Does a grain of salt have exactly equal amount of sodium and chlorine? Does a grain of salt have exactly equal amount of sodium and chlorine? I know that the formula unit of salt is 1:1, but for example if there is 10000000 atoms of sodium is it possible that that there is 9999999 atoms chlorine? Thanks.

Crystal solids, however small, have defects in them. These defects are basically irregularities in the arrangement of constituent particles. They are of two types:

Point defects
Line defects.

Point defects themselves are divided into other kinds of defects. Here, I'll explain only the relevant one.

One type of point defect is a vacancy defect in which some of the lattice sites in the crystal are vacant. This means that some atoms of Na or Cl or both are missing from their lattice sites.

So the answer to your question is, yes. That scenario is very much possible.


Is the surface of the planet Mercury made of copper minerals? I have been looking at photographs of the planet Mercury, and I notices that in most of the pictures I have seen, the rocks on the surface seem to be covered in blue in green. From this, is it correct to say that the surface of Mercury is mostly made up of copper deposits?

I think you probably fell into a trap concerning color-coding.

In reality, mercury is dark grey and mostly covered in silicate rocks/dust.

From a scientific point of view, the "real color" of an object like mercury is of little interest. What's more interesting, is it topography etc.
To make it easier to interpret data, man often uses color-coding (since our color-vision is quite ok) to communicate different information about such things.

Or ist just about beauty:
When you see pictures of nebulae like the pillars of creation you notice they are in a lot of different colors. This is because a lot of the radiation is not in the visible spectrum, and you can "color-code" it to make it interesting to look at.


Biphenyl esterification I was wondering that at optimum temperature can 6,6'-bis(hydroxymethyl)[1,1'-biphenyl]-2,2'-dicarboxylic acid show esterification reaction.

I know that the two phenyl groups aren't in the same plane so is esterification reaction still possible?

these type of things are possible,

esterification is possible, check the reference.


Are there any sodium salts that are insoluble in water? Are there any sodium salts (e.g. NaCl) that are insoluble in water? If so, what are their approximate $K_\mathrm{sp}$ values?

From here:

  Sodium tends to form water-soluble compounds, such as halides,
  sulfate, nitrate, carboxylates and carbonates. But examples of many
  insoluble sodium compounds such as sodium bismuthate ($\ce{NaBiO3}$),
  sodium octamolybdate ($\ce{Na2Mo8O25·4H2O}$), sodium thioplatinate
  ($\ce{Na4Pt3S6}$), and sodium uranate ($\ce{Na2UO4}$) is found in
  nature.

Other examples include:

Feldspars(aluminosilicates of sodium, potassium, calcium)
Sodium metaantimonate, $\ce{NaSbO3.7H2O}$ in the pyro form $\ce{Na2H2Sb2O7.H2O}$ has a solubility 0.7g/L.
Sodium metaphosphate, $\ce{NaPO3}$ has both soluble and insoluble form.


Understand Hexabromocyclododecane CAS numbers Hexabromocyclododecane has 16 possible stereo-isomers. So far so good.
I found that:

134237-50-6 is the CAS of alpha-Hexabromocyclododecan

134237-51-7 is the CAS of beta-Hexabromocyclododecan

134237-52-8 is the CAS of gamma-Hexabromocyclododecan

3194-55-6 is the CAS of a mixture of different Hexabromocyclododecan

Now, I have two more CAS which are a bit opaque to me:

678970-15-5

Pubchem says it's the same as 134237-50-6
However, ChemIDplus says it's alpha-Hexabromocyclododecane, (-)-

138257-18-8

Pubchem says it's the same as 134237-51-7
However, ChemIDplus says it's beta-Hexabromocyclododecane, (-)-

Question: Describe each of the CAS different chemicals or are 678970-15-5 and 138257-18-8 just synonyms. If they are different chemicals, what does (-)- mean here?

The (−)- in the chemical name refers to the optical activity of the compound, that is to tilt the plane of linerarly polarized light to the left (levo-rotatory), or to the right (dextro-rotatory) as determined with a polarimeter.  It may be either (−)- (minus sign) or (+)- (plus sign), as seen, for example, for two of the three forms of tartaric acid.  Compounds with stereogenic centers which do not tilt the plane of the polarization (e.g., meso-compounds) may be labeled by (±)-, too.
Similar to this, and equally used for molecules with stereogenic centers − especially carbohydrates  − is the the use of (D)-, (L)- or (DL)- which however relates to the spatial arrangement around the sterogenic atom of highest priority (there are rules how to determine this just from reading structure formula).  It may be used independently as well as simultaneously to the former system of (−)- and (+)-; but note the knowledge what to write within former system does not translate 1:1 into what to use in the later system.
As for the CAS number, this is a book-keeping number (see here); contrasting to a chemical name, it does not convey structural or even chemical meaning.  CAS for example systematically attributes more than one registry number to the same molecule if there is evidence that it crystallizes in different polymorphs.  This «already» happens with inorganic molecules like $\ce{TiO2}$ which may crystallize in the form of the mineral anatase (CAS 1317-70-0), or rutile (CAS 1317-80-2).  Thus, if your literature search is about the isolated molecule alone (say, molecular mass), it does not matter which entry you follow further.  Yet if you are interested in say the melting point of your compound or crystal structure, you have to treat the CAS numbers separate, like referring to different compounds.  Thus, the relationship between CAS number and a structure is not bijective.
(Occasionally, e.g. for some of the more substituted fullerenes, you may encounter many CAS numbers for one and the same compound in a database like Scifinder or Reaxys.  Possibly the difficulty to draw their formulae contributed to create whole sets of CAS numbers.)


Predicting pressure inside a container based on temperature I'm a mathematician and computer scientist, and for this particular problem I would benefit from some chemical expertise.

Suppose I fill a container up with some liquid propane. I believe this is done under very cold conditions so that the propane remains liquid while the container is being filled. It is then sealed off. Suppose now the inside of the container changes temperature. I would like to calculate the pressure that would be exerted on the walls of the container at any given internal temperature $T$.

Intuitively, I would think that the pressure would be a function of temperature and some initial conditions (how full is the tank initially)? If I filled up the tank to 95% capacity with liquid propane, and then heated the tank, I would expect the resulting pressure on the walls of the container to be much higher than if I filled up the tank to only 1% capacity and did the same thing.

However, the only resources I've been able to find so far relate to the vapour pressure, which is a function of only temperature and this seems incomplete.

Can anybody point me in the right direction on how to approach this problem? Some factors that seem to complicate things are:

The containers always seem to contain some level of liquid (due to internal pressure?), the rest is gas. Does the liquid have an effect on the pressure exerted on the walls? Can I just ignore it?
If, rather than fill up the tank with liquid at very cold temperatures, the tank is just directly filled with gas, how does this change things? My guess is that given the initial pressure and temperature you can predict the pressure at a different temperature, but I think the fact that some of the gas becomes liquid at some point is tripping me up and I'm not sure how to approach the problem.

The problem with answering this question is that propane is not remotely close to being a permanent gas at typical room temperature. This means, in practice, that it is easily liquefied at room temperature with a little pressure and you can't, therefore, use the simple gas law equation to work out the pressure exerted by a given amount of the gas in a vessel of fixed volume.

This means that the tank will fairly certainly contain some liquid (as a butane-filled cigarette lighter will at very modest pressures despite butane being a gas at normal temperatures and pressures).

You can, therefore, estimate the pressure inside the vessel as being the vapour pressure of propane at the given temperature (which ignores any other gas included in the vessel though this may be a good approximation as the filling process is likely to sweep out any other gas). The Wikipedia entry on propane has a convenient chart of the vapour pressure and you can look up the value for a given temperature. The amount of the vessel filled won't make much of a difference to this as long as there is some liquid and some gas in the vessel.

Because simple gas laws don't do a good job of predicting the vapour/gas equilibrium, this isn't a gas-law question and the best answer will always be empirical (see the Wikipedia page).


Hartree-Fock dipole moment I read this sentence in a paper (J. Phys. Chem. A 2014, 118 (20), 3678–3687. DOI: 10.1021/jp502475e):

  Calculations using Hartree–Fock theory systematically overestimated
  dipole moments and underestimate polarizabilities.

Is there any intuitive explanation that the HF approximation affects dipole moments and polarizabilities differently?

I originally wrote this as a comment, but I will post it as an answer as I'm fairly certain this is most of the answer.

In general, there are two distinct types of electron correlation which can be described. Exchange correlation, which results from the indistinguishability of electrons, and Coulomb correlation which correlates the motions of electrons by considering individual repulsions between electrons.

Hartree-Fock (HF) theory accounts for the exchange interaction, but does not have Coulomb correlation as HF is a mean-field theory. That is, each electron experiences the average potential of the other electrons.

So, there is a sort of imbalance here from the beginning. It is known that the presence of this exchange interaction, which only happens for electrons of the same spin, results in two phenomena called Fermi holes and Fermi heaps. The references in that Wikipedia page are quite good. Essentially, a Fermi heap results when you have electrons of opposite spin. Because they must have a spatially symmetric wavefunction, the ends up being a greater probability of finding both electrons at the same location. This sounds silly, but read the references in the wikipedia page and you'll see what I mean. A fermi hole results when you have two electrons of the same spin. These must have an antisymmetric spatial wavefunction, so there is no probability of finding them in the same location.

If we think about these two effects in terms of electrical repulsion, the Fermi hole is a lower energy configuration than the Fermi heap. Because we almost always have both spin-up and spin-down electrons in multi-electron systems, both of these phenomena will take place, and the overall electron density will be some kind of average of all of these. (This is only really a conceptual framework.) If we take a simple case of a homonuclear diatomic, then these heaps and holes will all average out and we will get a symmetric electron density. If, on the other hand, we take a heteronuclear diatomic, then we see that there will be a tendency for these Fermi heaps to form over the atom with the larger nuclear charge just based on Coulombic attraction.

From here, I believe it is fairly clear that Coulomb correlation will tend to make the effect of these Fermi heaps smaller, and hence the overall separation of charge will decrease rather than increase, which explains why HF tends to overestimate dipole moments. By the same reasoning, the electron density is already polarized due to this heaping effect, so the system is not likely to be as polarizable as if this electron density were allowed to relax as is the case when Coulomb correlation is included.

This is all a very conceptual way of thinking about this, but I don't know that the math describing HF will be all that much more useful.


Is Hydrolysis same as dissolution? I have book that contains the following paragraph:

  Due to high dielectric constant, it (water) has a very strong hydrating tendency. It dissolves many ionic compounds. However, certain covalent and some ionic compounds are hydrolysed in water.

Is this statement self-contradictory? Doesn't 'hydolysed in water' mean dissolved in water?

Hydrolysis is a broad chemical term that means reacting with the solvent water in some way. So for examples esters can be hydrolysed according to the following reaction:

$$\ce{H3CCOO-C2H5 + H2O -> H3CCOOH + HO-C2H5}$$
(Ethyl acetate and water react to form acetic acid and ethanol.)

Dissolution is the process of something being dissolved in water, intermixing with the water molecules so that each atom/ion/molecule of the substance is more or less completely surrounded by water molecules. This process is usually understood to happen without any reaction, thus the term hydrolysis is not used.

However, the above is often a simplification. For example, some water-free salts such as $\ce{AlCl3}$ or $\ce{CuSO4}$ will react with the water to form the corresponding hexaaquacomplexes $\ce{[Al(H2O)6]^3+ + 3 Cl-}$ or $\ce{[Cu(H2O)6]^2+ + SO4^2-}$. These are technically reactions, best exemplified in the aluminium example where the metal was originally coordinated by chlorido ligands and after dissolution is coordinated by aqua ligands.

