text:
Would it be possible to produce a non-toxic atmosphere using a coal fire and frozen oxygen?
A Pail of Air is a classic science fiction short story by Fritz Leiber from 1951 describing how a family survives on an Earth that has left its orbit around the sun, using desparate low-tech methods: a shelter isolated by many layers of blankets and a coal fire to thaw frozen oxygen gathered in the eponymous pails.
I am curious how realistic this scenario is from a basic chemistry and biology perspective. I think not for a number of reasons, but don't know enough to quantify these factors:

Would a coal fire produce enough energy to thaw the oxygen the fire itself uses?
Would the resulting atmosphere (at sufficient pressure for the oxygen to support human life) contain toxic levels of carbon dioxide?
...or carbon monoxide?
How does oxygen at lower pressures affect flammability? Wouldn't it all go up in a blaze at the drop of a hat?

Frozen oxygen? Oxygen turns to a gas at about the temperature that liquid nitrogen does, so a coal fire would melt it for sure. However, liquid oxygen is highly flammable and highly, highly explosive, so what I see happening with melting solid oxygen with a coal fire is a massive explosion.

The most important issue is probably not the compassion of the atmosphere but whether it is possible to contain an atmosphere at all at any reasonable pressure. Using blankets to hold anything likely to retain atmospheric pressure would be a bigger challenge than making sure there wasn't too much CO and CO2. If you could, then all the other issues would be a lot less important.

@matt_black: I disagree - maintaining pressure (especially when it's mostly oxygen and something like 0.15 atm might be sufficient) is a problem solvable through low-tech engineering, but the concerns I listed above could prove the method at the heart of the survival strategy fundamentally invalid.

Summary
Okay, so after doing some serious number crunching and searching for literature values all over the web, I've come to the following conclusion: Yes, it is possible to survive in these conditions!
Plausibility of Environment
First, let us examine the plausibility of the environment which is described in the short story. Outside the safety of the living space, cordoned off by 30-or-so blankets, lined with aluminium foil, there is a vacuum, and all of the gases in the atmosphere have frozen out and precipitated onto Earth's surface. The exception is helium, which still is liquid.
If we check the phase diagram of oxygen (note that in the figure below we would have to look at the pressure of 0.001), we see that indeed it is possible:

(Source)
Maintaining Pressure
The minimal pressure that humans need to survive (while breathing pure oxygen) is set at about 19 kPa, let's round that up to 20 kPa (0.2 atm). Let's furthermore also assume the space where this small family is living as about $3\times 3\times 2~\mathrm{m^3} = 18~\mathrm{m^3}$. Adding in the factor of a comfortable enough temperature (I pen that at $15~\mathrm{°C} = 288~\mathrm{K}$) we can use the ideal gas law: $$pV = nRT \Longleftrightarrow n=\frac{pV}{RT} = 150~\mathrm{mol\, \ce{O2}}$$
So we need $150~\mathrm{mol}$ of gas constantly in order to maintain pressurization at the desired temperature. Whether 30 heavy rugs really can do that I'll leave to an experimental physicist.
Oxygen Depletion
The fire isn't the only source of oxygen depletion, also the humans inhabiting the small space will use oxygen. Using this data, an average adult human being will consume $22.5~\mathrm{mol}$ $\ce{O2}$ per day. So we'll round this to $70~\mathrm{mol\, d^{-1}}$ for the whole family (2 adults, 2 kids). This is already corrected for oxygen exhalation, so the system generates an equal amount of $\ce{CO2}$.
Total Oxygen Balance and Required Heat
The only source of oxygen is the pail above the fire, in which the solid oxygen gets heated up to the boiling point (which, at 20 kPa is 77.1 K, see figure below; data source: NIST) and then further to ambient temperature.

Here some of the constants that I used for the following calculation: $c_\text{p} = 54~\mathrm{J\, mol^{-1}\, K^{-1}}$ for the liquid and $c_\text{p} = 29~\mathrm{J\, mol^{-1}\, K^{-1}}$ for the vapor phase with $\Delta H_\text{vap}(77.1~\mathrm{K},\; 20~\mathrm{kPa}) = 7.19~\mathrm{kJ\, mol^{-1}}$. The melting heat (at standard pressure, but let's just take this value from Wikipedia) is $0.445~\mathrm{kJ\, mol^{-1}}$, and the heat capacity of the solid I estimated at $25~\mathrm{J\, mol^{-1}\, K^{-1}}$. Initial temperature is $T_\text{i}=5~\mathrm{K}$, melting temperature (again for standard pressure, but we'll take it) $T_\text{m}=54~\mathrm{K}$, boiling point is given above and final temperature is $288~\mathrm{K}$.
The total heat needed for heating up $150~\mathrm{mol\, \ce{O2}}$ is $q_\text{tot, 150}=2.43~\mathrm{MJ}$.
Using the formula given in the paper Calculations of the Heat Release Rate by Oxygen Consumption for Various Applications I end up with $\dot{V} = 0.177~\mathrm{m^3\, d^{-1}}$, for a oxygen depletion coefficient of $0.8$ in a pure atmosphere of oxygen ($x_\mathrm{\ce{O2}}^0 = 1$, standard pressure is assumed in the formula). This is about 1% of the total volume in the cove, and is also only needed for the initial pressurization. However, since the inhabitants also produce $\ce{CO2}$ at non-negligible levels, and the concentration of said gas should stay under 2% for normal breathing and to avoid the feeling of dizziness, we need to up the oxygen generation rate.
So, to keep this whole system going with $x(\ce{CO2}) < 1.5~\%$, we need to generate 4700 mol $\ce{O2}$ every day (and this is without even looking at $\ce{CO2}$ generation by the fire).
It is inevitable that the exhaust gases from the fire should be led out via the chimney (that is also described in the story) without great mixing with the air in the room. But all in all the heat generation is possible (the oxygen needed to heat the 4700 mol to the temperature is only 46 mol), thus making it a survivable climate.
Conclusion
By the above estimation-based calculations and considering the limitations, I conclude that it would indeed be conceivable to survive in these conditions.

It took me some time (I've been pondering the answer for so long, I decided to calculate the values just to be sure), but finally my answer is here!

Awesome answer! So if I understand correctly, one would need to generate much more oxygen than the humans need to survive, mainly to keep the CO2 they breathe out below toxic levels. But that's very good for the plausibility of the scenario since there's going to be a lot of leakage anyway. This leaves flammability as a main concern, as well as the amount of coal required - over 37kg each day, over 13 tons per year. But I think the least realistic aspect may be something else - the construction of a makeshift pressure suit to haul in the oxygen.

As to the leakage... $\ce{CO2}$ has a higher density ($0.37~\mathrm{kg, m^{-3}}$ vs $0.27~\mathrm{kg, m^{-3}}$ at 20 kPa) so it would be most beneficial to position the leak at the bottommost point in the room.

Some sort of intensive hydroponics would help, converting additional CO2 to O2. Presumably they need one anyway in order to produce food.

Yes, it is possible to produce a non-toxic atmosphere using a coal fire and frozen oxygen.
I'll answer your questions last to first.
No, low pressure oxygen will not make flammability worse, fires require something to bond with oxygen (often carbon, causing carbon dioxide). Low pressure will make things less flammable as there is less oxygen to bond with.
Yes, using an open coal fire in a semi-enclosed space will gain toxic levels of carbon dioxide, carbon monoxide and sulfur dioxide (among a few others) quickly. However with a minor modification from a open fire to a fire in a old fashioned chimney pipe stove (venting outside the hut) would eliminate this.
Lastly, Definitely Yes, a coal fire will produce more than enough energy to thaw the oxygen it uses. To quote figures 1 pound of coal will produce roughly 12,500 btu's (British Thermal Units) Assuming the temperature of the frozen oxygen we are thawing is -455°F (the "temperature of deep space") and about 3 degrees below the freezing temperature of helium. To thaw 1 gallon of the frozen oxygen will require roughly 900 btu's (if your stove has a 80% efficiency) and produce 860 gallons of pure oxygen at 80°F and atmospheric pressure. Coal will consume 140% of it's weight in pure oxygen, so for 1 pound of coal 135 gallons of your oxygen will be consumed. Leaving you with 11,500 btu's extra heat and 725 gallons of oxygen. The average Human consumes 25 gallons of oxygen per day making this plenty to survive on.

That's just what I was looking for! Great to see that it would at least work in theory...

No, it would not be possible to use a coal fire and frozen oxygen to create a atmosphere able to sustain life for any prolonged amount of time.
Liquid oxygen, though not explosive or flammable by itself, would mix with the soot given off by the coal fire and create a highly explosive mixture.
Also, a coal fire gives off a toxic amounts of carbon dioxide, and potentially other gases. Not to mention the soot, which would quickly suffocate anyone breathing it in.

I think so too, but what I'm really looking for is concrete numbers concerning toxicity and energy budget. I see liquid oxygen and soot as a distraction from the core question, since ensuring that the fire produces a minimal amount of soot and that the oxygen is contained while it's liquid should not be too hard even with low-tech methods.

Sorry, I was thinking you were saying how they do it in the book.


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Looking for raw interferogram data / raw FID data from FT-IR / FT-NIR / FT-NMR
I'm trying to get my hands on some raw interferogram data / raw FID data from an FT-IR / FT-NIR / FT-NMR so I can run some tests using FFT with it (it needs to be real data). Here's a picture below of the data I'm looking for highlighted in yellow.  I'm looking for Alkali metals (Sodium, Potassium, but also Magnesium, Gold, Silver, Platinum and Palladium).  Any ideas where I can get these or if someone can send them to me as a text/data file?  I've searched NIST but everything is already converted

I think I may have a HCl-DCl FTIR. Let me check and I'll get back to you.

I don't unserstand if these are FT-IR spectra maybe take a look..

@GM - Those indeed are interferograms of a variety of organic compounds. RickT - I'm uncertain if the metals you list will have meaningful data using these techniques. For example, IR spectroscopy is all about the vibrations of covalent bonds. In order for a vibration to be IR active, it needs to be symmetry-allowed (usually by experiencing a change in dipole).

In The Journal of Chemical Education, published by the American Chemical Society, a rather recent article presented a related project, hosted by the University of Minnesota.  Intentionally it is about raw FIDs-as-obtained, out of the NMR spectrometer, aiming to teach correct data processing (including spectra with two orthogonal frequency scales). The article is sponsored by the authors (i.e. freely accessible), the database requires an e-mail to register.  Reference: J. Chem. Educ., 2013, 90 (7), pp 941–943; DOI: 10.1021/ed300787v

source: Stack Exchange

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What type of sensor technologies exist to determine the chemical composition of a polymer?
Let's say I have several pieces of trash and each is made with a different type of material like:

Plastic bottle : High Density Polyethylene (HDPE)
Cereal box : Paperboard & Epichlorohydrin
Shrink Wrap : Polyolefin
Kitchen utensil : Melamine-resin
Textiles : Urea-formaldehyde
Crown Cork Bottle Cap : Metal and Plastisol lining
Rock
Plasterboard
Coffee cup : Ceramic and Glaze (silica + metal oxides)

My objective is to build a system complete with sensors that automatically determines the material (chemical makeup) of each object. Are there material sensors currently on the market that can identify all of the materials that make up these objects?

Perhaps a XRF spectrometer.

