Is it "common practice in Fourier transform spectroscopy to multiply the measured interferogram by an apodizing function"? If so, why?

You usually post interesting questions, which appear to be deceptively simple yet they are very challenging. Modern day data acquisition and signal processing is so complicated that it is almost like a black-box. It is amusing when I ask electrical engineers some signal processing questions, they don't know the answers and when I ask mathematicians it is too much applied mathematics for them. I am not an FTIR expert either but as an analytical chemist, I can add some remarks. I think apodization is routinely done in order to decrease noise and oscillations when you do an inverse transform. If you do some mathematical operations in the frequency domain, when you do the inverse transform, the level of noise is crazily high. Of course, when you are looking at very small signals do not wish to have noise.

In general, sometimes I need to fit functions to the data to peak shaped functions. I always get a better fit once I digitally filter the data and fit later (moving average, Savitsky Golay, Hamming in the time domain and so on.) Noise is the biggest enemy of any analytical chemist or a spectroscopist.

With any smoothing process, either in the frequency domain or in the time domain, you tend to loose resolution. There is always a sweet spot for digital filtering or if you have heard the story of Goldilocks who entered the house of three bears..."When she had finished the porridge Goldilocks felt tired and went to find somewhere to sit down. The first chair she found was too big, the second chair was still too big, but the third chair felt just right." The same rule of thumb goes for digital filtering. Too little is useless, too much you loose all resolution and just the right filtering gives you the best signal to noise ratio.


Why are ionic compounds bad conductors of electricity in solid state?

Electric charge is transferred by physically moving charged particles around. In the case of an electric current moving through a wire (for example), the electrons are moving.

In an ionic compound, the ions are locked in place. They can move around a little bit, but there is not much translational motion - the ions stay in their places on the crystal lattice. In addition, the ions are "happy" with the number of electrons that they have. The ions formed in the first place by giving up or accepting electrons in order to minimize the overall potential energy of the system. If an anion were to transfer an electron back to a cation (for example) the energy of the system would increase, and so in general, transfer of electrons after the compound has formed is not favorable.

In solution or in a molten state, the ions themselves can move around - they become the charge carriers. In a solid, the ions can't move, and so electricity cannot be easily transferred.

You mentioned heat transfer - heat is the transfer of the kinetic energy of atoms and molecules. Heat can still be transferred (in some cases quite easily) in an ionic solid because, as you said, ions can vibrate about a mean position. When this happens they bump into their neighbors, which spreads the kinetic energy around.

In summary, ionic compounds don't conduct electricity very well because the charge carriers can't move through the crystal. They can conduct heat because the kinetic energy itself is the "heat carrier" - it can be transferred without moving ions too far from their mean positions.


How many hydride ions are available in NaBH4?

Practically speaking, a chemist will use one equivalent of sodium borohydride to reduce an aldehyde or ketone. However, theoretically only 0.25 equivalents are necessary. Borohydride donates a hydride to the carbonyl carbon, generating borane, which is a weaker reducing agent than borohydride. However, borane is a Lewis acid that can react with the alkoxide formed in the reduction. This gives a new borohydride, which can reduce a second molecule of the ketone. This process can continue until all of the hydrides are consumed from the boron center.

Keep in mind that sodium borohydride reductions are commonly performed in alcohol solvents, so the exact nature of these steps could be altered. The alkoxide could be protonated by solvent, and then the solvent's conjugate base could reactivate a neutral boron species.

Also keep in mind that while sodium borohydride isn't VERY basic, it is a little basic, so it will deprotonate alcohols slowly. It's that reaction that causes the practical requirement for more than the theoretical borohydride stoichiometry.


Part of the reason why there has been very little research done on organoastatine chemistry is the extreme radioactivity of astatine; the half-life of $\ce{^210At}$, the longest-lived isotope, is about 8 hours. However, the element can be incorporated into organic compounds (e.g., via a destannylation reaction).1

In organic chemistry, the prefix is "astato-", as in astatobenzene or 3-astatotamoxifen, both of which have been synthesized.1,2 This is analogous to "fluoro-", "chloro-", and so on. As a general term, astatine compounds can be referred to as astatides (e.g. aryl astatides, analogous to aryl fluorides, chlorides, etc.).1

The aforementioned synthesis of 3-astatotamoxifen is shown below.

As for inorganic nomenclature, astatine can be denoted with the prefix "astatido-"; this follows from the IUPAC Red Book for nomenclature of inorganic compounds.

Milius, R. A., W. H. McLaughlin, R. M. Lambrecht, A. P. Wolf, J. J. Carroll, S. J. Adelstein, and W. D. Bloomer. Appl. Radiat. Isot. 1986, 37 (8), 799-802.
Samson, G., and A. H. W. Aten. Radiochimica Acta 1970, 13 (4), 220-221.


The main use of $\ce{Hg}$$^{\text{2+}}$ is to polarize the electron cloud of the alkyne.
Else the attacking specie wont be able to approach the bond


Does aniline breakdown in groundwater produce benzene

Aniline is not easily converted to benzene in aqueous conditions. A google search turns up several papers describing the breakdown products of aniline in groundwater but benzene is not one of them. Methanogenic and denitrifying bacteria seem to play a key part in aniline breakdown, releasing methane and carbon dioxide as byproducts. The other main products seem to be polymerisation products resulting from condensation and oxidation reactions.

In answer to your question about detection; no you would not expect to see benzene as a breakdown product in a sample of groundwater that has been contaminated by aniline.

http://www.ncbi.nlm.nih.gov/pmc/articles/PMC1519315/


Applicability of Enthalpy-Heat relation for a non-isobaric process

Disclaimer: The below discussion only holds to be completely true for reversible processes and for ideal gases.
I think the problem is you're thinking of it the wrong way, that is, you first think that $ q= nC_p \Delta T$ and for constant pressure process $q_p= \Delta H$ , and hence, $ \Delta H = nC_p \Delta T$
In actuality, it is the other way around. Similar to how internal energy is always $nC_v \Delta T$, the enthalpy is always $nC_p \Delta T$ for an ideal gas. A simple proof is shown below:
$$ \Delta H = \Delta U + \Delta (PV)$$
For $n$ fixed moles of an ideal gas,
$$ U = n C_v \Delta T$$
and,
$$ \Delta (PV) = nR \Delta T$$
Hence,
$$ \Delta H =n\Delta T [  R + C_v]$$
Now, from Mayers relation,
$$ C_p  = C_v + R$$
And hence,
$$ \Delta H = n \Delta T C_p$$
So, as seen above, if you can accept that $\Delta U= nC_v \Delta T$ for an ideal gas always then you must be able to accept a similar statement for enthalpy

Regarding state variables:
Yes, there are indeed many states for an ideal gas which correspond to the same enthalpy since enthalpy is a function of the only temperature. The state variables of pressure and volume are irrelevant for enthalpy calculations unless you don't have the final and initial temperatures. In such a case, you can use the ideal gas law to find expression for the temperature at different states.

Note:
A discussion of proving $q_p =\Delta H$ can be found in this post
You may think the above proof is circular reasoning since regularly Mayers relation is proven by using $ \Delta H = nC_p \Delta T$. However, there exists another way to prove it by considering path functions. Have a look at the first twenty minutes of this lecture by MIT OCW.


Through experimentation, I've come up with satisfactory answers to my problems.

1) What is the best material to cast indium in? Silicone rubber (the kind used for cookware).

2) How can I recovery indium that has "wetted" a stainless steel pot? Put the pot on the stove and heat until the indium is molten. Use a silicone rubber spatula to scrape the indium off, let it harden on the spatula, then pick it off. You can also use a wet sponge as long as it's the kind used for cleaning solder irons (make sure it's a new clean sponge!). I folded it in half and held it with a large clamp. That gave me something to hold onto.

3) How can I recover indium that has "wetted" glass, broken glass, or is contaminated with other odd bits of material?

Number 3 is more involved, and I will break it down into steps:

1) Go through the material with a pair of forceps and discard any bits of glass that have no indium on them.

2) Place the material in Pyrex beaker.

3) Fill the beaker with mineral oil until the material is fully submerged (water may also work, but the oil will probably help keep the indium from re-wetting the glass bits).

4) Place the beaker on a hot plate and slowly heat up to about 160 degrees C (320 F). Periodically stir while heating (a screwdriver worked fine for me; I didn't use a magnetic stir rod because the broken glass might chew it up). Also, make sure your area is well ventilated; the oil will produce vapor.

5) Once you have reached 160 C, poke the indium and make sure it is molten. Increase temperature if it is not molten. Stir for about 30 seconds to break up the indium. Keep on heat; try to maintain ideal temperature.

6) Use a glass dropper to drop in 31.45% Hydrochloric acid (I used 4 drops total when the oil and material together was about 100mL). Add one drop at a time, and let the drops run down the inside of the beaker into the oil. (WARNING: This can result in splashing! Don't over heat the oil, add the acid drop-wise, and use proper safety gear: gloves, long sleeves/pants, face shield, and respirator with acid vapor cartridges). Adding the acid will react with the indium oxide layers and help break the indium free from other materials.

7) Stir vigorously with your implement of choice. The oil will become a nearly opaque brown with all the oxides. The stirring should also free up most of the indium from the glass bits.

8) When the oil seems saturated, thoroughly decant the oil into a waste receptacle (I used a mason jar). Save as much solid material as you can.

9) Fill the beaker with clean oil; do NOT begin stirring yet. Keep on heat.

10) Use your forceps to pick out the larger pieces of glass that seem mostly clean. Use another implement to scrape off little bits of indium from the glass bits as you are pulling them out of the oil.

11) Once most of the glass is removed, use strong stirring once again until the oil is mostly brown.

12) Using your stirring implement, try to push a good amount of the molten indium together into one blob on one side of the beaker. Now decant the oil into a waste receptacle.

13) Place beaker back on heat for a short time just to ensure the indium is molten. Make sure most of the material is in one corner of the beaker.

14) Use a broad-shaped implement (I used the flat head of a long "elevator bolt") to hold back most of the scrap material and pour off the indium from the beaker into a silicone rubber mold. The poured off indium should be MOSTLY clean. Repeat this step until you feel you cannot get anymore. Put the scrap material off to the side.

15) Once the clean blobs have hardened, dab them with a paper towel and cut them into tiny chunks with a pair of wire snips.

16) Take each individual chunk and crush it paper-thin using a flat metal implement and a flat metal surface (I used an elevator bolt and a piece of sheet metal). This step is to find any tiny bits of glass. Glass has quite a different "crunch" than the typical "crackling" of the indium's crystal structure. You will know when you find one. Fine tweezers and some sort of magnifying glass helps.

17) You can now re-melt the clean indium into one chunk. I use a stainless steel shot glass and pour it off into a silicone rubber mold. If it's still oily, wash it with Ronsonol/Zippo lighter fluid, then rinse it with water. I actually chose to keep this piece of indium separate from my other indium pieces that I KNOW are clean. You never know, there might be one last sand-grain sized piece of glass in there.

18) As for the scrap material, there's still a bit of dirty indium in it. You can decide how hard you want to work and continue salvaging it. Use tweezers to pick out as much "clean" glass as possible. Wash it with lighter fluid and rinse it with water. Perhaps re-melt it in a steel shot glass and try to pour a little bit more out of it. Try putting a blob of solder flux in with the molten material and get it to stick to a copper wire. It's all going to be loss anyway, so you may as well try different things.

