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Why the radius of noble gases is more than the halogens or the previous groups?
Noble gases have larger radii than that of halogens.
Sometimes it is greater than the radius of group I elements.
Why is it like that? When we talk about radii of noble gases, what type of radius is referred? Is it van der Waals or atomic?

From Ref. 1:

Unlike a ball, an atom doesn't have a fixed radius. The radius of atom
can only be found by measuring the distance between the nuclei of two
touching atoms and then halving that distance.

The answer lies in the way in which size measurements can be done:

i. Metallic/Covalent radius (left)
ii. Van der Waal radius (right)
Noble gases don't form (or don't easily form) bonds. Therefore their van der Waals radius is measured.
The rationale here is based on the fact that noble gases have very low chemical reactivity and their measured atomic radii are non-bonded. An explanation lies in the electronic configuration of noble gases. They have completely filled outer orbitals with increased repulsion.
In addition, following Hund's rule, in a set of degenerate orbitals, spin-paired electrons exhibit more repulsion compared to singly occupying orbitals which have parallel spins (and consenquently less repulsion).
This also implies that increased electron repulsion between the completely-filled orbitals (as seen in noble gases which all have spin paired electronic configuration) has an effect of increasing "van der Waals radius" as electrons tend to move as further away as possible to minimize repulsion.
However there is some disparity when comparing noble gases radii with group 1 elements (metallic radii) or halogens (covalent radii) because you are comparing two different unrelated properties, nevertheless it makes sense to compare radii between different noble gases.
References

Atomic and Ionic radius

Conceptual Chemistry

Your last statement is righr that we can't compare 2 differently observed things....but just suppose IF we have a way to compare their ORIGINOL radius(not by bonding but the distance btw the outer electron and nucleus of an isolated atom) then also will the noble gas atom radius is more than previous halogen ? (Ignoring bonding style )  how much extend the outer shell of noble gas can expand due to inter electronic repulsion?

Although its bit misleading to compare the van der waal radius of the noble gases which is mostly calculated for them but still ;
Since the attraction between the noble gas molecules is the weakest i.e the van der waal attractions so this causes only a touch or gap between two molecules of gap and as you may know , this distance divided by 2 shall give you radius.
In group 1 ionic radius are measured which have strong bond but still the largest in the period ( except the noble gas) cause here we see " the nuclear attraction or the $Z_{effective} $ " increases on moving left to right.
And hence atomic size decreases.
A bit more clarity about types of radius measurements would help : http://www.chemguide.co.uk/atoms/properties/atradius.html


text:
Relative stabilities of tropylium and cyclopropenyl carbocations
How can the stabilities of the tropylium and cyclopropenyl carbocations be compared?

Both are aromatic according to Hückel's rule. I think that to compare the stabilities, we must either count the number of resonance structures, or perhaps we could say that cyclopropenyl is less stable due to angle strain. Am I on the right track?

Both ions are indeed strongly stabilized by aromatic coupling, and salts containing both ions are known.  But, of course, aromaticity is not the only thing that determines stability.  Tropylium ion, with its lesser ring strain and more atoms over which to distribute the positive charge, forms ionic salts with more anions (e.g., ionic $\ce{C_7H_7Br}$ versus covalent $\ce{C_3H_3Br}$) and in that sense is more stable.
The ionic structure of some tropylium salts is documented in Wikipedia[1][2][3][4]

In 1891 G. Merling obtained a water-soluble bromine containing compound from a reaction of cycloheptatriene and bromine.[1] Unlike most hydrocarbyl bromides, this compound, later named tropylium bromide, is a water-soluble solid and is insoluble in hydrocarbons, chloroform, and ether. It is purified by crystallization from hot ethanol. Reaction with aqueous silver nitrate immediately gave a precipitate of silver bromide. The structure of tropylium bromide was deduced to be a salt, C7H7+Br–, by Doering and Knox in 1954 by analysis of its infrared and ultraviolet spectra.[2][3] The ionic structures of tropylium perchlorate (C7H7+ClO4–) and tropylium iodide (C7H7+I–) in the solid state have been confirmed by X-ray crystallography.[4] The bond length of the carbon-carbon bonds were found to be longer (147 pm) than those of benzene (140 pm) but still shorter than those of a typical single-bonded species like ethane (154 pm).

For examples of cyclopropenyl cation salts (e.g. $\ce{C3H3SbCl6}$) see Breslow and Groves[5].
References:

Merling, G. Ueber Tropin. Ber. Dtsch. Chem. Ges. 1891, 24 (2), 3108–3126. DOI: 10.1002/cber.189102402151.

von E.doering, W.; Knox, L. H. The Cycloheptatrienylium (Tropylium) Ion. J. Am. Chem. Soc. 1954, 76 (12), 3203–3206. DOI: 10.1021/ja01641a027.

Balaban, A. T.; Oniciu, D. C.; Katritzky, A. R. Aromaticity as a Cornerstone of Heterocyclic Chemistry. Chem. Rev. 2004, 104 (5), 2777–2812. DOI: 10.1021/cr0306790.

Kitaigorodskii, A. I.; Struchkov, Y. T.; Khotsyanova, T. L.; Vol'pin, M. E.; Kursanov, D. N. Crystal structures of tropylium perchlorate and iodide. Russ Chem Bull 1960, 9 (1), 32–36. DOI: 10.1007/BF01178699.

Ronald Breslow and John T. Groves (1970). "Cyclopropenyl cation. Synthesis and characterization". J. Am. Chem. Soc. 92 (4), 984–987.
https://doi.org/10.1021/ja00707a040

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Is it reasonable to use natural orbitals of an unrestricted HF calculation as a basis to start a CASSCF calculation for a radical
Let's say we want to run a CASSCF calculation of a radical, for example the cation of a neutral closed-shell molecule (therefore, we calculate an open-shell radical cation). MOLCAS, as well as probably every other program package out there, needs starting orbitals when performing a CASSCF calculation. For a neutral species, these are normally assumed to be the regular old Hartree-Fock orbitals. Usually, when dealing with a cation, it seems that the Hartree-Fock-SCF calculation is performed for the neutral species or the di-cationic species to yield the necessary HF starting-orbitals. They are in these cases the result of closed-shell calculations. MOLCAS also permits the use of a UHF calculation as the basis for a subsequent CASSCF calculation. Then, the natural orbitals of the unrestricted calculation are used.
Is the usage of the natural orbitals of an unrestricted HF calculation reasonable as a basis for a subsequent CASSCF calculation? If not, why? When is it a good idea / when is it a bad idea? What are the differences in using natural UHF orbitals in contrast to the SCF orbitals of the neutral or di-cationic species? What are the pitfalls?
Edit: I am aware (or at least it is my understanding) that conceptually it is no problem to perform the CASSCF using the natural UHF orbitals, as it could for example also be performed using only the alpha orbitals of the UHF calculation. My question is therefore aimed at finding out in which cases it is generally a good or bad idea to use the natural UHF orbitals.

Isn't this simply called UNO-CASSCF?

The best starting orbitals for a CASSCF calculation are optimised orbitals of another CASSCF (or RASSCF) calculation. That sounds a bit ridiculous, but this is probably the best way to figure out the active space.
At first you are probably choosing something quite small, like CAS(2,2) to CAS(4,4). For these calculations it barely matters what kind of starting orbitals you are choosing. The change will be negligible and probably depending heavily on the system you are using.
The advantage of calculating the cations by dismissing the radicals is often a good choice since you can run a very fast RHF calculation and the orbitals are symmetric, which is what you need for CAS.
Natural orbitals of an UHF calculation are also a quite sensible choice, because they already allow for fractional occupation and should also be symmetric. The only downside I can think of is, that this kind of start orbital generation takes a lot longer.
Similarly you can choose to run a ROHF calculation, if you are uncomfortable with too much cationic charge.
As you have said you can choose also an UHF calculation by throwing away one set. I would advise caution here. Sometimes - probably especially when CAS is necessary - spin contamination in UHF is significant. Therefore one set might not give you anything close and you could end up choosing the wrong CAS.
I am not aware of any major pitfalls except for the just mentioned choosing of a completely wrong active space.
In any case it is a good idea to give the CASSCF calculation a few cycles and see how the active space is developing. Sometimes you need to adjust the active space long before the calculation converges and then you will have already a much better guess than with your initial starting orbitals.
I personally use the following routine

R(O)HF [x+] -> CAS(x, ⌈½x⌉ +1)
            -> RAS(x+2n, ⌈½x⌉ +n +1)
            -> CAS(x+2n, ⌈½x⌉ +n +1)
            -> RAS(...)
            -> CAS(...) -> ...

to narrow down the active space and include all orbitals I want and need. In most cases I am not converging the intermediate calculations, except for the minimal CAS.
As you can see the procedure that follows after your start orbitals is much™ more time consuming, so it really does not matter.

No time to read it, so I'll just leave that here for later: Peter Pulay and Tracy P. Hamilton. UHF natural orbitals for defining and starting MC‐SCF calculations. J. Chem. Phys. 1988, 88, 4926.

Hartree-Fock Orbitals are Always the Wrong Choice for CAS Calculations
In Hartree-Fock (HF) theory, only the occupied orbitals contribute to the electronic energy. Due to the presence of Fock exchange, the Fock operator is different for occupied and virtual orbitals. For occupied orbitals, the Fock exchange exactly cancels the Coulomb interaction of the electron with itself. Hence, electrons in occupied orbitals experience the mean-field potential of the N-1 remaining electrons. Since virtual orbitals do not contribute to the potential, there exists no exchange term to cancel part of the Coulomb potential for these orbitals. Therefore, virtual orbitals experience the full potential due to the N electrons in the occupied orbital space. Compared to the occupied orbitals, these orbitals are relatively diffuse and high in energy. They rather correspond to anionic states and do not represent a good estimate to describe charge-neutral excited states. Consequently, HF virtual orbitals are not a reasonable starting point for CAS calculations, since their character can change dramatically once they become partially occupied.
Though it may seem counterintuitive in an ab-initio framework, Kohn-Sham (KS) DFT provides a much better starting point. Due to the lack of orbital-dependent Fock exchange in (semi-)local DFT, the occupied and virtual KS orbitals are generated in the same mean-field Coulomb and exchange potential. Thus, occupied and virtual orbitals are described on equal footing, i.e., both experience a mean-field potential from approximately N-1 electrons. Therefore, virtual KS orbitals represent a much better guess for charge-neutral excited states (cf. the reasonable performance of TD-DFT compared to CIS for vertical excitation energies). A comprehensive discussion of the band gap in Hartree-Fock theory, as well as in "exact" and approximate KS-DFT, is given in the following paper:
Baerends & Gritsenko & van Meer: The Kohn-Sham gap, the fundamental gap and the optical gap: the physical meaning of occupied and virtual Kohn-Sham orbital energies
There are alternative options, if one wants to avoid (semi-)local KS-DFT orbitals, which are inevitably affected by the so-called self-interaction error. For example, fractional occupation number (FON)-HF orbitals from an electronic temperature treatment serve as a good starting point, since the virtual orbitals become partially occupied. Such a treatment comes at no extra computational cost and is used, e.g., in FOMO-CASCI. Using natural transition orbitals (NTOs) form a CIS calculation is more physically motivated. While it is somewhat more elaborate, the additional computational cost is negligible compared to the subsequent CAS treatment. The higher quality for the initial set of orbitals will more than pay back through a faster and more robust convergence of the CAS procedure.

