diH2O, dH2O, and DI H2O. What do they mean? What do $\ce{diH_2O, dH_2O}$, and $\ce{DI H_2O}$ mean?

I'm not asking for a description of deionized water and distilled water. I'm asking what the three abbreviations/formulas mean.

In other words, I'm asking someone to fill in the blanks below.

$\ce{diH_2O = \_\_\_\_\_}$
$\ce{dH_2O = \_\_\_\_\_}$
$\ce{DI H_2O = \_\_\_\_\_}$

I don't think there is a commonly accepted standard. If I had to guess, I'd say more often than not $\mathrm{di}\ce{H2O}$ is used to mean deionized water, but I've nothing but my own experience to base that on. Sorry for the lack of references. Also, $\mathrm{dd}\ce{H2O}$ is a common abbreviation that used to mean "double-distilled water" but now is used to refer to highly pure water produced by laboratory purification systems such as those from Milli-Q or similar.

$\mathrm{DI}\ce{H2O}$ = deionized water
$\mathrm{di}\ce{H2O}$ = deionized water

Such water is often available from centralized water purification systems in buildings that host many chemical laboratories.  It isn't as pure as "$\mathrm{dd}\ce{H2O}$" (see below) but is subtanstially more pure than tap water.

$\mathrm{d}\ce{H2O}$ = distilled water
$\mathrm{dd}\ce{H2O}$ = doubly distilled water
$\mathrm{mq}\ce{H2O}$ = Milli-Q water

The brand name of a popular type of water purifier that gives purity similar to $\mathrm{dd}\ce{H2O}$


Concentration of gas in water: What is the time derivative of concentration toward steady state? The Henry's law gives the steady state concentration for a gas in water. But it does not tell about the speed for this to happen if the concentration is not in steady state. What is the time derivative of concentration of oxygen in gas?

The transient diffusion equation for the oxygen concentration is $$\frac{\partial C}{\partial t}=D\frac{\partial^2 C}{\partial x^2}$$where D is the diffusion coefficient.  Henry's law applies at the interface.


What are the GDB-13 criteria for "synthetically accessible organic molecules"? I'm trying to understand the QM datasets (QM7, QM8, QM9), however, in the description of QM7 the data is described as follows:

  [The] QM7 dataset, which is a subset of GDB-13 (a database of nearly 1 billion stable and synthetically accessible organic molecules) containing up to 7 heavy atoms C, N, O, and S. The 3D Cartesian coordinates of the most stable conformations and their atomization energies were determined using ab-initio density functional theory (PBE0/tier2 basis set)

What exactly does the phrase "stable and synthetically accessible organic molecules" mean? If I had to guess I would have said it means they can be synthesized in a lab, but I didn't want to guess.

What you have provided is a definition of all the terms involved. This is fine, but there remains the (rather interesting) question of exactly how this is defined for a molecule in silico, especially because your question asks specifically about the GDB-13 database. Of course, one could evaluate "synthetic accessibility" by trying to make it in the lab: if you succeed, then it is synthetically available, and if not, then it isn't. Likewise, one could measure "stability" by (potentially) making it and studying its thermodynamic properties.

However, for a database of 1 billion compounds, this is obviously unfeasible. So the designers of the database have to resort to using some heuristics, or rules, to determine whether a given molecular structure is "synthetically accessible" and "stable". The rules are outlined in the original paper in which the GDB-13 database was published,[1] as well as the first GDB paper from slightly earlier.[2] Both are very interesting papers and well worth a read if you are interested in cheminformatics in general.

Because all organic molecules have a carbon chain or ring as their backbone, the first step is to construct all possible frameworks of carbons and hydrogens using graph theory (up to a maximum number of carbons, of course). Here there is not yet any consideration as to what other atoms may possibly exist in a typical organic molecule. However, there are some computer-generated frameworks which are almost certainly impossible to make because of strain or other considerations (such as Bredt's rule). On top of that, it appears that they have intentionally removed three- and four-membered rings, which are stable but would have dominated the database, thus making it highly unrepresentative:[2]

  The vast majority of these graphs (99.8%) contained three‐ and four‐membered rings and was excluded to avoid generating a database consisting almost exclusively of such small rings.

The next step is to introduce heteroatoms, which can be done fairly straightforwardly because in organic chemistry atoms have very specific bonding patterns. Thus, for example, because carbon forms four bonds and nitrogen three, you could replace a $\ce{CH2}$ group in a molecule with $\ce{NH}$. This is not trivial to do in the lab, but is very easy for a computer:[2]

  [...] all possible atom‐type combinations by introducing carbon, nitrogen, oxygen, and fluorine (as a model halogen) at each node

although for GDB-13, they have ignored fluorines:[1]

  We also eliminated fluorine because it was rarely found and never considered in our group for synthesis in virtual-screening guided drug discovery applications of GDB-11.

and also added chlorines (replacing $\ce{OH}$ groups in molecules which have them) and sulfurs (replacing $\ce{O}$ atoms).

The problem with doing that exhaustively is that too many heteroatoms tends to make a molecule very unstable. For example, directly joining heteroatoms via single bonds is (generally) a good recipe for making explosive compounds. So, essentially all such molecules were removed. The authors found that most of this work could be very quickly automated by using an even simpler heuristic: simply removing any compound with a high heteroatom:carbon ratio.[1]

  Because most of the rejected molecules contained multiple heteroatoms, we reasoned that it might be possible to accelerate the database computation using a very fast “element-ratio” filter. Analysis of databases of known compounds suggested cutoff values of (N + O)/C < 1.0, N/C < 0.571, and O/C < 0.666

and specifically disallowed cases, which have to be filtered out in another step, included:[1]

  The following functional groups are discarded as too unstable to be considered: hemiacetals, hemi-aminals, aminals, acyclic imines, non-aromatic enols, orthoesters and analogs, carbamic acids, non-aromatic enamines (except acylated enamines and vinylogous enamines), beta-keto-carboxylic acids and beta imino-carboxylic acids, and all compounds containing both a primary or secondary amines and an aldehyde or ketone.

This is also helpful for preparing a database of "druglike" molecules because molecules with too many heteroatoms are very polar, which makes it almost impossible to diffuse across cell membranes (see e.g. https://en.wikipedia.org/wiki/Lipinski%27s_rule_of_five).

What you get at the end is probably not likely to be 100% "synthetically accessible". However, the filters applied mean that if you were to pick a random molecule from the database, there is an exceptionally good chance that you could make it in the lab if you so desired. Quoting from the authors one final time:[2]

  The database construction strategy chosen also ensures that the majority of GDB, although presently unknown, should be synthetically accessible.

References

Blum, L. C.; Reymond, J. 970 Million Druglike Small Molecules for Virtual Screening in the Chemical Universe Database GDB-13. J. Am. Chem. Soc. 2009, 131 (25), 8732–8733. DOI: 10.1021/ja902302h.
Fink, T.; Bruggesser, H.; Reymond, J. Virtual Exploration of the Small-Molecule Chemical Universe below 160 Daltons. Angew. Chem. Int. Ed. 2005, 44 (10), 1504–1508. DOI: 10.1002/anie.200462457.


Does the entropy of a chemical system increase in an endothermic reaction? Since entropy increases when temperature increases, the entropy of surroundings increases during an exothermic reaction. When I apply the same logic, it seems like the entropy of a chemical system increases in an endothermic reaction. Is this true?

Not necessarily.
Exo- and endothermic refer to the direction in which heat is exchanged between system and surroundings, not to changes in the entropy of the system, or of the temperature of either system or surroundings. Your logic though is partly correct: exothermic means the entropy of the surroundings increases, as it would do with an increase in T - the catch is that we usually assume that the surroundings has constant T, but that's because we say it has infinite heat capacity.
It is a recurrent subject of confusion to conflate changes in the entropy of the surroundings, of the system, and of the sum of both. Always keep the second law in mind. What determines whether a reaction will be spontaneous is the sum of the changes in entropy of the surroundings and the system.
When a reaction occurs reversibly, $\Delta S_{\mathrm{surr}} + \Delta S_{\mathrm{sys}}=0$.
When spontaneous, $\Delta S_{\mathrm{surr}} + \Delta S_{\mathrm{sys}}>0$.
When a reaction is exothermic,  $\Delta S_{\mathrm{surr}} >0$.
When a reaction is endothermic,  $\Delta S_{\mathrm{surr}} <0$.
Nothing about the thermicity (whether the reaction is endo or exothermic) tells you what happens to $\Delta S_{\mathrm{sys}}$, unless you already know whether the reaction is spontaneous or not (in which case the sum of the entropy changes is positive).


Potassium permanganate (VII) + glucose + sodium hydroxide =? I saw this experiment many times but still can't explain why.

The experiment is first prepare two solutions. The first one is to dissolve potassium permanganate crystal into distilled water to form potassium permanganate (VII) while the second one is to put sodium hydroxide and glucose (sugar) into distilled water.

Secondly, I pour the second solution into the first one. My observation is the purple solution will slowly turn into blue, then slowly turn into green and lastly turn into yellowish orange.

So, what really happen in the solution?

This is a redox reaction in which the permanganate ion is reduced and the glucose is oxidised. Potassium permanganate is usually used in acid solution and under these conditions is a very powerful oxidising agent in which the manganese is reduced from the oxidation state of +7 to +2 with a colour change from purple to colourless (actually extremely pale pink). In alkaline solution manganese is only reduced from +7 to +6 changing colour from purple to green. (The blue colour you mention is due to a mixture of purple and green). The green manganate(VI) ion is unstable and slowly disproportionates to manganate(VII) (purple) and manganese(IV) oxide which is brown. The manganate(VII) goes on to react as before until the brown manganese(IV) oxide is all that remains. In alkaline solution this tends to form a colloidal suspension which (if fairly dilute) can appear orange. The oxidation of the glucose forms no coloured productss.


What is considered room temperature in Celsius? Recently, I have seen room temperature listed at 20 °C and 25 °C. Which one is more accurate for use in chemistry problems?

Considering you've read Wikipedia article on room temperature (r.t.), you probably already have an idea that there is no uniformly defined value due to the subjective matter the term is defined upon.
There are numerous standards across industries defining temperature ranges, most of which lead to the value for r.t. of $\pu{(20 ± 5) °C}.$

In natural sciences there are several well-defined standard conditions describing the combination of exact values for temperature, pressure and humidity.
Room temperature is not among these, so ideally, an exact value must be specified alongside, if relevant, e.g. "the parameter has been determined at r.t. $(\pu{20 °C})$".

German Wikipedia article presents an observation that in physical chemistry and statistical physics in European region the typical value taken for r.t. is $\pu{20 °C},$ whereas in U.S.A. the majority of textbooks refer to r.t. as to $\pu{25 °C}.$

In nuclear chemistry, thermal neutron energy $E_0 = \pu{25.26 meV}$ is fixed as characteristic energy at r.t., which is a reference point for tabulated parameters such as neutron cross section.
Using $E = k_\mathrm{B}T$ relation, an exact value for r.t. can be determined, which is usually taken as $\pu{20.43 °C}$ (see e.g. [1, p.  11-173] and references therein).

