How does a combustion process produce carbon monoxide? My question is in the context of combustion heater safety- I want to identify the mechanism by which CO is produced, and hopefully have additional understanding, and clues to the warning signs, to identify unsafe stoves or installations. The primary safety measures will always be CO detectors and good ventilation.

The worst horror stories usually concern charcoal barbecues in confined spaces- solid carbon burning at about 900 °C. Is it a safe educated guess that the main combustion product here is CO, which fails to burn fully to CO2 in depleted oxygen levels? Or is a more complex process at work?

How does the yellow combustion of an oil or gas flame produce CO? Is it a simple matter of poor gas flow in the flue, and depleted oxygen in the feed air?

Thanks for any help

If combustion occurs in oxygen, (at least) the following reactions are relevant:
$$\ce{C + O2 -> CO2} $$
$$\ce{C + 1/2 O2 -> CO} $$

You can favor the first reaction by having more oxygen in the gas phase, since it required twice $\ce{O2}$ as the first.

However, one of the most important factors is the temperature: high temperature always favors the $\ce{CO}$ product, since reaction wants to proceed fast and react with any available oxygen.
You can clearly see these dependences in the Figure 6 of this article, at low temperatures the CO/CO2 ratio is low, but quickly increases by heating.

More detailed information is easily found in some classical papers, a full PhD thesis on it, or even calculations with modern quantum chemistry methods.


Difference between chemical potential and gibbs free energy I know that the chemical potential is the molar Gibbs free energy and that it is an intensive variable and that the gibbs free energy is extensive. However, they have identical units: $\mathrm{J~mol^{-1}}$. I don't understand why $G$ is extensive if it has $\mathrm{mol^{-1}}$ in its units.

I know that the chemical potential is the molar Gibbs free energy and that it is an intensive variable and that the gibbs free energy is extensive.

The chemical potential is related to the Gibbs energy in one of its definitions:

$$\mu_i = \left(\frac{dG}{dn_i}\right)_{T,P,N_{j\neq i}}$$

So it is the change in Gibbs energy when one component changes amount while the other components are of constant amount. The dimensions are energy per amount of substance (with units of e.g. J/mol).

The term "Gibbs energy" is used for a lot of things, and the dimensions could be energy, or energy per amount of substance. The Gibbs energy of reaction $\Delta_r G$, arguably the most important Gibbs energy for a chemist, is defined as:

$$\Delta_r G = \left(\frac{dG}{d\xi}\right)_{T,P}$$

where $\xi$ is the extent of reaction with dimensions amount of substance (i.e. amounts of all reactants and products change as $\xi$ changes, according to the stoichiometry specified in the reaction equation). As a consequence, the dimensions of chemical potential and Gibbs free energy of reaction are the same.

For other uses, the term "Gibbs energy" might refer to the Gibbs energy of a given system, and would have dimensions of energy. $G$, like the enthalpy $H$, does not have a defined reference point that is zero, so there is never a specific value associated with $G$. You can, however, use $\Delta G$ to describe the change in Gibbs energy from a defined state to another defined state. This $\Delta G$ would have dimensions of energy, and would be extensive (if you duplicate the system and look at the total, it would have a $\Delta G$ that is twice as large).

It would be easier if the symbols would indicate which quantities are per amount of substance (intensive) and which are not (extensive). Some textbooks use bold typeface for all molar quantities, but introductory textbooks are usually not rigid in terms of dimensions at all.

  I don't understand why G is extensive if it has $\pu{mol−1}$ in its units.

The Gibbs energy of reaction is not extensive. However, if you multiply all coefficients by a constant factor, you also have to multiply the Gibbs energy of reaction by the same factor. For example,

$$\ce{H2 + 1/2 O2 -> H2O}\ \ \ \ \Delta_r G^\circ = \pu{−237.13 kJ mol−1}$$

$$\ce{2H2 + O2 -> 2H2O}\ \ \ \ \Delta_r G^\circ = \pu{−474.26 kJ mol−1}$$

The first is equal to the Gibbs energy of formation of water, and the second avoids fractional coefficients. The two are different because the extent of reaction is linked differently to changes in the amount of the reactants and products.


You'd need to have some sort of ligand dictionary. For many PDB ligands, this exists, and you can look up the atom naming.

For example, the PDB ligand expo allow you to search for ligands by 3-letter code (e.g., HEM).

The PDBeChem also lets you find atom names based on 3-letter codes (the link is HEM).

However, as far as I know, there isn't (yet) any sort of pattern matching that would allow you to go through PDB files and automatically perceive your ligands, sort the atoms, and name them systematically.

If you find one, please let me know - it would be a great addition to programs like Open Babel


Why does mixing inks make subractive colors? I have little knowledge of color theory and chemistery. Recently I came to know that printers use CMYK colored inks and combine them in different proportions to achieve any color on print.

Let us suppose I fill white paper fully with red sketch then white light will fall on inked surface and all colors(wavelengths) except red will be absorbed and only red will be emitted/reflected. Good. Now I overfill that paper with green sketch. Now two things can happen:

Both the inks of sketches mix but do not interact molecully. Like two chemicals A and B mixed in water and both A and B are everywhere in water but they are still distinct molecules. A's and B's molecules don't react chemically.
Both the inks of sketches mix and interact molecuelly. Like HCL and Na2SO4 are mixed in water and they react molecually to make h2SO4 and NaCl.

Now if 1 is true then on any point on paper's surface I have two types of molecules, 1. that reflect red light and 2. that reflect green light. So when white light will fall on paper then from every point two colors will be reflected, red and green. When they'll reach someone's eye then the perceived color will be red+green = yellow. So inks should make additive color.

On the other hand if 2 is true then molecules of red and green ink will react chemically to make a new kind of compound. Now it is not necessary that the new kind of compound absorbs both red and green. It could emit/reflect any random color. So the paper could be of any color depending upon the chemical formula of inks used. So in this case too inks don't make subtractive colors.

Sadly in reality inks make subtractive color and the color of paper actually is dark brownish black not bright yellow. So in reality the mixed ink absorbs almost all colors including red and green which were remaining.

So my question is why does both red and green got subtracted from the mixed ink? In general why does mixture of inks produce subtractive color rather than additive?

The issue is that you're thinking about inks reflecting color, when in reality you should be thinking about the inks absorbing colors.

If the function of inks were to reflect colors, then by adding more and more ink (of a single color) you'd be able to make brighter and brighter colors, as more ink means more reflection. Instead, it's the white paper (rather than the ink) that's reflecting the light, and more and more ink (even of a single color) makes things darker because more and more light is being absorbed.

That's why mixing pigments/inks is a subtractive process - as you add inks, you increase the amount of light that's being absorbed.

So let's go to your example. Let's simplify things by considering only three wavelenghts of light: red, green and blue. Your red ink absorbs the green and blue light, whereas the green ink absorbs the red and blue light. Thus when you mix the two, each absorbs the light it would in isolation, and you have very small amount of blue light, and also very little red or green light, leaving you with a very dark, perhaps slightly yellowish-brown color.

Now consider subtractive inks. You have yellow ink, which absorbs just the blue light and you have magenta ink, which absorbs just the green light. You mix the two together, and the magenta ink absorbs the green light, the yellow ink absorbs the blue light, and the red light is free to be reflected, resulting in a red colored piece of paper from Y+M.

Generally speaking, modern inks (at least the standard, non-color changing variety) involve no chemical reactions in color production: mixing inks and pigments is solely a physical change based on absorption of light.


How to calculate molecular emission spectra? Is there free software that can calculate the fluorescence emission spectra of organic molecules?
Let's say, for example, rubrene

There is a bunch of quantum free quantum chemistry software:
https://en.wikipedia.org/wiki/List_of_quantum_chemistry_and_solid-state_physics_software
However, which one would be the most suitable for this purpose? Could this be done on a reasonable up-to-date laptop? Of course, computing times will depend on the number of atoms in the molecule and the desired precision. What would be approximate calculation times? In the order of hours, days or weeks?
I can imagine that this is most probably not a simple "draw molecule and push calculate-button"-task, but any hints where to further dig into are appreciated.

Addition after the question was put on hold:
This question certainly can fill a semester of a lecture on quantum or computational chemistry. In that sense, yes, it is broad.

However, instead of detailed instructions or input scripts, I was hoping for some rough guidance about the essential steps and the available software which can be recommended for that purpose (and equally important: which software can be excluded or cannot be recommended to avoid investing a lot of time and find out later that there are much better options). And of course, whether you need a super-computer or whether you can do it on a laptop within a reasonable amount of time.

As you mentioned, Gaussian can do it but is proprietary.

I believe ezSpectrum (https://github.com/iopenshell/ezSpectrum) might be able to do what you are looking at but I have not used it myself so far.

There used to be a program available called FCClasses which would run the Franck-Condon computations for you based on (I think) Gaussian output, but it seems like the author took it down, I can only find a front-end to it right now: https://github.com/jcerezochem/fcc_tools The FCClasses homepage is mentioned as http://www.pi.iccom.cnr.it/fcclasses on that page but, as mentioned above, no longer works.


Why atropisomers are called conformers? Changing one atropisomer to another requires bond breaking (in some cases the removal and reattachement of steric groups, according to my understanding) so how it is possible that according to IUPAC atropisomers are classified as conformers (conformational isomers)?
According to IUPAC:

Atropisomers - A subclass of conformers which can be isolated as
separate chemical species and which arise from restricted rotation
about a single bond.
Conformers - One of a set of stereoisomers, each of which is
characterized by a conformation corresponding to a distinct potential
energy minimum.
Conformation - The spatial arrangement of the atoms affording
distinction between stereoisomers, which can be inter-converted by
rotations about formally single bonds. Some authorities extend the
term to include inversion at trigonal pyramidal centres and other
polytopal rearrangements.

This definition of "conformation" seems vague to me, but it is often said that changing one conformation to another involves no bond-breaking and changing one configuration to another requires bond-breaking. This is the difference between them.
IUPAC gives examples of atropisomerism: e.g. (E)-cyclooctene (given in Gold Book "planar chirality" entry). This definitely should involve bond-breaking. Other IUPAC examples of atropisomerism include substituted ortho-biphenyls.

I think your notation of "Changing one atropisomer to another requires bond breaking (in some cases the removal and reattachement of steric groups, according to my understanding)" is not correct. To my knowledge, the free rotation of one single bond of an atropisomer is not restricted by connecting bond(s). I think you get confused by some literature explaining breaking a bond selectively directed to pair of atropisomers (e.g., Ref.1):

According to your post, IUPAC definition of atropisomers:

Atropisomers - A subclass of conformers which can be isolated as separate chemical species and which arise from restricted rotation about a single bond.

A variety of sources confirm this definition. For instance, Ref.2 and 3 described atropisomers as:

Atropisomers are stereoisomers (rotamers) resulting from hindered rotation about single bonds (having an energy barrier to rotation about a single $\sigma$ bond) usually due to steric hindrance. This energy barrier to rotation is high enough to allow the isolation of the conformers (rotamers).