I can't tell you when a UHF reference wave-function would be a good idea (maybe if you don't have ROHF), but I think it might be a BAD idea if UHF generates a significantly spin-contaminated wave-function (UHF wave-functions are not eigenfunctions of S^2).  Orbitals in your CASSCF active space will obviously be optimized, but the non-active occupied and external spaces will forever stay, in your case, UHF natural orbitals.
Two configuration MCSCF (TCSCF) is a tiny bit more expensive than UHF but results in S^2 eigenfunctions.  Also, I have had luck with MP2 natural orbitals (RHF/ROHF reference).

Are you entirely sure that using natural orbitals from a spin-contaminated UHF calculation will yield false results? Theoretically, all needed for CI calculations is a one-determinant-basis. This is then used for building CSFs (thereby assuring that they are eigenfunctions of S^2, no matter if the UHF-WF was, right?). Of course I agree with you that working with a severely spin-contaminated WF is probably a very bad idea. I am curious about the role of the CSFs when pluggin in spin-contaminated UHF natural orbitals.

Even if you have spin contamination in your system, natural orbitals are symmetric and provide a reasonable guess. "Orbitals in your CASSCF active space will obviously be optimized, but the non-active occupied and external spaces will forever stay, in your case, UHF natural orbitals." This is just wrong. You still need reference orbitals for TCSCF, how would you compute these? For systems with a MR character, MP2 is unreliable, expensive, and you still need a reference WF. And what's the point in the first place, you are just doubling your effort. (cc @mrnicegyu11)

For example, UHF can result in 'artifactual symmetry-broken' states, that are artificially stabilized by nonphysical symmetry-breaking (geometric distortions, the Jahn-Teller effect).  This means one could potentially be starting from the incorrect point group (a lower point-group).  This would be a very bad guess and require much trudging through phase space to even just get back to the correct higher-order point group.

You are right, MP2 natural orbitals for a small cas is probably a waste but I am intretsed in MRCI wavefunctions.

I'd like to hear if @Martin-マーチン agrees. I am aware of symmetry breaking in UHF calculations, but is this an issue for natural orbitals?

'Is this an issue of UHF' is a very system dependent question.  UHF is fine for a lot of systems and is very close to the RHF/ROHF result.  In other cases, UHF can variationally minimize to a symmetry-broken wave-function, possibly with an equilibrium geometry that is artificially distorted relative to nature.  Notice I am saying that not only will your orbitals be symmetry-broken, but UHF geometry optimization can lead to geometry distortions (the natural orbitals of this geometries wave-function would be no good to you).  Do you not have access to ROHF code?

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Relation between ∆(tetrahedral) and ∆(octahedral), ∆t = 4/9 ∆o
I remember my teacher saying that ∆(tetrahedral) being 4/9 times the ∆(octahedral) for the same ligands and the same central metal ion due to some analogy by number of ligands, but I do not seem to remember it.


text:
Why doesn't phosphorus form as many oxides as nitrogen?
Nitrogen has oxides like $\ce{N2O}$, $\ce{NO}$, $\ce{NO2}$, $\ce{N2O3}$, $\ce{N2O5}$, but as far as I know, phosphorus has probably only $\ce{P2O3}$ and $\ce{P2O5}$.
Why does nitrogen have so many oxides in the first place?
Also why doesn't phosphorus have the analogous oxides?

See https://en.wikipedia.org/wiki/Diphosphorus_tetroxide, also https://pl.wikipedia.org/wiki/Tlenki_fosforu

It is generally due to the larger size of phosphorus. Nitrogen and oxygen both are members of 2nd period.Hence the p$\pi$ - p$\pi$ bond formed between nitrogen and oxygen has larger area of overlap resulting in shorter bond length and increased stability. Phosphorus is a member of 3rd period so when 3p orbital of phosphorus overlaps with 2p orbital of oxygen, due to difference in energy and size the area of overlap decreases which leads to lesser stability.Hence it forms less oxides.

Also the oxides for phosphorus are somewhat special and I think they just have a high tendendy to form. Organic chemistry like the Wittig-reaction uses the oxophilia of phosphorus a lot.
If you assume from basic organic chemistry that 6-membered rings are something stable and a cyclopropane isn't that stable you can try to apply that to phosphorus as well. But it is a different element and you'll come up with a basic thumb rule that says that 5-membered rings are quite good for phosphorus while the more 3-membered rings you have the more unstable it will become. $\ce{P4}$, which is the smallest (stable) unit of phosphorus however consists of triangles, only so it has to be highly reactive. the bonding orbitals just can't bent that way to get the tetrahedral angles and the bonds are sometimes called 'bananoid' bonds for this reason. If oygen is present it will 'insert' into this $\ce{P-P}$ bond and act as a bridge for it that way you loose a lot of ring strain on the P atoms. Your tetrahedron has 6 of these edges and all of them are unstable so it will autoamtically form a $\ce{P4O6}$ or $\ce{P2O3}$ as you know it. And if you check on its structure you will see that it is only a tetrahedron where each edge is capped by an oxide. For the next step you will have to descide whether P is nucleophilic or electrophilic and this depends on the other partner in the reaction. With oxygen however it is further oxidized to form $\ce{P4O10}$ or $\ce{P2O5}$.
And phosphorus is just so reactive towards oxygen that this happens for all the possible positions on the tetrahedron therefore those two oxides are the most favored ones and will likely always form.
Nitrogen on the other hand has different structural motifs starting of with an $\ce{N2}$ molecule for example and that $\ce{N \tbond N}$ bond is really strong, too. So the way how it is oxidized is different from how phosphorus behaves.
I don't know if similar oxides exist for phosphorus as we know them for nitrogen and this was only a guess based on a phosphorus lecture I heard a while ago.


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Wavenumber range reversed in IR spectroscopy
I recently stumbled upon a result table from an IR spectroscopy, that is an image full of peaks. The goal was to decide which peak belongs to which group by looking up the intervals in a related table.
While the task itself is pretty straightforward, I can't explain to myself why the axis, as well as the intervals in the table, are "reversed" - meaning the higher values are on the left and the lower are on the right. In my opinion, this makes it much harder to quickly estimate if the given peak is in an interval or not:
e.g. for a number 2053, compare those two sets of intervals:
2100 - 2000
2070 - 2010
2050 - 2030

vs
2000 - 2100
2010 - 2070
2030 - 2050

The estimation seems to be easier with the second, at least for me.
Can someone provide a background on why the ordering is high-to-low and not low-to-high?
NOTE: I don't have any real background in the topic and so my view is rather naive.

Spectroscopy is unfortunately littered with lots of different conventions. It's very common for the horizontal axis to go from high wavenumber to low wavenumber. If you do a Google image search for "IR spectrum of <your favourite chemical>" you'll see what I mean. So, "2100-2000" actually goes from left to right on the actual spectrum.

It seems to come from the time when chart recorders were used, but so many spectra were recorded this way that when everything became computerised the convention was maintained.  Its the same for nmr.

@orthocresol I see, thanks. However, the question still holds - if the spectrum would go from low wavenumber to high, it would be easier to read the intervals, and I believe that's stronger than any other reason behind this.

@porphyrin That makes sense. Sounds like convention won over clarity.

The convention of peaks pointing down in IR (when plotted as %T, which is customary for structure elucidation) also traces to the physical design of early instruments.  I'm surprised that didn't catch your attention more than the reversed x-axis.

The peak convention does not make the task of reading the intervals harder, so that's why I didn't concern myself with it.

Sorry, I can't comment yet. I thought I'd add a historical perspective, which I don't know if it is true, but it occurred to me as a good reason for why this might be the case.
UV should historically preceed IR. In UV, the energies axis has been historically reported in wavelength (nm), which is inversely proportional to energy, meaning you have high energies on the left and low energies on the right.
It might be that for consistence with UV, the wavenumber axis has been reversed so to have higher energies on the left.
This is highly speculative, though.


text:
What will be the IUPAC name of this organic compound and why?

I cannot understand how to name this compound. The answer is 5-Hydroxycyclopent-2-enone but shouldn't the $\ce{-OH}$ group be given more priority than alkene?

The most important simplified criteria for the numbering of simple cyclic compounds are:

lower locants for suffixes
lower locants for multiple bonds
lower locants for prefixes
lower locants for substituents cited first as a prefix in the name

The corresponding actual wording in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows:

P-14.4 NUMBERING
When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority:
(…)
(c) principal characteristic groups and free valences (suffixes);
(…)
(e) saturation/unsaturation:
  (i) low locants are given to hydro/dehydro prefixes (…) and ‘ene’ and ‘yne’ endings;
  (ii) low locants are given first to multiple bonds as a set and then to double bonds (…);
(f) detachable alphabetized prefixes, all considered together in a series of increasing numerical order;
(g) lowest locants for the substituent cited first as a prefix in the name;
(…)

Therefore, when assigning locants to the compound that is given in the question, the lowest locant is given first to the suffix for the principal characteristic group according to (c), i.e. ‘…-1-one’. Then, a low locant is given to the double bond according to (e), i.e. ‘…-2-en-…’. Finally, a low locant is given to the prefix for the hydroxy substituent according to (f), i.e. ‘5-hydroxy…’. Thus, the complete name is 5-hydroxycyclopent-2-en-1-one.

That astonished me now, too. I checked on the IUPAC rules and it says that only the group with the highest priority is to mention all the others are just added alphabetically. So I guess you will take the highest priority + its corresponding carbon chain. This is like the molecule you start with and you can add additional functional groups which don't influence the priority anymore. (But as I said only a guess on this one or how I interpret that line in their rules).


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Why is the inside of a cell both more acidic and more negative than its environment?
In a very simplistic way of thinking, an acidic solution has more positively charged $\ce{H+}$ ions present, whereas a basic solution has more negatively charged $\ce{OH-}$ ions present.
Can that be simply reconciled with the fact that (1) the intracellular pH of about 6.8 is more acidic than the intercellular pH of about 7.3, and that (2) a typical cellular membrane potential is -40 to -80 mV?

Are you talking about mammalian cells?

@airhuff, I can accept that as a candidate for best answer if you post as an answer and if no one else weighs in

@Bob, I was thinking about human cells, so yes, sort of.