References

Haynes, W. M.; Lide, D. R.; Bruno, T. J. CRC Handbook of Chemistry and Physics: A Ready-Reference Book of Chemical and Physical Data, 97th ed.; Taylor & Francis Group (CRC Press): Boca Raton, FL, 2016. ISBN 978-1-4987-5429-3.


What is the crystal structure of Calcium Acetate? I tried conducting an experiment I saw at school with some materials I had at home. The experiment consisted of mixing a solution of Acetic Acid with Calcium Carbonate to yield Calcium Acetate: $$\ce{CaCO3 + CH3COOH -> Ca(CH3COOH)2 + CO3 + H2O}$$ I'm providing as much detailed in a summarized form to show what I did step-by-step if I did anything wrong. I didn't have any lab grade source of $\ce{CaCO3}$, but I found that chalk has a large concentration of $\ce{CaCO3}$, and I used that instead. I worked out a small calculator on Spreadsheets and found the appropriate quantities of each reactant. I then combined both of the ingredients and left the reaction to settle. After a couple days, I stirred and saw little to no fizzing, however, there was a lot of precipitate at the bottom of the beaker which I assumed to have been non-$\ce{CaCO3}$ residue from the chalk. I filtered the solution through a two coffee filters and let the solution evaporate. I should note that the solution was crystal clear after it was filtered. I came back a week later only to see a hard, styrofoam look-a-like on the sides of the beaker, and a layer of cloudy crystals at the base of the beaker. At the center of the base there's a crystal structure sprouting separately from the rest of the crystal layer.

I initially though perhaps my calculations or initial data gathered were incorrect but even after triple-checking them there was no error. Looking for possible structures of Calcium Acetate there was only one image that looked somewhat like the crystal sprouting at the center of the beaker, but still it doesn't explain what all the other substances are. Does Calcium Acetate have multiple crystal structures? Or has there been some sort of contamination from the chalk. Any help would really be appreciated:)

Too long for a comment, hence an answer:

What you are asking about is not crystal crystal structure, it's called crystal morphology. Crystal structure is a model describing atomic positions and overall symmetry within a unit cell, whereas crystal morphology is the appearance of the crystal on a macro-scale. And yes, crystal morphology is also affected by the conditions crystals are growing under (concentration and evaporation rate, in your case), so that the same substance can have crystals of different shape.
The formula for calcium acetate and hence the equation is wrong, the correct one would be $(\ce{OAc-} = \ce{CH3COO-})$
$$\ce{CaCO3 + 2HOAc → Ca(OAc)2 + H2O + CO2}$$
and, since your reaction's stoichiometry suggested 1:1 ratio between calcium carbonate and acetic acid, I suspect you didn't use enough acid leaving $\ce{CO3^2-}$ in the solution. Then, as the primary product – calcium acetate $\ce{Ca(OAc)2}$ – has crystallized (white edgy crystals on the bottom), the remaining calcium carbonate $\ce{CaCO3}$ started to precipitate or co-crystallize on the walls of the beaker (what you call a "styrofoam look-a-like" phase), resembling how calcite grows in nature (see this and other images of natural calcite on MineralMiner's FTP server):

Also note that there is no such thing as $\ce{CO3}$. It's either carbonate $\ce{CO3^2-}$ (solution) or carbon dioxide $\ce{CO2}$ (gas).
Regarding the the impurities from the chalk, usually there are parts of shells of marine animals and silicates (can be eliminated by filtering), also minor amounts of $\ce{MgCO3}$. However, if you took a chalk bar from the classroom, this is a different story as it's often nearly 50% $\ce{CaSO4}$ (I hope you didn't do that).


Can the Bohr atomic model be applied to sodium and other single valence electron species? Is $\ce{Na}$ a single electron species and can I apply Bohr's Model on it?
Please explain why not as it also contains one electron in its outermost shell.

One of the great successes of the Bohr model was the ability to explain the emission spectrum of hydrogen (and hydrogen-like monoelectronic ions), by allowing a simple theoretical derivation of the experimentally determined and remarkably accurate Rydberg formula:

$$\frac{1}{\lambda}=RZ^2 \bigg(\frac{1}{n_f^2}-\frac{1}{n_i^2}\bigg)$$

$\lambda$ is the wavelength of light emitted, $R$ is a proportionality constant known as the Rydberg constant (which actually varies slightly with changes in atomic mass), $Z$ is the amount of positive charge attracting the electron (the nuclear charge in multiples of $q=1.6022\times 10^{-19}\ \mathrm{C}$), and $n_f$ and $n_i$ are positive integers (called quantum numbers) which previously had no experimental interpretation, but which now arise from the postulates of the Bohr model, being attributed to the concept of electron shells.

After this great new insight into atomic physics, naturally one would like to investigate how generally applicable the model is. Given the success in describing the emission spectra of hydrogen, one of the first places to look would be the spectra for other elements. So here they are:

(Source - isn't it a beautiful image? Kudos to the designer! Open in a tab for gigantic resolution)

As you can see, most elements have rather elaborate spectra. After undoubtedly many frustrated attempts, it was found that the Bohr model, as accurate as it was for hydrogen, applied quite poorly to the rest of the elements, and no spectra could be reliably predicted.

However, the alkali metals stood out in that a modified version of the Bohr model had some success, producing a Rydberg-like formula, though salvaging it came at the cost of additional experimental parameter fitting, again with no theoretical background.

The formula for the energy of a electron in a hydrogen atom according to the Bohr model is:

$$E_n=-\frac{hcR}{n^2}=-\frac{13.6\ \mathrm{eV}}{n^2},\ \ \ \ n \in \mathbb{N} \ \ (n=1,2,3,4...)  $$

For the alkali metals, the formula had to be modified to:

$$E_n=-\frac{hcR}{n_{eff}^2}=-\frac{hcR}{(n-\delta_l)^2},\ \ \ \ n_{eff} \in \mathbb{R_+}  $$

To keep the same structure, the mismatch between theory and experiment is brushed into the quantum numbers $n$, which are replaced by effective quantum numbers $n_{eff}$ that are no longer positive integers. To recover integer quantum numbers, a value called the quantum defect ($\delta_l$) had to be introduced. This value is different for each value of $n$, and also varies with another parameter $l$. This may seem like an arbitrary mathematical attempt to salvage the equation, but while the values of $\delta_l$ had to be tabulated, it was found that they were relatively constant with changes in $n$, and that there were a limited number of different cases ($l$ was a parameter limited to only a few values). The resulting Rydberg-like equation is:

$$\frac{1}{\lambda}=R \bigg(\frac{1}{(n_f-\delta_{l_f})^2}-\frac{1}{(n_i-\delta_{l_i})^2}\bigg)$$

While not as tidy as the result from the Bohr model, at least the equation can be used to approximately explain the atomic emission spectra for the alkali metals. Details of an experimental procedure for chemistry undergrads to determine the emission spectrum of sodium and apply the modified Rydberg formula can be found here, along with theoretical background.

Nowadays, we know why the alkali metals seemed to be a better fit to the Bohr model than the rest of the elements. With finer knowledge about the details of the electronic structure of atoms, borne out of the full quantum mechanical model of the atom, we know that alkali metals share an important property with hydrogen; they all possess a single electron in their outermost shell, i.e. have a valence shell configuration of $(ns)^1$. This is the only electron which is important in electronic transitions in the near-ultraviolet, visible and infrared regions of the electromagnetic spectrum, as the core electrons of the alkali metals are held much more tightly to the nucleus and require high energy photons to excite. This creates a much simpler spectrum.

Furthermore, atoms with multiple electrons suffer from electron-electron repulsions, which are not taken into account at all in the Bohr model, and thus are a significant source of error. However, in the alkali metals the consequences of these repulsions are simplified, as all electrons are in core shells, except the single valence electron. For all other multielectronic elements, electron-electron repulsions in the valence shell preclude a simple application of the Bohr model.


Confusion in comparing melting point We have to compare melting points of $\ce{LiH}$, $\ce{NaH}$, $\ce{KH}$, $\ce{CsH}$. I know the melting point of an ionic compound is more than that of a covalent compound.

So according to Fajan's rule, ionic character increases in $\ce{LiH}$, $\ce{NaH}$, $\ce{KH}$, $\ce{CsH}$. As ionic character increases, melting point should also increase in $\ce{LiH}$, $\ce{NaH}$, $\ce{KH}$, $\ce{CsH}$.

But actually, the melting point decreases in $\ce{LiH}$, $\ce{NaH}$, $\ce{KH}$, $\ce{CsH}$. What could be the reason behind this?

You can get an estimate of the bond strength by using Coulomb's law and periodic trends for atomic radius. $$F = k{ {q_1q_2} \over r^2}$$The charge of all these alkali metals is the same in these compounds, so the only difference in bond strength comes from the the distance between the charges. The distance between the charges is related to each atoms' atomic radius and that is an increasing trend as you move down the alkali metals.

So, if distance is increasing, the force holding the solid together is decreasing, which would justify the trend $\ce{LiH}>\ce{NaH} >\ce{KH}> \ce{CsH}$ for melting point.


Can an element have a third energy shell without completely filling the second energy shell? I came across an electronic configuration diagram, it had three shells. The first shell had only one electron, the second had six and the third energy shell had five. An electron was also present outside all the shells.
Is this configuration possible?

Yes, it is possible. When an element is excited, some electrons from lower shell jumps to higher shells. For eg, in-ground state suppose an element has electronic configuration
$\ce{(1s)^2 (2s)^2 (2p)^5}$ . If we energize it, then some of the electrons jump up for a short period of time(till the energy is available) . The config becomes $\ce{(1s)^2 (2s)^1(2p)^6 }$.


What open-source software can produce 3D PDF representation of molecules? Most of the existing software I know (like VMD or Pymol) say that they can be used to create PDF files with 3D representations of molecules: actual 3D models, which you can rotate inside the Adobe Acrobat viewer. However, they all require the use of “Adobe Acrobat Pro 9 Extended” for its 3D capture feature. From what I can read, this feature does not exist any more in Acrobat X and was transferred to a separate product, Tetra4D, which doesn't work on Mac OS.

So, how can I produce PDF files including 3D molecular models using free (or better, open source) tools?

Disclaimer

This is not for the faint-hearted.
If you are not interested in using LaTeX, stop reading now, as this uses the media15 or movie9 packages to embed U3D in a pdfLaTeX generated PDF. This answer assumes basic familiarity with LaTeX. I have tested this with pdfLaTeX and it should work fine in XeLaTeX.
One of the tools  I have used (DAZ Studio) is apparently 'free for a limited time'. It is also not available on Linux. FOSS replacement suggestions are welcome.
I'm of the personal opinion that 3D in PDF is not as good as it sounds.