The only thing that changes is the immediate vicinity of the metal ion in these cases, so speaking of dissolution rather than hydrolysis (essentially the metal cation remains the same as do the corresponding anions) is often tolerated.


Database of 3D crystal models I'm looking for a database of very basic crystal models for all the various types - triclinic, monoclinic, orthorhombic, tetragonal, rhombohedral, hexagonal, and cubic.
A basic example would be a CAD file for a cube, to represent the cubic structure. I don't want like a packing model which shows the individual atoms because I'm trying to represent macroscopic crystals.
Any ideas where I could find what I'm looking for? Not a 3D graphic designer.

Joining @Karl's comment on «what do you want to do with them?» the work to present point groups is work already solved.  For off-line teaching, the freely accessible collection of the 32 point groups by Casas and Estop by (UB Barcelona) may suit your needs:

The model data are provided as interactive 3D .pdf files which a reader like Adobe Acrobat 9.0 may let you toggle on/off the axes / planes, etc.
Beside some background provided here, the authors outline the workflow and tools used (SolidWorks, Adobe Illustrator CS3, and Adobe Acrobat 9 Pro) in an accompanying publication (doi 10.1016/j.cageo.2013.09.004, paywall).  Later, the concept was extended to cover the space groups as to complement the International Tables of Crystallography, too (title page, paywall).


Why do we use coke instead of coal in order to reduce FeO as basic reaction in ironmaking process? Why is coke used as a reducing agent to reduce FeO to produce iron instead of coal.I admit that coke is carbonaceous but what is it that compels us to use coke instead of the naturally available coal?

The single most important factor is strength ( mechanical compressive ); coal is heated to make coke, the resulting coke is stronger than the original coal. Also, coke helps to make the charge of iron oxides and limestone more porous to permit gas flow up and droplets of liquid iron and slag down. The coke oven heating drives off volatiles from the coal, which are a valuable source of chemicals, and contain much of the undesirable $\ce{S}$ and $\ce{P}$, which would otherwise go into the pig iron. Minerals like silica (ash) remain in the coke but they are easily collected by the slag. A bit off topic, but powdered coke and gasses are blown into the bottom of the blast furnace to provide more $\ce{C}$, which makes modern blast furnaces much more productive than 60+ years ago.


How to predict one equilibrium constant given two others? $$\begin{alignat}{2}
\ce {CO2 (g) + 3H2 (g) \;& <=> CH3OH (g) + H2O (g)}\qquad&&{k_1=\;?} \\
\ce {CO (g) + H2O (g) \;& <=> CO2(g) + H2(g)}\qquad&&{k_2= 1.0\times10^5} \\
\ce {CO (g) + 2H2 (g) \;& <=> CH3OH(g)}\qquad&&{k_3= 1.4\times10^7}
\end{alignat}$$

What is the value of $k_1$? I have tried reversing equation two and then multiplying $k_2$ and $k_3$ together, but I am not sure how to work this problem out.

Remember that when you reverse the reaction the equilibrium constant changes. For the general gas-phase reaction $\ce{A(g) + B(g) <=> C(g) + D(g)}$ the equilibrium constant expression is $$K_\text{f} = \frac{p_\ce{C} p_\ce{D} }{p_\ce{A} p_\ce{B}}$$ (Strictly, it's activities, not partial pressures, but the principle is the same.)
The reversed reaction $\ce{C(g) + D(g) <=> A(g) + B(g)}$ has the equilibrium constant expression $$K_\text{r} = \frac{p_\ce{A} p_\ce{B} }{p_\ce{C} p_\ce{D}} = \frac{1}{K_\text{f}}$$

You are correct in multiplying the equilibrium constants to get the equilibrium constant for the combined reaction.


The synthesis of 1-phenylprop-1-ene from propiophenone I would like to know how to make 1-phenylprop-1-ene ($\ce{C6H5-CH=CH-CH3}$) from propiophenone ($\ce{C6H5-CO-CH2-CH3}$)?

I think carbonyl group ($\ce{C=O}$) of propiophenone is electron-withdrawing group because of Oxygen of carbonyl group.
So the carbonyl group of propiophenone will be attacked by nucleophile. And this will be oxidized to make double bonds.

But I don’t know the detail reaction mechanism to make phenylprop-1-ene from propiophenone.
Can anyone help me understand this reaction mechanism?

This can be done it 2 simple steps. You are not adding to the carbon count so all you need to do is operate on propiophenone.

Step 1 - reduce the carbonyl group to a secondary alcohol. Sodium borohydride in $\ce{MeOH}$ or $\ce{EtOH}$ will do this in high yield and the workup is straightforward.

Step 2 - dehydrate the secondary alcohol to give the double bond. As it is a benzylic alcohol, protonation of the $\ce{-OH}$ group and formation of the benzylic cation is strongly favoured and will give you the alkene you want. Catalytic para-toluenesulfonic acid in refluxing toluene (with water removal by Dean-Stark apparatus) will do this easily.


How are elements used in the human body (for example) encoded for in the genome? Given that bones require calcium and iron is used in red blood cells, how are these materials gathered by the developing organism? Is it possible that very small quantities used as "references" are somehow stored in the human egg? Or is there some way of encoding for an element in DNA without any amount of the element already available?

No. DNA is not a blueprint, it is an integral part of the machine. There aren't little people in your cells who can read an instruction set. DNA, together with a living cell, produces (actually is used to produce RNA, which goes into producing) specific proteins, some of which bind to e.g. iron ions and integrate them into the enzyme mechanisms. It's all about making the right protein (e.g. hemoglobin to bind iron) in the right place and time. A living cell is a reactor, and the genes are your catalyst.

What kind of proteins your DNA produces depends on the surroundings. If in a skin cell, it produces proteins that directly or indirectly are necessary to produce skin. If (in an embryo) close to the surface, but not yet a skin cell, it produces proteins that later help turn it into a skin cell. And so on.

If an alien got hold of just our DNA and RNA, he could never ever even find out how a human works or what he looks like.


Effective attenuation coefficient for polychromatic light The Beer-Lambert law gives the optical intensity of collimated light as a function of depth $z$ as:

$$I(z)=I_{0}\mathrm e^{-\gamma z},$$

where $\gamma = \alpha + \beta$ is the wavelength-dependent attenuation coefficient, with $\alpha$ and $\beta$ being the absorption and scattering coefficients respectively.

Suppose two completely different wavelengths $\lambda_1$ and $\lambda_2$ are present in the light beam. Is it possible to define the "effective attenuation coefficient" of the material $\overline{\gamma}$ for two or more wavelengths as an addition of the different attenuation coefficients?

That is,

$$\overline{\gamma}=p_{\lambda_1}\gamma_{\lambda_1}+p_{\lambda_2}\gamma_{\lambda_2},$$

where $p$ is the fraction of the beam which is of a given wavelength (indicated by the subscript). The mathematics seems to work out, but I have never seen this used in literature before. Any explanation is greatly appreciated.

The intensity transmitted at  wavelength $a$ where the extinction coefficient is $\epsilon_a$ is $I_a=I_{0a}\exp(-\epsilon_a L C)$ with path length $L$ and concentration $C$, and similarly for a second wavelength $b$. If, for clarity, we let $x=\epsilon l C $ then then $I_a=I_{0a}\exp(-x_a)$ and $I_b=I_{0a}\exp(-x_b)$. The total transmitted light is $I_t=I_a+I_b$ and the total initial amount $I_{0t}=I_{0a}+I_{0b}$. Thus the total transmittance ratio is

$$ \frac{I_t}{I_{0t}}=\frac{I_{0a}e^{-x_a}+I_{0b}e^{-x_b}}{I_{0a}+I_{0b}}$$

You want to make this equal to $I_{0ab}e^{-x_{ab}}$  but there seems no systematic way to do this and this confirms the fact that the simple  Beer-Lambert law only applies for monochromatic light.


What is the mechanism of oxidation of phenol to benzoquinone? I looked up the mechanism in a lot of organic chemistry books, including Clayden, Klein, Solomons, McMurry but I couldn't find the mechanism of oxidation of a phenol. I also looked it up in pharmacognosy books, but with no luck. All I could find in these books is the following scheme, with no mechanism:

It seems like other oxidizing agents can be used, such as hydrogen peroxide and silver oxide.

Unlike primary and secondary alcohols, phenol doesn't have an hydrogen atom attached to the carbon directly bound to oxygen. My guess is that the double bond of phenol will react with the oxidizing agent, but not sure of how this will work.

The reaction is used industrially to make hydroquinone (source: wikipedia), but often using hydrogen peroxide as the oxidation agent in the presence of a catalyst.
A study on catalytic hydroxylation of benzene comes to the following conclusion:

experimental evidences have allowed first discarding Fenton-like mechanisms and later proposing the plausible competition between an electrophilic aromatic substitution pathway with the alternative rebound (hydrogen abstraction) route.

The Fenton-like mechanism refers to addition of a free oxygen radical species to benzene, which they reject based on experiments with radical scavengers. Instead, they hypothesize formation of a radical oxygen:catalyst complex. This either reacts with benzene in an electrophilic substitution, or abstracts a hydrogen from benzene, which then rebounds and reacts with the oxygen still attached to the catalyst.
Although the study I am citing is for hydroxylation of benzene instead of phenol, the paper also talks about substituted benzenes.

[OP:] My guess is that the double bond of phenol will react with the oxidizing agent, but not sure of how this will work.

You can't pin down a single double-bond in an aromatic system. What you can say is that the existing hydroxyl group will steer the reaction to make an ortho- or para- product. Once hydroquinone is formed, it oxidizes quickly to benzoquinone under the oxidizing conditions.

[OP:] What is the mechanism of oxidation of phenol to benzoquinone?

For the reactions conditions given by the OP, I did not manage to find a mechanism. However, the mechanisms I discussed probably are sufficiently general that they apply in this special case as well. That is just a guess, though.


Cationic vacancies created when non-stoichiometric iron oxide is doped with silicon(IV) ions Problem

It is believed that non-stoichiometric compound $\ce{F_{0.93}O}$ forms by doping of $\ce{Fe^3+}$ ions in $\ce{FeO}$ crystal by replacement of $\ce{Fe^2+}$. Calculate the number of cationic vacancies if all of the $\ce{Fe^2+}$ ions are replaced by $\ce{Si^4+}$ ions in $\pu{0.1 mol}$ of $\ce{Fe_{0.93}O}$. Express your answer as a multiple of Avogadro's Number.

Answer stated by the book

 0.0465

My Solution
As the compound is electrically neutral I equated the net charges due to the cations and anions like this, assuming $x$ to be the fraction of $\ce{Fe^2+}$ in the compound.
$$[2x + 3(0.93-x)] - 2 = 0$$
This gave $x = 0.79$. Now, in $\pu{0.1 mol}$ of the compound there would be $\pu{0.079 mol}$ of $\ce{Fe^2+}$. As each silicon ion would replace two ions of $\ce{Fe^2+}$, one vacancy would be created for every replacement.
The number of replacements would be $\left(\frac{0.079}{2}\right)\times N_\mathrm{A} = 0.0395 N_\mathrm{A}$, and we would have as many vacancies. ($N_\mathrm{A}$ is Avogadro's Number)
Can anyone point out the error in my solution?