The key is that $\ce{H+}$ and $\ce{OH−}$ are not the only ions at play here.
The selective permeability of different cellular membranes for different anions (i.e. $\ce{Cl−}$) and cations (i.e. $\ce{K+, Na+, Ca^{2+}, Mg^{2+},}$ etc.) results in charge separation. Cellular membrane potentials are thereby created, with or without an intercellular / intracellular pH gradient.


text:
Order of the subshells in an electron configuration
According to what my chemistry teacher said, the energy of a subshell is given by the sum of its principal quantum number (n) and its second quantum number (l).
e.g.:

1s = n+l = 1+0 = 1
2s = n+l = 2+0 = 2
2p = n+l = 2+1 = 3
3s = n+l = 3+0 = 3

However, when it comes to make a graphical representation of an electron configuration (which follows an order of increasing energy), she doesn't put the subshells having the same energy in the same level, like in the following picture:

Shouldn't the subshells 2p and 3s (which both have an energy equals to 3), as well as 4p and 3d, be put in the same level, since they have the same energy?

Whilst the said concept is correct, as orbitals increase in number the less they are attracted to the centre of nuclei, consequently the further they move away the higher the energy they possess.
The 2s electrons spent more time closer to the nucleus (when compared to 2p electrons) and this reduces the energy of electrons in s orbitals, the nearer they get, the lower the energy. This is why the s orbital always has a slightly lower energy than p orbital at the same energy level.
The 2s, 3s .... orbitals get progressively away from the nucleus.

"This is the s orbital always has a slightly lower energy than p orbital at the same energy level"
But the subshells 2p and 3s, although they have the same energy, aren't in the same level. 3s is put upper than 2p. Why?

Well it is not entirely true that to say that a 2p orbital has the same energy level as the 3s orbital, (although theoretically it may be proven by calculations) because the 3s already is in an extra "shell" thus making it further away from nuclei and I have explained why orbitals further away from nuclei have a higher energy level.


text:
How many valid resonance forms can be constructed for aniline?

Based on the above picture, it seems that there are four. However, I have two questions:

Aren't the top right and bottom left forms simply mirror images of each other, and thus considered the same?
If not, would an alternate version of the top left form be a fifth structure, if the double and single bonds were flipped?

Your four structures are correct and you could also include your suggested fifth structure (your first and fifth structures are comparable to the resonance structures of benzene). While some of the structures are mirror images, we generally think of particular atoms as being fixed when writing out resonance structures. You can think of the carbons as having a number (which they technically would if you named the resonance structures IUPAC style), which leads to the "mirrored" structures not being identical since the double bonds are between different carbons.

(+1) for a nice answer but I must comment that I was taught that resonance structures that are mirror images of each other (aka identical to each other) are not counted separately. That might have been a different context though. Do you know of any other context where duplicates are not counted separately? Or is what you said applicable in all contexts? Thank you!

@Guarang Tandon I had always counted mirrored structures under the assumption that we weren't rotating the atoms of the original structure. I guess you wouldn't necessarily have to draw out these resonance structures if you remembered that the corresponding mirrored structure also exists, but drawing it out for cases like benzene makes the resonance more explicit.

The way I view it is that resonance is a superposition of (localised) wavefunctions, so for every wavefunction that has a negative charge on C2, you need a corresponding wavefunction with a negative charge on C6 with the same weight. These wavefunctions are symmetry-related, just like how C2 and C6 are symmetry-related, but they aren't the same.

@Tyberius: can I assume an isotopic carbon at C-6 (like we do in benzyne mech.) to explain this to myself?

I believe there are only three resonance structures for aniline. The upper right and lower left structures are the same. Flipping the single and double bonds within the ring (your suggested fifth structure) would not make another resonance structure because the ring is a benzene ring. The pi bonds forming the double bonds are actually delocalized. They are "equally shared" throughout the ring.

"upper right and lower left are the same" That's not true: if mirror images were the same, then ozone would have no resonance structures. Or acetate ion, for that matter.

"Flipping the single and double bonds within the ring (your suggested fifth structure) would not make another resonance structure because the ring is a benzene ring." This is equally wrong, because that would mean benzene cannot be described with resonance either.

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text:
Is paper or water the stationary phase in paper chromatography?
I've been looking at about 10-ish sources, and half of them say that paper is the stationary phase, and the other half say that water is the stationary phase. I understand that water is on the paper through humidity, but is there a clear stationary phase, or is it both cellulose and water (if that's possible)? Does the stationary phase depend on the humidity of the room?

The stationary phase in chromatography is the one that doesn’t move according to the eyes of a macroscopic (i.e. human researcher) observer. (That complicated way to put it was to prevent anybody raising any relativism arguments.)
Obviously, the paper does not move through the water but the water does through the paper. You should discard the five-ish sources that state water be the stationary phase as they obviously have no clue about chromatography.

@Jan everything depends on the frame of reference :P


text:
How to determine the speed of GLUT1 transporter using Michaelis-Menten kinetics?
I am reviewing a passage in my Lehninger Biochemistry Textbook (6th Edition), and the following passage has me very confused.
How can one sentence from this passage claim that when [S]out = Kt, the rate of uptake is 1/2 Vmax, but then go on to say that when the concentration in the blood is close to Kt that GLUT1 operates near Vmax?


text:
Do chemists use molecular orbital theory to explain physical characteristics of a compound (i.e. melting point, boiling point, etc)
When you think of a molecule as simple as methane CH4, you can see that it has four identical C-H bonds. Carbon has 6 electrons, 2 in 1s, 2 in 2s and 1 in 2px, 1 in 2py and 0 in 2pz. Interestingly, one of the electrons in 2 s goes to 2pz. How does methane form 4 identical C-H bonds with just 1 s and 3 p orbitals? The answer is that carbon uses hybridized atomic orbitals called sp3 that has 25% s-character and 75% p-character.
Molecular orbital theory can explain why we have triple bonds. For example, ethyne. The sigma bond formed by the sp-sp overlap along with the two perpendicular p-orbitals that are parallel with each other that form 2 pi bonds explains the triple bond.
I've also noticed that unpaired sp3 bonds like those in ammonia (NH3) are used to form bonds and paired sp3 are the lone pairs.
I was wondering how any of this knowledge is useful for someone wanting to find out about the physical properties of compounds? Do chemists think about these things?

Certainly the electronic structure of compounds are of interest when predicting or interpreting the physical properties - but not necessarily the molecular orbital theory in and by itself. The symmetry of compounds (and by extension this also means;) the symmetry of the bonds in the compound are vital to many physical properties. However, MO theory is mainly used to describe bonding in more detail, not to infer "macro"-properties FROM bonding - so I guess the answer is... "no and yes". Yes, chemists think about these things, at least the ones that work close to the quantum realm. No, MO theory is not really useful in the way you are describing - with one giant exception:
When we are talking about "possible" physical properties of not-yet-synthesized compounds, the answer is resoundingly "yes". Graphene, for instance, has more theoretically predicted qualities than have been measured. Super conductance, tensile strengths, diffusion rates... there is just so much exiting being suggested all a result of computer modeling on the quantum level from MO theory and the likes.

Yes, things like boiling points can be calculated today, but only with crude approximations because such calculations are very time consuming. Let's say chemists have been moderately successful with this as in what has been calculated so far is completely useless for anything other than academic amusement because of the massive error bars. People are working on it though.
This however has nothing to do with hybridization. Hybridization is a model concept that, at its best, describes qualitatively the geometry of very simple molecules. You are asking for properties that are the result of molecular interactions. Anyways, the idea of atomic orbitals mixing with each other and then forming bonds with each other is fundamentally wrong and not useful for anything quantitative I can think of. However, molecular orbitals are still a useful tool in the calculations mentioned earlier. Explaining this in detail goes beyond this question I fear.


text:
electroplating with different electrode and electrolyte
I wanted to make a school project on galvanic
 cells and I would like to know of I can make electroplating with electrolyte of $\ce{NaCl}$ to plate let's say a spoon with the sodium from the salt but with out sodium anode.
my view is that electroplating works when the positively charged $\ce{Na+}$ ions are reduced to $\ce{Na}$ at the cathode so why do we need the cathode to be sodium as long as it can reduce the sodium ions?

You can, but you need to do it in molten salt. Google Down's Process. Best avoided for school projects!

Before you start with your experiments, you might want to read about the concept of a standard electrode potential and have a look at some data, particulary for hydrogen, sodium and chlorine.
Can I electrolyze an aqueous solution of sodium chloride?

 Yes you can - but you'll be surprised!
 That is, unless you you have checked and understood the data in the link given above.

 You would then realize that instead of $\ce{Na+}$, protons will be reduced at the cathode while chlorine will be formed by anodic oxidation of $\ce{Cl-}$.

In short: No sodium metal will be formed here.

Can I produce sodium metal by electrolysis of sodium chloride?

 Yes, you can - in principle!

 But you probably can't in a school project. Electrolysis requires that your medium is electrically conductive. An aqueous $\ce{NaCl}$ solution is, but that won't help you. A block of solid sodium chloride, on the other hand is a lousy conductor. The situation changes dramatically if you can manage to melt the salt. Molten sodium chloride has a high electrical conductivity. The downside is the melting point of $\ce{NaCl}$ - it is around 800 °C. You could add some calcium chloride to lower the melting point of the mixture to ~ 500 °C, but that's still a lot. Probably a bit too hot for a school project.

So, does anybody use this sh*t?

   Yes, this is the established way to produce sodium metal using a special setup known as the Downs cell. But remember, this is in industry, and not suitable for a school project ;-)

But I could use a big spoon as an electrode and electroplate it?

 No, sorry! Under the conditions of the industrial electrolysis of molten sodium chloride, sodium is formed. But the melting point of sodium metal is just ~ 98 °C. It is a liquid with a lower density than the molten sodium chloride. The liquid sodium metal is collected above the (iron) cathode.