What doesn't work

VMD offers the ability to output as OBJ, which can then be processed with FOSS MeshLab into a U3D file that can be embedded into a PDF. This is sub-optimal for two reasons:

There is no way to tell MeshLab that the VMD-generated mesh is supposed to be Gouraud-shaded. This means that the hard facets of the mesh will be glaringly obvious in the pdf, even if you use an unwieldy degree of subdivision.
Colours are not preserved. Your model will appear in the PDF with a uniform grey texture.

As such, MeshLab is not suitable at this stage.

Preparing a model

VMD offers a number of decent export options to get our geometry out of .PDB or what have you and into a mesh format. I opted to use the VRML format to output a mesh of a ribbon representation of PDB entry 4DM9. I then imported this into Blender.

Blender has no reliable U3D export capabilities, however I can use it to assign Gouraud shading to all of the shapes and assign colours to them. I then use the blender export function to export the scene as a Wavefront .OBJ file, with attendant .MTL file.

Processing with DAZ Studio

I am indebted to the author of this article (John Nyquist), which indicated that DAZ Studio can convert Wavefront to U3D just fine, with preservation of textures. DAZ is a very specific program that seems to be heavily geared towards posing, clothing and rendering humanoid figures in the spirit of programs like Poser. However all we really need is the U3D export function and we can read in the structure just fine.

Sometimes, this is how I see myself:

Okay in all seriousness, don't put a human figure in or on your molecule without good reason. Export as U3D and then we can embed in pdfLaTeX.

pdfLaTeX and movie9

As a proof of concept, I just followed the minimal example given in the media9 manual for getting a 3D model into a pdfLaTeX-generated PDF. Here it is, being viewed in Adobe Reader X (I know, I know!). Bear in mind that very few readers can actually display 3D PDF, which is one of the reasons why I feel that it is not such a good idea for chemical visualisation.

It starts off zoomed way too far in because I did not set up any options for the import (however you can find out all about that in the media9 or movie15 package manuals) and is a bit laggy on account of having over 1 million faces (I went way overboard with subdivision), however the important thing is that it works.

Improvements welcome!

Obviously, this is a pretty Rube Goldbergesque route to getting a molecule into a PDF. There probably exist easier ways which should definitely be suggested.


Organic-catalysis microwaves When we perform a organic reaction without any addtional catalyst except that the reaction is carried out under microwave condition we can accelerate this reaction in comparison with one under heating only. So, microwaves can be considered a form of catalysis according to the concept of reaction kinetics?

Here is an example of an entire online GoogleBook dedicated to the topic Microwave Assisted Organic Synthesis, apparently based on some 2,000 papers.
On page 3 of the cited reference, notes to quote:

Interaction between microwave radiation and solutions of polar molecule polar molecules may be adequately described using classical models, which may be derived from Maxwell's equations.

And further:

Although this analysis has concentrated on the oscillating components of the electromagnetic field of the microwave radiation,  it is also possible to get interaction between the oscillating magnetic field and the magnetic dipoles in the sample. The high-heating rates for Fe304 may be attributed to such interactions.

With respect to benefits, per page 23 to quote:

Reaction rates are generally very high and yields in many cases can be greatly improved, as competing side reactions can be minimized...

I would also add that the use of microwave-based dielectric heating is, at times, more than just fast and efficient and heating, as it can, in the presence of activated carbon, for example, introduce surface-based radicals into the reaction mix. See, for example, Generation of hydroxyl radical in aqueous solution by microwave energy using activated carbon as catalyst and its potential in removal of persistent organic substances.
As such, I described the general use of this new heating technology as microwave-assisted synthesis, and not per se, as catalytic, albeit depending on the reaction mix, it can be (per last cited reference).


How do I determine the mass percent of Barium nitrate and Sodium acetate in an original unknown solution I am given $20 \mathrm {mL}$ of the unknown mixture of Barium Nitrate and Sodium Acetate dissolved in water. I am given the total mass of the two solids before they were put into the mixture.
I am given $0.200 \mathrm{M}$ of $\ce{NaOH}$, $0.200 \mathrm{M}$ of $\ce{Na2SO4}$, $0.200 \mathrm{M}$ of $\ce{NaCl}$ and ethanol to separate a precipitate.
I am learning about selective precipitation.

It occurred to me that you might have been asking how to complete this separation experimentally rather than conceptually, which is what the comments and previous answer all were pointing toward.  If that is the case, then once you've precipitated out the barium sulfate, you have to isolate it from all other substances in order to find its true mass.

The precipitate is a solid, so you could separate it from most of its contaminants by pouring the solution through a pre-weighed filter paper, making sure to wash the walls and bottom of the reaction vessel with distilled water and pouring this wash through the filter as well in order to collect all of the precipitate.

The filtrate would contain the bulk of the sodium acetate and sodium sulfate in the reaction mixture.  What remains in the filter paper would be barium sulfate and a small amount of reaction mixture, including the sodium salts.  To eliminate the sodium salts, add additional distilled water to the filter to wash the precipitate (and filter paper) clean.

Once you have a filter paper with clean barium sulfate, dry it thoroughly and weigh.  I don't believe that barium sulfate forms a hydrate, so the mass of the precipitate should all be barium sulfate.

Note: Barium salts are quite toxic, particularly highly soluble ones like barium nitrate.  Be careful using them and dispose of the product appropriately.


Hydrogen bonding in complex compounds? Does the water molecules in coordinated water (where water acts as ligand) form hydrogen bonding with other complex molecules? Will the water molecules inside the complex form intramolecular hydrogen bonds?
Consider the examples coordinated water and hydrogen bonded water molecules like $\ce{[Cr(H2O)6]^3+}$ and $\ce{CuSO4.5H2O}$.

In the generality the question cannot be answered with a definititve statement. You have to look at the cases. For example in the solid state structure  of CuSO$_4\cdot$5 H$_2$O, which is actually more like [CuSO$_4$(H$_2$O)$_4$]$\cdot$H$_2$O, the single crystal water (non-coordinating water) in deed forms hydrogen bonds to oxygen molecules of other coordinating water molecules. In general one could maybe predict, that coordinating water molecules for steric reasons are less likely forming hydrogen bonds to other coordinating water molecules: The oxygen points to the metal and the protons will point to outside so the likelihood of protons from coordinating water molecules  meeting oxygens of coordinating water molecules is small.


Vapour pressure of a liquid subjected to pressure I am following the derivation by Atkins in section 4B(c) (Physical transformations of pure substances; thermodynamic aspects of phase transitions), in which the authors derive the equation ${p = p_* * exp[V_m(liquid)* ΔP/RT]}$. The derivation starts by noting that chemical potentials of vapour and liquid are equal at equilibrium, $μ_{liquid} = μ_{gas}$, and follows by integration over the pressure to calculate the change in chemical potentials due to the added external pressure.

What is unclear to me is why the integration for the liquid phase is taken over the ambient pressure while the integration for the gas phase is taken over the vapour (NOT ambient) pressure. It seems counterintuitive to me because I would expect that $μ_{gas}$ is a function of the ambient pressure and therefore should be integrated over the ambient pressure. Am I missing something?

EDIT: I guess the question could be rephrased: in the definition of molar volume for gas, $V_m=(\frac{\partial \mu} {\partial P})_{T,n}$, is $P$ the partial or ambient pressure? More generally, does $\mu_{gas}(T,P)$ depend on the partial or ambient pressure?

Here's how it works.  The fugacity of the liquid at temperature T and total pressure P is equal to the equilibrium vapor pressure at T times the Poynting correction (assuming that the vapor phase behaves as an ideal gas mixture).  The fugacity of the same species in the vapor is equal to the partial pressure of the substance in the gas phase.  The vapor and liquid are at equilibrium at temperature T and total pressure P if the fugacity of the substance in the liquid is equal to the fugacity of the substance in the vapor phase.


Electronegativity and anion stability If two similar sized atoms have different electronegativity values, the atom with higher electronegativity holds the electrons more tightly, and thus is more stable. Example: hydroxide vs azanide (NH2-).

If two atoms have different sizes (down the periodic table), the larger atom has electrons more spread out and thus is more stable.
Example: fluoride vs iodide anion

To me, these two statements are contradictory. One says tight electron cloud means not reacting while the other says spread electron cloud means stability (less repulsion). Is one of the statements wrong? I'm sure I am missing something here. Thank you in advance for your help.

I believe that you are mixing up the concepts of electronegativity and of electron affinity.
The first statement you present is related to Electronegativity.

  If two similar sized atoms have different electronegativity values, the atom with higher electronegativity holds the electrons more tightly, and thus is more stable.

Yes. This is generally true.
Electronegativity establishes how well an element can pull electron density towards itself.

The second statement is about electron affinity.

  If two atoms have different sizes (down the periodic table), the larger atom has electrons more spread out and thus is more stable. Example: fluoride vs iodide anion

Generally true as well. Flouride anion is less stable than iodide anion due to the repulsion of electrons in the outermost octet of fluoride. But this rule is not absolute as chloride has the highest stability (in the gaseous state) though it is above bromine and iodine.

Edit:
Considering some confusion in the comments I shall put this here.

The electronegativity is generally not compared based on inter-electron repulsion because the shared pair doesn't completely enter the orbits of the electronegative atom but is shared and so electron density is between the two atoms and not concentrated on one atom for repulsion to take much effect.


Is there any reaction between amines and AgNO3? I found a question in a book about the reaction between $\ce{AgNO3}$ and aromatic/aliphatic amines, but I couldn't find any information about
it. What should be the expected reaction mechanism and products formed in this reaction?

I found this reaction between amines and silver nitrate:


Why does freezing point decrease when vapour pressure decreases in dilute solutions? I was studying colligative properties of dilute solutions, and one of the properties was freezing point depression. The confusion I had was that, if vapour pressure decreases, it should be easier to reach the freezing point (point where vapour pressure of solid and liquid are equal). So I thought the freezing point will increase, but freezing point actually decreases. Is there any logical explanation?

Consider graphs of the vapor pressure of the liquid and of the solid, both as a function of temperature. Intuitively, both graphs increase with temperature, and intersect at the freezing point. By lowering the vapor pressure of the liquid, we lower the corresponding vapor pressure curve and hence the temperature at the point of intersection.


Polarity of the nitrate ion We observe that nitrate ion has a trigonal planar geometry according to its lewis structure and the VSEPR theory. Its resonance hybrid consists of three contributing structure -three probable double bond organisations. According to my knowledge, we would expect that pi bond to be distributed equally over the molecule and thus, the dipole moments would cancel each other out. My question is, why such an ion would dissolve in water even though it is  non-polar?

Don't overthink it.

The oxygens have a partial negative charge in nitrate from resonance.  The protons in water have a partial positive charge from dipoles.  Positives attract negatives.

Molecule polarity becomes an issue when discussing neutral molecules interacting with neutral molecules.  Molecular polarity is one thing, and bond polarity is another.  Transformations / arrangements / alignments of molecules such that the dipoles (the things that make a molecule or bond 'polar') align to maximize interaction are certainly important for dissolution, but when a species is a bona fide free ion, water will tend to dissolve it.