You can identify many polymers without any clever sensor technologies at all but by following a number of simple physical tests. Sensor technology might help for some of the very fine distinctions, but if you can make a broad categorisation first with simple physical tests then that is what you should do.
And, of course, most plastic opponents are actually labelled with symbols to aid in recycling so you can often just look to identify the chemical nature of the polymer. For reference the symbols are shown below (V identifies chlorine-containing polymers like PVC):

The steps required to identify polymers, if you don't have a label, involve some simple physical tests. First identify whether the polymer is elastic or hard. Then identify whether it is a thermoplastic or a thermoset (by using a hot object like a soldering iron to see whether it remelts). Then a variety of tests can be applied: does it float in water or not? Does it burn and if so how does it behave and what does the flame look like?
One flowchart for this is available here as a pdf. Another simpler one is here also as a pdf but sufficient to distinguish the major classes of polymer. There are even interactive versions of such flowcharts such as this one (flash).
Included below is another version originally produced by the major, but now defunct, British chemical company ICI:

￼￼

Sorry. I realize I did not make my original question clear enough but I just edited it adding more information that should clarify it. In brief I am looking to create an automated material sensor system that can identify the chemical make up of any object that might be thrown out as trash (basically any object in existence). Are there material sensors available that one can purchase that can achieve this?

@max You may be being overambitious. the reason people are encouraged to separate their recycling into different groups is that a universal identification/separation system is not remotely cost-effective. If you only need to separate plastics or plastics plus paper, the techniques described above can work at a reasonable cost. Many actual recycling plants also use some human-driven separation which is also often cheaper than any available automated process.

I understand but an automatic material sensing system is a requirement for a technology I am designing. I just need to know if such a system can be purchased or if the necessary sensors to identify all the objects in my question actually exist and can be purchased. Thanks.

Well, it seems the question has emerged to How to design a polymer recycling plant? ;)
Rocks and pieces of plasterboard can be separated from the rest of material by streams of compressed air along or across perforated sheets of metal in a conveyor belt: they'll keep sitting there - like a rock ;)
Plastics should be milled down to flakes then. The mixture can be submitted to a swim-and-sink step: PE whould swim on water, PS and PVC will sink.
Whether it is technically useful to repeat this step using NaCl solutions of different concentration to achieve further separation is beyond my knowledge.
IR spectroscopy will provide a lot of information on the type of polymer measured. Aren't InGaAs line cameras/sensors used in this context? Pattern matching with stored spectra might be used to control streams of air to blow identified material off a conveyor belt.
I'm sorry that I can't provide more precise information, but I'm not an engineer. However, I wouldn't be astonished if full-fledged setups are commerically available nowadays.

I edited my question. Sorry for being rather vague before.

Haha. Yes a polymer recycling plant indeed! Thank you for you answer. I will look into some of the keywords you mentioned. Oh and where would be the best place online see if full-fledged material identification setups are commercially available? I am completely new to the realm of chemistry.

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Non-hardening adhesive
I am wondering if any of you know about glue/adhesives that do not harden over time after they are applied or glue/adhesives that take a really long time to harden after application. I need the glue to temporarily hold polymer sheets before they are sewed together.

Are the polymers hydrophilic? In that case it might be possible to just use water and rely on capillary forces to make the sheets stick together

I cannot use water. I have tried it since you suggested it. The polymer is hydrophilic, but it does not hold it well enough. Furthermore, there's the problem of evaporation.

Ok, too bad. But cool that you tried it!

What about water soluble adhesives? would that be an option? So you can easily clean the sheets after sewing

I think you could use some types of Butylrubber. If I'm not wrong from Isobutylene is derived the Post-it adhesive. However you can find Butyl tape easily in every big hardware store (especially nautical) or online.
EDIT
At this link you can find some more information this is an excerpt:

Tapes and flooring adhesives  Exxon butyl rubber can be used in
adhesive formulations for transparent tape, hot-melt pressure
sensitive adhesives, mastic for pipe wrap tape and vinyl floor tile
adhesives. It can also be used for semiconducting electrical splicing
tape. A rubber splicing tape must be self-fusing; i.e., when wrapped
upon itself, it must fuse to the underlying layer, forming a permanent
bond. The self-fusing is a result of compounding and eliminates the
need for application of a separate adhesive coat. These compounds have
very high tensile strengths and exhibit some degree of recovery or
elasticity. Electrical splicing tapes based on Exxon butyl rubber have
moisture resistance, outstanding aging resistance and good electrical
properties. Polyisobutylene elastomers have long been successfully
used in pressure sensitive adhesives both as primary elastomers and as
tackifiers and modifiers. Exxon chlorobutyl rubber can be used alone
or in blends with other polymers to achieve special properties in
pressure sensitive adhesives.

Thank you. However, I need to sew through the polymer and the adhesive. The needle and the "thread" that I am using aren't very resilient (which is why I need the glue to be non-hardening) . I am not sure whether the needle would be able to handle an extra layer of tape. Do you happen to know if there's a liquid, low-viscous form of butylrubber?

@user97554 I've try to find more information but I think very thin layer could be transparent..


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What is the chemical basis for superglues?
There are a variety of different types of glues, some derived from animal or natural materials and some designed by chemists. Some glues need to be activated before they harden and stick (many epoxies, for example) but others seem to stick very quickly without any obvious activation (like Loctite Superglue or The Original Super Glue).
What chemical reactions make these glues work?

This is a very general answer to your question.
According to the Chemistry World podcast transcript, Cyanoacrylate and the article The Chemistry of Superglue (Brubaker, 2000), the main ingredient cyanoacrylate, which is

a type of plastic known as an acrylic resin. It exists in the tube as single molecules in liquid form. But when they come into contact with water, the molecules react with hydroxide ions to form long polymer chains that set into a hard solid. As there is a thin layer of moisture covering most surfaces, cyanoacrylate can bond them together very effectively.

This also explains why superglue tends to harden very quickly when in contact with skin.
A diagram of cyanoacrylate polymerisation in the presence of water is shown below:

from the University of Rochester, who also state that asides from water, that the presence of

Weak bases (such as those on the proteins in skin) work even better

for polymerising the cyanoacrylate.

So glues like these don't actually 'dry'!


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What eco-friendly substance can I use to write and erase on a wall for UV glowing without detection in daylight?
I rent an office and want an alternative to a whiteboard and continual dry-erase marker purchases.
I can shut out the light and use the surface under UV lamps only, that's fine.
Commonly listed organic glowing elements are blood, urine and semen. None of which, I think would be untraceable by eyes (or nose) during the day!
The surface I would like to use contains typical white interior paint, applied last year.

I think that in a wall you can't write and erase so easily. You can write and then re-paint the surface in this case you can use Zinc White (make sure that what you use is $ZnO$). It was a common white pigment before $TiO_2$ discovery and has a strong fluorescence under UV light so if your wall is white (best $TiO_2$ painted, $CaCO_3$ may have some fluorescence) you will not notice $ZnO$ area in visible light but only under UV light.


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What are the key chemical structures of liquid crystals used in display technology?
The majority of current flat screen TVs and computer monitors are based on Liquid Crystal displays.
Liquid crystals are compounds that orient under an applied electric field and alter the polarisation of transmitted light. This can be used to build what are effectively valves that selectively transmit light.
So what are the typical chemical structures of the chemicals that have these properties and how do the structures relates to how they work?

This may be of interest http://scifun.chem.wisc.edu/chemweek/liqxtal/liqxtal.html


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What makes charcoal a good adsorbent to organic and non polar substances while alumina and silica are best for polar substances?
My question is, what makes charcoal a good adsorbent to organic and non polar substances while alumina and silica are best for polar substances?

related http://chemistry.stackexchange.com/questions/7033/why-is-charcoal-such-an-excellent-adsorbent

Charcoal (although activated carbon, a specially processed microporous charcoal product is better) is a material that is mostly carbon. The black allotrope of carbon is graphite, which is made of overlapping planar sheets of graphene. Graphite is very nonpolar (since it has only carbon atoms). Thus, nonpolar substances will adsorb. Most organic substances impurities are nonpolar substances (though not all, for example, sugars). Additionally, the graphene sheets are aromatic. The ring current generates a dipolar magnetic field and a quadrupolar electric field that attract other aromatic substances through $\pi$-stacking. Thus graphite is best at removing aromatic substances, which include all organic dyes. Carbon filtering can also remove contaminants like chlorine, although this is likely a reaction leading to chlorination of the surface.
Silica $(\ce{SiO2})$ and alumina $(\ce{Al2O3})$ are polar materials. They are polar for three reasons. 1) They are composed of polar $\ce{Si-O}$ and $\ce{Al-O}$ covalent bonds. 2) The geometries of these minerals especially around the oxygen atoms are not perfectly in-line with VSEPR predictions, so polarity is not canceled by symmetry (see the silica unit image below).

3) Both materials are hygroscopic and absorb/adsorb water from the atmosphere. The $\ce{Si-O-Si}$ and $\ce{Al-O-Al}$ covalent network on the surface is partially hydrolyzed so that the surfaces of these materials actually present $(\ce{-OH})$ groups, increasing the polarity of the surface. Polar substances are attracted to silica and alumina by dipole-dipole interactions and hydrogen bonding.
$$\ce{-Si-O-Si - + H2O -> -Si-OH + HO-Si -}$$


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Why is M1*V1 = M2*V2 only used for diluting solutions?
Why is $M_1V_1 = M_2V_2$ only used for diluting solutions? Why can it not be used to figure out how many moles to add to make a solution more dilute?

Because it is based on conservation of moles. $M_1V_1=\text{moles before dilution}=M_2V_2=\text{moles after dilution}$

As Satwik has stated in the comments, the equation relies on the conservation of moles principle.

$$\dfrac{mol_1}{V_1} \times {V_1} = \dfrac{mol_2}{V_2} \times {V_2}$$

Based on the above, given a constant volume of a solution, adding more moles of solute will only make the concentration increase, not decrease.  Adding in more solvent is the only answer in that case.


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SN1 or SN2, the kinetic isotope effect
I know that you can't always trust Wikipedia, so I just want to check something regarding their kinetic isotope effect page.
The first example implies that $\sideset{^{12}}\ C$ and $\sideset{^{13}}\ C$ can be used to determine whether a reaction is $S_N1$ or $S_N2$:

...In the nucleophilic substitution reaction of methyl bromide with
  cyanide, the kinetic isotope effect of the methyl carbon, in this case
  defined as $k_{12}/k_{13}$, was found to be 1.082 ± 0.008 ... The observed
  methyl carbon kinetic isotope effect is indicative of an $S_N2$
  mechanism."

No explanation appears to be offered for this inference. I know that the reaction will not proceed via $S_N1$ because of the unstable carbocation but (how) does the KIE datum predict an $S_N2$ mechanism?

Your quote ends in a reference to an article in Advances in Physical Chemistry. http://dx.doi.org/10.1016%2FS0065-3160%2806%2941004-2.  I agree that this is poor practice for Wikipedia, but normal for journal articles. The author of this this portion of the page was probably more sued to a different type of writing. The article I linked is behind a paywall for me, so I cannot say more about it, but your answer is probably there.