In summary, I suggest that you rethink your school project ;-)


text:
units of mole formula
I am a high school student.I am currently studying "moles" chapter in my textbook.I think my book is misleading me in regards with units of a formula.
The formula is,
NO.of moles= (Given mass of an element)/(Molar mass of element).
so,the units used here are,(Given mass of an element)=gram,g AND (Molar mass
of element)=g/mol
I think that the denominator of the formula is incorrect.The therm "Molar mass of element" should be replaced with "Gram Atomic Mass of the element". The thing which forces me to think so is the "Derivation of this formula".
The formula is derived in this way-
The mass of an atom of carbon 12, call it mC12,
  so,mC12=12 amu/atom
mC12=12 g/mol
thus, =>  12 amu/atom = 12 g/mol
=> 1 amu/atom= 1 g/mol
=> (1 amu /atom) * 1 mol = 1 g
let,1 mol = k (constant)
thus, 1 amu is directly proportional to 1 g
now let an element has atomic mass "x" amu.
thus, an element of mass "x" amu must have its gram atomic mass as x g (keeping k= 1 mol constant).
now, (x amu/atom) * 1 mol = x g
now, keeping q= x amu/atom constant,
we have,
1 mol is directly proportional to x g.
so,x g is directly proportional to 1 mol.
Now,by using the unitary method we have,
1 g = 1mol/x
now,let the given mass of an element is y g.
thus, y* 1 g= y * 1mol / x
y g= y/x mol
here, x= gram atomic mass of the given element.
      y=given mass of the element.
thus,the formula is.
No. of moles= (Given mass of the element)*1 mol/(gram atomic mass)
the units here are, (Given mass of the element)= gram
(gram atomic mass)= gram
So,i arrived at a result in which the denominator is "gram atomic mass"(g) not "molar mass"(g/mol).
Do you think my book is wrong?

You're vastly trying to overthink this. Because of changing definition a number of different terms have been used for the molar mass of an element. The book is simply try to save that if you weight out 3 grams of carbon then you have (3g)/(12 g per mole) = 0.25 moles of carbon.

Gram atomic mass is the molar mass multiplied by one mole. All you've done is replace $$\frac{\ce{g}}{\ce{mol}}$$ in the formula by $$\ce{mol}\cdot\frac{\ce{g}}{\ce{mol}}\cdot\frac{1}{\ce{mol}}$$
which can be done far quicker without the algebra contortion.


text:
How to predict which gas shows maximum ideal behavior at STP ? (given its critical temperature)
Suppose we know the Van Der Waals coefficients $a$ and $b$ for a set of gases in the real gas equation, then which is the correct method to predict which gas shows most ideal behavior at STP ?
Our professor said that we must check the critical temperature using the formula $$T_c=\frac{8a}{27Rb}$$ Then he said: "The gas for which $$|T_c-273|$$ is minimum will show the most ideal behaviour".
I sort of don't agree with this logic. Let us have a look at Andrew's curves:

We notice that greater is the temperature compared to critical temperature more is the ideal behaviour. So the correct method should be checking $(273-T_c)$. The gas which has maximum value of $(273-T_c)$ should show maximum ideal behaviour at $273 K$. Is my logic correct or not ? Why?

Its not clear why since a $T_c$ of $173$ and $373$ gives the same result. Presumably he means the one with the lowest $T_c$ in degrees Kelvin.

@porphyrin Yeah that is one problem. But the bigger problem is he said that difference between Tc and 273 should be minimum for ideal behavior. Shouldn't it be maximum for ideal behavior ?

Its just a conversion. Helium seem to have the lowest critical temp, $-268$ C or$ 5$ K, (hydrogen $33$ K, nitrogen $126$ K). Values seem to be mainly listed in centigrade not Kelvin, hence the $273$ is to convert. Can't see what $273$ has to do with it otherwise. Lowest temp (in K) should exhibit most ideal behaviour.

"Lowest temp (in K) should exhibit most ideal behaviour." <--- You mean the gas having lowest critical temperature shows most ideal behaviour at 273 K ? @porphyrin

I think you should read the question. "The most ideal behaviour at STP". If a gas has a high critical temperature, by the time it reaches STP, it would have deviated way too much from ideal behaviour.

If he did not use the absolute value Syntax, he would be correct.


text:
How do we explain shapes of molecules using molecular orbital theory or do we even do that?
How do we explain shapes of molecules using molecular orbital theory  or do we  even do that ?
While reading about the theories to explain chemical combination and stabilities of compounds I read about VBT and MOT and how when VBT failed to explain certain facts MOT was introduced but i do not know how do we explain shape of molecules using MOT as we do in case of VBT .

See Walsh diagrams for a start

There is some information about predicting the shapes of small molecules  in this answer. http://chemistry.stackexchange.com/questions/14981/why-does-bond-angle-decrease-in-the-order-h%E2%82%82o-h%E2%82%82s-h%E2%82%82se/57973#57973


text:
finding molar mass of a molecule
Why is the molar mass of a molecule defined as the the sum of molar mass of all the atoms present in the molecule?
For example, suppose we have to calculate the molar mass of $\ce{H_2O} $ molecule.
Then, $$\text{Molar mass of a } \ce{H_2O}\text{ molecule} = 2 \times \text{Molar mass of }\ce{H} + \text{Molar mass of }\ce{O}$$
= 2 * 1g/mol of H + 16g/mol of O
= 2g/mol of H + 16g/mol of O
As you can see we get two contrasting results - "1g/mol of H" and "2g/mol of H atoms)
How can both "1g" and "2g" form 1 mole of Hydrogen?
I think that the confusion is with the use of units.
I have represented one mole of H like this = 18 g/1 mole of H
But when I multiply with 2 then also I get the same denominator,the result I get is 36 g/1 mole of H
Why am I getting the same denominator (1 mole of H) ?
I have one more question
Why $6.022 \times 10^{23}$ atoms of different elements (constituents of molecule) form $6.022 \times 10^{23}$ molecules of the substance?

make your question more elaborate and you're right ? what doubt do you have ?

My question is that how both 1 g and 2 g form 1 mole of H?

I am still not clear with your language . Although 1 mole of H atom would weigh about 1g .

In your units manipulation you're missing that it is 2 atoms of hydrogen per molecule of water.

@MaxW why is  the relative atomic mass of an atom numerically equal to it's gram atomic mass?

@user42209 because if you are counting atoms they are the same thing (barring terminology).

Stoichiometry will decide how many moles of each element forms how many moles of the product. Hence, in your equation, 2 moles or 2x6.023x10^23 atoms of H will combine with 1 mole or 6.023x10^23 atoms of Oxygen to form 1 mole or 6.023x10^23 molecules of water. I hope this clears your confusion.

You can format mathematical and chemical expressions on Chemistry.SE using MathJax; this post contains further details.

@orthocresol thanks for letting me know! I was wondering how other people do the formatting here :)

No problem. There are a number of resources scattered on [meta], feel free to drop me a line if you need anything.

@MrityunjayGupta why is  the relative atomic mass of an atom numerically equal to it's gram atomic mass?

The mole is simply a way of counting things, but you have to be careful what you are counting. Molar mass is just a way of counting the mass of one mole of something.
If you are counting atoms then 3 moles of atoms (two moles of hydrogen atoms and one of oxygen atoms) come together to make one mole of water molecules. Moles are not additive because they count different things on the two sides of the equation (atoms on one side, molecules on the other). But mass is additive so the total mass of one mole of water is the sum of the masses of the three moles of its components. So the molar mass of water is about 18g/mole.
The molar counts (and masses) depend on what we are counting. If we count hydrogen and oxygen molecules then there would be 1 mole of hydrogen molecules (H2) and half a mole of oxygen molecules. The one mole of water would still weight the same if we count and weigh correctly.

why is  the relative atomic mass of an atom numerically equal to it's gram atomic mass?


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How is the position of a peak on an IR spectrum determined?
In an IR spectrum, all sources describe the position of a peak as a range. The ranges seem to be defined arbitrarily.  Take, for example, the IR spectrum of ethanol (which I can't upload for some reason); why is the $\ce{O-H}$ stretch defined as having a range of $3200$ to $3750~\mathrm{cm}^{-1}$. If it was up to me, I would say that the trough goes from $3100$ to $3650~\mathrm{cm}^{-1}$. And if the $\ce{C-H}$ stretch wasn't present, I would say that the $\ce{O-H}$ trough extends to below $3100~\mathrm{cm}^{-1}$.
Moreover, sometimes there are very specific values given. These also seem to be arbitrarily defined. How do I determine the exact specific value that corresponds to a peak?
And finally, what is the correct way to identify the position of a trough on an IR spectrum? Is it a range? Or do I have to give a specific value?
P.S. I know I asked three questions, but they all serve the overall purpose to help me define how a trough is defined within an IR spectrum.

If your instrument doesn't have a peak/trough picking routine then numerically differentiate the spectrum and look for zero crossings (where the gradient is zero), some smoothing may be necessary, then plot this together with the normal spectrum and so identify peaks and troughs.

Here is a link to a IR spectra of ethanol. http://webbook.nist.gov/cgi/cbook.cgi?ID=C64175&Type=IR-SPEC&Index=26#IR-SPEC

Phase transitions also have a huge effect on the spectra: crystalline phases are well ordered, thus all molecules have very similar surroundings and the bands are narrow (but you may be able to observe differences in the spectra in the volume of a bulk phase vs. surface). Liquid but pure substances have much broader bands (as different molecules have different environments), and also band position may change respectively. Gas phase typically has approximately no intermolecular forces, and the spectra again have very narrow bands (narrow enough to observe rotation in addition to vibration).

A liquid mixtur like ethanol + water has yet different band positions and band width, as again the molecular environment is different. Note: the IUPAC defines chemical species via "having different spectra" - so you can say that all those different spectra of ethanol correspond to different ethanol species - and you have to expect different spectra if you have different species (but not all bands / spectral ranges need to be affected).

Generally, the width of IR absorption peaks depends on the environment of the target molecule. Intermolecular forces alter vibrational modes.
If intermolecular attractions are fairly strong (like the hydrogen bonding present in liquid ethanol), the number and strength of interactions can vary greatly from molecule to molecule, causing the vibrational modes observed in an IR absorption spectrum to widen.
As for the reported width of the peak, the convention is to report the width at 1/2 the height of the peak.

Just to add to your last sentence- this convention is called "Full width at half max" (FWHM) and is a pretty common way to define peak widths for all kinds of spectra across different fields (this is also used in physics and astronomy, for example). See: https://en.wikipedia.org/wiki/Full_width_at_half_maximum

FWHM is only meaningful if you have one peak. If you have multiple overlapping peaks it is meaningless.

Point is that in IR spectra nice clear peaks are rare.

Also changes in intramolecular forces (conformation) change vibrational spectra.

In spectroscopy (and spectrophotometry), the position of a peak/trough is defined by the location of the tip of the peak or trough.  This is why you see such precise values given for their locations.
The ranges you refer to (e.g., $3200$ to $3750\pu{ cm^{-1}}$ for the $\ce{-OH}$ stretch) are the ranges of wavenumber values where the tip of the relevant peak/trough is usually found, and not the range of values over which the peak/trough spreads.