Your approach is correct, however, you forgot to account for the vacancies created when Fe3+ replaces Fe2+. First, calculate the cationic vacancies formed due to the replacement of Fe2+ by Fe3+.
Assuming  to be the fraction of Fe2+ in the compound, the fraction of Fe3+ in the compound is (0.93 - x).The total charge on x Fe2+ and (0.93 – x) Fe3+ should be equal to the charge due to O2- ions.
Hence, 2x + 3(0.93 – x) = 2
=> 2x + 2.79 – 3x = 2.00 or x = 0.79. So, fraction of Fe3+ ions in the compound is 0.14
Now, 2 Fe3+ ions replace 3 Fe2+ ions, so there's one vacancy created for every 2 Fe3+ ions. That means for 0.1 mole of F0.93O, the number of vacancies will be 0.007A due to Fe3+ replacing Fe2+ in FeO. This is the part you missed.
Now, as you mentioned, each Si4+ ion will replace two Fe2+ ions in the compound, so again, one vacancy per replacement. And, as you already solved, The number of replacements would be (0.079/2)A = 0.0395A. So, the total number of vacancies are 0.0395NA + 0.007NA = 0.0465NA.
Hope this helps!


Is the lattice point in ZnO crystal structure occupied by the molecule ZnO or the ion O^2-/Zn^2+? I learnt that ZnO has a hexagonal crystal system. I was expecting that cations or anions would occupy the lattice points in hcp while voids would be occupied by anions or cations respectively. But while googling the ZnO structure, I found this structure below which was confusing because of my notion about lattice point occupied only by cation/anion while voids occupied by its countercharged part.

But after close inspection of this above structure, I thought maybe the lattice point is occupied by the molecule ZnO itself while voids are left unoccupied as opposed to the cation/anion occuping the lattice point and anion/cation occuping the voids. Because only then,this structure is possible. Am I right in thinking so?

You are falling in the rather common mistake of confusing the crystal structure with the lattice. The lattice is a mathematical object (it does not exist as a real object), whereas the crystal structure is the convolution of the lattice with a basis (an atom or a collection of atoms).
In some cases atoms are distributed exactly at the same positions of the lattice points; for example Cu has a face centered cubic structure where copper atoms occupy the same positions of the face centered cubic lattice. In this case Cu at 0,0,0 position is your basis.
In your case neither the Zn atoms, nor O atoms occupy lattice positions. Your basis is contituted of Zn at 1/3,2/3,0 plus O at 1/3,2/3,z; 'attaching' this basis to each lattice point (taking into account the correct distances of both Zn and O atoms from each lattice point), you obtain the final structure.
In any case, the picture that you posted does not represent the real hexagonal structure; rather, I would that is the trigonal polymorph.


Openbabel: Generate random 3D geometry Is it possible with openbabel to generate random 3D geometries (i.e. conformers) of a molecule without taking energy into account? Ideally with the property of being truly random, so that after generating conformers like this for a sufficient amount of time, a good coverage of the conformational space is obtained. Because the --conformer options of openbabel takes energy into account, but I want to input the geometries into a qc program that then calculates the energies for the conformers.

Random conformer search generates random conformers

Open Babel, like many conformer sampling tools, has a set of common dihedral angles to sample. This minimizes the search space, e.g., only 3-4 likely cases for each rotatable bond.

The RandomConformerSearch method indeed generates conformers using a random sampling of dihedral angles. It then (optionally) runs a force field for some number of steps to minimize any steric clashes. No energies are considered.

If you're finding only one conformer output, that's because you're not using the --writeconformers flag:
http://open-babel.readthedocs.io/en/latest/3DStructureGen/multipleconformers.html


Why does urine smell like ammonia? Apparently, water or $\ce{H2O}$ combines in some way with nitrogen (from the air or elsewhere?) to form ammonia, or  $\ce{NH3}$. Where/how in the body is this $\ce{NH3}$ (or whatever nitrate) produced before it is excreted with urine?

All animals (including humans) consume significant amount of proteins. Proteins consist of aminoacids, which include amine nitrogen. This nitrogen traces back to mineral fertilizers humans use in agriculture. As proteins in our bodies are destroyed, the waste is utilized, formine urea (carbamide), so ammonia didn't cause too much trouble. However, urea is slowly hydrolized in water back into ammonia, so old urine stinks of (among other things) ammonia.


Do the units of the law of mass actions convey any meaning? Since the number that you end up with, when calculating the value of the law of mass action for a given reaction often also has a unit, I was wondering if there is any physical meaning to this given unit or if it is basically "useless"?

Example:

$$\ce{CH3COOH <=> CH3COO- + H+}$$

$$K_c = \frac{c(\ce{CH3COO-})\cdot c(\ce{H+})}{c(\ce{CH3COOH})}\qquad [K_c] = \pu{mol L-1}$$

Well, you can calculate difference in number of moles of gases and active liquid species on reactant and product side species using units of $K_c$ and $K_p$, respectively, as these are mostly used for homogeneous reactions.

For example, the unit of $K_c$ is $\text{(concentration)}^n$ if $n > 0$, more liquid or aqueous (ionic) species on product side.

Then, you can predict whether the product is formed or reactant when you increase or decrease external pressure or volume of a container containing a mixture in equilibrium. Temperature's effect would be only heat of reaction dependent.

Hint: write $K_c$ in terms of moles and volume to check effect of volume, write $K_p$ in terms of mole fractions and total pressure to check effect of external pressure.


How does the pyramidalization effect the bonding angles in a distorted octahedral environment? This is a follow up on "Generally Octahedral - Meaning?".

Since Dissenter asked in the comments to my answer:

  I'm not sure why the $\ce{F_{eq}-S-F'_{eq}}$ bond angles would be compressed, even if the $\ce{S}$ were out of the plane.

Since ron's answer already satisfies the original question about $\ce{BrSF5}$ and I did not want to clutter my answer even further, I am asking it here in a Q&A style.

How does the pyramidalization effect the bonding angles in a distorted octahedral environment?

So we are again talking about $\ce{BrSF5}$ but this post extends to all generally octahedral complexes/ molecules and even quadratic planar geometries. The main requirement is, that the molecule has a $C_{\mathrm{4v}}$ point group. (It may be extended to all non-linear $C_{n\mathrm{v}}\ (n>1)$ point groups.) First of all reconsider the molecular shape again.

Now admittedly the degree of pyramidalization is more than minimal, with less than $1^\circ$ deviation from the plane. However, this has a slight effect on the equatorial $\angle(\ce{F-S-F'})$ bond angle, purely to symmetric reasons.

Basic trigonometric considerations are enough to understand this phenomena. Consider the following scheme, where the peak of the pyramid is represented by $\mathbf{\ce{Y}}=\ce{S}$ and the base is span by $\mathbf{\ce{X_{eq}}}=\ce{F_{eq}}$, $\mathbf{\ce{X'_{eq}}}=\ce{F'_{eq}}$, $\mathbf{\ce{X''_{eq}}}=\ce{F''_{eq}}$ and for completeness $\mathbf{\ce{X'''_{eq}}}=\ce{F'''_{eq}}$.

So we have some basic quantities, that we can address, but in the end we can derive the connection of the two angles without caring about the other quantities, only knowing the point group.

$a$ is the $\mathbf{d}(\ce{F_{eq}\cdots{}F'_{eq}})$ distance
$d$ is the $\mathbf{d}(\ce{F_{eq}\cdots{}F''_{eq}})$ distance
$s$ is the $\mathbf{d}(\ce{S\cdots{}F_{eq}})$ distance
$h$ is the height of the pyramid and it refers to the out of plane shift of $\ce{S}$ compared to the $\ce{F_{eq}\cdots{}F'_{eq}\cdots{}F''_{eq}\cdots{}F'''_{eq}}$ plane
$h'$ is another virtual quantity, the angle bisector of $\angle(\ce{F_{eq}-S-F'_{eq}})$
$\alpha$ refers to the bonding angle $\angle(\ce{F_{eq}-S-F''_{eq}})$
$\beta$ refers to the bonding angle $\angle(\ce{F_{eq}-S-F'_{eq}})$

From Pythagoras's Theorem we can easily derive the following connection (square in the scheme):
\begin{align}
             && d^2 &= a^2 + a^2\\
  \therefore && d   &= \sqrt{2}\cdot a\tag{1}
\end{align}

So now we are going to use some trigonometric functions for the angles (look at the triangles):
\begin{align}
             && \sin\left(\frac{\alpha}{2}\right)
                    &= \frac{\frac12d}{s}\\
  \therefore && s   &= \frac{\frac12d}{\sin\left(\frac{\alpha}{2}\right)}\tag{2}
\end{align}

\begin{align}
             && \sin\left(\frac{\beta}{2}\right)
                    &= \frac{\frac12a}{s}\\
  \therefore && s   &= \frac{\frac12a}{\sin\left(\frac{\beta}{2}\right)}\tag{3}
\end{align}

Let's derive the connection between the angles by equalizing $(2)$ and $(3)$:
\begin{align}
             && \frac{\frac12d}{\sin\left(\frac{\alpha}{2}\right)}
                     &= \frac{\frac12a}{\sin\left(\frac{\beta}{2}\right)}\\
  \therefore && \sin\left(\frac{\beta}{2}\right)
                     &= \frac{a}{d}\sin\left(\frac{\alpha}{2}\right)\\
  \text{with (1)}\implies &&
                \sin\left(\frac{\beta}{2}\right)
                     &= \frac{1}{\sqrt{2}}\sin\left(\frac{\alpha}{2}\right)\\
  \equiv     && \sqrt{2} &= \frac{\sin\left(\frac{\alpha}{2}\right)}{\sin\left(\frac{\beta}{2}\right)}\tag{4}
\end{align}

From this you can see, that the degree of pyramidalization is codependent on the equatorial bond angles. When the pyramidalization increases, $\alpha\to0$, then the equatorial bond angle has to decrease, $\beta\to0$, i.e. a constant ratio.

In the special case of $\alpha = 180^\circ$ this simplifies to $\beta=90^\circ$. In all other cases you can derive the degree of pyramidalization from the equatorial bond angles and vice versa.

Transformation of $(4)$ yields a function that displays the equatorial bond angle in dependence of the pyramidalization, with $\alpha\in~]0^\circ;180^\circ]$ or $\alpha\in~]0;\pi]$.
$$\beta = 2\cdot\arcsin\left[\frac{1}{\sqrt{2}}\cdot\sin\left(\frac{\alpha}{2}\right)\right]$$

Addendum

The out of plane height can also be derived from Pythagoras's theorem (look at right triangle):
\begin{aligned}
                 && h^2 &= s^2 + \left(\frac{d}{2}\right)^2\\
  \therefore     && h   &= \sqrt{s^2 + \left(\frac{d}{2}\right)^2}\\
  \text{with (1)}&& h &= \sqrt{s^2 + \frac{1}{2}a^2}\\
\end{aligned}

This is usually a value that is given for domed complexes, so it is nice to know how to find it.


Oxidation number of oxygen atoms in oxygen difluoride My textbook says:

Oxygen is assigned an oxidation number of $+1$ in oxygen difluoride. The number assigned to each Oxygen atom will depend upon the bonding state of oxygen but this number would now be a positive figure only.

Can there be multiple bonding states for oxygen in oxygen difluoride?
What can be the oxidation state of each oxygen atom in oxygen difluoride?

Oxygen difluoride $\ce{OF2}$ has a single oxygen atom. You must mean dioxygen difluoride $\ce{O2F2}$.
The rules are the same as for hydrogen peroxide.
Both oxygen atoms in $\ce{O2F2}$  are equivalent, so their oxidation number is +1.


How can you qualitatively tell when the hydrolysis of aspirin is over? I'm doing an experiment on the effect of temperature on the time taken to produce acetic acid and salicylic acid. Are there any qualitative observations I can make about when the reaction has finished?
I am using a 0.01 M aspirin in pure water.

First I was thinking that maybe a chemical indicator could be used, however the pKa values of all three components are relatively close to each other (especially aspyrin and salicylic acid). It might still be a bit of a help to stick a pH electrode in the system and observe the change - however, this is not exactly something you can see easily.
Another thing that came to mind is that salicylic acid (as all phenols) will form a colorful complex with Fe(III). However, the appearance of the red-ish color will only tell you when the reaction has started, since it's borderline impossbile to observe that the solution is not getting any darker anymore. Again, if you have some machinery, you could stick your solution into a cuvette and put it in a photometer.
Apart from the two ideas above, I don't really see an easy way of observing anything.