The rate constant depends on the energy difference between the ground state of the reactants and the transition state. The ratio of the rate constants due to the kinetic isotope effect therefore depends on the difference in vibrational zero point energy between the initial state and the transition state. Vibrational ZPE depends on bond stiffness and reduced mass. The change in reduced mass from $\ce{^{12}C}$ to $\ce{^{13}C}$ causes the kinetic isotope effect, and the bond stiffness determines the size of the KIE. Bond stiffness is to a first approximation proportional to bond order.
In both $S_N1$ and $S_N2$, the initial state has a single bond between the methyl $\ce{C}$ and $\ce{Br}$  (as well as some $\ce{C-H}$ bonds which stay the same throughout the reaction).
For $S_N2$, the transition state has partial bonds between the methyl $\ce{C}$ and both $\ce{Br}$ and the $\ce{C}$ of $\ce{CN}$.
In the $S_N1$ transition state, the methyl $\ce{C}$ has only a partial bond to $\ce{Br}$.
Since the change in bond order is greater between initial state and $S_N1$ transition state than between ground state and $S_N2$ transition state, $S_N1$ will have a greater kinetic isotope effect. Measuring the KIE of an unknown reaction and comparing it to the KIEs of reactions with known mechanisms will therefore distinguish between $S_N1$ and $S_N2$.


text:
Why is CCSD(T)/aug-cc-pVTZ not a suitable method for the geometry optimisation of omeprazole?
I know that CCSD(T) with the aug-cc-pVTZ basis set for geometry optimization of omeprazole (5-methoxy-2-[(4-methoxy-3,5-dimethylpyridin-2-yl)methanesulfinyl]-1H-benzimidazole, the active substance in Losec) is not suitable using normal computers. The substance is:

(omeprazole)
Can someone explain to me what exactly makes these methods too expensive for this substance?

Is it just that the molecule is too damn large? I've seen simple CCSD on 6-atom systems run a full week and generate 750GB of scratch work.

@chipbuster If you're using a computer from the early 90s, a CCSD job could take that long.  However, a CCSD(T) job on 6 atoms can take just an hour with a double-zeta basis set...

@LordStryker wasn't my personal calculation, so I don't know. Maybe these guys were on ye olde pentium II, but more likely I'm mistaken. Thanks :)

@chipbuster I've run a CCSD(T)/aug-cc-pCVQZ frequency job on an 8 atom system that took about 3 months walltime (yes, core-valence basis set).  Insane in the membrane.

To transform the question into something simpler, let's assume CISD instead CCSD(T), which is cheaper and share the common expensive step.
The molecule has 182 electrons and 1545 basis functions in the basis you specified, that means approx. 1400 virtual orbitals.
Therefore, the number of double excitation $ij \rightarrow ab$ is approximately $180 \cdot 180 \cdot 1400 \cdot 1400 = 32400 \cdot 1960000 = 63504\cdot10^6$
Unfortunately, in each of this excitation you have to calculate integral of type $(ab|cd)$, and there are $1400^4$ of them.
Such a big problem is out of reach for traditional CCSD, but tractable with more approximate methods.

I'd also like to add to ssavec's response. In addition to the insane amount of flops that you have to carry out to run the computation, you also run into problems of storage for the computation. In a CCSD calculation you have to store single and doubly excited amplitudes. ssavec pointed out that there are ${6.3\cdot10^{10}}$ double excitations...assuming 8 bytes per double precision floating point, that would mean storing the doubles would require 450 GB of storage! (Not to mention the I/O between iterations...)

Both previous answers by ssavec and jjgoings accurately describe the problem for a single point calculation. Doing a geometry optimisation, you have to run the procedure multiple times until you find a minimum on the potential energy surfaces.
Since one point already requires insane resources to compute this, an optimisation at this level will take significantly longer. From experience and common usage of chemical calculations it is often not necessary to optimise a geometry at this level. Often much simpler and more approximate methods, like density functional approximations or perturbation theory yield good and reliable results.


text:
Can fluorine act as the central atom in interhalogen compounds?
Why can't fluorine be the central atom in inter-halogen compounds?
A $\ce{F-F}$ bond is weaker than a $\ce{F-X}$ bond and thus, fluorine should be happy to form inter-halogen compounds. But, why doesn't it act as the central atom?
In my textbook, the reason given is because of the high electronegativity of fluorine atoms. But, how does that affect fluorine being the central atom or not?
If fluorine is the central atom, it can draw electrons from other less electronegative halogens and be happy. Does this lead to instability of the compound?

Fluorine being smallest than all other halogens can't act as the center. Majority of Inter halogen compounds are formed by $Br$ and $I$, which combine with small size atoms such as fluorine, since it is possible to associate more small size atoms around a large central atom. Your text book is right, more electronegative atoms are smaller in size, as like fluorine.

Why can't fluorine be the central atom in inter-halogen compounds?

First off, fluorine can be at the "center of things."  Examples would include the strongly hydrogen bonded hydrofluoric acid

and the very relevant example of the trifluoride anion
$$\ce{[F-F-F]^-}$$
The trifluoride anion example is critical as it demonstrates that fluorine can be hypercoordinated (hypervalent).  Of course, in this last example we've "tricked" the fluorine into occupying the middle position, so your question remains (slightly altered), why doesn't fluorine occupy a central position more often?
In order to answer this question, two general cases need to be considered,

The anionic situation $\ce{[X-F-X]^-}$ vs.  $\ce{[X-X-F]^-}$
The cationic situation $\ce{[X-F-X]^+}$ vs. $\ce{[X-X-F]^+}$

where X is Cl, Br or I.
In the case of the anion, Lewis dot structures suggest a negative charge on the central atom.

However, computational analysis of the electron distribution in the triiodide anion (see p. 156) suggests that the central atom is positively charged (+0.27 electron), while the terminal atoms are negatively charged (-0.63 electrons on each terminal atom).  Assuming that the same general trend would be applicable to mixed trihalide anions, then in the case of $\ce{[X-F-X]^-}$ we would be placing positive charge on highly electronegative fluorine atom instead of the larger, more polarizable (e.g. more capable of bearing a positive charge) bromine, chlorine or iodine atom.
In the case of the cation, similar arguments come into play.  Again, calculations for a variety of trihalide cations (mixed and homogenous) suggest substantial positive charge on the central atom (this time in agreement with Lewis Dot structures). The same arguments made above would operate here to destabilize structures with a central fluorine atom. This analysis also found extremely long (less stabilizing) bonds resulted when fluorine (or the most electronegative of the 3 atoms) occupied the central position.
Arguments that don't explain the infrequent occurrence of fluorine at a central position in inter-halogen compounds include:

the lack of "d-orbitals"; as other's have commented, this is an old, invalid argument; the existence of $\ce{F3^-}$ demonstrates that fluorine is capable of hypercoordinate bonding without the need for d-orbital involvement
inefficient 2p-3p overlap; $\ce{FCl}$ has a higher bond strength and lower heat of formation than $\ce{Cl2}$

Wait, your analysis of the triodide ion's charges doesn't add up to -1.

In the anion there are 6 + 6 + 6 = 18 lone pair electrons, plus 4 bonding pair electrons for a total of 22.  In neutral fluorine there are 7 electrons; 3 x 7 = 21.  21 - 22 = -1.  Am I missing your point?

I mean when you say that the central atom in the trihalide ion triiodide is +0.27 and the peripheral atoms is -0.63 - that adds up to -0.99.

As it should for the anion with one unit of negative charge.

so the last 0.01 units of charge is just a rounding error?

Yes, just a rounding error.

why might the central interhalogen atom always have a positive charge? And does FCl actually exist; I cannot find any evidence on Google @ron.

Can't give a better answer then that's the lowest energy arrangement in these hypercoordinated halogens.  For ClF See http://en.m.wikipedia.org/wiki/Chlorine_monofluoride

Explanations that you mentioned in the end are still taught in many high schools :(

As a practical example there is (1) a fluorine-bridged iodine structure of $\ce{[R_\text{f} - I - F - I - R_\text{f}]-}$ composition with 5-centered 6-electron bond stabilized by fluorine in the middle:

Here is an isolated anionic part of tris(diethylamino)sulfonium bis((perfluorophenyl)iodo)fluoride, $\ce{F - I}$ distance is approx. 2.5 Å:

This is not exactly an interhalogen compound, but the closest known analogue I found with fluorine in the middle.
(1) Farnham, W. B.; Dixon, D. A.; Calabrese, J. C. J. Am. Chem. Soc. 1988, 110 (25), 8453–8461.DOI: 10.1021/ja00233a022

Several examples of species with fluorine in the middle of a chain are reported here. What they have in common: the fluorine is embedded in an environment of high or enhanced electronegativity, which makes fluorine in the center more energetically favorable.

The fluorine-iodine bridged structure in andselisk's answer. Iodine itself is not very electronegative, but here it is incorporated into a functional group that is rendered electron-withdrawing via both the ring of $sp^2$ carbon atoms and the attached terminal fluorines.

Gold heptafluoride, discussed here. This low-temperature compound is a difluorine complex, having a structure that may be rendered as $\ce{F5Au - F - F}$. The geometry at the bridging fluorine atom is bent. The $\ce{AuF5}$ fragment is so electronegative it imparts partial positive charges to the fluorine atoms in the $\ce{F2}$ ligand.

Caesium trifluoride, $\ce{Cs(F3)}$ (not yet experimentally known). This proposed high-pressure compound would contain a true polyhalide species, specifically $\ce{F3^-}$. It is calculated to be an intermediate stage towards other salts and complexes where the caesium becomes a $5p$ electron donor and assumes oxidation states greater than $+1$. See Ref. [1].

Reference
1.
Miao, M.S. "Caesium in high oxidation states and as a p-block element." Nat Chem. 2013 Oct; 5(10):846-52. https://doi.org/10.1038/nchem.1754. PMID: 24056341.

Fluorine can't be the central atom in inter-halogen compounds because it is an element from the period 2 in the periodic table. It can not have more than 8 valence electrons. And as it has 7 valence electrons, it can only form one bond. This is not the case of the other halogens $\ce{Cl, Br, I,...}$ where these elements are below period 2 in the periodic table. That means they can have an expanded octet. They can have more than 8 valence electrons. So they can be central atoms in inter-halogen compounds.
From an orbital viewpoint, we can attribute this to the fact that in the case of $\ce{Cl, Br,  I,...}$  the orbitals $\ce{3d}$ , $\ce{4d}$ and $\ce{5d}$  respectively are available for hybridization and can interact with the valence orbitals $\ce{ns, np}$ of these atoms. So these halogen atoms have the possibility to form more than one bond with other atoms. In the case of fluorine, the valence orbitals $\ce{2s, 2p}$ can not interact with $\ce{3d}$ due to the big difference of energy.

The use of d orbitals for hypercoordination is, as ron mentioned this in his answer, already successfully disproved. As this answer continues, the existence of $\ce{F3-}$ also disproves your one bond theory.

@martin my interpretation would still be fluorine forms just one net bond. However, this can be delocalized over multiple linkages.

@oscar yes, maybe, probably. Over the years and the further and farther I looked I lost sight of "the bond" and realised there are multiple interpretations, view points, opinions, etc. and no good way to reconcile any of that. However, on a purely mathematical level, the argument with d orbitals falls through - there simply is a simpler way to do it (pun intended). Therefore I remain with my vote on this answer: it's wrong.

fluorine is high electronegativity so when it combine with other halogen which is less electronegativity then fluorine can take electron from other atom due to this compound become unstable that's why  fluorine can't be central atom in interhalogen compound.

This makes little sense. If I understood your answer, you're implying that when fluorine reacts, an unstable compound is formed than the reactants, which is usually not the case

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text:
Why is gold sol used for intra-muscular injection?
In my textbook it is written that gold sol is used for intramuscular injection. But, actually what is its use? Why would people use such a precious element for injection. Can other colloids be used like silver sol?
Note:I posted this question to get a deeper understanding of what is given in my text about the use of gold sol in intramuscular injection. Googling it didn't help.