Taking IR spectroscopy as representative, since that's the technique you're asking about, the following is from LibreTexts:

[This equation] gives the frequency of light that a molecule will absorb, and gives the frequency of vibration of the normal mode excited by that light.
$$\nu={1\over2\pi}\left({k\over\mu}\right)^{1\over 2} \\ \nu = \text{ frequency in cm}^{-1} \\ k = \text{ force constant in N/cm} \\ \mu = \text{ reduced mass in kg}$$

Thus, in theory a given vibration corresponds to an exact, specific wavelength of light that would be absorbed when recording an IR spectrum.  However, due to effects like thermal excitation and solvation, we don't observe "infinitely narrow" absorption lines, but instead peaks of finite width. From that same LibreTexts reference:

In general, the width of infrared bands for solid and liquid samples is determined by the number of chemical environments which is related to the strength of intermolecular interactions such as hydrogen bonding. Figure 1 shows hydrogen bond in water molecules and these water molecules are in different chemical environments. Because the number and strength of hydrogen bonds differs with chemical environment, the force constant varies and the wavenumber differs at which these molecules absorb infrared light.
In any sample where hydrogen bonding occurs, the number and strength of intermolecular interactions varies greatly within the sample, causing the bands in these samples to be particularly broad. This is illustrated in the spectra of ethanol and hexanoic acid. When intermolecular interactions are weak, the number of chemical environments is small, and narrow infrared bands are observed.

Thus, if not for environmental effects, every absorption peak would be an absorption line. But, environmental effects cause the absorption to "spread" around that theoretical value. The peak is still considered to be located at that central point, though.
This helps to explain why, for example, the $\ce{-OH}$ stretch peak/band tends to be extraordinarily broad (lots of local variations in the hydrogen bonding environment), whereas the carbonyl $\ce{>C=O}$ stretch is often quite narrow (fewer environmental effects are able to perturb its frequency).

Thank you very much for that, but may you please provide a source to prove that? I have been trying to look for a source, but couldn't find one. You and Dan Burden seem to disagree on the fact that the range values are where the tip of the trough is to be found. Also, see the description of the IR spectrum for lactic acid in this link: http://www.chemguide.co.uk/analysis/ir/interpret.html

@Mathematician You're right, it is hard to find a source that says it explicitly. See my edit; is that argument enough?

@Mathematician I read the description of the lactic acid spectrum at that link as being fully consistent with my answer. I still could be wrong, though!

@Mathematician In Dan's answer, when he refers to "width" and "1/2 height of the peak," he's talking about the standard method for estimating the area of a peak (see here), and not about the range of wavenumbers where a peak is considered to be located.

I am a high school student, and your knowledge of chemistry seems much better than mine. Your source and deduction seem very logical. I trust that you are indeed correct, even more so supported by other questions that I've seen on this site before i asked this question. Lastly, just to make sure, take a look at question 26 on the paper in the next comment. Do you agree that the O-H stretch should be at around 3330 rather than at 3400?

http://qualifications.pearson.com/content/dam/pdf/A%20Level/Chemistry/2013/Exam%20materials/6CH04_01_que_20100127.pdf

Let us continue this discussion in chat.

The other effect to consider is that, say, a carbonyl group, has slightly different vibrational frequencies in an aldehyde, ketone, amide, or quinone etc. As well as solvent effects the chemical environment of the molecule changes the frequency. Most organic chemistry text books will have a list of what these changes are. Look for 'fingerprint' region.

@porphyrin Right, but doesn't chemical context typically shift the position of the peak, whereas environment usually just affects the width of the peak?

@hBy2Py: no, it doesn't matter whether the context change is intra- or intermolecular in order to shift the band position. Broadening is secondary to such shifts: it occurs if lots of slightly different contexts lead to slightly different shifting of a band. E.g. in aqeous solution many H-bridges do have varying strength, thus leading to varying weakening of, say, the -O-H bond in question. A similar but intra-molecular situation occurs with the amide I band of proteines: lots of amide groups have slightly varying contexts, thus the band is broad.


text:
What can be a harmless electrolytic additive for water for creating an oxy-hydrogen flame?
I was thinking of preparing a tiny oxy-hydrogen flame for soldering. I am going to use 6V DC and aluminum sheet as the electrode for this. Now, I have to add some electrolyte to water, that can be a salt, base or acid. But, if I add $\ce{NaCl}$ or $\ce{HCl}$, they will produce chlorine. If I add $\ce{NaOH}$, it will eat electrodes. Adding $\ce{H2SO4}$ may produce poisonous gases like $\ce{SO2}$ and $\ce{H2S}$. So, what can be a harmless/least harmful electrolyte for such purpose? Is $\ce{Na2CO3}$ a good choice?

More importantly, this is a very bad way to create a flame. It will require far more effort than just buying a butane or propane torch.

@matt_black Obviously butane or propane torch are good ways, probably the best. But, I was just thinking of creating a flame myself or at least trying to do so.

Electrolysis of water has become an increasingly common means of creating torches for soldering and welding, and there are commercial devices available as well as plenty of internet do-it-yourself sites. According to this site regarding the SafeFlame consortium:

"[...]the European Union-funded SafeFlame consortium has developed a
  torch system that generates a flame using nothing but H2O and
  electricity.   SafeFlame utilizes an electrical current to electrolyze
  ordinary water, separating it into hydrogen and oxygen gas. Those
  gases are then mixed and ignited as they exit the torch's nozzle. By
  fine-tuning the proportions of the two gases, different types of
  flames can be produced for different applications."

A commonly used electrolyte / electrode combination is $\ce{KOH}$ or $\ce{Na2CO3}$ with stainless steel electrodes. I believe this would be pretty corrosive toward aluminum, but stainless steel for such a setup should be cheap and easy to come by.


text:
Why aren't the R-side-chains of acidic and basic amino acids the site of polypeptide synthesis?
If polypeptides are synthesized by acid-base reactions between the carboxyl group of one amino acid and the amine group of another amino acid, what stops the formation of a peptide bond between the alpha amine group of one amino acid and the carboxyl group (that's in the side chain) of another amino acid?
For example, why doesn't the side chain of glutamic acid bond to the alpha amine of another amino acid (essentially creating a new branch at which polypeptide synthesis can occur)?

It can happen, but it's hardly the norm. See, for example, glutathione. As you may anticipate this reaction is catalysed by a different enzyme, not the ribosome.

Chemically
If you were making a small polypeptide you'd have a protecting group on the side-chain of glutamic acid, and preferably also an activating group on the carboxyl where you want to attach it. If you're just mixing glutamate with another amino acid you'll get a mixture as you expect (including all kinds of possible oligomers).
Biologically
Protein synthesis in the ribosome actually works the other way around (the amine-end is in the ribosome, tRNAs are attached to residues at the carboxyl side). So your expected error would be the continuation of protein synthesis at another alpha amine: the lysine sidechain. Apparently the ribosome is quite good at distinguishing this 4-carbon difference.
My 2 minute hypothesis is that this is one of the reasons why there are residues with shorter acid sidechains (glutamate and aspartate) while a shorter version of lysine is not used in biological protein synthesis.
How the tRNA synthases are so specific that they don't load the amino acid with the wrong carboxyl group is a different question, but tRNAs are extremely specific enzymes in many ways (they have no problem recognizing the difference between a methyl and a hydroxyl group).


text:
equilibrium state of a galvanic cell
At what conditions does equilibrium attain in the galvanic cell. What exactly happens at the equilibrium state of a galvanic cell redox reaction.
Consider the setup Zn|Zn2+||Cu2+|Cu,
   Where

Anode: $\ce{Zn -> Zn^2+ + 2e}$
Cathode: $\ce{Cu^2+ + 2e -> Cu}$
Final reaction: $\ce{Zn + Cu^2+  -> Zn^2+ + Cu}$

Here the electrons would flow from Anode to cathode through a metallic wire which connects the two half cells.
There comes a stage where the $\ce{Zn^2+}$ ions get piled up in anode compartment and depletion of $\ce{Cu^2+}$ ions in the cathode compartment. At one stage the reaction stops which is when the cell/battery power is all utilised.
But where does the equilibrium state come from?
Will the equilibrium reaction occur when the salt bridge is connected?
The reason I stressed more on equilibrium state is because of the equilibrium constant appearing in the Nernst equation. I'm pretty much in  a confused state reagarding the concept.
Any help would be greatly appreciated.

The nernst equation calculates at what voltage would the hypothetical cell be at equilibrium. At equilibrium, the current is zero. This voltage is then used to create a "ranking" of the (half) reactions so as to predict which will happen "first" or prioritized.

when the current becomes zero, how can the cell be at equilibrium, i mean no transfer of electrons takes place due to development of excess potential near respective electrodes.

Sure there is transfer of electrons. Just the same amount each direction. Equilibrium. The forward reaction rate equals the backward one.


text:
Cheap spectral analysis to analyze water quality
I need some guide about spectrometers
I saw a lot of DIY spectrometers to analyze light.
Does anyone try to use those DIY spectrometers to analyze water?
 To detect heavy metals? or a chemical? or any pollutant in the water?
In theory, a spectrometer can detect chemical, biological components and a lot of other things (I have no idea about the technical part)
My question is, making this homemade spectrometer using a webcam, is enough to detect heavy metals in the water ?

No really. UV-Vis doesn't have much specificity. In order to use UV-Vis you'd have to first isolate each metal then do an analysis. Also UV-Vis isn't extremely sensitive.

It seems that.. these guys made it, but there is  not much info about it
http://creative-technology.net/CTECH/Water_Canary.html

I've seen a number of these sorts of Vis spectrometers made. I've collected UV fluorescent rocks. Several folks were offering a spectrometer unit that hooked into  a PC. // The unit that Leo linked tests for "pathogens" not metal ions.

the  main goal is to find a cheap way to detect contaminants in the water,
given those limitations.... what kind of contaminant can I detect using a spectrometer made from a webcam and an old DVD. I don't want to measure it, i just need to know if there is something harmful in the water, I know is not a simple task

The phrase in your last comment,"I don't want to measure it, i just need to know if there is something harmful" is a key part of any analytical problem, and the root of much misunderstanding in that whatever you are measuring, there is almost certainly some of it there. A cup of the purest water will contain billions of Pb molecules. But because there are a trillion, trillion, trillion water molecules in that cup, the lead concentration is too low to measure, or cause harm. So one aspect of designing any analysis is defining how little you need to be able to see, aka the detection limit.

http://creative-technology.net/CTECH/Water_Canary.html
what about "pathogens"? What kind of pathogens? I'm just trying to understand the possibilities under those limits

@airhuff the detection limits will be the  maximum allowed  by the EPA or WHO for drinking water, below that is not relevant for the analysis


text:
What is the convention on color of molecular orbitals in publication figures?
I'm new to MO-visualization and I'm using the program Orca to generate visualizations of molecular orbitals of a system of interest. I've been able to produce some nice images in VMD but I'm not sure what they actually mean:

I've seen pictures like this before with blue/red blobs to represent molecular orbitals, but I don't know what (specifically) they are representing. My suspicion is blue=HOMO and red=LUMO, or blue = negative wavefunction and red = positive wavefunction, but I'd like to know if there is a universal standard.
For reference, this is the "Graphical Representations" panel in VMD for this system:

The colours represent the signal/phase of the molecular orbital, i.e. positive or negative.
Below is the HOMO (made with ChemCraft) of butadiene (PBE-D3BJ/def2-SVP with ORCA) as an example.