DISCLAIMER: I am not a professional! This procedure is not advisory. It is an account of my experiment and what worked for me.

If anyone has suggestions or constructive criticism, please feel free to comment. I know there are probably better ways to do this.


What is the reason for decomposition of layered cathode materials?

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


What is the CO2 content in the air in a compartment of air and water, as a function of temperature?

Simple, partial pressure of $\ce{CO2}$ is misleading.

$\ce{CO2}$ actually reacts with sea water:
$$\ce{H2O + CO2 -> H2CO3}$$
which then ionizes:
$$\ce{H2CO3 -> H+ + HCO3-}$$
which lowers the pH of the sea water (making it more acid).

In previous high $\ce{CO2}$ periods, $\ce{CaCO3}$ came out of sea water
making limestone and marble rock formations geologically. $\ce{CO2}$ at high pressure is responsible for other side effects, also killing
fish, and microbes, damaging coral reefs and generally messing up ocean life.


Carbocation Stability: Tertiary versus Secondary Benzyl for Hydrobromic Acid Addition

In the first instance, bromine will attach itself on the benzylic carbon as the benzylic carbocation formed in this case would be more stable than than the tertiary carbocation due to the mesomeric effect.

However in the second case, the benzylic carbonation formed would be destabilized due to the -M effect of the nitro group, like you pointed out. So bromine would favor the the tertiary carbocation which in this case would be more stable due to hyperconjugation and inductive effects.


Is the compressibility factor smaller or greater than 1 at low temperature and high pressure?

If we use the Van-der-Waals equation then a plot of Z vs reduced pressure can be constructed. By using reduced values, i.e. pressure and temperature relative to the critical values then a general plot can be made on which data from many different molecules can be superimposed. Look under sections dealing with the Law of Corresponding states in your textbook. From such a plot then it is possible to determine whether the attractive or repulsive part of the potential is the dominant one.

The figure shows such a plot of Z vs. reduced pressure $P_R$ at different reduced temperatures. The reduced temperature is $T_R=T/T_c$ where T is the actual temperature and $T_c$ the temperature at the critical point.

From the figure when $T_R \lt 2$ and $P_R \lt 6.5$ then $Z \lt 1$ and this indicates that a real gas (in so far as such is described by the Van-der-Waals eqn) has a lower pressure than an ideal gas and so this indicates that the molecules are more influenced by the attractive part of the potential than its repulsive part, which is due to their finite molecular volume. If the temperature is higher, above $T_R \approx 2$ (not shown), then Z is positive for all pressures, and similarly above $Z \approx 7$ for all temperatures.  In these cases repulsion dominates over attraction.


A solution by definition is a homogeneous mixture, so its "concentration" is same throughout the bulk. Let us have a look as to how one would prepare a solution of a given concentration, say in 1 L. We would weigh out a known weight of a solute and dissolve it  ~ 500 mL of solvent, with mechanical stirring or shaking. We would add more solvent until the total volume is 1 L. We would stir it sufficiently, again, until we believe that the stirring has made the solution uniform.

a) If it a colorless solid, like NaCl, there is no visual way to tell if the mixing is complete and the solution is uniform. You have to stir it very well.

b) A colored solid, like copper sulfate or iron sulfate is more or less visible if it is not homogenous. Different colors in the bulk.

If you are still not (mentally) satisfied, you can take portions of the solution from surface, middle and bottom. Measure their refractive index and they should be identical. If they are not not, the mixture needs more stirring.

Edits:
Some people may wish to differentiate microscopic vs. macroscopic homogeneity regarding the definition of solutions. IUPAC does not offer such a distinction. For an analytical chemist or any researcher interested in measuring or making known concentrations of solutions- a solution has to have identical concentration throughout the bulk. Ultimately, it will boil down to semantics or the problem of "how many angels can land on the head of the pin" - a debate which has no end! en.wikipedia.org/wiki/…


Spontaneous reaction between chromium(III) and aluminum

Yes, your rationalization here is correct. Chromium ions will oxidize aluminum container and deplete chromium while contaminating the solution with aluminum ions. It's exactly the same process, as for instance, copper nitrate with an iron paper clip experiment that we have all seen in a chemistry class at some point. But here you have an aluminum container instead of a paper clip :)


Explaining the oxidation states of palladium

Ions of the transition metals do not often occur "naked" the way one might think of a sodium ion. Of course, sodium ions cannot occur by themselves either, but let's compare the two for a moment.

  Sodium nitrate, $\ce{NaNO3}$

When $\ce{NaNO3}$ dissolves in water, the two ions dissociate.

$$\ce{NaNO3 -> Na+(aq) + NO3- (aq)}$$

Both ions are solvated by the solvent. The water molecules (which are polar) surround the ions. The partial charges due to the dipoles in the water molecules stabilize the charges on the ions. However, the interactions between the sodium ions and the water molecules are fleeting. These interactions are not strong enough to be consider "bonds" (there is no covalent electron sharing between water and sodium ion). Thus, we often consider sodium ions in solution (despite their solvation) to be "just" sodium ions.

  Palladium nitrate, $\ce{Pd(NO3)2}$

Transition metal ions are different. They tend to form complex ions by coordinating solvent molecules. These dative interactions are far stronger than the ion-dipole interactions in solvated sodium ions.

Palladium nitrate does dissociate, but the palladium ion picks up water molecules that it holds tightly. Thus, we do not worry about the electron configuration of $\ce{Pd^{2+}}$, since it never exists on its own. It is always part of a complex ion or coordination compound, and we care about the electron configuration of that complex ion.

$$\ce{Pd(NO3)2 + nH2O -> Pd(H2O)6^{2+}(aq) + 2NO3- (aq)}$$


Yes it is very strong, albeit not compared with Inorganic acids like HCl...

As far as organic acids are concerned, it is really one of the strongest ones... as its pKa is 0.3 [ more acidic than acetic, benzoic, formic, oxalic, trichloro-acetic acids...]

Possibility of intramolecular H-bonding, though exists, is offset reasonably as the negative charge of phenolic oxygen is very nicely delocalized over the 3 nitro groups. [In fact, if we compare the pKa's of phenol (9.98) with o-, m-, p- nitro phenols (7.23, 9.88, 7.15 respectively); we can see that H-bonding is not decreasing the acidic nature "upto greater extent"]

It is always possible that due to bulkier -NO2 groups, the repulsions betweens the LP of O atoms in -OH might be leading to the H of -OH group to be oriented far off from the range of effective H-bonding


In a substance do atoms actually touch each other?

Quantum mechanics fuzzes out the electrons into clouds of probability distribution (the motion of little electrons orbiting the nucleus is from an old outdated model).  In molecules these electronic clouds do not so much touch, rather they overlap.  See, for instance, https://en.wikipedia.org/wiki/Covalent_bond.


Since $\ce{ I_2}$ is a non-polar covalent molecule, it does not ionize in water. It  cannot also be soluble in the polar water.

$\ce{KI}$, which is a polar, ionic compound, will ionize and dissolve in water. When KI dissolves in water, it ionizes to $\ce{K^+}$ and $\ce{I^-}$.

The $\ce{I^-}$ will react with $\ce{ I_2}$ to form the complex ion $\ce{I_3^-}$. $\ce{I_3^-}$ being negatively charged will dissolve in water.


How can change in Gibbs free energy ever be positive?

You didn't do anything wrong here. This was a very good observation. This is one of the criteria of spontaneity, which I believe you haven't learnt yet.
As you correctly wrote, $\Delta G$ can only be negative or 0 for any reaction. This is actually a consequence of the second law of thermodynamics, which also goes on to say that in any spontaneous reaction the entropy change of the universe is always positive.
Essentially, the answer to your question heading is: in any spontaneous reaction, the Gibb's free energy change can never be positive. Since you took the conditions of the second law, which essentially explains spontaneity, you arrived at this result.


How to calculate the new concentration of a solution after adding more solute and convert it to ppm?

Let's think about it :

You have the formula of the mass concentration $C_m$:

$C_m = \frac{m}{V}$,

where $m$ is the mass in grams of solute and $V$ the volume in liters of the solution.

A first step for you would be to calculate the mass of solute already in the solution : your solution has a volume of $0.5~\mathrm m^3$, which is $V=500~\mathrm L$, with a massic concentration of $C_m=10~\mathrm{mg\cdot L^{-1}}$. Therefore, there is already a mass

$m=C_m~\times~V = 10\cdot10^{-3} \times 500 = 5~\mathrm g$

of salt in your solution.

To this mass, you add $2~\mathrm g$ of salt, so the mass of solute in your solution is now $m=5~\mathrm g + 2~\mathrm g = 7~\mathrm g$.

So, at end you have a mass $m=7~\mathrm g$ of salt.

I let you compute the mass of solution, knowing that $1~\mathrm L$ of water weights $1~\mathrm{kg}$.


TL;DR The cathode in an electrolytic process is considered to be
negative, so there is actually no contradiction. The cathode is a positive electrode in a galvanic cell.

There are different notations for the sign (±) of the cathode used in the literature, which are determined, in particular, by the nature of the process. A very broad definition of a cathode is that it is the electrode of some device connected to the negative pole of the current source.
For electrolysis it is commonly believed that the “−” cathode is the electrode on which the reduction process takes place, and the “+” anode is the one where the oxidation process takes place. When the cell works (for example, during copper refining), an external current source provides an excess of electrons (negative charge) at one of the electrodes — the cathode, where metal is reduced. On the other electrode, there is a lack of electrons and oxidation of metal takes place — this is the anode.
At the same time, during the operation of a galvanic cell (for example, copper–zinc), an excess of electrons (and a negative charge) on one of the electrodes is provided not by an external current source, but by the actual oxidation reaction of the metal (zinc dissolution). Following the definition given, this electrode is an anode. Electrons passing through the external circuit participate in the reduction of copper, so the cathode will be a positive electrode in this case.


What benefits do higher-rated NMR tubes provide?

A general rule is: crap goes in, crap comes out. A large-sample low-field 1D NMR at room temperature is usually only minimally affected by using a cheap NMR tube. There are important differences though and I’ll highlight a few.

The first distinction between prices is what the tube is constructed from: quartz obviously costs more than borosilicate. Why would a chemist ever use the more expensive quartz? You can heat/cool quartz faster (nice for thermal studies), the UV cutoff is lower (think 190 nm opposed to 320 nm) which is important for photolysis, you can work with quartz at higher temperatures (around 1300 °C instead of 250 °C), and the purity of quartz is better controlled than your typical Pyrex. There are different grades of quartz, fused and synthetic, and there are different grades of borosilicate, such as the high-quality Pyrex or the lower-quality Class B, each comes with its own limitations as far as purities are concerned and so forth.

Three more important parameters have to do with the manufacturing of your tube are: concentricity, camber and wall thickness. Lower quality tubes will tend to have less precision and accuracy over each of these parameters and as a result your sample may wobble while spinning (introducing problems such as modulation sidebands). A particularly bad tube can hit your RF coils and cause damage to your probe over time slowly or quickly if it is ignoring any reasonable standard – even more apparent for a tube at this level of “quality” is that it may be easier to break while acquiring your sample and we all should be aware of how much fun that is for everybody involved.

Shimming can deal with impurities present in the glass (such as ferric oxide) and increased impurities in the glass/inhomogeneities will result in taking longer to get a good shim. Time is money.