Nonetheless, the name, coined by German biochemist Richard Kuhn in application to a theoretical concept 1n 1933 (Wikipedia) is derived from Greek (a = not and tropos = turn; thus, atropos meaning "without turn"), which supported the definition. However, the atropisomerism was first detected in 6,6'-dinitro-2,2'-diphenic acid by Cristie, et al. in 1922 (Ref.4):

It is also important to know the nomenclature of atropisomers (Wikipedia):

Determining the axial stereochemistry of biaryl atropisomers can be accomplished through the use of a Newman projection along the axis of hindered rotation. The ortho, and in some cases meta substituents are first assigned priority based on Cahn–Ingold–Prelog priority rules (Ref.5 & Ref.6). One scheme of nomenclature in based on envisioning the helicity defined by these groups (Ref.7). Starting with the substituent of highest priority in the closest ring and moving along the shortest path to the substituent of highest priority in the other ring, the absolute configuration is assigned $P$ or $\Delta$ for clockwise and $M$ or $\Lambda$ for counterclockwise. Alternately, all four groups can be ranked by Cahn–Ingold–Prelog priority rules, with overall priority given to the groups on the "front" atom of the Newman projection. The two configurations are termed $R_\mathrm{a}$ and $S_\mathrm{a}$ in analogy to the traditional $R/S$ for a traditional tetrahedral stereocenter (Ref.8}:

Atropisomers are also found in nature. The main pigment of cotton seeds, gossypol is an atropisomer, which exists in both form of $(R_\mathrm{a})$- and $(S_\mathrm{a})$-conformations. Most commercial Upland (Gossypium hirsutum) cottonseeds have an $(R_\mathrm{a})$- to $(S_\mathrm{a})$-gossypol ratio of ~$2:3$, but some Pima (Gossypium barbadense) seeds have an excess of $(R_\mathrm{a})$-gossypol. Between two isomers, $(R_\mathrm{a})$-gossypol is more toxic and exhibits significantly greater anticancer activity than its $(S_\mathrm{a})$-atropisomer:

Two fun facts about atropisomers are (KU.edu):

Atropisomers are detectable by $\mathrm{NMR}$ if half lives exceed $\pu{10^{-2} s}$.
Atropisomers are isolatable if the half-life is above $\pu{10^3 s}$.

References:

Gerhard Bringmann, Thomas Hartung, “Atropo-enantioselective biaryl synthesis by stereocontrolled cleavage of configuratively labile lactone-bridged precursors using chiral H-nucleophiles,” Tetrahedron 1993, 49(36), 7891-7902 (https://doi.org/10.1016/S0040-4020(01)88014-5).
Alan R. Katritzky, Christopher A. Ramsden, John A. Joule, Viktor V. Zhdankin, In Handbook of Heterocyclic Chemistry, Third Edition; Elsevier Limited: Amsterdam, The Netherlands, 2010 (ISBN: 978-0-08-095843-9).
Jonathan Clayden,  “Atropisomerism,” Tetrahedron 2004, 60(20), 4335 (https://doi.org/10.1016/j.tet.2004.03.002).
George Hallatt Christie, James Kenner, "LXXI.—The molecular configurations of polynuclear aromatic compounds. Part I. The resolution of $\gamma$-6 : 6′-dinitro- and 4 : 6 : 4′ : 6′-tetranitro-diphenic acids into optically active components," Journal of the Chemical Society, Transactions 1922, 121, 614–620 (https://doi.org/10.1039/CT9222100614).
R. S. Cahn, Christopher Ingold, V. Prelog, “Specification of Molecular Chirality,”  Angew. Chem. Internat. Ed. Engl. 1966, 5(4), 385-415 (https://doi.org/10.1002/anie.196603851) and Corrigendum: Angew. Chem. Internat. Ed. Engl. 1966, 5(5), 511-511 (https://doi.org/10.1002/anie.196605111).
G. P. Moss, "Basic terminology of stereochemistry: IUPAC Recomendations 1996," Pure and Applied Chemistry 1996, 68(12), 2193-2222 ()(PDF).
http://goldbook.iupac.org/terms/view/H02763: IUPAC. Compendium of Chemical Terminology, 2nd ed. (the "Gold Book"). Compiled by A. D. McNaught and A. Wilkinson. Blackwell Scientific Publications, Oxford (1997). Online version (2019-) created by S. J. Chalk. ISBN 0-9678550-9-8. https://doi.org/10.1351/goldbook.
http://goldbook.iupac.org/terms/view/A00547: IUPAC. Compendium of Chemical Terminology, 2nd ed. (the "Gold Book"). Compiled by A. D. McNaught and A. Wilkinson. Blackwell Scientific Publications, Oxford (1997). Online version (2019-) created by S. J. Chalk. ISBN 0-9678550-9-8. https://doi.org/10.1351/goldbook.


Does acid anhydride give a positive iodoform test? I thought that acetic anhydride $\ce{CH3COOCOCH3}$  would give a positive Iodoform test because it has a $\ce{CH3-(CO)}-$ group.But it is given that this does not happen.

Is this true? If so, can anybody explain the reason? Thanks

According to this source, both acid halides and anhydrides can form enols reasonably readily and can be halogenated. I'd imagine, for anhydrides, this to be related to the potential for intramolecular hydrogen bonding in the enol form (as shown below):

This allows halogation at the alpha position by treatmeant with the halogen. I'd also imagine this process can be catalysed in anhydrous acidic conditions, avoiding hydrolysis.
The same cannot be said for base-facilitated enolate formation. As user55119 implied, the highly reactive $C=O$ is likely to be attacked before the alpha hydrogens (or even if deprotonation occurs first, proceed through an E1cB pathway due to the presence of a good leaving group; implied by Clayden Organic Chemistry), decomposing the anhydride instead of forming the enolate.
Enolate formation in basic conditions is usually required for multiple halogenation, and generally three substitutions are required to form a suitable leaving group for the haloform reaction to proceed. This is why it is a test for a methyl group next to a $C=O$ (as Mathew has said); as there are three alpha hydrogens.
Considering basic conditions are not an option, only mono-halogenation is feasible (perhaps even in acidic conditions) and the haloform test is made virtually impossible.


What happens to the oxygen when carbohydrates turn into hydrocarbons? According to the biogenic theory of hydrocarbon formation, hydrocarbons come from pressurized carbohydrates.

I am looking for an explanation of what happens to the oxygen in the process. How could the oxygen break away from the hydroxyl groups of molecules like glucose to form hydrocarbons?

The process likely involves pyrolysis, a generic term for the thermal degradation pathways of organic compounds.

In the case of carbohydrates, like glucose (shown below), the most common function groups are alcohols (R-OH). Alcohols undergo dehydration in the presence of strong acids, but the reaction can also be promoted by extreme heat in the absence of oxygen. The presence of oxygen leads to combustion (bad!).

In a dehydration elimination reaction, the OH on one carbon and the H on a neighboring carbon leave to form water. This usually follows an E1-like mechanism, involving a carbocation intermediate. In the geochemical version, the necessary acid could come from nearby slightly acidic mineral.

Protonation of the alcohol:
$$\ce{(CH3)2CH-OH + H+ -> (CH3)2CH-OH2+}$$
Loss of leaving group:
$$\ce{(CH3)2-CH-OH2+ -> (CH3)2CH+ + H2O}$$
Loss of proton (regeneration of acid):
$$\ce{CH3-CH+-CH2-{\bf{H}} -> CH3-CH=CH2 + {\bf{H+}}}$$

Overall:
$$\ce{(CH3)2CH-OH ->[\ce{H+}][\Delta] CH3-CH=CH2 + H2O }$$

In the case of carbohydrates, they tend to dehydrate to carbonyl compounds, which can then decompose further into hydrocarbons by reductive and radical pathways.

For example, fructose is converted into furan derivatives, like 5-hydroxymethylfurfural, by a series of dehydrations and rearrangements:

$$\ce{C6H12O6 ->[\ce{H+}][\Delta] C6H6O3 + 3H2O}$$

The image above was created by Wikipedia user V8rik and released under the GNU Free Documentation License and the Creative Commons Attribution-Share Alike 3.0 Unported License.


How to measure amount of oxygen (or carbon dioxide) in air? I'm not chemist, and I'm looking for easiest way to measure how much oxygen plant in closed container will produce per hour [can measure over time for increased accuracy].

My first idea was to measure percentage of oxygen in air with automotive lambda sensor.

However I don't expect good precision of measurement like this, and this is a little complicated (lambda sensor needs high temperature, so I need to heat it up).

I wonder if there is some chemical way to perform measurement like this.

This is untested, but I think it might work. You could light some fuel on fire inside the closed container and see how much of it burns (just put the cup of fuel on a kitchen scale) before it dies out and than back calculate how much oxygen was present. Preferably you should take a fuel that burns easily to make sure that the total combustion reaction is the only reaction that occurs. Also if you want to repeat the experiment every hour or so, you want to be sure that the fuel does not evaporate (substantially) because that might cause the entire container to light up when you try to burn the fuel.

I am not really sure which 'household' fuel would be a good option, but maybe one of the other people on this site have suggestions?!

You can probably use one of those candle lighters and stick the long side through a small hole in your container and than seal the rest of the hole to be able to light a fire inside, without having air flowing in or out.

Just be careful not to set the plant (or anything other than the fuel) on fire!


What would happen if you heat volatile/flammable material in a vacuum? Fire is an oxidizing chemical reaction, in the simplest sense it needs oxygen and fuel to burn in order to stay alight. But what reaction would you observe from a flammable substance like gasoline or lighter fluid in a vacuum?

If there's any trace of oxygen, and a spark, the vapors would burn like normal.

If there isn't any oxidising agent, it'll just vaporize and stay that way (maybe the vapors will attain equilibrium with the liquid if the vapor pressure is high enough, basically if there's enough liquid to start with).

That's mostly it... Nothing spectacular :)


What is the reason behind the 2% increase in silver metal composition just after melting the raw silver alloy (55% silver)? I determined the silver composition from an alloy obtained from a jeweler to be 55% and wanted to extract silver out of it. However, upon melting around 5 kgs of alloy, the silver composition went up to 57%. I noticed similar increase for another 2 times (different alloys of silver, i.e. different samples)

Could it be something related to oxidation/reduction of silver or any metal resulting in the increase in composition while in the furnace?

What are some of the analytical tests out there I could use to accurately determine the elemental composition and the compounds present in the original alloy vs the one just after melting it?

Zinc is used in some silver alloys; Zn would be easy to lose into dross or vaporization.


Formation of a bubble which is not a sphere Recently a friend of mine received a project from her college which was like -  "Bring a bubble which isn't spherical in class. ".  Could anyone please suggest how to prepare such a bubble sufficiently big and which can sustain enough. So that she could bring in class.  I confirmed that a soap film(like a cylinder) could also do.
Please help me in suggesting a method and the liquid we should use to develop this bubble

Surface tension will relax most or all bubbles into a
spherical shape around internal pressure.  So, one solution is to create a situation where surface tension is nonuniform:a bubble adhering to the wall of a glass. Because the glass is not part of the liquid's surface, that
face of the bubble needn't be spherical.

Another is to modulate the pressure the film contains.  Two bubbles in collision is the
easiest way: if the bubbles are the same internal pressure, they
will form a flat wall where they meet.  If they do NOT have the same
internal pressure, the wall curvature is concave toward the
center of the smaller bubble.