Figure 1: HOMO of butadiene (cutoff 0.03)
ChemCraft shows by default negative phase as blue and positive as red.
VMD seems to follow the same convention.
You can certainly change the colours if you want.
Below is how the controls for surfaces in ChemCraft look like:

Figure 2: surface controls in ChemCraft
Other programs may use different conventions, such as NBOView (yellow and blue).

The advantage of yellow and blue or red is that if you print them monochrome, you can still make out the phases. This is a lot harder for red and blue. But I'm not aware of any convention that is predominant. Black and white is also a good choice, as it matches most schematics. I personally use different colour schemes for doubly occupied, singly occupied and unoccupied orbitals.


text:
How far has "self-organizing chemistry" gone?
This is my first question here so, please, bear with me if I go out of the style or the scope of the SE section. I think I am quite new here, so I am prone to make mistakes.
I am a molecular biologist (not a chemist, that's why I really don't know where to look for references) who wants to know more about the field of very complex chemistry-I mean the possible real experimental designs producing an increase in the number, size and network topology (as a measure of organization) of the chemical system in question. My primarily research interests have been (and still are) in the fields of theoretical, system and synthetic biology with implications for the origins of life so I could say I'm pretty aware of all the progress going on with our attempts of replicating the processes occurring in the living cells and to synthesize "artificial living organisms" related to biology. But what I'm not very aware of (because I'm not a chemist and don't know all the possible places I could look for) are other chemical reaction systems which could possibly lead to complex self-organizing behaviour. By these words I mean reactions able to produce molecules of increased dimensions (both in size and molecular weight), increased number of different reaction products (e.g. reactions generating diversity) and reaction networks able to produce increased organization between the reactions participating in them (e.g. reactions the products of which can interact with each other and form new reaction networks among themselves and even their reaction products can repeat the process generating an increase in the relationships among the reactions of the network).
I am aware of complex self-catalyzing reactions like the Belousov-Zhabotinski's reaction which are far from equilibrium but my point here isn't "only" to generate a system out of equilibrium but to understand what chemicals ("real-world" chemicals we know exist, not some hypothetical compounds we design for our needs in computer database) could potentially form a network of self-organizing reactions which can replicate and enlarge itself if given enough "raw materials". I'm not talking here about replicating biological molecules, In fact the point of my question is can such "self-organizing" networks be created with chemicals that have nothing to od with biomolecules? Can we (even in theory) use some other class (or classes) of organic molecules that could form complex interactions among themselves able to produce a network able to replicate (and even diversify) itself given enough "raw materials", e.g. substrates? Have any such experiments been done (or at least proposed) for any class of molecules different than the biomolecules we know today?
In short, can we use any class (or classes) of organic molecules to design "artificial biochemistry" and has anybody in the chemical literature explored (or even only proposed) such options? Is there any papers (in journals specialized in only in chemistry) exploring (or even only just mentioning) the idea for any particular chemicals (or interactions among classes of chemicals) you are aware of? Has such research ever been done and if so which chemicals it explored? To be as concise as possible has any chemist in any journal ever tried to propose the idea of building an "artificial biochemistry"  of any kind?
Thank you very much.

Welcome to Chem.SE! This is an interesting topic and question, but it's unfortunately too broad and likely to be closed. Can you narrow your query down to one or more questions within this scope, that would be more specific? You could then post these as new questions.

Have a look at the Wilson & Meldrum web pages at Chemistry at Leeds University UK.  I'm sure that there are many others working in the areas  you are interested around the world but I only recall hearing these talks.


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First Order Degradation Constant (λ)(d^-1)
Can you please tell me the First Order Degradation Constant (λ)($d^{-1}$) for 1,3,5-Trimethylbenzene (mesitylene)?
If you don't have this value, can you please tell me which VOC in 35 Illinois Administrative Code 740. Appendix A. Table A. is "most similar" to 1,3,5-TMB that I may use its First Order Degradation Constant? Please provide an explanation. Thank you!

Hello and welcome to Chemistry.SE! At this point I recommend taking the short [tour] to better familiarize yourself with how this site works. Regarding your question, any additional information and context that you can give will help others to lead you to the answer you need.

For anyone trying to help with this, there is an online copy of this code with the "Table A" referred to in the OP available here.

By far the most similar compound(s) to 1,3,5-trimethylbenzene listed in Table A is "Total Xylenes", which is a mixture of  "1,2-dimethylbenzene", "1,3-dimethylbenzene" and "1,4-dimethylbenzene". I don't really know what your end goal is, but I would not necessarily expect these to have the same first order degradation constants (how close do you need?) as  1,3,5-Trimethylbenzene. As far as a single (not a mixture) compound listed in Table A, toluene ("1-methylbenzene") would be the most similar, but not very similar.

Based on a bit more than a cursory search, there doesn't seem to be a first order degradation constant readily available for 1,3,5-trimethylbenzene (mesitylene) in the open literature. This leaves us with your second option, the Title 35 Illinois Administrative Code 740$^{[1]}$, Table A (herein referred to as Table A).
By far the most similar compound to 1,3,5-trimethylbenzene that can be found in Table A, based on both chemical structure and physical properties, is the total xylenes (CAS # 1330-20-7).  This is actually a mixture of the three possible isomers of xylene.
Both the images below and all physical and chemical properties given for the total xylenes were taken from this Wikipedia page unless otherwise stated. The 1,3,5-trimethylbenzene image and all of its chemical and physical properties were taken from here.

1,3,5-trimethylbenzene (mesitylene)

In order: o-xylene, m-xylene, p-xylene
Selected properties:
Mesitylene: density = 0.86 ($\mathrm{20^oC}$), B.P. = $\mathrm{165^oC}$, M.P. = $\mathrm{-45^oC}$, and water solbility = 50 mg/L at $\mathrm{25^oC}^{[2]}$
Xylenes : density = 0.86 ($\mathrm{20^oC}$), B.P. = $\mathrm{138^oC}$, M.P. = $\mathrm{-47^oC}$, and water solbility = 110 mg/L at $\mathrm{25^oC}^{[2]}$
1) TITLE 35: ENVIRONMENTAL PROTECTION, SUBTITLE G: WASTE DISPOSAL, CHAPTER I: POLLUTION CONTROL BOARD PART 740, SITE REMEDIATION PROGRAM
2) Yalkowsky, S.H., He, Yan., Handbook of Aqueous Solubility Data: An Extensive Compilation of Aqueous Solubility Data for Organic Compounds Extracted from the AQUASOL dATAbASE. CRC Press LLC, Boca Raton, FL. 2003


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What are some notable results from the field of mathematical chemistry?
I recently found that there is actually a small field of research called mathematical chemistry. It seems that a lot of the research done in this field focuses on graph-theory as applied to chemistry, and topology as applied to defining the structure of molecules.
It's a bit disappointing to me that mathematical chemistry is not a larger field, but it also makes sense because for some reason chemistry does not seem to be as easily asbtracted into a mathematical framework as physics or biology. Note that I don't mean applying mathematical models to chemistry as we do this all the time. For instance, in mathematical physics, which is a very diverse field populated by people trained as both mathematicians and physicists, physical problems are reframed in a natural mathematical framework and solved accordingly. I think a lot of work on the Navier-Stokes equation would qualify as the type of work I mean.
So, is anybody aware of any notable results which have come from the field of mathematical chemistry? Furthermore, do any of these results indicate anything physical which we did not already know to be true? Also, an answer which addressed potential applications of some results of mathematical chemistry would be interesting.
I suppose I will say that I am looking for references as well as summaries so as to narrow the scope of the question.

While not exactly a discovery, the application of graph theory and related to (sub)structure search in tools like Scifinder is of great importance. Group theory of point groups, space groups as applied to spectroscopy and x-ray diffraction is also a very important tool.

Do you want to exclude quantum say for MO theory, and molecular dynamics? There has been quite a lot in the past so what about 'pencil and paper' theory ( I mean non computer based to get results) such as Marcus electron transfer, Forster energy transfer, radiationless transitions in molecules (Lin, Jortner, Siebrand) and ligand field /crystal field theory.

Those sound like reasonable answers to me. I had in mind more of the modern work which uses graph theory and some higher level applications of group theory. I suppose what I'm thinking of when saying mathematical chemistry is that the work is done similar to pure mathematics where a paper basically consists of presenting a proof along with some commentary.


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How to obtain the radial probability distribution function from a quantum chemical calculation?
In the book "Quantum Chemistry" by Ira Levine I found a plot of the radial probability distribution function of argon:

The figure appears in the context of the following section:

Electron densities calculated from Hartree–Fock wave functions are quite accurate. Figure 11.1 compares the radial distribution function of argon (found by integrating the electron density over the angles θ and φ and multiplying the result by r2) calculated by the Hartree–Fock method with the experimental radial distribution function found by electron diffraction. (Recall from Section 6.6 that the radial distribution function is proportional to the probability of finding an electron in a thin spherical shell at a distance r from the nucleus.) Note the electronic shell structure in Fig. 11.1. The high nuclear charge in 18Ar makes the average distance of the 1s electrons from the nucleus far less than in H or He. Thus there is only a moderate increase in atomic size as we go down a given group in the periodic table. Calculations show that the radius of a sphere containing 98% of the Hartree–Fock electron probability density gives an atomic radius in good agreement with the empirically determined van der Waals radius. [See C. W. Kammeyer and D. R. Whitman, J. Chem. Phys., 56, 4419 (1972).]

Source: Ira N. Levine. Quantum Chemistry 7th ed.; Pearson, 2014; page 294.
What kind of chemical software allows one to do this?
It is possible to do with Gaussian, and, of course, how is it possible?

The electron density can be calculated with maybe all the common quantum chemical software, however it is generally exported as a bunch of value on a grid (cube files), and this histogram is not directly calculated. To make this particular plot you most probably have to do the calculation yourself from the the calculated electron distribution.

@Greg Ok, I get .cube file for Argon like this http://rgho.st/8MSlxwyMZ
How can I process this file for relevant data?