How does the addition of acetone cause a shift from hexaaquacobalt(II) to tetrachloridocobaltate(II)? For this reaction, at equilibrium,

$\ce{[Co(H_2O)_6]^2+ + 4Cl^- <=> [CoCl_4]^2- + 6H_2O}$

$\ce{[Co(H_2O)_6]^2+}$ and $\ce{[CoCl_4]^2-}$ are red and blue in aqueous solution, respectively.

When acetone is then added, the solution changes from red to blue. What causes this?

The great problem is that the ability of a chemical substance to act in some way, is controlled by the activity of the substance and not by the concentration of it. Oftein people consider the concnetration in a first approximation where the activity coefficent of everything is assumed to be one. But for a good prediction or good undertsanding we need the activity coefficents.

for a species X

activity (a) = concentration x activity coefficent (gamma)

We need to understand that as the composition of the solvent changes or the temperture changes the activity coefficents will change.

I think that the answer relates to the relative dielectric constant of the medium, as you replace water with ethanol (or acetone) or increase the temperature then the relative dielectric constant decreases.

By lowering the relative dielectric constant you will make the coordination of the anions to the cations more favourable. I think that this is the reason why a solution of cobalt chloride in ethanol/water is thermochromic. It is pink when cold and blue when hot.

The replacement of the water with a less coordinating solvent will also help to shift the equilibrium in favour of the blue tetrachloro anionic complex.

If you look at the Debye-Huckel equation for the activity coefficient and make sure that you look at what is inside the A term, you will find that when the relative dielectric constant changes that the activity coefficients of ions will change. I think that the change of colour that you see is due to a change in activity coefficient.

The extended Debye-Huckel equation in its full glory is

$$\log (\gamma i) = \left(\frac{-Z_i^2 e^3}{4 \pi \left({\epsilon_0 \epsilon_r k_\mathrm B T}^{1.5}\right)}\right)\frac 1{1 + r_i(8 \pi n e^2/\epsilon_0 \epsilon_r k_\mathrm B T)^{0.5}}(n/2)^{0.5}$$

$Z_i$ is the charge on the ion of interest
$e$ is the electronic charge
$\epsilon_0$ is the vacuum permittivity
$\epsilon_r$ is the relative dielectric constant
$k_\mathrm B$ is Boltzmann’s constant
$T$ is the temperature in kelvin
$n$ is the number of ions in one cubic meter.
$r_i$ is the sum of the radii of the cation and anion which are in contact.

I have done the calculations for the activity coefficient of chloride in mixtures of water and acetone for a 500 μM solution of cobalt(II) chloride at 298 K in mixtures of water and acetone. As the acetone replaces the water the activity coefficient of the chloride goes down, I think that this will be due to the electrostatic interactions of the chloride anions with the cobalt cations (ion pairing):

I repeated the calculation for the activity coefficients of the cobalt ion and the $\ce{CoCl4^2-}$ ion. Bear in mind that the sum of the radii for $\ce{Co^2+}$ and $\ce{CoCl4^2-}$ will be greater than the sum of the radii for $\ce{Co^2+}$ and $\ce{Cl-}$.

Now if we consider the equation

$$K_\mathrm{eq} = a\frac{\ce{CoCl4^2-}}{a(\ce{Co^2+}){a\ce{(Cl-)^4}}}$$

Now if $K_\mathrm{eq}$ is a constant regardless of the solvent system, then as

$$a(\ce{Cl-}) = [\ce{Cl-}] \times \text{activity coefficient of } \ce{Cl-}$$

$$a(\ce{CoCl4^2-}) = [\ce{CoCl4^2-}] \times \text{activity coefficient of } \ce{CoCl4^2-}$$

$$a(\ce{Co^2+}) = [\ce{Co^2+}] \times \text{activity coefficient of } \ce{Co^2+}$$

Then if we have a new equation which is

$$\text{Random greek symbol} = \text{activity coefficient } \ce{CoCl4^2-} / {\text{activity coefficient } \ce{Co^2+} \times (\text{activity coefficient } \ce{Cl-})^4}$$

If we multiply "random greek symbol" with $K_\mathrm{eq}$' we will get a new equilibrium constant for the system.

($K_\mathrm{eq}$' is the $K$ value determined in water where it is an apparent constant - one calculated making the assumption that all activity coefficients are one)

We can get a new equilibrium constant $K$ (solvent composition) which we can plot against the fraction of the solvent mixture which is acetone. Here we have the graph.

Here K' is the true equilibrium constant which would be for a system in water with a zero concentration of cobalt chloride.

We can see as the acetone content increases that the cobalt cations are more likely to react with the chloride anions to form $\ce{CoCl4}$ dianions than they are in the aqueous system.

If you read the papers by people like Bromley[3]The Journal of Chemical Thermodynamics 1972, 4 (5), 669–673. who clearly expresses the ideas of Guggenheim you will find a specific ion interaction term in the activity coefficient equation (Bromley equation). I am sure that the value of the specific ion interaction constant for cobalt and chloride will increase as the water is replaced with acetone.

Sadly I very much doubt if the constants for water/acetone mixtures exist, most of the research effort has gone into aqueous only systems. You could make a lot of the measurements by doing UV/vis spectroscopy on solutions of chloride salts in acetone/water mixtures. But it will be a complex problem, I know that while sodium chloride is a strong electrolyte in water it is a weak one in acetone. I know a classic reaction in organic chemistry for the formation of alkyl iodides is to treat an alkyl bromide or alkyl chloride with sodium iodide in acetone. The larger anion size makes sodium iodide more soluble (and a stronger electroyte) in acetone than sodium chloride/sodium bromide.

If you look at the supporting information for [1], you will see at the end an example of thermochromism. Here mixtures of water, cobalt chloride and the lactic acid / choline chloride eutectic are shown both hot and cold.

This effect has also been reported with nickel chloride in deep eutectic solvents made from choline chloride by some chemists in China. Similar effects were seen for nickel by [2]. They have observed thermochromism which I think is due to a change in the coordination chemistry which in turn is induced by the temperature induced change in relative dielectric constant.

References:

Foreman, M. R. S. J.; Holgersson, S.; Mcphee, C.; Tyumentsev, M. S. Activity coefficients in deep eutectic solvents: implications for the solvent extraction of metals. New J. Chem. 2018, 42 (3), 2006–2012. DOI: 10.1039/c7nj03736h.
Gu, C.; Tu, J. Thermochromic behavior of chloro-nickel(II) in deep eutectic solvents and their application in thermochromic composite films. RSC Adv. 2011, 1 (7), 1220. DOI: 10.1039/C1RA00345C.
Bromley, L. A. Approximate individual ion values of β (or B) in extended Debye-Hückel theory for uni-univalent aqueous solutions at 298.15 K. The Journal of Chemical Thermodynamics 1972, 4 (5), 669–673. DOI: 10.1016/0021-9614(72)90038-9.


Formaldehyde from Urea? I was wondering if it is possible to make formaldehyde (1) from urea (2).

I noticed that they have a similar structure, but the two hydrogen atoms in formaldehyde are replaced with two amide groups in urea. Would it be possible to remove those groups and replace them with hydrogen?

This is more a proposition than an answer but after a lot of search on the net, in publications and also with the help of @orthocresol, @Jan and @jerepierre, I propose a reaction that will work. Also the question ask is really unusual, because it is obviously awfull to want to synthetize formaldehyde from urea.

You will find first in this "answer" the best way we found to make it. When I'll have time, I'll add all ideas I had and explain why it was not a good point.

A good question raised by this question is, "is it possible to "cut" the $\ce{C-N}$ bond to replace it by a $\ce{C-H}$ bond?" and if possible in one step. For the moment nobody knows.

The best way we found :

For the moment I can't explain you why I chose to make the N-alkylation but I know I got it in many of my organic chemistry lessons that this groupement will leave the molecule involving LAH so I wanted to do that. Also we are not at one more step from urea to get formaldehyde.

For the other part formaldehyde is much more reactive than the amide function because on the "diamide" both lone pair of nitrogen atoms are involve in mesomeric forms with the oxygen. Then the carbon is really not electrophilic and LAH will react then one the most reactive function in the reactor which will be the aldehyde.

And then the Formox process because it is use in the industry so I'm sure it will work.

After you will find ideas I had but that won't work too. I'll explain as much as as possible when it will be possible. Sometime it will be only because by experiments nobody have ever succeeded to do it.

This one won't work, one of us, I don't remember who didn't find anything in SciFinder about the possibility to make Eschweiler-Clarke reaction from an amide and especially urea.

This is really awfully awkward because you need formaldehyde itself to make the synthesis of formaldehyde. I was not thinking at that when I start but few minutes later it appears to me that I haven't realize before ^^

At the begining I thought it was great to use $\ce{HCl}$ in excess with $\ce{NaNO_2}$ and then get the carbonic acid $\ce{HO(CO)OH}$, because there is nice ways to reduce this acid and its derivatives but just one time. You can find informations in this publication. But when you have $\ce{H(CO)OH}$ I have really no idea of what to do, even $\ce{LiAlH_4}$ won't work. So at the moment I just have the first possible reaction.

Well also in terms of atoms economy using urea to synthetise formaldehyde is absolutely awkward too.

I hope you find some fun here, I keep thinking about it but it looks to be tricky. Have a good day.


Why does sp hybridised orbital have more directional character than p orbital? It states so in my textbook. But since s orbital is non directional, shouldn't sp hybridized orbital be less directional in character than p orbital?

Ignore everything else in image , this image is just for shape comparison.

As you can see in image that so orbital is a lobe with unsymmetrical shape this leads to its more directionality.

P orbital possess equal Electron Density on both sides of the orbital , but in case of sp orbital electron density isn't equal and that's the reason why sp is more directional than p .

That's also the reason for their different Electronegativity.


Hydrogen visible transitions When dealing with electrons and the electromagnetic spectrum, why are there only four visible transitions in the hydrogen atom?

This all falls back to the hydrogen absorption spectra. There are multiple "series" discovered by different scientists. The series in the visible region is the Balmer series.

Since the electronic energy levels of the hydrogen atom are quantized, there can only be certain discrete transitions.

As it turns out, if you solve the Balmer series (n=2), you get:

$$\frac{1}{\lambda} = \frac{4}{B}\left(\frac{1}{2^2} - \frac{1}{n^2}\right) = R_\mathrm{H}\left(\frac{1}{2^2} - \frac{1}{n^2}\right) \quad \mathrm{for~} n=3,4,5,...$$

You will then get lines at $\lambda$ = 656.3 nm, 486.1 nm, 434.1 nm, 410.2 nm...

Only a few of those fall in the visible. The Lyman series (n=1) occurs in the UV spectra, and the Paschen series (n=3) occurs in the IR.


What is the difference between polyethene and polyethane? The monomer in polyethene seems to drop the $\ce{C=C}$ bond and end up with a $\ce{C-C}$ bond. The polymer chain then looks like $\ce{[CH2-CH2]_{n}}$. If I were to look at that polymer chain I would probably call it polyethane not polythene (or I could even call it polymethane) because it doesn't have a double bond, so I wouldn't classify it as an alkene. Am I doing something wrong with my analysis here? Or is there some sort of rule about nomenclature that change when things become polymers and drop a functional bond (in this case the $\ce{C=C}$ bond)?