Change inchikey directly Preamble:
Now, I am aware that directly editing the inchikey is, generally speaking, a bad idea, but...

Can I just change the 2nd block of an inchikey to "UHFFFAOY" to get the correct main-layer-only version of the inchikey of that molecule?

example where this seems to be the case (from pubchem):
BSAIUMLZVGUGKX-BQYQJAHWSA-N = trans-2-nonenal
BSAIUMLZVGUGKX-FPLPWBNLSA-N = cis-2-nonenal
BSAIUMLZVGUGKX-UHFFFAOYSA-N = 2-nonenal

I have seen lots of other examples (100s if not 1000s) to support this hypothesis (both for cis/trans and for sterochemistry) and never a counter-example.

According to the InCHI FAQ, the 2nd block encodes "encodes stereochemistry and isotopic substitution information". So if you want to generate the corresponding InChIKey for the case where there is no stereo present nor isotopes specified, then it looks like this is possible. If you also want to force the neutral form, you could try setting the final character to 'N' (it may instead be P or O).

Your mileage may vary. To be honest, I wouldn't do this if I were you. I would generate a SMILES without isotopic or stereo information (or remove this programmatically), and then get the InChIKey of that. This would avoid any nasty surprises.


If you have the latest copy of cclib installed, it can extract both the energies and geometries and write the geometries to an XYZ trajectory file that you can open with Avogadro or VMD.

To print the HF or DFT energies to the screen,

$ ccget scfenergies admp.out

but be warned that the energies are in eV, not hartree.

To dump geometries to XYZ files, here's a simplified version of a script I use all the time:

#!/usr/bin/env python

"""cclib_extract_geom.py: Extract geometries from a quantum chemical
output file using cclib.
"""

from __future__ import print_function

import os.path

from cclib.io import ccopen, ccwrite

def getargs():
    """Get command-line arguments."""

    import argparse

    parser = argparse.ArgumentParser()

    parser.add_argument('outputfilename', nargs='+')

    parser.add_argument('--trajectory',
                        action='store_true',
                        help="""Extract all possible geometries into a trajectory-like file?""")
    parser.add_argument('--suffix')

    args = parser.parse_args()

    return args

if __name__ == '__main__':

    args = getargs()

    for outputfilename in args.outputfilename:

        job = ccopen(outputfilename)
        data = job.parse()

        stub = os.path.splitext(outputfilename)[0]
        if args.suffix:
            xyzfilename = ''.join([stub, '.', args.suffix, '.xyz'])
        else:
            xyzfilename = ''.join([stub, '.xyz'])

        ccwrite(data, outputdest=xyzfilename, allgeom=args.trajectory)

To write the energies in to a text file in hartree, you might add something like

        from cclib.parser.utils import convertor # put me at the top of the file
        with open(stub + '.energies', 'w') as fh:
            for energy in data.scfenergies:
                fh.write(str(convertor(energy, 'eV', 'hartree')) + '\n')


Are the particles of a colloid "solid"? I want to understand what are the particles of a colloid in the microscopic scale. According to Wikipedia:

A colloid is a phase separated mixture in which one substance of
microscopically dispersed insoluble or soluble particles is suspended
throughout another substance.

What I can't understand is how we can have a phase seperation when we have soluble particles? I looked for some pictures and I got more frustated. For example in the following picture Colloids:
the particles seem to be like ordinary atoms/molecules just scaled up. So how can this particles exhibit bulk properties and classified as "solid" particles? Is the picture a simplification? I mean should colloid particles be represented as clusters forming a phase (e.g. solid) that is invisible to human eye?

Colloidal solutions may contain particles made of liquids or of solids. The particles are liquid in substances like mayonnaise or moisturizer, whose particles are droplets of oil (in water) for mayonnaise or droplets of water (in oil) for moisturizer. They can be solid in substances like acrylic paints, where the particles are made of semi-solid paint dispersed in water. The great advantage is that these paints can be diluted with water. Fresh spots are easily removed. Colloidal gold does exist. The colloidal particles can even be gas in whipped cream, froth and foams.


Does an electrically conductive gas exist at room conditions? As I know, the gases are insulators, because their particles are mainly electrically neutral, and thus there is no movable charges in them.

But maybe it shouldn't be always so, for example if a gas has a significant proportion of charged particles, I think the electrical conductivity isn't impossible. If at least one of the electrons of its particles bounds very weakly, maybe even at room conditions is it possible to have enough low resistance to be considered as a conductor.

Am I right? Does a such gas already exist? Or there is some mechanism which prohibits this idea?

Extensions:

Maybe a such gas could be considered as plasma, but I think from this aspect we could see the plasma as gas as well.
I am thinking at around 1 atm pressure and 300 K temperature, in the gas, not around the box around it. Electrostatic discharge is also problematic, because in most gases it can happen only far from the room conditions and isn't an equilibrium state.
If there is none, I am looking for the gas which is the possible nearest to the these 3 conditions (300 K, 1 atm, conductive).

It depends on what you are prepared to consider a gas and what you are prepared to consider room conditions.

The gas inside all discharge lamps (fluorescent lamps and neon lamps in shop signs, for example) conducts electricity and the lamps work under normal room conditions. However, the gases are often at low pressure and are, strictly speaking, plasmas not room temperature neutral gases. Some lamps, especially high intensity ones like the xenon discharge lamps in floodlights) exist with high pressure gas when working so low pressure isn't the key.

And air itself will conduct electricity given the right conditions. Lightning couldn't exist if it didn't. Nor could you get the (much safer) discharges from Tesla Coils or Van de Graaff generators. But, again, the effect relies on the creation of a plasma from the breakdown of the neutral molecules in the gas.

So the answer is "yes" but only if you allow plasmas created from the medium to count as a normal gas.


How can I choose an indicator for determining the concentration of a base using its pKb values? I have a base $\ce{M(OH)_{3}}$ having 3 $\mathrm{p}K_{\mathrm{b}}$ values as,
$$
\mathrm{p}K_{\mathrm{b}_{1}} = 0.5\\
\mathrm{p}K_{\mathrm{b}_{2}} = 3.7\\
\mathrm{p}K_{\mathrm{b}_{3}} = 7.7\\
$$

If a strong acid is given as the titrant and phenolphthalein and methyl orange are given as possible indicators, how can I choose an indicator using above $\mathrm{p}K_{\mathrm{b}}$ values ?

Since:
\begin{align}
K_\mathrm{w} &=K_\mathrm{a} \cdot K_\mathrm{b}\\
K_\mathrm{a} &=\frac{K_\mathrm{w}}{K_\mathrm{b}}\\
\log{K_\mathrm{a}} &=\log K_\mathrm{w}-\log K_\mathrm{b}\\
K_\mathrm{w} &=1 \times 10^{-14}\\
\log{K_\mathrm{a}} &= -14-\log K_\mathrm{b}\\
-\log{K_\mathrm{a}} &=14+\log K_\mathrm{b}\\
-\log{K_\mathrm{a}} &=\mathrm{p}K_\mathrm{a}\\
-\log{K_\mathrm{b}} &=\mathrm{p}K_\mathrm{b}\\
\mathrm{p}K_\mathrm{a} &=14-\mathrm{p}K_\mathrm{b}
\end{align}

Using the above formula, the three $\mathrm{p}K_\mathrm{a}$ for the conjugate acid of the conjugate base are: $13.5~(\mathrm{p}K_\mathrm{a_1})$, $10.3~(\mathrm{p}K_\mathrm{a_2})$, and $6.3~(\mathrm{p}K_\mathrm{a_3})$, where $\mathrm{p}K_\mathrm{a_1}=14-\mathrm{p}K_\mathrm{b_1}$, etc.

An important equation to note is that the $\mathrm{pH}$ at the equivalence point is given by the equation: $$\mathrm{pH_\mathrm{eq}}=\frac{\mathrm{p}K_\mathrm{a1}+\mathrm{p}K_\mathrm{a2}}{2}$$

This means that two of the equivalence points exist at $$\mathrm{pH_\mathrm{eq_1}}=\frac{13.5+10.3}{2}=11.9$$ and $$\mathrm{pH_\mathrm{eq_2}}=\frac{10.3+6.3}{2}=8.3$$

The $\mathrm{pH}$ range for methyl orange is $3.0~-~4.4$ and the $\mathrm{pH}$ range for phenolphthalein is $8.2~-~10.0$. Phenolphthalein would be useful for detecting the second equivalence point $(\mathrm{pH_\mathrm{eq_2}})$, since this shift occurs within phenolphthalein's range.

The final equivalence point cannot be calculated from the information given, but given that methyl orange's color transition occurs in an acidic medium, and also the relatively basic $\mathrm{p}K_\mathrm{a_1}$, methyl orange can be eliminated as a possible indicator.


Identification of the organic compound An organic compound A $(\ce{C10H14O})$ exhibits following characteristics.
A) It reacts with metallic sodium to give a colorless odorless gas.
B) It is oxidized by $\ce{KMnO4}$ to benzoic acid.
C) It can be resolved.
D) It does not give precipitate with iodine in presence of $\ce{NaOH}.$
E) It changes the color of acidified potassium dichromate from orange to bluish
green, and gives a chiral compound.
How many of the following compounds satisfy all the five criteria from A to E?

The given answer is the compounds 4, 5, 10 satisfy the criteria but I think the compounds 5, 6, 12 satisfy the criteria.
From statement A, we come to know it's an alcohol (which is common to every compound).
From statement B, we come to know that it has a benzylic hydrogen (which rules out the compound 11).
From statement C, we come to know that it has a chiral carbon. (which rules out options 4, 7, 9).
From statement D, we come to know that $\ce{OH}$ group is not attached to a methyl group (which rules out options 3, 10).
For statement E, I guess potassium dichromate will oxidise 2° alcohol to ketone and 1° alcohol to carboxylic acid and the compound still remains chiral (which rules out options 1, 2, 8).
So where's the mistake?

The given answer is wrong in many ways:

4 did not satisfy both criteria C and E, because it is not a chiral compound to begin with.
10 did not satisfy the criteria D, because it does give a precipitate with iodine in presence of $\ce{NaOH}$.

That gives only 5 in given answer to satisfy all criteria. Let's eliminate given compounds in systematic order:

That leaves only 5 and 6 to satisfy all criteria.


Dichloromethane cleaning Is it safe to drink out of a bottle that use to contain Dichloromethane?  What should I use to clean a bottle of Dichloromethane?  Is there any chemicals that I could use to dissolve the Dichloromethane and make it safe to drink out of?

More info:

So if I had a glass bottle containing >= 99.5% Dichloromethane with amylene as a stabilizer and left out overnight is that safe? The bottle was used in a chemistry department so it is high quality, I would assume. Made from the company Sigma-Aldrich and in a Glass bottle.