Will this help?
Toxicity and cellular uptake of gold nanoparticles: what we have learned so far?
Gold has a long tradition as a disease-modifying antirheumatic drug in the treatment of rheumatoid arthritis and it seems to relieve pain and decreases swelling of the joints.

Sir, my question is about gold sol and not nano particles.

We are not taught anything about intra-molecular injection but,it's given in our chemistry textbook as an application of colloids.

You might want to read your last comment again. Sometimes our thoughts and our writing take different directions ;)

Gold nanoparticles are sometimes referred to as "colloidal gold" in mining industry, which I think matches the definition of a gold sol closely enough.


text:
Ozone catalysis
I know that catalysts don't change chemical equilibrium because they accelerate both the reactions in the same way.
I can't understand why CFC catalysts accelerate ozone destruction but not ozone formation.
Could you help me?

Would you mind to outline a mechanism for the formation of ozone under CFC catalysis?

One possibility to consider is that  ozone destruction could be thermodynamically favourable, but kinetically inhibited. this would then explain your doubt.

You make the assumption that the ozone concentration in the upper atmosphere is in equilibrium. It isn't.
$\ce{O3}$ is a much less stable molecule than $\ce{O2}$ (the heat of formation from $\ce{3/2 O2}$ is $143~\mathrm{kJ/mol}$) and the concentration at equilibrium would be very low. A significant concentration exists in the upper atmosphere because  $\ce{O3}$ is continuously formed from the reaction of $\ce{O2}$ and UV light. Also, though it is a highly reactive molecule, there are still kinetic barriers to its conversion back to $\ce{O2}$. There is no simple chemical route back to normal oxygen and the UV absorption of the $\ce{O2}$ molecule that creates ozone is different to the UV absorption of the ozone molecule, so the driving force of ozone creation doesn't destroy it.
CFCs release $\ce{Cl}$ atoms in the upper atmosphere via a photochemical reaction. The $\ce{Cl}$ atoms react with ozone in a variety of reactions the simplest of which are:
$$\ce{Cl + O3 -> ClO + O2}$$
and
$$\ce{ClO + O3 -> Cl + 2 O2}$$
So, strictly speaking, the $\ce{Cl}$ from CFCs is not directly catalysing the equilibrium between $\ce{O3}$ and $\ce{O2}$ but providing a new reaction pathway via several intermediates that, in a roundabout way, allows the "equilibrium" to be reestablished. Another way to think about this is that the presence of $\ce{Cl}$ facilitates a lower kinetic barrier route to ozone destruction. A lower kinetic barrier to the reaction allows the relative concentrations to get closer to the natural equilibrium. So even if there were a reverse pathway (which there isn't in these complex reactions) the result would be a much lower concentration of ozone.

Due to presence of UV, molecular oxygen breaks down into oxygen atoms. These oxygen atoms combine with molecular oxygen to form ozone. Now this ozone reacts with UV to form molecular oxygen again.

So why is ozone and molecular oxygen not in an equilibrium state as both forward and backward reactions are happening?


text:
Decomposition products of PET
On Wikipedia, it says that polyethylene terephthalate (PET) decomposes at temperatures above 350°C.
What are these decomposition products? I read further that acetaldehyde is produced, but is that the only decomposition product?
Also, what compounds are produced when PET is burned? ($\ce{+O2->?}$)

Since most PET used for a particular end use requirement the deliberate addition of other plastic additives are added that changes the nature of the toxic degradation products this includes cross contaminates from other polymers added to PET

This is a partial answer to your query.  According to the paper Degradation and Recyclability
of Poly (Ethylene Terephthalate)
 (Venkatachalam, 2012), there are a few degradation products, from the article:

Apart from aldehyde, $\ce{CO}$, $\ce{CO2}$, ethylene, benzene, biphenyl are also identified as
  degradation products

(The article also shows diagrms of the chemical processes of PTE degradation).
an important aspect to note is experiments/observations have

shown that rate of thermal degradation in the closed system is
  about three times higher than that in open system.

In regards to burning of PET, the products can vary according to the page EPB 433- Health and Environmental Effects of Burning Waste
Plastics
 (Saskatchewan Ministry of Environment) stating that

the composition of byproducts of plastic combustion as to their type and concentration depends
  on the combustion temperature and the flame residence time.

According to the Peiyu Plastics page (a reference from Wikipedia), if complete combustion occurs, then the products would be $\ce{CO2}$ and $\ce{H2O}$, incomplete combustion could produce soot and carbon monoxide.
Further reference:
Here is the MSDS for PET
A study of thermal decomposition and combustion products
of disposable polyethylene terephthalate (PET) plastic using
the High Resolution Fourier Transform Infrared
Spectroscopy, Selected Ion flow Tube Mass Spectrometry
and Gas Chromatography Mass Spectrometry  (Sonova et al. 2008).


text:
explain the reaction mechanism
$\ce{ROH + NaBr + H2SO4 -> RBr + NaHSO4 + H2O}$
What is exactly the reaction mechanism.
Is it $\ce{NaBr + H2SO4 -> NaHSO4 + HBr}$ and then
$\ce{HBr}$ reacts with $\ce{ROH}$?
If yes the why in the reaction of $\ce{NaBr}$ and $\ce{H2SO4}$ does $\ce{H2SO4}$ oxidise $\ce{NaBr}$?

HBr reacts with ROH?

Exactly. It's just a nucleophilic substitution.


text:
Ideal gas derivations
For an ideal gas, how can someone obtain this:
$$ \bar{V}dP = RdT- P \bar{dV}$$
from which it can de deducted, that for a constant temperature
$$ \frac{dP}{\bar{dV}} = -RT $$

This looks like the differential of the ideal gas equation. I'm going to use $V_m$ for the molar volume.
$$PV_m=RT$$
Now we convert each side to the differential. We keep $P$ and $V_m$ together since they are both variable. We can move $R$ out of the differential because it is a constant.
$$d\left(PV_m\right)=RdT$$
We can evaluate $d\left(PV_m\right)$ using the product rule:
$$d\left(PV_m\right) = PdV_m + V_mdP$$
Now:
$$PdV_m + V_mdP = RdT$$
Which is equivalent to your first equation.
**However, at constant temperature $dT=0$, so at constant temperature, the equation simplifies to:
$$PdV_m + V_mdP=0$$
$$V_mdP = - PdV_m$$
$$\frac{dP}{dV_m}=-\frac{P}{V_m}$$
Which is not equal to your second equation. The closest I can get is the following:
$$\frac{dP}{dV_m}=-\frac{P}{V_m}\cdot\frac{V_m}{V_m}=-\frac{PV_m}{V^2_m}=-\frac{RT}{V^2_m}$$
Which is what we would get by evaulating:
$$\left(\frac{\partial}{\partial V_m}\right)_T \left(P=\frac{RT}{V_m}\right)$$


text:
How to predict a COSY NMR spectra for an AMX spin system?
Does anyone know how to draw (predict) a hypothetical COSY NMR spectra for AMX spin system (the spin of A M X nuclei is 1/2)?

An AMX spin system represents three nuclei coupled to each other, with their signals separated by a large chemical shifts compared to the coupling between them (such that $\Delta \delta >> 3J$). This system is said to be first-order, and the A, M and X signals are each a doublet of doublets, and the couplings can be extracted by inspection.
The simple COSY spectrum of this will show the three core signals along the diagonal, with cross peaks appearing for all signals ie each signal will have two cross peaks. Below is a (very basic) schematic for an AMX COSY.

The appearance of the cross peaks will depend on the type of COSY experiment you are running, such as a COSY-45 or COSY-dqf etc, but this should give you the general picture for what you are asking.

Thank you for your answer! I get the concept. But how would the signals of A, M and X look if the spin of M and X nuclei is 1 (and the spin of A remains 1/2)? They wouldn't be doublet of doublets, right?

The quick answer as a comment (ask a new question if you need further clarification): Remember that splitting is described by the formula 2nI+1. For the system of A (n=1, I=1/2) M (n=1, I=1/2) and X (n=1, I=1/2), you would have the spectrum appearing as A: 9 line multiplet (9 lines of equal intensity which some describe as a triplet of triplets), and M and X each a 6 line multiplet (6  lines of equal intensity, which some describe as a doublet of triplets). The actual appearance though will depend on the coupling constants as you may get overlap of some of the lines.


text:
Can lye be extracted from bamboo ash?
I know that the best source of lye is hardwood. A brief search tells me that Bamboo is considered a hardwood of the monocotyledon variety. I've done a lot of searching to find this answer, but I can't seem to find if I can use the same process with bamboo ash as I would with, say, Apple ash, to extract lye.
My question is two-fold:
1) Is it possible to extract lye from bamboo ash?
and if so,
2) Would this lye be a quality lye, capable of being used to create bar soap?

Yes, it's possible but as pointed out in Bamboo: an overlooked biomass resource? Biomass and Bioenergy, Vol. 19, pages 229-244:

[bamboo] shares a number of desirable fuel characteristics with certain other bioenergy feedstocks, such as low ash content and alkali index

Potassium oxides are between 33-50% of the ash depending upon species and age of the plant.
Potassium oxides were between 0.16-0.58% of dry matter.
Sodium oxides were about a factor of 100 lower.

I know that the best source of lye is hardwood.

Is that really so?
In THE SODIUM, POTASSIUM AND PHOSPHORUS CONTENTS OF TREE SPECIES GROWN IN CLOSE STANDS, published in New Phytologist, 1958, 57(3), 273–284 (DOI), J. D. Ovington and H. A. I. Madgwick state:

[...] In nearly all cases, conifers contain greater weights of sodium, potassium
  and phosphorus than hardwoods of the same age grown under similar conditions. [...]

This still doesn't answer the question. Perhaps I could rephrase it to what would be the best source of natural lye? The caveats being it needs to be easily and quickly grown in a contained space (such as an in-door greenhouse with no natural lighting). It should also be capable of being used for heat, thus not burning too quickly.

I'm not going to pretend to be an arborial (arboral?) expert by any means, but while it refers to averages of certain elements, do those necessarily directly correspond to 1. purity of sodium hydroxide formed in the ash and 2. The ease of extracting said sodium hydroxide (possibly chemically bound or reacted to other compounds during the asking process)? I skimmed thru the paper and saw nothing in it that addressed these points. Again tho, not my field.

The answer above is more like a comment


text:
Intuitive understanding of standard electrode potential
I have read many books on standard electrode potential and they define the term in various ways but I just can’t seem to get an intuitive idea of it.
$$\ce{Cu^2+ + 2e- <=> Cu}$$
Standard electrode potential, $E= 0.34\ \mathrm{V}$
Does this mean that when $1\ \mathrm{C}$ of charge passes from the hydrogen half cell to copper half cell, $0.34\ \mathrm{J}$ of energy is released?
$$\ce{Zn^2+ + 2e- <=> Zn}$$
Standard electrode potential, $E= -0.76\ \mathrm{V}$
Does this mean that when $1\ \mathrm{C}$ of charge passes from the hydrogen half cell to zinc half cell, $0.76\ \mathrm{J}$ is absorbed?