The reason it is called polyethene (or polyethylene) is because it was produced from the monomer of ethylene. To produce polyethylene, a radical initiator is added to ethylene, which attacks one carbon and opens up the double bond leaving a radical at the end. This radical attacks another ethylene molecule, etc. The chain stops forming when it encounters another radical. So, from ethylene, we get polyethylene (even though it looks like polymethylene).


Find the amount of atoms for a given mass The atomic mass of two elements A and B is 40 and 80 respectively. If x g of A contains y atoms, how many atoms are present in 2x g of B?

I have tried to solve by assuming that

x g of A has y atoms
so x g of B has 2y atoms, because atomic weight of B is 2 times higher than A
so 2x g of B contains 4 y atoms.

But the answer key says it was wrong, the correct answer is y. What am I doing wrong?

For the elements A and B, the relation between the number of atoms n, the mass m, and the atomic mass M is given by

  $$n_A = \frac{m_A}{M_A}\quad,\quad n_B = \frac{m_B}{M_B}$$

Your assignment states:
$$m_B = 2\cdot m_A $$
$$M_B = 2\cdot M_A $$

Solve for $n_B$ ;-)

I seriously suggest to practise these calculations a lot. Ask your tutor for more worksheets or find a workbook (with solutions). Solve it alone, or work in groups. No matter how, you have to practise.

If you are in doubt, where to divide and where to multiply, remember:

Units are your friends!


How do people know HCN smells like almonds? I was told by my chemistry teacher that $\ce{HCN}$ smells like almonds. She then went on to tell a story about how some of her students tried to play a prank on her by pouring almond extract down the drain to make her think that they had inadvertently created $\ce{HCN}$ gas. She said that she knew that it wasn't $\ce{HCN}$ because if she had smelled the almond scent, then she would have already been dead.

I never asked her, but how do people know $\ce{HCN}$ smells like almonds if they would die before they knew what it smells like?

The odour threshold for hydrogen cyanide $(\ce{HCN})$ is in fact quite a bit lower than the lethal toxicity threshold. Data for $\ce{HCN}$ can be found in many places, but here and here are a couple of good references. That subset of the human population that can detect bitter almonds do so at a threshold of $0.58$ to $\pu{5 ppm}$. The lethal exposure dose is upwards of $\pu{135 ppm}$. That's a whole $\pu{100 ppm}$ range in which to detect and report the fragrant properties.


Is there a way to estimate the nutritional calorific value of a chemical The chemical in question is L Carvone. It is an ingredient of some flavors. Tastes and smells like spearmint.

What's a good way to estimate its calorific value? If any. I cannot see it as a carb / protein / fat etc.

Does it have a calorific value at all?

If carvone ($\ce{C10H14O1}$) is digested by the body using oxygen to generate carbon dioxide and water, you can estimate the heat of this chemical reaction:

$$\ce{C10H14O1 + aO2 -> xCO2 + yH2O}$$

After balancing the equation:

$$\ce{C10H14O1 + 13O2 -> 10CO2 + 7H2O}$$

The required enthalpy can be know from the formation heats of each molecule multiplied by the found factors.

A quick estimation

I just did this exercise using computational chemistry (B3LYP/6-31G(d) in Gaussian09) and found the following energies (in kcal/mol):

carvone: -291456.0217
oxygen: -94322.73333
carbon dioxide: -118326.709
water: -47947.30569

The change in energy of the reaction is then:

$$\begin{multline}10\times (\pu{-118327kcal/mol})+7\times (\pu{-47947kcal/mol})\\-1\times (\pu{-291456kcal/mol})-13\times (\pu{-94323kcal/mol}) = \pu{-1137 kcal/mol}\end{multline}$$

this is $\pu{-1137 kcal/mol}$ (exothermic, as expected).

Since molecular weight of carvone is $\pu{150.2 g/mol}$, we can convert to $\mathrm{kcal/g}$, the negative value indicates it releases heat:

$$\frac{\pu{-1137 kcal/mol}}{\pu{150.2 g/mol}} = \pu{-7.6 kcal/g}$$


When to regard water as a strong base Question:

Classify each of the following ions according to whether they react with water to give a neutral, acidic, or basic
solution:

As a general rule:

Strong acid + Strong base $\rightarrow$ Neutral solution
Strong acid + weak base $\rightarrow$ acidic solution
Weak acid + Strong base $\rightarrow$ Basic solution

According to my solutions manual.

a) $\ce{F-}$ is the conjugate base of a weak acid, therefore  the solution is basic.
b) $\ce{Br-}$ is the anion of strong acid, therefore the solution is neutral.
c) $\ce{NH4+}$ is the conjugate acid of a weak base, therefore the solution is acidic
d) $\ce{[K(H2O)6]+}$ is the conjugate base of a weak acid, therefore the soltuion is basic.
e) $\ce{SO3^2-}$ is the conjugate base of a weak acid, therefore the solution is basic
f) $\ce{[Cr(H2O)6]^3+}$ is an acidic solution, therefore the solution is acidic.

However, all these answers imply that water is acting as a strong base. How is this possible when water pH is $\sim 7$?

Question:

Classify each of the following ions according to whether they react with water to give a neutral, acidic, or basic
solution:
As a general rule:

Strong acid + Strong base $\rightarrow$ Neutral solution

Strong acid + weak base $\rightarrow$ acidic solution

Weak acid + Strong base $\rightarrow$ Basic solution

That's not true. The only time in which these rules are true is when you have an equimolar concentration of the acid and base in question, and the acids and bases must both be monoprotic and monobasic, or diprotic and dibasic, etc. Otherwise there are too many variables to consider for these guidelines to always ring true. Consider mixing 1.0 molar $\ce{H2SO4}$ with 0.10 molar $\ce{NaOH}$ ... I guarantee you that the pH of that solution is not "neutral."

d) $\ce{K(H2O)6+}$ is the conjugae base of a weak acid, therefore the soltuion is basic.

I didn't know that potassium hydrated to a significant extent in water; even so, what the solution manual has written down is not the conjugate base of hydrated potassium cation - the weak acid.

However, all these answers imply that water is acting as a strong
base. How is this possible when water pH is $\sim 7$?

How do these answers imply that water is acting as a strong base?
Also regarding how water can be a strong base in general - it just depends on your solvent system.


How to Calculate Lennard-Jones Parameters from TDDFT? I am trying to develop parameters for AMBER force fields and to calculate Lennard-Jones parameters for the $\ce{Fe^3+}$ ion. I did a deep search and found that the best way to do this with a high accuracy is to use TDDFT. I found many articles about this, for example (I can put more if there is need):

Marques, M. A. L.; Castro, A.; Malloci, G.; Mulas, G.; Botti, S. Efficient calculation of van der Waals dispersion coefficients with time-dependent density functional theory in real time: Application to polycyclic aromatic hydrocarbons. The Journal of Chemical Physics 2007, 127 (1), 014107 DOI: 10.1063/1.2746031.
Banerjee, A.; Autschbach, J.; Chakrabarti, A. Time-dependent density-functional-theory calculation of the van der Waals coefficient $C_6$ of alkali-metal atoms Li, Na, K; alkali-metal dimers $\ce{Li2}$, $\ce{Na2}$, $\ce{K2}$; sodium clusters $\ce{Na_n}$; and fullerene $\ce{C60}$. Phys. Rev. A 2008, 78 (3) DOI: 10.1103/PhysRevA.78.032704.

But, I couldn't understand the articles, because in some there are so many mathematical formulas, in others the method is not clear. So, to get some things clear before starting, I need to answer these questions (let's say I did my TDDFT calculation):

After performing TDDFT calculations, are there software or tutorials that I can use to calculate van der Waals parameters?
If there is no software or tutorial to calculate this, where can I start?

I'm basing my answer on T. Gould, T. Bucko, J. Chem, Theory Comput. 2016, 12, 3603-3613.

In your TDDFT calculation, what you want to compute is the frequency-dependent dipole polarizability, $\alpha_{X/Y}(i\omega)$, that determines the dispersion coefficient, $C_6$.
$$ C_{6,XY} = \frac{3}{\pi}\int d\omega \alpha_X(i\omega)\alpha_Y(i\omega)$$

They also propose some approximate relationships that depend only on the static dipole polarizabilities, $\alpha(0)$.
$$ C_{6,XY} \approx \Xi [\alpha_X(0)\alpha_Y(0)]^{0.73\pm 0.01} $$
where $\Xi = (1.5 \pm 0.1)[\text{hartree} \cdot a_0^{1.62} ]$.

I don't know any tutorials for this, but you could start trying to reproduce some of the numbers on these papers.


Difference between shells, subshells and orbitals What are the definitions of these three things and how are they related? I've tried looking online but there is no concrete answer online for this question.

Here's a graphic I use to explain the difference in my general chemistry courses:

All electrons that have the same value for $n$ (the principle quantum number) are in the same shell
Within a shell (same $n$), all electrons that share the same $l$ (the angular momentum quantum number, or orbital shape) are in the same sub-shell
When electrons share the same $n$, $l$, and $m_l$, we say they are in the same orbital (they have the same energy level, shape, and orientation)

So to summarize:

same $n$ - shell
same $n$ and $l$ - sub-shell
same $n$, $l$, and $m_l$ - orbital

Now, in the other answer, there is some discussion about spin-orbitals, meaning that each electron would exist in its own orbital. For practical purposes, you don't need to worry about that - by the time those sorts of distinctions matter to you, there won't be any confusion about what people mean by "shells" and "sub-shells." For you, for now, orbital means "place where up to two electrons can exist," and they will both share the same $n$, $l$, and $m_l$ values, but have opposite spins ($m_s$).


Burning different isotope combinations of hydrogen and oxygen, how much does the energy vary? In this comment I suggested that there would be a (very) small difference in energy released from burning a stoichiometric mixture of hydrogen and oxygen, depending on the isotopes, due to a slight difference in bond energy. The differences can be simply related to different masses of the nuclei, or to other quantum mechanical effects due to the large differences in nuclear spin and moments.

I'd like to repeat the calculation in this answer just to see the sensitivity to the isotopic shifts.

Approximately how large are these small differences in bond energies? Are the differences mostly due to the mass difference or due to the other nuclear properties?

Isotopes in question: $\ce{{}^1H, {}^2H, {}^3H$,  ${}^{16}O$, ${}^{17}O$, ${}^{18}O}$

Either the bond energies themselves, or just a link to a source where I can look it up myself would be fine.

Let me add to Paul's excellent answer with a simple thermodynamic calculation.

Hydrogen oxidation.
$$\ce{H2 + 0.5 O2 -> H2O(g)}$$
This reaction has a standard free energy change of about -229 kJ/mol.  Using $\Delta G^\circ = -RT \ln K$, I get the equilibrium constant for this reaction of $K=1.2\times10^{40}$.
Deuterium isotope exchange between water and hydrogen.

A 1976 paper gives the value for this reaction.

$$\ce{H2O(g) + HD -> HDO(g) + H2}$$
The value given is $K = 3.5$, which by the standards of isotope exchange reactions is very large. Using $\Delta G^\circ = - RT \ln K$, we can calculate that the standard free energy change of this reaction is $\Delta G ^\circ =$ -3.1 kJ/mol.
Add the above reactions.
$$\ce{HD + 0.5 O2 -> HDO(g)}$$
We can add the free energies of the reactions to get that $\Delta G = $-231 kJ / mol.

Thus, burning $\ce{HD}$ gas provides about 1.4% more free energy than burning $\ce{H2}$ gas. And this is by far the largest difference that we would observe for any common stable isotope.  For nearly any isotopic exchange reaction that involves other isotopes, such as $\ce{^{13}C}$ or $\ce{^{18}O}$, the relevant equilibrium constants will be $0.9 \lt K \lt 1.1$ or so.  Only deuterium vs. protium (i.e. $\ce{D}$ or $\ce{^2H}$ vs $\ce{^1H}$) would often have values outside that range.  Thus for isotopes like $\ce{^{13}C}$ or $\ce{^{18}O}$, the difference will be closer to 0.1%, or, as isotopic geochemists like to say, 1 ‰ (i.e. one per mille).