Basics of wave-mechanical model of atom Why is it that for the $\mathrm{d}$ subshell we have  $\ \ \mathrm{d}_{xy}, \ \ \mathrm{d}_{yz}, \ \ \mathrm{d}_{xz}, \ \ \mathrm{d}_{x^{2}-y^{2}},$ and $\mathrm{d}_{z^{2}}$ orbitals only?
Why aren't there $\mathrm{d}_{y^{2}-z^{2}}$ or $\mathrm{d}_{y^{2}}$ or other orbitals similar to that?
Perhaps it is because we have an empirical evidence that there are only 5 orbitals in $\mathrm{d}$-subshell, in that case, why are $\mathrm{d}_{x^{2}-y^{2}}$ and $\mathrm{d}_{z^{2}}$ orbitals named in this particular fashion only? (as the axes are arbitrary and there could have been a different choice of axes also)

These orbitals represent the angular part of the wavefunction. The solution obtained directly from solving the Schrödinger equation produces equations containing complex numbers so cannot be drawn on normal $xyz$ axes and are hard to visualise.
The angular part of the wavefunction is given by functions called Spherical Harmonics these usually are given the symbol $Y$ and require two quantum numbers $\ell$ and $m_z$ making $Y_\ell^{m_z}$.
What is done to make the orbitals real functions, as opposed to complex ones, is to take linear combinations of the spherical harmonics, thus choosing the $z$-axis then the $\mathrm{d}_{z^2} = Y_2^0 = N(3\cos^2(\theta)-1)$) where $N$ is the normalisation constant. In cartesian coordinates ($xyz$) rather than polar ones (which are in $r,\;\theta,\;\phi$), $\displaystyle Y^0_2= N\frac{3z^2-r^2}{r^2}$ so you can see where the $z^2$ comes from and $r$ is the radial distance.
The other orbitals are also linear combinations e.g. $\mathrm{d}_{xz}=(Y_2^{1}+Y_2^{-1})/\sqrt{2}$ and in cartesian coordinates $\displaystyle Y_2^1 \sim \frac{z(x+iy)}{r^2},\; Y_2^{-1}\sim \frac{z(x-iy)}{r^2}$ so you can see where the notation $zx$ comes from and by making a linear combination the sum is no longer a complex number ($i=\sqrt{-1}$).
For completeness $\mathrm{d}_{yz}=(Y_2^{1}-Y_2^{-1})/\sqrt{2},\; \mathrm{d}_{x^2-y^2}=(Y_2^{2}+Y_2^{-2})/\sqrt{2},\; \mathrm{d}_{xy}=(Y_2^{2}-Y_2^{-2})/\sqrt{2}$  and in cartesians $\displaystyle Y_2^{\pm2}\sim \frac{(x\pm iy)^2}{r^2}$.


How can I update a map/mesh and coordinates in a PyMol session file? I have a PyMol session file I was using to create my figures.  It contains a protein and ligand broken into several groups of atoms with a fair bit of overlap.  There are also a few meshes built from a density map for a few figures.

Unfortunately, I wasn't quite done with refinement, and now Refmac has given me shiny new PDB and MTZ files with subtly different coordinates and terms.  Is there a way to "update" my coordinates (despite the myriad groups, all the atoms still have the same name) with the new PDB?  With all the groups I made that are shown/hidden/colored differently for several scenes to produce all the figures, I'd rather not recreate it all manually.

I'm more pessimistic about the map/mesh, but I created them via a script so I can repeat that.  Should I also script the coordinate to figure stuff in the future if it's not finalized?

Until someone identifies an ‘update’ function in PyMol, I think the next best thing is to use scripting. (See the pymol wiki) It is an imperfect solution, but it may work for the situation presented in the original post if the session may be reproduced.
To begin capturing the commands in PyMol, including menu selections to a script file, select:
File -> Log File -> Open -> myscript.pml

When done with creating the display panels, select:
File -> Log File -> Close

The input data files and the script itself may then be updated or replaced. In a fresh PyMol session, execute the script with:
File -> Run Script -> myscript.pml

To test the above, I generated a PyMol session where I captured a script as above. I loaded the atomic coordinates of a small protein, a ligand and a 2mFobs-DFcalc electron density map. Then I displayed some panels along with mesh surface around the compound. The co-structure was then re-refined, thus generating modified atomic coordinates and electron density maps. I replaced the original files with the updates and executed the script in a fresh PyMol session. The display panels were updated accordingly.
I recommend the Advanced Scripting Workshop.


Drawing the condensed structure of ethylcyclobutane The IUPAC name for this is ethylcyclobutane. I know that the formula is $\ce{C6H12}$.

I have no problems drawing the skeletal structure, but I have a little difficulty drawing the condensed structure. I know this is basic, but I just started learning how to name them.

I tried it below. For $\ce{CHCH2CH3}$, am I right to write it in a straight chain?

$\ce{CH3CH2CH(CH2)3}$ is the best you can do.

If you want to draw the cycle, then you should make it pendant, drwing a bond from the ring to the ethyl group.


Why are chemistry droppers bad at holding bromine? Question in my chemistry test last week. The question was related to a previous question on bromine's density.

Since the standard dropper is designed to hold common liquid such as water or ethanol, the capillary size have some particular size. When it comes to bromine, the density is 3 times of water. In other words, the dropper need to hold a super heavy droplet, so you could expect that it won't be as efficient as usual. In addition, your previous question hinted on the density, so it is pretty obvious right?


How to determine at temperature and pressure to keep Cl2 a gas? I have chlorine tanks and they will be used industrially. When I use them, they will be exposed to all sorts of varying temperatures. I was wondering if there was any way that when I use them, I could figure out at what pressure I could release the $\ce{Cl2}$, depending on the temperature, so that it remains a gas?

I already know it only takes $\pu{6.8 atm}$ at $\pu{70^\circ C}$ to liquefy $\ce{Cl2}$ to a volume of $\pu{0.322812 L}$, and its boiling point is $\pu{-33.97 ^\circ C}$ at $\pu{1 atm}$.

What you need to know is the vapor pressure of $\ce{Cl2}$ as a function of temperature.  Just ensure that the release pressure of the chlorine is lower than the vapor pressure, and it will be a vapor.

I found one web site that gives a graph (and a literature reference) of $\ce{Cl2}$ vapor pressure as a function of temperature from -30 °F to 220 °F.  The references they give are:

  Reprinted from Chlorine Manual, the Chlorine Institute, 3rd ed. 1959. Data for graph from R. Kapoor and J.J. Martin, Thermodynamic Properties of Chlorine, Eng. Res. Inst. Univ. of Michigan, 1957.

At -30 °F they report $\ce{Cl2}$ vapor pressure is "0" psi (which has to be an approximation; the vapor pressure of all materials is non-zero at any non-zero temperature) and at 220 °F it is 587 psi.


Why is the pressure value altered in this gas law numerical? When I work through the following problem it is unclear to me why 0.00120 mmHg is converted to 0.00332 mmHg.

  What is the temperature of 0.00332 g of mercury in the gas phase with pressure of 0.00120 mmHg and a volume of 435 L?

I converted the grams of mercury to moles and then used the ideal gas law constant of 62.36 to solve for T. Is there any reason that the pressure of 0.00120 mmHg should be converted to 0.00332 mmHg?

Your tutor is a master of confusion :-D

Would the exercise be easier if the pressure was given in stones per square acre? ;-)

Write down the equation for the ideal gas law again. Pay attention to the units!

Don't let the mercury fool you ;-)


Is it relevant to pursue a master degree or a doctorate when applying to a job in the petroleum field? I have a few more questions of this type for an interview. ( University work )
Thanks in advance.

Professional careers in the petroleum industry can be very competitive, so a higher qualification such as masters degree could make you a more attractive candidate for a scientific or engineering position. Many companies nowadays specify post-graduate qualifications as necessary or desirable.

A PhD is generally preferred for research positions.


Chirality of heteroatoms Why is a phosphorus atom with three different substituents chiral, but nitrogen isn't? Nitrogen inverts fast, while phosphorus obviously much slower. How is that explained?

Typical amines ($\ce{NR3}$) are roughly $\ce{sp^3}$ hybridized.  A chiral amine must undergo nitrogen inversion in order to racemize. The transition state for nitrogen inversion is a planar $\ce{sp^2}$ structure.  So in order to invert the ligands attached to nitrogen we must move from an $\ce{sp^3}$ ground state with ca. 109 degree bond angles to an $\ce{sp^2}$ transition state with 120 degree bond angles.

On the other hand typical $\ce{PR3}$ compounds are roughly unhybridized with 90 degree bond angles in the ground state.  In order to racemize they must also invert at phosphorous which again requires an $\ce{sp^2}$ transition state with 120 degree bond angles.  Changing the geometry from 90 to 120 degrees in the case of phosphorous requires more energy that a distortion from 109 to 120 degrees in the case of nitrogen.

P.S. Put a chiral nitrogen in a 3-membered ring and chirality can be preserved because you've now made it so much more difficult to expand the 60 degree bond angle in the ring to 120 degrees.


In the lattice model, what's a nearest neighbor? I was reviewing my grad Thermodynamics textbook to go over a concept that I never learned well: the Lattice model.  One set of equations1 has:
\begin{align}
zN_1 &= 2N_{11} + N_{12}\\
zN_2 &= 2N_{22} + N_{12}\\
\end{align}
where $z$ is the coordination number, or the number of nearest neighbors (defined twice).

My trouble is a simple thought experiment:

pure solution of $1$
$z = 10$
$N = 100$

So then the first equation becomes $10{\times}100=2{\times}N_{11}+0$, so $N_{11}=500$. Five-hundred nearest pairs, right?  But if I actually had a solution of 100 species 1 each with 10 neighbors, then I'd expect (ignoring boundaries) to have about $100{\times}10=1000$ pairs of $1-1$.

I can't seem to figure out what is meant by $N_{11}$.  Could someone explain it?

1 J. Prausnitz, R. Lichtenthaler, and E. Gomes de Azevdeo. Molecular Thermodynamics of Fluid-Phase Equilibria, 3rd Edition, p.331, Eq.7-68.

A nearest neighbour in general terms is literally that: Find the closest atom of any given element, that is your nearest neighbour distance for that element in the lattice. The number of nearest neighbours for that element is the number of atoms that are at this distance from your starting atom.

$N_{11}$ is the number of nearest neighbour pairs of atoms of element 1, and $N_{12}$ is the number of nearest neighbour pairs between any atom of 1 and any atom of 2 etc. For the $N_{11}$ or $N_{22}$ case, you have to consider that each atom of 1 or 2 in the equation has a contribution to $zN$, hence the term being multiplied by 2 when viewed from the perspective of element 1 or 2.

In the case of your thought experiment, your expected value neglects edges (you can't just neglect edges!) so will be too high by a good margin.

The dissertation of Alexander Zhou1 covers these equations (p.34) and explains them slightly differently (and more clearly I think) than de With's textbook.2

Alexander Zhou. Explicit Non-Random Contribution in Lattice Fluid Equation of State. Explizite nicht zufällige Wahrscheinlichkeitsverteilungen in gitterbasierten Zustandsgleichungen. Doktor Ingenieur, Technische Fakultät der Universität Erlangen-Nürnberg, Nürnberg, 2013. See UB Friedrich-Alexander-Universität Erlangen-Nürnberg (FAU)
Gijsbertus de With. Liquid-State Physical Chemistry: Fundamentals, Modeling, and Applications; John Wiley & Sons: 2013. See Google Books


Why doesn't D block contraction affect the other P block groups 14 15 16 etc Usually we come across anomalies in the trends of periodic properties for example the size [Atomic radius value from Raymond Chang Chemistry 9th Edition] of Gallium (135 pm) is smaller than Aluminium (143 pm) as opposed to the basic trend that atomic size increases down the group. This is usually attributed to the fact that the poor shielding of nuclear charge by intervening D orbitals leads to a greater $Z_\mathrm{eff}$ hence the radius is less as opposed to the trend.