A lot of these things have lower tolerances in more complex experiments and at higher fields. It really does depend on your particular experiment and what you’re hoping to get out of it.


Actually, activity series works well in aqueous medium, while these reactions happening at high temperature will leave all waters long gone in vapours.

Well, here we go by (delta)G for the coupled reactions.
We know that a negative (delta)G amounts to SPONTANEOUS process.

The 2 reactions which need to be coupled must be such that sum of (delta)G1 for the reaction-1 and (delta)G2 for the reaction-2 is NEGATIVE.

Seeing the curve, we can conclude, Al can reduce all the metals listed except Mg, which of course could be above ~1600 degree celcius.

The conclusion: upto 1600 degrees Mg can reduce Al2O3, while above 1600 degrees Al can reduce MgO [Unlike what happens in aq. medium!]


The only place where a plane of symmetry may make sense would be vertically through the centre, through the spiro carbon. However, as the image by Tetrahydrocannabinol shows, these rings are not in the same plane (implied by the wedge and dash notation). Therefore, using that plane as a plane of symmetry would map each of the two rings into the void.

The only element of symmetry I can see in the compound is a $C_2$ axis (rotation by $180^\circ$). This axis would also bisect the spiro carbon, but it would map the oxygens onto each other. Having only a $C_2$ axis allows a compound to be chiral (point group $C_2$) and thus your compound is chiral.


What really happens in nature has higher priority than our believe what happens or should happen. See Dinitrogen_trioxide with the structure, bond lengths and angles.
Be aware both $\ce{NO}$ and $\ce{NO2}$ are radicals with an unpaired electron.
$\ce{N2O3}$ being( $\ce{ON-NO2}$ ) as 2 paired radicals is the direct analogy
to the dimer $\ce{N2O4}$ being $\ce{O2N-NO2}$, both being formed at low temperatures.


Like any other polymer decomposition process, the products of PTFE decomposition depends on the chemical species present while PTFE is undergoing the process and temperatures.

General process is this : Decomposition  is  initiated  by  random-chain
scission, followed by depolymerization. Termination is by
dis-proportionation. And all of this happens rapidly above 600 K (~326 C). There are no other conditions applied for the decomposition to take place hence can happen in dry or aqueous environments, which would give different by products.  Cooking below 200 C, it would be completely safe since the mass loss below 300 C is undetectable. Only above the glass transition temperature (~326 C) is the mass loss significant.

Decomposition products will depend on the environment and usually
Oxygen usually does not enter the cycle directly but through water, to give species like,

Carbonyl fluoride
Carbonyl difluoride

As expected other species like fluorinated alkanes and alkenes are obtained like,

the monomer, tetrafluoroethylene
Hexafluoroethane
Octafluorocyclobutane
Octafluoroisobutylene
Perfluoroisobutylene
Tetrafluoroethylene
and more.

However, the chemicals (species) listed above are for controlled lab experiments. What happens in real life cooking scenario is not (or cannot) be anticipated. But, it can be readily said that the above species can react with other chemicals in food to give fluorinated compounds that will be harmful for humans. As a precaution, one should not cook above 200 C on PTFE coated utensils to be completely safe.

Check this extremely detailed page for abstracts and toxicity remarks from research papers on PTFE decomposition on Fluoridealert.org where the conditions and level of toxicity is reported in an organized manner.

Polymer decompositions- check PTFE section for general comments.


How can I perform calculations including the relativistic approximation SC-ZORA and solvent approximation COSMO in Gaussian 09?

Gaussian can’t effectively treat relativistic effects (look up integral=dkh [1] and try its limited use. AFAIK you can only have 50 parameters in internal coordinates which is not very realistic for common molecules today.) and has no ZORA implemented. It is also not able to do COSMO PCM calculations. Instead one would use other PCM schemes, e.g. IEFPCM which is the default in Gaussian.

So can you perform your calculations with Gaussian 09? Unfortunately not.

Instead, you should change to other programs that fit your needs better. As it is free for academic use [2], you can have a look at ORCA and if it fits to your needs.

[1] http://gaussian.com/integral/
[2] read the licence carefully, it is quite strict


Why does this reaction not proceed by sn1/sn2?

The major reasons I would give for going with the SN1 products are:

The amount of steric hindrance on tertiary alkyl halides in this context highly limits the amount of second-order reactions that can occur (which require participation from two molecules).
EtOH is not a strong nucleophile in this context. In contrast, EtO- would be, but it's not present. Had EtOH been a strong nucleophile, SN2 would be a tad more likely, but a primary alkyl halide situation, for instance, is much more likely to favor SN2. Therefore, an SN2 product cannot form as a major product. However, considering both I AND III, the product is not formed through SN2 (unless you pick only III, in which case you assume I is not formed).
III is one product in the racemic mixture that would result from forming the tertiary carbocation (which is nearly planar around carbon 3, allowing for both retention and inversion of the stereochemistry). Also, the 20 degrees C does not convince me that elimination is a major product since lower temperature makes elimination more difficult. It is harder for the molecules to orient themselves quickly enough before all reactant is used up if the environment is cold. Therefore, the answer is probably I AND III, assuming that somehow the answer key is incorrect.

I also agree with orthocresol that product II is strange. I would expect by Zaitsev's rule that the new pi bond would form across carbons 3 and 4 in the main chain since his rule says that typically, the elimination occurs by acquiring a proton from the carbon with fewer protons, forming typically the more substituted alkene.


Why would water, and not saltwater, undergo electrolysis?

What is the "correct" answer

You are asking why the answer key has water and not saltwater as an answer. Decomposition is defined as:

  A reaction where a single compound breaks down into simpler compounds.

For salt water, the salt reacts at one electrode, and the water at the other, and you are making NaOH. For pure water, you are splitting water into the elements. Neither process matches the definition of decomposition that I quote exactly, but pure water is the better fit.

In your question, you write:

  I was thinking that water in itself is not conductive and therefore will not facilitate electrolysis by itself.

The half reactions could be written as:

$$\ce{2H2O + 2e- -> H2 + 2OH-}$$ and

$$\ce{H2O -> 1/2 O2 + 2H+ + 2e-}$$

So there are protons produced at one electrode and hydroxides at the other (You could also have protons be reduced at one electrode, and hydroxide oxidized at the other - the net effect is the same). The protons are very mobile in water and will quickly diffuse and neutralize the hydroxide. This takes care of the charge transport through the solution.

Is this a good question?

No, but most multiple choice questions aren't.


Can we break the covalent bonds of water molecules using resonance frequencies?

Yes this is possible. Your question does not state the source of the frequency, but in this internet article they have used laser light.

I am not sure at what frequency you need to irradiate water to break the O-H bond, but I would speculate that a good frequency would be close to the IR stretching frequencies of the O-H bond in water.


The key here is the concept of metallic bonding. or, the sharing of electrons between metal atoms like those in a copper wire (a typical conductor). In this type of atomic bond, electrons no longer belong to one atom or another; they're shared equally amongst all atoms.  As such, one electron truly does go bouncing around between the atoms in the direction of the electric current.

One atom is close enough to another in metallic-bonded substances that an electron is attracted just as much to one nucleus as the next one over, so it's not so "illogical" to think that an electron might "break free" of its original nucleus.

Chemical reactions occur between atoms, not electrons, and atoms in a metal are already metallic-bonded together, so a few stray electrons wouldn't do anything to create new reactions.  However, the electrons do behave oddly!  That's the beauty of metal conductors: their electrons are free to flow wherever an electric or magnetic field dictates, meaning that they can do some pretty cool things!

The nuclei of a conductor don't become ions because every time one electron leaves, another takes its place.  Think of people at a gas station on a busy day: every time one car fills up and leaves, another pulls in almost immediately.  Thus, the metal atoms never have time to become ions, because there's plenty of electrons to keep them company.

As (nearly) always, Wikipedia has a bunch of information on the subject.


It is difficult to give a general answer. A glass is an amorphous solid that may change its composition in a continuous way from one point to another. When melting brown bottles and colorless bottles in the oven of a glass factory, you may obtain a non homogeneous mixture, with regions more or less colored. It is obviously one and the same phase all over the mixture. But as the mixture is not homogeneous, its composition is not the same everywhere.


How is a reduction in the number of molecules associated with a decrease in Entropy?

Interpreting entropy as "disorder", which comes from thermodynamics applications, is only one way to interpret entropy. Entropy can also be described according to statistical mechanics as the statistical distribution of possible degenerate microstates:

$$S=k_B \ln\Omega$$

where $k_B$ is Boltzman's constant and $\Omega$ is the number of microstates. Thus it is a measure of uncertainty. The more equivalent/degenerate microstates there are, the less likely we know exactly which one of them the system is in, and thus, the higher the entropy. This probabilistic interpretation provides a more intuitive and satisfactory understanding of the relationship between number of molecules and entropy of the system.

  A decrease in the number of molecules leads to a decrease in the total number of degenerate microstates, which leads to a decrease in entropy. There are more degenerate ways to distribute N objects in M spaces than there are degenerate ways to distribute (N-1) objects in M spaces.

Let's look at a very simple example, in which we have three molecules $\ce{2H2 + O2}$ in a container with room for four molecules. How many ways can we fill four spaces with these three molecules? There are four spaces in which we could put the $\ce{O2}$ molecule:

$$\begin{pmatrix}
\ce{O2} & ... \\
... & ... \\
\end{pmatrix}
\begin{pmatrix}
... & \ce{O2} \\
... & ... \\
\end{pmatrix}\begin{pmatrix}
... & ... \\
\ce{O2} & ... \\
\end{pmatrix}
\begin{pmatrix}
... & ... \\
... & \ce{O2} \\
\end{pmatrix}
$$

How many ways can we put two $\ce{H2}$ molecules in one of these arrangements? In order to answer this, we need to temporarily distinguish one $\ce{H2}$ from the other $\ce{H2^{\circ}}$ (or to distinguish the one we place first from the one we place second) in order to generate all of the degenerate states.

$$\begin{pmatrix}
\ce{O2} & \ce{H2} \\
\ce{H2^{\circ}} & ... \\
\end{pmatrix}\begin{pmatrix}
\ce{O2} & \ce{H2} \\
... & \ce{H2^{\circ}} \\
\end{pmatrix}\begin{pmatrix}
\ce{O2} & \ce{H2^{\circ}} \\
\ce{H2} & ... \\
\end{pmatrix}\begin{pmatrix}
\ce{O2} & ... \\
\ce{H2} & \ce{H2^{\circ}} \\
\end{pmatrix}\begin{pmatrix}
\ce{O2} & \ce{H2^{\circ}} \\
... & \ce{H2} \\
\end{pmatrix}\begin{pmatrix}
\ce{O2} & ... \\
\ce{H2^{\circ}} & \ce{H2} \\
\end{pmatrix}$$

There are six arrangements of two $\ce{H2}$ molecules for each location of $\ce{O2}$ for a total of $\Omega=24$ arrangements. The entropy $S=k_B\ln(24)=4.39\times10^{-23}) \frac{\text{J}}{\text{K}}$. Note that this happened to be $\Omega=4\times3\times2=12$, or 4 ways of placing the first molecule, 3 ways of placing the second, and two ways of placing the third. We will use this shortcut next time we do this.