Only valence shell electrons participate in bonding then how is Cr in H2CrO4 forming 6 covalent bonds although it has only 1 valence electron? The structure of $\ce{H2CrO4}$ is as follows:

Shell configuration of chromium is 2,8,13,1 i.e. it has only 1 valence electron and only valence electron participate in bonding but Chromium in the above structure is forming 6 covalent bonds i.e. it is sharing 6 electrons. How's that possible ? How can it share 6 electrons when it only has 1 valence electron.

The same is the case for $\ce{Mn}$ in $\ce{H2MnO4}$ where it has only 2 valence electrons but is forming 6 covalent bonds and for $\ce{Mo}$ in $\ce{H2MoO4}$ where it has only 1 valence electron but forming 6 covalent bonds.

The electronic configuration of $\ce{Cr}$ is: $[\ce{Ar}]4s^2 3d^4$. Because of very little energy difference between 4s- and 3d- orbitals, it can lose 6 of it's valence electrons and thus form 6 covalent bonds.


Quaternary ammonium anti-microbials Also good against fungi, amoeba, and enveloped viruses,[7] quats are believed to act by disrupting the cell membrane.

Quaternary ammonium cleaners have long-alkyl substituent chains. Why is this so?

Does this have anything do with increasing the ability of the cleaners to penetrate the phospholipid bilayer, which has its non-polar portion oriented outward and its polar phosphate backbone oriented inside the cell? Does the alkyl portion dissolve into the lipid bilayer and wreck havoc?

What is the exact mechanism?

A phospholipid has a charged, hydrophilic phosphate head and a neutral, hydrocarbon hydrophobic tail.  In a phospholipid bilayer, two phospholipids combine tail-to-tail such that the two outer, or exposed, surfaces are comprised of the polar phosphate heads.  The bilayer can function as a cell membrane; it can keep what is in "in" and what is out "out".  It can also regulate critical cell functions such as pH and ion balance through the incorporation of proteins in the bilayer that can serve as ion pumps.  Clearly, any disruption to the bilayer could allow the cell contents to leak out and bad stuff to leak in, which would lead to either cell inactivation or cell death.

The mechanism of action of quaternary salts on lipid bilayers is still an active area of investigation, but here is a simplified overview of what is thought to be happening. .  The first step is thought to involve the electrostatic interaction of the positively charged ammonium ion with the exposed, negatively charged bilayer surface, followed by incorporation of the hydrocarbon chain of the ammonium ion into the bilayer. The electrostatic interaction affects membrane surface pressure which, in turn, disrupts cell function.  Shorter chain lengths (n=12-14) are more effective against gram-positive bacteria, while longer  chain lengths (n=14-16) are more effective against gram-negative bacteria.

Here's a reference to a recent thesis (2012) that studies the resistance of bacteria to quaternary ammonium compounds.  Section 1.2 (pages 2-5) provides a clear, concise summary of the latest thinking on the subject.


Are graphite intercalation compounds stable in water? I want to do an experiment with graphite sulfate, but I want to know if it will de-intercalate in a beaker of distilled water?

Thanks!

According to "Graphite sulphate – a precursor to graphene", Siegfried Eigler, Chem. Commun., 2015,51, 3162-3165:

  Graphite sulphate is used as a precursor to graphene for the first
  time. The positively charged graphene layers react with water to yield
  a processable graphene derivative.

This would seem to imply that it would indeed de-intercalate in a beaker of distilled water.


Does a wavefunction have to be an eigenfunction of an operator in order to obtain an expectation value? I am reviewing an intro quantum chemistry book and I came across the line "If $\hat{A}$ corresponds to a physical property of the system, whether or not $\psi$ is an eigenfunction of $\hat{A}$, the average value of that property is given by the expectation value of the operator." Doesn't $\psi$ have to be an eigenfunction of the operator in order to obtain an expectation value?

No, assuming that wavefunction is normalized, the expectation value $o$ of the operaror $\mathcal O $ is obtained as $\langle \Psi | \mathcal O | \Psi \rangle $.

EDIT

In response at your comment: I am not asking how to calculate the expectation value, I am asking why the wavefunction does not have to be an eigenfunction of the operator in order to calculate an expectation value

The fact that the expectation value is obtained with $\langle \Psi | \mathcal O | \Psi \rangle $ implies that there is no need that the wavefunction be an eigenfunction of the hermitian operator $\mathcal O$.

If it is an eigenfunction, then it is possible to get a simple expresion with:

$$ \langle \Psi | \mathcal O | \Psi \rangle = \langle \Psi | \mathcal O  \Psi \rangle =  \langle \Psi | \mathcal o  \Psi \rangle  = o  \langle \Psi | \mathcal   \Psi \rangle = o$$

If not the case, then you can not make the substiution  $\mathcal O  \Psi = o \Psi$. So, the application of the map $\mathcal O$ to $\Psi$ will be some function $\Omega$. Then, the expectation value will be  $\langle \Psi | \mathcal   \Omega \rangle$ that is an scalar that you can compute as
$$ \int d\mu \Psi^*\Omega$$


Does adding prefixes to SI units make them no longer SI units? Our chemistry professor said that "mg for mass, g/m3 for density, ms for time are all SI units".

I want to make sure now, even if we added a prefix to the unit, will it still be an SI unit?

Yes, when prefixes are used with coherent SI units, the resulting units are still SI units; however, they are no longer coherent.

  The metre per second, symbol m/s, is the coherent SI unit of speed. The kilometre per second, km/s, the centimetre per second, cm/s, and the millimetre per second, mm/s, are also SI units, but they are not coherent SI units.

BIPM: The International System of Units (SI)


Please explain the mystery of the equilibrium constant expression for chemical reactions Given a reversible chemical reaction, e.g.
$$\ce{CuCl4^2-(aq) <=> Cu^2+(aq) + 4Cl-(aq) + \text{heat}}$$
occurring in a medium, the equilibrium constant is defined as
$$K = \frac{[\ce{Cu^2+}][\ce{Cl-}][\ce{Cl-}][\ce{Cl-}][\ce{Cl-}]}{[\ce{CuCl4^2-}]}$$
where [X] is the molarity or molar concentration of X = amount$_X$ / volume.

Then, as you can read anywhere (e.g. google 'equilibrium constant equals 1' to see (1st hit) - 'The Equilibrium Constant, K - Chemistry LibreTexts' -

  If K is about equal to 1, the reaction will reach equilibrium as an
  intermediate mixture, meaning the amounts of products and reactants
  will be about the same.

Yet, if the 'amounts of products and reactants' are the same, that means to me the sum of the molarities of the products would equal the sum of the molarities of the reactants, but K = 1 means that the products of molarities are the same.

I can't attach any intuitive meaning to the product of the molarities.

Note:  the first answer is exactly what I needed.  This question was a pedagogical question - as noted above the first google hit gives a textbook with incorrect info on the equilibrium constant.  That had me confused.  Also, I note that my kid's 10 grade honors chemistry text defines the equilibrium constant as the ratio the product of the molarities of the products divided by the product of molarities of the reactants when the system is in equilibrium, and not as in answer 1 below as the ratio of forward and reverse reaction coefficients, and the latter is a much better definition.

I think the most intuitive way to think about this is to view it through a kinetic lens. That is, recall that at equilibrium, the rate of the forward reaction and the rate of the reverse reaction must be equal. When we write expressions for the rates of reactions, we multiply the reactants because the probability of collision is a function of the product of the two concentrations, not the sum. This only really makes sense for an elementary step, but the result applies to the whole reaction.

Let's consider the simpler reaction $\ce{A + B <=> C}$.

If there's is only one step in the reaction, the rate of the forward reaction will be given by $k_\mathrm{fwd}[A][B]$ and the rate of the reverse reaction by $k_\mathrm{rev}[C]$.

At equilibrium, these are equal, so

$$k_\mathrm{fwd}[\ce{A}][\ce{B}] = k_\mathrm{rev}[\ce{C}].$$

Rearranging,

$$\frac{k_\mathrm{fwd}}{k_\mathrm{rev}}=\frac{[\ce{C}]}{[\ce{A}][\ce{B}]},$$

and we see the product of the concentrations in the equilibrium constant.

You can also go the route of deriving the expression from the free energies and Boltzmann distribution, but I don't think that's nearly as intuitive as thinking in kinetic terms.


Shouldn't weak acids act as strong ones in alkaline solutions? A weak acid in water dissociates reversibly as follows:

$$\ce{XH <=> X- + H+}$$

A strong base is one which dissociates completely.

Now, let's assume we're doing acid-base titrations, adding a weak acid to a strong base, trying to arrive at the titration curve.

My question is mainly why the $\mathrm{pH}$ at the equivalence point isn't $7$. Intuitively, it makes sense, one dissociates fully while the other doesn't. But if you dig into it a bit deeper you realise that that answer is incomplete.

Bear with me. You add the weak acid to the strong base until you reach the equivalence point. The strong base dissociates completely while the acid establishes an equilibrium as described above. However, thinking about it you realise that at no point should this be possible. An $\ce{OH-}$ ion from the base quickly reacts with any H+ ions produced meaning that $[\ce{H+}]$ is always near zero, meaning that more and more of the weak acid needs to dissociate to establish equilibrium. However, the $\ce{H+}$ concentration will always be 0 as any hydrogen ions are snapped up meaning that the acid will completely dissociate before it has a chance to establish equilibrium, at which point all of the $\ce{OH-}$ ions will also have been neutralised. The $\mathrm{pH}$ of this resulting solution should in other words at that point be $7$, because the acid, even though it's weak, acts as a strong one in an alkaline environment.

This however I know isn't correct. What's the issue with my logic above?

While titrating weak acid with strong base, the titration does occur like a normal acid, because the base added quickly takes the dissociated $\ce{H^+}$ ion of the weak acid and thus the equilibrium of the acid is distorted. To maintain constant $\ce{K_c}$, the weak acid also produces $\ce{H^+}$ instantaneously, and thus its behaviour almost becomes like a strong acid if base is added fast(which is normal in titration) to maintain equilibium.But the $\ce{pH}$ at end point is not $7$ not due to the acid but the salt formed as a product of the titration.Suppose the weak acid($\ce{HA}$) is being titrated with a strong base(say for example,$\ce{NaOH}$), and both are taken in same equivalents. Then the reaction does go until all the acid (though weak) is exhausted.$$\ce{HA + NaOH -> NaA + H_2O}$$At t= o, say both $\ce{HA }$ and $\ce{NaOH }$ has $c$ equivalents each. So, at the end of the reaction, both $\ce{HA }$ and $\ce{NaOH}$ will be totally exhausted and $\ce{NaA }$ will be $c$ equivalents.Now, in the medium only $\ce{NaA }$ is present which will control the $\ce{pH}.$ But $\ce{NaA}$ is actually "SALT OF A WEAK ACID AND STRONG BASE". So it hydrolyses as,$$\ce{NaA + H_2O <=> HA + Na^+ + OH^-}$$So  you can see as $\ce{HA }$ is a weak acid ,it's conjugate base (stronger) $\ce{A^- }$ takes the proton of water and leaves only $\ce{OH^- }$, which in turn is the reason for incresing the alkalinity of the medium.That's why generally, $\ce{pH }$ is above 7 at the end point. By doing mathematical calculation, it can be easily shown that if the weak acid's $\ce{pK_a}$ is given and the final concentration of the salt is say, $\ce{ c' M}$. Then the $\ce{pH }$ of the medium can be calculated as ,$$\ce{pH = 1/2 pK_w + 1/2 pK_a + 1/2logc'}$$. Assuming the titration is being carried at $25$ degrees, then$$\ce{pH= 7 + 1/2pK_a +1/2logc' >  7 (generally) }$$ if your acid is tending to strong i.e. $\ce{pK_a}$ is very less positive, and $\ce{c'}$ is close to $1 $, the $\ce{pH}$ can be close to $7$, but $\ce{pK_a}$'s  are normally of the order of $3$ or $4$. So, generally $\ce{pH}$ exceeds $7$.