If that's the reason why don't we observe similar effects for other groups like group 15 having Phosphorus (110 pm) and Arsenic (120 pm) [Here the size doesn't increase much but is still significant as compared to changes taking place in group 13] , where the size actually increases even when there are D orbitals to intervene. Or in short why are the effects so profound for group 13 and not the other members?

There's another example of In (166 pm) and Tl (171 pm). Both belong to group 13. As compared to other groups these two members have the lowest increase in atomic radius.

As a property showing d-block contraction, the question cited atomic radius. However, d-block contraction can manifest itself in other properties as well:

$\ce{GeO}$ is described as much more stable than $\ce{SiO}$. $\ce{Sn(II)}$ compounds are described as reducing. [Riedel]
The pentachlorides of $\ce{P, As, Sb}$ show a kink in their stability: $\ce{PCl5}$ melts at 167°C, $\ce{AsCl5}$ decomposes at $-50\mathrm{°C}$, $\ce{SbCl5}$ melts at 4°C (and decomposes at 140°C) [HoWi]
The redox potentials in acidic solution for the (0)/(IV) oxyanions and oxides of chalkogens show a peak at $\ce{Se}$ [HoWi]:

\begin{align}
\ce{SO2} + \ce{4e-} + \ce{4H+} &\leftrightharpoons \ce{S} + \ce{2H2O} \quad &E^0 = +0.50 \ \mathrm{V} \\
\ce{SeO_3^{2-}} + \ce{4e-} + \ce{6H+} &\leftrightharpoons \ce{Se} + \ce{3H2O} \quad &E^0 = +0.74 \ \mathrm{V}\\
\ce{TeO2} + \ce{4e-} + \ce{4H+} &\leftrightharpoons \ce{S} + \ce{2H2O} \quad &E^0 = +0.57 \ \mathrm{V}
\end{align}

the oxidative strength (at both low and high pH) of halogen oxyacids follows the pattern $\ce{BrO4-} > \ce{H_5IO6} > \ce{ClO4-}$. [HoWi]

I think that stability of oxidation states, stability of comparable compounds, and redox potentials of similar compounds are valid examples of d-block contraction. Of course, other chemists may arrive at different conclusions.

Literature:
[Riedel] E Riedel, Anorganische Chemie, Vol. 6, de Gruyter (2004).
[HoWi] AF Holleman, E Wiberg, Lehrbuch der Anorganischen Chemie, Vol. 101, de Gruyter (1995).


What test can be used to differ amide and amine? For aliphatic amide and amine. I know we can use $\ce{NaOH}$ because amide will give off $\ce{NH3}$, but what will amine yield if we test it with $\ce{NaOH}$?

Easiest way is to use TLC(thin layer chromatography) plate and stain with ninhydrin. Ninhydrin colors primary and secondary amines and amides only with vigorous heating when amide bond breaks.


Boys function for Gaussian integrals in ab-initio calculations A few days ago I mentioned a problem with an Hartree-Fock program I am writing (HF using cartesian Pople's STO-3G basis set). I can reproduce overlap and kinetic integrals of some references for ($\ce{H2}$, $\ce{HeH+}$ and $\ce{H_2O}$). Things start to change (at the third/fourth decimal digits) in the nucleus-electron attraction integral. At the beginning I thought the error I get on the total energy was mainly caused by electron-electron repulsion integrals (as errors on nucleus-electron attraction seem minimal). But now I am quite convinced the problem comes from my calculation of the Boys function
$$
    F_n(x) = \int_0^1 t^{2n}e^{-xt^2}\, dt.
$$
A clue of this comes from the fact that for $\ce{H2}$ and $\ce{HeH+}$ everything works fine: in this case Boys function is only computed for $n = 0$.

In order to compute this function I used numerical integration routines of SciPy (Python library) or directly QUADPACK (I traduced my Python program into FORTRAN90). However, looking at the literature it seems that nobody compute Boys function via numerical integration. I tried to implement different versions (recursive, asymptotic values, ...) of Boys function but all of them eventually had problems (depending on the range of $x$ or on the value of $n$).

Since I am loosing too much time on this problem, which seems to be numerical, I was wondering if there is some standard implementation of Boys function that I can simply use in my code.

Do you know any Python or Fortran library containing a good (safe and efficient) implementation of Boys function? If not, there is a simple algorithm that works well in any range of $x$ and is not sensible to numerical errors?

I am not aware of any existing Fortran code for direct numerical quadrature of this problem, but it is worth pointing out that Mathematica can perform this integral symbolically:

Integrate[t^(2 n) Exp[-x t^2], {t, 0, 1}]

(* 1/2 x^(-(1/2) - n) (Gamma[1/2 + n] - Gamma[1/2 + n, x]) *)

where the $\Gamma$ function can be computed by exponentiating easy-to-find $\ln \Gamma$ functions, often called lngamma().  (One of them is the Euler function $\Gamma(z)$ and the other incomplete $\Gamma(a,z)$ function).

As a check:

With[
  {x = 1, n = 2},
  1/2 x^(-(1/2) - n) (Gamma[1/2 + n] - Gamma[1/2 + n, x])
  ] // N

(* 0.100269  *)

and numerical quadrature yields:

With[
 {x = 1, n = 2},
 NIntegrate[t^(2 n) Exp[-x t^2], {t, 0, 1}]
 ]

(* 0.100269  *)

One has to be a little bit careful, as the $\Gamma$ function is closely related to the factorial, and it can explode as the arguments get too large.  I would experiment and see whether you can get by with this analytic solution, which essentially takes you from numerical quadrature into evaluation of special functions.

The $\Gamma$ function approach is faster by over two orders of magnitude  than the quadrature approach.  This is critical when evaluating molecular integrals.  I have seen implementations where the Boys function is actually interpolated rather than evaluated, and there is discussion about the error in interpolation being almost insignificant if suitable interpolation basis functions are used.


Is there a name for this algorithm to calculate the concentration of a mixture of two solutions containing the same solute? There is an algorithm called "Mischungskreuz" (German for "x of mixing") that is sometimes taught as a shortcut to figure out the following problem:
You have two solutions that contain a solute at different concentrations $c_1$ and $c_2$. At what ratio $V_1/V_2$ do you have to mix them so that the mixture has the desired concentration $c_m$?
For example, let's say you want to make a 22% solution from a 35% and a 15% solution. You write the desired concentration in the center and the available concentrations at the left ends of the "x", and get the ratio of volumes on the right side of the "x" as shown below:

So in this case, mix 7 parts of 35% with 13 parts of 15% solution to get the desired 22%.
Source: https://de.wikipedia.org/wiki/Mischungskreuz
My questions are: Is this method taught outside of Germany, and is there a non-German (maybe English?) name for it?

This is a so-called "Pearson's square" or "Box method" of balancing ratios, originally used extensively in dairy industry (at least since 1920s judging from Google Books search).
Earlier the similar approach has been used in sugar industry by using "Cobenz diagrams" aka spider diagrams.

Widely popularized in Soviet books for analytical chemistry at least since 1940s (probably adapted from the German literature as many other tech novelties of that time were), also used in current Russian literature by the names "Метод креста" ("Cross method"); "Конверт Пирсона" ("Pearson's envelope") or "Диагональная схема правила смешения" ("Diagonal mixing rule scheme").


Quotable Literature for the Enthalpy of a Proton A proton obviously has no electronic energy, no vibrational and no rotational degrees of freedom. Therefore I think it is fair to assume, that $$H(\ce{H+})=\frac32\cdot \mathcal{R}\cdot T$$ holds for the translational degrees of freedom. I am looking for some quotable literature where they state that, or if that has not been stated anywhere, how to derive this.

I am doing quantum chemical calculations, i.e. gas phase. I define the proton affinities via
$$\text{PA}(\ce{M})=-\left(H(\ce{H+M})-\bigr[H(\ce{M})+H(\ce{H+})\bigr]\right)$$
for the reaction
 $$\ce{H+ + M -> H+M}.$$
This is respecting the (quite common) sign convention to show the energy released upon binding a proton.
I need a reliable source or at least a best-practice example(s) for $H(\ce{H+})$.

A quite complete procedure in the calculation of proton affinities from ab initio calculations is referenced in an article by H. Audier et al., Int. J. Mass Spectrom. 1996. It goes much further in the procedure than only on the calculation of the enthalpy of a proton (including corrections due to the basis set superposition error and so on), but confirms that you are right in assuming that the enthalpy of a proton is $3/2 RT$.


How does sulfur has atomicity 8 How does sulfur has atomicity 8, when its valence electrons are 6 like oxygen- ie even 2 sulfur atoms can share a pair of electrons and achieve the octet rule, why should it involve 8 atoms?

Octasulfur is just one of few different allotropic forms of sulfur. It happens to be the most stable form at room temperature, but different conditions favor different forms. Thus, for instance, sulfur vapor usually contains $\ce{S, S2, S4, S6, S8}$, and the higher the temperature, the lower is the amount of $\ce{S8}$ molecules since they break up into smaller allotropes of sulfur.

So, while two sulfur atoms can share a pair of electrons by forming a diatomic disulfur molecule with a double bond, so can also eight atoms, but by forming a puckered ring structure with few single bonds. Then it is basically the question of thermodynamic and kinetic stability, which form would dominate others at some particular conditions.


Does silver oxide react with hydrogen sulfide? Electronic connectors are often silver plated. However, silver tarnishes fairly quickly and heavily. There exists a widespread misconception that the tarnishing of silver contacts is harmless, because silver oxide has about the same conductivity as silver itself. The problem however is that silver does not oxidize under normal conditions. The tarnish on the contacts is not silver oxide, but silver sulfide that develops due to the presence of some hydrogen sulfide in the air. Unlike silver oxide, silver sulfide is not a conductor, but a semiconductor with various potential adverse effects for the connection.

I have come across a reference that suggests oxidizing silver contacts before using them in order to prevent the development of the silver sulfide layer. This implies that silver oxide does not react with hydrogen sulfide in the air under normal conditions. Is this claim correct?

EDITS

I have corrected the typo by changing "sulfur dioxide" to "hydrogen sulfide" in the title and body of the question. Thanks for pointing this out in the answer!

Silver sulfide is formed by Ag and hydrogen sulfide not sulfur dioxide. You need reducing conditions where the latter can be reduced to

$$\ce{SO2 +reduction -> H2S}$$

Actually, sulfur dioxide chemisorbs on ultraclean silver surface, however heating can remove it. So this is reversible sorption, as suggested by Lassiter [1].

Just note that Auger electron spectroscopy is done under extremely clean environment. There is no trace of water, oxygen or any other component! Real atmosphere is far more complicated and tons of photochemical reactions occur in the atmosphere. A typical indoor air has plenty of undesirable components. How does $\ce{SO2}$ react with Ag must be another story because we cannot avoid or control other factors.