The absolute electrode potential, i.e. the actual volts between electrode and electrolyte, is incredibly difficult to measure accurately.    Instead we simplify: declare the hydrogen electrode potential to be zero, when actually it's roughly +4.4V @25C deg.    Then measure other electrode potentials using a standard hydrogen electrode to contact the electrolyte.


text:
How to read a given chemical formula?
I'm trying to read this chemical formula but I'm a bit confused. I'm hoping that someone could help break this down for me to explain what the various parts mean. Someone purchased epigallocatechin in a purchase request and I found that it's synonymous with the following.
(−)-cis-2-(3,4,5-Trihydroxyphenyl)-3,4-dihydro-1(2H)-benzopyran-3,5,7-triol
I believe that the minus up front refers to the way that the polarity changes as light passes through the molecule. What does the 3,4,5 refer to and the 2?

The correct IUPAC name actually is (2R,3R)-2-(3,4,5-trihydroxyphenyl)-3,4-dihydro-2H-1-benzopyran-3,5,7-triol.

Klaus's answer has all the right explanations but I always find these things hard to follow without pictures, so here are some to clarify.
Note that in the pictures below we don't explicitly draw carbons or the hydrogens attached to them (we just mentally add enough hydrogens to make the number of bonds to each carbon equal 4).
First what is benzopyran?

The two numbers on the first structure tell us where the oxygen appears in the pyran ring and where the saturated carbon (with an extra hydrogen) appears. Both could be in different places. The second picture showns the numbering convention for each atom in the ring so we can keep track of where anything else is attached. The third picture shows what happens if we notionally add extra hydrogens on carbons 3 and 4 to give a fully saturated ring.
Now we add an OH (hydroxy) in certain positions:

The picture on the right shows how we number the parts of trihydroxy phenyl when it is attached to something else (it will be stuck on the 2-carbon on the benzopyran in the final compound).
Finally we have the epigallocatechin:

The "cis" indicates that the two hydrogens (mentally fill them in in positions 2 and 3) are on the same side of the benzopyran (the opposite side to the two wedge-like bonds on the OH and phenyl groups).
I hope that all makes some sort of sense. Note that none of the above bears any relation to how the molecules are made: this is all notional deconstruction for the purpose of explaining how the molecule is named.

In addition to what ssavec already outlined:
cis refers to the relative stereochemistry of the substituents at C2 and C3 of the benzopyran, which is the core structure (and thus name-giving) of the title compound.

3,4-dihydro-1(2H)-benzopyran

This tells us a lot!

Benzopyrans come in two isomers, which are distinguished by the position of the $\ce{O}$ atom in the pyrane ring. Apparently this is a 1-benzopyran, the $\ce{O}$ is close to the benzene ring.
Benzopyrans have a $\ce{C=C}$ double in the pyrane ring as well as a saturated centre, the one with the two $\ce{H}$ atoms. The position of the double bond is determined by the saturated centre. Exactly this is denoted with (2H).
As a matter of fact, our benzopyrane doesn't have a double bond ;) 3,4-dihydro denotes were it was, i.e. where a formal hydrogenation has taken place.

3,4,5 are the locants of phenolic OH on the trihydroxyphenyl substituent and 2- is its position on the 1-benzopyran.

Can you please expand on your explanation, for the asker's sake?

Also note that cis-… name would mean either of the two stereoisomers

and

(or their mixture), so the (−)-cis-… is to denote the single enantiomer. However the optical rotation orientation is not obvious from the structure without experimental data, so structure-only based name would better use R/S stereodescriptors denoting the absolute configuration of the provided structure (first one), (2R,3R)-…, but it's OK (but rather redundant) to add the optical rotation based descriptor as well, (−)-(2R,3R)-…
Also note the other, nicer, name from the linked page (stereochemistry omitted),
3,3’,4’,5,5’,7-hexahydroxyflavane, which is the semi-systematic flavonoid name. Flavane is 2-phenyl-3,4-dihydro-2H-1-benzopyran (note slight change around ‘H’ in contemporary preferred names). Unprimed numbers are for benzopyrane ring, primed ones for phenyl.

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text:
What's the difference between mass spectrometry, mass spectroscopy, and mass spectrography?
It seems that the terms mass spectrometry, mass spectroscopy, and mass spectrography are mostly used like synonyms. Is this correct? Or is there a difference?

According to the Encyclopaedia Britannica, mass spectrographs use(d) photographic plates or films for the detection of ions, as in the apparatus thought out by F. W. Aston in 1919.


text:
Do the compounds NaₓIrO₂ and NaOsO₂ exist?
Based on some physics background, I want to know whether have scientists synthesized these two materials: $\ce{Na_{$x$}IrO2}$ and $\ce{NaOsO2}$ in the lab, if so, what properties do they have?
I will be very appreciative if someone could recommend me some relevant articles.

Would you be willing to share your interest in these two materials? For example, is there a specific application in mind? If so, we might be able to suggest alternate materials should these two not exist.

I want to seek a material that has triangular lattice structure, strong spin-orbital coupling (e.g. 5d element) and large (or moderate) coulomb interaction between electrons. For example, $\ce{ABO2}$ type, $\ce{NaOsO2}$ compounds, where $\ce{A}$ and $\ce{B}$ are alkali and transition-metal ions, respectively. So far, I know $\ce{IrTe2}$ may be a one.
Do you mean that both $\ce{NaxIrO2}$ and $\ce{NaOsO2}$ do not exist?

Is there any information on iridium and osmium oxides in these low oxidation states?

For example, $\ce{Na2IrO3}$ and $\ce{Li2IrO3}$. But these two belong to the honeycomb lattice structure, and what about the triangular lattice structure?

Welcome to Chemistry.SE. Would you be willing to share your interest in these two materials? For example, is there a specific application in mind? If so, we might be able to suggest alternate materials should these two not exist.

Is there any information on iridium and osmium oxides in these low oxidation states?

@Ben Norris I want to seek a material that has triangular lattice structure, strong spin-orbital coupling (e.g. 5d element) and large(or moderate) coulomb interaction between electrons. For example, ABO$_2$ type(NaOsO$_2$) compounds, where A and B are alkali and transition-metal ions, respectively. So far, I know IrTe$_2$ may be a one. Do you mean that both Na$_x$IrO$_2$ and NaOsO$_2$ do not exist?Thank you very much.

@Klaus Warzecha For example, Na$_2$IrO$_3$ and Li$_2$IrO$_3$. But these two belong to the honeycomb lattice structure, and what about the triangular lattice structure? Thank you very much.

I don't know about those two compounds but 1. Potassium osmate is well known and researched. I couldn't find any corresponding sodium salt. 2. $\ce{Na2IrO3}$ is well known. In 2017, a corresponding copper salt was synthesized from it. Also, a  new sodium iridate of formula $\ce{NaIrO3}$ is discovered and synthesized from the former compound.

...and in a 2019 paper: "although neither experimental nor theoretical investigation on homologous derivatives NaIrO2 is available in the literature. However, the photocatalytic properties are not explored up to date"

For $\ce{Na_{x}IrO2}$ the current answer seems to be "no" it has not been made but it is discussed in the literature theoretically.
From Kitaev-Heisenberg models for iridates on the triangular, hyperkagome, kagome, fcc, and pyrochlore lattices:

However, no triangular lattice iridate with the relevant edge sharing octahedra structure is currently available; a compound of the
  type NaxIrO2 may or may not turn out to be stabilized.

and in Chiral spin liquid and topological superconductivity in the undoped and doped
triangular Kitaev-Heisenberg model:

In addition, a class of ABO2 (where A and B are alkali and transition metal ions, respectively) type transition metal compounds and the possible material NaxIrO2 may be potential candidates for realizing the triangular Kitaev-Heisenberg model.

(But that's your paper isn't it? I don't have any information beyond what you already know, so really this is a useless answer.)


text:
Reduction of terephthalic acid to an alcohol (or phenol?)
I want to convert terephthalic acid (TPA) to an alcohol (however I think the correct term would be "a phenol").

I've read that lithium tetrahydridoaluminate ($\ce{LiAlH4}$) can produce an alcohol from a carboxylic acid. And, according to Wikipedia, dicarboxylic acids (such as TPA) react around the same. So does $\ce{LiAlH4}$ reduce TPA to a phenol (or an other product)?

I want to make sure that you understand the different between two commonly confused molecules (their names don't help).
Phenol is an $\ce{-OH}$ group directly attached to an aromatic ring. Because of this connectivity, phenols exhibit different chemical behavior than other alcohols. In particular they are more acidic. Phenol can be abbreviated $\ce{PhOH}$, where the $\ce{Ph}$ represents the $\ce{C6H5}$ group of the aromatic ring. As a class of molecules, phenols can be represented generically as $\ce{ArOH}$, where $\ce{Ar}$ is used not for argon (there are no compounds of argon), but for generic arenes with any substitution.

Benzyl alcohol is a different molecule. In benzyl alcohol, there is a $\ce{CH2}$ group between the benzene ring and the $\ce{OH}$. Benzyl alcohol behaves like other alcohols. Benzyl alcohol is abbreviated $\ce{PhCH2OH}$.

Conversion of benzoic acid to phenol requires loss of a carbon atom.
$$\ce{Ph{\bf C}O2H -> PhOH + {\bf CO}}$$
Lithium aluminum hydride converts carboxylic acids to alcohols, but it does not cause the loss of the carbon atom. Lithium aluminum hydride will reduce benzoic acid to benzyl alcohol. If you want the benzyl alcohol derivative, stop here. If you want the phenol, read on.
$$\ce{PhCO2H} \overset{1. \ \ce{LiAlH4}}{\underset{2. \ \ce{ H2O}}{\ce{->}}}\ce {PhCH2OH}$$
To convert benzoic acid to phenol, we need a way to loose that carbon atom. One way is to oxidize benzyl alcohol to benzaldehyde using perhaps a Swern oxidation. Then benzaldehyde can be converted to phenyl formate using a Baeyer-Villiger Oxidation. The formate could be hydrolyzed in aqueous base.
$$\ce{PhCH2OH \overset{Swern}{->} PhCHO}$$
$$\ce{PhCHO \overset{Baeyer-Villiger}{->} PhOCHO}$$
$$\ce{PhOCHO}\overset{1.\ \ce{NaOH, \  H2O}}{\underset{2. \ \ce{H3O+}}{\ce{->}}} \ce{PhOH}$$
In practice, it would be easier to buy the phenol. In your case, you want hydroquinone.

Thanks for the very good answer - but is it also possible to remove two methanol molecules from the 1,4benzenedimethanol molecule to also yield benzene?

As nice and thoroughly described by Ben Norris, 1,4-(bishydroxymemethyl)benzene is available via reduction of the starting material with $\ce{LiAlH4}$. In order to minimize the amount of alanate used, one might consider to first convert the terephthalic acid to a corresponding ester.
If you're heading for the phenol, which doesn't make a lot of sense since hydroquinone is cheap as dirt, the following "strategy" might work:

brominate terephthalic acid to yield 2,5-dibromo terephthalic acid
heat the dibromo acid in an aqueous solution of sodium hydroxide, cupric bromide and 1,2-diaminoethane. You'd end up with 2,5-dihydroxy terephthalic acid, probably through a copper-catalyzed nucleophilic substitution
decarboxylate the 2,5-dihydroxy terephthalic acid (or its disodium salt) by heating with calcium oxide.