Now, let us run the reaction, converting hydrogen and oxygen to water.

$$\ce{2H2 + O2 -> 2H2O}$$

And let us repeat the exercise. How many ways can we fit two $\ce{H2O}$ molecules in four spots? There are 4 ways to place the first one, and 3 ways to place the second for each of those 4 placements of the first. Thus, $\Omega=4\times3=12$ and the entropy is $S=k_B\ln(12)=3.43\times10^{-23}) \frac{\text{J}}{\text{K}}$.

  Note that there is a subtle flaw in my explanation. The molecules themselves have an intrinsic entropy hidden in their structures as well. According to this site the standard molar entropies of $\ce{O2},\ \ce{H2},$ and $\ce{H2O}$ are (in$\frac{\text{J}}{\text{K}\cdot\text{mol}}$) 205.1, 130.7, and 188.8, respectively. Thus, we have an entropy change of reaction:

$$\Delta_rS^\circ=\sum S^\circ_{products} - \sum S^\circ_{reactants}$$ $$ \Delta_rS^\circ =\left(2\times188.8 - (205.1+2\times130.7)\right)\frac{\text{J}}{\text{K}\cdot\text{mol}}=-40\frac{\text{J}}{\text{K}\cdot\text{mol}}$$

By comparing standard entropies, we also get a decrease in entropy. Standard entropies are easier to work with because we are frequently working on the molar scale. If we changed our simplified example to be moles of $\ce{H2}, \ \ce{O2},$ and $\ce{H2O}$, in 4 moles of spaces we would have the original number of microstates be:

$$\Omega=(4N_A)(4N_A-1)(4N_A-2)=64N_A^3-48N_A^2+8N_A$$

When you plug in $N_A=6.022\times10^{23}$, all but the most staunch calculators and computer programs are going to give up. Wolfram Alpha can do it because it does so symbolically, and we get

$$\Omega=1.398\times10^{73}$$

$$S=8.314\frac{\text{J}}{\text{K}\cdot\text{mol}}\ln(1.398\times10^{73})=1.400\times10^3\frac{\text{J}}{\text{K}\cdot\text{mol}}$$

(At the molar scale we replace $K_B$ with $R$, the ideal as constant.)

Another apparent weakness of the statistical approach is that it assumes no phase change. As F'x mentioned, there is an entropy change associated with a phase change. Roughly, the number of available spaces increases or decreases.


Synthetic ruby - damaged by heat less than its melting point?

However, would there be any damage to its optical properties?

It is almost guaranteed. Corundum is incredibly sensitive to thermal shock and unless you can do this in a furnace that heats very slowly, it will fracture/crack.

As far as any other properties, your ruby should not see any change in other optical properties for such a short duration, but for longer durations (hours at high temperature) metals can diffuse into the ruby and cause color changes, but I doubt this is an issue for what you describe.


why does the gap between the energy levels of a hydrogen atom keep decreasing as we move further and further away from nucleus?

Whatever the radius, the allowed energy levels are proportional to $1/n^2$.
The first gap is proportional to $1/1^2 - 1/2^2 = 0.75$.
The second gap is proportional to $1/2^2 - 1/3^2 = 1/4 - 1/9 = 5/36 = 0.139$. This is smaller than $0.75$.
The third gap is proportional to $1/9 - 1/36 = 1/12 = 0.083$. It is still smaller.


What would be SMILES notation for a compound with delocalized bonding?

There are general problems representing delocalized bonding and organometallic compounds in cheminformatics (generally) and SMILES in specific.

To quote the Open SMILES standard:

  This simple mental model [of a connection table between atoms using valence bonds] has little resemblance to the underlying quantum-mechanical reality of electrons, protons and neutrons, yet it has proved to be a remarkably useful approximation of how atoms behave in close proximity to one another. However, the valence model is an imperfect representation of molecular structure, and the SMILES language inherits these imperfections. Chemical bonds are often tautomeric, aromatic or otherwise fractional rather than neat integer multiples. Delocalized bonds, bond-centered bonds, hydrogen bonds and various other inter-atom forces that are well characterized by a quantum-mechanics description simply don’t fit into the valence model.

Other interesting cases are agostic interactions, 3-center 2-electron bonds (e.g., diborane).

There are several suggestions for advancing beyond the typical connection-table valence bond view, including:

"Yet Another Representation of Molecular Structure" J. Chem. Inf. Comput. Sci., 1995, 35 (5), pp 787–802
"Overcoming the Limitations of a Connection Table Description:  A Universal Representation of Chemical Species" J. Chem. Inf. Comput. Sci., 1997, 37 (4), pp 705–714

That both papers were published 17+ years ago should be some indication of:

Chemists like the valence bond view of molecules, despite the limitations
How complicated bonding can be in non-organic chemistry
How far cheminformatics needs to go to properly handle organometallic and non-organic compounds

By the way... anyone interested in implementing such efforts should let me know.. Many people would be happy to help. :-)

UPDATE

I should point out that there are some practical solutions to representing such things. For example Alex Clark suggested recording zero-order bonds (and sometimes dummy atoms):

"Accurate Specification of Molecular Structures: The Case for Zero-Order Bonds and Explicit Hydrogen Counting" J. Chem. Inf. Model., 2011, 51 (12), pp 3149–3157


What has old Potassium metabisulfite turned into?

According to my research, I found the reaction of oxidation of potassium metabisulfite but the products of this reaction does not seem to match with your experimental observations.

  The jar has been standing with a a loose fitting lid for thirty years in a cool and dry place.

This means that your old  potassium metabisulfite has been subjected to atmospheric oxidation in a cool and dry place and thus the temperature of the surrounding did not exceed the room temperature.

$$\ce{2K2S2O5 + O2 ->[298K] 2K2SO4 + 2SO2}$$

(source)

  The jar contains a dry substance now like matte flakes. The substance does not react visibly to hydrochloric acid and it does not smell of sulfur dioxide.

Yes, potassium sulfate has a flaky, crystalline appearance. Potassium sulfate does not react with hydrochloric acid. But one of the product is sulfur dioxide and the flakes smell of sulfur dioxide.

  The substance does not dissolve visibly in cold or boiling water. A
  few flakes poured in water remain on the bottom seemingly unaltered.

Potassium sulfate is soluble in water: 120 g/L (25 °C).

  When heated to red heat in a test tube, it does not melt, the coarse
  powder is not sintered. No water is given off. Some sulfur is
  sublimated, a yellow deposit in the cold part of the test tube and a
  smell of sulfur. Not much sulfur is given off, the volume of the
  original substance is not visibly altered.

Potassium sulfate does not decompose even at high temperatures. But if the surrounding has carbon monoxide, it gets reduced to potassium sulfide which has a yellow-brown appearance when impure.

Summarizing as a table,


Spectroscopy and Excitation of Electrons

I assume it is a homework question...
You may want to into the concept of blackbody radiation, as an explanation. It is discussed in most PChem / Quantum Chem books in the first chapters. Feel free to update your question according to.

Generally, we do not say that photons are excited. Photons do not have ground and excited states in the same chemical sense as electrons. Excited states (of e.g. electron) may emit photons during de-excitation, but those are the excited state of the electron (before the photon even existed) not the photon.

Also, we generally do not say photon decay: technically you can argue that absorption is something like that, but it is definitely not a common way to phrase it, and obviously absorption is absorption, so it will not emit any radiation.

Note: After this answer, the question was changed to talk about electronic excitation instead of "photonic excitations", therefore a large part of my answer is irrelevant.


In the photoelectric effect, why we use momentum to calculate kinetic energy?

Relativity
Since electrons are travelling at near the speed of light, we need to use relativity.
In relativity, the correct equation relating energy to mass and momentum is the energy-momentum relation:

$E^2 = (m_0 c^2)^2 + (p c)^2$
where:

$E$ denotes the total energy of the particle.
$m_0$ denotes the (rest) mass of the particle.
$c$ denotes the speed of light.
$p$ denotes the momentum of the particle.

The momentum of a particle can be found from its (rest) mass and its velocity:

$\displaystyle p = \gamma m_0 v$
where:
$\displaystyle \gamma = \frac {1} {\sqrt{1 - \left({ \frac v c }\right)^2}} = \left({1 - \left({ \frac v c }\right)^2}\right)^{-0.5}$

Note that $p = m v$ is only true when $0 \lt v \ll c$, when $\gamma \approx 1$.
Kinetic energy
When the particle is not moving, $p = 0$, so $E = m_0 c^2$ (mass-energy equivalence) is denoted the rest  energy.
Substituting the definition of momentum gives:

$\displaystyle E^2 = m_0^2 c^4 + \gamma^2 m_0^2 v^2 c^2 = m_0^2 c^4 \left({\frac {c^2} {c^2 - v^2}}\right) = m_0^2 c^4 \gamma^2$

Now, the kinetic energy is defined as the total energy minus the rest energy:

$E_{\mathrm k} = E - m_0 c^2$
$\displaystyle E_{\mathrm k} = m_0 c^2 \left({\gamma - 1}\right)$

Classical approximation: $\displaystyle E_\mathrm k = \frac 12 m_0 v^2$
Now, the classical formula $\displaystyle E_\mathrm k = \frac 12 m_0 v^2$ is an approximation of $E_\mathrm k$ when $0 \lt v \ll c$:

$E_{\mathrm k}$
$\displaystyle = m_0 c^2 \left({\gamma - 1}\right)$
$\displaystyle = m_0 c^2 \left({\left({1 - \left({ \frac v c }\right)^2}\right)^{-0.5} - 1}\right)$
$\displaystyle \approx m_0 c^2 \left({1 - (-0.5) \left({ \frac v c }\right)^2 - 1}\right)$
$\displaystyle = \frac 1 2 m_0 v^2$

However, as I have mentioned, it only works when $v$ is very small, which is not the case for electrons.
Another approximation: $\displaystyle E_{\mathrm k} = \frac {p^2} {2 m_0}$

$E_{\mathrm k}$
$\displaystyle = \sqrt{m_0^2 c^4 + p^2 c^2} - m_0 c^2$
$\displaystyle = m_0 c^2 \sqrt{1 + \frac {p^2} {m_0^2 c^2}} - m_0 c^2$
$\displaystyle \approx m_0 c^2 \left({1 + \frac {p^2} {2 m_0^2 c^2}}\right) - m_0 c^2$
$\displaystyle = \frac {p^2} {2 m}$

Comparison
Now, for reasons unknown, the second approximation is actually better than the first:

(For convenience reasons, $c$ is took to be $1$ in the graph. It can be seen that the second approximation (blue) goes to infinity with the real kinetic energy (black) as $v \to c$, while the first approximation (red) does not.)
Perhaps this is the reason why the second approximation is used instead of the first.
Note that it is still an approximation.

Another reason is that, $\displaystyle \nu = \frac c \lambda$, but that is $\nu$ (Greek letter nu) for frequency not the $v$ (Latin letter vee) for velocity. The former has a unit of $\mathrm{s^{-1}}$ while the latter has a unit of $\mathrm{m~s^{-1}}$.


Mechanism of arene side chain oxidation by permanganate

Some general information on side-chain oxidation in alkylbenzenes is available at Chemguide:

  An alkylbenzene is simply a benzene ring with an alkyl group attached
  to it. Methylbenzene is the simplest alkylbenzene.

  Alkyl groups are usually fairly resistant to oxidation. However, when they are attached to a benzene ring, they are easily oxidised by
  an alkaline solution of potassium manganate(VII) (potassium
  permanganate).