Coming to the second part of the query: If we have surface layer of silver oxide, will it prevent sulfide formation. In principle, possibly yes, because $\ce{Ag2O}$ is decent oxidizing agent. The moment traces of $\ce{H2S}$ come in contact with the oxide, it will reduce the oxide to elemental silver. This is my personal speculation.

As per Franey et al. [2]:

  Polycrystalline silver has been exposed to the atmospheric gases $\ce{H2S}$, $\ce{OCS}$, $\ce{CS2}$ and $\ce{SO2}$ in humidified air under carefully controlled laboratory conditions. $\ce{OCS}$ is shown to be an active corrodant while $\ce{CS2}$ is quite inactive. At room temperature, the rates of sulfidation by $\ce{H2S}$ and $\ce{OCS}$ are comparable, and are more than an order of magnitude greater than those of $\ce{CS2}$ and $\ce{SO2}$. It appears that $\ce{OCS}$ is the principal cause of atmospheric sulfidation of silver except near sources of $\ce{H2S}$ where high concentrations may render the latter gas important. At constant absolute humidity, the sulfidation rate of
  silver by both H2S and OCS decreases from 20 to 40 °C and then increases to 40 to 80 °C.

So may hydrogen sulfide may not be a major culprit!

References

Lassiter, W. S. Interaction of Sulfur Dioxide and Carbon Dioxide with Clean Silver in Ultrahigh Vacuum. J. Phys. Chem. 1972, 76 (9), 1289–1292. https://doi.org/10.1021/j100653a011.
Franey, J. P.; Kammlott, G. W.; Graedel, T. E. The Corrosion of Silver by Atmospheric Sulfurous Gases. Corrosion Science 1985, 25 (2), 133–143. https://doi.org/10.1016/0010-938X(85)90104-0.


Why oxygen leaks from tires faster than nitrogen? In critical applications tires are inflated using pure nitrogen (link).

  Nitrogen molecules have a more difficult time escaping through the
  microscopic spaces that exist between a tire's rubber molecules. [...]
  Nitrogen reduces the loss of tire pressure due to permeation through
  rubber over time by about 1/3. (source)

According to WebElements, nitrogen bond length (109.76 pm) is smaller than oxygen bond length (120.74 pm), so why nitrogen permeation through rubber is said to be lower than oxygen?

The rate of oxygen, $\ce{O2}$, permeation through tire rubbers is about 3.4 times faster than nitrogen.  The reference for this is "Permeability Properties of Plastics and Elastomers" by L. W. McKeen.

The reason is that $\ce{O2}$ is a smaller molecule than $\ce{N2}$.  The size of a molecule is not determined by only the bond lengths between atoms but the size of the electron cloud around the atoms.  The kinetic diameter of $\ce{O2}$ is 0.346 nanometers and $\ce{N2}$ is 0.364 nanometers.  So, $\ce{N2}$ is larger than $\ce{O2}$ even though the $\ce{N-N}$ bond length is shorter than the $\ce{O-O}$ bond length.


Why is a singlet state called singlet and a triplet state called triplet? I kind of get the idea of singlet and triplet states. But why are they called singlet and triplet (what is the single and what is the triple in these cases)?

I feel that I am missing something obvious!

The terms arose back in the early days of quantum physics when spectral lines that were expected to be singlets were actually observed to be more complex (doublets, triplets, etc.).

An electron can have a spin quantum number of $+\frac12$ or $-\frac12$.  For a system that exists as a singlet, all spins are paired and the total spin for the system is $S=0$.  If we have a single electron, $S=\frac12$.  If we have a triplet system with 2 unpaired electrons, $S=1$.  Also associated with these electrons/ systems is a spin-angular momentum vector, $L$.  Quantum mechanics tells us that $L$ is allowed to have $2S+1$ distinct values.  So

for a singlet, $L = 2(0) + 1 = 1$     only one state exists (singlet)

for a doublet, $L = 2\left(\frac12\right) + 1 = 2$   two states exist (doublet)

for a triplet, $L = 2(1) + 1 = 3$     three states exist (triplet)


Relative strengths of methyl, primary, secondary, and tertiary C–H bonds Which of the highlighted C–H bonds in the following compounds is weakest? (R = generic alkyl group)

I personally thought that the answer was 1 but it is given as 4 in the answer key. I didn't understand how it is so.

$\ce{X–H}$ bond dissociation energies usually correspond inversely to $\ce{X^.}$ radical stability, so another way of looking at it is to recall that the stability of radicals increases in the order methyl < primary < secondary < tertiary.

The root cause of this is hyperconjugation (from $\sigma_\ce{C–H}$ to singly filled $\mathrm{p}$ orbital on carbon), analogous to the stabilisation of carbocations. So, it is just the second side of the same coin, really.


Physical significance of magnitude of ΔG (change in Gibbs free energy) of a reaction In thermodynamics, we know that if $\Delta G$ of a reaction is negative, the reaction is spontaneous.
Suppose we are given with two spontaneous reactions ($\Delta G\lt0$) with corresponding magnitudes of $\Delta G$. Can we extract any other information from these magnitudes?

It is only a rule of thumb that a negative value for $\Delta G^0$ of a reaction means that a reaction is spontaneous.  If you start out with pure reactants (and no products present), all reactions are spontaneous, irrespective of the $\Delta G^0$.  A negative value for $\Delta G^0$ only signifies that the equilibrium constant is greater than 1.


Remove formalin from food using chemical found in kitchen In some parts of the world food is stored with formalin so it looks fresh forever! This is astonishing, but true (See refs here). As formalin is very bad for health and damages kidneys and liver very quickly, I am looking for a way to remove it.

When I buy fish, I know it has formalin in it. Every fruit I bought has formalin. I remember I bought 2 kilgrams of apples once. And it lasted for 1 month, looking fresh whole time. Then I threw it away.

My question is: how can you remove formalin from food? (Especially fish and fruit). It's best if we can use a chemical found in the kitchen. If not, I'd think one that is found at a chemical store.

While the idea of filtering formaldehyde from food might sound naive, the motivation is understandable and the problem apparently real.

I've found at least one publication on the topic:

T Yeasmin, *M S Reza, M N A Khan, F H Shikha and M Kamal

PRESENT STATUS OF MARKETING OF FORMALIN TREATED FISHES IN
DOMESTIC MARKETS AT MYMENSINGH DISTRICT IN BANGLADESH

Int. J. BioRes. 2010, 1, 21-24

From the abstract:

  [...]Study showed that formalin was not detected in any
  fish produced locally, but was detected in the imported ones of catla and rohu ranging from 0.5% to 1% which was
  sold in different markets of Mymensingh with comparatively lower price than those produced locally. The shelf life of
  the locally produced fish was much longer than those of imported fish which may be related to the loss of shelf life
  during its transport from importing countries. [...]

Concerning the initial question:
I'm afraid but i don't see any chance to remove formaldehyde from food without turning it to another toxic cocktail.

But since formaldehyde was mostly used to increase the shelf time of imported fish, buying from markets predominantly selling local products might be a way--if available and affordable.


How tartaric acid gives Tollen's test? I wanted to know how Tollen's test is given by tartaric acid. Google searches say that it belongs to some "Hydroxyl" group. I know that it oxidises aliphatic and aromatic aldehydes along with α-hydroxy ketones and some other compounds such as alkynes and it's mechanism.
Can someone help me out in the mechanism of oxidation or provide some more detail on this reaction?

Maxted[1] reported in 1926 that the reaction of tartaric acid with silver produces oxalic acid, formic acid, CO2 and water as products and that the stoichiometry is 6 moles of silver reduced per tartartic acid. He proposed the following possible net reactions:

$\ce{C4H6O6 + 6 Ag+ + 2H2O -> 2(C2H2O4) + 6 Ag + 6 H+}$

$\ce{C4H6O6 + 6 Ag+ + 2H2O -> C2H2O4 + CO2 + HCO2H + 6 Ag + 6 H+}$

$\ce{C4H6O6 + 6 Ag+ + 2H2O -> 2(HCO2H) + 2CO2 + 6 Ag + 6 H+}$

For the mechanism, he concluded that initial decarboxylation to 2-hydroxy-3-oxo-propionate or splitting into two glyoxylic acid molecules were both possibilities. The glyoxylate could be oxidized to oxalic acid or oxidatively decarboxylated to formic acid, while the 2-hydroxy-3-oxo-propionate could split into glyoxylic acid and formic acid.

I'm sure someone must have done more recent mechanistic work on this to pin it down further, but that gives a pretty good idea of the possibilities.

[1] Maxted, D.R. (1926) CCLXXXVIII-The Oxidation of Tartartic Acid by Solutions of Silver Salts; J. Chem Soc. 129:2178.


Is there a way to make a liquid cold by a chemical reaction? I'm looking for a way to combine two substances in a chemical reaction that uses a lot of energy, therefore absorbing heat, to make a substance cold.

I'm thinking about a two-chamber thermos that in the middle chamber is, for example, juice and in the outside chamber will be placed the other substances. When these substances react, is it possible that they absorb heat, making the juice colder?

Ammonium thiocyanate and barium hydroxide octahydrate when mixed will draw a huge quantity of heat. This reaction often used as a chemistry demo of endothermic reactions where the chemicals are mixed in a beaker that's placed on a wet piece of wood. The wood freezes to the beaker.


Why does alcohol crack acrylic (plexiglass)? I have seen videos and discussions about alcohol cracking acrylic, and waterblock manufacturers reminding users to not use alcohol in cooling systems with plexiglass acrylic, but why is this so?

The question proposed as duplicate only asked for the effect, but here I already know the effect. I am asking for the reason for the effect, which is not answered in the answer to the 'duplicate' question.

If a solvent causes an amorphous polymer to craze, this is because the polymer is principally soluble in it. It diffuses into the surface, but without instantly dissolving it, as the polymer chains are still stuck in the glassy network of the substrate. The polymer swells (like a fruit gum in water), but the volume increase can only happen perpendicular to the surface.

The polymer chains thereby end up oriented towards the surface, weakening the structure. When the solvent dries off again, the chains adhere to each other again, but cannot reentangle in time. The surface has become thicker now, but as it is still the same amount of polymer, the additional volume is made up by the cracks and crazes appearing in it.


Number of Stereoisomers for Terpin Terpin (4-(2-hydroxypropan-2-yl)-1-methylcyclohexan-1-ol) has two $\mathrm{sp^3}$ carbons with different substituents on a cycloalkane, so it has two cis-/trans-isomers, and from what I was taught, is a stereoisomer.

The way I was taught to find stereoisomers was to count the number of double bonds that can be E/Z and the stereocentres, and do $2^n.$ There are no stereocentres, and no double bonds, so the number of stereoisomers would only be 1 (only one molecule). This disagrees with the cis-/trans-isomers number which would be 2.

I can't find anything online about these kinds of cases. Is this general rule of counting stereocentres and double bonds with E/Z not apply in every case? And is this a case that happens a lot that when I'm solving problems, I should always keep this in mind?

The simple rules don't work with complicated examples. There are non-chiral molecules with "stereocenters", and chiral molecules lacking "stereocenters". The textbook example for your question is 1,4 dichloro cyclohexane (or 1-chloro 4-bromo cyclohexane to get even closer to your example):

William Reusch explains this and other cyclohexane cases here: https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Chirality/Stereoisomers/Stereoisomerism_in_Disubstituted_Cyclohexanes

  The way I was taught to find stereoisomers was to count the number of double bonds that can be E/Z and the stereocentres, and do $2^n$.