You'll get hydroquinone, and unless your arene was labeled, you won't notice that you haven't replaced the $\ce{COOH}$ with $\ce{OH}$ but actually changed the positions of the substituents.
But all of this is nothing but a mind game ;)


text:
What is an equivalent?
I want to know what is an equivalent and how is it different from equivalent mass. How to calculate and why is it used in chemistry for knowing concentration?

In a simple way, you can consider the literal meaning of equivalent. A molecules of $\ce{H2SO4}$ is equivalent to two molecules of $\ce{HCl}$, in the sense of how much hydrogen ions released by them. Not much to get confused about, it's rather easy to understand.

In reaction stoichiometry, it is the amount of one substance that reacts with one mole of another substance.
Equivalent mass is molar mass divided by the n factor which isn't always dependent on the other reactant. As a rule of thumb "One equivalence of one reactant reacts completely with one equivalence of another."
These concepts of equivalence are extremely important in acid-base titration to determine the neutralisation point.
$xM_1V_1 = yM_2V_2$
here $M_i$ is molarity.
For a relation in terms of equivalence use normality.
NOTE: Normality is a measure of concentration that is equal to the gram equivalent weight per liter of solution. Gram equivalent weight is a measure of the reactive capacity of a molecule.
As an example a solution containing $\pu{1M}$ $\ce{H2SO4}$ is actually $\pu{2N}$ as each molecule dissociates to give 2 $\ce{H+}$ ions.
I hope that clears it.

Equivalent concept is an archaic unit of measurement that was used in chemistry and the biological sciences in the era before researchers knew how to determine the chemical formula for a compound.
An equivalent (symbol: officially equiv; unofficially but often Eq) is the amount of a substance that reacts with (or is equivalent to) an arbitrary amount of another substance in a given chemical reaction. The mass of an equivalent is called its equivalent weight.
In a more formal definition, the equivalent is the amount of a substance needed to do one of the following:

react with or supply one mole of hydrogen ions ($\ce{H+}$) in an acid-base reaction.
react with or supply one mole of electrons in a redox reaction.

I know that it is a bit difficult to understand by the means of only definitions, so let me give you some intuition about it.
The basic idea of equivalent concept is that,

$1$ eqiv of a substance reacts with exactly $1$ equiv of another substances to produce $1$ equiv of each product.

Let us start with the example of Acid-Base reaction, so for acids we define one equivalent as the amount of the acid (in moles) that supplies one mole of $\ce{H+}$ ions and for base we define it as the amount of base (in moles) that supplies 1 mole of $\ce{OH-}$.
For Example:-

$1$ mole $\ce{HCl}$ supplies $1$ mole $\ce{H+}$ ions, so $1$ equiv of $\ce{HCl}$ is equal to $1$ mole $\ce{HCl}$.
Similarly, $1$ equiv of $\ce{H2SO4}$ = $1/2$ mole $\ce{H2SO4}$, $1$ Eq $\ce{NaOH= 1}$ mole $\ce{NaOH}$,
$1$ equiv $\ce{Al(OH)3}$ = $1/3$ mole $\ce{Al(OH)3}$.

Now if you look at the above closely, you will notice that since 1 equiv of acid supplies 1 mole $\ce{H+}$ ions and 1 equiv of base supplies 1 mole $\ce{OH-}$ ions so 1 equiv of an acid should completely react with 1 equiv of base to produce 1 equiv of each product ( eg. 1 mole $\ce{H2O}$, since its a neutralization reaction).
In a similar manner we can define equivalent for all other types of reactions, such that the main idea i.e. "1 eqiv of a substance reacts with exactly 1 equiv of another substances to produce 1 equiv of each product", remains unaltered.
Now lets define equivalent for other things,

Element
The equivalent of an element can be defined as the amount of the element that reacts with or supplies 1 mole electron in a reaction, so if you try a bit you can conclude that 1 equiv of an element is equal to 1/n mole of element where n is the valency of the element. (NOTE:- the equivalent of a substance can be different for different reactions, eg. 1 equiv of $\ce{Fe=}$ 1/2 mole $\ce{Fe}$ if it under goes to an oxidation state $\ce{Fe^2+}$ and it will be 1/3 mole if it undergoes to an oxidation state $\ce{Fe^3+}$).

Ions
Similar to element equivalent of ions can be defined as 1/n mole of ion where n is the charge on ion (try to understand it yourself using concept of electrons used earlier).

Salt
A salt have an equal number of cations and anions, so we need to define an equivalent in such a way that, 1 equiv of salt supplies 1 mole of cations (with +1 charge or 1/n moles of cation with +n charge) and anions (with -1 charge or 1/n moles of anion with -n charge) each so that these cations and anions react completely with another one mole of anions and cations (having -1 and +1 charge) that may have been supplied 1 equiv of another salt/ substance.
Now if you think a bit harder you can get to the point that one equiv of a salt is equal to 1/n mole of salt where n is the total number of charge either on cationic part or on anionic part (eg. in NaCl, total cationic/anionic charge = +1 so, 1 eq of NaCl= 1 mole NaCl, now if you pay your attention, you can notice that 1 mole NaCl will give 1 mole Na+ and 1 mole Cl- which is in accordance to what we said earlier.)
Some more examples for your help: 1 equiv of $\ce{AgCl2= 1/2 AgCl2}$ as it will supply 1/2 mole of $\ce{Ag^2+}$A and 1/2 mole of (Cl-)2, in simpler terms it will give one mole of positive charge and one mole of negative charge.
1 equiv of AlCl3= 1/3 moles of AlCl3.

Oxidizing agent/ Reducing agent
Now if you have come this far without dozing off, so you should be able to define that 1 equiv of an Oxidizing/ Reducing agent is equal to 1/n mole of that agent where n is the number of electrons gained or lost.

Now lets get to the question " Why is equivalent concept is used in chemistry for knowing concentration? ", well you might yourself have got the answer that it is because it is much simpler to apply as you do not need the Stoichiometry of the reaction going on, equivalent concept works fine without it.
And also to know the equivalent mass you simply need to find the mass of one equivalent of a substance using mole concept.
So that was all, I knew. Lets end this with a quote " Science is nothing but simple logic which is often not so simple."

@Tusher. $\ce{AgCl2}$ does not exist. You should have used $\ce{CaCl2}$ instead


text:
What is the pH of 1M Glycerol?
If the pKa of Glycerol is 14.15. How do you calculate the pH for it?
I assume that the Henderson-Hasselbalch derivative that works for weak acids and bases is not applicable here.

The answer is approximately 6.88. Therefore a $1\  M$ solution of glycerol in water will be ever so slightly acidic (considering that the hydroxyls are much weaker bases than acids, i.e. that the equilibrium constant for the reaction $\ce{C3H7O2OH + H2O <=> C3H7O2O^- + H3O^+}$ is much larger than the constant for $\ce{C3H7O2OH + H2O <=> C3H7O2OH2^+ + OH^-}$, which may or may not be true. Inclusion of the second equilibrium will drive the true pH even closer to 7)
For very weak or very dilute acids/bases, solving a dissociation problem is a bit more tricky because the self-dissociation equilibrium of water must be taken into account. It is often a valid approximation to forget about it, as water has only a small tendency to self-dissociate and so its effect on equilibrium concentrations of $\ce{H^+_{(aq)}}$ and $\ce{OH^{-}_{(aq)}}$ is often negligible, but this is not your case; glycerol in water seems to be about as weak an acid as water itself.
The best way to solve equilibrium problems is to set up a large equation including all factors by combining smaller equations. We need to make an equation that takes into account all the species in the medium ($\ce{H^+_{(aq)}}$, $\ce{OH^{-}_{(aq)}}$, $\ce{C3H7O2O^{-}_{(aq)}}$ and $\ce{C3H7O2OH_{(aq)}}$ {for convenience I shall label the latter two $\ce{A^-}$ and $\ce{HA}$}) and the related equilibrium constants ($K_a$ and $k_w$). The smaller equations are found by balancing charges (one equation) and the amount of matter of each substance (one or more equations depending on the problem).
Charge balance: $[H^+]=[OH^-]+[A^-]$ (I)
Matter balance: $[HA]+[A^-]=C_{acid}=1\ M$ (II)
The equations for the equilibrium constants are:
$K_a=\frac{[H^+][A^-]}{[HA]}=10^{-14.15}$ (III)
$k_w=[H^+][OH^-]=1\times 10^{-14}$ (IV)
To solve the problem, we need to play with the equalities until we get a single equation with a single variable. It's convenient to obtain an equation in $[H^+]$ because we are able to find the pH directly by applying $-log$ to the answer. Notice that
$[OH^-]=\frac{k_w}{[H^+]}$ (V)
$[HA]=\frac{[H^+][A^-]}{K_a}$ (VI)
Substituting (VI) in (II) yields, after some algebra:
$[A^-]=\frac{C_{acid}K_a}{[H^+]+K_a}$ (VII)
Now insert (V), (VI) and (VII) in (I) so that you get:
$[H^+]=\frac{k_w}{[H^+]}+\frac{C_{acid}K_a}{[H^+]+K_a}$
A little bit of persistence will get you to the following polynomial:
$[H^+]^3+K_a[H^+]^2-(C_{acid}K_a+k_w)[H^+]-K_ak_w=0$
Replacing the values of $K_a$, $k_w$ and $C_{acid}$, the only positive root for the equation is $[H^+]=1.30688\times 10^{-7}$, which after applying the antilogarithm results in $pH = 6.88$.

Correct, Henderson-Hasselbalch is for when you have a buffered solution of both an acid and its conjugate base.
Here you want to imagine dissolving one mole of Glycerol in one liter of water. When we say that the pKa of glycerol is 14.15 we mean that the Ka for the reaction:
$$C_3H_7O_2OH \Rightarrow C_3H_7O_2O^- + H^+$$
is $10^{-14.15}$. This is a very small number, which indicates that while glycerol dissolved in water does produce some $H^+$, it likely won't decrease the pH very much. So here you would expect a pH slightly less than 7.
To calculate this value, I'd construct what many call an "ICE chart"
Example 2 in this document seems similar to the problem you're working on.

I'm not sure whether ICE tables are capable of solving for equilibria concentrations for very dilute or very weak acids/bases, as one needs to simultaneously take into account the dissociation of the acid/base and the self-dissociation of water. It's best to learn the more general way of solving such problems, which involves the usage of charge and mass balance equations. See here and here for details.

I need more help here. I have no chemistry knowledge. How would I calculate the pH of glycerol based on its pKa? Thanks.

I know a formula relating $K_a$,dissociation constant A and concentration C for weak acids and bases as $K_a$=$A^2$C.This may help you out.

I also know a formula but applying this formula gives me odd readings.


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SN2 nucleophilic attack with alpha unsaturation
The following two questions prompted me to ask this question.

Determine the order of the following compounds according to reactivity in Williamson Ether Synthesis.
  a. $\ce{CH3CH2CH2Br}$
  b. $\ce{CH2=CHCH2Cl}$

Which of these represent the ether($\ce{CH2=CHCH2OC3H7}$) on forming peroxide in air
  a. $\ce{CH2=CHCH2OCH(OOH)CH2CH3}$
  b. $\ce{CH2=CH-CH(OOH)OC3H7}$

In both the case, I think the question is the relative ease of nucleophilic attack on carbon with $\alpha$-unsaturation. In the former, the bromine is a better leaving group. But, the presence of $\alpha$-unsaturation in the chloro compound, changes the order to b>a. In the latter, the attack of oxygen depends on the most stable electrophilic centre.
My question:-
Is $\alpha$-unsaturation favourable for nucleophilic attack? Relevant factors would be the increased electrophilicity of the carbon attached to the $sp^2$ hybridised ethylinic carbon and the repulsion of the negative nucleophile by the pi electrons.