What acid will dissolve iron carbonate? Issue: removing through dissolution a thin (2-4 mm) iron carbonate crust covering fossilized bone without damaging the bone.  Microblasting with conventional baking soda ineffective; removal of iron carbonate with mini-pneumatic air chisel also removes top layer of bone with the iron carbonate crust. Any solutions would be appreciated.

Two issues arise immediately: 1) dissolving a variable thickness of iron carbonate (probably also of varying composition) and 2) not affecting the fossilized bone (which also probably has varying composition). A third issue might be that the iron carbonate seems to be either very tightly bonded to the bone, or perhaps even part of it by being part of the first substitution layer.

If the fossil is known to be mostly silica, acetic acid (vinegar) could work. Best method would be to scrub the fossil under the surface of the solution with frequent examination to see where additional scrubbing is appropriate, and where not.

If the fossil is largely calcium carbonate, recourse could be to oxidize the ferrous to ferric using a) calcium nitrite solution or b) hydrogen peroxide (H2O2). Calcium nitrite is a commercial corrosion inhibitor for rebar steel in concrete, and functions to oxidize ferrous to ferric at pH 11-12. The product at high pH is ferric oxide (hydrated). The aim in this process is to chemically attack the iron carbonate and turn it into something else that is not attached to the fossil. Scrubbing under the liquid surface (using rubber gloves!!) could turn the iron carbonate to a powdery product that is not attached to the fossil, at high enough pH to avoid dissolving calcium compounds. The chemical reaction is:
2 Fe++  +  2 NO2-  +  2 OH-  -->  Fe2O3  +  H2O  +  2 NO.
The NO will oxidize in air to NO2 which forms nitric and nitrous acids in water, and is toxic, but so is the nitrite, so you have to be very careful not to breathe the vapors or leave the oxidizing solution on your skin.

Perhaps a safer method would be to use hydrogen peroxide. This reaction is well-known, but is usually done in acid; a pH of 5 should be safe enough for the fossil, even if it is carbonate-based. Doing the reaction in acid keeps the ferric iron soluble, but hydrogen peroxide is also reactive in alkaline solution and a high pH might be more suitable for carbonate substrates. The reaction is:
H2O2  + 2 H+  +  2 Fe++  -->  2 H2O  +  2 Fe+++.
There is the added advantage that using excess peroxide causes foaming which could help remove particles from the fossil. Foaming causes mist which should not be breathed, and rubber gloves should be worn because H2O2 will eat your skin if you leave it on you for any significant period.


What is the final molar concentration (molarity) of the nitrate ions in the resulting solution? A chemist mixes 16.3 mL of 1.23 M sodium nitrate with 6.28 mL 0.823 M solution of calcium nitrate. What is the final molar concentration (molarity) of the nitrate ions in the resulting solution?

What I know: I know I can add the volumes to use it to find the final molarity. After I multiply both volume and molarity and get amount of substance, I multiply by the ratio to get the amount of product and use those amounts to calculate molarity. But I don't know how to isolate the amounts of nitrate from the overall amount of product. Can someone explain how I do that?

First, when solving problems like this one, you have to make sure that both salts possess good solubility and that no double salt is precipitating:

$$
\begin{align}
\ce{NaNO3 &<=>> Na+ + NO3-}\label{rxn:R1}\tag{R1}\\
\ce{Ca(NO3)2 &<=>> Ca^2+ + 2NO3-}\label{rxn:R2}\tag{R2}
\end{align}
$$

Both nitrates are indeed well soluble in water (equilibrium is shifted to the far right), and no poorly soluble side-product is formed, e.g. we are likely to have a true solution consisting of hydrated cations $\ce{Na+}$, $\ce{Ca^2+}$ as well as $\ce{NO3-}$ anions.

As you already suggested, one can indeed use total amounts of substances $n_i$ and the preservation of volumes $V_i$ to find the final molar concentration on nitrate (here indices $1$ and $2$ refer to initial sodium and calcium nitrate solutions, respectively):

$$c(\ce{NO3-}) = \frac{n_1(\ce{NO3-}) + n_2(\ce{NO3-})}{V_1 + V_2}\label{eqn:1}\tag{1}$$

According to the stoichiometry (see \eqref{rxn:R1} and \eqref{rxn:R2}):

$$n_1(\ce{NO3-}) = n(\ce{NaNO3}) = c(\ce{NaNO3})\cdot V_1\tag{2.1}$$

$$n_2(\ce{NO3-}) = 2\cdot n(\ce{Ca(NO3)2}) = 2\cdot c(\ce{Ca(NO3)2})\cdot V_2\tag{2.2}$$

Finally, we can plug all the known quantities back to \eqref{eqn:1}:

$$
\begin{align}
c(\ce{NO3-}) &= \frac{c(\ce{NaNO3})\cdot V_1 + 2\cdot c(\ce{Ca(NO3)2})\cdot V_2}{V_1 + V_2}\tag{3}\\
  &= \frac{\pu{1.23 mol L-1}\cdot\pu{16.3 mL} + 2\cdot\pu{0.823 mol L-1}\cdot\pu{6.28 mL}}{\pu{16.3 mL} + \pu{6.28 mL}}\\
  &= \pu{1.35 mol L-1}
\end{align}
$$


Synthesis Golf VII: (+)-Clopidogrel (+)-Clopidogrel (sold as Plavix) is a common pharmaceutical used to help combat the risk of heart disease. The World Health Organisation lists Plavix on its list of 'essential medicines' due to its broad efficacy, low toxicity and low cost (less than 1USD/month in the developing world).

Many industrial routes rely on racemic synthesis and resolution, however syntheses of a single isomer have been reported.

The molecule makes an interesting target as it requires the synthesis of an unusual heterocyclic core, as well as the challenging stereocentre which is prone to racemisation.

Some ground rules:

the synthesis must include formation of the heterocyle (the whole thing, both rings), and a way for setting the stereocentre
you can start from anything you like (subject to the above condition), but it must be available in the Sigma Aldrich catalogue

This was a tricky synthesis. There were several issues. One is that the aryl glycine is possibly easy to epimerize, though the ortho substitution may help prevent that. One other major issue is that the thiophene is reducing and not compatible with any oxidative conditions and probably not compatible with electrophiles.

The amino acid moiety can be synthesized in several ways to set the stereogenic center. One is epoxide ring opening, but here I opted for an asymmetric Strecker reaction.

The six member ring was created via a [4+2] cycloaddition. This is probably the sketchiest portion of the synthesis.

The thiophene moiety was created using Paal-Knorr reaction with Lawesson reagent (B).

Detailed route:

imine formation
asymmetric Strecker with 5% PHTA (partially hydrolyzed titanium alkoxide) and 5% ligand (L). 1 eq tBuOH, tol. (Org. Lett., 2010, 12, 264.) Also, possible with Hoveyda type ligand: (JACS, 2000, 122, 2657.) This exact substrate is reported with good ee.
Hydrolysis with HCl. Similar route reported in (JACS, 1999, 121, 4284.) Though there, the authors Boc-protect at the end. We'll attempt to trap a different way...
formalin, toluene, pTsOH. reflux. Product imine is A.
LDA; TMSCl on methyl vinyl ketone to make electron rich diene.
Hopefully, a little heat will coax this [4+2] cycloaddition with A to take place.
Tsuji-Trost allylation is established for these types of TMS-protected enol ethers. Conditions are $\ce{Pd_{3}(dba)_{2}}$, DPPE, allyl methyl carbonate, THF. (Chem Lett., 1983, 1325.) We'll carry both diastereomers forward.
$\ce{O3}$; DMS.
Lawesson reagent (B).


Why is the rate of disappearance negative? Rate of disappearance is given as $-\frac{\Delta [A]}{\Delta t}$ where $\ce{A}$ is a reactant. However, using this formula, the rate of disappearance cannot be negative.

$\Delta [A]$ will be negative, as $[A]$ will be lower at a later time, since it is being used up in the reaction. Then, $[A]_{\text{final}} - [A]_{\text{initial}}$ will be negative. Therefore, the numerator in $-\frac{\Delta [A]}{\Delta t}$ will be negative.

$\Delta t$ will be positive because final time minus initial time will be positive.

This means that $-\frac{\Delta [A]}{\Delta t}$ will evaluate to $(-)\frac{(-)}{(+)} = (-) \cdot (-) =(+)$

However, we still write the rate of disappearance as a negative number. Also, if you think about it, a negative rate of disappearance is essentially a positive rate of appearance. The reactants disappear at a positive rate, so why isn't the rate of disappearance positive?

Reaction rates are generally by convention given based on the formation of the product, and thus reaction rates are positive. So, for the reaction:

$$\ce{A->B}$$

$$\text{Rate} = \frac{\Delta[\ce{B}]}{\Delta t}$$

To ensure that you get a positive reaction rate, the rate of disappearance of reactant has a negative sign:

$$\text{Rate} = -\frac{\Delta[\ce{A}]}{\Delta t}=\frac{\Delta[\ce{B}]}{\Delta t}$$


What are the primary, secondary and tertiary physicochemical properties of ionic liquids? In the following sentence

  ..the ability to modulate the individual components presents an advantageous framework for tuning secondary and tertiary physicochemical characteristics without sacrificing the primary function of the ionic liquids.

What do they mean by secondary and tertiary physicochemical characteristics?
Can somebody give example of this categorization of the physicochemical property?

The definition of generations of physicochemical properties can be attributed to Hugh et al. [1] as recapitulated by e.g. Ferraz et al. [2]:

  The search for new and different ILs has led to the progressive development and application of three generations of ILs: 1) The focus of the first generation was mainly on their unique intrinsic physical and chemical properties, such as density, viscosity, conductivity, solubility,and high thermal and chemical stability. 2) The second generation of ILs offered the potential to tune some of these physical and chemical properties, allowing the formation of “task-specific ionic liquids” which can have application as lubricants, energetic materials (in the case of selective separation and extraction processes), and as more environmentally friendly(greener) reaction solvents, among others. 3) The third andmost recent generation of ILs involve active pharmaceutical ingredients (API), which are being used to produce ILs with biological activity.

The passage quoted in the OP stems from an article by Zakrewsky et al. [3] and refers to a study on a third generation IL. Third generation characteristics would in this case include those granting the IL antibacterial, nontoxic and transdermal delivery properties, that is allowing solubilization of drug compounds, minimizing damage to human cells, and disupting bacterial biofilms. In this case, "secondary" properties were not targeted, but these might include (as per the above quote) reduced environmental toxicity.

References

Hough, W. L.; Smiglak, M.; Rodríguez, H.; Swatloski, R. P.; Spear, S. K.; Daly, D. T.; Pernak, J.; Grisel, J. E.; Carliss, R. D.; Soutullo, M. D.; et al. The Third Evolution of Ionic Liquids: Active Pharmaceutical Ingredients. New J. Chem. 2007, 31 (8), 1429–1436. https://doi.org/10.1039/B706677P.
Ferraz, R.; Branco, L. C.; Prudêncio, C.; Noronha, J. P.; Petrovski, Ž. Ionic Liquids as Active Pharmaceutical Ingredients. Chem. Med. Chem. 2011, 6 (6), 975–985. https://doi.org/10.1002/cmdc.201100082.
Zakrewsky, M.; Lovejoy, K. S.; Kern, T. L.; Miller, T. E.; Le, V.; Nagy, A.; Goumas, A. M.; Iyer, R. S.; Sesto, R. E. D.; Koppisch, A. T.; et al. Ionic Liquids as a Class of Materials for Transdermal Delivery and Pathogen Neutralization. PNAS 2014, 111 (37), 13313–13318. https://doi.org/10.1073/pnas.1403995111.