  Methylbenzene is heated under reflux with a solution of potassium
  manganate(VII) made alkaline with sodium carbonate. The purple colour
  of the potassium manganate(VII) is eventually replaced by a dark brown
  precipitate of manganese(IV) oxide.

  The mixture is finally acidified with dilute sulfuric acid.

  Overall, the methylbenzene is oxidised to benzoic acid.

  Interestingly, any alkyl group is oxidised back to a -COOH group on
  the ring under these conditions. So, for example, propylbenzene is
  also oxidised to benzoic acid.

Regarding the mechanism, a Ph.D. student at the University of British Columbia did his doctorate on the mechanisms of permanganate oxidation of various organic substrates.1 Quoting from the abstract:

  It was found that the most vigorous oxidant was the permanganyl ion ($\ce{MnO3+}$), with some contributing oxidation by both permanganic acid ($\ce{HMnO4}$) and permanganate ion ($\ce{MnO4-}$) in the case of easily oxidized compounds such as alcohols, aldehydes, or enols.

The oxidation of toluene to benzoic acid was one of the reactions investigated, and a proposed reaction mechanism (on pp 137–8) was as follows. In the slow step, the active oxidant $\ce{MnO3+}$ abstracts a benzylic hydrogen from the organic substrate.

  $$\begin{align}
\ce{2H+ + MnO4- &<=> MnO3+ + H2O} & &\text{(fast)} \\
\ce{MnO3+ + PhCR2H &-> [PhCR2^. + HMnO3+] & &\text{(slow)}} \\
\ce{[PhCR2^. + HMnO3+] &-> PhCR2OH + Mn^V} & &\text{(fast)} \\
\ce{PhCR2OH + Mn^{VII} &-> aldehyde or ketone} & &\text{(fast)} \\
\ce{aldehyde + Mn^{VII} &-> benzoic acid} & &\text{(fast)} \\
\ce{ketone + Mn^{VII} &-> benzoic acid} & &\text{(slow)} \\
\ce{5 Mn^V &-> 2Mn^{II} + 3Mn^{VII}} & &\text{(fast)}
\end{align}$$

The abstraction of a benzylic hydrogen atom is consistent with the fact that arenes with no benzylic hydrogens, such as tert-butylbenzene, do not get oxidised.

Reference

Spitzer, U. A. The Mechanism of Permanganate Oxidation of Alkanes, Arenes and Related Compounds. Ph.D. Thesis, The University of British Columbia, November 1972. DOI: 10.14288/1.0060242.


Changing the order of the reaction with change in concentration

You would change the reaction speed, but not the order.

The order of the reaction can be changed by reagent concentration, if it is result of multiple reaction steps with different orders. By variation of concentrations you can manipulate the relative reaction speed,
as the order is often ruled by the slowest reaction.

Let suppose the 2 step reaction
$$\begin{align}
\ce{2A + B &-> C} \\
\ce{A + A &->[slow] D} \\
\ce{B + D &->[fast] C} \\
\end{align}$$

For small $c_A$, the reaction speed
$$\frac{dc_C}{dt}=k_{low A}.{c_A}^2$$.

For high $c_A$, the first reaction becomes faster then the second.

$$\frac{dc_C}{dt}=k_{high A}.{c_A}^2.c_B$$.

You can also change the effective order of reaction, if some reactant is in about constant abundance and is not explicitly considered.

E.g. The speed of the reversed esterification
$$\ce{R1COOR2 + H2O -> R1COOH + R2OH}$$ is the reaction kinetics of the 2nd order. But in abundance of water, it is effectively a reaction of the 1st order:
$$\frac{dc_{\ce{R1COOR2}}}{dt}=-k_1.c_{\ce{R1COOR2}}$$.


What general uses do beta-lactam functional groups have?

β-lactams, also known as 2-azetidinones, have experienced a resurgence of interest in the last decade or so, with a number of new applications being exploited. Of course, antibiotic activity is well-known, but I hope you don't limit your acceptance of drug design just to this class of chemical.

β-Lactams have found use in a number of fields of medical application, inlcuding gene activation, enzyme inhibition and cholesterol absorption inhibition, and recently there has been growing interest in the use of these molecules as synthetic templates for accessing other target moelcules.

There are lots of sources of references if you go digging, but some key reviews that may interest you include:

TETRAHEDRON Volume 56, Issue 31, Pages 5553-5742 (July 2000). The entire issue is concerned with developments in β-lactam chemistry;
EUROPEAN JOURNAL OF MEDICINAL CHEMISTRY  Volume 45, Issue 12, Pages 5541-5560   (DEC 2010) "2-Azetidinone - A new profile of various pharmacological activities"  gives some very good insights into new medical applications of this class of compound, including anticancer, antiparkinson, antidiabetic, antiinflammatory activities. I think they can also do the washing up and buy your groceries.
TETRAHEDRON  Volume 59, Issue 39, Pages 7631-7649 (Sep 2003) "Recent progress in the synthesis and chemistry of azetidinones" gives a pretty good summary of synthetic approaches to these molecules.
Alcaide and Almendros do a lot of work using 2-azetidones as reactive synthons in catalytic cyclisations. A recent review by them is

ACCOUNTS OF CHEMICAL RESEARCH  Volume 47, Issue 3, Pages 939-952 (Mar 2014) "Gold-Catalyzed Cyclization Reactions of Allenol and Alkynol Derivatives", but they have a heap of other papers also.

Another example, probably somewhat similar, of their use as a reactive synthon is
BULLETIN OF THE KOREAN CHEMICAL SOCIETY  Volume 28, Issue 1, Pages 17-28 (2007) "Indium and gallium-mediated addition reactions" - I have to admit I only read the abstract of this paper, but they appear to use azetidones as synthons in metal-mediated syntheses of a range of compounds.

Finally, the August edition of Chemical Reviews Vol 114, Issue 16 looks at small heterocycles in synthesis and has some interesing papers, inparticular:

CHEMICAL REVIEWS Volume 114, Issue 16, Pages 7930-7953 "Chemical Synthesis of beta-Lactams: Asymmetric Catalysis and Other Recent Advances". This is a very good paper for you, I think.
CHEMICAL REVIEWS Volume 114, Issue 16, Pages 8257-8322 "Four-Membered Ring-Containing Spirocycles: Synthetic Strategies and Opportunities". An interesting application of lactams in spiro molecule formation.
CHEMICAL REVIEWS Volume 114, Issue 16, Pages 7954-8015 "Synthesis and Reactivity of 2-(Carboxymethyl)aziridine Derivatives". This looks at lactams being used as synthons for a range of molecules, here focussing on aziridines.

I hope some of these point you in the direction of what you are looking for.


What does pO2 of blood mean and why do we use it?

There is a good explanation in Relating oxygen partial pressure,
saturation and content: the
haemoglobin–oxygen dissociation
curve Breathe 2015; 11: 194–201

  The partial pressure of oxygen (also known
  as the oxygen tension) is a concept which often
  causes confusion. In a mixture of gases, the
  total pressure is the sum of the contributions
  of each constituent, with the partial pressure of
  each individual gas representing the pressure
  which that gas would exert if it alone occupied
  the volume. In a liquid (such as blood), the partial
  pressure of a gas is equivalent to the partial
  pressure which would prevail in a gas phase in
  equilibrium with the liquid at the same temperature.
  With a mixture
  of gases in either the gas or
  liquid phase, the rate of diffusion of an individual
  gas is determined by the relevant gradient of its
  partial pressure, rather than by its concentration.
  While in a gas mixture, the partial pressure
  and concentration of each gas are directly proportional,
  with oxygen in blood the relationship
  is more complex because of its chemical combination
  with haemoglobin. This allows blood
  to carry an enormously greater concentration
  (content) of oxygen than, for example, water
  (or blood plasma). Measurement of $p_\ce{O_2}$, therefore,
  does not give direct information about the
  amount of oxygen carried by blood.

So blood $p_\ce{O_2}$ does not correspond to a particular concentration of oxygen, because the concentration of haemoglobin can vary, and most of the oxygen is bound to the heme iron.

$P_\ce{O_2}$ is the partial pressure of oxygen in a hypothetical gas phase which would make the blood oxygen and gas phase oxygen be in equilibrium.


You've correctly identified the $C_3$ axis that contains the $\ce{C-Cl}$ bond.  Methyl chloride also has 3 planes of symmetry that contain the $C_3$ axis and each of the $\ce{C-H}$ bonds.  Since these planes contain the $C_3$ axis they are referred to as $\sigma_{v}$ planes of symmetry.  There are no other symmetry elements in methyl chloride so it belongs to point group $C_{3v}$ (n $\sigma v$ planes plus a $C_n$ axis).

There is no $C_{nd}$ point group.  There is a $D_{nd}$ point group which contains a $C_n$ axis, n $\sigma_{v}$ planes (so far methyl chloride fits the description), but also n $C_2$ axis.  Methyl chloride does not contain any $C_2$ axes, so it does not belong to the $D_{nd}$ point group.


How to classify the hydride shift in the following diazacyclopenta[cd]pentalene system?

It took me a while to see what was actually happening. Initially, I was focused on a [1,5] suprafacial sigmatropic proton shift followed by a diaza-electrocyclic ring opening reaction. However, after drawing the corresponding orbitals I realised that I would have a eight-electron antiaromatic π system, in which a proton shift would need to occur antarafacially — not possible.[1]

It still is clear that it is some kind of pericyclic reaction, and going through the list we can exclude everything except for a (retro-) group transfer reaction for which the (retro-) ene reaction is the most well-known example. (We are transforming one σ bond into one π bond which excludes all the others.)

It took me a while to actually see the retro-ene system in the substrate, but finally I did, and I took the liberty of highlighting it in red in the scheme below. Note that you have to ignore the carbon-nitrogen bond which belongs to the lowest five-membered ring from the ene-system. It is not taking part, if anything it is scaffolding.

Scheme 1: Retro-ene reaction of the title compound with the pericyclic six-membered ring system highlighted. Dashed bonds represent those that are broken/formed within the reaction.

Counting electrons, I realised that my initial assumption of a proton shift was wrong and that it is indeed a hydride shift.

Notes:

[1] [1,n]-rearrangements are classified as suprafacial or antarafacial depending on whether the group is on the same side or the other side of a π system after migration. Technically, there is a further difference between retention and inversion, but hydrogen cannot rearrange under inversion so that is moot. The general rule for proton transfers is: If the number of bonds is $4n + 1$, the mechanism is suprafacial, for $4n + 3$ it is antarafacial.


How is the change in Gibb's free energy a condition for spontaneity for reactions not happening under constant pressure conditions?

Gibbs energy have an another equation which is:-

So according to this equation spontaneity of a reaction depends both on the enthalpy change as well as entropy change.
We all know that a system attains stability as the system's energy is lowered that is it has to release energy. If in a reaction energy is released enthalpy change is negative.
We also know that a reaction is favoured in the direction of increased chaos or a positive entropy change.

So from this we can see that Enthalpy change is negative and entropy change is positive for a spontaneous reaction to take place and the overall equation of

becomes negative that is Gibbs free energy must always be negative for a spontaneous reaction.....Can be applied to any reactions.
Hope this helps.


What are the differences between the HMQC and HSQC experiments?

You get a better resolution in the indirect dimension with an HSQC compared to an HMQC. The HSQC signal is a singlet while the HMQC is usually a multiplet as you see certain couplings on it. So generally you should prefer the HSQC over the HMQC.