As a note, the peroxide forming reaction in #2 is a radical reaction substituting H for OOH, so arguments involving nucleophilicity and electrophilicity are not explicitly relevant since these concepts are defined for ionic reactions.

As Ben Norris said, the aspects of nucleophilic and electrophilic attacks are not relevant in the second case.
Talking about the compound given in question 1(b), the first step of nucleophilic attack is opposed by the conditions you mentioned, i.e., of the electron withdrawal effect of the double bond and replusion of nucleophile by $\pi$ electron cloud.
But talking in terms of yield of the product, it is necessary to consider two things:

Chlorine is a good leaving group.
The stability of the carbocation intermediate formed by $SN_1$ mechanism (steric factors don't favour $SN_2$). The carbocation formed will be very stable due to conjugation with the double bond. Hence, the yield of the product is not much low.

While solving the above question, we should look at the stability of carbocation intermediate formed,which is a more dominating effect. Though Bromine is a better leaving group, and there is decrease in tendency for nucleophilic attack due to $\alpha $-unsaturation; both factors which support higher nucleophilic attack for the first compound, stabilisation of carbocation by resonance (conjugation) is highly stabilising, and this shadows the opposing factors.

Thank you for the answer.Your arguments are logically satisfying. But can you provide some reference to support the claims.

Stability of compounds/intermediates is compared using Internal energy(IE) of the compound and activation energy(AE) of the reaction.Any system tries to minimise IE,to become stable.High AE makes it difficult for a reaction to proceed,so such an intermediate isn't stable.In the 2nd compound there's a delocalised pi electron cloud,due to resonance.Resonance is known to decrease the IE of a compound significantly.Stable reaction intermediates imply low AE.Couldn't give you direct references,you can find the data regarding AE and resonance energy of compounds in detailed textbooks and Wikipedia.

Whenever we compare stability, it's all about thermodynamics.That's why detail organic chemistry textbooks mention activation energies,yield of each possible product,all energies supplied and evolved during a reaction;while explaining a reaction mechanism.


text:
Thermodynamics Problems!

A cylinder, 0.5 m long and of cross section $S$, closed at each end is fitted with a smooth thermally insulating piston, so as to contain gas in each end A and B. Initially, the piston is in equilibrium at a distance of 0.2 m from one end of the cylinder, and the gas in each compartment is at temperature 27 °C. The temperature of the smaller volume of gas A is now increased to 177 °C, while that of gas in the larger volume B is maintained at 27 °C. The piston moves through a distance of $X$ until a position is obtained.
1) Write an equation relating the initial and final pressure, volume and temperature of the gas in A.
2) Write an equation relating the initial and final pressure, volume and temperature of the gas in B.
3) Hence find the new position of equilibrium.

P.S. This is not my assignment or homework so you could feel free to explain.

This question reads like a homework or textbook problem, even if it was not assigned. Our homework policy extends to self-study problems that you have found.

The volume $V$ of a cylinder is given as $V=\pi r^2 h$ where $h$ is the height and $r$ the radius. Since the cross-section of the cylinder is $S$, the radius is given as $r=S/2$.
Now, we know that the piston is at equilibrium in the initial position, and the equilibrium in this case is $p_\text{A} = p_\text{B}$ for the pressures of the corresponding gases.
We also know the ideal gas law, which, when adding the information above (and assuming the gases A and B behave ideally) is: $$pV = nRT \Longleftrightarrow p = \frac{nRT}{V} = \frac{nRT}{\pi (S/2)^2 x} = \frac{4nRT}{\pi S^2 x}$$
For the distance $x$ of the piston to one wall.
For gas A we have the initial and final conditions:
$$ p_\text{i}^{(\text{A})} = \frac{4n^{(\text{A})} RT_\text{i}}{\pi S^2 x_\text{i}} \; \text{and} \;\; p_\text{f}^{(\text{A})} = \frac{4n^{(\text{A})} RT_\text{f}}{\pi S^2 x_\text{f}} $$
And for B (with a total height of the cylinder of $h$):
$$ p_\text{i}^{(\text{B})} = \frac{4n^{(\text{B})} RT_\text{i}}{\pi S^2 (h-x_\text{i})} \; \text{and} \;\; p_\text{f}^{(\text{B})} = \frac{4n^{(\text{B})} RT_\text{i}}{\pi S^2 (h-x_\text{f})} $$
We know that the initial and final pressures were the same (equilibrium condition):
$$ \frac{4n^{(\text{A})} RT_\text{i}}{\pi S^2 x_\text{i}} = \frac{4n^{(\text{B})} RT_\text{i}}{\pi S^2 (h-x_\text{i})} \Longleftrightarrow \frac{n^{\text{(A)}}}{x_\text{i}} = \frac{n^{\text{(B)}}}{h - x_\text{i}}$$
$$ \frac{4n^{(\text{A})} RT_\text{f}}{\pi S^2 x_\text{f}} = \frac{4n^{(\text{B})} RT_\text{i}}{\pi S^2 (h-x_\text{f})} \Longleftrightarrow \frac{n^{\text{(A)}}T_\text{f}}{x_\text{f}} = \frac{n^{\text{(B)}}T_\text{i}}{h-x_\text{f}}$$
We can now solve both equations for $\frac{n^\text{(A)}}{n^\text{(B)}}$ and equate them:
$$ \frac{n^\text{(A)}}{n^\text{(B)}} = \frac{x_\text{i}}{h-x_\text{i}} \; \text{and} \;\; \frac{n^\text{(A)}}{n^\text{(B)}} = \frac{x_\text{f}T_\text{i}}{T_\text{f}(h-x_\text{f})} $$
$$ \frac{x_\text{i}}{h-x_\text{i}} = \frac{x_\text{f}T_\text{i}}{T_\text{f}(h-x_\text{f})}$$
Solving for $x_\text{f}$ gives us the final equation:
$$ x_\text{f} = \frac{\frac{hx_\text{i}T_\text{f}}{T_\text{i}(h-x_\text{i})}}{1 + \frac{x_\text{i}T_\text{f}}{T_\text{i}(h-x_\text{i})}}$$
For the values provided by you I get $x_\text{f} = 0.25~\mathrm{m}$, which is exactly in the middle.


text:
How to synthesize t-Boc protected derivative of PVA
I've been toying with different potential recipes for DIY photoresist. My current favored design is a chemically amplified resist based on PVA (poly(vinyl alcohol)). For those who are unaware, a CAR works like this: a polymer is functionalized with a protecting group, and is mixed with a photoacid generator (PAG), which as the name suggests, generates acid upon exposure to light. The acid then cleaves the protecting group (in this case, t-Boc), rendering the polymer soluble in the developer. Basically, my question is how to functionalize PVA with the t-Boc group.
References are scant, but I did find one, which is unfortunately extremely vague. It says to react fully hydrolyzed PVA with di-tert-butyl dicarbonate (boc anhydride) and sodium hydride in N-methylpyrrolidone. However, the paper itself says that this reaction is not very predictable, and the authors are (were?) looking for better options.
Not being satisfied, I decided to lookup the reaction used for functionalizing poly(hydroxystyrene), which is the polymer used in the original CAR, APEX. Unfortunately, I wasn't really able to find anything on this either, save for even more vague references.
Anyhow, I have looked up information on how amino acids are protected with t-Boc, but the problem here is that I'm not sure if that would even work on PVA.
Does anyone have any idea how I would I go about doing this? Even more preferable would be a method not involving dangerous/explosive and for that matter, expensive substances such as Sodium hydride, or dangerous solvents (although NMP is not really that bad).
I did find one option  that simply involves reacting amino acids directly with molten boc anhydride. Is it likely this procedure would work with PVA?

Interesting, and i like the part with the photoacid.
I can't read the full details of the cited article (paywall), but since the authors aren't very pleased with the functionalization of the PVA, how about a change in the order of events:

prepare or buy the monomer (vinyl tBOC)
polymerize it according to this article (or a better source)

Hi! Well, I had actually read the article you just linked. The problem is that polymerizing vinyl tBOC involves some rather dangerous chemicals, and tricky things like nitrogen atmosphere. Although, I will look into it more. Thanks for the suggestion.

My bad, I forgot that you mentioned DIY! In the lab, with a working fume hood and proper waste management (!) I would mind to give all this a try. However, I would NOT do it at home in my appartment.

Yes, hah, not exactly something one would want to do at home. However, slightly adjunct to this question, but I did find an interesting patent which details a non chemically amplified photoresist using PVA, that doesn't use hexavalent chromium (you can use dichromate salts to make a very simple one).


text:
Alternatives to Ring Stands
I'm trying to find an alternative to simple ring stands to hold a simple electrochemical cell.
The issue is that it is difficult to adjust the height to the right position to within even a couple inches. If you try to screw the stand in it can drop or lean, and I need to be able to adjust it to within at least an inch without it falling down.
I use a micrometer to adjust an electrode coming in from the top, and I don't have a lot of play to push down.
It is also not easy to adjust the stand from under a few hood, due to the nature of the screw.
It would also be nice if the clamp could be adjusted along the y-axis to move it closer or farther from the stand for different situations.

Let me guess: You used the "breaking-bad" tag because it's bad when the precious glassware breaks? :D

Surprisingly we've never broken glassware using this method. It's extremely bad when you go through the trouble to set everything up and then you mess up the junction by adjusting the stand just a little bit too high, before even starting the experiment. :)

What about puting the whole stand on the lab-lift, hopefully there will be also some precision version.

This seems like a really good idea! Are there ones made with about 6in, 15cm or so in the z-axis? I'm finding ones with like 3mm z-axis movement. It doesn't have to be too precise as the micrometer on the top has a few cm of play.

After looking through some catalogs, I found some and I think this will be the simplest method. Thanks!

I'm glad I could offer you some of the organic chemistry tricks ;) As for the y-axis movement, use those:
http://www.sigmaaldrich.com/labware/labware-products.html?TablePage=102186322 and
http://www.sigmaaldrich.com/catalog/product/aldrich/z243620

On the more improvised/DIY side, you may be able to get away with something like a gooseneck clamp of the sort commonly used for microphones; something like this:

(from here)
Obviously they're not really designed to be exposed to hazardous chemicals, though, and you'll want to make sure that it can handle the weight you need to put on it without bending further.

This seems pretty interesting, I'll go ahead and order one and see how it works out. I'll probably try having the stand pre-adjusted then just clip/unclip it from a mount to get it ready. I'm hoping others will have more answers as well, Thanks!

Does your university have a workshop with some precision mechanics, who can handle a milling machine and a lathe to process metals and teflon?
Talk to the guys and explain them what you need. If they know their job (and I'm certain they do) they'll figure out all technical details and build it.

Our machine shop doesn't really do that much, and is understaffed, so that would probably take a year and we still wouldn't get it. I was hoping there was just something available to buy off the shelf maybe with some grooves in the stand instead of just being flat

Unfortunately, I don't know about lab suppliers in the US. For CV and SWV only used electrochemical cells from Metrohm (as far as I remember) where the lid was fixed to a stand and the glass cell body scewed to it. The position of the working electrode was adjusted via a teflon screw (with a drill-hole along the longer axis).