Safety regulations in every lab I have heard of strictly disallow mixing of containers of any sort that contain, have contained or will contain chemicals with containers of any sort that contain, have contained or will contain food, beverages or anything else intended for human consumption. You should never drink out of a bottle that had contained chemicals, you should never store food or drinks and chemicals together and you should not consume anything where there are chemicals. Your question has a premise that should never happen due to safety regulations.

With that gotten out of the way the answer is quick and dirty: It’s not the dichloromethane that is your prime point of concern. Dichloromethane boils at $40~\mathrm{^\circ C}$ and if you keep a bottle that contained it open and lying down overnight at room temperature, there are no traces of dichloromethane in harmful concentrations left. But unless you really go out of your way to buy $99.999~\%$ purity dichloromethane, it will contain impurities which generally do not evaporate as easily, i.e. which will stay in the bottle. Since no bottle of dichloromethane is designed for drinking out of it, there are no safety tests performed with these impurities or additives and thus you cannot assume it to be harmless even if dichloromethane is gone.

You should not read this answer to say that bottles of highest-purity dichloromethane be safe to drink out of. Never mix food or beverages and chemicals. Never, ever!


Difference between intercalated lithium and metallic lithium (Li-ion batteries) I have a question about Li-ion batteries. I sometimes see the two terms - intercalated lithium and metallic lithium - used when talking about this type of batteries and I was wondering what is the difference between the two?

I understand that the intercalation means that the Li cations are transferred from one electrode to another and that is why it is called intercalated lithium (or am I wrong?). But what about metallic lithium? What does that mean?

I found this on Wikipedia:

  Li-ion batteries use an intercalated lithium compound as one electrode material, compared to the metallic lithium used in a non-rechargeable lithium battery

But what makes them different? In other words, what are the differences between intercalated lithium and metallic lithium? Can anyone help me to understand this?

Intercalation is the addition of lithium ions into a host material without significantly changing the host's structure. In my research, I will often prepare test cells where the cathode is an intercalation material, like $\ce{CoO2}$, which can accomodate (or "intercalate") layers of lithium ions between the oxygen layers in the crystal lattice to form $\ce{LiCoO2}$. It's not that there are layers of cathode material surrounding lithium, but rather there's only 1 material, but lithium ions can move into and out of it.

$$ \ce{CoO2 + Li+ + e^- -> LiCoO2} $$

I mostly use lithium metal for the anode because it's easy to work with and has a stable potential, making it a good reference electrode:

$$ \ce{Li^+ + e^- -> Li_{(s)}} $$

In this setup we have an intercalation cathode ($\ce{LiCoO2}$), and a lithium metal anode. The major downside is that when you cycle lithium metal, it forms dendrites which can puncture the separator and short-circuit the cell, leading to catastrophic failure. To solve this problem, commercial cells use graphite anodes, which can intercalate lithium ions:

$$ \ce{C6 + Li+ + e- -> LiC6} $$

This solves the dendrite problem, but makes the battery less efficient. Since dendrite formation is only a problem with repeated cycling, non-rechargeable ("primary") batteries just use lithium metal instead. Lots of researchers have tried to solve "the dendrite problem" for decades, with little success.

The diagram below shows a lithium ion battery with an intercalation cathode on the right: the green parts are lithium, the blue parts are transition metals (Ni, Co, etc.) and the red spheres are oxygen. The anode on the left is the graphite intercalation anode, but you could replace it with a metallic lithium anode.


How to predict the outcome of a crossed Cannizzaro reaction? My question is, how do we predict the products of a cross-Cannizzaro disproportionation in presence of a concentrated base? i.e, How to judge which of the aldehydes (both don't have any alpha hydrogen) is reduced and which one is oxidized?

If we are asked

$$\ce{CH3-O-C6H4-CHO   + HCHO ->[OH^- (conc)] ?}$$

Case: if we see acc to RDS, which is hydride transfer, we must see which compound is better hydride donor. Then it will be $\ce{CH3-O-C6H4-CHO}$. This will get oxidized, and $\ce{HCHO}$ reduced.

Case: if we see which is a better nucleophilic substrate, $\ce{HCHO}$ will get oxidised, and $\ce{CH3-O-C6H4-CHO}$, reduced.

Now my problem is, which of the above cases is correct? i.e, do we see which one of the substrates is better $\ce{H-}$ donor, or which one of them is a better nucleophilic substrate?

Please advise which case is correct, and how to approach such problems.

To answer this, first let us look at the mechanism of the Cannizzaro reaction in general. It is depicted in scheme 1 as a cyclic transition state. The cycle itself isn’t as important, the realisation that one aldehyde is present as a (deprotonated) hydrate after the action of $\ce{OH-}$ is.

Scheme 1: General mechanism of a Cannizzaro reaction.

In your reaction, you are reacting formaldehyde with a (substituted) benzaldehyde. The question boils down to ‘can we predict, which aldehyde will form a hydrate mor easily?’ And the answer is: we can!

$$\begin{array}{cc}\hline
\text{carbonyl} & \text{hydrate ratio} \\ \hline
\ce{H2CO} & 99~\% \\
\ce{MeCHO} & 60~\% \\
\ce{Me2CO} & < 1~\% \\ \hline\end{array}$$

We have a clear tendency for formaldehyde to almost exclusively form hydrates under standard conditions. Thus, we can conclude that formaldehyde will alwys be the hydrate partner, end up oxidised and reduce its reaction partner as shown in scheme 2.

Scheme 2: Cannizzaro reduction of an aldehyde with formaldehyde as sacrificial aldehyde.


Synthesis of a spirocycle I am trying to come up with a synthesis of 7-hydroxyspiro[4.5]decan-1-one;

I think I could use these starting reagents below;

Could I potentially generate the enolate of the cyclic ketone starting material and it would attack the double bond of the alcohol? If this is correct, how would the ring close to create the spirocycle? Thanks.

Mesylate the alcohol, react that with the enolate, epoxidise the double bond then reform the enolate.


Using baking soda paste (alkaline) to "neutralize" honeybee venom (acidic). Is this scientifically sound? I recall reading (courtesy: my 9th grade Chemistry text-book, chapter Acids, Bases and Salts), something along the lines of:

  Bee venom is acidic. This is why the area of skin (stung by a bee) is treated (first-aid) by the topical application of a sodium bicarbonate paste. This method works by neutralization of the acidic venom.

The problems with this theory? I see three:

Firstly,

A honeybee's stinger can grow to be as long as 3mm. If bee venom were acidic, then how would the baking-soda paste neutralize it (entirely) by simply applying it on the surface?

Secondly,

Wikipedia provides the composition of honeybee venom;

  Melittin

  Apamin

  Adolapin

  Phospholipase A2

  Hyaluronidase

  Histamine

  Dopamine and Noradrenaline

  Tertiapin

As far as I know...none of these are actual acids, instead, they're pretty much all proteins. Which appears to seriously contradict the popular "Bee venom is acidic" notion.

Thirdly,

Some sources (online) suggest that application of vinegar (which is acidic) also relieves the pain. Additionally, even meat tenderizers (proteases), toothpaste and wet aspirin tablets seem to do the trick.

So I'll break up my question into two bits (to hopefully, make it easier to answer):

1) Is honeybee venom acidic? If so, what substance present in it is responsible for the "acidic" nature of the venom?

2) Why does applying baking-soda (paste) relieve the pain? Does it somehow denature/"neutralize" the venom?

Honeybee venom is acidic. It contains formic acid, also known as $\ce{HCOOH}$. However,
Rubbing baking soda probably isn't really reducing the pain because

The formic acid isn't causing the pain
The venom is injected under tissue
The ratio of venom to baking soda is way out of proportion

For more information, visit this website.


What red and blue area represent in a orbital software? I have a software in my android device named hydrogen atom. It show orbital of hydrogen we have to set the quantum numbers and probability of finding particle. But what the blue and red colour area in below pictures represent.

To answer this question, we need to start from the beginning.

Movement of an electron is described by wave function. The best layman description for it is that wave function is a quantum equivalent of trajectory. It isn't exactly trajectory, but it serves as equivalent for quantum mechanics.
A trajectory repeating itself and wrapped around an object is called an orbit. A quantum mechanical equivalent is a wave function, remaining constant in time. Such a function for hydrogen atom is called orbital.
Wave function is a function. It means, that it can be positive and negative. (Well, in case of a stationary function, see below). It usually doesn't matter in simple chemistry, but once we start talking about orbital interactions it becomes important. In short, orbitals interact with formation of orbitals with reduced energy as long as lobes with same sign overlap. Thus, it is important to know sign of the wave function in lobes of the orbitals.
So, whenever you see an orbital depicted using two colors, it means two different signs of the wave function. Note, since math, it is not important which color depicts what sign. If one inverses sign of all lobes simultaneously, the result will depict the same system, actually.
Side note 1: the surfaces depicted in the picture likely are isosurfaces (i.e. the function in consideration has same value in all surface points) of square of wave function that enclose volume containing some fixed amount of electron density (usually somewhere around 90-95%). There is no good way to depict wave function/orbital in 3 dimensions really.
Side note 2: wave function has well defined sign only in case of stationary system (i.e. unchanged with time). Whenever we talk about system evolving in time, we need to consider arcane math of complex numbers.


Why are hydrogen bonds in an antiparallel beta sheet stronger than those in parallel beta sheets? Beta sheets are illustrated as such in most diagrams, where:

In an antiparallel β-sheet, the polypeptide strands are arranged such that a $\ce{C=O}$ and an $\ce{NH}$ from adjacent strands face each other, and the $\ce{H}$ forms an H-bond with the $\ce{O}$, with $\ce{C=O**HN}$ all lying on the same plane.
In a parallel β-sheet, the strands are literally parallel, and so the H-bond formed has to form a sort of angle, where the $\ce{C=O}$ and $\ce{NH}$ no longer lie on the same plane (when viewed from the perspective of the diagram).

Most explanations explain the stronger bond strength by stating the obvious from the diagram: that the "hydrogen bonds are aligned directly opposite each other, making for stronger and more stable bonds." (taken from chem libretexts, https://chem.libretexts.org/Bookshelves/Biological_Chemistry/Supplemental_Modules_(Biological_Chemistry)/Proteins/Protein_Structure/Secondary_Structure%3A_-Pleated_Sheet)
This seems intuitive, but my issue arises when considering some orbital geometry: in a peptide bond, the $\ce{C}$, $\ce{N}$ and $\ce{O}$ have orbitals with ${sp^2}$ hybridization, meaning that bond pairs/lone pairs are organized in a trigonal planar orientation around each of those atoms. As a consequence of this, one can also observe that all the atoms illustrated in the diagram should lie on the same plane, making things rather easy to visualize.
However, if one considers that the 2 lone pairs and one bond pair in the $\ce{O}$ atoms are orientated in a trigonal planar fashion, then shouldn't the ideal, strongest, H-bond is orientated at an angle (specifically $\pu{{60}^\circ}$) relative to the $\ce{C=O}$ bond? Now, if I observe the diagram again, it seems this $\pu{{60}^\circ}$ H-bond angle is better fulfilled in the parallel strand than in the antiparallel one. So shouldn't the H-bond in the parallel strand, theoretically, be stronger?
A possible explanation would be that the bond length in parallel strands would be longer, thereby weakening it. But that's not how most textbooks explain it, so what am I missing out here?