You established that this rule does not work here. If you added an oxo group to one of the ring carbons (turning it into a ketone and making the two branches of the cyclohexane different from one another), you would get the expected pattern: two stereocenters, four stereoisomers.

I found a more exact formulation of the rule on another one of William Reusch's pages:

  [William Reusch] As a general rule, a structure having $n$ chiral centers will have $2^n$ possible combinations of these centers. Depending on the overall symmetry of the molecular structure, some of these combinations may be identical, but in the absence of such identity, we would expect to find $2^n$ stereoisomers.

Source: https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Chirality/Stereoisomers/Compounds_with_Several_Stereogenic_Centers

So what about 1-chloro 4-bromo cyclohexane or terpin? If you use the following definition of stereocenter, both have two stereocenters:

  [Gamini Gunawardena] If the interchange of two ligands on an atom in a molecule results in a stereoisomer of the molecule, the atom is called a stereocenter or stereogenic center.

Source: https://chem.libretexts.org/Ancillary_Materials/Reference/Organic_Chemistry_Glossary/Stereocenter

You have 4 combinations, but there are pairs of identical molecules when you consider the two chair conformations inter-converting.

Other examples where the simple $2^n$ rule does not work are the following:

Meso compounds: Stereochemistry of meso compounds
Axial chirality: M/P nomenclature for axial-chirality
Planar chirality: Why is trans-cyclooctene chiral?
Molecular knots: Molecular knots absolute configuration


Why is it so hard to turn plastics into something useful or decompose it? Even after so many years, why haven't we yet come up with a solution to the plastic problem ?

Why are plastics so hard to decompose via chemical or physical processes into something harmless or useful ?

Most plastics are reusable . The problem is wastes are mixes of various different plastics and other materials . The cost to reliably separate them into specific materials is greater than the cost to make new plastic.


Finding original concentration w%/w% A question from a grade 9 admission exam:
When 1 kg of salt is added to a solution of salt
and water, the solution becomes 33 1/3% salt by
mass. When 1 kg of water is added to the new solution, the resulting solution is 30% by mass. The percentage of salt in the original solution is...
I tried:
(Salt orig +1)/(Water orig + salt orig + 1) = 33 1/3. (S + 1)/(W+S+2) = 30. Which means 33 1/3*(W+S+1)=30*(W+S+2); 33 1/3 W+33 1/3 S +33 1/3 = 30W + 30S + 60; 3 1/3 W + 3 1/3 S = 26 1/3; S+W = 79. Putting that into the 1st eq gives S+1 / 80 = 33 1/3 meaning S= 26 2/3. Solving for W = 79-26 2/3 = 52 1/3. So the original concentration = 26 2/3 / (52 1/3 + 26 2/3) = 33 3/4%. But that's not one of the multiple choice answers (I did get further this time!).


SCF (Roothan-Hall) Method, why Density Matrix? I've been aksing myself, when I try to variationally minimize a Hartree-Fock energy by using the Roothan equations, what I do is to set a Fock-Matrix, then transform it, then diagonalize it so I can read the energy from the diagonal elements and then I test for convergence. But why is usually the density matrix compared to the previous run and not just the sum of the diagonal elements? I know in the end they both have to be evaluated but just for testing for any changes, pretty much any parameter, which changes with the Fock-Matrix and has some relationship with the energy could be used as a criteria for convergence. Or where do I see something wrong here?

When doing the SCF calculation taking the RMSD of the density matrix compared to taking the RMSD of the diagonal of the density matrix takes about the same with respect to the total calculation time. The comparrison of the density matrices is not the time consuming step in the SCF calcultion. When you take the RMSD of the total density matrix you also "keep an eye" on the off diagonal elements, this is a more rigid test for convergence.


Plotting ΔG vs Temp for Electrochemical Reaction I’m doing an electrochemistry lab that requires me to construct various galvanic cells in an attempt to solve for $\Delta H$, $\Delta G$, and $\Delta S$ values of an unknown. During one part of the experiment, I had to heat a cell to a temperature of 70 °C and record its voltage in 15 degree increments until the temperature reached 10 °C.

The cell I am using to perform this step on is represented by the equation:

$\ce{Zn + Sn^2+ -> Zn^2+ + Sn}$

After performing the procedure, I’m getting concerned that my data may be inaccurate. Theoretically, I was using 0.1 M concentrations of each solution in the cells. According to the Nernst equation, it seems that the voltage or potential of the cell should not be affected by a change in temperature if the concentrations are equal because $\ln(0.1/0.1) = 0$.

However, according to my professor, I should be seeing a linear relationship when I plot $\Delta G$, calculated from the voltage measurements I took, vs the temperature in K.

My question is, theoretically, what should happen to the voltage as I heat the cell, assuming that the concentrations of the solutions were off? I’ve calculated the $\Delta S$ value based on the standard values listed for these entities and I’m getting a value of −85.18 J/(K mol).

Calculation: $\Delta S = (-112.1 + 51.55) - (41.63 + -17.0) = -85.18\ \mathrm{J/(mol\ K)}$

Based off of this result, it would appear that the slope of this plot must come out to be a positive value because $m = - \Delta S$. Does this mean that I should be seeing an increase in voltage as temperature decreases? Regarding the $\Delta H$ of this reaction, which is represented by the Y-intercept (b) of the equation of the best fit line for my data, it appears I must also have a negative value of −145090 (in units of J) based upon the calculation I did for this, also using standard values.

Calculation: $\Delta H = (-153.89 + 0) - (0 + -8.8) = -145.09\ \mathrm{kJ/mol}$

I’ve tried to be as detailed as possible in describing by question, and hopefully it makes sense. I’m just getting confused on what the data should look like. Unfortunately, its not possible for me to redo the experiment. Please note that this is for my own understanding, I will not submit the assignment with fixed data. I appreciate any assistance that anyone can offer me.

My question is, theoretically, what should happen to the voltage as I heat the cell, assuming that the concentrations of the solutions were off?

This question is tough to answer because I don't know what you mean by "off".  So let me ask (and answer) a few related questions that I think will help you.

What should happen to voltage as a function of temperature in an electrochemical cell

The electrical potential of electrochemical cells varies with temperature, even when the cell is at standard conditions.  A 1985 NIST document explains the situation well.  Only the standard hydrogen electrode has a standard potential that does not vary with temperature.  (The SHE potential is defined as 0 volts at every temperature.)  For other cells, usually the assumption of a linear variation with temperature is usually very good.  That is,

$${\cal E}_T^\circ = {\cal E}_{298}^\circ  + (T-298.15)\cdot \frac{d{\cal E}_{298}^\circ}{dT}$$

Note that this variation is completely unrelated to the Nernst equation.  The same reference gives values for $\frac{d{\cal E}_{298}^\circ}{dT}$ for zinc(II)/zinc metal of 0.119 milliVolts per Kelvin and for tin(II)/tin metal -0.32 milliVolts per Kelvin.

Don't be confused by the $T$ in the Nernst equation

The Nernst equation is

$$ {\cal E}= {\cal E}^\circ + \frac{RT}{nF}\ln Q$$

It does contain a $T$ in there, but don't be mislead into thinking that the $T$ you see is the sole means by which a cell potential ${\cal E}$ can vary with temperature.  ${\cal E}^\circ$ also varies with temperature (see above).

  According to the Nernst equation, it seems that the voltage or potential of the cell should not be affected by a change in temperature if the concentrations are equal because ln(0.1/0.1)=0.

Thus, hopefully my answer makes clear that this statement is wrong, because ${\cal E}^\circ$ varies with temperature.


Acids, Bases and Salts? I'm a 7th grader, and this is what is printed in my Chemistry textbook:

An Acid is a chemical compound that is sour in taste and is soluble in water.
A base is a chemical compound that is bitter in taste.
A salt is a compound that is form alongside water when an Acid and a Base reacts.

Come on. That can't be what Acids and Bases are! What is an Acid, a Base and a Salt?

I will try to make the answer as non-technical as possible, to make it understandable for you.

Some substances are composed of atoms, like $\ce{H}, \ce{O}, \ce{C}, \ce{Fe}$, etc. Each of these atoms actually exist out of even smaller particles, called protons, neutrons and electrons. Protons are positively charged ($+1$), neutrons are, as you might have guessed, neutral, and electrons are negatively charged ($-1$).

The protons and the neutrons are together in the core of the atom, the electrons are in a cloud around it. Because atoms are not charged, an atom has equally many protons as neutrons. A picture will help: (Source: User:Fastfission at Wikipedia)

However, in some cases, those atoms take an extra electron or give a electron to another atom. They are then called ions. For example, a normal natrium atom $\ce{Na}$ has 11 protons and 11 electrons, but a natrium ion $\ce{Na^+}$ has 11 protons and 10 electrons. Similiarly, a normal chloride atom $\ce{Cl}$ has 17 protons and 17 electrons, but a chloride ion $\ce{Cl^-}$ has 17 protons and 18 electrons.

Now we know enough to define what a salt is: A salt is a praticle that is is composed of one or more positively charged ions and one ore more negatively charged ions. For example $\ce{Na^+Cl^-}$, more commonly just written as $\ce{NaCl}$, is common salt, i.e. the kind of salt people do on some of their food (at least in some countries).

There are two special ions, $\ce{H+}$ and $\ce{OH-}$. If you have a liquid, then these two are inside. Now, we call something an acid if there is more $\ce{H+}$ inside than $\ce{OH-}$, and we call something a base if there $\ce{OH-}$ inside than $\ce{H+}$.

I'd also like to note that the book definition is dangerous: An Acid is a chemical compound that is sour in taste. Acids, or chemical compounds in general, are of course not things you should taste to test what they are! They might be poisonous.


Claisen rearrangement in substituted ring This is an example of Claisen rearrangement, property of allyl vinyl/phenyl ethers.

I believe that the allylic migration should take place to the position ortho to both $\ce{-CH3}$ and $-\ce{OR}$ because:

There's more electron density there due to the presence of electron donating group $\ce{-CH3}$

and I read this in March under Claissen rearrangement:

  Since the Claisen rearrangement doen't involve ions, it shouldn't greatly depend on the presence or absence of substituent groups on the ring. This is the case. Electron donating groups increase the rate and electron withdrawing groups decrease it, but the effect is small.

So I expect the product to be 2-[(2E)-but-2-en-1-yl]-3-methylphenol:

However my teacher made us write the product to be 2-[(2Z)-but-2-en-1-yl]-5-methylphenol:

His rationalisation was that the carbanion like species attacking the allyl group is more stable at the place away from methyl group.

But my question is that if this reaction is single step, followed by tautomerism why should it even depend on the carbanion like species which isn't even formed.

Another great example of how it's important to look at actual evidence. There have been various studies on Claisen rearrangement selectivity, here's a combined experimental/theoretical one on meta-substituted ethers specifically: Gozzo, F. C.; Fernandes, S. A.; Rodrigues, D. C.; Eberlin, M. N.; Marsaioli, A. J. Regioselectivity in Aromatic Claisen Rearrangements. J. Org. Chem. 2003, 68 (14), 5493–5499. DOI: 10.1021/jo026385g.

An important screenshot is shown here:

Based on this article at least, it turns out that for your specific reaction, it's likely that the regioisomers would be formed in equal amounts. (See allyl aryl ether 1). We further see that sterics are not as significant of a consideration as electronics; allyl aryl ether 6 actually prefers the "more sterically hindered" product 14 to 1.

