in here:

I get an interactive image that looks like this:

To me, the second image doesn't look nice, and I prefer the first one. However, I cant seem to find a way to get the image of first type for any molecule after converting from SMILES. I want to draw the structure of, say for example, Bismabenzene? How do I produce a graphical image for that if it's not available on Wikipedia?

You can generate the geometry for your molecule of choice by drawing it in Avogadro. More often than not, you want to refine the geometry through a force field calculation.

Avogadro supports multiple ways of visualization, which all can be configured according to your preferences. A ball-and-stick model, which you seem to prefer, is included.

If you want to generate an animation from the input file, QuteMol might be one option.

The interactive visualization on the wikipedia page of benzene is generated using JSmol, which is a JavaScript implementation of Jmol.

If you are only interested in online tools, MolView could be an option.


Are there any molecules with delta bonds in their ground states? When looking at excited states of molecules, $\delta$ bonds are relatively common, but I've never come across a molecule with a $\delta$ bond in its ground state.

Are there molecules with $\delta$ bonds in their ground states?

What about even higher angular momentum states ($\phi$ and $\gamma$ bonds)?

Chromium(II) acetate has a Cr-Cr quadruple bond - and hence a $\delta$ bond - in its ground state. This sort of "paddlewheel" configuration, with four carboxylic acids coordinating a pair of metal ions, is reasonably common (see also copper(II) acetate, for instance), although the extent to which the metals can be considered covalently bonded varies.


Details of Boltzmann distribution derivation I have 2 questions regarding the derivation of the formula which calculates the probability of molecules having a particular amount of kinetic energy $E_x$ in a system of $N$ molecules. It states that:
$$p(E_x) = \frac{\exp(-E_x/kT)}{\sum_{i=0}^\infty \exp(-E_i/kT)}$$
This formula is initially derived from finding the maximum number of permutations $\Omega$ with $N$ molecules distributed over $n$ energy compartments within a system according to:
$$\Omega = \frac{N!}{n_1!n_2!n_3!...n_i!}$$
Question 1. It is explained that one must find the maximum number of permutations by differentiating this permutation function and solving for $0$. However, I can already see from the formula that the maximum number of permutations is achieved if the denominator is equal to 1 (it can't be less), which means that there should be 1 molecule in each energy compartment $n_i$. Why isn’t it possible to reason the maximum that way? I'm aware this would need $N$ number of energy compartments resulting in some having very improbable high energy levels, but can't the difference in energy levels between the compartments be very small?

Question 2.  Suppose the molecules only possess translational kinetic energy. I understand that translational kinetic energy is continuous: it does not have discrete energy levels like rotational or vibrational kinetic energy. Since it is continuous, there are no energy compartments with exact energy values. The number of possible energy compartments within such a system should therefore be infinite and one could only derive probability densities. How is it possible nonetheless to calculate a limited number of permutations using a limited number of energy compartments? Does each energy compartment cover a certain range of energy levels?