This is gladly not a big problem as MultiWFN (open source) can do that for you. I'll guide you through the process from start to finish. My current setup is Gaussian 09 Rev D.01 and MultiWFN 3.3.8; there are newer versions out there with which this should work, too.
Run the calculation, for example ar.com at the HF/STO-3G level of theory, but anything else will also be fine:

%chk=ar.chk
#P HF/STO-3G
gfinput gfoldprint iop(6/7=3)
symmetry(loose)
pop=full pop=nbo6

title card

0 1
Ar    0.0  0.0  0.0

This will generate a checkpoint file ar.chk which needs to be formatted:
formchk -3 ar.chk ar.fchk

Now fire up your MultiWFN program, you should get a prompt similar to the following:

 Multiwfn -- A Multifunctional Wavefunction Analyzer (for Linux)
 Version 3.3.8, release date: 2015-Dec-1
 Project leader: Tian Lu  (Beijing Kein Research Center for Natural Sciences)
 Citation of Multiwfn: Tian Lu, Feiwu Chen, J. Comp. Chem. 33, 580-592 (2012)
 Multiwfn official website: http://Multiwfn.codeplex.com
 Multiwfn official forum (in Chinese): http://bbs.keinsci.com
 Bug reporting, question and suggestion, please contact: Sobereva@sina.com
 ( The number of threads:  2   Current date: 2017-03-09   Time: 14:46:57 )

 Input file path, for example E:\Haiyore!Nyaruko-san\Nyaruko.wfn
 (Supported types: .wfn/.wfx/.fch/.31/.chg/.pdb/.xyz/.cub/.grd/.molden, etc.)
 Hint: To reload the file last time used, simply input the letter "o". Input such as ?miku.fch can open miku.fch in the same folder of the file last time used

If you loaded MultiWFN from the directory you're files are located, simply put in ar.fchk. It will load the file and output some basic information and take you to the "Main function menu":

 Please wait...
 Loading various information of the wavefunction
 Loading basis-set definition...
 Loading orbitals...
 Converting basis function information to GTF information...
 Generating overlap matrix...
 Generating SCF density matrix...

 Total/Alpha/Beta electrons:     18.0000      9.0000      9.0000
 Net charge:     0.00000      Expected multiplicity:    1
 Atoms:      1,  Basis functions:      9,  GTFs:     27
 Total energy:  -521.222880802727 Hartree,   Virial ratio:  2.01704078
 This is a restricted single-determinant wavefunction
 Orbitals from 1 to     9 are occupied
 Title line of this file: title card
 Formula: Ar1
 Molecule weight:        39.94800

 Loaded ar.fchk successfully!

                    ------------ Main function menu ------------

For the sake of sanity I will not copy every menu screen here, just simply what you have to put in. On this screen the radial distribution function is hidden in the "Other functions (Part 2)" and there it is option 5, also see the manual chapter 3.200.5 on page 175 (v.3.3.8). So choose
200 // Other functions (2)

and then
5 // RDF

You should get the following menu:

   ====== Plot radial distribution function for a real space function ======
 -1 Exit
 0 Calculate radial distribution function and its integration curve
 1 Select real space function, current:   1
 2 Set sphere center, current    0.0000    0.0000    0.0000 Ang
 3 Set lower and upper limit of radial distance, current:   0.0000   5.0000 Ang
 4 Set the number of integration point in each shell, current:  2030
 5 Set the number of radial points, current:   500

Here you can select a couple of options. The electron density is already preselected, but you can change it in option 1. You'll get a table with all available functions.
The only thing you might want to adjust are the limits, so choose
3 // Limits
0.0,2.0

and calculate the RDF:
0 // Calculate RDF

Choose 1 to plot and view the function. Or choose
3 // Save graph

or choose
5 // Export to RDF.txt

for post-procesing (with gnuplot).
You can shorten the journey if you prepare a file like this mwfn.in

ar.fchk
200 // Other functions p2
5   // RDF menu
1   // Select real space fucntion
1   // Electron density
2   // Set sphere centre
0.0,0.0,0.0
3   // Set limits
0.0,2.0
0   // Calculate RDF
3   // Save graph
5   // export to text file

and pipe it into the program:
multiwfn < mwfn.in

The native graphing capabilities of MultiWFN don't allow for customisation, so you might want to run it through gnuplot. Here is a file which gives you the graph below, I am using gnuplot 4.4, but later versions are more feature rich.

# General Settings
set terminal svg enhanced font "Arial,12" size 800,600
set tics font "Arial, 9"
set output "ar-rdf.svg"
unset key
#or set key top right Left # etc
set title 'Argon Radial Distribution Function HF/STO-3G'
#
# X-Axis: radius / angstrom
set xlabel 'r / Å'
set xtics in scale 0.6, 0.4 nomirror
set xtics 0,1
set mxtics 2
set xrange [0:2]
# Y-Axis: RDF/a.u.
set ylabel "RDF / a.u." rotate left
set ytics  in  scale 0.6, 0.4 nomirror
set ytics 0,10
set mytics 2
set yrange [0:25]
#
#
# PLOT!
plot \
     "RDF.txt"   using ($2):($3) title "HF/STO-3G"   with lines lt 1 axes x1y1

This is the resulting graph:

This is incredibly useful.

Tested with Gaussian 16 B.01 and Multiwfn 3.6.0 and it worked the same. Same for gnuplot 4.6 patchlevel 6.

hi, Gaussian is a must here? may I know if I dont have Gaussian, is there any way (open source software) combined with multiwfn to get the RDF ? @Martin-マーチン

@Jiadong probably not. Multiwfn understands quite a few packages, essentially you'll need a wave function. Please consult their manual, I'm not active in the field anymore, so I can't really help.


text:
Reaction of phosphonates with conjugated carbonyl compounds
I was wondering whether the phosphonate ion would do a Michael addition on to the double bond rather than attacking the $\ce{C=O}$ bond to form the alkene product as it usually does (e.g. in a Horner–Wadsworth–Emmons reaction).

Something to do with HSAB?


text:
For Penicillin Binding proteins, why is the Enzyme-peptide complex less stable than an enzyme-β-lactam complex?
I'm trying to figure this out. I cannot find any publications that go into good detail about the chemistry of PBP inhibition by β-lactam antibiotics.
PBPs cross-link adjacent pentapeptides to form peptidoglycan. My understanding is that in most bacteria, this pentapeptide has a terminal D-Ala-D-Ala dipeptide. The enzyme removes the terminal D-Ala and forms a D-Ala-Enzyme complex, where the enzyme is linked to the rest of the peptide via ester bond to the D-Ala.
However, when PBPs bind to β-lactams, they also for a β-lactam-ester-enzyme complex. However, this complex is more stable so the β-lactam remains in the active site. Why? What's different about the β-lactam ester? Or is it perhaps steric bulk that prevents the enzyme from being freed?
Any good publications are also very welcome.


text:
How to choose a UV lamp for photoisomerization of 4,4'-azobenzene dicarboxylic acid?
I have prepared a thin film containing 4,4'-azobenzene dicarboxylic acid and I want to study the photoresponsive behaviour of them. For the trans-cis isomeration , I used a 365 nm UV LED with irradiation intensity varied in the range of 5-50 mW/cm². Not seeing any isomeration I have the doubt if I am using the right lamp.  Articles which have prepared thin films containing 4,4'-azobenzene dicarboxylic acid have used other types of lamps such as 200 W Xenon-doped mercury lamp(using a 355 nm cut off filter), 500 W high pressure mercury lamp or 400 W high pressure mercury lamp. I also took the UV-Visible spectroscopy of my films and noticed that the absorption of trans azobenzene-4,4'-dicarboxylic acid is centered at 330 nm.
Am I using the right lamp?

Related: http://chemistry.stackexchange.com/questions/31730

@FelipeSchneider Using MathJax only for exponents can lead to problems. It might look okay for you, but it might break on different devices. Instead of 5-50 mW/cm$^2$ use 5-50 mW/cm², or wrap everything in MathJax $\pu{5-50 mW/cm^2}$.

If you use a high pressure Xe  lamp (and probably high pressure  Hg) you will also need a water filter (with quartz windows to transmit UV) to remove infra-red so that you don't fry your sample then followed by a uv window filter,  such as Schott glass UG1, UG5 or UG11 or something similar  depending on exactly what wavelengths you need.

Water would absorb all the UV.

I'd probably try a more powerful lamp. Plain low-pressure mercury tubes have their emission maximum at 254 nm, which isn't what you want here. But these tubes also come in various coated variants with emission maxima around 300 or 365 nm.
A reasonable choice would be a Rayonet RPR-3000A or a Luzchem LZC-UVB lamp in a proper housing.


text:
What should be the pKa and IUPAC name of this compound?

Note that this is not squaric acid as it has no double bond between the carbons holding the hydroxy groups.

Too bad. Then it won't be stabilized in the anionic form the same way as squaric acid, and hence would be less acidic.

@IvanNeretin but the alpha protons will still be labile ... in water I would imagine that there are 3 sp2 carbons the oxygens of which share one proton.  It would be interesting to see a geometry optimised molecular model.

What I want to know in particular is its pKa, 'cause I want to know whether it will form effervescence of CO2 with NaHCO3.

I should be able to get you an optimized geometry and I might be able to figure out a pKa. I'm practicing how to use Gaussian anyway so it should be useful to me to try it on a real example.

The preferred IUPAC name for this compound is 3,4-dihydroxycyclobutane-1,2-dione. If you have any detailed questions about this name, you might want to consider asking a separate question (see also our guidelines on how to ask nomenclature questions).

I know this isn't exactly an answer, but I think it might help in reaching a qualitative understanding. These are the geometry optimized structures of the protonated and singly deprotonated form of the molecule (I could not obtain a theoretical pKa; it is a much more involved calculation than I anticipated). I would expect the pKa to be fairly low, as the deprotonated structure would seem to be fairly stable due to hydrogen bonding.
(Images were obtained using GaussView. Optimized structures were from Gaussian Optimization calculation using APFD, Basis set 6-311G+(2d,p)).

You might also want to consider the trans-configuration of the molecule.

@Martin That would probably help in the long to demonstrate if there is different behavior, but at this point I'm still stuck on getting a pKa value I can be confident in for this cis-configuration.

I guess if you change the structure of the molecule to get both the carbonyl grops opp to each other...(by Lobry De Bryn Van akelstein rearrangement) the hydroxyl carbon would be acidic enough to give a ppt with $\ce{NaHCO3}$

This is the sort of thing that I was thinking about in my comment above, " but the alpha protons will still be labile" etc ...


text:
Why are thermoluminescent materials mostly mostly based on calcium sulfate rather than on zinc sulfide or calcium fluoride?
I am doing a research in thermo-luminiscent topic and over a part of phosphor called calcium sulfate? The question my mentor asked to me is that why mostly people choose calcium sulfate to do research on and why not other phosphor parts like zinc sulfide or calcium fluoride? Is there something special about calcium sulfate or it is easy to do research over it because he said that he observed in his career that mostly people let it teachers or professors or students will do research over calcium sulfate and not on zinc sulfide or calcium fluoride.

In what kind of research? Could you be more specific?

The question is wrong in its premises, and you should be able to recognize it as such. This world is wide, wide indeed; by no means is it limited to one, or two, or three, or even four compounds. Many are used in research. Then again, there is another possibility: maybe you are telling us only a half of the question?