The article concludes that "the preferred conformation of the reactants ... with no exception, was found to lead to the major isomeric product.... The regioselectivity for the Claisen rearrangement of meta-substituted allyl aryl ethers is mainly influenced, as expected, by the electronic effects of the meta substituent, which governs the relative stability of conformers, transition states, and intermediate to final products.


Why did my milled powders catch fire? I milled a small amount of the sample ($75\%$ Al, $15\%$ Cu, $10\%$ Ni) mixed with toluene for 5 hours after which I set it on butter paper to dry. Then after 13 minutes I switched on the fan in the room, at which point my powder caught fire.

Could you please explain why my powder would have caught fire?

They caught fire because partly because they were milled in toluene and partly because they were such fine powders. As you probably know aluminum forms a passivating oxide on the surface. When you mill the powder, you increase the surface area which will oxidize when exposed to air. You milled it in toluene so oxidation did not occur during and after you were milling as long as it was covered in toluene.

$$\require{cancel}\begin{align}\\
\ce{Al (s) + O2 (g) &->  Al2O3 (s)} \qquad \qquad{\mathrm{\Delta H} = \pu{-1675.7 kJ mol-1}}\\
\ce{Al (s, toluene) + O2 (g) &\cancel{\ce{->}} Al2O3 (s)}
\end{align}$$
Once you filtered off the toluene parts and let it sit the toluene started evaporating and the aluminum started oxidizing. Once you turned the fan on you allowed the reaction to happen much faster to the point that it could not cool as it oxidized and the powder ignited. This is similar to smoldering coals, they are not glowing, but hot and still combusting. When you blow air on them the rush of oxygen will make them hot enough to glow again. This is essentially what turning the fan on did to your powder.


Volume changes when mixing DMSO and water Since I know that a mixture of 500ml $\ce{H_2O}$ and 500ml ethanol gives less than a liter of mixture I wonder how strong this effect is when mixing DMSO and $\ce{H_2O}$. I'd assume it's nonzero - but is it negligible?

If we assume that the density of the mixture is just the average of the individual densities, would get 1 litre of a mixture with a density of 1.04635 kg/l.
However, measured density (1, table 1) is 1.0682.

We thus expect to see a volume that is 0.9795 times the added individual volumes.

In this case: 0.9795 litres.

LeBel, R. G.; I. Goring, D. A. Density, Viscosity, Refractive Index, and Hygroscopicity of Mixtures of Water and Dimethyl Sulfoxide. J. Chem. Eng. Data, 1962, 7 (1), 100–101


How come the same reaction can be balanced in two distinct ways? I faced this reaction which can be balanced in $2$ distinct ways which are not multiples of each other.

$$\begin{align}\ce{6XeF4 + 12H2O &-> 2XeO3 + 24HF + 4Xe + 3O2}\tag{1}\\[0.4em]
\ce{4XeF4 + 8H2O &-> 2XeO3 + 16HF + 2Xe + O2}\tag{2}\end{align}$$

The first reaction is as it is given in the books and the second reaction was balanced by me.

I sense that only a mechanism can resolve this. I found a similar question on googling, but the answer is unsatisfactory. (It's Yahoo Answers. What can you expect?)

I also found two independent reactions here which on adding produce equation $(2)$. I am confused as to what exactly happens.

As per Nicolau's suggestion, :

$$\{(1-2)\times\color\red{4}\}-2$$

gives

$$2\ce{XeO3}\to2\ce{Xe}+\ce{3O2}$$

Thus, I conjecture that the second and third reaction differ by the fact of decomposition of $\ce{XeO3}$

$2$ can be converted into $1$ by multiplying by $\color \red{\frac 3 2}$ and decomposing one of the $\ce{XeO3} $ molecules.

Wikipedia also says that $\ce{XeO3}$ is unstable.


What do atoms look like? A professor of mine noted that when he was in school, microscopes weren't powerful enough to resolve certain things (I forgot what it was). But current microscopes are powerful enough.
Extrapolating from this trend, he hypothesized that by the time we were his age, we would be able to see atoms.

Wait. Seeing atoms? Do we expect to see discrete little dots for electrons (I don't think so)? What exactly can we expect to "see," if anything at all?

Yes, we can definitely see atoms in modern microscopes.  I'll try not to get deep into the subject of "seeing", but I will note that most people say they are seeing when they use their eyes to observe objects.  That is, they are using an optical system (eye) consisting of a lens and a detector (retina) and some image processing carried out by a computer (brain).  Thus, any instrument which uses some sort of detector and a computer to turn the signals of that detector into an image we feel comfortable looking at qualifies to me as "seeing in a microscope."  If you're comfortable with that then read on.

My favorite microscope which is capable of seeing atoms is the transmission electron microscope (TEM).  Using a high quality TEM, we can easily resolve objects around an angstrom in size, which is the distance between atoms.  To image a single atom, you also need to have a sample which is only a single atom thick at some point.  This is tough to accomplish but can be done.  Today, images of graphene provide arguably the best "single atom" TEM images.  A seminal paper is here (paywall).  You can also search the internet for graphene images such as this.

In the links above, the images show the shapes of atoms as little blobs, and to a large extent the blobs are more a description of the microscope's ability to focus rather than the shape of the atom itself.  However, the atomic force microscope (AFM) excels at seeing the electrons at the surface of an atom.  With AFM, researchers are even able to image the chemical bonds of molecules sitting on a substrate.  In this case, the image shows the bonds as rods and the atom centers are not really visible.  Chemistry textbooks often show the "shapes" of atoms by drawing the electron orbitals, so in a sense these rods are the shapes of these atoms too.

This just points up that the "shape" of atoms is a non-trivial subject, which is very dependent on how you look at them.


What is the reason for decomposition of layered cathode materials? Articles like this claim oxygen can evolve from layered metal oxide cathode materials when charging the Li-ion battery.
Is it the high positive potential given to the cathode while charging pulling the electrons from the chemical bonds of the cathode materials and thus cause the decomposition to evolve oxygen?
If not, what is the cause of the evolution of oxygen?

What is the " ... Article like this ..." that you write at the very beginning of your post ?
Then, try to be clear in using the word "cathode". A cathode is the electrode where a reduction happens, whatever its working mode (cell or electrolysis). So when recharging a Li-ion battery, the $\ce{Li+}$ ion are going to the cathode to be reduced : reduction happens on the Li-graphite electrode (cathode). And no Oxygen can be formed on this electrode. Oxygen may be produced by electrolysis at the other electrode made of Cobalt and Manganese oxides. These oxides are oxidized to a higher number of oxidation. But of course, these oxides are not vrery stable, they may lose part of their oxygen as $\ce{O2}$.
To summarize, the Lithium electrode is always the negative pole, whatever its working mode. The Lithium atom works as an anode when the cell is working (producing $\ce{Li+}$ ions and electrons), and as a cathode when the cell is being recharged. The other electrode is always the positive pole, whatever its working mode. Its metallic oxides are reduced when the cell is producing current (cathode). When being recharged, this electrode is an anode, as its oxides are oxidized to a higher oxidation number.


In the nonideal solution, why can't one of the solute be positive deviation and the other be negative deviation? In the nonideal solution, why can't one of the solute be positive deviation and the other be negative deviation?
Let's say A and B makes a nonideal solution.
The moleculat force between A and B means A-B.
Let's say A-A > A-B > B-B.
Then A would evaporate more than in the pure state, and B would evaporate less than in the pure state.

For multiple component systems we use activity coefficient $\gamma$ in modified Roult's law as:
$$K_i=\gamma \frac{P_i^\star}P $$

and $\gamma$ is defined in the terms of $f$ : fugacity or fictitious pressure for a real gas.

You should read further about "Vapour Liquid Equilibrium" for multiple components.

Maybe:  http://www.journalrepository.org/media/journals/BJAST_5/2012/Dec/1355741742-Olaleke312012BJAST1789.pdf. You can see there is no reason why it can't have deviation in multiple directions.


Why hydrogen peroxide sizzles in the bathroom sink? Extremely short question. I pour a bit of hydrogen peroxide in my bathroom sink and it starts to sizzle, almost the same way it does on the open wound. Why? What elements in there it reacts with?

Hydrogen peroxide contains an peroxide linkage and it decompses rapidly to give oxygen.

But when substances like detergent come in its contact in a bathroom (as in the question) it not only  catalyses the reaction but if it was a foam making detergent it would make the decomposition process more evident and visible as oxygen bubble would produce effervescence in foam.


Impact of Pressure on equilibrium of Tin allotropes The pressure at the bottom of the Mariana Trench in the Pacific Ocean is $1090$ bar. What temperature will the two allotropes of tin be at equilibrium? Assume that the molar volume, energy, and entropy change does not vary with temperature.
Relevant data: Density of white tin: $7.287$ g/mL; Density of grey tin: $5.766$ g/mL
At $1$ bar: The enthalpy change from white tin to grey tin is $-2.016$ kJ/mol and the entropy change is $-7.04$ J/mol K.

My attempt: Since the two species are in equilibrium, I thought first to use the Clausius Clapeyron equation. Assuming that the molar volume is independent with temperature, I calculated the molar volume change to be $4.26\cdot 10^{-6} \frac{\pu{m^3}}{\pu{mol}}$ from white tin to grey tin. Now, I substituted these values into the equation to find that $$\frac{dP}{dT}=\frac{\Delta S}{T\Delta V} =-\frac{1.6\cdot 10^7}{T}$$
upon integration, I find that $$P-P^\circ = 1.6\cdot 10^7 \ln(\frac {T^\circ}{T}).$$
Substituting the temperature of equilibrium at one bar from the data given I find that the temperature is $286.4 \pu{ K},$ thus $T^\circ = 286.4\pu{K}$ and $P^\circ = 10^5\pu{Pa}.$ Thus at $P = 1090\cdot 10^5\pu{Pa},$ $T = 0.317\pu{K}.$ However, the temperature given in the answer is much higher ($220.35$ K).  The given answer was derived from a completely different method not involving the Clausius Clapeyron equation.
Hence I have the following question:
Question: Why is it not valid to use the Clausius Clapeyron equation in this scenario? (Or did I make a mistake in my derivation that caused my values to be off?)

First (as you already did) you need to determine the equilibrium temperature at $1\ \mathrm{bar}$. This happens to be when  $\Delta G^\circ = 0$, so that $T^\circ_{trans}= \frac{\Delta H^\circ}{\Delta S^\circ}=286.36\ \mathrm K$.

Next you can apply the Clausius equation in integrated form

$$ T = T^\circ_{trans} + (P-P^o) \frac{\Delta V_m^\circ}{\Delta S^\circ} $$

I obtain the following value for $\Delta V_m^\circ$:

$\Delta V_m^\circ = (1/5.766-1/7.287)\times \pu{ 118.71\times10^-6 m3/mol} = \pu{4.2973\times10^-6 m3/mol}$

Plugging this and the values you provide for $\Delta P$ ($1089\ \mathrm{bar}$) and the standard entropy changes $\Delta S_m^\circ$ gives $T=\pu{219.89 K}$. Not exactly what you say is the right answer, but, closer.


Can atomic number change the type of electromagnetic radiation being emitted when a electron drop from a energy level to another? Without getting into too much/any math, I just wanna talk about intuitions if possible.