What you say is a good idea, but is not quite correct because we must maximize the distribute subject to two constraints. I will reiterate the derivation following along with McQuarrie's Statistical Mechanics.
Constraints in the Ensemble:
The constraints I mention are actually the properties that determine the ensemble we are working with:
$$
\sum_ja_j=A
$$
$$
\sum_j\epsilon_ja_j=E
$$
The first constraint says that sum over all states of the population numbers for each state $j$, given by $a_j$, must equal the total number of members in the system $A$. This is the number of molecules if you like.
The second constraint says that the energy of each state, $\epsilon_j$, multiplied by the population of that state, $a_j$, must equal the total energy, $E$, of the system when the sum is taken over all $j$. These constraints are so intuitive physically that it is easy to forget they even have to be enforced.
By the way, we are not actually constraining the total energy. We only constrain that whatever the energy is, it can be found from the second constraint. In fact, in McQuarrie, this derivation is undertaken for the canonical (NVT) ensemble.
Method of the Most Probable Distribution:
Now, as you say, the distribution which we primarily observe will be the one which maximizes the number of ways to arrange the members of the ensemble. Notice I say primarily observe. This is discussed below. This number of ways is given by the equation you write down,
$$
W(\mathbf{a})=\frac{A!}{\prod_ka_k!}
$$
There are many possible distributions that satisfy the constraints we have given. Given this indeterminate (but probably infinite) number of ensembles which satisfy our constraints, the probability of being found in state $j$ is given by,
$$
P_j=\frac{\bar{a}_j}{A}=\frac{1}{A}\frac{\sum_{\mathbf{a}}a_j(\mathbf{a})W(\mathbf{a})}{\sum_{\mathbf{a}}W(\mathbf{a})}
$$
Notice that we are now summing over the vector of all the quantum states for each ensemble, $\mathbf{a}$, because we want to find $\bar{a}_j$, the average population of quantum state $j$ in each different ensemble satisfying our constraints.
This is clearly not very useful to us because there are so many possible ensembles we will never be able to do something with them. We can, however, take advantage of the form of $W(\mathbf{a})$. Namely, by just playing around with the numbers, you will find that the multinomial coefficients are very strongly peaked at their maximum when all of the $a_j$'s are large. Fortunately, this is exactly the regime we live in because we are thinking about something on the order of $10^{23}$ particles, so the population of states will be very large.
This means we can make an approximation, which is formally exact in the limit as $A\rightarrow\infty$, that $\bar{a}_j/A=a_j^*/A$ where $a_j^*$ is the $a_j$ in the distribution which maximizes $W(\mathbf{a})$. So, the fact we are going to maximize this distribution comes about through an approximation, albeit one which becomes exact in the limit near where we are working.
So, we can simplify our equation for the probabilities to,
$$
P_j=\frac{1}{A}\frac{a_j^*W(\mathbf{a^*})}{W(\mathbf{a^*})}=\frac{a^*_j}{A}
$$
Method of Lagrange Multiplies:
We want to find $a_j^*$ which means we must perform a maximization subject to the two constraints we listed at the beginning. Maximizing something subject to constraints does not seem to be easy, but there is a very nice method for doing it called the method of Lagrange multipliers. Essentially, we introduce some undetermined coefficients to the maximization equation so that the constraints are included, but in reality we are just adding zero to the equation.
What I mean by this is that I can rewrite the constraints as follows:
$$
\alpha\left(\sum_ja_j-A\right)=0
$$
$$
\beta\left(\sum_j\epsilon_ja_j-E\right)=0
$$
where $\alpha$ and $\beta$ are the so-called Lagrange multipliers. Then, we can simply maximize our distribution as normal:
$$
\frac{\partial}{\partial a_k}\left(\ln W(\mathbf{a^*})-\alpha\left(\sum_ja_j-A\right)-\beta\left(\sum_j\epsilon_ja_j-E\right)\right)=0
$$
It really isn't obvious that this should maximize the system while satisfying the constraints, so you will have to read elsewhere why this works, but the fact is that it does work.
Notice, I have actually played a trick. Instead of maximizing $W(\mathbf{a^*})$, I am maximizing $\ln W(\mathbf{a^*})$. This is because $W(\mathbf{a^*})$ has factorials in it, so by taking the logarithm we can employ Stirling's approximation. This is also more or less exact in the limit of a large number of particles, which we always satisfy. The reason we can do this is because the logarithm of a function will have a maximum at the same coordinate as the function itself and the logarithm is smooth such that will not somehow create another maximum or something which would spoil the trick.
After taking the derivative, we lose the constant and every term in the sum where $k\ne j$. So, we have,
$$
-\ln a_j^*-\alpha-1-\beta E_j=0
$$
Or, after rearranging and solving for $a_j^*$:
$$
a_j^*=e^{-(\alpha+1)}e^{-\beta E_j}
$$
Putting it All Together:
So, all that remains is to solve for the undetermined mutlipliers and plug it all back into our probability statement, $P_j=a_J^*/A$.
This can be done very simply (and we actually don't care about the mutliplier $\alpha$). If we take the sum over all $j$ of the equation we wrote for $a_j^*$, we realize that the left side is equal to $A$ due to our first constraint and we can rearrange to write $\alpha$ in terms of $\beta$:
$$
e^{\alpha+1}=\frac{1}{A}\sum_je^{-\beta E_j}
$$
Then, we can find the final probabilities from,
$$
P_j=\frac{a_j^*}{A}=\frac{e^{-(\alpha+1)}e^{-\beta E_j}}{A}=\frac{A}{\sum_je^{-\beta E_j}}\frac{e^{-\beta E_j}}{A}
$$
We finally arrive at the final probability distribution which defines Boltzmann statistics:
$$
P_j=\frac{e^{-\beta E_j}}{\sum_je^{-\beta E_j}}
$$
Thus, the only one of these Lagrange multipliers we care about is $\beta$. We have not actually found what this multipliers is yet, but I will stop here as it should be no surprise that the multiplier is given by $\beta=1/k_bT$.