Why are hydrogen bonds in an antiparallel beta sheet stronger than those in parallel beta sheets?

They are probably not. The difference is small, and depends on sequence context. Also, the diagrams do not reflect the typical conformation of the backbone in beta sheets. To complicate matters, most sheets are mixed rather than purely parallel or anti-parallel, and many are irregular (different length of strands, bulges, one strand part of two sheets etc).

"hydrogen bonds are aligned directly opposite each other, making for stronger and more stable bonds."

No, the geometry around the hydrogen bond is fairly similar. Here is a depiction of structures from an ab initio study showing the geometry:

[OP ..] shouldn't the ideal, strongest, H-bond is orientated at an angle (specifically 60∘) relative to the C=O bond?

Based on analyzing known structures with a single hydrogen bond for a given acceptor, the hydrogen lies in the plane of the carbonyl group but has a fairly smooth distribution with many examples having a 180 degree angle (carbonyl carbon, carbonyl oxygen, hydrogen).

A possible explanation would be that the bond length in parallel strands would be longer, thereby weakening it. But that's not how most textbooks explain it, so what am I missing out here?

The ab initio study talks about the primary effect (having a hydrogen bond), secondary effects (repulsion between partial charges of oxygen atoms, repulsion between partial charge of hydrogen atoms, C-H...O hydrogen bond in the parallel sheet) and tertiary effects (induced dipoles due to formation of hydrogen bonds). You have to consider all of these; as the differences in hydrogen bonding geometry are subtle, the secondary and tertiary effects will become significant.


Solutions of Group 1 and Group 2 metals in Ammonia When we add Group-1 and Group-2 metals to liquid ammonia, they dissolve to form metal cations and solvated electrons.
$$\ce{M + NH3(liq) -> M+ + e-}$$
Now, when the G-1 solutions evaporate, we get the metal back, but in the case of G-2 solutions (except Beryllium), they give a metal complex, $\ce{[M(NH3)6]}$.

"Evaporation of the ammonia from solutions of Group 1 metals yields the metal, but with Group 2 metals evaporation of ammonia gives hexammoniates $\ce{[M(NH3)6]}$ of the metals. These slowly decompose to give amides."
Page No. 335, Chapter 11-'Group 2 - the alkaline earth elements', Concise Inorganic Chemistry

and,

"The alkaline earth metals except beryllium form similar solutions [to alkali metals] readily, but upon evaporation, a solid 'ammoniate', $\ce{[M(NH3)x]}$ is formed."
Page No. 249, Chapter 9-'Chemistry in Aqueous and Nonaqueous Solvents', Inorganic Chemistry: Principles of Structure and Reactivity

What is the reason for this contrasting behaviour, and do all Group 2 metals(except beryllium) exhibit such a property?
Source:
1. Lee, J.D.; Concise Inorganic Chemistry; Wiley-Blackwell; Dec. 1998
2. Huheey, James E., Keiter, Ellen A., Keiter, Richard L.; Inorganic Chemistry: Principles of Structure and Reactivity, 4th Edition; Pearson Education Inc.; 1993

As per P. Bahadur's Objective Chemistry 1st Year Program for IIT-JEE and All Other Engineering Entrance Examinations, under the chapter "s-Block Elements - The Alkaline Earth Metals", under Solubility in Liquid Ammonia, it reads:

  Like Alkali Metals, alkaline earth metals also dissolve in liquid ammonia and form coloured solutions. When such a solution is evaporated, hexa-ammoniate, $\ce{M(NH3)6}$ is formed, which slowly decomposes to amides.
  $$\ce{M(NH3)6 -> M(NH2)2 + 4NH3 +H2}$$

(I noticed that it is almost exactly the same as JD Lee's Concise Inorganic Chemistry text.)

So far, it doesn't specify whether this metal $\ce{M}$ is only Beryllium or is the representation of the other alkaline earth metals as well... Since this is the case, I can safely assume that this applies to all the alkaline earth metals, and not just beryllium.

Another point mentioned in the book by Bahadur, under the chapter "s-Block Elements - The Alkali Metals", under Complex Ion Formation is that the Group 1 metals, i.e., the alkali earth metals have a larger size, low nuclear charge and hence don't tend to form complexes too easily. On the other hand, the Alkaline Earth metals are smaller, have higher nuclear charge, and hence, have a greater tendency to accept electrons, forming complexes. In Bahadur's Book, under "s-Block Elements - The Alkaline Earth Metals", under Formation of Complexes, he writes that:

  However, $\ce{Be^{2+}}$ on account of smaller size forms many complexes such as $\ce{BeF3-}$, $\ce{BeF4^{2-}}$ and $\ce{Be(H2O)4^{2+}}$. Of the others, only $\ce{Mg}$ and $\ce{Ca}$ show much tendency to form complexes in solution and these are usually with oxygen donor ligands.

$\ce{NH3}$ is a ligand, and hence acts as a powerful complexing agent. This is why the $\ce{NH3}$ coordinates with these metals, forming hexa-ammoniates.

Source:

Bahadur, P.; Objective Chemistry 1st Year Programme for IIT-JEE & All Other Engineering Entrance Examinations; GRB Books; 2016
Lee, J.D.; Concise Inorganic Chemistry; Wiley-Blackwell; Dec. 1998


What does oscillator strength have to do with state character? I was reading S.H.A.R.C. Manual (Surface Hopping in the Adiabatic
Representation Including Arbitrary Couplings) and came across the following paragraph:

  "Each state is colored, with one color as contour and another at the core of the line. The contour color represents the total spin expectation value of the state. The core color represents the oscillator strength of the state with the lowest state. See figure 7.3 for the relevant color code. Note that by definition the “oscillator strength” of the lowest state with itself is exactly zero, hence the lowest state is also light grey. This dual coloring allows for a quick recognition of different types of states in the dynamics, e.g. singlets vs. triplets or nπ ∗ vs. ππ ∗ states."

How can one recognize nπ ∗ vs. ππ ∗ from this kind of colour scheme? What does oscillator strength have to do with state character?

This does not seem to make any sense to me either unless I'm missing something very obvious. Suffice it to say that spin changing transitions are generally strongly forbidden, unless paramagnetic electron spin or heavy atoms make them allowed, and that spin 'forbidden-ness' is stronger than symmetry 'forbidden-ness'. Symmetry forbidden transitions can be made allowed by 'intensity borrowing' (Herzberg-Teller) involving a higher electronic state & a vibration of suitable symmetry, as in benzene.  When both spin & symmetry are forbidden transitions are very weak indeed. $n\pi ^*$ transitions are often symmetry forbidden (e.g. in carbonyls) and so weak compared to $\pi \pi ^*$. Other than these general rules you just have to look at the individual molecules to work out what is going on, it might be quicker than trying to understand the diagram.


What is the mechanism of the reaction of a ketone with sodium nitrite/ hydrogen chloride? What is the mechanism of reaction of a ketone with $\ce{NaNO2/HCl}$?
Can $\ce{NaNO2/HCl}$ be considered as freshly prepared $\ce{HONO}$ here? I am confused about how the reaction will proceed.

$\ce{NaNO2/HCl}$ is as you suggest a source of $\ce{HONO}$. This will be protonated by $\ce{HCl}$ to give the nitrosonium ion $\ce{NO+}$ (in equilibrium with nitrosyl chloride). This strongly electrophilic species will react with the enol of the ketone to nitrosate the alpha position. This nitroso compound then rearranges to give the alpha oxime.

Image from: Emilia Iglesias and Isabel Brandariz, A further study of acetylacetone nitrosation. Org. Biomol. Chem. 2013, 11, 1059-1064. DOI: 10.1039/C2OB26073E; available as pdf from pdfs.semanticscholar.org.


Bimolecular and Isomerization reactions Is it possible to describe a reaction of the
$$\ce{A + B <=> AB}$$
exactly using the solution of an isomerization reaction?
$$\ce{A <=> B}$$

I just can not figure it out the equations seem so similar but the solution in my script is so convoluted that I can not prove they are equal.

Equations for top reaction:
$${\partial c_\ce{AB} \over \partial t} = -k_\mathrm{off} c_\ce{AB} + k_\mathrm{on} c_\ce{A} c_\ce{B}$$
and for the bottom:
\begin{align}
{\partial c_\ce{A} \over \partial t} &=
  -k_{\ce{A -> B}} c_\ce{A} + k_{\ce{B -> A}} c_\ce{B}\\
{\partial c_\ce{B} \over \partial t} &=
   k_{\ce{A -> B}} c_\ce{A} - k_{\ce{B -> A}} c_\ce{B}\\
\end{align}

Could we not just represent $\ce{A + B <=> AB}$ as a new reaction with $\ce{C $=$ A + B}$ and $\ce{D $=$ AB}$ and thus use the isomerization results on $\ce{C <=> D}$?

I could write a set of equations for the $ A+B \rightleftharpoons AB $ reaction :
$${\partial c_{AB} \over \partial t} = -k_{off} c_{AB} + k_{on} c_A c_B$$
$${\partial c_{A}c_{B} \over \partial t} = +k_{off} c_{AB} - k_{on} c_A c_B$$

which looks suspiciously like the $$A \rightleftharpoons B$$  reaction.

A starting point might to prove they can be used to express each other would be to use the equilibrium relation
$$k_d = { c_\ce{A} c_\ce{B} \over c_\ce{AB}}$$
with the relation for the total number of species A $c_\mathrm{total} = c_\ce{A} + c_\ce{AB}$ to get :
$$c_\ce{A} c_\ce{B} = k_d (c_\mathrm{total} -c_\ce{A})$$

TL;DR; Addition in a chemical reaction does not mean mathematical addition, and translating this operation to a rate expression makes no physical sense.

Elaboration:
Your idea $\ce{C = A + B}$ changes the order of reaction, and has no
physical basis.

The product of two concentrations, $\ce{c_{a}}$ and $\ce{c_{b}}$ in the equation:
$${\partial c_\ce{AB} \over \partial t} = -k_\mathrm{off} c_\ce{AB} + k_\mathrm{on} c_\ce{A} c_\ce{B}$$
has the physical interpretation that $\ce{A}$ and $\ce{B}$ must collide for reaction to occur, and is hence a second order reaction.

Your proposed solution would change that to first order in a fictional component, $\ce{C}$, and leads you astray.

To help clarify, consider the following reaction:
$$\ce{Br^{\cdot} + Br^{\cdot} <=> Br2}$$

In this case A would be $\ce{Br^{\cdot}}$ and B would be $\ce{Br^{\cdot}}$, and thus your C would be $\ce{[Br^{\cdot}] + [Br^{\cdot}] = 2 [Br^{\cdot}]}$.  The rate expressions would therefore be first order in $\ce{Br^{\cdot}}$  as opposed to second order.