What is the significance of enthalpy definition not under constant pressure? Under constant pressure (and only PV work), change in enthalpy can be shown to equal change in heat. If enthalpy is not meaningful outside of constant pressure, then why not use heat instead? So I think it must be meaningful outside of constant pressure scenarios, but I couldn't figure it out myself. Can someone help?

There are many processes that take place at constant pressure, such as the change of phase from a liquid to a vapor for a pure substance at constant temperature.  Also, the heats of chemical reactions and heats of formation of compounds are typically cataloged at constant temperature and pressure.  So the enthalpy comes into play in both these instances.

The first law of thermodynamics for a closed system is expressed conveniently in terms of the change in internal energy of the system.  However, for an open system (in which mass is flowing in and out) operating at steady state, the first law is more conveniently expressed in terms of the enthalpy, not the internal energy.  And, of course, all industrial applications of thermodynamics involving continuous processes are operating at steady state.

Chet


Most Comprehensive Chemistry Book(s)? Although I am now taking a pretty fundamental chemistry course, I am really looking to gain an extremely comprehensive knowledge of the subject, specifically, in inorganic and organic chemistry. I am looking for two texts, keeping in mind that the most important thing is comprehensiveness and the most information about the most topics, not ease of readability. I have the time and dedication to do it, but I really don't want to start with a book that doesn't cover everything and I have to fill in the holes as I go or individually seek out topics it misses.

For the inorganic chemistry I recommend Holleman-Wiberg's text, also exists in english. It is reasonably readable and very thick and contains a lot.
Holleman-Wiberg's Inorganic Chemistry. If you prefer, the original is in German and updated very often.

For the organic chemistry, besides being thick also very nicely readable,  Organic Chemistry by Clayden, Greeves and Warren

Besides that, you would probably need some nice general chemistry book, but I have no good personal experience in this area.


What are the best chemical liquids to clean metallic surfaces? I'm just a simple man curious about things. I've no big understanding of chemistry. I had this question today in mind. What chemicals could I use to achieve the cleanest metallic surface?

For example, I could first start by using isopropyl alcohol to remove all dirt and oils. But there are other liquids such as acetone and acids.

If you were a chemist, how would you clean a metallic surface?

Assuming clean means anti-pathogenic, I would use Hydrogen Peroxide "oxyclean" since it can "clean" metals by removing pathogens. $\ce{H2O2}$ is marketed in various strengths. 1% dilution can kill up to 99.99% bacteria in 30 seconds.

Any butyl based cleaning product is a very good degreaser but it may harm rubber products.

Check here and here for more.


In atomic diffusion in Au/Cu metal contact, where do the atoms come from ? (need melting temperature to break bonding in lattice ?) It is known that Au metal is diffused when it is contacted with copper metal. And Au metal have lattice which consist of atoms. If so, may I think that atom is diassociated from lattice by heating ? If so, it needs to break bond between atoms before diffuion ?
I think that it needs melting temperature to break bonding in lattice.

Yes, it takes energy to get an atom off a lattice site. However, as an activated process with an energy barrier, Arrhenius' equation shows you that even with that requirement, some number of atoms will be off their lattice sites in thermal equilibrium, making vacancies (empty sites) and interstitial atoms.

This was realized some time ago by Simmons and Balluffi. They investigated many metals, using precision measurements of the change in length of a bar of material in a furnace, coupled with x-ray diffraction to determine the lattice constant at the furnace temperature. The result is you can back out how many lattice sites are vacant from the difference in the change of length actually measured vs what you would expect from the change in lattice spacing.

The linked article focuses on gold, but has some data on other fcc metals as well. Just at the melting point of gold (not yet melted), the concentration of vacancies is about $7 \times 10^{-4}$, or 0.07% of lattice sites are vacant. The activation energy is given as 0.94 eV.

Bottom line - there are native populations of point defects even at room temperature, and those drive diffusivities. Certain point defects or defect complexes may be more mobile than others, and the concentration of a given defect will vary with temperature.


Why is 3,6-dimethylpiperazine-2,5-dione optically active in spite of having a centre of symmetry? I learnt in my chemistry class that if a molecule has an element of symmetry (either a plane of symmetry or a centre of symmetry), it will not be chiral and hence will not show optical activity.

My professor told me that in alanine anhydride the carbon atoms with methyl groups would be non-planar, hence it will be non-symmetrical:

But if I consider one of the above-mentioned carbon atom above the plane of the ring, and the other below the plane of ring, a centre of symmetry will still exist. Where am I going wrong?

There is no evidence (e.g. aromaticity) to support your claim that the compound is planar. In fact, crystalographic studies [1] revealed existence of both cis- and trans-3,6-dimethylpiperazine-2,5-dione. Corresponding CCDC numbers are LCDMPP01 and TRDMPP01, respectively — feel free to open the links and play around with the 3D structures in JSmol to understand the difference.

According to Sletten, in cis (ʟʟ) modification the ring is puckered resembling a skewed boat conformation, whereas the ring in trans (ᴅʟ) molecule is nearly planar:

Figure 1a. cis-3,6-dimethylpiperazine-2,5-dione (drawn from LCDMPP01)

Figure 1b. trans-3,6-dimethylpiperazine-2,5-dione (drawn from TRDMPP01)

References

Sletten, E. Conformation of Cyclic Dipeptides. The Crystal and Molecular Structures of Cyclo-D-Alanyl-L-Alanyl and Cyclo-L-Alanyl-L-Alanyl (3,6-Dimethylpiperazine-2,5-Dione). J. Am. Chem. Soc. 1970, 92 (1), 172–177. DOI: 10.1021/ja00704a028. (Open Access)


What is the difference between temperature and thermodynamic temperature? My course book says that kelvin is the unit of thermodynamic temperature, then what is the difference between temperature and thermodynamic temperature.

Temperature is more of a relative notion--heat flows from higher temperature to lower temperature, but that's about it. The zero in a temperature scale is meaningless; it is an arbitrary convention. Usually, we measure temperature in Celsius, Fahrenheit, Rankine, etc (note that Kelvin measures temperature as well--but it also measures thermodynamic temperature)

Thermodynamic temperature is such that zero temperature is absolute zero (unlike Celsius, where -273.15°C is absolute zero). It is related to the energy (specifically the kinetic energy) of the molecules of a substance-- for example, in a gas, kinetic energy per molecule is $\frac12k_BT$ (rotational energy, etc can be obtained if you know the type of molecule--but temperature is principally a measure of kinetic energy). It also has strong ties with entropy.

We generally don't use "temperature" in thermodynamics, thermodynamic temperature is more convenient. Temperature is only useful in equations involving temperature difference--whereas equations involving the absolute value of temperature (Arrhenius equation in kinetics, entropy, Maxwell distribution equation, etc) demand thermodynamic temperature.


Open Source tools to draw a set of 2d molecular graphs from sdf file I have a sdf file for around 50 small-molecules for which I am doing structure-activity study. I would like to draw 2d graphs of all the molecules such that the image would fit into a single journal page. For example, in this paper, page number 3, they have listed 2d molecular structure. I have attached the image file from the same paper. I have software tools like MarvinSketch which give me 2d or 3d graph for a given sdf file. But I want the integrated image like shown below suitable for publication.

An SDF file is just a bunch of MOL entries in a single file; $$$$ serves as the separator for the individual entries.

Splitting into individual files AND simultaneous conversion to graphic file formats (PNG, SVG) is possible with Open Babel, either directly on the command line or through scripts/programmes using the bindings available for various programming languages, such as Pybel for Python.

To my knowledge, the direct generation of PostScript files - probably the nicest option when using $\LaTeX$ - is not possible, so you might want to used SVG as an alternate vector-based format.

obabel -isdf lotsofstructures.sdf -m -osvg -xC -xu -xd -O basename.svg

will do the job.

-isdf specifies the input file format, -m does the splitting, -osvg specifies the output file format, and -O basename.svg specifies the prefix and extension for the individual numbered output files that are generated. -xC prevents that terminal carbon atoms are printed explicitly with attached hydrogen atoms. Surely a matter of taste, but I wouldn't want to have lots of $\ce{-CH3}$ in the drawings. -xu prevents the use of colour for heteroatoms and generates a black and white drawing. -xd suppressed the name of the compound to be written and displayed in the SVG file.

This should work for SDF files with 2D and 3D coordinates, but in the latter case I'd suggest to check it carefully; I've seen 3D -> 2D transformations on some molecules that went wrong.


Explaining the result of experiment---Aluminum foil heated in boiling KMnO4 solution I've recently been thinking back to an impromptu "experiment" I performed almost exactly 3 years ago. The setup was as follows:

  A concentrated solution of KMnO$_4$ was produced by dissolving 2 grams of
  Condy's crystals in 250 ml of water. This was subsequently poured into
  an Erlenmyer flask containing a strip of aluminum foil, and the entire
  system was then heated at high on a stove until the water came to a boil. As the water evaporated, the solution lost the intensity of its violet
  hue, turning a light pink-violet instead. The stove was then turned off, and the aluminum strip was removed, having turned a golden color.

My questions are two:

What, precisely, was the chemical reaction(s) that occurred?
How much of a risk did the steam from the boiling solution pose in
   terms of potential acute manganese inhalation toxicity---and what are the potential long-term effects? Unfortunately, this
   experiment was performed on an ordinary kitchen stove and not in a
   fume hood (I'm kicking myself to this day).

My guess as to the former is that MnO$_2$ was produced, hence the observed color changes of the solution and foil. I am not sure whether this occurred via heat decomposition of the MNO$_4^-$ (which seems to occur at temperatures of >240$^\circ$C, and thus should not have been able to take place in boiling water) or oxidization of the aluminum (having somehow penetrated the surface layer of aluminum oxide). As for the latter, welp.

Thanks in advance!

Potassium permanganate solution tends to decompose with time, especially those which are not acidified. Just let an aqueous solution sit in a glass bottle and it will leave a brown film of tenacious manganese dioxide on the surface at room temperature. If water has any organic material, it will become brown and tap water is not pure water. It decomposes on the skin as well leaving a brown stain. You Al surface is no different, with a thin deposit of manganese dioxide, it may appear yellowish-golden.

Long time ago potassium permanganate was added to water wells to disinfect water. Of course it would decompose with all the organics. If you are well after three years, what is the problem?


Whats the effect of pressure on water electrolysis of sea water with increase in depth? I’m currently working on an underwater robot and was hoping to use the gas bubbles produced by water electrolysis for buoyancy control. The operating depths range from 200–1000 m.

The higher pressure itsels doent make it much more difficult. But...

For water electrolysis you need high overpotential, preferably large electrodes with small electrode gap between them. You want clear, filtered water and no fishes electrocuted, burning down the whole submarine due to an accidental short.

And you will need plenty of gas for buoyancy control, especially at $200{-}1000\ \mathrm{m}$ depths, therefore you need a lot of electricity:

According to Faraday's law $$m=(Q/F)(M/z)$$ where $F = 96485\ \mathrm{C/mol}$ you can make like $4.5\ \mathrm{l}$ of hydrogen gas (1 atm, 25 °C) with a $10\ \mathrm{A}$ current in 1 hour! That is like $0.04\ \mathrm{l}$ at $1000\ \mathrm{m}$ ($100\ \mathrm{atm}$) of a potentially explosive gas.