If atomic increase, the attraction increase, and consequently the potential energy increase. More energy leads to higher frequency, different radiations.

I guess the reason if this diagram is universal is that the gap between 1 and 2 is too large too that the difference in energy cannot simply be compensated by adding more protons.

Absolutely, the differences in potential energies will be different for every element. This is because the potential energies will be different for every element. Notice that the potential energy for each energy level can be determined from the ground state. Let $E_g$ be the potential energy of the ground state (with the sign). Then for energy level $n$ the potential energy is $\frac{E_g}{n^2}$. Thus Differences between energy levels $n$ and $m$ can be written as $E_g\left|\frac{1}{n^2}-\frac{1}{m^2}\right|$.

Clearly it will be different for a different ground state energy which is precisely what the number of protons affects.

Note that the only important part of the mathematics is that all differences rely on the ground state potentials.


Derivation of Gibbs free energy formula Okay, so I'm trying to derive an important formula which states that -
                             ∆G(r) = ∆G° + RTln( RQ )
where ∆G(r) is the instantaneous rate of change of Gibbs energy with respect to moles of reagent consumed (am I right on this?) I'm hight confused because I have 'almost' derived it but due to some conceptual difficulty I'm not able to get the exact formula!

NOTE:
(1) Please ignore any non expansion work done on the system and tell my mistake as simply as possible because I have just started thermodynamics.

(2) The line above G represent molar Gibbs energy of substance. The small letters on the side of the symbol of reactent are the stochiometric coefficients.

(3) Many times I see variations of this formula, instead of ∆G(r) they simply use ∆G, there is a difference between two. Is this formula valid if ∆G is used instead of ∆G(r) in left hand side of equation? Also don't see any 'sense' in using ∆G because if ∆G is there then we need the initial and final quantities of the substances in the reaction but in the reaction quotient we plug the instantaneous activities of reactents and products.

I'm getting the above equation but a stochiometric coefficient is also included in this ! Obviously it is wrong. Kindly tell me my mistake.

I think you are using the first approach that I described.  So, here is the analysis based on the first approach.  Suppose you have the following numbers of moles of A, B, C, and D in the reaction mixture at some time: $n_A$, $n_B$, $n_C$, and $n_D$.

And the total pressure of the mixture (in bars) is held constant at P.  So the partial pressure of species A is $$p_A=\frac{n_A}{n_A+n_B+n_C+n_D}P$$And the partial molar free energy of species A is $$\bar{G}_A=G^0_A+RT\ln{\left(\frac{n_A}{n_A+n_B+n_C+n_D}P\right)}$$where $G^0_A$ is the free energy of pure species A at the reference pressure of 1 bar.  Similar equations can be written for the other species.  The total free energy of the mixture at any point in time during the reaction is:  $$G=n_A\bar{G}_A+n_B\bar{G}_B+n_C\bar{G}_C+n_D\bar{G}_D$$
Now, let the concentrations of the chemical species in the reaction mixture be expressed as follows:
$$n_A=n_{A0}-a\xi$$
$$n_B=n_{B0}-b\xi$$
$$n_C=n_{C0}+c\xi$$
$$n_A=n_{D0}+d\xi$$
where the subscript 0 refers to the initial state and $\xi$ is the conversion.  To find the equilibrium condition, you want to find the value of $ \xi $ that minimizes the free energy of the solution mixture.  To do that, you have a calculus problem to solve.


Can a bond be electrophilic or nucleophilic? The indicated bond is:

A) Nucleophilic because it is electron-deficient.
B) Electrophilic because it is electron-rich.
C) Nucleophilic because it is electron-rich.
D) Electrophilic because it is electron-deficient.

The answer is D. I'm arguing with my friend because of this particular question.
I argued that there may be an error in this question. The arrow is pointing toward a carbon atom, so the question should be "The indicated carbon atom is:". But, if you take the question as it is, I answered C since I think that the "indicated bond" is the double bond, and is therefore nucleophilic and electron-rich. If it is not the double bond, then what is the indicated bond in this question? And is it possible for a single bond to be electrophilic?

Bonds can indeed be considered to be nucleophilic. For example, in the electrophilic aromatic substitution mechanism an electrophile interacts with benzene’s π system. By analogy, we can also see this as a nucleophilic electron pair (situated in the HOMO) interacting with the electrophile’s LUMO.

However, since bonds are considered to be the bonding electrons which by definition carry a negative charge, bonds can only be nucleophiles or non-nucleophilic; never electrophiles.

Bonds also cannot be electron-deficient. A bond in a Lewis structure is made up of two electrons, full stop. In MO theory, we may no longer consider the bond localised but we still have an electron pair in a certain bonding orbital and thus never electron-deficient.

I suspect the book made an error here and wanted to point at the carbon atom so the text should be as you suggest.


Why is the water reduction of oxygen favored in acidic conditions? The standard reduction potential of diatomic oxygen in acidic conditions is +1.23 volts. However, the standard reduction potential of diatomic oxygen in basic conditions is only +0.40 volts.

Why is the reduction of diatomic oxygen by acidic media more favored than the reduction of oxygen by water that is basic?

I can think of a few reasons:

1) We are likely going to be forming negative two oxidation state oxygen anions when we oxidize diatomic oxygen. Oxygen adopts an oxidation state of negative two except in peroxides, which aren't very stable, and I don't think the oxygen radical will be very stable in solution.

Can someone provide a less fluffy reason that we're going to be forming negative two oxidation state oxygen atoms? I'd imagine that the electronegativity and electron affinity of oxygen are two reasons it might favor the negative two oxidation state over the negative one oxidation state in solution. I.e. why doesn't this occur to a significant extent?

$\ce{O2 + 2e^- -> 2O^-}$

Or this?

$\ce{O2 + 2e^- -> HOOH}$

2) From the above we can deduce that water is the product; water is pretty stable in either acidic or basic medium - i.e. it won't react to a large extent with either hydronium or hydroxide ion. Plus the oxygen in water has a -2 oxidation state. So if we look at our skeleton half reaction:

$\ce{O2 + 4e^- -> 2H2O}$

We realize that we need a source of protons. And the proton source in acidic medium is likely hydronium ion.

However, in basic medium, the proton source is water. I'd imagine that water is a poorer proton source than hydronium ion.

So would this account for the difference in standard reduction potential of oxygen in media that vary in pH?

I understand that voltage is an extension of K through the Nernst equation; is what I'm seeing here basically that the protonation of a base ($\ce{O^2-}$) is more favored in acidic rather than basic media? Could there also be kinetic factors? In other words, why is this statement true?

Also:

1) What's the mechanism for the reduction of diatomic oxygen?

2) Why is the reduction of protons at high pH less favored than the reduction of protons at low pH? My first guess: you don't find as many protons in basic solutions. But then, protons don't exist by themselves anyway. So what can the proton source be? I'm guessing that in basic pH, there doesn't exist much hydronium ion. So the only two sources of protons are water and hydroxide anion. Heterolytically cleaving the hydroxide anion to form the highly basic oxide anion doesn't seem thermodynamically favorable. Cleaving water to form hydroxide anions ... are we running into a Le Chatlier's Principle effect?

ETA: Or should I be considering proton activities? Is the activity of the proton less in basic solution than in acidic solution?

If you look at the half equations, you will see when you add them together, you get 1.23 V, 1.22 V, and 1.23 V.

As the mixture becomes more basic, it is more difficult for the oxygen to grab $H^+$ ions and easier for $H_2$ to lose electrons to generate $H^+$ ions.

The mechanism probably involves these combinations: $O_2, H_2O_2, H^+, HO_2^-, H_3O_2^-, H_2O, and\ HO^-$.


Methyl Iodide synthesis I was trying to determine whether I could synthesize methyl iodide by combining methanol, potassium iodide and sulphuric acid together... The way I see it, the sulphuric acid will protonate the OH instead of the I forming a very stable leaving group of water. Then the iodine ions would donate an electron pair to the carbon. You would be left with the salt Potassium bisulfate.... Then to remove my methyl iodide, just distil.

Will this work?

So what I said in the comment still holds: it could work (particularly in combination with a Dean-Stark apparatus to remove excess water), but there is no way to know for sure unless someone tries it out.

The more usual (and early) method is described in Organic Syntheses and utilizes the reaction of methanol with iodine in the presence of phosphorous

I don't know the exact reaction mechanism (if anyone has a clue, please feel free to edit!) but what happens is that $\ce{PI_3}$ is generated in situ from the elemental iodine and phosphorous and reacts with methanol to yield the methyl iodide and phosphorous acid

$\ce{3 CH3OH + PI3 -> 3 CH3I + H3PO3}$

Phosphorous acid does not act as nucleophile and so the problem of a competing backward reaction, which you would have in your proposed synthesis, is largely prevented.

I'd also like to add some caution: the reaction described in Organic Syntheses is not suited for beginners or in amateur setting. The chemicals involved are nasty: highly toxic, highly flammable, alkylating and pyrophoric. Don't try this at home :)


Racemization of ketones in presence of catalytic acid Question:
Which of the following ketones racemise in aqueous solution containing acidic or basic impurities?

My answer:
A and B should racemise in aqueous solution, for the carbonyl groups in A and B contain alpha-hydrogens, leading to the possibility of enolisation, hence racemization. Compound C can't form an enol, hence shouldn't racemise.

It's very clear that compound C doesn't racemise, my confusion is about A and B.

My logic (stability of carbanion) says that A and B should both racemise, however, the answer I have is only B.

I agree that the carbanion formed by A would be less stable than that formed by B, but not so much so to ignore the possibility of enolisation and racemization.

Could someone please help? Is there anything obvious that I'm missing? or is it related to data? A good explanation would help. Thanks a lot.

You are right that the racemization occurs through the enol form.

The chiral center is on the carbon attached to the chlorine atom. In B, that carbon participates in the enol form, but in A not. Therefore, in A the enol is formed but there is no racemization.


Why does N(OH)3 not exist? The hydrolysis of Group-15 tetrahalides gives us their hydroxides via a $\ce{S_{N^2}}$ mechanism:
$$\ce{PCl3 + 3H2O -> H3PO3 + 3HCl}$$
Nitrogen Trichloride, however, yields a different set of products, not forming the trihydroxide as expected.
$$\ce{NCl3 + 3H2O -> NH3 + 3HOCl}$$

A little bit of research led me to this exact same question (with 2 unsatisfactory answers) and this article, which claims that it exists, but I cannot read it. Why does $\ce{N(OH)3}$ not exist?

Another thought along similar lines is the existence of gem-diols in general, which are quite unstable, as well as $\ce{C(OH)4}$, which does not seem to exist as well. Is this an abnormal property of period 2 elements? All the other elements in group 14 and 15 form stable hydroxides whereas the first members do not seem to do so.

Although the acid and its salts are unknown, orthonitrite esters have been found, and from a natural source at that [1].  The orthonitrite function, which is trivalent rather than monovalent as nitrate would be, is incorporated into a macrocycle that hinders its decomposition.

The picture below is from Ref. 1.

Reference:

1.
Leanne Murray,* Tang K. Lim,* Graeme CurrieB and Robert J., "Aplidites (A-G): Macrocyclic Orthonitrites from
an Australian Tunicate, Aplidium sp.", Aust. J. Chem., 1995, 48, 1253-1266.Link