So, I believe the above answers your question 1. That is, you have to worry about the constraints associated with you ensemble so what you mention does not quite work.
As to your second question, it is not quite true that translational energy is continuous. Pretty much everyone pretends it is, but kinetic energy is quantized. It is, however, very rare that the boundaries a particle is confined to are sufficiently small (and impermeable) that quantization of the kinetic energy is noticeable. Also, because there are many of these states which are closely spaced, one must reach very low temperatures for the quantization to be important.
What I have described above is the origin of Boltzmann statistics (not Maxwell–Boltzmann, which is often confused with it). This is a common confusion because of the famous Maxwell–Boltzmann distribution of velocities. I only bring this up because the Maxwell–Boltzmann distribution of velocities makes exactly the approximation which you describe by pretending the kinetic energies are continuous.
That is, if we simply plug the kinetic energy of a particle $E=1/2mv_x^2$ into our final result (and note the deonominator is the partition function), we get,
$$
P_{v_x}=\frac{e^{-\frac{mv_x^2}{2kT}}}{Q}
$$
Then, the partition function, $Q$, is just a normalization constant. Because the states of different kinetic energies are so closely spaced, we can elevate $P_{v_x}$ to a function, $P(v_x)$, and integrate over all velocities to determine the probability of finding a particle with a velocity between $v_x$ and $v_x+dv_x$. I will leave this as an exercise. Notice this is just an integral over a gaussian function.


Protons on aromatic rings in NMR Recently, I have started to learn about nuclear magnetic resonance (NMR) in school and something which I cannot seem to reconcile is the fact that all aromatic protons on any substituted benzene ring would give the same chemical shift. This is very, very strange... Clearly, the aromatic protons are in different chemical environments due to the fact that they are different distances away from the substituents from the benzene ring. I understand that the strong delocalisation in the aromatic ring may tend to "even out" the electron distribution across all the aromatic $\ce {C-H}$ bonds. However, I would not think that this effect would result in all the aromatic protons having the same chemical shift. Could anyone propose a explanation to this peculiarity?

The following image was taken from p. 279 of Clayden, Warren & Greeves (2012):

It also seems to suggest the equivalence of aromatic protons despite them being of different physical distance from the substituent. Here's another one from p. 278:

If strong delocalisation and aromaticity really is the effect responsible for this apparent chemical equivalence of the aromatic protons, then would we also observe such equivalent signals in other aromatic heterocycles or even in polyaromatic hydrocarbons?

Reference

Clayden, J., Greeves, N., & Warren, S. (2012). Organic Chemistry (2nd ed.). New York : Oxford University Press Inc.

This is true that all the aromatic protons come near the region having $\delta$: $\pu{7\!-\! 8 ppm}$. But you can always easily find examples where all the aromatic protons doesn't have the same chemical shift (which definitely include your examples). In fact the cases where all the aromatic protons have exactly similar chemical shift is hardly found.