@Ivan Neretin  Dear sir, I am doing a research in thermoluminiscent topic and over a part of phosphor called calcium sulphate? The question my mentor asked to me is that why mostly people choose calcium sulphate to do research on and why not other phosphor parts like zns or calcium fluoride? Is there something special about calcium sulphate or it is easy to do research over it because he said that he observed in his career that mostly people let it teachers or professors or students will do research over calcium sulphate and not on zns or calcium fluoride.

To my knowledge, pure(!) calcium sulfate does not show thermoluminescence.
It is however true that it can be turned into thermoluminescent materials by doping with rare metals, such as dysprosium or terbium.
According to a conference contribution, $\ce{CaSO4:Dy}$ is prepared from calcium sulfate and dysprosium oxide in concentrated sulfuric acid at temperatures between 200 and 300 °C.
While calcium sulfate will survive these reaction conditions, other base materials, such as calcium fluoride or zinc sulfide, will rather not.


text:
Rusting on an iron material
I was wondering if someone could tell me why do we write down 'x' in front of $\ce{H2O}$ in the reaction that takes place to form rust on an iron material. I searched for it and went through many websites but couldn't get a specific answer so please someone help me.

"x" stands for "dunno".

@IvanNeretin more accurately, ‘x’ stands for ‘unknown value’

Rust consists of hydrated iron(III) oxides Fe2O3·nH2O, iron(III) oxide-hydroxide (FeO(OH), Fe(OH)3.
The rusting of iron is an electrochemical process that begins with the transfer of electrons from iron to oxygen. The rate of corrosion is affected by water and accelerated by electrolytes. The key reaction is the reduction of oxygen:
$$\ce{O2 + 4 e- + 2H2O -> 4OH-}$$
The following redox reaction also occurs in the  presence of water and is crucial to the formation of rust:
$$\ce{2Fe2+ + 0.5 O2 <=> 2 Fe3+ + O2−}$$
Additionally, the following multistep acid-base reactions affect the course of rust formation:
$$\ce{Fe2+ + 2 H2O <=> Fe(OH)2 + 2H+}$$
$$\ce{Fe3+ + 3 H2O <=> Fe(OH)3 + 3 H+}$$
as do the following dehydration equilibria:
$$\ce{Fe(OH)2 <=> FeO + H2O}$$
$$\ce{Fe(OH)3 <=> FeO(OH) + H2O}$$
$$\ce{2 FeO(OH)  <=> Fe2O3 + H2O}$$
From the above equations, it is also seen that the corrosion products are dictated by the availability of water and oxygen. With limited dissolved oxygen, iron(II)-containing materials are favoured, including FeO and black lodestone (Fe3O4). High oxygen concentrations favour ferric materials with the nominal formulae Fe(OH)3-xOx/2.
Thus you can actually calculate the value of x (water of hydration) in the compound. Hope this works
Credits: Chemical formula of rust


text:
Cis/trans Proline vs. Double bond character
I am learning about proteins and we have learned that the C-N bond is rigid because of it's partial double-bond character. We have also learned that proline is commonly found in beta turns (type 1) because it can isomerize from trans to cis (which helps make the 180 degree turn).
I don't understand how these 2 things fit together--how can it change from trans to cis if it can't rotate around the C-N bond?
I found this which talks about the double bond being broken (in this case, it would be the partial double-bond, I guess) and then reforming...is that what is going on here?

The change from cis to trans (called isomerization) is not spontaneous. There are proteins called prolyl isomerases (see Wikipedia Prolyl Isomerase) that assist in the isomerization process. This happens when the protein is being folded into its final secondary and tertiary structures.


text:
Why is naphthalene less stable than benzene according to per benzene ring?
Why is benzene more stable than naphthalene according to per benzene ring. Which results in a higher heat of hydrogenation (i.e. energy released on hydrogenation) of benzene than naphthalene according to per benzene ring, i.e. if we hydrogenate only one benzene ring in each. I think it should be opposite.
Both are aromatic in nature both have delocalised electrons but naphthalene has more resonance structures and more delocalisation so overall it must be more stable for a single ring.
May someone help?

What do you mean by stability?

Stability means thermodynamic stability ie enthalpy of formation .

Which source tells you benzene is more stable than naphthalene?

Stability is a relative concept, this question is very unclear. Please also add the source (quote and cite) that gave you this idea.

counting resonance structures is a poor way to estimate aromaticity or the energy involved.

A short answer is: you're forgetting that a double bond is shared between the two rings in naphthalene.
A long answer is given below.
In terms of heat of hydrogenation
Naphthalene can be hydrogenated to give tetralin.
Further hydrogenation gives decalin.

Scheme 1: hydrogenation of naphthalene.
On the other hand, the hydrogenation of benzene gives cyclohexane.

Scheme 2: hydrogenation of benzene.
In an old report it reads (Sherman, J. Oil Soap (1939) 16: 28):

If benzene could be simply regarded as cyclohexatriene with no interaction between the double bonds, the heat of hydrogenation to form cyclohexane should be just three times the heat of hydrogenation of cyclohexene to form cyclohexane.
  From Table I, this is $3 \times -28.6 = -85.8$ kcal/mol.
  The experimental value is $-49.8$ kcal/mol.
  Hence benzene is more stable by an amount $-49.8 -(-85.8) = 36.0$ kcal/mol than it would be if it were completely unsaturated in character (with no interaction between the double bonds).
  From the theoretical viewpoint, this extra stability of benzene (and other aromatic compounds) is shown to be a consequence of the fact that the normal state of the molecule is not that corresponding to either Kekulé structure but is a sort of combination of the two.

This $36.0$ kcal/mol may be regarded as a direct measure of aromaticity and is cited in many places as resonance energy.
In particular, the resonance energy for naphthalene is $61$ kcal/mol.
This value is shifted by around $5 \times -28.6 = -143.0$ kcal/mol (five double bonds) from the actual heat of hydrogenation of naphthalene.
This means the heat of hydrogenation of naphthalene would be somewhere around $61 - (-143.0) = -82$ kcal/mol.
The actual value from NIST (to cis-decalin) is $-318$ kJ/mol, or $-76$ kcal/mol.
So it costs $-49.8$ kcal/mol to hydrogenate benzene to cyclohexane but only $-76$ kcal/mol to hydrogenate naphthalene to cis-decalin, less than twice a benzene.
Furthermore, part of this energy is due to the resonance energy, which is $36.0$ kcal/mol for benzene, but only $61$ kcal/mol for naphthalene, again less than twice a benzene.
What did we forget?
Naphthalene rings are fused, that is, a double bond is shared between two rings.
This makes the above comparisons unfair.
A better comparison would be the amounts of resonance energy per $\pi$ electron.
Benzene has 6 $\pi$ electrons, which makes its resonance energy equal to $36.0 / 6 = 6$ kcal/mol per $\pi$ electron.
Naphthalene (10 $\pi$ electrons) shows a remarkably similar value: $61 / 10 = 6.1$ kcal/mol.
Even comparison of heats of hydrogenation per double bond makes good numbers.
For benzene (hypothetically three double bonds) it costs $49.8 / 3 = 16.6$ kcal/mol to hydrogenate each double bond.
For naphthalene it would be $-76 / 5 = 15.2$ kcal/mol, again very similar.
What about tetralin?
Again NIST comes to our rescue.
The most recent value for the heat of hydrogenation from naphthalene to tetralin is $-125$ kJ/mol, or $-29.9$ kcal/mol (Scheme 1, first arrow).
As a per double bond value it gives us $-29.9 / 2 = -15.0$ kcal/mol, close to both benzene and naphthalene (all of them differ by less than $1.6$ kcal/mol).
In terms of electronic structure

Both are aromatic in nature both have delocalised electrons but naphthalene has more number of $\pi$ bonds and hence more resonance structures and more delocalisation so overall it must be more stable.

I believe the highlighted sentence tells it all.
As you said, delocalisation is more significative in naphthalene.
A simple model for delocalisation is the particle in a box (in this case, more of a rectangle).
The $n$-th energy level for an electron confined in a single dimensional box (line segment) is
$$E_n = \frac{n^2 h^2} {8 m_e L^2}\text{,}$$
where $m_e$ is the mass of the electron, $h$ is the Planck's constant and $L$ is the length of the line segment.
So energy decreases with the square of the length of the confinement.
Results are analogous for other dimensions.
From this simple model, the more confined an electron, the higher will be its energy.
It is thus perfectly reasonable to rationalise that more delocalisation results in more stability.
Given its larger delocalisation, it seems rational that the energy levels of naphthalene have less energy in comparison to benzene.

So you're saying that in benzene there is more delocalisation?

No, I mean energy levels of naphthalene have less energy due to its greater delocalisation.
I have edited the answer to make it clearer.

So if they have less energy that means they are more stable.

In a thermodynamical sense, as you stated, yes.

I was making a confusion earlier today, sorry. I corrected my answer. The conclusion is the same.

I am still incredibly confused which kind of stability we are talking about. W.r.t. the energy levels outlined by you, I agree. But if we look at $\ce{C6H6 + H2}$ versus $\ce{C10H8 + H2}$ it is going to be a different story. I think the question still is very unclear.

I tried to address the issue in an edit, mostly focused in the comparison between heat of hydrogenation and stability (resonance energy).

This is a good answer. However saying that naphthalene (though I don't remember the specific case, says a PAH) is less aromatic per ring makes sense considering the rings chemistry.


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Solid aluminum powder
I am curious if there is a way to dissolve or break down solidified aluminum powder. I have tried to break it apart but it is a solid chunk and I need to figure out how I can break it down to 150 pound pieces.
Can a block of hardened aluminum powder be physically or chemically broken down back to a soft form.

What do you mean by "solidified aluminium powder" ?

Can you modify your question? If you have a solid chunk, it doesn't make sense that you're trying to dissolve a powder.

Luke - I have a couple of questions: 1 Are you trying to get rid of it or trying to use it for an application? 2 How is this 150 mass organized i.e. in a too heavy to lift bucket or bag or ... ?  And then for physically breaking it apart, you could try rock grinders - will not powder the material but will make moving it easier.

If you had a sealed container of aluminum powder that became exposed to air, your prospects for turning your current material back into aluminum powder are grim.
Your block probably appears like something that became clumpy when exposed to humidity (table salt, sugar, etc.), and that you aught to be able to just break it apart with a little force to get the powdery stuff again.
Unfortunately, aluminum reacts with the oxygen in air to form a very hard oxide layer, which is likely what is now binding the formerly pure aluminum powder together. In other words, you don't just have aluminum "sticking" to aluminum, but you've formed strong chemical bonds, actually creating a new compound altogether (aluminum oxide) There is likely no simple solution for returning your 150 pound blocks of mixed aluminum / aluminum oxide back into pure aluminum powder.