Other comments:

You also forgot to include the other reactions for the bimolecular case:
$${\partial c_\ce{A} \over \partial t} = {\partial c_\ce{B} \over \partial t} = k_\mathrm{off} c_\ce{AB} - k_\mathrm{on} c_\ce{A} c_\ce{B}$$Omitting these may be leading to some confusion.
Little k is a rate constant.  Equilibria are described with big K.  It is unclear what $\ce{k_{d}}$ means in your question.


Why weren't the moles of oxygen calculated the same way as C? Combustion analysis of a $\rm1.500~g$ sample of ascorbic acid yields $4.023\rm~g$ of $\ce{CO2}$ and $\rm0.96~g$ of $\ce{H2O}$. What is the empirical formula of ascorbic acid?

Actually my question is that in many solutions I saw it's solved like that:

In $\ce{CO2}$ the moles of C is calculated as $$\frac{4.023}{44.02}=0.9139\rm~mass=1.0976~g$$

and the $\ce{H2O}$ is solved with getting out the H moles with same way that C was solved and the mass is
$$\rm m~H=0.10761~g$$
 the mass of O will be $$\rm 1.500-(0.10761~g-1.0976~g)=0.29489~g~moles=0.1842$$
My question is why didnt we use the same procedure we did with H and C to compute the moles with O? Why did we make it in the last step?

You have done combustion analysis, so you have introduced a large amount of oxygen to the ascorbic acid sample.  Because of this, you can't know how much of the oxygen in the $\ce{CO2}$ and $\ce{H2O}$ you produced actually came from the acid sample and how much came from the atmosphere.

But you do know that the mass of your ascorbic acid is the mass of all the carbon plus the mass of all the hydrogen, which you calculated, plus the mass of oxygen.  The difference in mass between what you have calculated and what you started with is the mass of oxygen that was in the ascorbic acid.

If there were any other elements in your sample, such as sulphur or nitrogen, those amounts would also have to be calculated before finally working out how much mass you haven't accounted for which must have been oxygen.


Ratios of produced gases in water electrolysis I just did a little water electrolysis experiment at home by immerging two aluminum electrodes in water with a small amount of table salt. The voltage of the battery was 6 V.

What surprised me is that I was able to collect a much bigger volume of gas at the electrode connected to the - than to the + (about a factor 10 by eye). I would have expected a ratio of 1/2 (2 times more hydrogen than oxygen).

Why did I get less gas than expected at the electrode connected to the +?
Was the oxygen consumed by oxydising the electrode? (It appeared quite damaged at the end.)
How can I get a ratio of 1/2?

We know from stoichiometry that you did produce twice as many moles of $\ce{H}$ at the negative electrode as $\ce{O}$ at the positive electrode. However, we don't know a few important things:

What were the actual volumes of the gases that were finally collected? Volume is notoriously hard to estimate by eye, unless you are comparing the levels in two identical containers. So, we'll take your word for it that it was "more", but we won't place any bets on the factor of 10.
What happened to the oxygen and hydrogen after the water split? Did it react?
What else was in the water that could have reacted? We know there is $\ce{NaCl}$, but maybe there are other dissolved compounds as well?

Question 1 can't really be answered here - the best we can do for now is assume that the volume of hydrogen was 2-10 times larger than that of oxygen. To find out for sure, you would need to conduct another experiment using graduated collection vessels.
Question 2 we can partially answer. If the electrode was corroded, then something reacted - either chlorine, oxygen, or some other more complicated set of reactions. We can get a rough estimate of the volume of oxygen this might consume by assuming it was oxidized by oxygen. If that 0.1 g of the electrode was oxidized, and that it was oxidized into $\ce{Al(OH)3}$, then at 1 atm and 298 K it would consume roughly 137 mL of $\ce{O2}$:
$\ce{Al^3+ + 3OH^- -> Al(OH)3}$
$0.1 \space \rm{g \space Al} * \frac{1\rm{\space mol \space Al}}{27\rm{\space g \space Al}} * \frac{3 \space mol \space O}{1 \space mole \space Al} * \frac{1 \space mol \space \ce{O2}}{2 \space mol \space O} = 0.0056 \space mol \space \ce{O2}$
$PV = nRT$ (oxygen is close enough to ideal under normal lab conditions)
$V = \frac{nRT}{P} = \frac{(0.0056 \space \rm{mol} \space \ce{O2})(0.08206 \space L \space atm \space mol^{-1}K^{-1})(298 \space K)}{1 \space \rm{atm}} = 137 \space \rm{mL}$
$\ce{Al(OH)3}$ is a likely candidate for a reaction product because of the white flakes you observed - aluminum hydroxide is insoluble in water. Aluminum oxide ($\ce{Al2O3}$) is also a potential candidate, and would consume 68 mL of $\ce{O2}$ under the same conditions.
So, assuming 0.1 g is a reasonable estimate for the mass of corroded aluminum, it seems very possible that what happened is the oxygen reacted with the aluminum electrode, reducing your yield of oxygen gas.
How can you get around this? Well, my five minutes of google research did not turn up any good candidates for inert electrodes. Usually graphite electrodes are recommended for other electrolysis setups, but they will react with oxygen as well. I have seen stainless steel recommended, but I would think that gold/platinum/silver are probably your best bets.
Question 3 is also hard to answer. We know that the $\ce{NaCl}$ will react and produce hydroxide ion (further supporting our hypothesis for aluminum hydroxide formation) at low concentrations, and chlorine gas at higher concentrations. If it were producing chlorine gas, you would know - it would be a pale yellow/green and it would make you sick as soon as you smelled it. So it is unlikely that your concentration of table salt was high enough for that to be the competing reaction. However, it would contribute to the aluminum oxidation. The easiest way to rule this out is to not use table salt. I have seen baking soda (sodium bicarbonate) recommended instead, and that would probably work. However, it might still produce hydroxide in solution. What about other unknown substances? If you used tap water, it is likely that there is additional chloride, possibly fluoride, and probably various magnesium and calcium salts. All of these would react during the hydrolysis, and would affect yields. The way to eliminate them is to buy or make distilled or deionized water.
Summary:

Why did I get less gas than expected at the electrode connected to the +?
Was the oxygen consumed by oxidising the electrode? (It appeared quite damaged at the end.)

The answer is most likely yes, the oxygen reacted with the aluminum. I would bet that it formed aluminum hydroxide, which then precipitated into the white flakes you observed.

How can I get a ratio of 1/2?

You might not be able to get it to exactly 1:2, but if you use sodium bicarbonate or a weak acid (vinegar would probably work) instead of salt as an electrolyte, and an inert electrode - like gold, or platinum, you can probably get close. Keep in mind - hydrogen gas is usually the desired product in hydrolysis. Oxygen is cheap if you don't need it to be pure! And even pure oxygen is relatively cheap compared to other gases.


How do you change the cis-trans isomerism of compounds? I would like to know a general procedure for the synthesising cis isomers from their trans counterparts and vice versa.

In general, any setup that gives enough energy to the compound can isomerize it. This can be done by heating it, irradiating it with visible or UV light, or by using a catalyst (reducing the energy required for transition rather than supplying more energy). Excitation of electrons from the $\pi$ to the $\pi^*$ (bonding orbital to anti-bonding orbital) allow free rotation about the $\sigma$ bond. Relaxation back to the $\pi$ orbital locks the configuration. The specific procedures depend on the compound of interest.

FYI, photoisomerization (light-induced isomerization) is the principle on which sight is based.


Covalent bond nature in electronegativity gap Does a large difference in electronegativity mean the covalent bond is weaker?

In a covalent bond between two atoms of different electronegativities, the bonding electrons are pulled towards the more electronegative electron. This makes the bond polar (eg. $HCl$).

In a diatomic molecule, the electronegativity is the same so the electrons are equally attracted to either atom. Does this make diatomic covalent bonds stronger than the former? And if so, as the difference in electronegativity increases, does the strength of the covalent bond decrease?

Strength of covalent bond depends on many factors, one of the key factors is "extent of overlapping of orbitals". For e.g., if you consider $\ce{H2}$ molecule, there is a head on overlapping of $\mathrm{1s-1s}$ orbitals involving two hydrogen atoms, but in the case of, say, $\ce{HCl}$, there is a head on overlapping of $\mathrm{1s-3p_z}$ orbital ($\mathrm{1s}$ of hydrogen, and $\mathrm{3p_z}$ of chlorine), and we all know that $\mathrm{p-s>s-s}$, that is, overlapping between a $\mathrm{p}$-orbital and a $\mathrm{s}$-orbital is stronger than $\mathrm{s}$ and  $\mathrm{s}$ orbital, and stronger the overlapping more is the covalent bond energy.

So, it largely depends on the type of orbitals involved and the nature of overlapping.

Check this to learn more about overlapping.


How does a Pt and reference electrode measure the cell potential in a potentiometric titration? I'm trying to perform a potentiometric titration to determine the redox potential of a $\ce{Fe^3+|Fe^2+}$ couple. To do this, I am using an $\ce{Ag|AgCl}$ electrode connected to an inert $\ce{Pt}$ electrode, which is submerged in a solution containing $\ce{Fe^3+}$ ions. I then titrate it with $\ce{Co^2+}$, which reduces it to $\ce{Fe^2+}$. The reference and inert electrode are connected with a standard voltmeter.

What I don't understand is why the inert electrode measures the redox potential of the $\ce{Fe^3+|Fe^2+}$ couple. As per my understanding, there is an electron flow from the reference electrode into the inert electrode, which would get accepted into the Fe solution, but now, the transfer of electrons is only between the $\ce{Co^2+}$ and $\ce{Fe^3+}$ ions, so what happens to that electron? Can someone explain exactly how the redox potential of the $\ce{Fe^3+|Fe^2+}$ couple gets measured, with reference to electron transfer between the reference and inert electrodes?

It is a very good question. Your question has nothing to do with current being drawn or not or a reference electrode etc. Imagine a single Pt electrode dipped in an Fe(II) solution and forget about the presence of cobalt. Imagine that a metal is a source (sea) of electrons.

The moment you dip the electrode, instantaneous exchange of electrons takes place between the metal and the reducible or oxidizable ions. Electrons can be released by Fe(II) to form Fe(III) and similarly electrons can be gained by Fe(III) to form Fe(II). This exchange rate is dependent on the concentration ratios of Fe(II) and Fe(III).

As a result of this exchange, the Pt metal develops a charge. It is very surprising that you only need to remove a very very very small number of electrons to develop a measurable surface charge. Hence, there is no perceptible change in the concentration of Fe ions. If platinum electrode develops a charge, the interface (solution electrode boundary) develops an opposite charge until equilibrium is achieved.