One advantage of the HMQC is that it uses significantly fewer pulses than the HSQC. This means that errors in the pulses don't accumulate as much as in the HSQC, you'll lose less signal if your pulses are wrong than in the HSQC.

If you had trouble setting up an HSQC, that could mean that the probe wasn't tuned and matched properly for your sample, or that your pulse lengths were wrong. The proton pulse should be determined on each sample individually; the $\ce{^13C}$ and $\ce{^15N}$ pulses should be set up correctly already. If you're not using a probe with automatic tuning and matching you should also do that manually before the measurement.

There are a lot more HSQC and HMQC variants that are used, e.g. Sofast-HMQC or TROSY sequences that are used for larger molecules like proteins. Another interesting variant you should consider is the multiplicity-edited HSQC, where $\ce{CH2}$ signals will have the opposite sign from $\ce{CH}$ or $\ce{CH3}$ signals.


What is the mechanism of the uncatalysed epoxide opening with alcohols?

Because of its special chemical struture Ethyelene oxide (EO) is highly reactive material thus easily participate in chemical reactions which includes polymerization. Mechanism of opening of EO's ring in acidic and alkaline medium is given below

If you see in both cases a molecule with alcoholic group ($\ce{-OH}$) is formed.If the epoxyethane is in excess, the reaction can continue and could lead to polymerization.

A good example is the reaction between EO and water

As explained in wikipedia ethoxylation is an industrial process to make surfactant from EO. Surfactants are chemical molecule which contain both hydrophobic groups (their tails) and hydrophilic groups (their heads).

The end result of reaction with EO and alchol is a long chain molecule with these hydrophobic ($\ce{-R}$) and hydrophilic ($\ce{-OH}$) group


You don't really need to do anything to get LPG vapour. It is a gas a room temperature and pressure so as long as you just want the clean gas not mixed with air (which would be dangerous as the mixture will be flammable and potentially explosive), all you need to do is allow the output of the canister to vent into a balloon.

LPG is mostly propane and butane which liquefy under mild pressure. Butane will liquefy at about 220 kPa at normal room temperature (this is a little over twice atmospheric pressure). This is what happens in a typical cigarette lighter like the one illustrated. There is no "wick" involved just a valve that allows the gas to vent to the atmosphere and a flint to create a spark and ignite it (which you can avoid doing by careful handling of the lighter activating just the valve and not the flint). Large cylinders are similar and it might be possible to vent directly into a balloon just by carefully controlling the exit valve on the cylinder.

It is possible that standard supplies of LPG for household use also contain stinky additives to make leaks easy to detect, but otherwise the gas should be similar to that found in a cigarette lighter.


Which compound is responsible for the borax bead test?

A little chemical detective work gives a plausible answer.

Say the metal is cobalt and it's oxidation state is +2.  Then, with the cobalt calcined to the oxide, you have two possible acid-base reactions:

$\ce{CoO + 2 NaBO2 -> Co(BO2)2 + Na2O}$

$\ce{CoO + B2O3 -> Co(BO2)2}$

Both reactions look good, but upon further review the first one gives something we recognize as a powerful base, more likely to be a reactant than a product.  Moreover,  a quick check here reveals that boric oxide has a low enough melting point to liquefy in the flame, favoring the second reaction kinetically.

So the second reaction above is really the only plausible one.


Determining the nature of common pH indicators (methyl orange) using Le Chatlier's principle

Methyl Orange is the common name for Sodium 4-[(4-dimethylamino)phenyldiazenyl]benzenesulfonate. So the compound is a salt or a conjugate base. It is the sodium salt for the indicator since that form is more soluble than the acid form.

The pKa is 3.47 in aqueous solution, so at pH values of 3.1 or lower the molecule is protonated.

At pH values greater than 4.4 all the molecules in solution are deprotonated as the sulfonate anion.

Linked two images from ChemWiki.

The confusion is caused by rather poor wording in a question.

I'd prefer the wording to be:

      ... In aqueous solution methyl orange indicator behaves like a weak acid. Letting HX represent the protonated form of the indicator, it dissociates as follows:

It is splitting hairs, but methyl orange is the name of the sodium salt, not the protonated form of the indicator.


What is the meaning of the phase state label (cr,l) appearing in ATcT thermochemical data tables?

The nearest I came to an explanation is an article (Ref. 1) by authors of the database explaining that, in the case of $\ce{H2O (cr,l)}$, the notation (cr,l) refers to the condensed state. Also, a link from the above data page explains that "cr=crystal" and "l=liquid" (thanks MaxW for hinting at the solution).

A further update: it finally dawned on me that the website sometimes provides enthalpy data at two different temperatures ($\pu{0 K}$ and $\pu{298.15 K}$) in one line, so evidently the lower T refers to the crystalline state, the higher to the liquid state. But it remains somewhat confusing notation. In the case of an aqueous solution of $\ce{(NH4)NO3}$, it would seem to make sense to use the notation "(aq)" for the liquid state. Evidently "(l)" is synonymous with "(aq)" in the case of solutes in aqueous solution, and "(cr)" refers to a frozen dilute aqueous solution at $\ce{0 K}$. But this remains confusing, since in this table, the last entry describes the following reaction: $$\ce{(NH4)NO3 (cr,l) → [NH4]+ (aq) + [NO3]- (aq) }$$ Here a distinction is made between the hydrated (or aqueous) ions and the species $\ce{(NH4)NO3 (cr,l)}$, whatever it may be. It seems unlikely that $\ce{(NH4)NO3}$ can exist as a molten salt. It seems (cr,l) is used to signify a condensed phase of the pure substance, whichever makes sense under the given conditions. In the above reaction this is presumably the crystalline solid, and the reaction represents solubilization.

References

Ruscic et al., J. Phys. Chem. A, Vol. 108, No. 45, 2004. See the caption to Table 1.


How to dialkylate a cyclohexene, and convert a carbonyl to an alkene?

You are right about using the Gilman in a 1,4-Michael addition. The enolate created from this can then be used in the nucleophilic attack of a haloalkane. Then all that's left to do is a Wittig reaction, swapping the carbonyl oxygen for a carbon. The full stepwise synthesis would be:


How is it possible, that the hydrolysis of ethers proceed via a SN2 mechanism?

While the mechanism may look simple in the way you described it there, it doesn’t work exactly like that.

First, you are in the presence of (diluted) $\ce{H2SO4}$ so aside from extremely short-lived accidental self-deprotonation of water, $\ce{OH-}$ ions will not be around. Therefore, $\ce{OH-}$ cannot act as an attacking group, whatever the mechanism.

Second, an alcoholate is one of the worst leaving groups in $\mathrm{S_N}$ reactions one can imagine. The only case where you will frequently see leaving alcoholates is for epoxide openings. But that’s not a problem here, we have $\ce{H2SO4}$. We can protonate the ether oxygen so that the leaving group will now be an alcohol.

Third, since $\ce{OH-}$ cannot be the attacking species, we need to choose from the nucleophiles present. Either $\ce{HSO4-/SO4^2-}$ or $\ce{H2O}$ can attack the $\alpha$ carbon (in $\mathrm{S_N2}$ or $\mathrm{S_N1}$ manner). In the former case, a later attack of $\ce{H2O}$ to liberate the sulphate needs to be assumed. Both attacks work, but neither nucleophile is strong, so the driving force must be the liberated alcohol.

Fourth, the attacking $\ce{H2O}$ will create a protonated alcohol. This needs to deprotonate to give the free alcohol. Thereby the proton we used in second (step 1 of the mechanism, if you wish) is reliberated and therefore the reaction is acid-catalysed.

Finally, one needs to consider the possibility of equilibrium. The reaction can proceed in both directions. Usually by the choice of solvent one can direct it into one or the other. Hydrolysis is usually achieved with a large surplus of water, driving the reaction towards the alcohols by Le Chatelier’s principle. The reverse reaction would be triggered by removal of water or high concentrations of either alcohol.


What is the appropriate way of writing the Iron/thiocyanate equilibrium?

The second representation is more accurate because in water there can be no ``naked'' $\ce{Fe^3+}$ ion; as it goes into solution, iron combines with water to form an ion consisting of
a $\ce{Fe}$ atom surrounded by six molecules of water. It is the oxygens that form bonds with the iron atom, in a roughly octahedral arrangement: two oxygen
atoms in each coordinate direction, one on either side. The reaction
consists of the replacement of one of these coordinated water
molecules by the $\ce{SCN-}$ ion; now the iron has bonds to 5 oxygens
and 1 nitrogen atom.

The addition of $\ce{H2O}$ to the formula doesn't mean we are
diluting the solution with water.  It just means we're letting the
water that's  present anyway play a role in the equation, to reflect reality.

It's not clear to me what they are asking in (ii).  The colors change
dramatically when you mix $\ce{[Fe(H2O)6]^3+}$ (pale yellow) with $\ce{SCN-}$
(colorless); the result is deep red. Perhaps some context will help.  Read the discussion of the reaction in the text to see what they say about color intensity.

(iv) Again, what is the context?  If the text is about the spectrospic
measuring of the equilibrium constant, low
concentrations ensure that the reation obeys the ideal law.  In
concentrated solutions the law doesn't work quite right unless you
replace concentrations by activities, something that is only
indirectly measurable (it's whatever makes the formula work).

(vi) Again, we need to know the context.

No, not all equilibrium reactions need to be in a closed vessel. An
aqueous reaction that only takes a short time (so that for example
water won't evaporate appreciably, messing up the concentration) is
just fine in a test tube.


Fulfilling the request from @Mockingbird, there is a quote from (1, p. 90)

  Surprisingly, [1.1.1]propellane is somewhat more stable to thermal
  decomposition than the next larger propellane, [2.1.1]propellane,
  indicating a reversal in the trend of increased reactivity with
  increased strain. To understand this observation, it is important to
  recognized that the energy of both the reactant and intermediate
  influence the rate of unimolecular reactions that lead to
  decomposition. In the case of propellanes, homolytic rupture of the
  central bond is expected to be the initial step in decomposition. This
  bond rupture is very endothermic for [1.1.1]propellane. Because
  relatively less strain is released in the case of [1.1.1]propellane
  than in the [2.1.1]- and [2.2.1]-homologs, [1.1.1]propellane is
  kinetically most stable.(2)

  Another manifestation of the relatively small release of strain
  associated with breaking the central bond comes from MP4/6-31G∗
  calculations on the energy of the reverse ring closure.(3)

  The thermal decomposition of [1.1.1]propellane has been studied both
  experimentally and by computation(4)

The book chapter further discusses reactivity of [1.1.1]propellane, if OP is really interested in this area, I would suggest to grab the book in the library or focus on the papers (2-4) the book references to.

(1) Carey, F. A.; Sundberg, R. J. Advanced organic chemistry. Part A. Structure and Mechanisms, 5th ed.; Springer: New York, 2007.
(2) Wiberg, K. B. Angew. Chem. Int. Ed. Engl. 1986, 25 (4), 312–322. DOI: 10.1002/anie.198603121
(3) Adcock, W.; Binmore, G. T.; Krstic, A. R.; Walton, J. C.; Wilkie, J. J. Am. Chem. Soc. 1995, 117 (10), 2758–2766. DOI: 10.1021/ja00115a011
(4) Jarosch, O.; Walsh, R.; Szeimies, G. J. Am. Chem. Soc. 2000, 122 (35), 8490–8494. DOI: 10.1021/ja994043v


A lot of non-systematic chemical names don't describe the composition of the compound, they describe what the compound is synthesised from. That is the case here since EDTA is derived from ethylene by adding two units of the diamine.