Yeah that sounds like a much better setup, and not too hard to machine together, it's a shame our university lacks the ability to do something like that. I'll definitely keep that in mind when thinking about a better way to proceed.


text:
How to find how much there is of a compound given the following information?
The detailed version of the question:
If we burn a strip of magnesium in air, there will be two main products, magnesium oxide and magnesium
nitride.  If we then treat that mixture with water, the nitride ion is converted to ammonia.  After boiling away
the water and ammonia, we would be left with only magnesium oxide.  Knowing that air is approximately
78% nitrogen and 20% oxygen, we might expect that the mixture that forms in the burning process would be
a similar distribution of magnesium nitride and magnesium oxide.  In fact, the ratio is closer to 75% of the
nitride to 25% of the oxide.  Assuming that the reactions are 100% efficient (i.e., no loss of magnesium),
how much magnesium oxide was formed initially, if the final product contains 3.22 grams of magnesium
oxide?
After thinking about this for a while and working it out, it seems like the other information provided was irrelevant to finding the amount of MgO. I believe since the reactions are 100% efficient, that the final amount of 3.22 grams of MgO would be exactly the same for the initial amount of MgO, which was left over at the part when ammonia and water was evaporated. However, I am not sure; I would like to know, do the given percentages of nitrogen and oxygen given affect the unknown amount of MgO?

This seems like it might be a homework question or have come from a textbook or similar source. Check out our [homework policy] so you can improve this question. Primarily, what work have you done and where are you stuck?

Maybe I'm just too tired right now, but didn't you mention that the absolute amount of magnesium oxide doesn't change during aqueous workup?
Isn't thus the amount of 3.22 g $\ce{MgO}$ exactly what was formed initially?
Wouldn't it be more reasonable to let the students calculate how much magnesium metal was burned initially, taking the $\ce{N2}$/$\ce{O2}$ ratio in air and the stoichiometry of the magnesium nitride into account? ;)

Hm, it just might be that my professor was adding extra information for distraction. I had thought the same with the final amount of 3.22 g of MgO being the same as the initial, but I was not sure with all of the other information provided.

The magnesium nitride hydrolyzes to Mg(OH)2 then dehydrates to MgO.  If formation is 3:1, then 4x = 3.22 g MgO.  Solve for x.  All the fun is in the footnotes.


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What can I use as a sacrificial electrode to stop the oxidation of aluminium?
For a solution of chromimum phosphate when putting in to solution aluminum there is removal of the aluminum.
It is possible to prevent a galvanic cell, stop the etching of the aluminum, by putting in a sacrificial anode.
In this situation I have access to metals like tantalum, titanium, tungsten and copper.
Which one of these would stop the oxidation of the aluminum.
I have slightly familiar with electronegativity and I understand that aluminum is 1.6.  I just do not know if you want some thing with a higher or lower electronegativity.
Thanks

I think that you are interested in the corrosion protection of aluminum. You have basically two choices depending on your experiment setup.
You can use an impressed current protection or a sacrificial anode method. The sacrificial anode method uses a more reactive metal (less noble) in contact with the metal requiring protecting. The less noble metal reacts first within corrosive environment.
The impressed current protection uses electric current to charge the metal requiring protecting with excess electrons. It changes the electrode potential in the negative direction until the metal is not reacting.
If you want to use the sacrificial anode method look in the table where you are looking for metal that is positioned higher than your aluminium (is more electronegative - so non of the mentioned metals).
More useful information can be found here CORROSION PROTECTION OF METALS
 or here metals.about.

Based upon the table you provided it seems that Magnesium would be a candidate, I am not an expert at this so I was wondering if you agree basically confirming my interpretation of the table.

Yes, your interpretation of the table is correct. But be aware that magnesium is quite reactive metal. So at the begging of your experiment, try it first with small quantity of sample.


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What is the pKa Range for weak acids and bases?
Am I correct in assuming that the following is true?

$\mathrm pK_\mathrm a < 3$ is for a strong acid
$3 < \mathrm pK_\mathrm a < 7$ is for a weak acid
$7 < \mathrm pK_\mathrm a < 11$ is for a weak base
$\mathrm pK_\mathrm a > 11$ is for a strong base

$\mathrm pK_\mathrm a$ is negative log of the acid dissociation constant ($K_\mathrm a$).
$$\mathrm pK_\mathrm a = -\log K_\mathrm a$$
Acid dissociation constant is the equilibrium constant of the dissociation of ions of an acid in an aqueous solution. Consider a weak acid $\ce{HA}$. Weak acids do not dissociate completely in aqueous solution. The equilibrium for the dissociation of such acids can be expressed as
$$ \ce{HA + H2O <=> H3O+ + A-}$$
The equilibrium constant for this reaction(dissociation) will be
$$K_\mathrm a = \frac{\ce{[H3O+][A^{-}]}}{\ce{[HA]}}$$
It can be seen that the greater is the extent of dissociation, the greater will be the value of $K_\mathrm a$. Or, the stronger is the acid, the greater will be its $K_\mathrm a$.
Since, $\mathrm pK_\mathrm a$ is negative log of $K_\mathrm a$, it's values will be greater for weaker acid.
There is no sharp boundary between weak and strong acid. Wikipedia defines strong acids as acids which ionize completely in aqueous solution. So, an acid, say $\ce{HA}$ , is said to be strong if one mole of this acid dissociates in aqueous solution to give one mole of $\ce{H+}$ and one mole of $\ce{A-}$. Now $K_\mathrm a$ of such an acid will be 1/0.
Usually, $\mathrm{pH}$ is used to measure the acid strength, which is negative log of $\ce{H+}$ ion concentration. $$\mathrm{pH} = -\log\ce{[H+]}$$
The concentration of $\ce{H+}$ ions in water is $10^{-7}$ (It has been found out experimentally). Hence, its $\mathrm{pH}$ is $-\log(10^{-7}) = 7$.
For acids dissolved in pure $\ce{H2O}$, the concentration of $\ce{H+}$ ions, $\ce{[H+]} > 10^{-7}$. So, $\mathrm{pH} < 7$ for acids and $\mathrm{pH} > 7$ for bases.
What you are referring to may actually be $\mathrm{pH}$ and in that case the ranges you've given are maybe correct. But once again there is no clear difference between weak and strong.
There is a simple relation between $\mathrm{pH}$ and $\mathrm pK_\mathrm a$,
$$\mathrm{pH} = \frac{1}{2}[\mathrm pK_\mathrm a - \log c]$$
where $c$ is concentration of the acid. You can derive this relation using Ostwald's dilution law.
Similarly, we have $K_\mathrm b$, $\mathrm pK_\mathrm b$ and $\mathrm{pOH}$ for bases which are basic analogues of $K_\mathrm a$, $\mathrm pK_\mathrm a$ and $\mathrm{pH}$. These variables show the strength of a base.
Conclusion

The greater is the value of $K_\mathrm a$, the stronger will be the acid and the weaker its conjugate base.

The greater is the value of $\mathrm pK_\mathrm a$, the weaker will be the acid and the stronger its conjugate base.

The greater is the value of $\mathrm{pH}^*$, the weaker will be the acid and the stronger will be the base. For acids, $\mathrm{pH} < 7$ and for bases, $\mathrm{pH} > 7$.

The greater is the value of $K_\mathrm b$, the stronger will be the base and the weaker its conjugate acid.

The greater is the value of $\mathrm pK_\mathrm b$, the weaker will be the base and the stronger its conjugate acid.

The greater is the value of $\mathrm{pOH}^*$, the weaker will be the base and the stronger will be the acid. For bases, $\mathrm{pOH} < 7$ and for acids, $\mathrm{pOH} > 7$.

$^*$ of a solution made by adding a defined amount of acid or base to pure water.

It should be noted that the final expression for pH is valid only when the dissociation constant of the acid is small and the autodissociation of water can be neglected.

Not quite, what you are describing is the pH scale, although the two terms are related, according to the Columbia University document Acidity, Basicity and $pK_a$, $pK_a$ is

It turns that that the $pK_a$ of an acid is the $pH$ at which it is exactly half dissociated

The full derivation is on the document (and a bit long to post here with explanations).
But another explanation is provided in the document $pH$ and $pK_a$, where they state

$pK_a$ tells you if a given molecule is going to either
give a proton to water at a certain $pH$, or remove a
proton

So, to answer your question, in terms of $pK_a$, strong and weak acids and bases can be defined by the following table of examples:

Image source: Strength of Acids and Bases, a key point is that there is no clear boundary that defines strong from weak acids or bases.
They conclude with the following very generalised rule for $pK_a$:

For acids: the stronger the acid, the smaller the $pK_a$
For bases: the stronger the base, the larger the $pK_a$

It's interesting that the table shown here propagates the view that the $pK_a$ of water is 15.7. It should be understood that this is wrong, that value (for thermodynamic consistency with the other tabulated values) should be 14.0.

What is the pKa Range for weak acids and bases?

Bases don't have a $\mathrm{p}K_\mathrm{a}$. To discuss the strength of a base, you can look at the $\mathrm{p}K_\mathrm{a}$ of the conjugate acid, though.

Am I correct in assuming that the following is true?

$\mathrm pK_\mathrm a < 3$ is for a strong acid
$3 < \mathrm pK_\mathrm a < 7$ is for a weak acid

To be called a strong acid in an aqueous system, the acid has to dissociate completely, or at least almost completely. An acid with a $\mathrm{p}K_\mathrm{a}$ of, say, 2 added to neutral water at a concentration of, say, 1 mol/L would not dissociate much. In fact, approximately 90% of it would be in the acid form (protonated) and only approximately 10% would be present as the conjugate base (deprotonated).
It is not clear where the separation between strong and weak acids should be. Usually, it is sufficient to memorize that $\ce{HCl, H2SO4 and HNO3}$ are common strong acids, and their $\mathrm{p}K_\mathrm{a}$ values are usually given as lower than zero. If an acid were to dissociates completely, the equilibrium constant would not be defined, and the $\mathrm{p}K_\mathrm{a}$ value would not be defined.

Am I correct in assuming that the following is true?

$7 < \mathrm pK_\mathrm a < 11$ is for a weak base
$\mathrm pK_\mathrm a > 11$ is for a strong base

No, because bases don't have a $\mathrm{p}K_\mathrm{a}$. It is true that as the $\mathrm{p}K_\mathrm{a}$ of the conjugate acid increases, the conjugate base is stronger. There is nothing special about an acid with a $\mathrm{p}K_\mathrm{a}$ of 7. Whether the $\mathrm{p}K_\mathrm{a}$ of an acid is 6.9, 7.0 or 7.1, it is still an acid.
So what can I learn from the $\mathrm{p}K_\mathrm{a}$ of an acid?
One use of the $\mathrm{p}K_\mathrm{a}$ is to define the pH-range a buffer composed of that acid and its conjugate base would have. The simplest buffer (one part acid and one part conjugate base) will have a pH approximately equal to the $\mathrm{p}K_\mathrm{a}$ value of the acid. Similarly, a pH indicator will switch colors when the pH is near the $\mathrm{p}K_\mathrm{a}$ of the conjugate acid form (the protonated form) of the indicator. The underlying reason is the same: when $\mathrm{p}K_\mathrm{a}$ = pH, the concentrations of protonated and deprotonated species (conjugate acid and base) will be almost equal (exactly equal for ideal solutions).

You write "Bases don't have a pKa
. To discuss the strength of a base, you can look at the pKa
of the conjugate acid, though." <-- What about the pKb of the base?