The main idea is the difference between the terms Chemical Equivalence & Magnetic Equivalence, which is often misinterpreted. In the first example you have given, all the protons don't have the same chemical shifts and this is seen from the multiplet pattern as well. What you can obviously predict is that the two ortho protons are chemically equivalent, and the two meta protons are also chemically equivalent, and the para proton is a separate different type of proton. For observing these differences, what you need to do is to record the NMR at high frequency (e.g., $500$ or $\pu{600 MHz}$). There you can see the three different peaks clearly. Now in this case only, the two ortho protons are also magnetically equivalent as they couple with all the other protons equally, i.e. the $^3\!J_{H-H}$ and $^4\!J_{H-H}$  values are same for the two ortho protons, but in general this may not be the case at all (which is simply to consider another different para substitution at the ring, then the two ortho protons will become magnetically non-equivalent and therefore will have different chemical shift which is observable). Similar is the case with meta potons i.e. they are also magnetically equivalent. So those two protons will have same chemical shift. But ortho, meta, and para protons will definitely have different chemical shifts in your first example. If you zoom in or record the spectra in a high frequency machine, you should see a doublet of doublet peak (due to $^3\!J_{H-H}$ and $^4\!J_{H-H}$) for both ortho and meta protons, and a triplet of triplet for the para proton.

In the second example also, the two protons ortho to $\ce{-N(CH3)2}$ are chemically as well as magnetically equivalent as they both couple with the remaining proton in the same way. So, applying the similar logic you should see a doublet peak (due to $^4\!J_{H-H}$ only) for those two protons and a triplet peak for the remaining one and these two peaks should be at two different chemical shifts.

To encourage you more here are some spectra from the book Introduction to Spectroscopy which emphasize on aromatic regions and consider different spin-spin splitting and show how complicate a structure can be:

The above spectra depicts the differences in the spectra between ortho, meta, and para substitution in the benzene ring. The below two are some typical examples of complicated spectra due to spin-spin splitting as well as magnetic non-equivalence.

From these examples you can see that the aromatic protons can definitely come at distinct regions and what you need is to just record them in high field NMR machine and zoom into that region.


Is there a nomenclature for linear alkanes with n carbons? Is there a common nomenclature (not necessarily IUPAC) for naming linear alkanes with $n$ carbons without resorting to Greek/Latin numerical prefixes (hex-, hept-, etc.)?

For instance, is there a name for the higher linear alkane of $\mathrm{50}$ carbon atoms other than n-pentacontane? ("Not necessarily IUPAC" is meant in the same sense that $\alpha$, $\beta$, and $\gamma$ nomenclature  don't follow IUPAC rules but are nevertheless often used to designate relative positions of carbon atoms)

This does not answer you question, just mentions an alternative name that does have that count arabic numeral in the name, but is not eliminating the greek/latin prefixes. One could use IR-7.4 INORGANIC CHAINS AND RINGS nomenclature that would be impractical, e.g. the mentioned n-pentacontane would become dohectahydridopentacontacarby[50]catena or maybe even 1,1,1,2,2,3,3,4,4,5,5,6,6,7,7,8,8,9,9,10,10,11,11,12,12,13,13,14,14,15,15,16,16,17,17,18,18,19,19,20,20,21,21,22,22,23,23,24,24,25,25,26,26,27,27,28,28,29,29,30,30,31,31,32,32,33,33,34,34,35,35,36,36,37,37,38,38,39,39,40,40,41,41,42,42,43,43,44,44,45,45,46,46,47,47,48,48,49,49,50,50,50-dohectahydridopentacontacarby[50]catena (not to confuse with catenanes).


Obtaining 100% pure metal While reading my book everywhere I found that 99.9% pure metal is obtained. Is it possible by any process that we get 100% pure metal on industrial level?

The only way to get a "100% pure" sample of an element is by mass spectroscopy, or something similar, which acts on each atom (or ion, actually) one at a time.

So you could get a sample of pure rubidium of a few dozen atoms. Caveat: "100 % pure" applies only to those atoms actually observed... there could be other elements in the vicinity that have not been observed.