It isn't that uncommon in chemical names. Polyethylene, for example, is (simplifying only slightly) a long chain of carbons connected by single bonds. But it is made by polymerising ethylene. Hence has a name that doesn't describe what its structure is but where it came from.


Do the mechanism (SN1 or SN2 ) which reaction follow depends more upon intermediate stability than on substrate and nucleophilic?

It will follow SN1 as the intermediate carbocation formed is resonance stabilised.


Why is hypochlorite a stronger oxidizing agent than other oxychlorides?

First, hypochlorite ($\ce{ClO−}$) is not a strong oxidizing agent compared to other chloride oxyanions, at least in acidic medium. Under standard conditions in acidic medium, chlorous acid ($\ce{HClO2}$) is the best oxidizing agent among them due to the largest positive number for standard electrode potential:

$$\begin{align}
\ce{HClO2 + 2H+ + 2 e- &<=> HClO + H2O} &\quad E^\circ &= \pu{1.645 V} \label{rxn:1}\tag{1}\\
\ce{HClO2 + 3 H+ + 3 e- &<=> 1/2 Cl2 + 2 H2O} &\quad E^\circ &= \pu{1.628 V} \label{rxn:2}\tag{2}\\
\ce{HClO2 + 3 H+ + 4 e- &<=>  Cl– + 2 H2O} &\quad E^\circ &= \pu{1.570 V} \label{rxn:3}\tag{3}\\
\end{align}$$

Keep in mind that in the reduction half-reaction \eqref{rxn:1}, $\ce{HClO2}$ reduced to $\ce{HClO}$ with its largest positive standard electrode potential $(E^\circ = \pu{1.645 V})$ among chlorine oxoacids. For comparison, we should look at half-reactions of which each of these two reagents reducing to the same product. Thus, look at half-reactions \eqref{rxn:2} versus \eqref{rxn:4}, and \eqref{rxn:3} versus \eqref{rxn:5} where $\ce{HClO2}$ shows its superiority over $\ce{HClO}$:

$$\begin{align}
\ce{HClO + H+ + e- &<=> 1/2 Cl2 + H2O} &\quad E^\circ &= \pu{1.611 V} \label{rxn:4}\tag{4}\\
\ce{HClO + H+ + 2 e- &<=>  Cl– + H2O} &\quad E^\circ &= \pu{1.482 V} \label{rxn:5}\tag{5}\\
\end{align}$$

Perchlorate ($\ce{ClO4-}$) and chlorate ($\ce{ClO3-}$) ions, on the other hand, are also good oxidizing agents, but not as strong as $\ce{HClO2}$ or $\ce{HClO}$, yet closed. This fact is clear when compared to the half-reactions where they directly reduce to $\ce{Cl2}$ or $\ce{Cl-}$ similar to $\ce{HClO2}$ and $\ce{HClO}$:

$$\begin{align}
\ce{ClO4- + 8H+ + 7e- &<=> 1/2 Cl2 + 4H2O} &\quad E^\circ &= \pu{1.39 V} \label{rxn:6}\tag{6}\\
\ce{ClO4- + 8H+ + 8e- &<=> Cl– + 4H2O} &\quad E^\circ &= \pu{1.389 V} \label{rxn:7}\tag{7}\\
\ce{ClO3- + 6H+ + 5e- &<=> 1/2 Cl2 + 3H2O} &\quad E^\circ &= \pu{1.47 V} \label{rxn:8}\tag{8}\\
\ce{ClO3- + 6H+ + 6e- &<=> Cl– + 3H2O} &\quad E^\circ &= \pu{1.451 V} \label{rxn:9}\tag{9}\\
\end{align}$$

However, the standard reduction potentials OP is mentioning is for different reduction half-reactions $\ce{ClO3-}$ and $\ce{ClO4-}$ involved (\eqref{rxn:12} and \eqref{rxn:10}, respectively):

$$\begin{align}
\ce{ClO4- + 2H+ + 2e- &<=> HClO3 + H2O} &\quad E^\circ &= \pu{1.189 V} \label{rxn:10}\tag{10}\\
\ce{ClO3- + 2H+ + e- &<=> ClO2 + H2O} &\quad E^\circ &= \pu{1.152 V} \label{rxn:11}\tag{11}\\
\ce{ClO3- + 3H+ + 2e- &<=>  HClO2 + H2O} &\quad E^\circ &= \pu{1.214 V} \label{rxn:12}\tag{12}\\
\end{align}$$

Accordingly, when consider Latimer diagrams $(\ce{ClO4- -> ClO3- -> HClO2 -> HClO -> Cl2 -> Cl-}),$ you'd see, $\ce{HClO2}$ cannot oxidizes $\ce{ClO3-},$ although it is the strongest oxidizing agent. However, $\ce{HClO2}$ can oxidize another species, which $\ce{ClO3-}$ cannot oxidize.

Note that all reduction potential values are from: http://depa.fquim.unam.mx/amyd/archivero/TablasdepotencialesREDOX_26700.pdf (Petr Vanẏsek), which cited the following references:

G. Milazzo, S. Caroli, V. K. Sharma, Tables of Standard Electrode Potentials; Wiley, Chichester, 1978.
A. J. Bard, R. Parsons, and J. Jordan, Standard Potentials in Aqueous Solutions; Marcel Dekker, New York, 1985 (Series: Monographs in electroanalytical chemistry and electrochemistry).
S. G. Bratsch, “Standard Electrode Potentials and Temperature Coefficients in Water at $\pu{298.15 K}$,” J. Phys. Chem. Ref. Data 1989, 18(1), 1-21 (https://doi.org/10.1063/1.555839).


Is sedimentation coefficient constant and why?

Sedimentation due to Gravity

The sedimentation coefficient of a particle falling in a fluid due to gravity is easier to understand. It is the ratio of the settling velocity ($v_{t}$) to the acceleration causing the settling($a_{g}$).

$$s=\frac{v_{t}}{a_{g}}$$

In fluids on earth $a_{g}$ is acceleration due to gravity which is known to be constant.

$v_{t}$ , which is also referred to as terminal velocity, is constant.  It is constant because the force driving the acceleration of the particle through a fluid ($F_{g}$) is equal in magnitude and in an opposite direction to the drag force ($F_{d}$).  Since these two forces cancel out the net force ($F_{n}$) is 0.

$$F_{g} - F_{d} = F_{n} = 0$$

There is no net acceleration $a_{n}$ since $F_{n}=m_{p}a_{n}=0$.  The particle mass ($m_{p}$) cannot be zero so there is no net acceleration of the particle, thus the terminal velocity is constant.

A ratio of two constants are constant so the sedimentation coefficient is constant.

Sedimentation due to Centrifugal Acceleration

Radial acceleration, $a_{r}$, equals the angular velocity $\omega$ squared times the radius, $r$.

$$a_{r} =\omega^{2}r$$

If the angular velocity is constant, which it is for most of centrifugation, and the radius is constant, then the radial acceleration, like the acceleration due to gravity, is constant.

The sedimentation velocity in centrifugation, just as in gravity settling, is constant.  It is constant because the drag force of the particle in the fluid is balanced by the centrifugal forces generated by the centrifuge. A centrifuge can generate forces several times higher than gravity so settling occurs much faster during centrifugation.

Again, the ratio of two constants are constant so the sedimentation coefficient is constant.

The other equation looks like it is a derivative of Stoke's law, which is used to determine the drag force of a particle in a fluid.


Here, limiting radius ratio is for geometry of this complex is square planar because $\ce{CN-}$ is a strong field ligand and causes pairing of unpaired electrons. So we can relate that $\ce{Ni^2+}$  occupies a square planar void of the $\ce{CN-}$ lattice. The square planar void is a cross section of the octahedral void.


Is beryllium difluoride covalent or ionic?

There is a very relevant article by Gillespie Covalent and Ionic Molecules: Why Are BeF2 and AlF3 High Melting Point Solids whereas BF3 and SiF4 Are Gases? J. Chem. Educ., 1998, 75 (7), p 923.

According to the article, the charge on Be is +1.81 and the charges on the Fs are -0.91. (citing to his earlier article Reinterpretation of the Lengths of Bonds to Fluorine in Terms of an Almost Ionic Model Inorg. Chem., 1997, 36 (14), pp 3022–3030)

So while nothing is completely covalent or ionic, BeF2 is extremely toward ionic.


How to synthesize tetrabutyl titanate from titanium dioxide powder?

Treat with a chlorination reagent the titania to form titanium tetrachloride. You could do this with coke and chlorine in an electric arc furnace. Then distill the titanium tetrachloride. Next react it with a mixture of sodium butoxide and butanol at a low temperture.

Next vacuum distill the product to obtain pure titanium butoxide


Composites of Immiscible Solids - Aluminum and Lead Example

You are correct that solid Al (fcc) and Pb (fcc) are (surprisingly for both being fcc) not miscible in each other (well, there is always some miscibility as driven by entropy). Heck, there is even an miscibility gap in the liquid phase as well. Al and Pb really don't like each other.

So, as you cool the liquid from above 1418C (the top of the liquid miscibility gap), you will phase separate in the liquid phase. At the aluminum melting point, the aluminum will solidify, and the Pb will remain liquid. When the Pb melting temperature is reached, the Pb will solidify. The exact specifics of the solidification will depend on how it is done (the kinetics), not the thermodynamics. This means you can't predict the final micro-structure without  details of the process.

The final result will be solid Al and Pb, likely intermixed. It is not an alloy, but a two-phase mixture. I would not call it a composite as I would generally reserve that name for actual composite materials.


How does oxygen fight odors? (Deodorant ad)

In my estimation, this is basically marketing nonsense. If you examine the list of ingredients, the primary active agent is invariably some aluminum or aluminum-zirconium salt, which act as anti-perspirants primarily by the mechanical actions of shrinking pores (due to their astringency) and clogging ducts in the apocrine glands.

Listed among the inactive ingredients is hydrogen peroxide, $\ce{H2O2}$. At sufficient concentrations, $\ce{H2O2}$ can be a potent oxidizing agent, and probably the source of the oxygen the ad refers to. It might indeed be able to neutralize some odors by oxidizing the molecules which cause them, thereby transforming those molecules into odorless ones. However, as it's listed among the inactive ingredients, it's almost certainly present in very low concentration. This is to be expected, since prolonged skin contact will invariably cause chemical burns unless the mixture is very dilute. Furthermore, much of the odor of sweat is the result of metabolic byproducts from skin-dwelling bacteria, chief among them certain short-chain carboxylic acids that are already maximally oxidized. The only malodorous substances in sweat that can be neutralized by oxidation, as far as I can tell, are certain amines and thiols. That said, I doubt the concentration of $\ce{H2O2}$ is sufficient to be of any use here. $\ce{H2O2}$ does also have anti-septic properties, so perhaps its purpose is bacteriostatic, but there I'm just speculating. Ultimately, I'm uncertain of its purpose in the formulation, but I'm extremely dubious that it helps substantially in eliminating odor.


Genuine confusion about the equilibrium constant

I gather that the equilibrium constant K = 1 implies that, at equilibrium, neither the forward nor the backward reactions are thermodynamically favoured.

There is nothing special about K = 1. At equilibrium, no matter the value of K, there is no net reaction. Forward and backward reactions proceed at the same speed.