Now given the fact that Pt electrode is charged i.e., if we connect Pt to another electrode, we can sense the potential difference between the Pt electrode and another electrode, which can be any other electrode. Now that you wish to have a stable. For that purpose you need an electrode whose potential is fixed i.e. a reference electrode.


Why are two C-O single bonds more stable than a C=O double bond (hemiacetal)? In my organic chemistry book it says that the formation of cyclic hemiacetals is energetically favorable in comparison to the open chain aldehyde as two C-O single bonds are more stable than a C=O double bond. Why is this the case?
To be more precise, I am specifically talking about monosaccharides.

I quote from "Organic Chemistry" by Clayden, Greeves, Warren (I am translating from German so it's not 100 % accurate):

  [Translated from German] The change of enthalpy that goes along with the reaction results from the change of bonds: In this case a C=O double bond becomes two C-O single bonds, and those C-O single bonds are slightly more stable than the C=O double bond, resulting in  $\Delta$H being slightly positive.

a C=O double bond becomes two C-O single bonds, and those C-O single bonds are slightly more stable than the C=O double bond resulting in ΔH being slightly positive.

The tables I found (e.g. OpenStax Chemistry, Wikipedia) have the C=O bond with roughly twice the bond dissociation energy compared to the C-O bond, and in all cases the C=O bond is slightly more stable than the two single bonds. So the statement in the textbook is surprising. As these values are aggregates derived from large sets of molecules, their values will vary depending on which set of molecules they are based on.

If you go from less stable bonds (easier to break, smaller bond dissociation energy) to more stable bonds (harder to break, larger bond dissociation energy), the enthalpy of reaction should be negative (exothermic), not positive as claimed in the statement above.

For completeness, these are not the only changes. In addition, the O-H bond of the hydroxyl group is broken, and the O-H bond of the hemiacetal is formed. If you are estimating bond dissociation energies from a simple table, however, this will just cancel out because it is treated as the same kind of bond.

Estimating enthalpy of reaction with bond energies

Bond energies give you a rough estimate of reaction enthalpy, but will not pick up subtle differences between cases that are treated equally using this estimation. Here are two other examples where a C=O double bond is replaced by two C-O bonds, and the counts of all the other bonds are equal:

Hydration of a ketone, e.g. acetone and water reacting to form 2,2 propanediol. According to this web site, the reaction is slightly endothermic (about +6 kcal/mol). On the other hand, formaldehyde is mostly hydrated in water, so ketones and aldehydes behave slightly differently.
Butanone isomerizing to tetrahydrofurane. This is a hypothetical reaction, but NIST data suggests that it is also endothermic, again different than the estimate from bond dissociation energies.

So first, you have to look at the experimental evidence, and then you can try to rationalize it with a simple model, which sometimes will give the expected answer and sometimes will not.

"energetically favorable" vs major species

Just because a reaction is exothermic does not mean the product will be the major species. Not all monosaccharides form rings, and mixtures of ketones and alcohols typically do not form hemiacetals as major species. There are other energetic factors  (e.g. ring strain), and entropic factors (e.g. intramolecular vs intermolecular, conformational degrees of freedom) that also contribute to making one or the other species more prevalent.

  Why are two C-O single bonds more stable than a C=O double bond (hemiacetal)?

This is what we observe. A simple model like adding up bond energies seems to support it, but maybe that is just luck. Computational methods are sufficiently sophisticated to predict that certain monosaccharides form cyclic hemiacetals. I'm not sure this answers the question of "why?", though.


Composition and pH of water produced by air conditioners I would like to know if there are specific studies or research regarding the condensed water that is generated by air conditioners.

Many people feel that this water is acid and corrosive but could not find anything that confirms or denies this belief.

If someone could send me a link or document with a reliable scientific research would help me.

Pure condensed water is neutral (pH 7). However, water in contact with air is slightly acidic (approximately pH 5.6) due to dissolved carbon dioxide $\ce{(CO2)}$. Remarkably, the same applies to rainwater.

Furthermore, water condensed from air is not pure. The condensing water collects small amounts of various airborne substances (soluble gases and vapours, dust, and even radioactive particles from radon decay). (Note the large volume of air that is passed through the air conditioner while a certain amount of water is condensed.) Usually, these substances do not significantly affect the pH of the water. However, when the air contains large amounts of pollutants, such as sulfur dioxide or nitrogen oxides (e.g. produced in industrial processes), the condensed water can become more acidic (like acid rain, which can have pH values of about 4).

By way of comparison, typical values for black coffee are about pH 5.0–5.1, and for orange juice about pH 3.3–4.2.


At what temperature does tautomerization stop? The enol and keto forms are said to be tautomers of each other, and the interconversion of the two forms involves the movement of an alpha hydrogen and the shifting of bonding electrons.

At what temperature does tautomerization stop? (E.g I can obtain both the enol and keto form without them interconverting)

To make this simple, let's just assume the molecule of choice is: acetaldehyde

In addition there is no tunneling of the two tautomers. Lastly it is known the the two tautomers of acetaldehyde are stable and do not convert at 11 K with argon matrix trapping. What's the highest temperature that do not interconvert?

Tautomerisation, as an observable process by NMR, is entirely system dependent, and there are many cases where non-exchanging or very slow-exchanging systems are observed at room temperature and higher. As always, equilibrium is both sterically and electronically controlled. Some systems require very low temperatures (liquid nitrogen and liquid helium temperatures) to slow the exchange processes, and of course, going to higher magnetic fields will also improve the case of observing individual exchanging species.

Acetaldehyde is a very small molecule; this would require extremely low temperatures; as you have said 11K allows for observation without interconversion.

A lot of this sounds like a homework question .... you might want to be more specific about what your own research is contributing towards answering your own questions.


Nucleophilic Addition of Hydrohalic Acids Hydrogen cyanide shows nucleophilic addition at carbonyl compounds' carbon to form a cyanohydrin. Alcohols, similarly, form acetals or ketals. Will hydrogen chloride - or hydrohalic acids in general - show a similar reaction to form alpha halohydrins?

As you noted, hydrogen cyanide will react with carbonyl compounds to form cyanohydrin and alcohols to form ketals/acetals. The list can be expanded to include aminals after attack of a nitrogen nucleophile, or thioacetals after sulphur attacks.

So why does it not happen with hydrogen halides? Well, in the case of the above mentioned, the reverse reaction is usually not favoured. In fact, ketals are used as protective groups for ketones (usually the reaction with ethan-1,2-diol or propan-1,3-diol); and propan-1,3-dithiol is similarly used for umpolung reactions. Likewise, the cyanohydrin has a new carbon-carbon bond formed, and these are relatively stable. To remove the ketal protective group, a rather strong Brønsted acid is often required along with an excess of water.

But in the case of a halide, that can just easily diffuse away again as a happy little halide; carbonyl double bond will be reformed. In fact, this reaction can be (and is!) used in exactly the opposite direction: liberate a hydroxide on the same carbon as a halide, and the halide will be eliminated to create the double bond.


How to get ionic radii for coordination number 12? I am studying $\ce{ABX3}$ perovskites, and I would like to calculate Goldschmidt tolerance factors for them. The $\ce{A}$ sites in these materials have a coordination number $12$.

The Shannon's paper (R. D. Shannon, Revised effective ionic radii and systematic studies of interatomic distances in halides and chalcogenides. Acta. Cryst. A 32, 751 (1976)) provides values for ionic radii for the coordination number $12$ only for a few elements, but doesn't do it for most other elements.

Are there any sources, where I can get ionic radii for $\text{C.N.} = 12$ for all elements?

The commonly used method of obtaining ionic radii for higher coordination numbers (C.N.) is to extrapolate values from the Shannon's scale [1] using the relationship between ionic radius and coordination number proposed by Zachariasen [2]:

  ... the bond lengths $D(N_1)$ and $D(N_2)$ for cation coordination numbers $N_1$ and $N_2$ were related as follows

  $$D(N_2)=D(N_1)(N_2A_1/N_1A_2)^{1/n} \tag{1}$$

  where $n + 1$ is the exponent in the Born repulsion term of the lattice energy and $A_1/A_2$ is the ratio of the Madelung constants. The quantity $A_1/A_2$ was taken to be $0.972$ for $N_2/N_1 = 12/9 = 8/6 = 4/3$ and $0.927$ for $N_2/N_1 = 12/8 = 9/6 = 6/4 = 3/2$.

Jia [3] used Zachariasen's formula and published effective ionic radii for all lanthanides with C.N. $6$ to $12$.

Using the same approach, selected ionic radii for the 12-coordinated $\ce{A}$-sites in oxo-perovskites $\ce{ABO3}$ [4] as well as halide perovskites $\ce{ABX3}$ [5] have been determined.

References

Shannon, R. D. Revised Effective Ionic Radii and Systematic Studies of Interatomic Distances in Halides and Chalcogenides. Acta Cryst. A 1976, 32 (5), 751–767. https://doi.org/10.1107/S0567739476001551.
Zachariasen, W. H. Bond Lengths in Oxygen and Halogen Compounds of d and f Elements. Journal of the Less Common Metals 1978, 62, 1–7. https://doi.org/10.1016/0022-5088(78)90010-3.
Jia, Y. Q. Crystal Radii and Effective Ionic Radii of the Rare Earth Ions. Journal of Solid State Chemistry 1991, 95 (1), 184–187. https://doi.org/10.1016/0022-4596(91)90388-X.
Li, C.; Soh, K. C. K.; Wu, P. Formability of $\ce{ABO3}$ Perovskites. Journal of Alloys and Compounds 2004, 372 (1), 40–48. https://doi.org/10.1016/j.jallcom.2003.10.017.
Li, C.; Lu, X.; Ding, W.; Feng, L.; Gao, Y.; Guo, Z. Formability of $\ce{ABX3}$ ($\ce{X = F, Cl, Br, I}$) Halide Perovskites. Acta Cryst. B 2008, 64 (6), 702–707. https://doi.org/10.1107/S0108768108032734.


Using baking powder to make baked food crispy The recipe I use for chicken wings achieves a fried texture by dusting the wings with baking powder and then baking at $220^\circ\pu C$ ($425^\circ \pu F$). The chicken is also dusted with salt and other seasonings, but I assume they're not chemically relevant.  What chemical process is going on, here?

I'd assumed it was some kind of reaction between the baking powder and the skin of the chicken, though I had no idea what. Then, one time, there was a stray boneless, skinless chicken thigh in with the wings I bought, and that also became crispy on the outside when cooked this way. I've also heard of people using the same technique on oven fries.

Not exactly an academic source, but a few articles on Serious Eats suggest that baking powder's gas releasing reaction forms small bubbles of protein on the surface that add surface area that can quickly dehydrate (a prerequisite for crispiness) and solidify.
It's also a pretty common ingredient in things like batter for fried chicken, where I imagine the expanding bubbles do a similar thing—allow easier removal of water from the batter by increasing the surface area so it can crisp up.

The pH change may also help break down the skin, if applied in advance.