Practically, though, there is no way to prevent a single atom of hydrogen, for example, from eventually contaminating your "100 %" pure sample.


pH of very dilute acids From doing some research on the site I have found that many people have posted about solutions of $\ce{HCl}$ where $[\ce{HCl}] = \pu{1e-8 mol dm-3}$

Here they are able to deduce

$$[\ce{H+}] = \frac{[\ce{HCl}]_0}{2} + \sqrt{\frac{[\ce{HCl}]_0^2}{4} + K_\mathrm{w}}$$

where $[\ce{HCl}]_0$ is the initial concentration of  $\ce{HCl}$ dissolved.

Can we make a generalisation of this and say if $\ce{H_xA}$ is a strong acid that will dissociate fully when dissolved in water? Then if $[\ce{H_xA}]_0 < \sqrt{K_\mathrm{w}}$ then the $\mathrm{pH}$ of solution can be expressed by

$$\mathrm{pH} = -\log_{10} \left(\frac{x[\ce{H_xA}]_0}{2} + \sqrt{\frac{(x[\ce{H_xA}]_0)^2}{4} + K_\mathrm{w}}\right)$$

Your formulation us correct for an acid with one dissociable proton ($x=1$) below pH 7, and if you assume ideal solution behavior it even works beyond your upper bound.  Typically you should take $x=1$ because, mostly, even most strong acids dissociate one proton.  However...

One limitation, however, is when you have an acid with two or more dissociable protons.  If the second proton dissociates before you get to pH 7, then you need to add that into your calculation (and then it gets a lot more complicated).  Among common acids sulfuric acid is the one that will do this; in the limit of extremely dilute solutions (0.001 molar or less, roughly), the pH of a sulfuric acid solution will be about 0.3 unit lower than your formula with $x=1$ because of that second dissociable proton.


Is barium sulfate soluble in dilute mineral acid? I have read that $\ce{BaSO4}$ is soluble in $\ce{HCl}$ , but I am not sure whether this works for dilute acid as well.

The reaction is -
$\ce{BaSO4 + 2 HCl ->BaCl2 + H2SO4}$

The key point is to have an acid with such a high $\ce{H+}$ activity that in large extent shifts the below reaction to the left :
$$\ce{  HSO4-  + H2O <=> SO4^2- + H3O+ } \\ K_\mathrm{a2}=1.0 \cdot  10^{-2}$$

Then activity of sulphate ions from dissolution of $\ce{BaSO4}$ is decreased by the equilibrium reaction and the solubility increases.

$$\begin{align}
K_\mathrm{s}&=a_{\ce{Ba^2+}}\cdot a_{\ce{SO4^2-}}=1.1\times 10^{-10}\\
\frac{K_\mathrm{a2}}{a_{\ce{H+}}}&=\frac{a_{\ce{SO4^2-}}}{a_{\ce{HSO4^-}}}\\
\end{align}$$

The  effective solubility product
$$K_\mathrm{s,ef}=K_\mathrm{s} \cdot    \frac{a_{\ce{H+}}}{K_\mathrm{a2}}$$

Note that despite activity coefficients {gamma} decrease with increasing ionic strength, for concentrated acids this tendency reverses and raises very rapidly

[HCl]  Gamma
0.0005 0.975
0.01 0.904
0.1 0.796
1 0.809
2 1.01
5 2.38
10 10.44
12 17.25

Therefore, $a_{\ce{H+}}$ of 10 M HCl = 104.4.

$$K_\mathrm{s,ef}=K_\mathrm{s} \cdot    \frac{104.4}{K_\mathrm{a2}}\\=1.1\cdot 10^{-10}\cdot 1.044\cdot 10^4=1.148\cdot 10^{-6} \mathrm{~ mol^2 L^{-2}}$$

Solubility would be then $ M \cdot \sqrt{K_\mathrm{s,ef}}=233.4 \cdot 0.001072=0.25\mathrm{~g/L}$

Conclusion: The $\ce{BaSO4}$ solubility in concentrated $\ce{HCl}$ is limited, in diluted one is very low. The solubility in concentrated sulphuric acid would be much higher, as $a_{\ce{H+}}$ would be very high.