  But if pressure of the system increases and this shifts the position of equilibrium, the value of K remains to be 1.

Yes, even a reaction with K = 1 can react. Just imagine you start with some reactant and no product. This situation is not at equilibrium. The reaction will proceed in the forward direction until Q = K. Then, you can disturb the system, changing Q, and there will be a net reaction again, changing concentrations until Q reaches K again.

  Please can I have a simple and comprehensive explanation of what the value of K actually tells us?

You said "please", so I will try for a short, simple and comprehensive answer. The value of K allows you to calculate equilibrium concentrations, no matter which concentrations you started with. For a lower K, the product equilibrium concentrations will be lower than for a higher K (same initial concentrations). The opposite is true for reactant equilibrium concentrations.

It should be clear that the equilibrium concentrations for a given K are not always the same. The simplest example is doubling initial concentrations (while keeping the volume constant). If the equilibrium concentrations were equal for both cases (initial concentrations doubled or not), it would go against the conservation of matter. (We doubled the amount of species present initially, so we can't have the "old" amount of equilibrium species).


Refractive Index (RI) variation with Temperature & disagreement between two standards

The refractive index of a pure liquid varies with the wavelength of the light (this dispersion causes the splitting of white light into rainbow colours) and with the density and thus temperature of the liquid (this causes the schlieren patterns when heating a liquid on a hot plate).

The section “Index of Refraction of Liquids for Calibration Purposes”, in CRC Handbook of Chemistry and Physics, 90th Edition (CD-ROM Version 2010), David R. Lide, ed., CRC Press/Taylor and Francis, Boca Raton, FL. lists values for several organic liquids at various wavelengths and temperatures.

Reference values close to the values given in the question can be found for toluene:

$$\textbf{Index of refraction of toluene}\\
\begin{array}{lll}
\hline
&\lambda/\mathrm{nm} &T=20\ \mathrm{^\circ C} &T=25\ \mathrm{^\circ C} &T=30\ \mathrm{^\circ C} \\
\hline
&667.81 &1.49180 &1.48903 &1.48619 \\
&656.28 &1.49243 &1.48966 &1.48682 \\
&589.26 &1.49693 &1.49413 &1.49126 \\
&546.07 &1.50086 &1.49803 &1.49514 \\
&501.57 &1.50620 &1.50334 &1.50041 \\
&486.13 &1.50847 &1.50559 &1.50265 \\
&435.83 &1.51800 &1.51506 &1.51206 \\
\hline
\end{array}$$

The uncertainty of the refractive index is estimated as $\pm0.00003$.

For example, the refractive index measured for toluene using the characteristic wavelength of a sodium-vapour lamp $(\lambda\approx589\ \mathrm{nm})$ at $T=20\ \mathrm{^\circ C}$ is $1.49693$. When increasing the temperature, the refractive index is decreased to $1.49413$ $(-0.19\ \%)$ at $T=25\ \mathrm{^\circ C}$ and to $1.49126$ $(-0.38\ \%)$ at $T=30\ \mathrm{^\circ C}$.

The range of values given in the question corresponds to a decrease between $-0.26\ \%$ and $-0.52\ \%$ when the temperature is increased from $T=20\ \mathrm{^\circ C}$ to $T=27\ \mathrm{^\circ C}$.


Is there any danger of forming explosive reaction products during the electrodeposition in Diethylene glycol?

Diethylne glycol reacts readily with perchloricacid, however the oxidation and hazard potential of perchloric acid depends on its concentration in side the solvent.

If concentrations are high enough and more than 10 percent perchloric acid formation is expected with 100 percent reaction yield, then it can be explosive.

What actually happens that as soon as diethylene gylcol meets with perchloric acid, diethylene glycol decomposes and some dangerous peroxides formed. These are very reactive to water, so if you can provide very dry and inert atmosphere it should be safe but you should give extra care to make sure that you do not have humidity and water both in your chemicals and in the atmosphere.

If you are going to do the experiment and use milimolar range chemicals then just do it, because you will not be able to form enough perchloricacid to get problems.


Directions & sites are said to be crystallographically equivalent, because there are symmetry operators that relate them.  So

learn about simpler symmetry operations (proper axes, mirror planes, centres of inversion) or advanced ones (like screw axes, glide planes, etc.)
learn how to recognize them in every-day's life.  It may be helpful for crystallography, and in other fields too: IR / Raman spectroscopy, for example, not only for understanding (descriptive), but for forcasting (predictive).  Tutorials like this represent an entry.


What is Redox reaction? Is it really possible?

I think you mean Redox, which is a portmanteu of Reduction-Oxidation. An easy example of redox is the Haber process (nitrogen and hydrogen form ammonia)

$$\ce{N2 + 3 H2 <=> 2 NH3}.$$

In the Haber process the nitrogen is reduced (the oxidation state changes from $0$ to $-3$), and the hydrogen is oxidized (the oxidation state changes from $0$ to $+1$). In that redox reaction, both oxidation and reduction take place, as you said.

There's no oxidation without reduction and vice-versa. The electrons lost in oxidation have to go somewhere! If they wouldn't, then there's no reason for the reaction to occur in the first place.


At boiling point, is the total pressure twice the atmospheric pressure?

Pressure is not additive

If you mix two gases of different pressure, the resulting gas does not have a pressure equal to the sum of the two pressures. Treating the gases as ideal gases, the pressure will be the average of the two pressures, weighted by the volumes. What is additive is the partial pressure of a gas mixtures, where the sum is the total pressure.

Closed container with constant volume

To simplify things, let's place a closed bottle filled with the liquid of interest into the container, adjust the pressure to 1 atm, and close the container. If we heat the container, the pressure will go up just because the gas temperature rises. If we open the bottle, some of the liquid will evaporate until the partial pressure is equal to the vapor pressure. Again, the container will have a pressure higher than atmospheric pressure. So the liquid will not boil when the temperature reaches the normal boiling point because the pressure will be above atmospheric pressure at that point, for the two reasons mentioned.

Closed container with variable volume

If the experiment is done in a closed, variable volume container that expands into the surrounding kept at 1 atm, the volume will increase as the temperature is raised (because the gas expands, and because the vapor pressure increases with increasing temperature). When the temperature reaches boiling temperature (slightly above, actually), the liquid will boil until it is all gone, and the volume will increase further during this boiling process. The partial pressure of the liquid will never reach 1 atm (because there are other gases present), so the vapor pressure will remain higher than the partial pressure throughout the boiling process.

  Does the total pressure become twice the atmospheric pressure at the surface?

No. Either there is room to expand, and the gas produced by boiling will push away the other gas (and then quickly mix with it), or we keep the volume constant, and then we don't call it boiling (because the total pressure exceeds 1 atm, so the liquid will not boil. However, when you heat a liquid to the boiling point and let the system reach equilibrium, the partial pressure of that substance will be 1 atm, and if any other gases are present, the total pressure will be higher. How much higher depends on the starting conditions. If our starting condition is a completely filled closed bottle, with inert gas (e.g. nitrogen or air) already at the boiling point of the liquid and at 1 atm, when we open the bottle the total pressure will reach 2 atm.


Number of atomic and molecular orbitals in different Hartree-Fock references

Straight answer to your question: whatever method you use, RHF, ROHF or UHF, your calculations will use the same number of basis functions (7 contracted functions for your water example). How will they be used and combined, however, is different. In your comment on the question you talk about molecular orbitals - it's important to keep these conceptually separate from the basis set you are using to construct the orbitals.

How many functions do you use for an orbital in ROHF?
In general, UHF/ROHF approaches use a separate molecular orbital for each electron, so you would use basis functions twice to describe an unrestricted pair compared to a restricted pair.

In other words, basis functions aren't "split" between the electrons when treated as an unrestricted pair; rather, two orbitals are calculated using the same basis functions, one corresponding to the alpha electron and another for the beta electron.

So, if you compare UHF with RHF, you'll be using the same basis functions to construct twice the number of molecular orbitals.

Let's take an example to illustrate it: $\mathrm{O_2}$.

For singlet $\mathrm{O_2}$ ($\mathrm{S=0}$), there are 8 alpha and 8 beta electrons. Using STO-3G, we'll have 10 contracted functions, those corresponding to $\mathrm{1s}$, $\mathrm{2s}$ and 3 x $\mathrm{2p}$ for each oxygen atom, each of them composed of 3 primitive gaussians.

In RHF, we can combine these 10 basis functions in a single Slater determinant, producing 10 molecular orbitals. The first 8 of these will be doubly occupied, the last 2 will be vacant.
In ROHF, we have exactly the same as for RHF, because ROHF is exactly the same as RHF for molecules with $\mathrm{S=0}$.
In UHF, we use the same 10 basis functions, but we use them to build two sets of 10 molecular orbitals - 10 for the alpha electrons and 10 for the beta electrons - in a single Slater determinant. We occupy them with 8 alpha electrons and 8 beta electrons, 1 electron per orbital, with 2 alpha and 2 beta orbitals left vacant.

For triplet $\mathrm{O_2}$ ($\mathrm{S=1}$), there are 9 alpha and 7 beta electrons. We have the same basis functions as before.

In UHF, we do the same as before, and we also get 10 alpha and 10 beta molecular orbitals in a single Slater determinant. Occupation changes, though, so now we have 9 occupied and 1 vacant alpha orbitals and 7 occupied and 3 vacant beta orbitals.
In ROHF, and since now we have an unequal number of alpha and beta electrons, we have a more complex construction of the Fock operator - one that is not unique. Essentially, we'll calculate alpha and beta electrons separately under the hood, and then combine them; and we'll be treating full, partially full and empty orbitals as three coupled sets. We'll produce 10 orbitals, of which we'll doubly fill the first 7 and partially fill the following 2, leaving 1 unoccupied.

In conclusion: the number of basis functions in all cases is the same: 10 contracted functions comprising 30 primitive gaussians. However, these basis functions are used a different number of times for RHF, UHF and ROHF; RHF uses them once in a simple Slater determinant, UHF uses them twice to form a Slater determinant with a two sets of orbitals, one for alpha and the other for beta electrons, and ROHF uses them twice but ends up producing a single set of full, partially or non-occupied orbitals that share their spatial part.


Work done by gas while expanding against constant external pressure?

In a reversible process, the pressure of the gas is uniform throughout, and the force per unit area that the gas is exerting on the piston face is equal to its pressure, as determined from its equation of state.  So the work is equal to the force integrated over the displacement, or, equivalently, the uniform gas pressure integrated over change in volume.

In an irreversible process, the pressure of the gas is not uniform throughout, but, at the piston face, the force per unit area that the gas exerts on the piston face must match the force per unit area that the piston face exerts on the gas, by Newton's third law.  But, in this case, because the gas pressure is not uniform throughout (and also because viscous stresses contribute to the force per unit area of the piston), the force per unit area at the piston face cannot be determined from the equation of state.  However, if the force per unit area at the piston face is imposed by specifying external conditions on the system, the work that the gas does can still be calculated as the integral of the external pressure integrated over the volume change.

So, to summarize, in both reversible and irreversible processes, the force per unit area exerted by the gas on the piston face is equal to the externally applied pressure and the work is equal to the external pressure integrated over the volume change.  However, in the case of a reversible process, the uniform gas pressure can be determined from the equation of state of the gas, but, for an irreversible process, it cannot (and needs to be determined by controlling the external pressure in some way).