How do I prepare a 5N HCl solution? I am trying to make  5N $\ce{HCl}$ solution. So far I understood that you need to get the molecular weight (36.5) and divide it by the number of $\ce{H+}$ that react in the acid-base reactions. The problem is that my $\ce{HCl}$ stock is 37 %. How should I take this into account?

Hydrogen chloride in its standard state is a gas. We usually do not make solutions of $\ce{HCl}$ by the method of dissolving a known mass of the the solute in enough solvent to make the appropriate volume of the specific concentration since gaseous substances are difficult to weigh (and weight is one of the most direct ways we have to determine mass in the lab).

Hydrochloric acid (aqueous solutions of $\ce{HCl}$) is commercially available in several concentrations, usually $1\ \text{N}$, $0.1\ \text{N}$. and "concentrated".

"Concentrated" simply means that you have in that bottle the highest concentration of $\ce{HCl}$ water that is stable. For $\ce{HCl}$, a concentrated solution is 36-38% by mass. If you know the density of the solution (should be on the bottle), you can convert % by mass into moles of $\ce{HCl}$ per liter of solution (you also need the molar mass of $\ce{HCl}$).

For example, if you had a solution that was 20% $\ce{HCl}$ by mass with a density of $1.098 \ \text{g/mL}$, you can determine molarity by converting mass of $\ce{HCl}$ into moles and mass of solution into liters:

$$20\% \ \ce{HCl} \ \text{by mass}=\frac{20 \text{ g }\ce{HCl}}{100\ \text{g solution}}$$
$$20 \text{ g }\ce{HCl} \times \frac{1 \text{ mol }\ce{HCl}}{36.46 \ \text{g }\ce{HCl}}=0.5485\ \text{mol }\ce{HCl}$$
$$100\ \text{g solution}\times \frac{1\ \text{mL solution}}{1.098\ \text{g solution}}\times \frac{1\text{ L}}{1000\ \text{mL}}=0.09107\ \text{L solution}$$
$$\frac{0.5485\ \text{mol }\ce{HCl}}{0.09107\ \text{L solution}}=6.023 \ \text{M}$$
However, other folks have the done the math for you. The Wikipedia article of hydrochloric acid lists the concentrations (and densities) for several $\ce{HCl}$ solutions as a function of % by mass. This listing of concentrated reagents lists $37.2\% \ \ce{HCl}$.

With a concentration for $37.2\% \ \ce{HCl}$, you can proceed to work through the dilution process using $c_1 v_1 = c_2 v_2$.


What is the atomic radius of chlorine? I'm trying to study for chemistry and there's a question in my book about identifying an element based on its atomic radius. But when I looked up the value for chlorine, I found different answers from different sources:

79 pm
175 pm (from the Princeton Review)
0.97 Å
175 pm (van der Waals radius?)
79 pm

What is the correct value?

There is no such thing as unified atomic radius. An atomic radius is a class consisting of van der Waals radii $R_\mathrm{vdW}$ (steric interactions), covalent radii $R_\mathrm{cov}$, and ionic radii $R_\mathrm{i}$ (and some other as well).

From the recent edition of CRC Handbook [1, p. 9-57]:

  $$
\begin{array}{llrr}
\hline
\text{Element} & \text{Symbol} & R_\mathrm{vdW}~(Å) & R_\mathrm{cov}~(Å)\\
\hline
\text{Chlorine} & \ce{Cl} & 1.75 & 1.00\\
\hline
\end{array}
$$

Ionic radii are given for the crystallographic data separately [1, p. 12-12]:

  $$
\begin{array}{llr}
\hline
\text{Ion} & \text{C.N.} & R_\mathrm{i}~(Å)\\
\hline
\ce{Cl^{+5}} & 3~\text{(pyramidal)} & 0.12\\
\ce{Cl^{+7}} & 4 & 0.08\\
\hline
\end{array}
$$

Reference

Haynes, W. M.; Lide, D. R.; Bruno, T. J. CRC handbook of chemistry and physics: a ready-reference book of chemical and physical data.; 2017; Vol. 97. ISBN 978-1-4987-5429-3.